From 5fc61e814a98ab47d389c0d8ab89c73618d002cb Mon Sep 17 00:00:00 2001 From: A-Larsen Date: Tue, 13 Oct 2026 20:33:47 -0400 Subject: [PATCH 1/1] first commit --- Algrebra/Angles.adoc | 23 + Algrebra/Angles.html | 1097 +++++++++ Algrebra/Division.adoc | 30 + Algrebra/Division.html | 1145 +++++++++ Algrebra/Exponents.adoc | 152 ++ Algrebra/Exponents.html | 1317 +++++++++++ Algrebra/Fractions.adoc | 81 + Algrebra/Fractions.html | 1183 ++++++++++ Algrebra/Logarithms.adoc | 244 ++ Algrebra/Logarithms.html | 1477 ++++++++++++ Algrebra/Polynomials.adoc | 255 ++ Algrebra/Polynomials.html | 1568 +++++++++++++ Algrebra/README | 1 + Algrebra/Radicals.adoc | 138 ++ Algrebra/Radicals.html | 1272 ++++++++++ Algrebra/Sequences.adoc | 111 + Algrebra/Sequences.html | 1219 ++++++++++ Algrebra/Terminology.adoc | 44 + Algrebra/Terminology.html | 1149 +++++++++ Algrebra/Triangulation.adoc | 12 + Algrebra/Triangulation.html | 1082 +++++++++ Algrebra/complex_numbers.adoc | 11 + Algrebra/complex_numbers.html | 1082 +++++++++ Algrebra/index.adoc | 22 + Algrebra/index.html | 1088 +++++++++ Algrebra/linear_inequalities.adoc | 16 + Algrebra/linear_inequalities.html | 1084 +++++++++ Algrebra/rules.adoc | 70 + Algrebra/rules.html | 1163 ++++++++++ Calculus/Asymptotes.adoc | 27 + Calculus/Asymptotes.html | 1113 +++++++++ Calculus/Derivatives.adoc | 511 ++++ Calculus/Derivatives.html | 1902 +++++++++++++++ Calculus/Integrals.adoc | 432 ++++ Calculus/Integrals.html | 1685 ++++++++++++++ Calculus/Intermediate-Value-Theorem.adoc | 21 + Calculus/Intermediate-Value-Theorem.html | 1098 +++++++++ Calculus/Limits.adoc | 570 +++++ Calculus/Limits.html | 1796 ++++++++++++++ Calculus/Mean-Value-Theorem.adoc | 43 + Calculus/Mean-Value-Theorem.html | 1131 +++++++++ Calculus/Spherical-Coordinates.adoc | 17 + Calculus/Spherical-Coordinates.html | 1094 +++++++++ Calculus/index.adoc | 16 + Calculus/index.html | 1070 +++++++++ Calculus/scarp | 4 + Geometry/Angles-And-Sides.adoc | 53 + Geometry/Angles-And-Sides.html | 1145 +++++++++ Geometry/Circle-Theorems.adoc | 114 + Geometry/Circle-Theorems.html | 1341 +++++++++++ Geometry/Geometric-Proportions.adoc | 92 + Geometry/Geometric-Proportions.html | 1253 ++++++++++ Geometry/Lines.adoc | 17 + Geometry/Lines.html | 1086 +++++++++ Geometry/Triangles.adoc | 27 + Geometry/Triangles.html | 1115 +++++++++ Geometry/index.adoc | 14 + Geometry/index.html | 1064 +++++++++ Graphics/3D-Objects.adoc | 18 + Graphics/3D-Objects.html | 1087 +++++++++ Graphics/Quanternions.adoc | 13 + Graphics/Quanternions.html | 1085 +++++++++ Graphics/Rasterization.adoc | 18 + Graphics/Rasterization.html | 1105 +++++++++ Graphics/index.adoc | 12 + Graphics/index.html | 1058 +++++++++ Graphing/Parent-Functions.adoc | 144 ++ Graphing/Parent-Functions.html | 1370 +++++++++++ Graphing/index.adoc | 10 + Graphing/index.html | 1052 +++++++++ Linear-Algrebra/Graphing.adoc | 27 + Linear-Algrebra/Graphing.html | 1109 +++++++++ Linear-Algrebra/Matrix-Math.adoc | 110 + Linear-Algrebra/Matrix-Math.html | 1218 ++++++++++ Linear-Algrebra/Matrix-Operations.adoc | 290 +++ Linear-Algrebra/Matrix-Operations.html | 1478 ++++++++++++ Linear-Algrebra/Matrix-Types.adoc | 161 ++ Linear-Algrebra/Matrix-Types.html | 1287 +++++++++++ Linear-Algrebra/Properties.adoc | 138 ++ Linear-Algrebra/Properties.html | 1264 ++++++++++ Linear-Algrebra/SCRAP | 11 + Linear-Algrebra/System-of-Equations.adoc | 786 +++++++ Linear-Algrebra/System-of-Equations.html | 2056 +++++++++++++++++ Linear-Algrebra/Terminology.adoc | 152 ++ Linear-Algrebra/Terminology.html | 1344 +++++++++++ Linear-Algrebra/Vector-Math.adoc | 188 ++ Linear-Algrebra/Vector-Math.html | 1320 +++++++++++ Linear-Algrebra/Vector-Rules.adoc | 51 + Linear-Algrebra/Vector-Rules.html | 1172 ++++++++++ Linear-Algrebra/index.adoc | 18 + Linear-Algrebra/index.html | 1076 +++++++++ Makehtml | 83 + Physics/Ball-Bouncing.adoc | 37 + Physics/Ball-Bouncing.html | 1164 ++++++++++ Physics/Ball-Dynamic-Collision.adoc | 280 +++ Physics/Ball-Dynamic-Collision.html | 1633 +++++++++++++ Physics/Ball-Static-Collision.adoc | 146 ++ Physics/Ball-Static-Collision.html | 1383 +++++++++++ Physics/Basics.adoc | 274 +++ Physics/Basics.html | 1649 +++++++++++++ Physics/Collision.adoc | 43 + Physics/Collision.html | 1194 ++++++++++ Physics/index.adoc | 14 + Physics/index.html | 1064 +++++++++ Probability/Pascals-Triangle.adoc | 13 + Probability/Pascals-Triangle.html | 1085 +++++++++ .../Permutations-and-Combinations.adoc | 12 + .../Permutations-and-Combinations.html | 1082 +++++++++ Probability/index.adoc | 11 + Probability/index.html | 1055 +++++++++ README.adoc | 14 + Triginometry/Basics.adoc | 81 + Triginometry/Basics.html | 1277 ++++++++++ Triginometry/Centers.adoc | 23 + Triginometry/Centers.html | 1104 +++++++++ Triginometry/Identities.adoc | 101 + Triginometry/Identities.html | 1327 +++++++++++ Triginometry/index.adoc | 12 + Triginometry/index.html | 1058 +++++++++ ...748c8d76a4b154138a91a12caf43c78a.png.cache | 1 + diagram/index.adoc | 9 + diagram/index.html | 1046 +++++++++ docinfo.conf | 3 + favicon.png | Bin 0 -> 2147 bytes index.adoc | 17 + index.html | 1074 +++++++++ notes.adoc | 28 + rendered.html | 854 +++++++ 128 files changed, 80466 insertions(+) create mode 100644 Algrebra/Angles.adoc create mode 100644 Algrebra/Angles.html create mode 100644 Algrebra/Division.adoc create mode 100644 Algrebra/Division.html create mode 100644 Algrebra/Exponents.adoc create mode 100644 Algrebra/Exponents.html create mode 100644 Algrebra/Fractions.adoc create mode 100644 Algrebra/Fractions.html create mode 100644 Algrebra/Logarithms.adoc create mode 100644 Algrebra/Logarithms.html create mode 100644 Algrebra/Polynomials.adoc create mode 100644 Algrebra/Polynomials.html create mode 100644 Algrebra/README create mode 100644 Algrebra/Radicals.adoc create mode 100644 Algrebra/Radicals.html create mode 100644 Algrebra/Sequences.adoc create mode 100644 Algrebra/Sequences.html create mode 100644 Algrebra/Terminology.adoc create mode 100644 Algrebra/Terminology.html create mode 100644 Algrebra/Triangulation.adoc create mode 100644 Algrebra/Triangulation.html create mode 100644 Algrebra/complex_numbers.adoc create mode 100644 Algrebra/complex_numbers.html create mode 100644 Algrebra/index.adoc create mode 100644 Algrebra/index.html create mode 100644 Algrebra/linear_inequalities.adoc create mode 100644 Algrebra/linear_inequalities.html create mode 100644 Algrebra/rules.adoc create mode 100644 Algrebra/rules.html create mode 100644 Calculus/Asymptotes.adoc create mode 100644 Calculus/Asymptotes.html create mode 100644 Calculus/Derivatives.adoc create mode 100644 Calculus/Derivatives.html create mode 100644 Calculus/Integrals.adoc create mode 100644 Calculus/Integrals.html create mode 100644 Calculus/Intermediate-Value-Theorem.adoc create mode 100644 Calculus/Intermediate-Value-Theorem.html create mode 100644 Calculus/Limits.adoc create mode 100644 Calculus/Limits.html create mode 100644 Calculus/Mean-Value-Theorem.adoc create mode 100644 Calculus/Mean-Value-Theorem.html create mode 100644 Calculus/Spherical-Coordinates.adoc create mode 100644 Calculus/Spherical-Coordinates.html create mode 100644 Calculus/index.adoc create mode 100644 Calculus/index.html create mode 100644 Calculus/scarp create mode 100644 Geometry/Angles-And-Sides.adoc create mode 100644 Geometry/Angles-And-Sides.html create mode 100644 Geometry/Circle-Theorems.adoc create mode 100644 Geometry/Circle-Theorems.html create mode 100644 Geometry/Geometric-Proportions.adoc create mode 100644 Geometry/Geometric-Proportions.html create mode 100644 Geometry/Lines.adoc create mode 100644 Geometry/Lines.html create mode 100644 Geometry/Triangles.adoc create mode 100644 Geometry/Triangles.html create mode 100644 Geometry/index.adoc create mode 100644 Geometry/index.html create mode 100644 Graphics/3D-Objects.adoc create mode 100644 Graphics/3D-Objects.html create mode 100644 Graphics/Quanternions.adoc create mode 100644 Graphics/Quanternions.html create mode 100644 Graphics/Rasterization.adoc create mode 100644 Graphics/Rasterization.html create mode 100644 Graphics/index.adoc create mode 100644 Graphics/index.html create mode 100644 Graphing/Parent-Functions.adoc create mode 100644 Graphing/Parent-Functions.html create mode 100644 Graphing/index.adoc create mode 100644 Graphing/index.html create mode 100644 Linear-Algrebra/Graphing.adoc create mode 100644 Linear-Algrebra/Graphing.html create mode 100644 Linear-Algrebra/Matrix-Math.adoc create mode 100644 Linear-Algrebra/Matrix-Math.html create mode 100644 Linear-Algrebra/Matrix-Operations.adoc create mode 100644 Linear-Algrebra/Matrix-Operations.html create mode 100644 Linear-Algrebra/Matrix-Types.adoc create mode 100644 Linear-Algrebra/Matrix-Types.html create mode 100644 Linear-Algrebra/Properties.adoc create mode 100644 Linear-Algrebra/Properties.html create mode 100644 Linear-Algrebra/SCRAP create mode 100644 Linear-Algrebra/System-of-Equations.adoc create mode 100644 Linear-Algrebra/System-of-Equations.html create mode 100644 Linear-Algrebra/Terminology.adoc create mode 100644 Linear-Algrebra/Terminology.html create mode 100644 Linear-Algrebra/Vector-Math.adoc create mode 100644 Linear-Algrebra/Vector-Math.html create mode 100644 Linear-Algrebra/Vector-Rules.adoc create mode 100644 Linear-Algrebra/Vector-Rules.html create mode 100644 Linear-Algrebra/index.adoc create mode 100644 Linear-Algrebra/index.html create mode 100755 Makehtml create mode 100644 Physics/Ball-Bouncing.adoc create mode 100644 Physics/Ball-Bouncing.html create mode 100644 Physics/Ball-Dynamic-Collision.adoc create mode 100644 Physics/Ball-Dynamic-Collision.html create mode 100644 Physics/Ball-Static-Collision.adoc create mode 100644 Physics/Ball-Static-Collision.html create mode 100644 Physics/Basics.adoc create mode 100644 Physics/Basics.html create mode 100644 Physics/Collision.adoc create mode 100644 Physics/Collision.html create mode 100644 Physics/index.adoc create mode 100644 Physics/index.html create mode 100644 Probability/Pascals-Triangle.adoc create mode 100644 Probability/Pascals-Triangle.html create mode 100644 Probability/Permutations-and-Combinations.adoc create mode 100644 Probability/Permutations-and-Combinations.html create mode 100644 Probability/index.adoc create mode 100644 Probability/index.html create mode 100644 README.adoc create mode 100644 Triginometry/Basics.adoc create mode 100644 Triginometry/Basics.html create mode 100644 Triginometry/Centers.adoc create mode 100644 Triginometry/Centers.html create mode 100644 Triginometry/Identities.adoc create mode 100644 Triginometry/Identities.html create mode 100644 Triginometry/index.adoc create mode 100644 Triginometry/index.html create mode 100644 diagram/diag-ditaa-md5-748c8d76a4b154138a91a12caf43c78a.png.cache create mode 100644 diagram/index.adoc create mode 100644 diagram/index.html create mode 100644 docinfo.conf create mode 100644 favicon.png create mode 100644 index.adoc create mode 100644 index.html create mode 100644 notes.adoc create mode 100644 rendered.html diff --git a/Algrebra/Angles.adoc b/Algrebra/Angles.adoc new file mode 100644 index 0000000..6f6e7a8 --- /dev/null +++ b/Algrebra/Angles.adoc @@ -0,0 +1,23 @@ += Algrebra | Complex numbers +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} + +== Perpendicular Lines + +You can tell if two rays are perpendicular to one another, through a few ways. + +=== Negative reciprocals + +The first way is if the two rays of slopes that are negative reciprocals of each +other. + +=== Multiplying Slopes + +Similar to the above method, if you multiplied both of these slopes and got +stem:[1] you would know the rays are perpendicular + diff --git a/Algrebra/Angles.html b/Algrebra/Angles.html new file mode 100644 index 0000000..c3d8bf7 --- /dev/null +++ b/Algrebra/Angles.html @@ -0,0 +1,1097 @@ + + + + + + + +Algrebra | Complex numbers + + + + + + + + + + + + + + + + + + +
+
+

Perpendicular Lines

+
+
+

You can tell if two rays are perpendicular to one another, through a few ways.

+
+
+

Negative reciprocals

+
+

The first way is if the two rays of slopes that are negative reciprocals of each +other.

+
+
+
+

Multiplying Slopes

+
+

Similar to the above method, if you multiplied both of these slopes and got +\(1\) you would know the rays are perpendicular

+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Division.adoc b/Algrebra/Division.adoc new file mode 100644 index 0000000..8df4b44 --- /dev/null +++ b/Algrebra/Division.adoc @@ -0,0 +1,30 @@ += Algrebra | Division +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Division Rules + +[%header, cols="1,1"] +|=== +| Divisor | Rule Description +| 1 | Every integer is divisible by 1. +| 2 | A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8). +| 3 | A number is divisible by 3 if the sum of its digits is divisible by 3. +| 4 | A number is divisible by 4 if the last two digits form a number that is divisible by 4. +| 5 | A number is divisible by 5 if its last digit is 0 or 5. +| 6 | A number is divisible by 6 if it is divisible by both 2 and 3. +| 7 | To check divisibility by 7, double the last digit, subtract it from the rest of the number, and see if the result is divisible by 7. +| 8 | A number is divisible by 8 if the last three digits form a number that is divisible by 8. +| 9 | A number is divisible by 9 if the sum of its digits is divisible by 9. +| 10 | A number is divisible by 10 if it ends in 0. +|=== + +== Long Division +* https://www.youtube.com/watch?v=_FSXJmESFmQ[The Organic Chemestry Tutor | Long division with polynomials] +* https://www.youtube.com/watch?v=FxHWoUOq2iQ[The Organic Chemestry Tutor | Synthetic division of polynomials] diff --git a/Algrebra/Division.html b/Algrebra/Division.html new file mode 100644 index 0000000..47ca4ff --- /dev/null +++ b/Algrebra/Division.html @@ -0,0 +1,1145 @@ + + + + + + + +Algrebra | Division + + + + + + + + + + + + + + + + + + +
+
+

Division Rules

+
+ ++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
DivisorRule Description

1

Every integer is divisible by 1.

2

A number is divisible by 2 if its last digit is even (0, 2, 4, 6, 8).

3

A number is divisible by 3 if the sum of its digits is divisible by 3.

4

A number is divisible by 4 if the last two digits form a number that is divisible by 4.

5

A number is divisible by 5 if its last digit is 0 or 5.

6

A number is divisible by 6 if it is divisible by both 2 and 3.

7

To check divisibility by 7, double the last digit, subtract it from the rest of the number, and see if the result is divisible by 7.

8

A number is divisible by 8 if the last three digits form a number that is divisible by 8.

9

A number is divisible by 9 if the sum of its digits is divisible by 9.

10

A number is divisible by 10 if it ends in 0.

+
+
+
+

Long Division

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Exponents.adoc b/Algrebra/Exponents.adoc new file mode 100644 index 0000000..902bab2 --- /dev/null +++ b/Algrebra/Exponents.adoc @@ -0,0 +1,152 @@ += Algrebra | Exponents +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Exponent Rules + +[cols="^.^1a, ^.^1a, <.^2a", frame="none", grid="rows", options="header"] +|=== +| Rule Name | Math (LaTeX) | Details + +| Product Rule +| +[stem] +++++ +a^m \cdot a^n = a^{m+n} +++++ +| [%collapsible] +==== +When multiplying two powers with the same base, keep the base and add the exponents. +==== + +| Quotient Rule +| +[stem] +++++ +\frac{a^m}{a^n} = a^{m - n} +++++ +| [%collapsible] +==== +When dividing two powers with the same base, subtract the exponent of the denominator from the exponent of the numerator. +==== + +| Power of a Power +| +[stem] +++++ +(a^m)^n = a^{m \cdot n} +++++ +| [%collapsible] +==== +To raise a power to a power, multiply the exponents together. +==== + +| Power of a Product +| +[stem] +++++ +(ab)^m = a^m \cdot b^m +++++ +| [%collapsible] +==== +Each factor inside the parentheses is raised to the power. +==== + +| Power of a Quotient +| +[stem] +++++ +\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} +++++ +| [%collapsible] +==== +Both the numerator and the denominator are raised to the power. (Note: $b \neq 0$). +==== + +| Zero Exponent +| +[stem] +++++ +a^0 = 1 +++++ +| [%collapsible] +==== +Any non-zero base raised to the power of zero is exactly 1. +==== + +| Negative Exponent +| +[stem] +++++ +a^{-m} = \frac{1}{a^m} +++++ +| [%collapsible] +==== +A negative exponent represents the reciprocal of the base raised to the positive power. +==== +|=== + +== Odd vs Even exponents and negative coefficients + +=== Even Exponents +[stem] +++++ +(-x)^4 +++++ + +In this case this has a coefficients of negative one. so it is equavilant to + +[stem] +++++ +x^4 +++++ + +=== Odd Exponents + +[stem] +++++ +3(-x)^3 +++++ + +This is equivalant to + +[stem] +++++ +-3x^3 +++++ + + +== Multiplying Negative Exponents + +=== When the bases are the same + +[stem] +++++ +a^{-n} \cdot a^{-m} = a^{-(n+m)} = 1 \div a^{n+m} +++++ + +=== When the bases are different and the negative powers are the same + +[stem] +++++ +a^{-n} \cdot b^{-n} = (a \cdot b)^{-n} = 1 \div (a \cdot b)^n +++++ + +=== When the bases and the negative powers are different + +[stem] +++++ +a^{-n} \cdot b^{-m} = (a^{-n}) \cdot (b^{-m}) +++++ + +=== Exponents of exponents + +== Links +- https://www.cuemath.com/numbers/multiplying-exponents/[cuemath | multiplying exponents] +- https://www.cuemath.com/algebra/exponent-rules/[cuemeth | exponent rules] diff --git a/Algrebra/Exponents.html b/Algrebra/Exponents.html new file mode 100644 index 0000000..1172f93 --- /dev/null +++ b/Algrebra/Exponents.html @@ -0,0 +1,1317 @@ + + + + + + + +Algrebra | Exponents + + + + + + + + + + + + + + + + + + +
+
+

Exponent Rules

+
+ +++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Rule NameMath (LaTeX)Details
+

Product Rule

+
+
+\[a^m \cdot a^n = a^{m+n}\] +
+
+Details +
+
+

When multiplying two powers with the same base, keep the base and add the exponents.

+
+
+
+

Quotient Rule

+
+
+\[\frac{a^m}{a^n} = a^{m - n}\] +
+
+Details +
+
+

When dividing two powers with the same base, subtract the exponent of the denominator from the exponent of the numerator.

+
+
+
+

Power of a Power

+
+
+\[(a^m)^n = a^{m \cdot n}\] +
+
+Details +
+
+

To raise a power to a power, multiply the exponents together.

+
+
+
+

Power of a Product

+
+
+\[(ab)^m = a^m \cdot b^m\] +
+
+Details +
+
+

Each factor inside the parentheses is raised to the power.

+
+
+
+

Power of a Quotient

+
+
+\[\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}\] +
+
+Details +
+
+

Both the numerator and the denominator are raised to the power. (Note: $b \neq 0$).

+
+
+
+

Zero Exponent

+
+
+\[a^0 = 1\] +
+
+Details +
+
+

Any non-zero base raised to the power of zero is exactly 1.

+
+
+
+

Negative Exponent

+
+
+\[a^{-m} = \frac{1}{a^m}\] +
+
+Details +
+
+

A negative exponent represents the reciprocal of the base raised to the positive power.

+
+
+
+
+
+
+

Odd vs Even exponents and negative coefficients

+
+
+

Even Exponents

+
+
+\[(-x)^4\] +
+
+
+

In this case this has a coefficients of negative one. so it is equavilant to

+
+
+
+\[x^4\] +
+
+
+
+

Odd Exponents

+
+
+\[3(-x)^3\] +
+
+
+

This is equivalant to

+
+
+
+\[-3x^3\] +
+
+
+
+
+
+

Multiplying Negative Exponents

+
+
+

When the bases are the same

+
+
+\[a^{-n} \cdot a^{-m} = a^{-(n+m)} = 1 \div a^{n+m}\] +
+
+
+
+

When the bases are different and the negative powers are the same

+
+
+\[a^{-n} \cdot b^{-n} = (a \cdot b)^{-n} = 1 \div (a \cdot b)^n\] +
+
+
+
+

When the bases and the negative powers are different

+
+
+\[a^{-n} \cdot b^{-m} = (a^{-n}) \cdot (b^{-m})\] +
+
+
+
+

Exponents of exponents

+ +
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Fractions.adoc b/Algrebra/Fractions.adoc new file mode 100644 index 0000000..99ede46 --- /dev/null +++ b/Algrebra/Fractions.adoc @@ -0,0 +1,81 @@ += Algrebra | Fractions +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Dividing Fractions +When dividing two fractions, you can flip the second fraction and solve as if +they were being multiplied to get your answer + +[stem] +++++ +\frac{1}{2} \div \frac{2}{1} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} +++++ + +== Rationalizing the denominator +Rationalizing is the process of multiplying a surd with another similar surd, to +result in a rational number. The surd that is used to multiply is called the +rationalizing factor (RF). + +In math, a conjugate refers to a pair of binomials that have the same terms but +opposite signs between them. + +Here we can multiply the top and bottom of the fraction by the conjugate to +simplify the denomitor +[stem] +++++ +\frac{10}{6+\sqrt{5}} * frac{6-\sqrt{5}}{6-\sqrt{5}} = +\frac{10(6-\sqrt{5})}{(6+\sqrt{5})(6-\sqrt{4})} = +\frac{10(6-\sqrt{5})}{36-\sqrt{25}} = +\frac{10(6-\sqrt{5})}{36-5} = +\frac{10(6-\sqrt{5})}{31} +++++ + +== Canceling in fractions + +=== Example 1 +[stem] +++++ +\frac{x^5y^2}{x^3} = \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x \cdot z} +++++ + +[stem] +++++ +\frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot y \cdot y}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot z} +++++ + +[stem] +++++ +\frac{x^2y^2}{z} +++++ + +=== Example 2 + +[stem] +++++ +\frac{a^2b^3(xy - pq)^4(c + d)}{ab^4z(xy - pq)^3} +++++ + +[stem] +++++ +\frac{a(xy - pq)(c + d)}{bz} +++++ + +=== When you can't cancel + +The stem:[+ 1] makes you unable to cancel. + +[stem] +++++ +\frac{a^2b^3(xy - pq)^4(c + d) + 1}{ab^4z(xy - pq)^3} +++++ + +=== Links +* https://www.youtube.com/watch?v=5j8a75aHaSE[The Oganic Chemestry Tutor | Rationalizing the denominator] +* https://www.cuemath.com/numbers/rationalize-the-denominator/[cue math | rationalizing the denominator] +* https://www.mathsisfun.com/algebra/conjugate.html[mathisfun | conjugate] diff --git a/Algrebra/Fractions.html b/Algrebra/Fractions.html new file mode 100644 index 0000000..dc78df7 --- /dev/null +++ b/Algrebra/Fractions.html @@ -0,0 +1,1183 @@ + + + + + + + +Algrebra | Fractions + + + + + + + + + + + + + + + + + + +
+
+

Dividing Fractions

+
+
+

When dividing two fractions, you can flip the second fraction and solve as if +they were being multiplied to get your answer

+
+
+
+\$\frac{1}{2} \div \frac{2}{1} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\$ +
+
+
+
+
+

Rationalizing the denominator

+
+
+

Rationalizing is the process of multiplying a surd with another similar surd, to +result in a rational number. The surd that is used to multiply is called the +rationalizing factor (RF).

+
+
+

In math, a conjugate refers to a pair of binomials that have the same terms but +opposite signs between them.

+
+
+

Here we can multiply the top and bottom of the fraction by the conjugate to +simplify the denomitor

+
+
+
+\$\frac{10}{6+\sqrt{5}} * frac{6-\sqrt{5}}{6-\sqrt{5}} = +\frac{10(6-\sqrt{5})}{(6+\sqrt{5})(6-\sqrt{4})} = +\frac{10(6-\sqrt{5})}{36-\sqrt{25}} = +\frac{10(6-\sqrt{5})}{36-5} = +\frac{10(6-\sqrt{5})}{31}\$ +
+
+
+
+
+

Canceling in fractions

+
+
+

Example 1

+
+
+\$\frac{x^5y^2}{x^3} = \frac{x \cdot x \cdot x \cdot x \cdot x}{x \cdot x \cdot x \cdot z}\$ +
+
+
+
+\$\frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot y \cdot y}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot z}\$ +
+
+
+
+\$\frac{x^2y^2}{z}\$ +
+
+
+
+

Example 2

+
+
+\$\frac{a^2b^3(xy - pq)^4(c + d)}{ab^4z(xy - pq)^3}\$ +
+
+
+
+\$\frac{a(xy - pq)(c + d)}{bz}\$ +
+
+
+
+

When you can’t cancel

+
+

The \$+ 1\$ makes you unable to cancel.

+
+
+
+\$\frac{a^2b^3(xy - pq)^4(c + d) + 1}{ab^4z(xy - pq)^3}\$ +
+
+
+
+

Links

+
+ +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Logarithms.adoc b/Algrebra/Logarithms.adoc new file mode 100644 index 0000000..63dd55d --- /dev/null +++ b/Algrebra/Logarithms.adoc @@ -0,0 +1,244 @@ += Algrebra | Logarithms +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== What is a Logarithm? + +[stem] +++++ +2^3 = 8 +++++ + +Is the same as + +[stem] +++++ +\log_{3}8 = 3 +++++ + +Can be an inverse of an exponential function + +[.center] +image::logorithm-exponent-inverse-graph.png[] + +== Properties + +[%header, cols="1,2a", options="header"] +|=== +| Property Name | Formula + +| **Logarithm of 1** +| [stem] +++++ +\log_c 1 = 0 +++++ + +| **Logarithm of the Base** +| [stem] +++++ +\log_c c = 1 +++++ + +| **Product Property** +| [stem] +++++ +\log_c(ab) = \log_c a + \log_c b +++++ + +| **Quotient Property** +| [stem] +++++ +\log_c\left(\frac{a}{b}\right) = \log_c a - \log_c b +++++ + +| **Power Property** +| [stem] +++++ +\log_c a^b = b \log_c a +++++ + +| **Change of Base Formula** +| [stem] +++++ +\log_a b = \frac{\log_c b}{\log_c a} +++++ + +| **Inverse Property (Base to Log)** +| [stem] +++++ +\log_a a^b = b +++++ + +| **Inverse Property (Log to Base)** +| [stem] +++++ +a^{\log_a b} = b +++++ +|=== + +== Common Logarithms (Base 10) + +The **Common Logarithm** is a logarithm with base 10. It is frequently used in +science and engineering (for example, in the pH scale or Richter scale +calculations). + +When a logarithm is written without an explicit base, such as stem:[\log x], it +is generally assumed to be the common logarithm (base 10). + +=== Standard Notation +[stem] +++++ +\log x = \log_{10} x +++++ + +=== Common Values +[cols="1,2a", options="header"] +|=== +| Expression | Result +| stem:[\log 10] | 1 +| stem:[\log 100] | 2 +| stem:[\log 1] | 0 +| stem:[\log 0.1] | -1 +|=== + +NOTE: In some advanced mathematics and computer science contexts, stem:[\log x] + may refer to the natural logarithm (base stem:[e]). Always check your + specific field's conventions! + +== Natural Logarithms (Base stem:[e]) + +The **Natural Logarithm** is a logarithm with the base stem:[e], where stem:[e \approx 2.71828]. It is widely used in physics, biology, and economics to describe growth and decay. + +Instead of writing stem:[\log_e], we use the specific notation **stem:[\ln]**. + +=== Standard Notation +[stem] +++++ +\ln x = \log_{e} x +++++ + +=== Fundamental Properties +[cols="1,2a", options="header"] +|=== +| Expression | Result +| stem:[\ln e] | 1 +| stem:[\ln 1] | 0 +| stem:[\ln e^x] | stem:[x] +| stem:[e^{\ln x}] | stem:[x] +|=== + +[IMPORTANT] +==== +The relationship between the exponential function and the natural log is inverse: +[stem] +++++ +y = e^x \iff x = \ln y +++++ +==== + +=== Properties of the Natural Logarithm (\ln) + +[cols="1,2a", options="header"] +|=== +| Name/Type | Property & Condition + +2+^| *Fundamental Values* + +| Log of 1 +| +[stem] +++++ +\ln 1 = 0 +++++ + +| Log of base e +| +[stem] +++++ +\ln e = 1 +++++ + +2+^| *Algebraic Properties* + +| Product Property +| +[stem] +++++ +\ln(xy) = \ln x + \ln y \quad (x, y > 0) +++++ + +| Quotient Property +| +[stem] +++++ +\ln(x/y) = \ln x - \ln y \quad (x, y > 0) +++++ + +| Power Property +| +[stem] +++++ +\ln(x^y) = y \ln x \quad (x > 0) +++++ + +| Root Property +| +[stem] +++++ +\ln(\sqrt[y]{x}) = \frac{\ln x}{y} \quad (x > 0, y \neq 0) +++++ + +2+^| *Comparison and Limits* + +| Monotonicity +| +[stem] +++++ +\ln x < \ln y \quad (0 < x < y) +++++ + +| Limits +| +[stem] +++++ +\lim_{x \to 0^{+}} \ln x = -\infty, \quad \lim_{x \to \infty} \ln x = \infty +++++ + +2+^| *Important Specific Limits* + +| At x=0 +| +[stem] +++++ +\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1 +++++ + +| Limit Form +| +[stem] +++++ +\lim_{\alpha \to 0} \frac{x^{\alpha} - 1}{\alpha} = \ln x \quad (x > 0) +++++ + +2+^| *Inequalities* + +| Standard +| +[stem] +++++ +\frac{x - 1}{x} \le \ln x \le x - 1 \quad (x > 0) +++++ + +| Alternate +| +[stem] +++++ +\ln(1 + x^\alpha) \le \alpha x \quad (x \ge 0, \alpha \ge 1) +++++ +|=== diff --git a/Algrebra/Logarithms.html b/Algrebra/Logarithms.html new file mode 100644 index 0000000..db449c1 --- /dev/null +++ b/Algrebra/Logarithms.html @@ -0,0 +1,1477 @@ + + + + + + + +Algrebra | Logarithms + + + + + + + + + + + + + + + + + + +
+
+

What is a Logarithm?

+
+
+
+\[2^3 = 8\] +
+
+
+

Is the same as

+
+
+
+\[\log_{3}8 = 3\] +
+
+
+

Can be an inverse of an exponential function

+
+
+
+logorithm exponent inverse graph +
+
+
+
+
+

Properties

+
+ ++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Property NameFormula

Logarithm of 1

+
+\[\log_c 1 = 0\] +
+

Logarithm of the Base

+
+\[\log_c c = 1\] +
+

Product Property

+
+\[\log_c(ab) = \log_c a + \log_c b\] +
+

Quotient Property

+
+\[\log_c\left(\frac{a}{b}\right) = \log_c a - \log_c b\] +
+

Power Property

+
+\[\log_c a^b = b \log_c a\] +
+

Change of Base Formula

+
+\[\log_a b = \frac{\log_c b}{\log_c a}\] +
+

Inverse Property (Base to Log)

+
+\[\log_a a^b = b\] +
+

Inverse Property (Log to Base)

+
+\[a^{\log_a b} = b\] +
+
+
+
+
+

Common Logarithms (Base 10)

+
+
+

The Common Logarithm is a logarithm with base 10. It is frequently used in +science and engineering (for example, in the pH scale or Richter scale +calculations).

+
+
+

When a logarithm is written without an explicit base, such as \(\log x\), it +is generally assumed to be the common logarithm (base 10).

+
+
+

Standard Notation

+
+
+\[\log x = \log_{10} x\] +
+
+
+
+

Common Values

+ ++++ + + + + + + + + + + + + + + + + + + + + + + + + +
ExpressionResult

\(\log 10\)

+

1

+

\(\log 100\)

+

2

+

\(\log 1\)

+

0

+

\(\log 0.1\)

+

-1

+
+
+ + + + + +
+ + +In some advanced mathematics and computer science contexts, \(\log x\) + may refer to the natural logarithm (base \(e\)). Always check your + specific field’s conventions! +
+
+
+
+
+
+

Natural Logarithms (Base \(e\))

+
+
+

The Natural Logarithm is a logarithm with the base \(e\), where \(e \approx 2.71828\). It is widely used in physics, biology, and economics to describe growth and decay.

+
+
+

Instead of writing \(\log_e\), we use the specific notation \(\ln\).

+
+
+

Standard Notation

+
+
+\[\ln x = \log_{e} x\] +
+
+
+
+

Fundamental Properties

+ ++++ + + + + + + + + + + + + + + + + + + + + + + + + +
ExpressionResult

\(\ln e\)

+

1

+

\(\ln 1\)

+

0

+

\(\ln e^x\)

+

\(x\)

+

\(e^{\ln x}\)

+

\(x\)

+
+
+ + + + + +
+ + +
+

The relationship between the exponential function and the natural log is inverse:

+
+
+
+\[y = e^x \iff x = \ln y\] +
+
+
+
+
+
+

Properties of the Natural Logarithm (\ln)

+ ++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Name/TypeProperty & Condition

Fundamental Values

Log of 1

+
+\[\ln 1 = 0\] +
+

Log of base e

+
+\[\ln e = 1\] +
+

Algebraic Properties

Product Property

+
+\[\ln(xy) = \ln x + \ln y \quad (x, y > 0)\] +
+

Quotient Property

+
+\[\ln(x/y) = \ln x - \ln y \quad (x, y > 0)\] +
+

Power Property

+
+\[\ln(x^y) = y \ln x \quad (x > 0)\] +
+

Root Property

+
+\[\ln(\sqrt[y]{x}) = \frac{\ln x}{y} \quad (x > 0, y \neq 0)\] +
+

Comparison and Limits

Monotonicity

+
+\[\ln x < \ln y \quad (0 < x < y)\] +
+

Limits

+
+\[\lim_{x \to 0^{+}} \ln x = -\infty, \quad \lim_{x \to \infty} \ln x = \infty\] +
+

Important Specific Limits

At x=0

+
+\[\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1\] +
+

Limit Form

+
+\[\lim_{\alpha \to 0} \frac{x^{\alpha} - 1}{\alpha} = \ln x \quad (x > 0)\] +
+

Inequalities

Standard

+
+\[\frac{x - 1}{x} \le \ln x \le x - 1 \quad (x > 0)\] +
+

Alternate

+
+\[\ln(1 + x^\alpha) \le \alpha x \quad (x \ge 0, \alpha \ge 1)\] +
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Polynomials.adoc b/Algrebra/Polynomials.adoc new file mode 100644 index 0000000..bb6c7c5 --- /dev/null +++ b/Algrebra/Polynomials.adoc @@ -0,0 +1,255 @@ += Algrebra | Polynomials +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Standard Form of a Polynomial +[stem] +++++ +P(x) = a_{n}X^n + a_{n-1}X^(n-1) + a_{n-2}X^(n-2) + ... + a_{1}X + a_{0} +++++ + +[highlight]#stem:[a_{n},a_{n-1},a_{n-2}]# are coefficients of +[highlight]#stem:[x^n,x^(n-1),x^(n-2)]# and constant term respectively and it +should belong to real number (stem:[\exists\mathbb{R}]). + +== Types of Polynomials +=== Binomial +[stem] +++++ +ax^m+bx^n +++++ +=== Quadradic Equation + +.Standard Form +[stem] +++++ +ax^2 + bx + c +++++ + +.Vertex Form +[stem] +++++ +y = a(x-h)^2 + k +++++ + +.Intercept form +[stem] +++++ +y = a(x-p)(x - q) +++++ + +=== Line Slope-Intercept Form +[stem] +++++ +y = mx + b +++++ +m:: slope +b:: y-intercept + +=== Line https://www.youtube.com/watch?v=yoHs1h5qtuQ[Point-Slope Form] +I you have one point of the line and a slope you can figure out more points on +the line with this formula + +[stem] +++++ +(y - y_{1}) = m(x - x_{1}) +++++ +m:: slope +stem:[x_{1},y_{1}]:: any pont on the line + + +== Zeros of Polynomial (Roots of the equation) +Values of the variable such that the polynomial equals 0. Often represented as +stem:[\alpha], stem:[\beta], and y respectively. + +[.center] +image::polynomial-roots.png[] + +WARNING: This image is not related to the below equation + +[.inline] +* *Positive real root*: is a positive number (like stem:[1], stem:[5], or + stem:[\sqrt{2}]) that, when plugged in for stem:[x], makes stem:[P(x) = 0]. + +[.inline] +* *Negative real root*: is a negative number (like stem:[-1], stem:[-3], or +stem:[-\pi]) that makes stem:[P(x) = 0]. + +[.inline] +* *Imaginary roots* (often part of Complex roots, written as stem:[a \pm bi]) + are solutions that involve the imaginary unit stem:[i] (stem:[\sqrt{-1}]). You + cannot plot an imaginary number on a standard Cartesian stem:[(x,y)] grap + +.example equation +[stem] +++++ +f(x) = x^3 - 2x^2 - x + 2 +++++ + +=== Finding the number of total real zeros +The degree of the polynomial represents the total number of real zeros. + +=== Finding the number of positive real zeros +Descrates Rule of Signs:: + The number of positive real zeros is either equal to the number of sign + changes, or less a positive even integer + +According To *Descartes Rule of signs* The numnber of positive real zeros could +be 2, because there were to sign changes, or 0 because 2 - 2 is zero and 2 is a +positive even integer. + +=== Finding the number of negative real zeros +First look at the equation as if you added a negative symbol to x. + +[stem] +++++ +f(-x) = (-x)^3 - 2(-x)^2 - (-x) + 2 \ +\ = -z^3 - 2x^2 + x + 2 +++++ + +The number of negative real zeros, is either equal to the number of sign +changes, or less than a positive even integer in the equation. + +There is one sign change so the answer is one. One is the only answer because +one minus two would be negative one and you cannot have a negative number of +negative real zeros. + +=== Finding the number of imaginary real zeros +In this case if we look at the chart, we know that there has to be three total +real zeros. If we fall into the situation where there are 0 positive real zeros +than there would have to be 2 imaginary real zeros. + +[%header,cols="1,1,1,4"] +|=== +|Pos | Negative | Imaginary | Total +|2 +|1 +|0 +|3 + +|0 +|1 +|2 +|3 +|=== + +== Factoring Trinomials +=== Leading Coeficient Of One +1. Find two numbers that when multiplied together equal the constant term + but when added together equal the first degree term. + +2. Take the two numbers from step one and put it in the multiplication of two + binomials in the form of stem:[(x+a)(x+b)], where `a` is the first number + number and `b` is the second term. + +Example: +[stem] +++++ +x^2 + 5x + 6 = 0 +++++ + +answer:: stem:[(x+3)(x+2)] + +=== Leading Coeficient Other Than One +1. Multiply the leading coeficient by the constant. + +2. Find two numbers that when multiplied equal the number from step 1, but when + added equals the first degree term. + +3. Replace the first degree term with the two numbers from step 2 added + together, and make sure they are multiplied by the same variable of the first + degree term. + +4. Factor until you get two binimals mulitiplied together + +Example: +[stem] +++++ +2x^2 - 3x - 2 +++++ + +1. stem:[2 * -2 = 4] + +2. stem:[-4 + 1 = -3\ and\ -4 * 1 = -4] + +3. stem:[2x^2 - 4x + 1x -2] + +4. stem:[2x(x-2) + 1(x - 2)] -> stem:[(x-2)(2x+1)] + +== Quadradic Trinomial Formula +[stem] +++++ +x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} +++++ + +== Discriminant +The part under the square root of the Quadradic Formula. + +[stem] +++++ +D = b^2-4ac +++++ + +stem:[D\gt0]:: 2 real solutions, factorable if perfect square +stem:[D=0]:: 1 real solution +stem:[D<0]:: 2 imaginary solutions + +== Important Factoring +=== Difference of Squares +[stem] +++++ +a^2-b^2 = (a+b)(a-b) +++++ + +=== Factoring Perfact cubes +[stem] +++++ +A^3 + B^3 = (A+B)(A^2-AB+B^2) +++++ + +[stem] +++++ +A^3 - B^3 = (A-B)(A^2+AB+B^2) +++++ + +== Standard Form to Vertex Form +1. Take the first degree term and take half of it squared added to both sides +2. factor + +example: +[stem] +++++ +y = x^2 + 6x - 5 +++++ + +1. stem:[y+3^2=x^2+6x+3^2-5] +2. stem:[y + 9 = (x + 3)^2-5] -> -9 from both sides -> stem:[y=(x+3)^2-14] + +== Perfect Square Trinomials +[stem] +++++ +a^2 + 2ab + b^2 +\ +or +\ +a^2 - 2ab + b^2 +++++ + +[.center] +image::Perfect-Square-Trinomial.png[] + +== Links +- https://www.desmos.com/calculator[Desmos Graphing] +- https://www.cuemath.com/algebra/perfect-square-trinomial/[perfect square trinomials] +- https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-functions-equations/x2f8bb11595b61c86:vertex-form/v/vertex-form-intro[vertex-form] +- https://www.youtube.com/watch?v=2MKigAgPZMQ[completing the square] +- https://www.cuemath.com/algebra/polynomials/[cuemath polynomials] +- https://www.cuemath.com/algebra/standard-form-of-quadratic-equation/[quadradic eqautions] +- https://www.cuemath.com/algebra/zeros-of-polynomial/[zerso of polynomial] +- https://www.youtube.com/watch?v=YaU5JTe3cPU[OCT Descartes Rules of Signs] diff --git a/Algrebra/Polynomials.html b/Algrebra/Polynomials.html new file mode 100644 index 0000000..1f8dfa2 --- /dev/null +++ b/Algrebra/Polynomials.html @@ -0,0 +1,1568 @@ + + + + + + + +Algrebra | Polynomials + + + + + + + + + + + + + + + + + + +
+
+

Standard Form of a Polynomial

+
+
+
+\$P(x) = a_{n}X^n + a_{n-1}X^(n-1) + a_{n-2}X^(n-2) + ... + a_{1}X + a_{0}\$ +
+
+
+

\$a_{n},a_{n-1},a_{n-2}\$ are coefficients of +\$x^n,x^(n-1),x^(n-2)\$ and constant term respectively and it +should belong to real number (\$\exists\mathbb{R}\$).

+
+
+
+
+

Types of Polynomials

+
+
+

Binomial

+
+
+\$ax^m+bx^n\$ +
+
+
+
+

Quadradic Equation

+
+
Standard Form
+
+\$ax^2 + bx + c\$ +
+
+
+
Vertex Form
+
+\$y = a(x-h)^2 + k\$ +
+
+
+
Intercept form
+
+\$y = a(x-p)(x - q)\$ +
+
+
+
+

Line Slope-Intercept Form

+
+
+\$y = mx + b\$ +
+
+
+
+
m
+
+

slope

+
+
b
+
+

y-intercept

+
+
+
+
+
+

Line Point-Slope Form

+
+

I you have one point of the line and a slope you can figure out more points on +the line with this formula

+
+
+
+\$(y - y_{1}) = m(x - x_{1})\$ +
+
+
+
+
m
+
+

slope

+
+
\$x_{1},y_{1}\$
+
+

any pont on the line

+
+
+
+
+
+
+
+

Zeros of Polynomial (Roots of the equation)

+
+
+

Values of the variable such that the polynomial equals 0. Often represented as +\$\alpha\$, \$\beta\$, and y respectively.

+
+
+
+polynomial roots +
+
+
+ + + + + +
+ + +This image is not related to the below equation +
+
+
+
    +
  • +

    Positive real root: is a positive number (like \$1\$, \$5\$, or +\$\sqrt{2}\$) that, when plugged in for \$x\$, makes \$P(x) = 0\$.

    +
    • +
+
+
+
    +
  • +

    Negative real root: is a negative number (like \$-1\$, \$-3\$, or +\$-\pi\$) that makes \$P(x) = 0\$.

    +
    • +
+
+
+
    +
  • +

    Imaginary roots (often part of Complex roots, written as \$a \pm bi\$) +are solutions that involve the imaginary unit \$i\$ (\$\sqrt{-1}\$). You +cannot plot an imaginary number on a standard Cartesian \$(x,y)\$ grap

    +
    • +
+
+
+
example equation
+
+\$f(x) = x^3 - 2x^2 - x + 2\$ +
+
+
+

Finding the number of total real zeros

+
+

The degree of the polynomial represents the total number of real zeros.

+
+
+
+

Finding the number of positive real zeros

+
+
+
Descrates Rule of Signs
+
+

The number of positive real zeros is either equal to the number of sign +changes, or less a positive even integer

+
+
+
+
+

According To Descartes Rule of signs The numnber of positive real zeros could +be 2, because there were to sign changes, or 0 because 2 - 2 is zero and 2 is a +positive even integer.

+
+
+
+

Finding the number of negative real zeros

+
+

First look at the equation as if you added a negative symbol to x.

+
+
+
+\$f(-x) = (-x)^3 - 2(-x)^2 - (-x) + 2\$ +\$\ = -z^3 - 2x^2 + x + 2\$ +
+
+
+

The number of negative real zeros, is either equal to the number of sign +changes, or less than a positive even integer in the equation.

+
+
+

There is one sign change so the answer is one. One is the only answer because +one minus two would be negative one and you cannot have a negative number of +negative real zeros.

+
+
+
+

Finding the number of imaginary real zeros

+
+

In this case if we look at the chart, we know that there has to be three total +real zeros. If we fall into the situation where there are 0 positive real zeros +than there would have to be 2 imaginary real zeros.

+
+ ++++++ + + + + + + + + + + + + + + + + + + + + + + +
PosNegativeImaginaryTotal

2

1

0

3

0

1

2

3

+
+
+
+
+

Factoring Trinomials

+
+
+

Leading Coeficient Of One

+
+
    +
  1. +

    Find two numbers that when multiplied together equal the constant term +but when added together equal the first degree term.

    +
    2. +
  2. +

    Take the two numbers from step one and put it in the multiplication of two +binomials in the form of \$(x+a)(x+b)\$, where a is the first number +number and b is the second term.

    +
    3. +
+
+
+

Example:

+
+
+
+\$x^2 + 5x + 6 = 0\$ +
+
+
+
+
answer
+
+

\$(x+3)(x+2)\$

+
+
+
+
+
+

Leading Coeficient Other Than One

+
+
    +
  1. +

    Multiply the leading coeficient by the constant.

    +
    2. +
  2. +

    Find two numbers that when multiplied equal the number from step 1, but when +added equals the first degree term.

    +
    3. +
  3. +

    Replace the first degree term with the two numbers from step 2 added +together, and make sure they are multiplied by the same variable of the first +degree term.

    +
    4. +
  4. +

    Factor until you get two binimals mulitiplied together

    +
    5. +
+
+
+

Example:

+
+
+
+\$2x^2 - 3x - 2\$ +
+
+
+
    +
  1. +

    \$2 * -2 = 4\$

    +
    2. +
  2. +

    \$-4 + 1 = -3\ and\ -4 * 1 = -4\$

    +
    3. +
  3. +

    \$2x^2 - 4x + 1x -2\$

    +
    4. +
  4. +

    \$2x(x-2) + 1(x - 2)\$ → \$(x-2)(2x+1)\$

    +
    5. +
+
+
+
+
+
+

Quadradic Trinomial Formula

+
+
+
+\$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\$ +
+
+
+
+
+

Discriminant

+
+
+

The part under the square root of the Quadradic Formula.

+
+
+
+\$D = b^2-4ac\$ +
+
+
+
+
\$D\gt0\$
+
+

2 real solutions, factorable if perfect square

+
+
\$D=0\$
+
+

1 real solution

+
+
\$D<0\$
+
+

2 imaginary solutions

+
+
+
+
+
+
+

Important Factoring

+
+
+

Difference of Squares

+
+
+\$a^2-b^2 = (a+b)(a-b)\$ +
+
+
+
+

Factoring Perfact cubes

+
+
+\$A^3 + B^3 = (A+B)(A^2-AB+B^2)\$ +
+
+
+
+\$A^3 - B^3 = (A-B)(A^2+AB+B^2)\$ +
+
+
+
+
+
+

Standard Form to Vertex Form

+
+
+
    +
  1. +

    Take the first degree term and take half of it squared added to both sides

    +
    2. +
  2. +

    factor

    +
    3. +
+
+
+

example:

+
+
+
+\$y = x^2 + 6x - 5\$ +
+
+
+
    +
  1. +

    \$y+3^2=x^2+6x+3^2-5\$

    +
    2. +
  2. +

    \$y + 9 = (x + 3)^2-5\$ → -9 from both sides → \$y=(x+3)^2-14\$

    +
    3. +
+
+
+
+
+

Perfect Square Trinomials

+
+
+
+\$a^2 + 2ab + b^2 +\$ +\$or +\$ +\$a^2 - 2ab + b^2\$ +
+
+
+
+Perfect Square Trinomial +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/README b/Algrebra/README new file mode 100644 index 0000000..6b69212 --- /dev/null +++ b/Algrebra/README @@ -0,0 +1 @@ +https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab[3Blue 1Brown] diff --git a/Algrebra/Radicals.adoc b/Algrebra/Radicals.adoc new file mode 100644 index 0000000..6983325 --- /dev/null +++ b/Algrebra/Radicals.adoc @@ -0,0 +1,138 @@ += Algrebra | Radicals +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} + +== Core Properties of Radicals + +[cols="^.^1a, ^.^1a, <.^2a", frame="none", grid="rows", options="header"] +|=== +| Rule Name | Math (LaTeX) | Details + +| Product Rule +| +[stem] +++++ +\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b} +++++ +| +[%collapsible] +==== +The root of a product is equal to the product of the roots. This is essential for simplifying radicals. +==== + +| Quotient Rule +| +[stem] +++++ +\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} +++++ +| +[%collapsible] +==== +The root of a fraction is equal to the root of the numerator divided by the root of the denominator. (Note: stem:[b \neq 0]). +==== + +| Root of a Power +| +[stem] +++++ +\sqrt[n]{a\vphantom{[}^m} = (\sqrt[n]{a})^m +++++ +| +[%collapsible] +==== +You can take the power before or after taking the root; the result remains the same. +==== + +| Fractional Exponent +| +[stem] +++++ +\sqrt[n]{a\vphantom{[}^m} = a^{\frac{m}{n}} +++++ +| +[%collapsible] +==== +This is the "Bridge" rule. It allows you to convert any radical into a power, making it easier to solve using standard exponent laws. +==== + +| Root of a Root +| +[stem] +++++ +\sqrt[m]{\sqrt[n]{a}} = \sqrt[m \cdot n]{a} +++++ +| +[%collapsible] +==== +When nesting radicals, you multiply the indices together. For example: + +[stem] +++++ +\sqrt{\sqrt[3]{x}} = \sqrt[6]{x} +++++ + +==== + +| Inverse Property +| +[stem] +++++ +(\sqrt[n]{a})^n = a +++++ +| +[%collapsible] +==== +Raising an stem:[n]-th root to the stem:[n]-th power cancels the radical out. +==== +|=== + +== Breaking down roots with Prime factors + +[stem] +++++ +\sqrt{255} +++++ + +[stem] +++++ +\begin{array}{c} + 225 \\ + \land \\ + \bigcirc \kern-0.7em 5 \quad \quad 45 \\ + \hspace{1.3cm} \land \\ + \hspace{1.4cm} \bigcirc \kern-0.7em 5 \quad \ 9 \\ + \hspace{2.2cm} \land \\ + \hspace{2.3cm} \bigcirc \kern-0.7em 3 \quad \bigcirc \kern-0.7em 3 +\end{array} +++++ + + +[stem] +++++ +\sqrt{5 * 5 * 3 * 3} +++++ + +take a number out of each pair of number + +NOTE: If there wasn't numbers under the square root that didn't have pairs, they + would be left under the square root. + +[stem] +++++ +5 * 3\sqrt{1} +++++ + +[stem] +++++ +\ = 15 +++++ + +== Rules + +* No negatives under roots. diff --git a/Algrebra/Radicals.html b/Algrebra/Radicals.html new file mode 100644 index 0000000..dc9e6e4 --- /dev/null +++ b/Algrebra/Radicals.html @@ -0,0 +1,1272 @@ + + + + + + + +Algrebra | Radicals + + + + + + + + + + + + + + + + + + +
+
+

Core Properties of Radicals

+
+ +++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Rule NameMath (LaTeX)Details
+

Product Rule

+
+
+\[\sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}\] +
+
+Details +
+
+

The root of a product is equal to the product of the roots. This is essential for simplifying radicals.

+
+
+
+

Quotient Rule

+
+
+\[\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}\] +
+
+Details +
+
+

The root of a fraction is equal to the root of the numerator divided by the root of the denominator. (Note: \(b \neq 0\)).

+
+
+
+

Root of a Power

+
+
+\[\sqrt[n]{a\vphantom{[}^m} = (\sqrt[n]{a})^m\] +
+
+Details +
+
+

You can take the power before or after taking the root; the result remains the same.

+
+
+
+

Fractional Exponent

+
+
+\[\sqrt[n]{a\vphantom{[}^m} = a^{\frac{m}{n}}\] +
+
+Details +
+
+

This is the "Bridge" rule. It allows you to convert any radical into a power, making it easier to solve using standard exponent laws.

+
+
+
+

Root of a Root

+
+
+\[\sqrt[m]{\sqrt[n]{a}} = \sqrt[m \cdot n]{a}\] +
+
+Details +
+
+

When nesting radicals, you multiply the indices together. For example:

+
+
+
+\[\sqrt{\sqrt[3]{x}} = \sqrt[6]{x}\] +
+
+
+
+

Inverse Property

+
+
+\[(\sqrt[n]{a})^n = a\] +
+
+Details +
+
+

Raising an \(n\)-th root to the \(n\)-th power cancels the radical out.

+
+
+
+
+
+
+

Breaking down roots with Prime factors

+
+
+
+\[\sqrt{255}\] +
+
+
+
+\[\begin{array}{c} + 225 \\ + \land \\ + \bigcirc \kern-0.7em 5 \quad \quad 45 \\ + \hspace{1.3cm} \land \\ + \hspace{1.4cm} \bigcirc \kern-0.7em 5 \quad \ 9 \\ + \hspace{2.2cm} \land \\ + \hspace{2.3cm} \bigcirc \kern-0.7em 3 \quad \bigcirc \kern-0.7em 3 +\end{array}\] +
+
+
+
+\[\sqrt{5 * 5 * 3 * 3}\] +
+
+
+

take a number out of each pair of number

+
+
+ + + + + +
+ + +If there wasn’t numbers under the square root that didn’t have pairs, they + would be left under the square root. +
+
+
+
+\[5 * 3\sqrt{1}\] +
+
+
+
+\[\ = 15\] +
+
+
+
+
+

Rules

+
+
+
    +
  • +

    No negatives under roots.

    +
    • +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Sequences.adoc b/Algrebra/Sequences.adoc new file mode 100644 index 0000000..6504fc0 --- /dev/null +++ b/Algrebra/Sequences.adoc @@ -0,0 +1,111 @@ += Algrebra | Sequences +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Arithmetic Sequence +Has a common difference. + +The below has a common difference of 4: +[stem] +++++ +3,7,11,15,19,23,27 +++++ +=== Find Specific Index + +[stem] +++++ +a_{n} = a_{1}+(n-1)d +++++ + +=== Find Partial Sum + +[stem] +++++ +Sn = (\frac{a_{1}+a_{n}}{2})n +++++ + +== Geometric Sequence +Has a common ratio. + +The below has a common ratio of 2: +[stem] +++++ +3,6,12,24,48,96,192 +++++ + +=== Find Specific Index +[stem] +++++ +a_{n} = a_{1}(r)^{n-1} +++++ + +=== Find Geometric Mean + +The goemetric mean of two numbers is another number between the two such that +there's a common ration between (1) the frist number and the mean and (2) the +mean and the second number. The goemetric mean of 4 and 9 is 6, because the +ration stem:[\frac{4}{6}] is the same as the ratio stem:[\frac{6}{9}]. Both +ratios are equal to stem:[\frac{2}{3}]. + +image::geometric-mean-and-geometric-sequences.png[] + +The goemetric mean will get you the number in between a and b in the geometric +sequence + +==== Formulas + +.standard notation + +[stem] +++++ +G = \sqrt[n]{x_1 \cdot x_2 \cdot \dots \cdot x_n} +++++ + +.product notation + +[stem] +++++ +G = \left( \prod_{i=1}^{n} x_i \right)^{\frac{1}{n}} +++++ + + +In many technical applications (like computer science or complex statistics), it +is easier to calculate the geometric mean by taking the average of the +logarithms. This prevents "overflow" when multiplying many large numbers +together: + +.Logarithmic Form +[stem] +++++ +\ln G = \frac{1}{n} \sum_{i=1}^{n} \ln x_i +++++ + +which can be written as: + +.Logarithmic Form v2 +[stem] +++++ +G = \exp \left( \frac{1}{n} \sum_{i=1}^{n} \ln x_i \right) +++++ + +.When to use it + +While the arithmetic mean (the standard average) is used for +additive data, the geometric mean is specifically used for: + + +=== Find Partial Sum +[stem] +++++ +Sn = \frac{a_{1}(1-r^n)}{1-r} +++++ + +== Links + +* https://www.youtube.com/watch?v=xt07TExQNVg[Organic Chemistry tutor | calculating geometric mean] diff --git a/Algrebra/Sequences.html b/Algrebra/Sequences.html new file mode 100644 index 0000000..5c247f6 --- /dev/null +++ b/Algrebra/Sequences.html @@ -0,0 +1,1219 @@ + + + + + + + +Algrebra | Sequences + + + + + + + + + + + + + + + + + + +
+
+

Arithmetic Sequence

+
+
+

Has a common difference.

+
+
+

The below has a common difference of 4:

+
+
+
+\[3,7,11,15,19,23,27\] +
+
+
+

Find Specific Index

+
+
+\[a_{n} = a_{1}+(n-1)d\] +
+
+
+
+

Find Partial Sum

+
+
+\[Sn = (\frac{a_{1}+a_{n}}{2})n\] +
+
+
+
+
+
+

Geometric Sequence

+
+
+

Has a common ratio.

+
+
+

The below has a common ratio of 2:

+
+
+
+\[3,6,12,24,48,96,192\] +
+
+
+

Find Specific Index

+
+
+\[a_{n} = a_{1}(r)^{n-1}\] +
+
+
+
+

Find Geometric Mean

+
+

The goemetric mean of two numbers is another number between the two such that +there’s a common ration between (1) the frist number and the mean and (2) the +mean and the second number. The goemetric mean of 4 and 9 is 6, because the +ration \(\frac{4}{6}\) is the same as the ratio \(\frac{6}{9}\). Both +ratios are equal to \(\frac{2}{3}\).

+
+
+
+geometric mean and geometric sequences +
+
+
+

The goemetric mean will get you the number in between a and b in the geometric +sequence

+
+
+

Formulas

+
+
standard notation
+
+\[G = \sqrt[n]{x_1 \cdot x_2 \cdot \dots \cdot x_n}\] +
+
+
+
product notation
+
+\[G = \left( \prod_{i=1}^{n} x_i \right)^{\frac{1}{n}}\] +
+
+
+

In many technical applications (like computer science or complex statistics), it +is easier to calculate the geometric mean by taking the average of the +logarithms. This prevents "overflow" when multiplying many large numbers +together:

+
+
+
Logarithmic Form
+
+\[\ln G = \frac{1}{n} \sum_{i=1}^{n} \ln x_i\] +
+
+
+

which can be written as:

+
+
+
Logarithmic Form v2
+
+\[G = \exp \left( \frac{1}{n} \sum_{i=1}^{n} \ln x_i \right)\] +
+
+
+
When to use it
+

While the arithmetic mean (the standard average) is used for +additive data, the geometric mean is specifically used for:

+
+
+
+
+

Find Partial Sum

+
+
+\[Sn = \frac{a_{1}(1-r^n)}{1-r}\] +
+
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Terminology.adoc b/Algrebra/Terminology.adoc new file mode 100644 index 0000000..dcd325d --- /dev/null +++ b/Algrebra/Terminology.adoc @@ -0,0 +1,44 @@ += Algrebra | Terminology +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +[%header, cols="1,2a"] +|=== +| Name | Definition and Example +| Surd +| Irrational numbers that cannot be expressed as exact fractions or + terminating decimals, typically represented in square root form. They are used + for precise calculations where exact values are necessary. + +[stem] +++++ +\sqrt{4}, \sqrt[3]{8}, \sqrt{25} +++++ + +| Exact Fraction +| An exact fraction is a fraction that can be expressed as a decimal without + repeating or terminating + +[stem] +++++ +\frac{1}{2}, \frac{3}{4} +++++ + +NOTE: Fractions with denominators that are powers of 2 or 5 (like 4, 5, 8, 10, + etc.) typically have exact decimal representation + +| Irrational Numbers +| A real number that *cannot* be written as a simple fraction + +image::rational-irational-numbers-example.png[] + +|=== + +== Links +* https://www.cuemath.com/numbers/surds/[cuemath | surds] diff --git a/Algrebra/Terminology.html b/Algrebra/Terminology.html new file mode 100644 index 0000000..80c599b --- /dev/null +++ b/Algrebra/Terminology.html @@ -0,0 +1,1149 @@ + + + + + + + +Algrebra | Terminology + + + + + + + + + + + + + + + + + + +
+
+
+ ++++ + + + + + + + + + + + + + + + + + + + + +
NameDefinition and Example

Surd

+

Irrational numbers that cannot be expressed as exact fractions or + terminating decimals, typically represented in square root form. They are used + for precise calculations where exact values are necessary.

+
+
+
+\[\sqrt{4}, \sqrt[3]{8}, \sqrt{25}\] +
+

Exact Fraction

+

An exact fraction is a fraction that can be expressed as a decimal without + repeating or terminating

+
+
+
+\[\frac{1}{2}, \frac{3}{4}\] +
+
+
+ + + + + +
+ + +Fractions with denominators that are powers of 2 or 5 (like 4, 5, 8, 10, + etc.) typically have exact decimal representation +
+

Irrational Numbers

+

A real number that cannot be written as a simple fraction

+
+
+
+rational irational numbers example +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/Triangulation.adoc b/Algrebra/Triangulation.adoc new file mode 100644 index 0000000..9374586 --- /dev/null +++ b/Algrebra/Triangulation.adoc @@ -0,0 +1,12 @@ += Algrebra | Triangulation +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Links +* https://www.youtube.com/watch?v=nIRBApwA-OY&t=14s[YouTube | triangulator with trigonometry] diff --git a/Algrebra/Triangulation.html b/Algrebra/Triangulation.html new file mode 100644 index 0000000..d471210 --- /dev/null +++ b/Algrebra/Triangulation.html @@ -0,0 +1,1082 @@ + + + + + + + +Algrebra | Triangulation + + + + + + + + + + + + + + + + + + +
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/complex_numbers.adoc b/Algrebra/complex_numbers.adoc new file mode 100644 index 0000000..aa1bf78 --- /dev/null +++ b/Algrebra/complex_numbers.adoc @@ -0,0 +1,11 @@ += Algrebra | Complex numbers +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} + +== Links +* https://www.youtube.com/watch?v=hPhhxwgk0A0[HackRF Michael Ossmann, Complex Numbers] diff --git a/Algrebra/complex_numbers.html b/Algrebra/complex_numbers.html new file mode 100644 index 0000000..3432205 --- /dev/null +++ b/Algrebra/complex_numbers.html @@ -0,0 +1,1082 @@ + + + + + + + +Algrebra | Complex numbers + + + + + + + + + + + + + + + + + + +
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/index.adoc b/Algrebra/index.adoc new file mode 100644 index 0000000..f9e59dd --- /dev/null +++ b/Algrebra/index.adoc @@ -0,0 +1,22 @@ += ./Algrebra +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Polynomials.html;hb=HEAD[Polynomials] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Division.html;hb=HEAD[Division] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Triangulation.html;hb=HEAD[Triangulation] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/linear_inequalities.html;hb=HEAD[linear_inequalities] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Radicals.html;hb=HEAD[Radicals] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Sequences.html;hb=HEAD[Sequences] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Logarithms.html;hb=HEAD[Logarithms] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Angles.html;hb=HEAD[Angles] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Exponents.html;hb=HEAD[Exponents] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/complex_numbers.html;hb=HEAD[complex_numbers] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/rules.html;hb=HEAD[rules] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Terminology.html;hb=HEAD[Terminology] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/Fractions.html;hb=HEAD[Fractions] diff --git a/Algrebra/index.html b/Algrebra/index.html new file mode 100644 index 0000000..04b7495 --- /dev/null +++ b/Algrebra/index.html @@ -0,0 +1,1088 @@ + + + + + + + +./Algrebra + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Algrebra/linear_inequalities.adoc b/Algrebra/linear_inequalities.adoc new file mode 100644 index 0000000..2310a59 --- /dev/null +++ b/Algrebra/linear_inequalities.adoc @@ -0,0 +1,16 @@ += Algrebra | Linear Inequalities +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Multiplying or dividing by negative numbers + +image::linear_inequalities.png[] + +When you multiply or divide by a negative number when solving for linear +inequalities make sure to change the direction of the inequality sign. diff --git a/Algrebra/linear_inequalities.html b/Algrebra/linear_inequalities.html new file mode 100644 index 0000000..686835c --- /dev/null +++ b/Algrebra/linear_inequalities.html @@ -0,0 +1,1084 @@ + + + + + + + +Algrebra | Linear Inequalities + + + + + + + + + + + + + + + + + + +
+
+

Multiplying or dividing by negative numbers

+
+
+
+linear inequalities +
+
+
+

When you multiply or divide by a negative number when solving for linear +inequalities make sure to change the direction of the inequality sign.

+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Algrebra/rules.adoc b/Algrebra/rules.adoc new file mode 100644 index 0000000..e823dcc --- /dev/null +++ b/Algrebra/rules.adoc @@ -0,0 +1,70 @@ += Algrebra | Rules +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Properties + +.Commutative Property of Addition + +[stem] +++++ +a + b = b + a +++++ + +.Commutative Property of Multiplication + +[stem] +++++ +a * b = b * a +++++ + +.Associative Property of Addition + +[stem] +++++ +(a + b) + c = a + (b + c) +++++ + +.Associative Property of Multiplication + +[stem] +++++ +(a * b) * c = a * (b * c) +++++ + +.Distributive Properties of Addition and Multiplication + + +[stem] +++++ +a * (b + c) = a * b + a * c +++++ + +and + +[stem] +++++ +(a + b) * c = a * c + b * c +++++ + + +image::https://o.quizlet.com/3gRs.XX5diEtgknc3VGaNA_b.png[] + +=== Links +* https://www.analyzemath.com/algebra/rules_algebra.html + +== Order of Operations (PEMDAS) +Please excuse my dear aunt sally + +- Parentheses +- Exponents +- Multiply +- Divide +- Add +- Subtract diff --git a/Algrebra/rules.html b/Algrebra/rules.html new file mode 100644 index 0000000..2d67026 --- /dev/null +++ b/Algrebra/rules.html @@ -0,0 +1,1163 @@ + + + + + + + +Algrebra | Rules + + + + + + + + + + + + + + + + + + +
+
+

Properties

+
+
+
Commutative Property of Addition
+
+\$a + b = b + a\$ +
+
+
+
Commutative Property of Multiplication
+
+\$a * b = b * a\$ +
+
+
+
Associative Property of Addition
+
+\$(a + b) + c = a + (b + c)\$ +
+
+
+
Associative Property of Multiplication
+
+\$(a * b) * c = a * (b * c)\$ +
+
+
+
Distributive Properties of Addition and Multiplication
+
+\$a * (b + c) = a * b + a * c\$ +
+
+
+

and

+
+
+
+\$(a + b) * c = a * c + b * c\$ +
+
+
+
+3gRs.XX5diEtgknc3VGaNA b +
+
+
+

Links

+
+ +
+
+
+
+
+

Order of Operations (PEMDAS)

+
+
+

Please excuse my dear aunt sally

+
+
+
    +
  • +

    Parentheses

    +
    • +
  • +

    Exponents

    +
    • +
  • +

    Multiply

    +
    • +
  • +

    Divide

    +
    • +
  • +

    Add

    +
    • +
  • +

    Subtract

    +
    • +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Asymptotes.adoc b/Calculus/Asymptotes.adoc new file mode 100644 index 0000000..8b3a8a4 --- /dev/null +++ b/Calculus/Asymptotes.adoc @@ -0,0 +1,27 @@ += Calculus | Asymptotes +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Solving for Vertical Asymptotes + +[stem] +++++ +f(x) = frac{x+4}{(x+4)(x-8)} +++++ + +* `-4` is removable +* `8` is non-removable + +In this function there would be a hole at `-4` + +== Solving for Horizontal and Slant Asymptotes + +== Links + +* https://youtu.be/kvhpu1TkSjI?si=sZV82P255DY-4mhS[The Orgainic Chemistry Tutor | Horizontal Asymptotes and Slant Asymptotes of Rational Functions] diff --git a/Calculus/Asymptotes.html b/Calculus/Asymptotes.html new file mode 100644 index 0000000..bbef716 --- /dev/null +++ b/Calculus/Asymptotes.html @@ -0,0 +1,1113 @@ + + + + + + + +Calculus | Asymptotes + + + + + + + + + + + + + + + + + + +
+
+

Solving for Vertical Asymptotes

+
+
+
+\$f(x) = frac{x+4}{(x+4)(x-8)}\$ +
+
+
+
    +
  • +

    -4 is removable

    +
    • +
  • +

    8 is non-removable

    +
    • +
+
+
+

In this function there would be a hole at -4

+
+
+
+
+

Solving for Horizontal and Slant Asymptotes

+
+ +
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Derivatives.adoc b/Calculus/Derivatives.adoc new file mode 100644 index 0000000..44fcdb4 --- /dev/null +++ b/Calculus/Derivatives.adoc @@ -0,0 +1,511 @@ += Calculus | Derivatives +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Derivative Formula + +[stem] +++++ +\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a) +++++ + +[.center] +image::Derivative-Formula.png[] + +[.center] +image::3Blue1Brown-Limit-definition.png[] + +Keep in mind that that stem:[\frac{df}{dx}] or stem:[\frac{d}{dx}] is just the +short had version of this definition. + +image::short-hand-derivative.png[] + +== Basic Rules + +=== Constant Rule +[stem] +++++ +\frac{d}{dx} [c] = 0 +++++ + +=== Power Rule +[stem] +++++ +\frac{d}{dx} [x^n] = nx^(n-1) +++++ + +=== Product Rule +[stem] +++++ +\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) +++++ + +=== Quotient Rule +[stem] +++++ +\frac{d}{dx} [\frac{f(x)}{g(x)}] = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} +++++ + +=== Chain Rule +[stem] +++++ +\frac{d}{dx} [f(g(x))] = f'(g(x)) * g'(x) +++++ + +== Points of Interest + +[.center] +image::points-of-interest-derivatives.png[] + +=== Concaving + +NOTE: Down looks like a frown. Up looks like a cup. + +Whenever a function is concave _down_, it's derivative (and slope) are +_decreasing_; werever a function is concave up, it's derivative (and slope) are +_increasing_. + +=== Points + +[cols="1,1,2a", options="header"] +|=== +| Concept | Key Points | Mathematical Description + +| **Local Maxima** +| b, g +| * <- Slope is positive +* -> Slope is negative + +| **Inflection Points** +| c, e, h +| * The point at which the slope changes. +* Steepest or least steep points in their immediate neighborhoods. + +| **The Corner** +| j +| * No tangent line means no slope. +* No slope means no derivative. +* The derivative is undefined at this point. + +| **Critical Points & Numbers** +| b, d, g, i, j +| * **Critical Points:** Where the derivative is either zero or undefined. +* **Critical Numbers:** The x-values of these critical points. +|=== + +Local Extrma:: + local maximums and minimums + +:stem: latexmath + +== Finding Critical numbers + +Find the critical numbers of +[stem] +++++ +f(x) = 3x^5 - 20x^3 +++++ + +1. Find the first derivative of f using the power rule. + +[stem] +++++ +f(x) = 3x^5 - 20x^3 +\ +f'(x) = 15x^4 - 60x^2 +++++ + +[start=2] +2. Set the derivative equal to zero and solve for x. + +[stem] +++++ +\begin{aligned} + & 15x^4 - 60x^2 = 0 \\ + & 15x^2(x^2 - 4) = 0 \\ + & 15x^2(x+2)(x-2) = 0 \\ + & 15x^2 = 0\ \ or\ \ x + 2 = 0\ \ or\ \ x - 2 = 0 \\ + & x = 0\ \ or\ \ x = -2\ \ or\ \ x = 2 +\end{aligned} +++++ + +These three X-Values are crtical number of f. + +WARNING: Additional critical numbers could exists if the first derivate were + undefined at some x-values, but that is not the case here. + +[.center] +image::Finding-Critical-Numbers.png[] + +=== The first derivative test + +Take a number line and put down the critical numbers you found above 0, -2, and +2. + +[.center] +image::critical-numbers-number-line.png[] + +Pick a value from each region, plug it into the first dierivate and note whether +your result is positive or negative + +[stem] +++++ +\begin{aligned} + & f'(x) = 15x^4 - 60x^2 \\[10pt] + & f'(-3) = 15(-3)^4 - 60(-3)^2 = 15 \cdot 81 - 60 \cdot 9 = 675 \\ + & f'(-1) = 15(-1)^4 - 60(-1)^2 = 15 - 60 = -45 \\ + & f'(1) = 15(1)^4 - 60(1)^2 = 15 - 60 = -45 \\ + & f'(3) = 15(3)^4 - 60(3)^2 = 15 \cdot 81 - 60 \cdot 9 = 675 +\end{aligned} +++++ + +Mark were the positives and negatives are. + +[.center] +image::critical-numbers-number-line-2.png[] + +* *local maximum*: -2 +* *local minimum*: 2 + +Lastly, plug in the x values + +[stem] +++++ +\begin{aligned} + & f(x) = 3x^5 - 20x^3 \\[15pt] + & f(-2) = 3(-2)^5 - 20(-2)^3 = 64 \\ + & f(2) = 3(2)^5 - 20(2)^3 = -64 +\end{aligned} +++++ + +The local max is located at stem:[(-2,64)] and the local min is at +stem:[(2,-64)]. + +=== The second derivative test + +* *positive value*: concave up + +* *negative value*: concave down + +* *zero value*: _inconclusive_. Could be concave up, concave down, or there + could be an inflection point there. + +Find the second derivative and then plug in the critical numbers you found + +* *-2*: concave down +* *2*: concave up +* *0*: inconclusive + +[stem] +++++ +\begin{aligned} + f(x) &= 3x^5 - 20x^3 \\[5pt] + f'(x) &= 15x^4 - 60x^2 && \text{(power rule)} \\ + f''(x) &= 60x^3 - 120x && \text{(power rule)} \\[15pt] + f''(-2) &= 60(-2)^3 - 120(-2) = -240 \\ + f''(0) &= 60(0)^3 - 120(0) = 0 \\ + f''(2) &= 60(2)^3 - 120(2) = 240 +\end{aligned} +++++ + +* *-2*: concave down +* *2*: concave up + +WARNING: The second derivative test is no help where the first derivative is + undefined. + +=== Examples + +[%collapsible] +==== +Find the local exteme of stem:[g(x) = 2x - 3x^{2/3} + 4] + +[.center] +image::derivative-figure-11-5.png[] + +1. *Find the first derivative of g.* + ++ +[stem] +++++ +\begin{aligned} + g(x) &= 2x - 3x^{2/3} + 4 \\[5pt] + g'(x) &= 2 - 2x^{1/3} && \text{(power rule)} \\ +\end{aligned} +++++ + +2. *Set the derivative equal to zero and solve* + ++ +[stem] +++++ +\begin{aligned} + 2 - 2x^{-1/3} &= 0 \\ + -2x^{-1/3} &= -2 \\ + x^{-1/3} &= 1 \\ + (x^{-1/3})^{-3} &= 1^{-3} \\ + x &= 1 +\end{aligned} +++++ + ++ +Thus 1 is a critical number + + +3. *Determine whether the first derivative is undefined for any x-values* + ++ +stem:[2x^{-1/3}] equals latexmath:[\frac{2}{\sqrt[3\]{x}}], you'd have +latexmath:[\frac{2}{0}] which is undefined. + + ++ +stem:[0] is another critical number. + +4. *Plot the critical numbers on a number line, and then use the first* + *derivative test to figure out the sign of each region.* + ++ +You can use -1, 0.5, and 2 as test numbers: + ++ +[stem] +++++ +\begin{aligned} + g'(x) &= 2 - 2x^{-1/3} \\[15pt] + g'(-1) &= 4 && \text{(pos.)} \\ + g'(0.5) &\approx -0.52 && \text{(neg.)} \\ + g'(2) &\approx 0.41 && \text{(pos.)} +\end{aligned} +++++ + ++ +[.center] +image::figure-11-6-critical-points-graph.png[] + ++ +Because the derivative of _g_ switches from positive to negative at 0 there's a +local max there. And becaause the first derivative switches from negative to +positive atr 1, there's a local min at stem:[x = 1]. + +5. *Plug the critical numbers into g to obtain the function values (the* + *heights) of these two local extrema.* + ++ +[stem] +++++ +\begin{aligned} + g(x) = 2x - 3x^{2/3} + 4 \quad & g(0) = 4 \\ + & g(1) = 3 +\end{aligned} +++++ + +So, there's a #local max at (0,4)# and a #local min at (1,3)#. You're done. + +You could have used the second derivative test instead of the first derivative +test in Step 4. First, you need the second derivative of latexmath:[g], which +is, as you know, the derivative of its first derivative: + +[latexmath] +++++ +\begin{aligned} + g'(x) &= 2 - 2x^{-1/3} \\ + g''(x) &= \frac{2}{3}x^{-4/3} +\end{aligned} +++++ + +Evaluate the second derivative at latexmath:[x = 1] (the critical number from +Step 2): + +[latexmath] +++++ +g''(1) = \frac{2}{3} +++++ + +:stem: latexmath + +Because latexmath:[g''(1)] is positive, you know that latexmath:[g] is concave +up at latexmath:[x = 1] and, therefore, that there's a local min there. + +At the other critical number, latexmath:[x = 0] (from Step 3), the first +derivative is undefined. The second derivative test is no help where the first +derivative is undefined, so you've got to use the first derivative test for that +critical number. + +==== + +:stem: + +== Derivatives to Memorize + +=== Logrithmic Derivative + +[.inline] +- stem:[\frac{d}{dx}\[\ln(u)\] = \frac{u'}{u}] + +=== Exponential Derivative + +[.inline] +- stem:[\frac{d}{dx}\[e^u\] = e^u * u'] + + +=== Trigonometric Derivatives + +[.inline] +- stem:[\frac{d}{dx}\[\sinx\] = \cosx] + +[.inline] +- stem:[\frac{d}{dx}\[\cosx\] = -\sinx] + +[.inline] +- stem:[\frac{d}{dx}\[\tanx\] = \sec^2x] + +[.inline] +- stem:[\frac{d}{dx}\[\secx\] = \secx\tanx] + +[.inline] +- stem:[\frac{d}{dx}\[\cotx\] = -\csc^2x] + +[.inline] +- stem:[\frac{d}{dx}\[\cscx\] = -\cscx\cotx] + +=== Inverse Trigonometric Derivatives + +[.inline] +- stem:[\arcsin(x) = \frac{1}{\sqrt{1 - x^2}}\ \text{(for |x| < 1)}] + +[.inline] +- stem:[\arccos(x) = \frac{-1}{\sqrt{1 - x^2}}\ \text{(for |x| < 1)}] + +[.inline] +- stem:[\arctan(x) = \frac{1}{1 - x^2}\ \text{(for all x)}] + +[.inline] +- stem:[\arccot(x) = \frac{-1}{1 + x^2}\ \text{(for all x)}] + +[.inline] +- stem:[\arcsec(x) = \frac{1}{|x| \sqrt{x^2 - 1}}\ \text{(for |x| > 1)}] + +[.inline] +- stem:[\arccsc(x) = \frac{-1}{|x| \sqrt{x^2 - 1}}\ \text{(for |x| > 1)}] + +:stem: latexmath + +== Examples + +.Problem 1 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{Find the equation of the tangent line to the curve } x^3 + 4xy^2 + y^3 = 107 \text{ at the point } (2, 3) \\ + & \text{using implicit differentiation.} \\[10pt] + & \frac{d}{dx} \left[ x^3 + (4x)(y^2) + y^3 = 107 \right] \\[10pt] + & 3x^2 + 4(y^2) + 4x(2y) \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0 \\[10pt] + & 3(2)^2 + 4(3)^2 + 4(2)(2 \cdot 3) \frac{dy}{dx} + 3(3)^2 \frac{dy}{dx} = 0 \\[10pt] + & 12 + 36 + 8(6) \frac{dy}{dx} + 27 \frac{dy}{dx} = 0 \\[10pt] + & 48 \frac{dy}{dx} + 27 \frac{dy}{dx} = -48 +\end{aligned} +++++ + +NOTE: Whenever you differentiate a *stem:[y]* variable with respect to + *stem:[x]* you need to add a stem:[\frac{dy}{dx}] in front of it + +[stem] +++++ +\begin{aligned} + & m_{\text{tan}} = \frac{dy}{dx} = \frac{-48 \div 3}{75 \div 3} = -\frac{16}{25} \\[15pt] + & y - y_1 = m(x - x_1) \\[10pt] + & \boxed{y - 3 = -\frac{16}{25}(x - 2)} +\end{aligned} +++++ + +==== + +.Problem 2 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} \qquad f(x) = \sin x \\[15pt] + & f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\[15pt] + & f'(x) = ? \qquad f'(x) = \boxed{\cos x} +\end{aligned} +++++ + +NOTE: Notice how the limit is similar to the formal definition of a derivative +==== + +.Critical numbers +[%collapsible] +==== +[stem] +++++ +\text{10. Identify all intervals where } f(x) \text{ is increasing given } f(x) = x^3 + \frac{3}{2}x^2 - 36x - 9. +++++ + +[stem] +++++ +\begin{aligned} + & f'(c) = 0 \quad f'(x) = 3x^2 + \frac{3}{2}(2x) - 36(1) + 0 \\ + & f'(x) = 3x^2 + 3x - 36 = 0 \\ + & 0 = 3(x^2 + 1x - 12) \qquad \begin{aligned} c &= -4 \\ c &= 3 \end{aligned} \\ + & 0 = 3(x + 4)(x - 3) +\end{aligned} +++++ + +Now plug in a number for each section of this number line, that will give you +whether numbers are increasing or decreasing in that section. + +[.center] +image::critical-numbers-example-1.png[] + +[stem] +++++ +\text{Inc } \boxed{(-\infty, -4) \cup (3, \infty)} +++++ + +You can identify your local maximum and minimums like this + +[.center] +image::local-minimum-and-maximum.png[] + +==== + +.Related Rates 1 +[%collapsible] +==== +[.center] +image::water-cylinder.png[] + +[stem] +++++ +\begin{aligned} + & \text{9. Water is flowing into a cylinder with a diameter of } 6\text{ft and a height of } 10\text{ft.} \\ + & \text{If the height of the water in the cylinder is increasing at } 3\text{ft/min, at what} \\ + & \text{rate is the volume of the water in the cylinder changing?} \\[10pt] + & V = Ah \qquad \frac{dh}{dt} = +3 \text{ ft/min} \\ + & [V = (\pi R^2)h] \frac{d}{dt} \\[10pt] + & \frac{dV}{dt} = \pi R^2 (1) \frac{dh}{dt} \\[10pt] + & \frac{dV}{dt} = \pi(3)^2(3) \\[10pt] + & \frac{dV}{dt} = 27\pi \text{ ft}^3\text{/min} +\end{aligned} +++++ + + +NOTE: Constants don't get derived but variables do. A constant is something that + doesn't change while a variable is changing. +==== + + +== Links +- https://www.derivative-calculator.net/[Derivative Calculator] +- https://www.youtube.com/playlist?list=PLTEBkBE3HYK9XpUFla9Th86wPfR1XxFlQ[The Organic Chemistry Tutor | Related Rates] +- https://www.youtube.com/watch?v=WmBzmHru78w[The Organic Chemistry Tutor | Calculus 1 Final Exam] diff --git a/Calculus/Derivatives.html b/Calculus/Derivatives.html new file mode 100644 index 0000000..9e9823e --- /dev/null +++ b/Calculus/Derivatives.html @@ -0,0 +1,1902 @@ + + + + + + + +Calculus | Derivatives + + + + + + + + + + + + + + + + + + +
+
+

Derivative Formula

+
+
+
+\$\lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a)\$ +
+
+
+
+Derivative Formula +
+
+
+
+3Blue1Brown Limit definition +
+
+
+

Keep in mind that that \$\frac{df}{dx}\$ or \$\frac{d}{dx}\$ is just the +short had version of this definition.

+
+
+
+short hand derivative +
+
+
+
+
+

Basic Rules

+
+
+

Constant Rule

+
+
+\$\frac{d}{dx} [c] = 0\$ +
+
+
+
+

Power Rule

+
+
+\$\frac{d}{dx} [x^n] = nx^(n-1)\$ +
+
+
+
+

Product Rule

+
+
+\$\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\$ +
+
+
+
+

Quotient Rule

+
+
+\$\frac{d}{dx} [\frac{f(x)}{g(x)}] = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\$ +
+
+
+
+

Chain Rule

+
+
+\$\frac{d}{dx} [f(g(x))] = f'(g(x)) * g'(x)\$ +
+
+
+
+
+
+

Points of Interest

+
+
+
+points of interest derivatives +
+
+
+

Concaving

+
+ + + + + +
+ + +Down looks like a frown. Up looks like a cup. +
+
+
+

Whenever a function is concave down, it’s derivative (and slope) are +decreasing; werever a function is concave up, it’s derivative (and slope) are +increasing.

+
+
+
+

Points

+ +++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
ConceptKey PointsMathematical Description

Local Maxima

b, g

+
    +
  • +

    ← Slope is positive

    +
    • +
  • +

    → Slope is negative

    +
    • +
+

Inflection Points

c, e, h

+
    +
  • +

    The point at which the slope changes.

    +
    • +
  • +

    Steepest or least steep points in their immediate neighborhoods.

    +
    • +
+

The Corner

j

+
    +
  • +

    No tangent line means no slope.

    +
    • +
  • +

    No slope means no derivative.

    +
    • +
  • +

    The derivative is undefined at this point.

    +
    • +
+

Critical Points & Numbers

b, d, g, i, j

+
    +
  • +

    Critical Points: Where the derivative is either zero or undefined.

    +
    • +
  • +

    Critical Numbers: The x-values of these critical points.

    +
    • +
+
+
+
+
Local Extrma
+
+

local maximums and minimums

+
+
+
+
+
+
+
+

Finding Critical numbers

+
+
+

Find the critical numbers of

+
+
+
+\[f(x) = 3x^5 - 20x^3\] +
+
+
+
    +
  1. +

    Find the first derivative of f using the power rule.

    +
    2. +
+
+
+
+\[f(x) = 3x^5 - 20x^3 +\ +f'(x) = 15x^4 - 60x^2\] +
+
+
+
    +
  1. +

    Set the derivative equal to zero and solve for x.

    +
    2. +
+
+
+
+\[\begin{aligned} + & 15x^4 - 60x^2 = 0 \\ + & 15x^2(x^2 - 4) = 0 \\ + & 15x^2(x+2)(x-2) = 0 \\ + & 15x^2 = 0\ \ or\ \ x + 2 = 0\ \ or\ \ x - 2 = 0 \\ + & x = 0\ \ or\ \ x = -2\ \ or\ \ x = 2 +\end{aligned}\] +
+
+
+

These three X-Values are crtical number of f.

+
+
+ + + + + +
+ + +Additional critical numbers could exists if the first derivate were + undefined at some x-values, but that is not the case here. +
+
+
+
+Finding Critical Numbers +
+
+
+

The first derivative test

+
+

Take a number line and put down the critical numbers you found above 0, -2, and +2.

+
+
+
+critical numbers number line +
+
+
+

Pick a value from each region, plug it into the first dierivate and note whether +your result is positive or negative

+
+
+
+\[\begin{aligned} + & f'(x) = 15x^4 - 60x^2 \\[10pt] + & f'(-3) = 15(-3)^4 - 60(-3)^2 = 15 \cdot 81 - 60 \cdot 9 = 675 \\ + & f'(-1) = 15(-1)^4 - 60(-1)^2 = 15 - 60 = -45 \\ + & f'(1) = 15(1)^4 - 60(1)^2 = 15 - 60 = -45 \\ + & f'(3) = 15(3)^4 - 60(3)^2 = 15 \cdot 81 - 60 \cdot 9 = 675 +\end{aligned}\] +
+
+
+

Mark were the positives and negatives are.

+
+
+
+critical numbers number line 2 +
+
+
+
    +
  • +

    local maximum: -2

    +
    • +
  • +

    local minimum: 2

    +
    • +
+
+
+

Lastly, plug in the x values

+
+
+
+\[\begin{aligned} + & f(x) = 3x^5 - 20x^3 \\[15pt] + & f(-2) = 3(-2)^5 - 20(-2)^3 = 64 \\ + & f(2) = 3(2)^5 - 20(2)^3 = -64 +\end{aligned}\] +
+
+
+

The local max is located at \((-2,64)\) and the local min is at +\((2,-64)\).

+
+
+
+

The second derivative test

+
+
    +
  • +

    positive value: concave up

    +
    • +
  • +

    negative value: concave down

    +
    • +
  • +

    zero value: inconclusive. Could be concave up, concave down, or there +could be an inflection point there.

    +
    • +
+
+
+

Find the second derivative and then plug in the critical numbers you found

+
+
+
    +
  • +

    -2: concave down

    +
    • +
  • +

    2: concave up

    +
    • +
  • +

    0: inconclusive

    +
    • +
+
+
+
+\[\begin{aligned} + f(x) &= 3x^5 - 20x^3 \\[5pt] + f'(x) &= 15x^4 - 60x^2 && \text{(power rule)} \\ + f''(x) &= 60x^3 - 120x && \text{(power rule)} \\[15pt] + f''(-2) &= 60(-2)^3 - 120(-2) = -240 \\ + f''(0) &= 60(0)^3 - 120(0) = 0 \\ + f''(2) &= 60(2)^3 - 120(2) = 240 +\end{aligned}\] +
+
+
+
    +
  • +

    -2: concave down

    +
    • +
  • +

    2: concave up

    +
    • +
+
+
+ + + + + +
+ + +The second derivative test is no help where the first derivative is + undefined. +
+
+
+
+

Examples

+
+Details +
+
+

Find the local exteme of \(g(x) = 2x - 3x^{2/3} + 4\)

+
+
+
+derivative figure 11 5 +
+
+
+
    +
  1. +

    Find the first derivative of g.

    +
    +
    +\[\begin{aligned} + g(x) &= 2x - 3x^{2/3} + 4 \\[5pt] + g'(x) &= 2 - 2x^{1/3} && \text{(power rule)} \\ +\end{aligned}\] +
    +
    +
    2. +
  2. +

    Set the derivative equal to zero and solve

    +
    +
    +\[\begin{aligned} + 2 - 2x^{-1/3} &= 0 \\ + -2x^{-1/3} &= -2 \\ + x^{-1/3} &= 1 \\ + (x^{-1/3})^{-3} &= 1^{-3} \\ + x &= 1 +\end{aligned}\] +
    +
    +
    +

    Thus 1 is a critical number

    +
    +
    3. +
  3. +

    Determine whether the first derivative is undefined for any x-values

    +
    +

    \(2x^{-1/3}\) equals \(\frac{2}{\sqrt[3]{x}}\), you’d have +\(\frac{2}{0}\) which is undefined.

    +
    +
    +

    \(0\) is another critical number.

    +
    +
    4. +
  4. +

    Plot the critical numbers on a number line, and then use the first +derivative test to figure out the sign of each region.

    +
    +

    You can use -1, 0.5, and 2 as test numbers:

    +
    +
    +
    +\[\begin{aligned} + g'(x) &= 2 - 2x^{-1/3} \\[15pt] + g'(-1) &= 4 && \text{(pos.)} \\ + g'(0.5) &\approx -0.52 && \text{(neg.)} \\ + g'(2) &\approx 0.41 && \text{(pos.)} +\end{aligned}\] +
    +
    +
    +
    +figure 11 6 critical points graph +
    +
    +
    +

    Because the derivative of g switches from positive to negative at 0 there’s a +local max there. And becaause the first derivative switches from negative to +positive atr 1, there’s a local min at \(x = 1\).

    +
    +
    5. +
  5. +

    Plug the critical numbers into g to obtain the function values (the +heights) of these two local extrema.

    +
    +
    +\[\begin{aligned} + g(x) = 2x - 3x^{2/3} + 4 \quad & g(0) = 4 \\ + & g(1) = 3 +\end{aligned}\] +
    +
    +
    6. +
+
+
+

So, there’s a local max at (0,4) and a local min at (1,3). You’re done.

+
+
+

You could have used the second derivative test instead of the first derivative +test in Step 4. First, you need the second derivative of \(g\), which +is, as you know, the derivative of its first derivative:

+
+
+
+\[\begin{aligned} + g'(x) &= 2 - 2x^{-1/3} \\ + g''(x) &= \frac{2}{3}x^{-4/3} +\end{aligned}\] +
+
+
+

Evaluate the second derivative at \(x = 1\) (the critical number from +Step 2):

+
+
+
+\[g''(1) = \frac{2}{3}\] +
+
+
+

Because \(g''(1)\) is positive, you know that \(g\) is concave +up at \(x = 1\) and, therefore, that there’s a local min there.

+
+
+

At the other critical number, \(x = 0\) (from Step 3), the first +derivative is undefined. The second derivative test is no help where the first +derivative is undefined, so you’ve got to use the first derivative test for that +critical number.

+
+
+
+
+
+
+
+

Derivatives to Memorize

+
+
+

Logrithmic Derivative

+
+
    +
  • +

    \$\frac{d}{dx}\[\ln(u)] = \frac{u'}{u}\$

    +
    • +
+
+
+
+

Exponential Derivative

+
+
    +
  • +

    \$\frac{d}{dx}\[e^u] = e^u * u'\$

    +
    • +
+
+
+
+

Trigonometric Derivatives

+
+
    +
  • +

    \$\frac{d}{dx}\[\sinx] = \cosx\$

    +
    • +
+
+
+
    +
  • +

    \$\frac{d}{dx}\[\cosx] = -\sinx\$

    +
    • +
+
+
+
    +
  • +

    \$\frac{d}{dx}\[\tanx] = \sec^2x\$

    +
    • +
+
+
+
    +
  • +

    \$\frac{d}{dx}\[\secx] = \secx\tanx\$

    +
    • +
+
+
+
    +
  • +

    \$\frac{d}{dx}\[\cotx] = -\csc^2x\$

    +
    • +
+
+
+
    +
  • +

    \$\frac{d}{dx}\[\cscx] = -\cscx\cotx\$

    +
    • +
+
+
+
+

Inverse Trigonometric Derivatives

+
+
    +
  • +

    \$\arcsin(x) = \frac{1}{\sqrt{1 - x^2}}\ \text{(for |x| < 1)}\$

    +
    • +
+
+
+
    +
  • +

    \$\arccos(x) = \frac{-1}{\sqrt{1 - x^2}}\ \text{(for |x| < 1)}\$

    +
    • +
+
+
+
    +
  • +

    \$\arctan(x) = \frac{1}{1 - x^2}\ \text{(for all x)}\$

    +
    • +
+
+
+
    +
  • +

    \$\arccot(x) = \frac{-1}{1 + x^2}\ \text{(for all x)}\$

    +
    • +
+
+
+
    +
  • +

    \$\arcsec(x) = \frac{1}{|x| \sqrt{x^2 - 1}}\ \text{(for |x| > 1)}\$

    +
    • +
+
+
+
    +
  • +

    \$\arccsc(x) = \frac{-1}{|x| \sqrt{x^2 - 1}}\ \text{(for |x| > 1)}\$

    +
    • +
+
+
+
+
+
+

Examples

+
+
+Problem 1 +
+
+
+\[\begin{aligned} + & \text{Find the equation of the tangent line to the curve } x^3 + 4xy^2 + y^3 = 107 \text{ at the point } (2, 3) \\ + & \text{using implicit differentiation.} \\[10pt] + & \frac{d}{dx} \left[ x^3 + (4x)(y^2) + y^3 = 107 \right] \\[10pt] + & 3x^2 + 4(y^2) + 4x(2y) \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0 \\[10pt] + & 3(2)^2 + 4(3)^2 + 4(2)(2 \cdot 3) \frac{dy}{dx} + 3(3)^2 \frac{dy}{dx} = 0 \\[10pt] + & 12 + 36 + 8(6) \frac{dy}{dx} + 27 \frac{dy}{dx} = 0 \\[10pt] + & 48 \frac{dy}{dx} + 27 \frac{dy}{dx} = -48 +\end{aligned}\] +
+
+
+ + + + + +
+ + +Whenever you differentiate a \(y\) variable with respect to + \(x\) you need to add a \(\frac{dy}{dx}\) in front of it +
+
+
+
+\[\begin{aligned} + & m_{\text{tan}} = \frac{dy}{dx} = \frac{-48 \div 3}{75 \div 3} = -\frac{16}{25} \\[15pt] + & y - y_1 = m(x - x_1) \\[10pt] + & \boxed{y - 3 = -\frac{16}{25}(x - 2)} +\end{aligned}\] +
+
+
+
+
+Problem 2 +
+
+
+\[\begin{aligned} + & \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} \qquad f(x) = \sin x \\[15pt] + & f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\[15pt] + & f'(x) = ? \qquad f'(x) = \boxed{\cos x} +\end{aligned}\] +
+
+
+ + + + + +
+ + +Notice how the limit is similar to the formal definition of a derivative +
+
+
+
+
+Critical numbers +
+
+
+\[\text{10. Identify all intervals where } f(x) \text{ is increasing given } f(x) = x^3 + \frac{3}{2}x^2 - 36x - 9.\] +
+
+
+
+\[\begin{aligned} + & f'(c) = 0 \quad f'(x) = 3x^2 + \frac{3}{2}(2x) - 36(1) + 0 \\ + & f'(x) = 3x^2 + 3x - 36 = 0 \\ + & 0 = 3(x^2 + 1x - 12) \qquad \begin{aligned} c &= -4 \\ c &= 3 \end{aligned} \\ + & 0 = 3(x + 4)(x - 3) +\end{aligned}\] +
+
+
+

Now plug in a number for each section of this number line, that will give you +whether numbers are increasing or decreasing in that section.

+
+
+
+critical numbers example 1 +
+
+
+
+\[\text{Inc } \boxed{(-\infty, -4) \cup (3, \infty)}\] +
+
+
+

You can identify your local maximum and minimums like this

+
+
+
+local minimum and maximum +
+
+
+
+
+Related Rates 1 +
+
+
+water cylinder +
+
+
+
+\[\begin{aligned} + & \text{9. Water is flowing into a cylinder with a diameter of } 6\text{ft and a height of } 10\text{ft.} \\ + & \text{If the height of the water in the cylinder is increasing at } 3\text{ft/min, at what} \\ + & \text{rate is the volume of the water in the cylinder changing?} \\[10pt] + & V = Ah \qquad \frac{dh}{dt} = +3 \text{ ft/min} \\ + & [V = (\pi R^2)h] \frac{d}{dt} \\[10pt] + & \frac{dV}{dt} = \pi R^2 (1) \frac{dh}{dt} \\[10pt] + & \frac{dV}{dt} = \pi(3)^2(3) \\[10pt] + & \frac{dV}{dt} = 27\pi \text{ ft}^3\text{/min} +\end{aligned}\] +
+
+
+ + + + + +
+ + +Constants don’t get derived but variables do. A constant is something that + doesn’t change while a variable is changing. +
+
+
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Integrals.adoc b/Calculus/Integrals.adoc new file mode 100644 index 0000000..a31ede8 --- /dev/null +++ b/Calculus/Integrals.adoc @@ -0,0 +1,432 @@ += Calculus | Integrals +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Funamental Theorem of Calculus + +image::fundemental-theorem-of-calculus-v1.png[width=80%] + +.version 1 +[stem] +++++ +A_{f}(x) = \int_{s}^{x} f(t)\ dt +++++ + +stem:[A_{f}(x)]:: + The area under the curve at point `x` for funtion `f()` + +`integral s`:: + The starting point + +`integral x`:: + The end point + +`dt`:: + An infinitely small width of each section + +// image::fundemental-theorem-of-calculus-v2-minus.png[width=75%] + +.version 2 +[stem] +++++ +\int_{a}^{b} f(x) \, dx = F(b) - F(a) +++++ + +`F()`:: + The antiderivative of `f()` +`x`:: + Represents all the values in the interval stem:[[a, b]] + + +.example +[%collapsible] +==== + +Find the area under the graph between 2 and 3: + +[.center] +image::fundemental-theorem-of-calculus-v2-example.png[] + +[stem] +++++ +\int_{2}^{3} (x^2+1)dx = F(3) - F(2) +++++ + +the antiderivative of stem:[x^2 + 1] is stem:[\frac{1}{3}x^3 + x] + +stem:[F(3) - F(2)] can be written as stem:[ \left[ \frac{1}{3}x^3 + x +\right\]_2^3 ], and thus, + +[stem] +++++ +\begin{aligned} +\int_{2}^{3} (x^2+1)dx &= \left[ \frac{1}{3}x^3 + x \right]_2^3 \\ +&= \left( \frac{1}{3} \cdot 3^3 + 3 \right) - \left( \frac{1}{3} \cdot 2^3 + 2 \right) \\ +&= 12 - 4\frac{2}{3} \\ +&= 7\frac{1}{3} +\end{aligned} +++++ +==== + +== Power Rule + +1. Bring the power in front where it will multiply the rest of the derivative + ++ +[stem] +++++ +5x^4 \rightarrow 4 * 5x^4 +++++ + +2. Reduce the power by one and simplify + ++ +[stem] +++++ +4 * 5x^4 \rightarrow 4 * 5x^3 = 20x^3 +++++ + ++ +thus, stem:[y'] = stem:[20x^3] + +== Reverse Power Rule + +1. Increase the power by one. + ++ +The 3 becomes a 4. + ++ +[stem] +++++ +20x^3 \rightarrow 20x^4 +++++ + +2. Divide by th enew power and simplify + ++ +[stem] +++++ +20x^4 \rightarrow \frac{20}{4}x^4 = 5x^4 +++++ + ++ +WARNING: The reverse power rule does not work for a power of negative one. + + +== "u" substitution + +find + +[stem] +++++ +\lim 2x \cos(x^2) dx +++++ + +. *Set _u_ equal to the argument of the main function.* + ++ +The argument of stem:[\cos(x^2)] is stem:[x^2], so you set stem:[u] equal to +stem:[x^2]. + +. *Take the derivative of _u_ with respect to _x_.* + ++ +[stem] +++++ +u = x^2\ \text{so}\ \frac{du}{dx} = 2x +++++ + +. *Solve for _dx_.* + ++ +[stem] +++++ +\begin{aligned} + & \frac{du}{dx} = \frac{2x}{1} \\[10pt] + & du = 2xdx\ \text{(cross multiplication)} \\ + & \frac{du}{2x}\ = dx\ \text{(dviding both sides by 2x)} +\end{aligned} +++++ + +. *Make substitutions* + ++ +In stem:[\int 2x \cos(x^2)dx], u takes the place of stem:[x^2] and +stem:[\frac{du}{2x}] takes the place of stem:[dx]. So now you've got +stem:[\int 2x\cos u \frac{du}{2x}]. The two stem:[2x]s cancel, giving you +stem:[\int \cos\ u\ du]. + +. *Anitdifferentiate using the simple reverse rule.* + ++ +[stem] +++++ +\lim \cos\ u\ du = \sin\ u + C +++++ + +. *Substitute stem:[x^2] back in for _u_, coming full circle.* + ++ +_u_ equals stem:[x^2], so stem:[x^2] goes in for the _u_: + ++ +[stem] +++++ +\lim \cos\ u\ du - \sin(x^2) + C +++++ + ++ +That's it. So stem:[\lim\ 2x\ cos(x^2)dx = \sin(x^2) + C] + +=== Examples + +.Example 1 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{8. Evaluate the integral shown below:} \\[10pt] + & \int 2x \sqrt{3x^2 + 5} \, dx \qquad \begin{aligned} u &= 3x^2 + 5 \\ du &= 6x \, dx \\ \frac{du}{6x} &= dx \end{aligned} \\[15pt] + & \int 2x \sqrt{u} \, \frac{du}{6x} \\[15pt] + & \frac{1}{3} \int u^{1/2} \, du = \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} + C \\[15pt] + & = \frac{2}{9} u^{3/2} + C = \boxed{\frac{2}{9} (3x^2 + 5)^{3/2} + C} +\end{aligned} +++++ +==== + + +== Integration by Parts + +[stem] +++++ +\int u \, dv = uv - \int v \, du +++++ + +NOTE: You can check "calculus for dumies" chapter 16 + +//// +[stem] +++++ +\begin{array}{|c|c|} +\hline +\rule[-1.2em]{0pt}{3em} \quad u \quad & \quad v \quad \\ \hline +\rule[-1.2em]{0pt}{3em} \quad du \quad & \quad dv \quad \\ \hline +\end{array} +++++ +//// + +[stem] +++++ +\begin{array}{ccc} +\begin{array}{c} \text{diff.} \\ \downarrow \end{array} & +\begin{array}{|c|c|} +\hline +\rule[-1.2em]{0pt}{3em} \quad u \quad & \quad v \quad \\ \hline +\rule[-1.2em]{0pt}{3em} \quad du \quad & \quad dv \quad \\ \hline +\end{array} & +\begin{array}{c} \text{int.} \\ \uparrow \end{array} +\end{array} +++++ + +[stem] +++++ +u \cdot v - \int v \cdot du +++++ + +[.center] +image::integration-by-parts-7.png[] + +=== Picking your "u" and "dv" + + +You can use thsi mnemonic to pick you're "u", anything that isn't you're "u" is +you're "dv". + +.LIATE mnemonic +[cols="1,2,2a", options="header"] +|=== +| Letter | Function Type | Example + +| **L** +| Logarithmic +| +[stem] +++++ +\ln(x), \log(x) +++++ + +| **I** +| Inverse Trigonometric +| +[stem] +++++ +\arctan(x), \arcsin(x) +++++ + +| **A** +| Algebraic +| +[stem] +++++ +x^2, 5x^2 + 3 +++++ + +| **T** +| Trigonometric +| +[stem] +++++ +\cos(x), \sin(x) +++++ + +| **E** +| Exponential +| +[stem] +++++ +e^x, 10^x +++++ +|=== + +.example +[%collapsible] +==== + +Integrate this + +[stem] +++++ +\int \arctan(x)dx +++++ + +You see that there are no logarithmic functions in arctan(x)dx, but +there is an inverse trigonometric function, arctan(x). So that’s your u. +Everything else is your dv, namely, plain old dx. + +[stem] +++++ +u = archtan(x) \\ +dv = dx +++++ + +[stem] +++++ +\begin{aligned} +\text{diff.} \downarrow \quad +\begin{array}{|wc{7em}|wc{7em}|} +\hline +\rule[-1.5em]{0pt}{4em} \arctan(x) & x \\ \hline +\rule[-1.5em]{0pt}{4em} \frac{1}{1+x^2} \, dx & dx \\ \hline +\end{array} +\hspace{1em} \uparrow \text{int.} +\end{aligned} +++++ + +showing integration from dx to x + +[stem] +++++ +\int 1 dx \\ +\int x^0 dx = +x^1 = x +++++ + +Do the seven thing. + +[.center] +image::integration-by-parts-7.png[] + +[stem] +++++ +\int udv = uv - \int vdu \\ +\int \arctan(x)dx = x \arctan(x) - \int x \cdot \frac{1}{1 + x^2}dx +++++ + +Now you can finish this problem by integrating +stem:[\int x \cdot \frac{1}{1 + x^2}dx]with the substitution method, setting +u=1+x2. + +NOTE: You can check "calculus for dumies" chapter 15 + +WARNING: _u_ in u=1+x2 has nothing to do with the integration-by-parts _u_. + +solution + +[stem] +++++ +\int \arctan(x)dx = x\arctan(x) - \frac{1}{2}\ln(1 + xx^2) + C. +++++ +==== + +== Basic Antiderivative Formulas + +[.inline] +- stem:[\int x^n\ dx = \frac{x^n+1}{n+1} + C\ (n \neq -1)] + +[.inline] +- stem:[\int \sin x\ dx = -\cos x + C] + +[.inline] +- stem:[\int \frac{dx}{x} = \ln|x| + C] + +[.inline] +- stem:[\int e^x\ dx = -\cos x + C] + +[.inline] +- stem:[\int a^x\ dx = \frac{1}{\ln a} a^x + C] + +[.inline] +- stem:[\int \sin x\ dx = -\cos x + C] + +[.inline] +- stem:[\int \cos x\ dx = \sin x + C] + +[.inline] +- stem:[\int \sec^2 x\ dx = \tan x + C] + +[.inline] +- stem:[\int \csc^2 x\ dx = - \cot x + C] + +[.inline] +- stem:[\int \sec x \tan x\ dx = \sec x + C] + +[.inline] +- stem:[\int \csc x \cot x\ dx = -\csc x + C] + +[.inline] +- stem:[\int \frac{dx}{x\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C] + +[.inline] +- stem:[\int \frac{dx}{a^2 - x^2} = \frac{1}{a}\arctan \frac{x}{a} + C] + +[.inline] +- stem:[\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a} \text{arcsec} \frac{|x|}{a} + C] + +NOTE: remember stem:[\frac{d}{dx}] means derivative which is similar to the the + change in y divided by the change in x (stem:[\frac{\Delta Y}{\Delta X}] + rise over run), except the scale of stem:[\frac{d}{dx}] is instantaneous, + where the scale of stem:[\frac{\Delta Y}{\Delta X}] is between two + distinct points on a line. + +== Examples + +.Example 1 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \int \frac{4x^5 + x^4 - 3x^2}{x^2} \, dx = \int (4x^3 + x^2 - 3) \, dx \\[15pt] + & = \frac{4x^4}{4} + \frac{x^3}{3} - 3x + C \\[15pt] + & = \boxed{x^4 + \frac{1}{3}x^3 - 3x + C} +\end{aligned} +++++ +==== diff --git a/Calculus/Integrals.html b/Calculus/Integrals.html new file mode 100644 index 0000000..7ee89f9 --- /dev/null +++ b/Calculus/Integrals.html @@ -0,0 +1,1685 @@ + + + + + + + +Calculus | Integrals + + + + + + + + + + + + + + + + + + +
+
+

Funamental Theorem of Calculus

+
+
+
+fundemental theorem of calculus v1 +
+
+
+
version 1
+
+\[A_{f}(x) = \int_{s}^{x} f(t)\ dt\] +
+
+
+
+
\(A_{f}(x)\)
+
+

The area under the curve at point x for funtion f()

+
+
integral s
+
+

The starting point

+
+
integral x
+
+

The end point

+
+
dt
+
+

An infinitely small width of each section

+
+
+
+
+
version 2
+
+\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\] +
+
+
+
+
F()
+
+

The antiderivative of f()

+
+
x
+
+

Represents all the values in the interval \([a, b\)]

+
+
+
+
+example +
+
+

Find the area under the graph between 2 and 3:

+
+
+
+fundemental theorem of calculus v2 example +
+
+
+
+\[\int_{2}^{3} (x^2+1)dx = F(3) - F(2)\] +
+
+
+

the antiderivative of \(x^2 + 1\) is \(\frac{1}{3}x^3 + x\)

+
+
+

\(F(3) - F(2)\) can be written as \( \left[ \frac{1}{3}x^3 + x +\right]_2^3 \), and thus,

+
+
+
+\[\begin{aligned} +\int_{2}^{3} (x^2+1)dx &= \left[ \frac{1}{3}x^3 + x \right]_2^3 \\ +&= \left( \frac{1}{3} \cdot 3^3 + 3 \right) - \left( \frac{1}{3} \cdot 2^3 + 2 \right) \\ +&= 12 - 4\frac{2}{3} \\ +&= 7\frac{1}{3} +\end{aligned}\] +
+
+
+
+
+
+
+

Power Rule

+
+
+
    +
  1. +

    Bring the power in front where it will multiply the rest of the derivative

    +
    +
    +\[5x^4 \rightarrow 4 * 5x^4\] +
    +
    +
    2. +
  2. +

    Reduce the power by one and simplify

    +
    +
    +\[4 * 5x^4 \rightarrow 4 * 5x^3 = 20x^3\] +
    +
    +
    +

    thus, \(y'\) = \(20x^3\)

    +
    +
    3. +
+
+
+
+
+

Reverse Power Rule

+
+
+
    +
  1. +

    Increase the power by one.

    +
    +

    The 3 becomes a 4.

    +
    +
    +
    +\[20x^3 \rightarrow 20x^4\] +
    +
    +
    2. +
  2. +

    Divide by th enew power and simplify

    +
    +
    +\[20x^4 \rightarrow \frac{20}{4}x^4 = 5x^4\] +
    +
    +
    + + + + + +
    + + +The reverse power rule does not work for a power of negative one. +
    +
    +
    3. +
+
+
+
+
+

"u" substitution

+
+
+

find

+
+
+
+\[\lim 2x \cos(x^2) dx\] +
+
+
+
    +
  1. +

    Set u equal to the argument of the main function.

    +
    +

    The argument of \(\cos(x^2)\) is \(x^2\), so you set \(u\) equal to +\(x^2\).

    +
    +
    2. +
  2. +

    Take the derivative of u with respect to x.

    +
    +
    +\[u = x^2\ \text{so}\ \frac{du}{dx} = 2x\] +
    +
    +
    3. +
  3. +

    Solve for dx.

    +
    +
    +\[\begin{aligned} + & \frac{du}{dx} = \frac{2x}{1} \\[10pt] + & du = 2xdx\ \text{(cross multiplication)} \\ + & \frac{du}{2x}\ = dx\ \text{(dviding both sides by 2x)} +\end{aligned}\] +
    +
    +
    4. +
  4. +

    Make substitutions

    +
    +

    In \(\int 2x \cos(x^2)dx\), u takes the place of \(x^2\) and +\(\frac{du}{2x}\) takes the place of \(dx\). So now you’ve got +\(\int 2x\cos u \frac{du}{2x}\). The two \(2x\)s cancel, giving you +\(\int \cos\ u\ du\).

    +
    +
    5. +
  5. +

    Anitdifferentiate using the simple reverse rule.

    +
    +
    +\[\lim \cos\ u\ du = \sin\ u + C\] +
    +
    +
    6. +
  6. +

    Substitute \(x^2\) back in for u, coming full circle.

    +
    +

    u equals \(x^2\), so \(x^2\) goes in for the u:

    +
    +
    +
    +\[\lim \cos\ u\ du - \sin(x^2) + C\] +
    +
    +
    +

    That’s it. So \(\lim\ 2x\ cos(x^2)dx = \sin(x^2) + C\)

    +
    +
    7. +
+
+
+

Examples

+
+Example 1 +
+
+
+\[\begin{aligned} + & \text{8. Evaluate the integral shown below:} \\[10pt] + & \int 2x \sqrt{3x^2 + 5} \, dx \qquad \begin{aligned} u &= 3x^2 + 5 \\ du &= 6x \, dx \\ \frac{du}{6x} &= dx \end{aligned} \\[15pt] + & \int 2x \sqrt{u} \, \frac{du}{6x} \\[15pt] + & \frac{1}{3} \int u^{1/2} \, du = \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} + C \\[15pt] + & = \frac{2}{9} u^{3/2} + C = \boxed{\frac{2}{9} (3x^2 + 5)^{3/2} + C} +\end{aligned}\] +
+
+
+
+
+
+
+
+

Integration by Parts

+
+
+
+\[\int u \, dv = uv - \int v \, du\] +
+
+
+ + + + + +
+ + +You can check "calculus for dumies" chapter 16 +
+
+
+
+\[\begin{array}{ccc} +\begin{array}{c} \text{diff.} \\ \downarrow \end{array} & +\begin{array}{|c|c|} +\hline +\rule[-1.2em]{0pt}{3em} \quad u \quad & \quad v \quad \\ \hline +\rule[-1.2em]{0pt}{3em} \quad du \quad & \quad dv \quad \\ \hline +\end{array} & +\begin{array}{c} \text{int.} \\ \uparrow \end{array} +\end{array}\] +
+
+
+
+\[u \cdot v - \int v \cdot du\] +
+
+
+
+integration by parts 7 +
+
+
+

Picking your "u" and "dv"

+
+

You can use thsi mnemonic to pick you’re "u", anything that isn’t you’re "u" is +you’re "dv".

+
+ + +++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
Table 1. LIATE mnemonic
LetterFunction TypeExample

L

Logarithmic

+
+\[\ln(x), \log(x)\] +
+

I

Inverse Trigonometric

+
+\[\arctan(x), \arcsin(x)\] +
+

A

Algebraic

+
+\[x^2, 5x^2 + 3\] +
+

T

Trigonometric

+
+\[\cos(x), \sin(x)\] +
+

E

Exponential

+
+\[e^x, 10^x\] +
+
+
+example +
+
+

Integrate this

+
+
+
+\[\int \arctan(x)dx\] +
+
+
+

You see that there are no logarithmic functions in arctan(x)dx, but +there is an inverse trigonometric function, arctan(x). So that’s your u. +Everything else is your dv, namely, plain old dx.

+
+
+
+\[u = archtan(x) \\ +dv = dx\] +
+
+
+
+\[\begin{aligned} +\text{diff.} \downarrow \quad +\begin{array}{|wc{7em}|wc{7em}|} +\hline +\rule[-1.5em]{0pt}{4em} \arctan(x) & x \\ \hline +\rule[-1.5em]{0pt}{4em} \frac{1}{1+x^2} \, dx & dx \\ \hline +\end{array} +\hspace{1em} \uparrow \text{int.} +\end{aligned}\] +
+
+
+

showing integration from dx to x

+
+
+
+\[\int 1 dx \\ +\int x^0 dx = +x^1 = x\] +
+
+
+

Do the seven thing.

+
+
+
+integration by parts 7 +
+
+
+
+\[\int udv = uv - \int vdu \\ +\int \arctan(x)dx = x \arctan(x) - \int x \cdot \frac{1}{1 + x^2}dx\] +
+
+
+

Now you can finish this problem by integrating +\(\int x \cdot \frac{1}{1 + x^2}dx\)with the substitution method, setting +u=1+x2.

+
+
+ + + + + +
+ + +You can check "calculus for dumies" chapter 15 +
+
+
+ + + + + +
+ + +u in u=1+x2 has nothing to do with the integration-by-parts u. +
+
+
+

solution

+
+
+
+\[\int \arctan(x)dx = x\arctan(x) - \frac{1}{2}\ln(1 + xx^2) + C.\] +
+
+
+
+
+
+
+
+

Basic Antiderivative Formulas

+
+
+
    +
  • +

    \(\int x^n\ dx = \frac{x^n+1}{n+1} + C\ (n \neq -1)\)

    +
    • +
+
+
+
    +
  • +

    \(\int \sin x\ dx = -\cos x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \frac{dx}{x} = \ln|x| + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int e^x\ dx = -\cos x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int a^x\ dx = \frac{1}{\ln a} a^x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \sin x\ dx = -\cos x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \cos x\ dx = \sin x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \sec^2 x\ dx = \tan x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \csc^2 x\ dx = - \cot x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \sec x \tan x\ dx = \sec x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \csc x \cot x\ dx = -\csc x + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \frac{dx}{x\sqrt{a^2 - x^2}} = \arcsin \frac{x}{a} + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \frac{dx}{a^2 - x^2} = \frac{1}{a}\arctan \frac{x}{a} + C\)

    +
    • +
+
+
+
    +
  • +

    \(\int \frac{dx}{x\sqrt{x^2 - a^2}} = \frac{1}{a} \text{arcsec} \frac{|x|}{a} + C\)

    +
    • +
+
+
+ + + + + +
+ + +remember \(\frac{d}{dx}\) means derivative which is similar to the the + change in y divided by the change in x (\(\frac{\Delta Y}{\Delta X}\) + rise over run), except the scale of \(\frac{d}{dx}\) is instantaneous, + where the scale of \(\frac{\Delta Y}{\Delta X}\) is between two + distinct points on a line. +
+
+
+
+
+

Examples

+
+
+Example 1 +
+
+
+\[\begin{aligned} + & \int \frac{4x^5 + x^4 - 3x^2}{x^2} \, dx = \int (4x^3 + x^2 - 3) \, dx \\[15pt] + & = \frac{4x^4}{4} + \frac{x^3}{3} - 3x + C \\[15pt] + & = \boxed{x^4 + \frac{1}{3}x^3 - 3x + C} +\end{aligned}\] +
+
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Intermediate-Value-Theorem.adoc b/Calculus/Intermediate-Value-Theorem.adoc new file mode 100644 index 0000000..87bf21a --- /dev/null +++ b/Calculus/Intermediate-Value-Theorem.adoc @@ -0,0 +1,21 @@ += Calculus | Intermediate Value Theorem +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Intermediate Value Theorem + +Suppose stem:[f] is continuous on stem:[[a,b]], and stem:[N], a number b/t +stem:[f(a)] and stem:[f(b)], where stem:[f(a)\ \ne\ f(b)], then there exists a +number stem:[c] in stem:[(a,b)] s.t stem:[f(c) = N]. + +[.center] +image::Intermediate-Value-Theorem.png[] + +== Links +* https://www.youtube.com/watch?v=9wEHwFrUyOU[TOCT | Intermediate Value Theorem] diff --git a/Calculus/Intermediate-Value-Theorem.html b/Calculus/Intermediate-Value-Theorem.html new file mode 100644 index 0000000..fb5f5e8 --- /dev/null +++ b/Calculus/Intermediate-Value-Theorem.html @@ -0,0 +1,1098 @@ + + + + + + + +Calculus | Intermediate Value Theorem + + + + + + + + + + + + + + + + + + +
+
+

Intermediate Value Theorem

+
+
+

Suppose \(f\) is continuous on \([a,b\)], and \(N\), a number b/t +\(f(a)\) and \(f(b)\), where \(f(a)\ \ne\ f(b)\), then there exists a +number \(c\) in \((a,b)\) s.t \(f(c) = N\).

+
+
+
+Intermediate Value Theorem +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Limits.adoc b/Calculus/Limits.adoc new file mode 100644 index 0000000..2ee024a --- /dev/null +++ b/Calculus/Limits.adoc @@ -0,0 +1,570 @@ += Calculus | Limits +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Delta, Epsilon Definition of a Limit + +[stem] +++++ +\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x \in \mathbb{R} +\text{ with } 0 < |x - a| < \delta, \text{ we have } |f(x) - L| < \epsilon +++++ + +[.center] +image::Epsilon-Delta-Definition-of-a-Limit.png[] + + + +== Limits to Memorize + +[.inline] +- stem:[\lim_{x \to 0^+} \frac{1}{x} = \infty] + +[.inline] +- stem:[\lim_{x \to 0^-} \frac{1}{x} = -\infty] + +[.inline] +- stem:[\lim_{x \to \infty} \frac{1}{x} = 0] + +[.inline] +- stem:[\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x}{\sin x} = 1] + +[.inline] +- stem:[\lim_{x \to 0} \frac{\cos x - 1}{x} = 0] + +[.inline] +- stem:[\lim_{x \to \infty} (\frac{1 + 1}{x})^x = e \approx 2.718] + +NOTE: 1 is the placeholder for any constant (any number) + +== When Limits exist + +[.center] +image::when-limits-exist.png[] + +.Left Side Limit +[%collapsible] +==== +When approaching the limit from the left + +[stem] +++++ +\lim_{x \to 3^{-}} +++++ + +you get stem:[1], even though there is a hole their. That is because it is only +"as it is approaching 3". +==== + +.Right side limit +[%collapsible] +==== +When approaching the limit from the Right + +[stem] +++++ +\lim_{x \to 3^{+}} +++++ + +you get stem:[1], +==== + +.Regular Limit +[%collapsible] +==== +When approaching the limit from both sides + +[stem] +++++ +\lim_{x \to 3} +++++ + +the answer is +DNE (Does not exist). That is because you cannot have two answers for a limit +==== + +== Bounded Function and Limit Proof with Squeeze Theorom + +[stem] +++++ +\begin{aligned} + & \text{Prove that if } g \text{ is bounded and } \lim_{x \to c} f(x) = 0, \text{ then } \\ + & \lim_{x \to c} (f(x)g(x)) = 0. \\[10pt] + & \underline{\text{Proof}} \text{ Spse. } g \text{ is bounded by say } M \text{ and spse. } \\ + & \lim_{x \to c} f(x) = 0. \qquad (|g(x)| \leq M) \\[10pt] + & \text{Then } 0 \leq |f(x)g(x)| = |f(x)| \cdot |g(x)| \leq |f(x)| \cdot M \\[10pt] + & \text{Then } \lim_{x \to c} 0 = 0, \quad \lim_{x \to c} |f(x)| \cdot M = |0| \cdot M = 0 \\[10pt] + & \text{Thus by squeeze, } \lim_{x \to c} |f(x)g(x)| = 0 \\[10pt] + & \text{So } \lim_{x \to c} f(x)g(x) = 0. \quad \square +\end{aligned} +++++ + +NOTE: Whenever you're trying to show something is equal to zero if you take the + absolute value and show that's zero, then whatever is inside is also zero + +== Solving Limits + +=== Plug in the number + +Try to plug in the number into the limit. If you can solve it than that's +you're answer. Always try this first. + +.example +[%collapsible] +==== +we plug 1 into the limit + +[stem] +++++ +\lim_{x \to 1} (3x + 2) = 3 \cdot 1 + 2 = 5 +++++ + +In this case we can solve so the solution is stem:[5] +==== + +If this doen't work then move on to the other steps + +=== Other Methods + +.Difference of squares +[%collapsible] +==== + +Using the difference of squares + +[stem] +++++ +\require{cancel} +\begin{aligned} + & \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \\ + & \lim_{x \to 3} \frac{\cancel{(x - 3)}(x + 3)}{\cancel{x - 3}} \\ + & \lim_{x \to 3} (x + 3) = 3 + 3 = 6 +\end{aligned} +++++ + +NOTE: You only get rid of the limit sign when you take the limit (plug in the + number). +==== + +.Difference of cubes +[%collapsible] +==== + +[stem] +++++ +\lim_{x \to 2} \frac{x^3 - 8}{x-2} +++++ + +[stem] +++++ +\require{cancel} +\begin{aligned} + & a^3 - b^3 = (a - b)(a^2 + ab + b^2) \\[10pt] + & \lim_{x \to 2} \frac{\cancel{(x - 2)}(x^2 + 2x + 4)}{\cancel{x - 2}} \\[10pt] + & \lim_{x \to 2} (x^2 + 2x + 4) = 2^2 + 2(2) + 4 \\ + & \phantom{\lim_{x \to 2} (x^2 + 2x + 4)} = 4 + 4 + 4 \\ + & \phantom{\lim_{x \to 2} (x^2 + 2x + 4)} = 12 +\end{aligned} +++++ +==== + +.Absolute Value 1 +[%collapsible] +==== + +This is actually a piecewise function + +NOTE: This piecewise function works is the absolute value is in the numerator or + denominator + +[stem] +++++ +\begin{aligned} + & \text{Find } \lim_{x \to 3^-} \frac{|x - 3|}{x - 3} \\[10pt] + & \text{Sol. } |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} +\end{aligned} +++++ + +Because stem:[x] is approaching from the left, stem:[x] must be less then three, +so stem:[x-3 \lt 0]. Looking at the piecewise function we see that we can +substute the top of the fraction with stem:[-(x-3)]. + +[stem] +++++ +\require{cancel} +\begin{aligned} + \lim_{x \to 3^-} \frac{-(x - 3)}{x - 3} & = \lim_{x \to 3^-} \frac{-\cancel{(x - 3)}}{\cancel{x - 3}} \\ + & = \lim_{x \to 3^-} (-1) \\ + & = -1 +\end{aligned} +++++ + +==== + +.Absolute Value 2 +[%collapsible] +==== + +[stem] +++++ +\lim_{x \to -6} \frac{2x + 12}{|x + 6|} \\[10pt] +++++ + +We can use the distributive property + +[stem] +++++ +\lim_{x \to -6} \frac{2(x + 6)}{|x + 6|} +++++ + +The piecewise function + +[stem] +++++ +\begin{aligned} + |x| &= \begin{cases} + x, & \text{if } x \geq 0 \\ + -x, & \text{if } x < 0 + \end{cases} \\[15pt] + |x + 6| &= \begin{cases} + x + 6, & \text{if } x + 6 \geq 0 \implies x \geq -6 \\ + -(x + 6), & \text{if } x + 6 < 0 \implies x < -6 + \end{cases} +\end{aligned} +++++ + +To solve this we are going to approach from the left and right. First we are +going to to approach from the left, so we'll use stem:[-(x + 6)]. + +NOTE: Whenever you have an absoluate value and you're solving for a normal limit + you want to use one sided limits, you take the limit from the left first + and then the limit from the right. If the limits are the same then that'll + be the limit and the limit exists other with it does not exist (DNE). + +[stem] +++++ +\require{cancel} +\begin{aligned} + \lim_{x \to -6^-} \frac{2(x + 6)}{-(x + 6)} &= \lim_{x \to -6^-} \frac{2\cancel{(x + 6)}}{-\cancel{(x + 6)}} \\[10pt] + &= \lim_{x \to -6^-} (-2) \\[10pt] + &= -2 +\end{aligned} +++++ + +Now we'll solve the right side. Looking at are peicewise function we can use +stem:[x + 6] for our denominator because we are approaching form the right. + +[stem] +++++ +\require{cancel} +\begin{aligned} + \lim_{x \to -6^+} \frac{2(x + 6)}{x + 6} &= \lim_{x \to -6^+} \frac{2\cancel{(x + 6)}}{\cancel{x + 6}} \\[10pt] + &= \lim_{x \to -6^+} (2) \\[10pt] + &= 2 +\end{aligned} +++++ + +The limit from the left is stem:[-2] and the limimt from the right is stem:[2] +so the limit doesn not exist. + +[stem] +++++ +\lim_{x \to -6} \frac{2x + 12}{|x + 6|} \\[10pt] = DNE +++++ + +==== + +.Limit of rational function example 1 +[%collapsible] +==== +[stem] +++++ +\require{cancel} +\begin{aligned} + & \lim_{x \to 0} \frac{x^2 + 3x}{x} \\ + & \lim_{x \to 0} \frac{\cancel{x}(x + 3)}{\cancel{x}} \\ + & \lim_{x \to 0} (x + 3) = 0 + 3 \\ + & \phantom{\lim_{x \to 0} (x + 3)} = 3 +\end{aligned} +++++ +==== + +.Limit of rational function example 2 +[%collapsible] +==== +[stem] +++++ +\require{cancel} +\begin{aligned} + & \lim_{x \to 2} \frac{2 - x}{x^2 - 4} \\ + & \lim_{x \to 2} \frac{-1(x - 2)}{(x - 2)(x + 2)} \\ + & \lim_{x \to 2} \frac{-1\cancel{(x - 2)}}{\cancel{(x - 2)}(x + 2)} \\ + & \lim_{x \to 2} \frac{-1}{x + 2} = \frac{-1}{2 + 2} \\ + & \phantom{\lim_{x \to 2} \frac{-1}{x + 2}} = -\frac{1}{4} +\end{aligned} +++++ +==== + +.Limit by rationalizing +[%collapsible] +==== + +We Multiple by the conjugate to get the radical on the bottom. +stem:[\sqrt{x + 3} - \sqrt{3}] is the difference of squares. + +* stem:[x + 3] is a +* stem:[3] is b + +[stem] +++++ +\require{cancel} +\begin{aligned} + & \text{Compute } \lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x} \\[10pt] + & \text{Sol. } \lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x} \cdot \left( \frac{\sqrt{x+3} + \sqrt{3}}{\sqrt{x+3} + \sqrt{3}} \right) \\[10pt] + & = \lim_{x \to 0} \frac{(x + 3) - 3}{x(\sqrt{x+3} + \sqrt{3})} \\ + & = \lim_{x \to 0} \frac{\cancel{x}}{\cancel{x}(\sqrt{x+3} + \sqrt{3})} \\ + & = \lim_{x \to 0} \frac{1}{\sqrt{x+3} + \sqrt{3}} = \frac{1}{\sqrt{0+3} + \sqrt{3}} \\ + & = \frac{1}{2\sqrt{3}} +\end{aligned} +++++ +==== + +.Limit by rationalizing 2 +[%collapsible] +==== +[stem] +++++ +\require{cancel} +\begin{aligned} + & \lim_{u \to 2} \frac{\sqrt{4u+1} - 3}{u - 2} \\[10pt] + & \text{Sol. } \lim_{u \to 2} \frac{\sqrt{4u+1} - 3}{u - 2} \cdot \left( \frac{\sqrt{4u+1} + 3}{\sqrt{4u+1} + 3} \right) \\[10pt] + & = \lim_{u \to 2} \frac{(4u + 1) - 9}{(u - 2)(\sqrt{4u+1} + 3)} \\ + & = \lim_{u \to 2} \frac{4u - 8}{(u - 2)(\sqrt{4u+1} + 3)} \\ + & = \lim_{u \to 2} \frac{4\cancel{(u - 2)}}{\cancel{(u - 2)}(\sqrt{4u+1} + 3)} \\ + & = \lim_{u \to 2} \frac{4}{\sqrt{4u+1} + 3} = \frac{4}{\sqrt{4(2)+1} + 3} \\ + & = \frac{4}{3 + 3} = \frac{4}{6} = \frac{2}{3} +\end{aligned} +++++ +==== + +.Limit of a piecewise function +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & f(x) = \begin{cases} x^2 - 4x + 6, & \text{if } x < 2 \\ -x^2 + 4x - 2, & \text{if } x \geq 2 \end{cases} \\[15pt] + & \text{Find } \lim_{x \to 2} f(x) +\end{aligned} +++++ + +We solve for the left and right. + +[stem] +++++ +\begin{aligned} + & \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 - 4x + 6) = 2^2 - 4(2) + 6 = 2 \\ + & \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (-x^2 + 4x - 2) = -(2^2) + 4(2) - 2 = 2 \\ + & \therefore \lim_{x \to 2} f(x) = 2 +\end{aligned} +++++ + +The left and right are equal so we get a solution. + +==== + +.Limit of a Difference Quotient with Delta x +[%collapsible] +==== +[stem] +++++ +\lim_{\Delta x \to 0^+} \frac{\frac{2}{x + \Delta x} - \frac{2}{x}}{\Delta x} +++++ + +Get the LCD + +[stem] +++++ +\begin{aligned} + & \frac{x \cdot 2}{x \cdot (x + \Delta x)} - \frac{2 \cdot (x + \Delta x)}{x \cdot (x + \Delta x)} \\[15pt] + & = \frac{2x - 2(x + \Delta x)}{x(x + \Delta x)} +\end{aligned} +++++ + +Distribute. + +NOTE: Dividing fractions is the same as multiplying by the reciprical + +[stem] +++++ +\require{cancel} +\begin{aligned} + & \lim_{\Delta x \to 0^+} \frac{\frac{2x - 2(x + \Delta x)}{x(x + \Delta x)}}{\Delta x} \\ + & = \lim_{\Delta x \to 0^+} \frac{\frac{\cancel{2x - 2x} - 2\Delta x}{x(x + \Delta x)}}{\Delta x} \\ + & = \lim_{\Delta x \to 0^+} \frac{\frac{-2\Delta x}{x(x + \Delta x)}}{\Delta x} \\ + & = \lim_{\Delta x \to 0^+} \frac{-2 \cancel{\Delta x}}{(x + \Delta x)x} \cdot \frac{1}{\cancel{\Delta x}} \\ + & = \lim_{\Delta x \to 0^+} \frac{-2}{(x + \Delta x)x} \\ + & = \frac{-2}{(x + 0)x} = \frac{-2}{x^2} +\end{aligned} +++++ + +==== + +.Squeeze Theorem Example 1 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{Find } \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) \\[10pt] + & -1 \leq \cos\left(\frac{1}{x}\right) \leq 1 \\ + & -x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2 \\[10pt] + & \lim_{x \to 0} (-x^2) = 0, \quad \lim_{x \to 0} x^2 = 0. \text{ So by the} \\ + & \text{squeeze thm, } \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0. +\end{aligned} +++++ +==== + +.Squeeze Theorem Example 2 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{Prove } \lim_{x \to 0} x^4 \cos\left(\frac{17}{x}\right) = 0. \\[10pt] + & \underline{\text{Proof}} \qquad (-1 \leq \cos x \leq 1) \\[10pt] + & -x^4 \leq x^4 \cos\left(\frac{17}{x}\right) \leq x^4 \\[10pt] + & \lim_{x \to 0} (-x^4) = 0, \quad \lim_{x \to 0} x^4 = 0 \\[10pt] + & \text{Thus by the squeeze thm,} \\ + & \lim_{x \to 0} x^4 \cos\left(\frac{17}{x}\right) = 0 +\end{aligned} +++++ +==== + +.Squeeze Theorem Example 3 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{Prove } \lim_{x \to 1} (x-1)^4 \sin\left(\frac{1}{x-1}\right) = 0 \\[10pt] + & \underline{\text{Proof}} \\[10pt] + & -1 \leq \sin\left(\frac{1}{x-1}\right) \leq 1 \\ + & -(x-1)^4 \leq (x-1)^4 \sin\left(\frac{1}{x-1}\right) \leq (x-1)^4 \\[10pt] + & \lim_{x \to 1} -(x-1)^4 = -(1-1)^4 = -0^4 = 0 \\ + & \lim_{x \to 1} (x-1)^4 = (1-1)^4 = 0^4 = 0 +\end{aligned} +++++ +==== + +.Limit sandwich +[%collapsible] +==== + + +[stem] +++++ +\begin{aligned} + & \text {Find } \lim_{x \to 0}(x \sin \frac{1}{x}) +\end{aligned} +++++ + +Because the range of the sine funciton is from negaitve 1 to positive 1, +whenever you multiply a number by the sine of anything, the result either stays +the same distance from zero or gets closer to zero. Thus +stem:[x \sin \frac{1}{x}] will never get above stem:[|x|] or below stem:[-|x|]. + +[.center] +image::limit-sandwich.png[] + +[stem] +++++ +\begin{aligned} + & f(x) \ge g(x) \ge h(x) \\[10pt] + & \text{Because } \lim_{x \to 0} f(x) = \lim_{x \to 0} h(x) = 0 \\[10pt] + & \text{it follows that } \\[10pt] + & \lim_{x \to 0} g(x) = 0 +\end{aligned} +++++ + +==== + +.Trig Example 1 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{Find the limit} \\ + & \lim_{x \to 0} \frac{\sin(3x)}{2x} \qquad \boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1} \\[15pt] + & \underline{\text{Sol}} \quad \frac{3}{1} \cdot \frac{1}{2} \left( \lim_{x \to 0} \frac{\sin(3x)}{3x} \right) = \frac{3}{2} \cdot 1 = \frac{3}{2} +\end{aligned} +++++ +==== + +.Trig Example 2 +[%collapsible] +==== +[stem] +++++ +\text{Find } \lim_{x \to 0} \frac{\sin(4x)}{x} \qquad \boxed{\begin{aligned} \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \\ \text{match!} \end{aligned}} \\[15pt] +++++ + +You can solve this problem in two different ways + +.way 1 +[stem] +++++ +\begin{aligned} + & \underline{\text{Sol}} \quad 4 \left( \lim_{x \to 0} \frac{\sin(4x)}{4x} \right) \\[10pt] + & = 4 \cdot 1 = \boxed{4} +\end{aligned} +++++ + +.way 2 +[stem] +++++ +\begin{aligned} + & 4 \cdot \lim_{x \to 0} \frac{\sin(4x)}{4x} \qquad , \quad \begin{aligned} &\text{let } t = 4x \\ &\text{Note as } x \to 0, t \to 0 \end{aligned} \\[15pt] + & 4 \left( \lim_{t \to 0} \frac{\sin t}{t} \right) = 4 \cdot 1 = \boxed{4} +\end{aligned} +++++ +==== + +.Trig Example 3 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \text{Find } \lim_{x \to 0} \frac{\sin(3x)}{\sin(2x)} \qquad \boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1} \\[15pt] + & \text{Sol} \\ + & \lim_{x \to 0} \frac{3x \cdot \frac{\sin(3x)}{3x}}{2x \cdot \frac{\sin(2x)}{2x}} = \frac{3}{2} \lim_{x \to 0} \frac{\left( \frac{\sin(3x)}{3x} \right)}{\left( \frac{\sin(2x)}{2x} \right)} = \frac{3}{2} \cdot \frac{1}{1} = \boxed{\frac{3}{2}} +\end{aligned} +++++ +==== + +.Trig Example 7 +[%collapsible] +==== +[stem] +++++ +\begin{aligned} + & \lim_{x \to 1} \frac{\sin(x^2 - 1)}{x - 1} \cdot \frac{x + 1}{x + 1} \\[15pt] + & \lim_{x \to 1} \frac{(x + 1) \sin(x^2 - 1)}{x^2 - 1} \\[10pt] + & \lim_{x \to 1} (x + 1) \cdot \frac{\sin(x^2 - 1)}{x^2 - 1} \\[10pt] + & \left( \lim_{x \to 1} (x + 1) \right) \left( \lim_{x \to 1} \frac{\sin(x^2 - 1)}{x^2 - 1} \right) \\[10pt] + & = (1 + 1) \cdot 1 = 2 \cdot 1 = 2 +\end{aligned} +++++ +==== + +== Links + +* https://www.youtube.com/watch?v=kfF40MiS7zA&t=292s[3Blue1Brown Limits | Essence of calculus] diff --git a/Calculus/Limits.html b/Calculus/Limits.html new file mode 100644 index 0000000..05e4d7a --- /dev/null +++ b/Calculus/Limits.html @@ -0,0 +1,1796 @@ + + + + + + + +Calculus | Limits + + + + + + + + + + + + + + + + + + +
+
+

Delta, Epsilon Definition of a Limit

+
+
+
+\[\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } \forall x \in \mathbb{R} +\text{ with } 0 < |x - a| < \delta, \text{ we have } |f(x) - L| < \epsilon\] +
+
+
+
+Epsilon Delta Definition of a Limit +
+
+
+
+
+

Limits to Memorize

+
+
+
    +
  • +

    \(\lim_{x \to 0^+} \frac{1}{x} = \infty\)

    +
    • +
+
+
+
    +
  • +

    \(\lim_{x \to 0^-} \frac{1}{x} = -\infty\)

    +
    • +
+
+
+
    +
  • +

    \(\lim_{x \to \infty} \frac{1}{x} = 0\)

    +
    • +
+
+
+
    +
  • +

    \(\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x}{\sin x} = 1\)

    +
    • +
+
+
+
    +
  • +

    \(\lim_{x \to 0} \frac{\cos x - 1}{x} = 0\)

    +
    • +
+
+
+
    +
  • +

    \(\lim_{x \to \infty} (\frac{1 + 1}{x})^x = e \approx 2.718\)

    +
    • +
+
+
+ + + + + +
+ + +1 is the placeholder for any constant (any number) +
+
+
+
+
+

When Limits exist

+
+
+
+when limits exist +
+
+
+Left Side Limit +
+
+

When approaching the limit from the left

+
+
+
+\[\lim_{x \to 3^{-}}\] +
+
+
+

you get \(1\), even though there is a hole their. That is because it is only +"as it is approaching 3".

+
+
+
+
+Right side limit +
+
+

When approaching the limit from the Right

+
+
+
+\[\lim_{x \to 3^{+}}\] +
+
+
+

you get \(1\),

+
+
+
+
+Regular Limit +
+
+

When approaching the limit from both sides

+
+
+
+\[\lim_{x \to 3}\] +
+
+
+

the answer is +DNE (Does not exist). That is because you cannot have two answers for a limit

+
+
+
+
+
+
+

Bounded Function and Limit Proof with Squeeze Theorom

+
+
+
+\[\begin{aligned} + & \text{Prove that if } g \text{ is bounded and } \lim_{x \to c} f(x) = 0, \text{ then } \\ + & \lim_{x \to c} (f(x)g(x)) = 0. \\[10pt] + & \underline{\text{Proof}} \text{ Spse. } g \text{ is bounded by say } M \text{ and spse. } \\ + & \lim_{x \to c} f(x) = 0. \qquad (|g(x)| \leq M) \\[10pt] + & \text{Then } 0 \leq |f(x)g(x)| = |f(x)| \cdot |g(x)| \leq |f(x)| \cdot M \\[10pt] + & \text{Then } \lim_{x \to c} 0 = 0, \quad \lim_{x \to c} |f(x)| \cdot M = |0| \cdot M = 0 \\[10pt] + & \text{Thus by squeeze, } \lim_{x \to c} |f(x)g(x)| = 0 \\[10pt] + & \text{So } \lim_{x \to c} f(x)g(x) = 0. \quad \square +\end{aligned}\] +
+
+
+ + + + + +
+ + +Whenever you’re trying to show something is equal to zero if you take the + absolute value and show that’s zero, then whatever is inside is also zero +
+
+
+
+
+

Solving Limits

+
+
+

Plug in the number

+
+

Try to plug in the number into the limit. If you can solve it than that’s +you’re answer. Always try this first.

+
+
+example +
+
+

we plug 1 into the limit

+
+
+
+\[\lim_{x \to 1} (3x + 2) = 3 \cdot 1 + 2 = 5\] +
+
+
+

In this case we can solve so the solution is \(5\)

+
+
+
+
+

If this doen’t work then move on to the other steps

+
+
+
+

Other Methods

+
+Difference of squares +
+
+

Using the difference of squares

+
+
+
+\[\require{cancel} +\begin{aligned} + & \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \\ + & \lim_{x \to 3} \frac{\cancel{(x - 3)}(x + 3)}{\cancel{x - 3}} \\ + & \lim_{x \to 3} (x + 3) = 3 + 3 = 6 +\end{aligned}\] +
+
+
+ + + + + +
+ + +You only get rid of the limit sign when you take the limit (plug in the + number). +
+
+
+
+
+Difference of cubes +
+
+
+\[\lim_{x \to 2} \frac{x^3 - 8}{x-2}\] +
+
+
+
+\[\require{cancel} +\begin{aligned} + & a^3 - b^3 = (a - b)(a^2 + ab + b^2) \\[10pt] + & \lim_{x \to 2} \frac{\cancel{(x - 2)}(x^2 + 2x + 4)}{\cancel{x - 2}} \\[10pt] + & \lim_{x \to 2} (x^2 + 2x + 4) = 2^2 + 2(2) + 4 \\ + & \phantom{\lim_{x \to 2} (x^2 + 2x + 4)} = 4 + 4 + 4 \\ + & \phantom{\lim_{x \to 2} (x^2 + 2x + 4)} = 12 +\end{aligned}\] +
+
+
+
+
+Absolute Value 1 +
+
+

This is actually a piecewise function

+
+
+ + + + + +
+ + +This piecewise function works is the absolute value is in the numerator or + denominator +
+
+
+
+\[\begin{aligned} + & \text{Find } \lim_{x \to 3^-} \frac{|x - 3|}{x - 3} \\[10pt] + & \text{Sol. } |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} +\end{aligned}\] +
+
+
+

Because \(x\) is approaching from the left, \(x\) must be less then three, +so \(x-3 \lt 0\). Looking at the piecewise function we see that we can +substute the top of the fraction with \(-(x-3)\).

+
+
+
+\[\require{cancel} +\begin{aligned} + \lim_{x \to 3^-} \frac{-(x - 3)}{x - 3} & = \lim_{x \to 3^-} \frac{-\cancel{(x - 3)}}{\cancel{x - 3}} \\ + & = \lim_{x \to 3^-} (-1) \\ + & = -1 +\end{aligned}\] +
+
+
+
+
+Absolute Value 2 +
+
+
+\[\lim_{x \to -6} \frac{2x + 12}{|x + 6|} \\[10pt]\] +
+
+
+

We can use the distributive property

+
+
+
+\[\lim_{x \to -6} \frac{2(x + 6)}{|x + 6|}\] +
+
+
+

The piecewise function

+
+
+
+\[\begin{aligned} + |x| &= \begin{cases} + x, & \text{if } x \geq 0 \\ + -x, & \text{if } x < 0 + \end{cases} \\[15pt] + |x + 6| &= \begin{cases} + x + 6, & \text{if } x + 6 \geq 0 \implies x \geq -6 \\ + -(x + 6), & \text{if } x + 6 < 0 \implies x < -6 + \end{cases} +\end{aligned}\] +
+
+
+

To solve this we are going to approach from the left and right. First we are +going to to approach from the left, so we’ll use \(-(x + 6)\).

+
+
+ + + + + +
+ + +Whenever you have an absoluate value and you’re solving for a normal limit + you want to use one sided limits, you take the limit from the left first + and then the limit from the right. If the limits are the same then that’ll + be the limit and the limit exists other with it does not exist (DNE). +
+
+
+
+\[\require{cancel} +\begin{aligned} + \lim_{x \to -6^-} \frac{2(x + 6)}{-(x + 6)} &= \lim_{x \to -6^-} \frac{2\cancel{(x + 6)}}{-\cancel{(x + 6)}} \\[10pt] + &= \lim_{x \to -6^-} (-2) \\[10pt] + &= -2 +\end{aligned}\] +
+
+
+

Now we’ll solve the right side. Looking at are peicewise function we can use +\(x + 6\) for our denominator because we are approaching form the right.

+
+
+
+\[\require{cancel} +\begin{aligned} + \lim_{x \to -6^+} \frac{2(x + 6)}{x + 6} &= \lim_{x \to -6^+} \frac{2\cancel{(x + 6)}}{\cancel{x + 6}} \\[10pt] + &= \lim_{x \to -6^+} (2) \\[10pt] + &= 2 +\end{aligned}\] +
+
+
+

The limit from the left is \(-2\) and the limimt from the right is \(2\) +so the limit doesn not exist.

+
+
+
+\[\lim_{x \to -6} \frac{2x + 12}{|x + 6|} \\[10pt] = DNE\] +
+
+
+
+
+Limit of rational function example 1 +
+
+
+\[\require{cancel} +\begin{aligned} + & \lim_{x \to 0} \frac{x^2 + 3x}{x} \\ + & \lim_{x \to 0} \frac{\cancel{x}(x + 3)}{\cancel{x}} \\ + & \lim_{x \to 0} (x + 3) = 0 + 3 \\ + & \phantom{\lim_{x \to 0} (x + 3)} = 3 +\end{aligned}\] +
+
+
+
+
+Limit of rational function example 2 +
+
+
+\[\require{cancel} +\begin{aligned} + & \lim_{x \to 2} \frac{2 - x}{x^2 - 4} \\ + & \lim_{x \to 2} \frac{-1(x - 2)}{(x - 2)(x + 2)} \\ + & \lim_{x \to 2} \frac{-1\cancel{(x - 2)}}{\cancel{(x - 2)}(x + 2)} \\ + & \lim_{x \to 2} \frac{-1}{x + 2} = \frac{-1}{2 + 2} \\ + & \phantom{\lim_{x \to 2} \frac{-1}{x + 2}} = -\frac{1}{4} +\end{aligned}\] +
+
+
+
+
+Limit by rationalizing +
+
+

We Multiple by the conjugate to get the radical on the bottom. +\(\sqrt{x + 3} - \sqrt{3}\) is the difference of squares.

+
+
+
    +
  • +

    \(x + 3\) is a

    +
    • +
  • +

    \(3\) is b

    +
    • +
+
+
+
+\[\require{cancel} +\begin{aligned} + & \text{Compute } \lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x} \\[10pt] + & \text{Sol. } \lim_{x \to 0} \frac{\sqrt{x+3} - \sqrt{3}}{x} \cdot \left( \frac{\sqrt{x+3} + \sqrt{3}}{\sqrt{x+3} + \sqrt{3}} \right) \\[10pt] + & = \lim_{x \to 0} \frac{(x + 3) - 3}{x(\sqrt{x+3} + \sqrt{3})} \\ + & = \lim_{x \to 0} \frac{\cancel{x}}{\cancel{x}(\sqrt{x+3} + \sqrt{3})} \\ + & = \lim_{x \to 0} \frac{1}{\sqrt{x+3} + \sqrt{3}} = \frac{1}{\sqrt{0+3} + \sqrt{3}} \\ + & = \frac{1}{2\sqrt{3}} +\end{aligned}\] +
+
+
+
+
+Limit by rationalizing 2 +
+
+
+\[\require{cancel} +\begin{aligned} + & \lim_{u \to 2} \frac{\sqrt{4u+1} - 3}{u - 2} \\[10pt] + & \text{Sol. } \lim_{u \to 2} \frac{\sqrt{4u+1} - 3}{u - 2} \cdot \left( \frac{\sqrt{4u+1} + 3}{\sqrt{4u+1} + 3} \right) \\[10pt] + & = \lim_{u \to 2} \frac{(4u + 1) - 9}{(u - 2)(\sqrt{4u+1} + 3)} \\ + & = \lim_{u \to 2} \frac{4u - 8}{(u - 2)(\sqrt{4u+1} + 3)} \\ + & = \lim_{u \to 2} \frac{4\cancel{(u - 2)}}{\cancel{(u - 2)}(\sqrt{4u+1} + 3)} \\ + & = \lim_{u \to 2} \frac{4}{\sqrt{4u+1} + 3} = \frac{4}{\sqrt{4(2)+1} + 3} \\ + & = \frac{4}{3 + 3} = \frac{4}{6} = \frac{2}{3} +\end{aligned}\] +
+
+
+
+
+Limit of a piecewise function +
+
+
+\[\begin{aligned} + & f(x) = \begin{cases} x^2 - 4x + 6, & \text{if } x < 2 \\ -x^2 + 4x - 2, & \text{if } x \geq 2 \end{cases} \\[15pt] + & \text{Find } \lim_{x \to 2} f(x) +\end{aligned}\] +
+
+
+

We solve for the left and right.

+
+
+
+\[\begin{aligned} + & \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 - 4x + 6) = 2^2 - 4(2) + 6 = 2 \\ + & \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (-x^2 + 4x - 2) = -(2^2) + 4(2) - 2 = 2 \\ + & \therefore \lim_{x \to 2} f(x) = 2 +\end{aligned}\] +
+
+
+

The left and right are equal so we get a solution.

+
+
+
+
+Limit of a Difference Quotient with Delta x +
+
+
+\[\lim_{\Delta x \to 0^+} \frac{\frac{2}{x + \Delta x} - \frac{2}{x}}{\Delta x}\] +
+
+
+

Get the LCD

+
+
+
+\[\begin{aligned} + & \frac{x \cdot 2}{x \cdot (x + \Delta x)} - \frac{2 \cdot (x + \Delta x)}{x \cdot (x + \Delta x)} \\[15pt] + & = \frac{2x - 2(x + \Delta x)}{x(x + \Delta x)} +\end{aligned}\] +
+
+
+

Distribute.

+
+
+ + + + + +
+ + +Dividing fractions is the same as multiplying by the reciprical +
+
+
+
+\[\require{cancel} +\begin{aligned} + & \lim_{\Delta x \to 0^+} \frac{\frac{2x - 2(x + \Delta x)}{x(x + \Delta x)}}{\Delta x} \\ + & = \lim_{\Delta x \to 0^+} \frac{\frac{\cancel{2x - 2x} - 2\Delta x}{x(x + \Delta x)}}{\Delta x} \\ + & = \lim_{\Delta x \to 0^+} \frac{\frac{-2\Delta x}{x(x + \Delta x)}}{\Delta x} \\ + & = \lim_{\Delta x \to 0^+} \frac{-2 \cancel{\Delta x}}{(x + \Delta x)x} \cdot \frac{1}{\cancel{\Delta x}} \\ + & = \lim_{\Delta x \to 0^+} \frac{-2}{(x + \Delta x)x} \\ + & = \frac{-2}{(x + 0)x} = \frac{-2}{x^2} +\end{aligned}\] +
+
+
+
+
+Squeeze Theorem Example 1 +
+
+
+\[\begin{aligned} + & \text{Find } \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) \\[10pt] + & -1 \leq \cos\left(\frac{1}{x}\right) \leq 1 \\ + & -x^2 \leq x^2 \cos\left(\frac{1}{x}\right) \leq x^2 \\[10pt] + & \lim_{x \to 0} (-x^2) = 0, \quad \lim_{x \to 0} x^2 = 0. \text{ So by the} \\ + & \text{squeeze thm, } \lim_{x \to 0} x^2 \cos\left(\frac{1}{x}\right) = 0. +\end{aligned}\] +
+
+
+
+
+Squeeze Theorem Example 2 +
+
+
+\[\begin{aligned} + & \text{Prove } \lim_{x \to 0} x^4 \cos\left(\frac{17}{x}\right) = 0. \\[10pt] + & \underline{\text{Proof}} \qquad (-1 \leq \cos x \leq 1) \\[10pt] + & -x^4 \leq x^4 \cos\left(\frac{17}{x}\right) \leq x^4 \\[10pt] + & \lim_{x \to 0} (-x^4) = 0, \quad \lim_{x \to 0} x^4 = 0 \\[10pt] + & \text{Thus by the squeeze thm,} \\ + & \lim_{x \to 0} x^4 \cos\left(\frac{17}{x}\right) = 0 +\end{aligned}\] +
+
+
+
+
+Squeeze Theorem Example 3 +
+
+
+\[\begin{aligned} + & \text{Prove } \lim_{x \to 1} (x-1)^4 \sin\left(\frac{1}{x-1}\right) = 0 \\[10pt] + & \underline{\text{Proof}} \\[10pt] + & -1 \leq \sin\left(\frac{1}{x-1}\right) \leq 1 \\ + & -(x-1)^4 \leq (x-1)^4 \sin\left(\frac{1}{x-1}\right) \leq (x-1)^4 \\[10pt] + & \lim_{x \to 1} -(x-1)^4 = -(1-1)^4 = -0^4 = 0 \\ + & \lim_{x \to 1} (x-1)^4 = (1-1)^4 = 0^4 = 0 +\end{aligned}\] +
+
+
+
+
+Limit sandwich +
+
+
+\[\begin{aligned} + & \text {Find } \lim_{x \to 0}(x \sin \frac{1}{x}) +\end{aligned}\] +
+
+
+

Because the range of the sine funciton is from negaitve 1 to positive 1, +whenever you multiply a number by the sine of anything, the result either stays +the same distance from zero or gets closer to zero. Thus +\(x \sin \frac{1}{x}\) will never get above \(|x|\) or below \(-|x|\).

+
+
+
+limit sandwich +
+
+
+
+\[\begin{aligned} + & f(x) \ge g(x) \ge h(x) \\[10pt] + & \text{Because } \lim_{x \to 0} f(x) = \lim_{x \to 0} h(x) = 0 \\[10pt] + & \text{it follows that } \\[10pt] + & \lim_{x \to 0} g(x) = 0 +\end{aligned}\] +
+
+
+
+
+Trig Example 1 +
+
+
+\[\begin{aligned} + & \text{Find the limit} \\ + & \lim_{x \to 0} \frac{\sin(3x)}{2x} \qquad \boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1} \\[15pt] + & \underline{\text{Sol}} \quad \frac{3}{1} \cdot \frac{1}{2} \left( \lim_{x \to 0} \frac{\sin(3x)}{3x} \right) = \frac{3}{2} \cdot 1 = \frac{3}{2} +\end{aligned}\] +
+
+
+
+
+Trig Example 2 +
+
+
+\[\text{Find } \lim_{x \to 0} \frac{\sin(4x)}{x} \qquad \boxed{\begin{aligned} \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \\ \text{match!} \end{aligned}} \\[15pt]\] +
+
+
+

You can solve this problem in two different ways

+
+
+
way 1
+
+\[\begin{aligned} + & \underline{\text{Sol}} \quad 4 \left( \lim_{x \to 0} \frac{\sin(4x)}{4x} \right) \\[10pt] + & = 4 \cdot 1 = \boxed{4} +\end{aligned}\] +
+
+
+
way 2
+
+\[\begin{aligned} + & 4 \cdot \lim_{x \to 0} \frac{\sin(4x)}{4x} \qquad , \quad \begin{aligned} &\text{let } t = 4x \\ &\text{Note as } x \to 0, t \to 0 \end{aligned} \\[15pt] + & 4 \left( \lim_{t \to 0} \frac{\sin t}{t} \right) = 4 \cdot 1 = \boxed{4} +\end{aligned}\] +
+
+
+
+
+Trig Example 3 +
+
+
+\[\begin{aligned} + & \text{Find } \lim_{x \to 0} \frac{\sin(3x)}{\sin(2x)} \qquad \boxed{\lim_{x \to 0} \frac{\sin x}{x} = 1} \\[15pt] + & \text{Sol} \\ + & \lim_{x \to 0} \frac{3x \cdot \frac{\sin(3x)}{3x}}{2x \cdot \frac{\sin(2x)}{2x}} = \frac{3}{2} \lim_{x \to 0} \frac{\left( \frac{\sin(3x)}{3x} \right)}{\left( \frac{\sin(2x)}{2x} \right)} = \frac{3}{2} \cdot \frac{1}{1} = \boxed{\frac{3}{2}} +\end{aligned}\] +
+
+
+
+
+Trig Example 7 +
+
+
+\[\begin{aligned} + & \lim_{x \to 1} \frac{\sin(x^2 - 1)}{x - 1} \cdot \frac{x + 1}{x + 1} \\[15pt] + & \lim_{x \to 1} \frac{(x + 1) \sin(x^2 - 1)}{x^2 - 1} \\[10pt] + & \lim_{x \to 1} (x + 1) \cdot \frac{\sin(x^2 - 1)}{x^2 - 1} \\[10pt] + & \left( \lim_{x \to 1} (x + 1) \right) \left( \lim_{x \to 1} \frac{\sin(x^2 - 1)}{x^2 - 1} \right) \\[10pt] + & = (1 + 1) \cdot 1 = 2 \cdot 1 = 2 +\end{aligned}\] +
+
+
+
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Mean-Value-Theorem.adoc b/Calculus/Mean-Value-Theorem.adoc new file mode 100644 index 0000000..fdae546 --- /dev/null +++ b/Calculus/Mean-Value-Theorem.adoc @@ -0,0 +1,43 @@ += Calculus | Mean Value Theorem +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +[.center] +image::Mean-Value-Theorem.png[] + +== Formal Definition + +If f is continuous on the closed interval [a, b] and differentiable on the open +interval (a,b) then tehre exists at least on number c in (a,b) such that + +[stem] +++++ +f\prime(c) = \frac{f(b) - f(a)}{b - a} +++++ + +NOTE: you can look at page 168 of calculus for dummies for a good explaination + +== Plain-English Definition + +There must be at least one point between a and b where the slope of the tangent +is the same as the slope of the secant line from a to be. + +Imagine that you grab the secant line connecting (a, f(a)) and (b, f(b)), and +then you slide it up, keeping it parallel to the original secant line. Can you +see that the two points of intersection between this sliding line and the +function -- the two points that begin at (a, f(a)) and (b, f(b)) -- will +gradually get closer and closer to each other until they come together at (c, +f(c))? If you raise the line any further, you break away from the function +entirely. At this last point of intersection, (c, f(c)), the sliding line +touches the function at a single point and is thus tangent to the function +there, and it has the same slope as the original secant line. + +[.center] +image::mean-value-theorem-visual.png[] + diff --git a/Calculus/Mean-Value-Theorem.html b/Calculus/Mean-Value-Theorem.html new file mode 100644 index 0000000..fb4c828 --- /dev/null +++ b/Calculus/Mean-Value-Theorem.html @@ -0,0 +1,1131 @@ + + + + + + + +Calculus | Mean Value Theorem + + + + + + + + + + + + + + + + + + +
+
+
+
+
+Mean Value Theorem +
+
+
+
+
+

Formal Definition

+
+
+

If f is continuous on the closed interval [a, b] and differentiable on the open +interval (a,b) then tehre exists at least on number c in (a,b) such that

+
+
+
+\$f\prime(c) = \frac{f(b) - f(a)}{b - a}\$ +
+
+
+ + + + + +
+ + +you can look at page 168 of calculus for dummies for a good explaination +
+
+
+
+
+

Plain-English Definition

+
+
+

There must be at least one point between a and b where the slope of the tangent +is the same as the slope of the secant line from a to be.

+
+
+

Imagine that you grab the secant line connecting (a, f(a)) and (b, f(b)), and +then you slide it up, keeping it parallel to the original secant line. Can you +see that the two points of intersection between this sliding line and the +function — the two points that begin at (a, f(a)) and (b, f(b)) — will +gradually get closer and closer to each other until they come together at (c, +f(c))? If you raise the line any further, you break away from the function +entirely. At this last point of intersection, (c, f(c)), the sliding line +touches the function at a single point and is thus tangent to the function +there, and it has the same slope as the original secant line.

+
+
+
+mean value theorem visual +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/Spherical-Coordinates.adoc b/Calculus/Spherical-Coordinates.adoc new file mode 100644 index 0000000..2eb91c5 --- /dev/null +++ b/Calculus/Spherical-Coordinates.adoc @@ -0,0 +1,17 @@ += Calculus | Spherical Coordinates +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Links + +* https://www.youtube.com/watch?v=jBcNOkwiS6k[YouTube | inegrating spherical coordinates] +* https://www.youtube.com/watch?v=w79nZGzWMyI[YouTube | deriving spherical coordinates] +* https://mathworld.wolfram.com/SphericalCoordinates.html[Spherical coordinates] +* https://www.youtube.com/watch?v=R50_LFbJ3aM +* https://www.youtube.com/watch?v=Ex_g2w4E5lQ[spherical coordinates animation] diff --git a/Calculus/Spherical-Coordinates.html b/Calculus/Spherical-Coordinates.html new file mode 100644 index 0000000..11b8a72 --- /dev/null +++ b/Calculus/Spherical-Coordinates.html @@ -0,0 +1,1094 @@ + + + + + + + +Calculus | Spherical Coordinates + + + + + + + + + + + + + + + + + + +
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Calculus/index.adoc b/Calculus/index.adoc new file mode 100644 index 0000000..c43a138 --- /dev/null +++ b/Calculus/index.adoc @@ -0,0 +1,16 @@ += ./Calculus +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Derivatives.html;hb=HEAD[Derivatives] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Intermediate-Value-Theorem.html;hb=HEAD[Intermediate-Value-Theorem] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Spherical-Coordinates.html;hb=HEAD[Spherical-Coordinates] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Integrals.html;hb=HEAD[Integrals] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Asymptotes.html;hb=HEAD[Asymptotes] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Limits.html;hb=HEAD[Limits] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/Mean-Value-Theorem.html;hb=HEAD[Mean-Value-Theorem] diff --git a/Calculus/index.html b/Calculus/index.html new file mode 100644 index 0000000..2f99bde --- /dev/null +++ b/Calculus/index.html @@ -0,0 +1,1070 @@ + + + + + + + +./Calculus + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Calculus/scarp b/Calculus/scarp new file mode 100644 index 0000000..b40828d --- /dev/null +++ b/Calculus/scarp @@ -0,0 +1,4 @@ +math sorcerer Calculus 1 +Ended at Calculus 1 with the Math Sorcerer Section 1.13 +Ended at Calculus 1 with the Math Sorcerer Section 3.21 +on section 5 diff --git a/Geometry/Angles-And-Sides.adoc b/Geometry/Angles-And-Sides.adoc new file mode 100644 index 0000000..f988c57 --- /dev/null +++ b/Geometry/Angles-And-Sides.adoc @@ -0,0 +1,53 @@ += Algrebra | Angles and Sides +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Transversals +When two parallel lines are cut by a transveresal, alternate interior angles are +formed. + +If two side of a triangle are congruent, the angles opposite those sides are +also congruent. + +== Triangles +All of the angles of a triangle equal 180 degrees. This image is a simple proof +of that. + +image::180-degree-triangle-proof.png[] + +In an equalateral triangle the angle bisector is always perpendicular to the +opposite angle + +There *cannot* be two abtuse angles in a triangle. + + +== Definitions + +bisector:: + divides an angle into two congruent angles. + +subtraction postulate:: + if equal quantities are subtracted from equal quantities, the differences + are equal. + +transversal:: + a line that passes through two lines in the same plane at two + distinct points in the geometry. + +circumscribed:: + a figure within another so that their boundries touch but do not intersect + +inscribed:: + a figure within another so that their boundries touch but do not intersect. + +circumcenter:: + a center of a circle when the circle is circumscribed around the figure. + +incenter:: + a center of a circle when the circle is inscribed in the figure. diff --git a/Geometry/Angles-And-Sides.html b/Geometry/Angles-And-Sides.html new file mode 100644 index 0000000..94b8cce --- /dev/null +++ b/Geometry/Angles-And-Sides.html @@ -0,0 +1,1145 @@ + + + + + + + +Algrebra | Angles and Sides + + + + + + + + + + + + + + + + + + +
+
+

Transversals

+
+
+

When two parallel lines are cut by a transveresal, alternate interior angles are +formed.

+
+
+

If two side of a triangle are congruent, the angles opposite those sides are +also congruent.

+
+
+
+
+

Triangles

+
+
+

All of the angles of a triangle equal 180 degrees. This image is a simple proof +of that.

+
+
+
+180 degree triangle proof +
+
+
+

In an equalateral triangle the angle bisector is always perpendicular to the +opposite angle

+
+
+

There cannot be two abtuse angles in a triangle.

+
+
+
+
+

Definitions

+
+
+
+
bisector
+
+

divides an angle into two congruent angles.

+
+
subtraction postulate
+
+

if equal quantities are subtracted from equal quantities, the differences +are equal.

+
+
transversal
+
+

a line that passes through two lines in the same plane at two +distinct points in the geometry.

+
+
circumscribed
+
+

a figure within another so that their boundries touch but do not intersect

+
+
inscribed
+
+

a figure within another so that their boundries touch but do not intersect.

+
+
circumcenter
+
+

a center of a circle when the circle is circumscribed around the figure.

+
+
incenter
+
+

a center of a circle when the circle is inscribed in the figure.

+
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Geometry/Circle-Theorems.adoc b/Geometry/Circle-Theorems.adoc new file mode 100644 index 0000000..738e0f3 --- /dev/null +++ b/Geometry/Circle-Theorems.adoc @@ -0,0 +1,114 @@ += Geometry | Circle Theorems +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + + +== Formulas + +* Circumfrerence = stem:[2\pir] +* Diameter = stem:[2r] +* Area = stem:[\pir^2] + +== Lines + +[.center] +image::lines-of-circle.png[] + +Chords:: + Line segment that connects two endpoints on a circle. If a chord passes + through the center of a circle its known as a diameter + +Tangent Line:: + Touches just one point at a curce. + +Secant Line:: + A line the passes through two points of a curve. + +Radius:: + half of diameter + +== Central Angle +[stem] +++++ +m\angle ACB = mACB +++++ +// [stem] +// ++++ +// \angle ACB = 50^\circ +// \ +// mACB = 50^\circ +// ++++ + +[.center] +image::Central-Angle.png[] + +== Inscribed Angle +[stem] +++++ +m\angle ABC = \frac{1}{2}m \overarc{AC} +++++ + +[.center] +image::Inscribed-Angle.png[] + +== Tangent-Chord Angle +[stem] +++++ +m\angle ABC = \frac{1}{2}(m\overarc{AB}) +++++ + +[.center] +image::Tangent-Chord-Angle.png[] + +== Unit Circle +y:: sine +x:: cosine +image::Unit-Circle.png[] + +== Standard Circle Formula + +[stem] +++++ +(x -h)^2 + (y - k)^2 =r^2 +++++ + +`h` and `k` are the x and y coordinates of the center of the circle. + +== Drawing a Circle in C + +Without Standard Circle Formula: +[source, c] +---- +void drawCircle(SDL_Renderer *ren, float radius, SDL_Point center, uint32_t color) +{ + + float x = 0; + float y = 0; + for(float i = 0; i < 2 * M_PI; i += 0.001f) + { + x = cos(2.0f * M_PI * i); + y = sin(2.0f * M_PI * i); + SDL_Rect point = { + (x * radius) + center.x, + (y * radius) + center.y, + 1, + 1 + }; + SDL_SetRenderDrawColor(ren, 0, 0, 0, 255); + SDL_RenderFillRect(ren, &point); + } +} +---- + +== Rotation + +image:Rotation.png[] + +== Links +- https://www.desmos.com/geometry[Desmos Geometry] diff --git a/Geometry/Circle-Theorems.html b/Geometry/Circle-Theorems.html new file mode 100644 index 0000000..0500e5c --- /dev/null +++ b/Geometry/Circle-Theorems.html @@ -0,0 +1,1341 @@ + + + + + + + +Geometry | Circle Theorems + + + + + + + + + + + + + + + + + + + +
+
+

Formulas

+
+
+
    +
  • +

    Circumfrerence = \$2\pir\$

    +
    • +
  • +

    Diameter = \$2r\$

    +
    • +
  • +

    Area = \$\pir^2\$

    +
    • +
+
+
+
+
+

Lines

+
+
+
+lines of circle +
+
+
+
+
Chords
+
+

Line segment that connects two endpoints on a circle. If a chord passes +through the center of a circle its known as a diameter

+
+
Tangent Line
+
+

Touches just one point at a curce.

+
+
Secant Line
+
+

A line the passes through two points of a curve.

+
+
Radius
+
+

half of diameter

+
+
+
+
+
+
+

Central Angle

+
+
+
+\$m\angle ACB = mACB\$ +
+
+
+
+Central Angle +
+
+
+
+
+

Inscribed Angle

+
+
+
+\$m\angle ABC = \frac{1}{2}m \overarc{AC}\$ +
+
+
+
+Inscribed Angle +
+
+
+
+
+

Tangent-Chord Angle

+
+
+
+\$m\angle ABC = \frac{1}{2}(m\overarc{AB})\$ +
+
+
+
+Tangent Chord Angle +
+
+
+
+
+

Unit Circle

+
+
+
+
y
+
+

sine

+
+
x
+
+

cosine

+
+
+Unit Circle +
+
+
+
+
+
+
+
+

Standard Circle Formula

+
+
+
+\$(x -h)^2 + (y - k)^2 =r^2\$ +
+
+
+

h and k are the x and y coordinates of the center of the circle.

+
+
+
+
+

Drawing a Circle in C

+
+
+

Without Standard Circle Formula:

+
+
+
+
void drawCircle(SDL_Renderer *ren, float radius, SDL_Point center, uint32_t color)
+{
+
+    float x = 0;
+    float y = 0;
+    for(float i = 0; i < 2 * M_PI; i += 0.001f)
+    {
+        x = cos(2.0f * M_PI * i);
+        y = sin(2.0f * M_PI * i);
+        SDL_Rect point = {
+            (x * radius) + center.x,
+            (y * radius) + center.y,
+            1,
+            1
+        };
+        SDL_SetRenderDrawColor(ren, 0, 0, 0, 255);
+        SDL_RenderFillRect(ren, &point);
+    }
+}
+
+
+
+
+
+

Rotation

+
+
+

Rotation

+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Geometry/Geometric-Proportions.adoc b/Geometry/Geometric-Proportions.adoc new file mode 100644 index 0000000..1f89317 --- /dev/null +++ b/Geometry/Geometric-Proportions.adoc @@ -0,0 +1,92 @@ += Geometry | Geometric Proportions +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + + +In geometry, proportional relationships appear whenever shapes are "scaled" +versions of each other. In Calculus—specifically Related Rates—these are used to +eliminate variables before differentiating. + +== 1. Similar Triangles (The "Trough" or "Shadow" Rule) +This applies when one triangle is nested inside another or when triangles are +created by a single line of sight. + +* **The Proportion:** + +[stem] +++++ +\frac{\text{base}_1}{\text{height}_1} = \frac{\text{base}_2}{\text{height}_2} +++++ + +* **Common Use:** Calculating the length of a shadow cast by a person walking + away from a street lamp, or water levels in a triangular trough. + +== 2. Conical Proportions (The "Water Tank" Rule) +Because a cone has a uniform slope, any cross-section of the liquid forms a +triangle similar to the triangle of the full cone. + +* **The Proportion:** + +[stem] +++++ +\frac{r}{h} = \frac{R}{H} +++++ + +**Note:** _r_ and _h_ are the liquid's radius/height; _R_ and _H_ are the tank's + total dimensions. + +== 3. Circles and Sectors (The "Pizza" Rule) +The relationship between a slice's angle, its arc length, and its area is +proportional to the whole circle. + +* **The Proportion:** + +[stem] +++++ +\frac{\text{Arc Length}}{2\pi r} = \frac{\text{Sector Area}}{\pi r^2} = \frac{\text{Central Angle}}{360^\circ} +++++ + +== 4. Direct Square/Cube Law (The "Scaling" Rule) +This describes how dimensions relate to surface area and volume as an object scales. + +* **Area Proportion:** + +[stem] +++++ +\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2 +++++ + +* **Volume Proportion:** + +[stem] +++++ +\frac{V_1}{V_2} = \left(\frac{s_1}{s_2}\right)^3 +++++ + +== 5. Chord-Segment Proportion +When two chords intersect inside a circle, the products of their segments are +proportional. + +* **The Proportion:** + +[stem] +++++ +a \cdot b = c \cdot d +++++ + +== Summary Reference Table + +[cols="1,2,2", options="header"] +|=== +| Shape | Relationship | Common Application +| **Triangle** | base / height = constant | Troughs, Shadows, Ladders +| **Cone** | radius / height = constant | Funnels, Sand piles, Water tanks +| **Circle** | arc length / radius = theta | Rotational motion, Gears +| **Sphere** | Volume is proportional to r^3 | Balloons, Melting snowballs +|=== diff --git a/Geometry/Geometric-Proportions.html b/Geometry/Geometric-Proportions.html new file mode 100644 index 0000000..a308683 --- /dev/null +++ b/Geometry/Geometric-Proportions.html @@ -0,0 +1,1253 @@ + + + + + + + +Geometry | Geometric Proportions + + + + + + + + + + + + + + + + + + +
+
+
+
+

In geometry, proportional relationships appear whenever shapes are "scaled" +versions of each other. In Calculus—specifically Related Rates—these are used to +eliminate variables before differentiating.

+
+
+
+
+

1. Similar Triangles (The "Trough" or "Shadow" Rule)

+
+
+

This applies when one triangle is nested inside another or when triangles are +created by a single line of sight.

+
+
+
    +
  • +

    The Proportion:

    +
    • +
+
+
+
+\[\frac{\text{base}_1}{\text{height}_1} = \frac{\text{base}_2}{\text{height}_2}\] +
+
+
+
    +
  • +

    Common Use: Calculating the length of a shadow cast by a person walking +away from a street lamp, or water levels in a triangular trough.

    +
    • +
+
+
+
+
+

2. Conical Proportions (The "Water Tank" Rule)

+
+
+

Because a cone has a uniform slope, any cross-section of the liquid forms a +triangle similar to the triangle of the full cone.

+
+
+
    +
  • +

    The Proportion:

    +
    • +
+
+
+
+\[\frac{r}{h} = \frac{R}{H}\] +
+
+
+

Note: r and h are the liquid’s radius/height; R and H are the tank’s + total dimensions.

+
+
+
+
+

3. Circles and Sectors (The "Pizza" Rule)

+
+
+

The relationship between a slice’s angle, its arc length, and its area is +proportional to the whole circle.

+
+
+
    +
  • +

    The Proportion:

    +
    • +
+
+
+
+\[\frac{\text{Arc Length}}{2\pi r} = \frac{\text{Sector Area}}{\pi r^2} = \frac{\text{Central Angle}}{360^\circ}\] +
+
+
+
+
+

4. Direct Square/Cube Law (The "Scaling" Rule)

+
+
+

This describes how dimensions relate to surface area and volume as an object scales.

+
+
+
    +
  • +

    Area Proportion:

    +
    • +
+
+
+
+\[\frac{A_1}{A_2} = \left(\frac{s_1}{s_2}\right)^2\] +
+
+
+
    +
  • +

    Volume Proportion:

    +
    • +
+
+
+
+\[\frac{V_1}{V_2} = \left(\frac{s_1}{s_2}\right)^3\] +
+
+
+
+
+

5. Chord-Segment Proportion

+
+
+

When two chords intersect inside a circle, the products of their segments are +proportional.

+
+
+
    +
  • +

    The Proportion:

    +
    • +
+
+
+
+\[a \cdot b = c \cdot d\] +
+
+
+
+
+

Summary Reference Table

+
+ +++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
ShapeRelationshipCommon Application

Triangle

base / height = constant

Troughs, Shadows, Ladders

Cone

radius / height = constant

Funnels, Sand piles, Water tanks

Circle

arc length / radius = theta

Rotational motion, Gears

Sphere

Volume is proportional to r^3

Balloons, Melting snowballs

+
+
+
+ + + + + \ No newline at end of file diff --git a/Geometry/Lines.adoc b/Geometry/Lines.adoc new file mode 100644 index 0000000..35e28d1 --- /dev/null +++ b/Geometry/Lines.adoc @@ -0,0 +1,17 @@ += Geometry | Lines +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Perpendicular Lines +Perpendicular lines have negative reciprocal slopes + +[.center] +image::Perpendicular-Angle.jpg[] + +The product of the two slopes should be equal to stem:[-1]. diff --git a/Geometry/Lines.html b/Geometry/Lines.html new file mode 100644 index 0000000..d36515e --- /dev/null +++ b/Geometry/Lines.html @@ -0,0 +1,1086 @@ + + + + + + + +Geometry | Lines + + + + + + + + + + + + + + + + + + +
+
+

Perpendicular Lines

+
+
+

Perpendicular lines have negative reciprocal slopes

+
+
+
+Perpendicular Angle +
+
+
+

The product of the two slopes should be equal to \$-1\$.

+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Geometry/Triangles.adoc b/Geometry/Triangles.adoc new file mode 100644 index 0000000..8b82a83 --- /dev/null +++ b/Geometry/Triangles.adoc @@ -0,0 +1,27 @@ += Geometry | Triangles +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Types of Triangles + +Scalene Triangle:: + A triangle whose all sides are unequal and all angles have different + measures. + +Scalene Right Triangle:: + Same as above but with a right angle. + + +== Similar Triangles + +image::similar-triangles.png[] + +== Links +* https://www.youtube.com/watch?v=bGe-d09Nc_M +* https://www.cuemath.com/geometry/similar-triangles/[cuemath | similar triangles] diff --git a/Geometry/Triangles.html b/Geometry/Triangles.html new file mode 100644 index 0000000..01092ab --- /dev/null +++ b/Geometry/Triangles.html @@ -0,0 +1,1115 @@ + + + + + + + +Geometry | Triangles + + + + + + + + + + + + + + + + + + +
+
+

Types of Triangles

+
+
+
+
Scalene Triangle
+
+

A triangle whose all sides are unequal and all angles have different +measures.

+
+
Scalene Right Triangle
+
+

Same as above but with a right angle.

+
+
+
+
+
+
+

Similar Triangles

+
+
+
+similar triangles +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Geometry/index.adoc b/Geometry/index.adoc new file mode 100644 index 0000000..5ac7919 --- /dev/null +++ b/Geometry/index.adoc @@ -0,0 +1,14 @@ += ./Geometry +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Geometry/Circle-Theorems.html;hb=HEAD[Circle-Theorems] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Geometry/Triangles.html;hb=HEAD[Triangles] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Geometry/Angles-And-Sides.html;hb=HEAD[Angles-And-Sides] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Geometry/Lines.html;hb=HEAD[Lines] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Geometry/Geometric-Proportions.html;hb=HEAD[Geometric-Proportions] diff --git a/Geometry/index.html b/Geometry/index.html new file mode 100644 index 0000000..792522a --- /dev/null +++ b/Geometry/index.html @@ -0,0 +1,1064 @@ + + + + + + + +./Geometry + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Graphics/3D-Objects.adoc b/Graphics/3D-Objects.adoc new file mode 100644 index 0000000..30d0b32 --- /dev/null +++ b/Graphics/3D-Objects.adoc @@ -0,0 +1,18 @@ += Algrebra | 3D Objects +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Plane + +=== Formula + +[stem] +++++ +ax + by + cz = 0 +++++ diff --git a/Graphics/3D-Objects.html b/Graphics/3D-Objects.html new file mode 100644 index 0000000..03e853c --- /dev/null +++ b/Graphics/3D-Objects.html @@ -0,0 +1,1087 @@ + + + + + + + +Algrebra | 3D Objects + + + + + + + + + + + + + + + + + + +
+
+

Plane

+
+
+

Formula

+
+
+\$ax + by + cz = 0\$ +
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Graphics/Quanternions.adoc b/Graphics/Quanternions.adoc new file mode 100644 index 0000000..e7134dd --- /dev/null +++ b/Graphics/Quanternions.adoc @@ -0,0 +1,13 @@ += Quanternions | Graphics +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Links +* https://www.youtube.com/watch?v=xoTqBqQtrY4[Covenant Turtle | Basic Intro for Quanternions for 3D rotations] +* https://youtu.be/-zsnHbQyRnc?si=xPlKN15Kt6GyioXz[Wumbo | how quanternions visualize 3d space] diff --git a/Graphics/Quanternions.html b/Graphics/Quanternions.html new file mode 100644 index 0000000..f379967 --- /dev/null +++ b/Graphics/Quanternions.html @@ -0,0 +1,1085 @@ + + + + + + + +Quanternions | Graphics + + + + + + + + + + + + + + + + + + +
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Graphics/Rasterization.adoc b/Graphics/Rasterization.adoc new file mode 100644 index 0000000..30aa289 --- /dev/null +++ b/Graphics/Rasterization.adoc @@ -0,0 +1,18 @@ += Graphics | Rasterization +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + + +== Interpolation +NOTE: This is helpful outside of rasterization + +image::interpolation.png[] + +== Links +- https://www.youtube.com/watch?v=PahbNFypubE&t=398s[Bisqwit | Texture mapping.. | interpolation] diff --git a/Graphics/Rasterization.html b/Graphics/Rasterization.html new file mode 100644 index 0000000..3e4f84a --- /dev/null +++ b/Graphics/Rasterization.html @@ -0,0 +1,1105 @@ + + + + + + + +Graphics | Rasterization + + + + + + + + + + + + + + + + + + +
+
+

Interpolation

+
+
+ + + + + +
+ + +This is helpful outside of rasterization +
+
+
+
+interpolation +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Graphics/index.adoc b/Graphics/index.adoc new file mode 100644 index 0000000..95ab256 --- /dev/null +++ b/Graphics/index.adoc @@ -0,0 +1,12 @@ += ./Graphics +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Graphics/Rasterization.html;hb=HEAD[Rasterization] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Graphics/3D-Objects.html;hb=HEAD[3D-Objects] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Graphics/Quanternions.html;hb=HEAD[Quanternions] diff --git a/Graphics/index.html b/Graphics/index.html new file mode 100644 index 0000000..61d0286 --- /dev/null +++ b/Graphics/index.html @@ -0,0 +1,1058 @@ + + + + + + + +./Graphics + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Graphing/Parent-Functions.adoc b/Graphing/Parent-Functions.adoc new file mode 100644 index 0000000..9c5539c --- /dev/null +++ b/Graphing/Parent-Functions.adoc @@ -0,0 +1,144 @@ += Parent Functions +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Constant Functions +[.example] +[stem] +++++ +y = c +++++ + +Domain:: stem:[(-\infty,\infty)] +Range:: stem:[C] (constant) + +[.center] +image::Constant-Functions.png[] + + +== Linear Functions +[.example] +[stem] +++++ +y = x +++++ + +Domain:: stem:[(-\infty,\infty)] +Range:: stem:[(-\infty,\infty)] + +[.center] +image::Linear-Functions.png[] + +== Quadradic Functions +[.example] +[stem] +++++ +y = x^2 +++++ + +Domain:: stem:[(-\infty,\infty)] +Range:: stem:[[0,\infty)] + +[.center] +image::Quadradic-Functions.png[] + +== Cubic Functions +[.example] +[stem] +++++ +y = x^3 +++++ + +Domain:: stem:[(-\infty,\infty)] +Range:: stem:[(-\infty,\infty)] + +[.center] +image::Cubic-Functions.png[] + +== Absolute Value Functions +[.example] +[stem] +++++ +y = |x| +++++ + +Domain:: stem:[(-\infty,\infty)] +Range:: stem:[[0,\infty)] + +[.center] +image::Absolute-Value-Functions.png[] + +== Radical Functions +=== Square Root +[.example] +[stem] +++++ +y = \sqrt{x} +++++ + +Domain:: stem:[(0,\infty)] +Range:: stem:[[0,\infty)] + +[.center] +image::Radical-Square-Root-Functions.png[] + +=== Cubed Root +[.example] +[stem] +++++ +y = \sqrt[3]{x} +++++ + +Domain:: stem:[(-\infty,\infty)] +Range:: stem:[(-\infty),\infty)] + +[.center] +image::Radical-Cubed-Root-Functions.png[] + +== Exponential Functions +[.example] +[stem] +++++ +y = b^x +++++ + +Horizontal Asymbtote::: 0 +Domain:: stem:[(\infty,\infty)] +Range:: stem:[(0,\infty)] + +[.center] +image::Exponential-Functions.png[] + +== Logoritmic Functions +[.example] +[stem] +++++ +y = log_{b} x +++++ + +Vertical Asymbtote:: 0 +Domain:: stem:[(0,\infty)] +Range:: stem:[(-\infty,\infty)] + +[.center] +image::Logorithmic-Functions.png[] + +== Reciprocal Functions +[.example] +[stem] +++++ +y = \frac{1}{x} +++++ + +Vertical Asymptote:: 0 +Domain:: stem:[(-\infty, 0) \cup (0, \infty)] +Range:: stem:[(-\infty, 0) \cup (0, \infty)] + +[.center] +image::Reciprocal-Functions.png[] diff --git a/Graphing/Parent-Functions.html b/Graphing/Parent-Functions.html new file mode 100644 index 0000000..8cc4d29 --- /dev/null +++ b/Graphing/Parent-Functions.html @@ -0,0 +1,1370 @@ + + + + + + + +Parent Functions + + + + + + + + + + + + + + + + + + +
+
+

Constant Functions

+
+
+
+\[y = c\] +
+
+
+
+
Domain
+
+

\((-\infty,\infty)\)

+
+
Range
+
+

\(C\) (constant)

+
+
+
+
+
+Constant Functions +
+
+
+
+
+

Linear Functions

+
+
+
+\[y = x\] +
+
+
+
+
Domain
+
+

\((-\infty,\infty)\)

+
+
Range
+
+

\((-\infty,\infty)\)

+
+
+
+
+
+Linear Functions +
+
+
+
+
+

Quadradic Functions

+
+
+
+\[y = x^2\] +
+
+
+
+
Domain
+
+

\((-\infty,\infty)\)

+
+
Range
+
+

\([0,\infty)\)

+
+
+
+
+
+Quadradic Functions +
+
+
+
+
+

Cubic Functions

+
+
+
+\[y = x^3\] +
+
+
+
+
Domain
+
+

\((-\infty,\infty)\)

+
+
Range
+
+

\((-\infty,\infty)\)

+
+
+
+
+
+Cubic Functions +
+
+
+
+
+

Absolute Value Functions

+
+
+
+\[y = |x|\] +
+
+
+
+
Domain
+
+

\((-\infty,\infty)\)

+
+
Range
+
+

\([0,\infty)\)

+
+
+
+
+
+Absolute Value Functions +
+
+
+
+
+

Radical Functions

+
+
+

Square Root

+
+
+\[y = \sqrt{x}\] +
+
+
+
+
Domain
+
+

\((0,\infty)\)

+
+
Range
+
+

\([0,\infty)\)

+
+
+
+
+
+Radical Square Root Functions +
+
+
+
+

Cubed Root

+
+
+\[y = \sqrt[3]{x}\] +
+
+
+
+
Domain
+
+

\((-\infty,\infty)\)

+
+
Range
+
+

\((-\infty),\infty)\)

+
+
+
+
+
+Radical Cubed Root Functions +
+
+
+
+
+
+

Exponential Functions

+
+
+
+\[y = b^x\] +
+
+
+
+
Horizontal Asymbtote
+
+

0

+
+
+
Domain
+
+

\((\infty,\infty)\)

+
+
Range
+
+

\((0,\infty)\)

+
+
+
+
+
+
+
+
+Exponential Functions +
+
+
+
+
+

Logoritmic Functions

+
+
+
+\[y = log_{b} x\] +
+
+
+
+
Vertical Asymbtote
+
+

0

+
+
Domain
+
+

\((0,\infty)\)

+
+
Range
+
+

\((-\infty,\infty)\)

+
+
+
+
+
+Logorithmic Functions +
+
+
+
+
+

Reciprocal Functions

+
+
+
+\[y = \frac{1}{x}\] +
+
+
+
+
Vertical Asymptote
+
+

0

+
+
Domain
+
+

\((-\infty, 0) \cup (0, \infty)\)

+
+
Range
+
+

\((-\infty, 0) \cup (0, \infty)\)

+
+
+
+
+
+Reciprocal Functions +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Graphing/index.adoc b/Graphing/index.adoc new file mode 100644 index 0000000..12074b0 --- /dev/null +++ b/Graphing/index.adoc @@ -0,0 +1,10 @@ += ./Graphing +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Graphing/Parent-Functions.html;hb=HEAD[Parent-Functions] diff --git a/Graphing/index.html b/Graphing/index.html new file mode 100644 index 0000000..ad064c6 --- /dev/null +++ b/Graphing/index.html @@ -0,0 +1,1052 @@ + + + + + + + +./Graphing + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Linear-Algrebra/Graphing.adoc b/Linear-Algrebra/Graphing.adoc new file mode 100644 index 0000000..b5439cf --- /dev/null +++ b/Linear-Algrebra/Graphing.adoc @@ -0,0 +1,27 @@ += Linear Algrebra | Graphing +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +=== Three Equations + +== Exactly One Soltion (Independent) + +In a System of three equations, if three planes intersect at a single unique +point then there is only one soution. (like the corner of a room where two walls +meet the ceiling). + +== Span + +In linear algebra, the span of a set of vectors is the collection of all +possible vectors you can create by scaling and adding them together (their +linear combinations). + +== Links + +* https://math.libretexts.org/Courses/University_of_California_Irvine/Linear_Algebra_(Math_3A_UCI)/01%3A_Systems_of_Linear_Equations/1.01%3A_Systems_of_Linear_Equations/1.1.01%3A_Lines_Planes_and_Space diff --git a/Linear-Algrebra/Graphing.html b/Linear-Algrebra/Graphing.html new file mode 100644 index 0000000..bcb2d87 --- /dev/null +++ b/Linear-Algrebra/Graphing.html @@ -0,0 +1,1109 @@ + + + + + + + +Linear Algrebra | Graphing + + + + + + + + + + + + + + + + + + +
+
+

Three Equations

+ +
+
+

Exactly One Soltion (Independent)

+
+
+

In a System of three equations, if three planes intersect at a single unique +point then there is only one soution. (like the corner of a room where two walls +meet the ceiling).

+
+
+
+
+

Span

+
+
+

In linear algebra, the span of a set of vectors is the collection of all +possible vectors you can create by scaling and adding them together (their +linear combinations).

+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Matrix-Math.adoc b/Linear-Algrebra/Matrix-Math.adoc new file mode 100644 index 0000000..ef106ac --- /dev/null +++ b/Linear-Algrebra/Matrix-Math.adoc @@ -0,0 +1,110 @@ += Linear Algrebra | Matrix Math +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + + +== Matrix Multiplication + +[stem] +++++ +A \otimes B = C +\\ +A = +\left[\begin{array}{a} +a_{11} & a_{12} \\ +a_{21} & a_{22} \\ +\end{array}\right] + +B = +\left[\begin{array}{b} +b_{11} & b_{12} & b_{13} \\ +b_{21} & b_{22} & b_{23} \\ +\end{array}\right] +\\ +C = +\\ +\left[\begin{array}{cc} +(a_{11} * b_{11}) + (a_{12} * b_{21}), & (a_{11} * b_{12}) + (a_{12} * b_{22}), & (a_{11} * b_{13}) + (a_{12} * b_{23}) \\ +(a_{21} * b_{11}) + (a_{22} * b_{21}), & (a_{21} * b_{12}) + (a_{22} * b_{22}), & (a_{21} * b_{13}) + (a_{22} * b_{23}) +\end{array}\right] +++++ + + +.Alternative Matrix Multiplication +[%collapsible] +==== +[.center] +image::alternative-matrix-multiplication.png[] +==== + +=== Demension Rules + +[stem] +++++ +B = p \times q +\\ +A = m \times n +++++ + +stem:[n] has to be equal to stem:[p]. The demensions of stem:[C] is stem:[m \times q] + +== Matrix and Scalar Multiplication + +[stem] +++++ +k \left[\begin{array}{cc} +4 & 3 \\ +9 & 3 \\ +\end{array}\right] = +\left[\begin{array}{cc} +k4 & k3 \\ +k9 & k3 \\ +\end{array}\right] +++++ + +== Matrix addition + +When you add or subtract matrices they need to be the same dimensions + +[stem] +++++ +\left[\begin{array}{cc} +4 & 3 & 5 \\ +2 & 1 & -4 \\ +1 & 2 & 2 \\ +\end{array}\right] + + +\left[\begin{array}{cc} +1 & 2 & 3 \\ +4 & 5 & -6 \\ +8 & 4 & 2 \\ +\end{array}\right] = + +\left[\begin{array}{cc} +4 + 1 & 3 + 2 & 5 + 3 \\ +2 + 4 & 1 + 4 & -4 + -6 \\ +1 + 8 & 2 + 4 & 2 + 2 \\ +\end{array}\right] +++++ + +== Three-Dimensional Transformations + +[.center] +image::3d-vector-transformation.png[] + +[.center] +image::3d-vecctor-transformation2.png[] + +== Links +* https://www.youtube.com/watch?v=pJzmiywagfY[Khan Academy | Cross Product] +* https://www.youtube.com/watch?v=eu6i7WJeinw[3Blue 1 Brown | Cross Product] +* https://www.youtube.com/watch?v=Ip3X9LOh2dk[3Blue 1 Brown | Determinant] +* https://youtu.be/XkY2DOUCWMU?si=GHdQZPG1qktUV1f2[3Blue 1Brown | Matrix Multiplication] +* https://www.youtube.com/watch?v=rHLEWRxRGiM&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab&index=6[3Blue 1Brown | Three-dimensional linear transofrmation] +* https://interactivetextbooks.tudelft.nl/linear-algebra/Chapter5/DeterminantsGeometric.html diff --git a/Linear-Algrebra/Matrix-Math.html b/Linear-Algrebra/Matrix-Math.html new file mode 100644 index 0000000..300184c --- /dev/null +++ b/Linear-Algrebra/Matrix-Math.html @@ -0,0 +1,1218 @@ + + + + + + + +Linear Algrebra | Matrix Math + + + + + + + + + + + + + + + + + + +
+
+

Matrix Multiplication

+
+
+
+\[A \otimes B = C +\\ +A = +\left[\begin{array}{a} +a_{11} & a_{12} \\ +a_{21} & a_{22} \\ +\end{array}\right] + +B = +\left[\begin{array}{b} +b_{11} & b_{12} & b_{13} \\ +b_{21} & b_{22} & b_{23} \\ +\end{array}\right] +\\ +C = +\\ +\left[\begin{array}{cc} +(a_{11} * b_{11}) + (a_{12} * b_{21}), & (a_{11} * b_{12}) + (a_{12} * b_{22}), & (a_{11} * b_{13}) + (a_{12} * b_{23}) \\ +(a_{21} * b_{11}) + (a_{22} * b_{21}), & (a_{21} * b_{12}) + (a_{22} * b_{22}), & (a_{21} * b_{13}) + (a_{22} * b_{23}) +\end{array}\right]\] +
+
+
+Alternative Matrix Multiplication +
+
+
+alternative matrix multiplication +
+
+
+
+
+

Demension Rules

+
+
+\[B = p \times q +\\ +A = m \times n\] +
+
+
+

\(n\) has to be equal to \(p\). The demensions of \(C\) is \(m \times q\)

+
+
+
+
+
+

Matrix and Scalar Multiplication

+
+
+
+\[k \left[\begin{array}{cc} +4 & 3 \\ +9 & 3 \\ +\end{array}\right] = +\left[\begin{array}{cc} +k4 & k3 \\ +k9 & k3 \\ +\end{array}\right]\] +
+
+
+
+
+

Matrix addition

+
+
+

When you add or subtract matrices they need to be the same dimensions

+
+
+
+\[\left[\begin{array}{cc} +4 & 3 & 5 \\ +2 & 1 & -4 \\ +1 & 2 & 2 \\ +\end{array}\right] + + +\left[\begin{array}{cc} +1 & 2 & 3 \\ +4 & 5 & -6 \\ +8 & 4 & 2 \\ +\end{array}\right] = + +\left[\begin{array}{cc} +4 + 1 & 3 + 2 & 5 + 3 \\ +2 + 4 & 1 + 4 & -4 + -6 \\ +1 + 8 & 2 + 4 & 2 + 2 \\ +\end{array}\right]\] +
+
+
+
+
+

Three-Dimensional Transformations

+
+
+
+3d vector transformation +
+
+
+
+3d vecctor transformation2 +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Matrix-Operations.adoc b/Linear-Algrebra/Matrix-Operations.adoc new file mode 100644 index 0000000..724df74 --- /dev/null +++ b/Linear-Algrebra/Matrix-Operations.adoc @@ -0,0 +1,290 @@ += Linear Algrebra | Matrix Operations +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Determinants + +=== Solving 2x2 matrices + +[stem] +++++ +\det(A) = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc +++++ + +[.center] +image::helpful-2x2-matrix-determinant-diagram.png[] + +=== Solving 3x3 matrices + +[stem] +++++ +\det \begin{pmatrix} \begin{bmatrix} +a & b & c \\ +d & e & f \\ +g & h & i +\end{bmatrix} \end{pmatrix} = a \det \begin{pmatrix} \begin{bmatrix} +e & f \\ +h & i +\end{bmatrix} \end{pmatrix} - b \det \begin{pmatrix} \begin{bmatrix} +d & f \\ +g & i +\end{bmatrix} \end{pmatrix} + c \det \begin{pmatrix} \begin{bmatrix} +d & e \\ +g & h +\end{bmatrix} \end{pmatrix} +++++ + +[.center] +image::rule-of-sarrus-3x3-matrices.png[] + +[.center] +image::six-tetrahedra-parallepiped.png[] + +=== Two-Dimensions + +Determinants determine the scale of the area of how much the unit parallelogram is +scaled. + +* *Positive Determinant*: stem:[\hat{i}] is right of stem:[\hat{j}]. + +* *Negative Determinant*: stem:[\hat{i}] is left of stem:[\hat{j}]. + +NOTE: You can think of negative determinants as inverting the oriantation of + space. (i.e. a reflection) + +=== Three-Dimensions + +Determinants determine the scale of the volume of how much the unit +parallelepiped is scaled into. A parallelpiped is prism whose bases are +parallelograms. (e.i. warped cubed). + +[.center] +image::3d-determinant-area.png[] + +.Right hand | Positive determinant | unchanged orientation +[.center] +image::right-hand-rule2.png[] + +.Left hand | Negative determinant, changed orientation +[.center] +image::right-hand-rule-flipped.png[] + +== Eigenvalues and Eigenvectors + +== Transposing a matrix + +.Matrix Transpose Property +[stem] +++++ +A = \begin{bmatrix} +a & b & c \\ +d & e & f +\end{bmatrix}, \quad +A^T = \begin{bmatrix} +a & d \\ +b & e \\ +c & f +\end{bmatrix} +++++ + +== Getting an Identity Matrix with Multiplication + +When a matrix is multiplied by an inverse matrix the result is an identity +matrix + +== Row Operations to get a Row-Equivalent Matrix + +To change a Matrix to a Row-Equivalent Matrix, you can: + +* Interchange any two rows + +* Multiply all the elements in a row by a real number (not 0). + +* Add multiples of the elements of one row the the elements of another row. + +== Row Notation Examples + +[.inline] +* *stem:[R_1 \leftrightarrow R_4]*: replace one row with the other + +[.inline] +* *stem:[\frac{1}{5}R_3 \leftrightarrow R_3]*: One-fith of each element in row 3 + becomes the new row 3. + +[.inline] +* *stem:[-5R_1 + R_2 \rightarrow R_2]*: Negative 5 times each element in row 1 + added to the elements in row 2 produces a new row 2. + +== Getting Inverse Matrix + +=== 2x2 Method + +[stem] +++++ +M = \left[\begin{array}{cc} +a & b \\ +c & d \\ +\end{array}\right] +++++ + +[stem] +++++ +M^{-1} = +\left[\begin{array}{cc} +\frac{d}{ab - bc} & \frac{-b}{ad - bc} \\ +\frac{-c}{ad - bc} & \frac{a}{ad - bc} \\ +\end{array}\right] +++++ + +If stem:[ad - bc] is zero, then the matrix has no inverse + +Many square matrices have inverses. When a matrix and its inverse are multiplied +together, in either order, the result is an identity matrix. Matrix inverses are +used to solve problems involving systems of equations and to perform matrix +_division_. + +=== Row Reduction Method for Square Matrices + +Create an identity matrix on the right. We go throught the row operations to +transform the left-hand matrix into an identity matix. When You're finished, the +matrix on teh right is the original matrix's inverse. + +.example +[example%collapsible] +==== + +.Row Operation #3 +[stem] +++++ +D = \left[ +\begin{array}{ccc|ccc} + 1 & 2 & 0 & 1 & 0 & 0 \\ +-1 & 3 & 3 & 0 & 1 & 0 \\ + 0 & 1 & 1 & 0 & 0 & 1 \\ +\end{array} +\right] +++++ + +.Row Operation #2 +[stem] +++++ +R_1 + R_2 \rightarrow R_2 \left[ +\begin{array}{ccc|ccc} + 1 & 2 & 0 & 1 & 0 & 0 \\ + 0 & 5 & 3 & 1 & 1 & 0 \\ + 0 & 1 & 1 & 0 & 0 & 1 \\ +\end{array} +\right] +++++ + +.Row Operation #3 +[stem] +++++ +-R_2 + R_1 \rightarrow R_2 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & -\frac{6}{5} & \frac{3}{5} & -\frac{2}{5} & 0 \\ + 0 & 1 & \frac{3}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ + 0 & 0 & \frac{2}{5} & -\frac{1}{5} & -\frac{1}{5} & 1 \\ +\end{array} +\right] +++++ + +.Row Operation #2 +[stem] +++++ +\frac{5}{2}R_3 \rightarrow R_2 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & -\frac{6}{5} & \frac{3}{5} & -\frac{2}{5} & 0 \\ + 0 & 1 & \frac{3}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ + 0 & 0 & 1 & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ +\end{array} +\right] +++++ + +.Row Operation #3 +[stem] +++++ +\frac{6}{5}R_3 + R_1 \rightarrow R_1 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & 0 & 0 & -1 & 0 \\ + 0 & 1 & \frac{3}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ + 0 & 0 & 1 & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ +\end{array} +\right] +++++ + +.Row Operation #3 +[stem] +++++ +\frac{6}{5}R_3 + R_1 \rightarrow R_1 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & 0 & 0 & -1 & 3 \\ + 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2} & -\frac{3}{2} \\ + 0 & 0 & 1 & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ +\end{array} +\right] +++++ + +You now have the identity matrix on the left and the inverse matrix +(stem:[D^{-1}]) on the right. stem:[D * D^{-1}] and stem:[D^{-1} * D] both equal +the identity matrix. + +==== + +== Shear + +=== Horizontal Shear + +[.center] +image::horizontal-shear.png[] + +[stem] +++++ +\left[\begin{array}{cc} +x'\\ +y'\\ +\end{array}\right] = +\left[\begin{array}{cc} +1 & k\\ +0 & 1\\ +\end{array}\right] +\left[\begin{array}{cc} +x\\ +y\\ +\end{array}\right] +++++ + +* stem:[x' = x + ky] + +* stem:[y' = y] + +=== Vertical Shear + +[.center] +image::vertical-shear.png[] + +[stem] +++++ +\left[\begin{array}{cc} +x'\\ +y'\\ +\end{array}\right] = +\left[\begin{array}{cc} +1 & 0\\ +k & 1\\ +\end{array}\right] +\left[\begin{array}{cc} +x\\ +y\\ +\end{array}\right] +++++ + +* stem:[x' = x] + +* stem:[y' = y + kx] diff --git a/Linear-Algrebra/Matrix-Operations.html b/Linear-Algrebra/Matrix-Operations.html new file mode 100644 index 0000000..c26f236 --- /dev/null +++ b/Linear-Algrebra/Matrix-Operations.html @@ -0,0 +1,1478 @@ + + + + + + + +Linear Algrebra | Matrix Operations + + + + + + + + + + + + + + + + + + +
+
+

Determinants

+
+
+

Solving 2x2 matrices

+
+
+\[\det(A) = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc\] +
+
+
+
+helpful 2x2 matrix determinant diagram +
+
+
+
+

Solving 3x3 matrices

+
+
+\[\det \begin{pmatrix} \begin{bmatrix} +a & b & c \\ +d & e & f \\ +g & h & i +\end{bmatrix} \end{pmatrix} = a \det \begin{pmatrix} \begin{bmatrix} +e & f \\ +h & i +\end{bmatrix} \end{pmatrix} - b \det \begin{pmatrix} \begin{bmatrix} +d & f \\ +g & i +\end{bmatrix} \end{pmatrix} + c \det \begin{pmatrix} \begin{bmatrix} +d & e \\ +g & h +\end{bmatrix} \end{pmatrix}\] +
+
+
+
+rule of sarrus 3x3 matrices +
+
+
+
+six tetrahedra parallepiped +
+
+
+
+

Two-Dimensions

+
+

Determinants determine the scale of the area of how much the unit parallelogram is +scaled.

+
+
+
    +
  • +

    Positive Determinant: \(\hat{i}\) is right of \(\hat{j}\).

    +
    • +
  • +

    Negative Determinant: \(\hat{i}\) is left of \(\hat{j}\).

    +
    • +
+
+
+ + + + + +
+ + +You can think of negative determinants as inverting the oriantation of + space. (i.e. a reflection) +
+
+
+
+

Three-Dimensions

+
+

Determinants determine the scale of the volume of how much the unit +parallelepiped is scaled into. A parallelpiped is prism whose bases are +parallelograms. (e.i. warped cubed).

+
+
+
+3d determinant area +
+
+
+
+right hand rule2 +
+
Figure 1. Right hand | Positive determinant | unchanged orientation
+
+
+
+right hand rule flipped +
+
Figure 2. Left hand | Negative determinant, changed orientation
+
+
+
+
+
+

Eigenvalues and Eigenvectors

+
+ +
+
+
+

Transposing a matrix

+
+
+
Matrix Transpose Property
+
+\[A = \begin{bmatrix} +a & b & c \\ +d & e & f +\end{bmatrix}, \quad +A^T = \begin{bmatrix} +a & d \\ +b & e \\ +c & f +\end{bmatrix}\] +
+
+
+
+
+

Getting an Identity Matrix with Multiplication

+
+
+

When a matrix is multiplied by an inverse matrix the result is an identity +matrix

+
+
+
+
+

Row Operations to get a Row-Equivalent Matrix

+
+
+

To change a Matrix to a Row-Equivalent Matrix, you can:

+
+
+
    +
  • +

    Interchange any two rows

    +
    • +
  • +

    Multiply all the elements in a row by a real number (not 0).

    +
    • +
  • +

    Add multiples of the elements of one row the the elements of another row.

    +
    • +
+
+
+
+
+

Row Notation Examples

+
+
+
    +
  • +

    \(R_1 \leftrightarrow R_4\): replace one row with the other

    +
    • +
+
+
+
    +
  • +

    \(\frac{1}{5}R_3 \leftrightarrow R_3\): One-fith of each element in row 3 +becomes the new row 3.

    +
    • +
+
+
+
    +
  • +

    \(-5R_1 + R_2 \rightarrow R_2\): Negative 5 times each element in row 1 +added to the elements in row 2 produces a new row 2.

    +
    • +
+
+
+
+
+

Getting Inverse Matrix

+
+
+

2x2 Method

+
+
+\[M = \left[\begin{array}{cc} +a & b \\ +c & d \\ +\end{array}\right]\] +
+
+
+
+\[M^{-1} = +\left[\begin{array}{cc} +\frac{d}{ab - bc} & \frac{-b}{ad - bc} \\ +\frac{-c}{ad - bc} & \frac{a}{ad - bc} \\ +\end{array}\right]\] +
+
+
+

If \(ad - bc\) is zero, then the matrix has no inverse

+
+
+

Many square matrices have inverses. When a matrix and its inverse are multiplied +together, in either order, the result is an identity matrix. Matrix inverses are +used to solve problems involving systems of equations and to perform matrix +division.

+
+
+
+

Row Reduction Method for Square Matrices

+
+

Create an identity matrix on the right. We go throught the row operations to +transform the left-hand matrix into an identity matix. When You’re finished, the +matrix on teh right is the original matrix’s inverse.

+
+
+example +
+
+
Row Operation #3
+
+\[D = \left[ +\begin{array}{ccc|ccc} + 1 & 2 & 0 & 1 & 0 & 0 \\ +-1 & 3 & 3 & 0 & 1 & 0 \\ + 0 & 1 & 1 & 0 & 0 & 1 \\ +\end{array} +\right]\] +
+
+
+
Row Operation #2
+
+\[R_1 + R_2 \rightarrow R_2 \left[ +\begin{array}{ccc|ccc} + 1 & 2 & 0 & 1 & 0 & 0 \\ + 0 & 5 & 3 & 1 & 1 & 0 \\ + 0 & 1 & 1 & 0 & 0 & 1 \\ +\end{array} +\right]\] +
+
+
+
Row Operation #3
+
+\[-R_2 + R_1 \rightarrow R_2 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & -\frac{6}{5} & \frac{3}{5} & -\frac{2}{5} & 0 \\ + 0 & 1 & \frac{3}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ + 0 & 0 & \frac{2}{5} & -\frac{1}{5} & -\frac{1}{5} & 1 \\ +\end{array} +\right]\] +
+
+
+
Row Operation #2
+
+\[\frac{5}{2}R_3 \rightarrow R_2 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & -\frac{6}{5} & \frac{3}{5} & -\frac{2}{5} & 0 \\ + 0 & 1 & \frac{3}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ + 0 & 0 & 1 & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ +\end{array} +\right]\] +
+
+
+
Row Operation #3
+
+\[\frac{6}{5}R_3 + R_1 \rightarrow R_1 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & 0 & 0 & -1 & 0 \\ + 0 & 1 & \frac{3}{5} & \frac{1}{5} & \frac{1}{5} & 0 \\ + 0 & 0 & 1 & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ +\end{array} +\right]\] +
+
+
+
Row Operation #3
+
+\[\frac{6}{5}R_3 + R_1 \rightarrow R_1 \left[ +\begin{array}{ccc|ccc} + 1 & 0 & 0 & 0 & -1 & 3 \\ + 0 & 1 & 0 & \frac{1}{2} & \frac{1}{2} & -\frac{3}{2} \\ + 0 & 0 & 1 & -\frac{1}{2} & -\frac{1}{2} & \frac{5}{2} \\ +\end{array} +\right]\] +
+
+
+

You now have the identity matrix on the left and the inverse matrix +(\(D^{-1}\)) on the right. \(D * D^{-1}\) and \(D^{-1} * D\) both equal +the identity matrix.

+
+
+
+
+
+
+
+

Shear

+
+
+

Horizontal Shear

+
+
+horizontal shear +
+
+
+
+\[\left[\begin{array}{cc} +x'\\ +y'\\ +\end{array}\right] = +\left[\begin{array}{cc} +1 & k\\ +0 & 1\\ +\end{array}\right] +\left[\begin{array}{cc} +x\\ +y\\ +\end{array}\right]\] +
+
+
+
    +
  • +

    \(x' = x + ky\)

    +
    • +
  • +

    \(y' = y\)

    +
    • +
+
+
+
+

Vertical Shear

+
+
+vertical shear +
+
+
+
+\[\left[\begin{array}{cc} +x'\\ +y'\\ +\end{array}\right] = +\left[\begin{array}{cc} +1 & 0\\ +k & 1\\ +\end{array}\right] +\left[\begin{array}{cc} +x\\ +y\\ +\end{array}\right]\] +
+
+
+
    +
  • +

    \(x' = x\)

    +
    • +
  • +

    \(y' = y + kx\)

    +
    • +
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Matrix-Types.adoc b/Linear-Algrebra/Matrix-Types.adoc new file mode 100644 index 0000000..9e40998 --- /dev/null +++ b/Linear-Algrebra/Matrix-Types.adoc @@ -0,0 +1,161 @@ += Linear Algrebra | Matrix Types +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Square Matrices + +[cols="1,2a", options="header"] +|=== +| Matrix Type | Definition & Example + +| **Upper Triangular** +| All the elements below the main diagonal are 0s. + +[stem] +++++ +\begin{bmatrix} +4 & 3 & 5 \\ +0 & 1 & -4 \\ +0 & 0 & 2 +\end{bmatrix} +++++ + +| **Lower Triangular** +| All the elements above the main diagonal are 0s. + +[stem] +++++ +\begin{bmatrix} +7 & 0 & 0 & 0 \\ +7 & 8 & 0 & 0 \\ +5 & 2 & 3 & 0 \\ +5 & -4 & -9 & 1 +\end{bmatrix} +++++ + +| **Diagonal** +| All entries above and below the main diagonal are 0s. + +[stem] +++++ +\begin{bmatrix} +4 & 0 & 0 & 0 \\ +0 & 3 & 0 & 0 \\ +0 & 0 & 1 & 0 \\ +0 & 0 & 0 & 2 +\end{bmatrix} +++++ +|=== + +== Addative Identity Matrix + +The identity matix needs to be the same dimensions as the original. + +[stem] +++++ +\left[\begin{array}{cc} +4 & 3 & 5 \\ +2 & 1 & -4 \\ +1 & 2 & 2 \\ +\end{array}\right] \to + +\left[\begin{array}{cc} +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +\end{array}\right] +++++ + +== Multiplicative Identity Matrix + +One's are on the diagonal from top left to bottom right. Every other element is +zero + +[stem] +++++ +\left[\begin{array}{a} +1 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 1 \\ +\end{array}\right] +++++ + +== Reduced Row Echelon Form and Rown Echelon Form + +[stem] +++++ +\begin{array}{cc} + \text{Reduced Row Echelon Form} & \text{Row Echelon Form} \\ + \begin{bmatrix} + 1 & 0 & 0 & 3 & 4 \\ + 0 & 1 & 0 & 0 & 2 \\ + 0 & 0 & 1 & -1 & 1 \\ + 0 & 0 & 0 & 0 & 0 + \end{bmatrix} & + \begin{bmatrix} + 1 & 3 & 0 & 2 & -3 \\ + 0 & 1 & 4 & 0 & 3 \\ + 0 & 0 & 1 & 1 & 4 \\ + 0 & 0 & 0 & 0 & 0 + \end{bmatrix} +\end{array} +++++ + +=== Reduced Row Echelon Form + +When a matrix is in the _reduced row echelon form_, it has the following +characteristics: + +[.inline] +* If the matrix has any rows whose elements are all 0x, those rows are at the + bottom of the matrix + +[.inline] +* Reading from left to right, the first element of a row that is _not_ a 0 is + always a 1. This first 1 is called the _leading_ 1 of its row + +[.inline] +* In any row, the leading 1 is always to the right of and below any leading 1 in + a row above. + +[.inline] +* If a column contains a leading 1, then all the entries above and below that + leading 1 are 0s. + +=== Row Echelon Form + +The same as roe Echelon form except it doesn't require 0s above and below each +leading 1. + +== Augmented matrix + +An augmented matrix is a compact way of representing a system of linear +equations by combining the coefficients of the variables and the constant terms +into a single matrix. + +[stem] +++++ +\begin{cases} +\begin{alignat*}{6} +-x & {}-{} & y & {}-{} & z & = 3 \\ +4x & {}+{} & 5y & & & = -6 \\ + & & y & {}-{} & 3z & = 5 +\end{alignat*} +\end{cases} + +\qquad % + +\left[ +\begin{array}{ccc : c} +-1 & -1 & -1 & 3 \\ + 4 & 5 & 0 & -6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right] +++++ diff --git a/Linear-Algrebra/Matrix-Types.html b/Linear-Algrebra/Matrix-Types.html new file mode 100644 index 0000000..5f3a6b5 --- /dev/null +++ b/Linear-Algrebra/Matrix-Types.html @@ -0,0 +1,1287 @@ + + + + + + + +Linear Algrebra | Matrix Types + + + + + + + + + + + + + + + + + + +
+
+

Square Matrices

+
+ ++++ + + + + + + + + + + + + + + + + + + + + +
Matrix TypeDefinition & Example

Upper Triangular

+

All the elements below the main diagonal are 0s.

+
+
+
+\[\begin{bmatrix} +4 & 3 & 5 \\ +0 & 1 & -4 \\ +0 & 0 & 2 +\end{bmatrix}\] +
+

Lower Triangular

+

All the elements above the main diagonal are 0s.

+
+
+
+\[\begin{bmatrix} +7 & 0 & 0 & 0 \\ +7 & 8 & 0 & 0 \\ +5 & 2 & 3 & 0 \\ +5 & -4 & -9 & 1 +\end{bmatrix}\] +
+

Diagonal

+

All entries above and below the main diagonal are 0s.

+
+
+
+\[\begin{bmatrix} +4 & 0 & 0 & 0 \\ +0 & 3 & 0 & 0 \\ +0 & 0 & 1 & 0 \\ +0 & 0 & 0 & 2 +\end{bmatrix}\] +
+
+
+
+
+

Addative Identity Matrix

+
+
+

The identity matix needs to be the same dimensions as the original.

+
+
+
+\[\left[\begin{array}{cc} +4 & 3 & 5 \\ +2 & 1 & -4 \\ +1 & 2 & 2 \\ +\end{array}\right] \to + +\left[\begin{array}{cc} +0 & 0 & 0 \\ +0 & 0 & 0 \\ +0 & 0 & 0 \\ +\end{array}\right]\] +
+
+
+
+
+

Multiplicative Identity Matrix

+
+
+

One’s are on the diagonal from top left to bottom right. Every other element is +zero

+
+
+
+\[\left[\begin{array}{a} +1 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 1 \\ +\end{array}\right]\] +
+
+
+
+
+

Reduced Row Echelon Form and Rown Echelon Form

+
+
+
+\[\begin{array}{cc} + \text{Reduced Row Echelon Form} & \text{Row Echelon Form} \\ + \begin{bmatrix} + 1 & 0 & 0 & 3 & 4 \\ + 0 & 1 & 0 & 0 & 2 \\ + 0 & 0 & 1 & -1 & 1 \\ + 0 & 0 & 0 & 0 & 0 + \end{bmatrix} & + \begin{bmatrix} + 1 & 3 & 0 & 2 & -3 \\ + 0 & 1 & 4 & 0 & 3 \\ + 0 & 0 & 1 & 1 & 4 \\ + 0 & 0 & 0 & 0 & 0 + \end{bmatrix} +\end{array}\] +
+
+
+

Reduced Row Echelon Form

+
+

When a matrix is in the reduced row echelon form, it has the following +characteristics:

+
+
+
    +
  • +

    If the matrix has any rows whose elements are all 0x, those rows are at the +bottom of the matrix

    +
    • +
+
+
+
    +
  • +

    Reading from left to right, the first element of a row that is not a 0 is +always a 1. This first 1 is called the leading 1 of its row

    +
    • +
+
+
+
    +
  • +

    In any row, the leading 1 is always to the right of and below any leading 1 in +a row above.

    +
    • +
+
+
+
    +
  • +

    If a column contains a leading 1, then all the entries above and below that +leading 1 are 0s.

    +
    • +
+
+
+
+

Row Echelon Form

+
+

The same as roe Echelon form except it doesn’t require 0s above and below each +leading 1.

+
+
+
+
+
+

Augmented matrix

+
+
+

An augmented matrix is a compact way of representing a system of linear +equations by combining the coefficients of the variables and the constant terms +into a single matrix.

+
+
+
+\[\begin{cases} +\begin{alignat*}{6} +-x & {}-{} & y & {}-{} & z & = 3 \\ +4x & {}+{} & 5y & & & = -6 \\ + & & y & {}-{} & 3z & = 5 +\end{alignat*} +\end{cases} + +\qquad % + +\left[ +\begin{array}{ccc : c} +-1 & -1 & -1 & 3 \\ + 4 & 5 & 0 & -6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right]\] +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Properties.adoc b/Linear-Algrebra/Properties.adoc new file mode 100644 index 0000000..016ba69 --- /dev/null +++ b/Linear-Algrebra/Properties.adoc @@ -0,0 +1,138 @@ += Linear Algrebra | Properties +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Properties + +[cols="^.^1, ^.^1, ^.^1, <.^2", frame="none", grid="rows", options="header"] +|=== +^| *Property* ^| *Parts* ^| *Definition* ^| *Details* + +| *Conditionally Commutative* | matrices +| stem:[A B \neq B A] +| Matrix multiplication is generally *not* commutative. Exceptions include the + identity matrix stem:[I], the inverse stem:[A^{-1}], or a zero matrix. + +| *Associative* | matrices +| stem:[(AB)C = A(BC)] +| Matrix multiplication is associative. As long as the order stays the same, you +can group them however you like. + +| *Commutative* | scalar and matrices +| stem:[kA = Ak] +| Scalar multiplication is commutative—you can place the scalar stem:[k] in +front or behind the matrix. + +| *Mixed Associative* | Scalar and Matrices +| stem:[(kA)B = k(AB)] +| You can multiply a matrix by a scalar first, or multiply the matrices first + and then apply the scalar. + +| *Mixed Associative* | scalars +| stem:[(kl)A = k(lA)] +| You can multiply the two scalars together first, or apply them to the matrix + one at a time. + +| *Commutative* | matrices +| stem:[A + B = B + A] +| Matrix addition is fully commutative. The order of operands does not change + the resulting sum. + +| *Associative* | matrices +| stem:[(A + B) + C = A + (B + C)] +| When adding three or more matrices, the grouping does not matter. + +| *Distributive* | matrices +| stem:[A(B + C) = AB + AC] +| Matrix multiplication distributes over matrix addition. Note: stem:[A] must + remain on the left side for both products. + +| *Distributive* | scalar and matrix +| stem:[k(A + B) = kA + kB] +| A scalar multiplied by the sum of two matrices is equal to the sum of the + scalar multiplied by each matrix individually. + +| *Distributive* | scalar and matrix +| stem:[(k + l)A = kA + lA] +| A matrix multiplied by the sum of two scalars is equal to the sum of the + matrix multiplied by each scalar individually. + +| *Distributive* | scalar and matrix +| stem:[(A + B)k = Ak + Bk] +| Because scalar multiplication is commutative stem:[Ak = kA], distributing a + scalar from the right side works exactly the same as distributing from the + left. + +| *Multiplicative Property* | determinants +| stem:[\det(M_1 M_2) = \det(M_1)\det(M_2)] +| The determinant of a product of matrices is the product of their determinants. + This holds even though matrix multiplication itself is not commutative. +|=== + + +=== Multiplicative Identity Matrix + +Identity matrices are only communative if it is a square matrix. + +When you do not have a square matrix the demensions of the identity matrix is +different depending on the order. + +[stem] +++++ +D = \left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] +\\ +D * I = \left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] * +\left[\begin{array}{cc} +1 & 0 \\ +0 & 1 \\ +\end{array}\right] = +\left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] +\\ +I * D = +\left[\begin{array}{cc} +1 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 1 \\ +\end{array}\right] * +\left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] = +\left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] +++++ + +== Invertable Matrices + +* A matrix with a row of 0s or a column of 0s isn't invertible -- it has no + inverse. + +== Singular Matrix + +A matrix that that doesn't have a inverse + +== Non-Singular Matrix + +A matrix that that has a inverse diff --git a/Linear-Algrebra/Properties.html b/Linear-Algrebra/Properties.html new file mode 100644 index 0000000..d0336be --- /dev/null +++ b/Linear-Algrebra/Properties.html @@ -0,0 +1,1264 @@ + + + + + + + +Linear Algrebra | Properties + + + + + + + + + + + + + + + + + + +
+
+

Properties

+
+ ++++++ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
PropertyPartsDefinitionDetails

Conditionally Commutative

matrices

\(A B \neq B A\)

Matrix multiplication is generally not commutative. Exceptions include the + identity matrix \(I\), the inverse \(A^{-1}\), or a zero matrix.

Associative

matrices

\((AB)C = A(BC)\)

Matrix multiplication is associative. As long as the order stays the same, you +can group them however you like.

Commutative

scalar and matrices

\(kA = Ak\)

Scalar multiplication is commutative—you can place the scalar \(k\) in +front or behind the matrix.

Mixed Associative

Scalar and Matrices

\((kA)B = k(AB)\)

You can multiply a matrix by a scalar first, or multiply the matrices first + and then apply the scalar.

Mixed Associative

scalars

\((kl)A = k(lA)\)

You can multiply the two scalars together first, or apply them to the matrix + one at a time.

Commutative

matrices

\(A + B = B + A\)

Matrix addition is fully commutative. The order of operands does not change + the resulting sum.

Associative

matrices

\((A + B) + C = A + (B + C)\)

When adding three or more matrices, the grouping does not matter.

Distributive

matrices

\(A(B + C) = AB + AC\)

Matrix multiplication distributes over matrix addition. Note: \(A\) must + remain on the left side for both products.

Distributive

scalar and matrix

\(k(A + B) = kA + kB\)

A scalar multiplied by the sum of two matrices is equal to the sum of the + scalar multiplied by each matrix individually.

Distributive

scalar and matrix

\((k + l)A = kA + lA\)

A matrix multiplied by the sum of two scalars is equal to the sum of the + matrix multiplied by each scalar individually.

Distributive

scalar and matrix

\((A + B)k = Ak + Bk\)

Because scalar multiplication is commutative \(Ak = kA\), distributing a + scalar from the right side works exactly the same as distributing from the + left.

Multiplicative Property

determinants

\(\det(M_1 M_2) = \det(M_1)\det(M_2)\)

The determinant of a product of matrices is the product of their determinants. + This holds even though matrix multiplication itself is not commutative.

+
+

Multiplicative Identity Matrix

+
+

Identity matrices are only communative if it is a square matrix.

+
+
+

When you do not have a square matrix the demensions of the identity matrix is +different depending on the order.

+
+
+
+\[D = \left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] +\\ +D * I = \left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] * +\left[\begin{array}{cc} +1 & 0 \\ +0 & 1 \\ +\end{array}\right] = +\left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] +\\ +I * D = +\left[\begin{array}{cc} +1 & 0 & 0 \\ +0 & 1 & 0 \\ +0 & 0 & 1 \\ +\end{array}\right] * +\left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right] = +\left[\begin{array}{cc} +3 & 4 \\ +2 & -1 \\ +-5 & 6 \\ +\end{array}\right]\] +
+
+
+
+
+
+

Invertable Matrices

+
+
+
    +
  • +

    A matrix with a row of 0s or a column of 0s isn’t invertible — it has no +inverse.

    +
    • +
+
+
+
+
+

Singular Matrix

+
+
+

A matrix that that doesn’t have a inverse

+
+
+
+
+

Non-Singular Matrix

+
+
+

A matrix that that has a inverse

+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/SCRAP b/Linear-Algrebra/SCRAP new file mode 100644 index 0000000..826848e --- /dev/null +++ b/Linear-Algrebra/SCRAP @@ -0,0 +1,11 @@ +3Blue 1Brown linear algrebra series: +https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab + +problems +https://www.math.purdue.edu/~zhan4306/spring_2022_la/sec_1_9.pdf +https://math.libretexts.org/Courses/University_of_California_Irvine/Linear_Algebra_(Math_3A_UCI)/02%3A_Linear_Transformations_and_Matrix_Algebra/2.07%3A_Determinants_and_Volumes/2.7.01%3A_Determinants_for_Area_and_Volume + +3d graphing calculator +https://c3d.libretexts.org/CalcPlot3D/index.html + + diff --git a/Linear-Algrebra/System-of-Equations.adoc b/Linear-Algrebra/System-of-Equations.adoc new file mode 100644 index 0000000..f1f6f59 --- /dev/null +++ b/Linear-Algrebra/System-of-Equations.adoc @@ -0,0 +1,786 @@ += Linear Algrebra | System of Equations +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Solving + +=== Two Equations + +The tools or methods for solving systems of equations algebraically are + +* *Elimination*: You eliminate one of the variables in teh system to make it a + simpler system, while preserving the original relationship between variables + +* *Back Substitution*: You use back substitution after you've found a value for + one of the variables and tehn solve for the values of the other variables. + +.back substitution +[example%collapsible] +==== +[stem] +++++ +\begin{cases} +\begin{alignat*}{4} +2x & {}+{} & 3y & = 23 \\ +4x & {}-{} & 3y & = 1 +\end{alignat*} +\end{cases} +++++ + +[stem] +++++ +\begin{array}{r} +\begin{alignat*}{4} +2x & {}+{} & 3y & = 23 \\ +2x & {}+{} & 3y & = 1 \\ +\hline +6x & & & 24 \\ +x & {}={} & \frac{24}{4} & = 4 +\end{alignat*} +\end{array} +++++ + +[stem] +++++ +\begin{gather*} +2(4) + 3y = 23 \\ +8 + 3y = 23 \\ +3y = 15 \\ +y = 5 +\end{gather*} +++++ +==== + +.Elimination +[example%collapsible] +==== +[stem] +++++ +\begin{cases} +\begin{alignat*}{4} +3x & {}+{} & 5y & = 12 \\ +4x & {}-{} & 3y & = -13 +\end{alignat*} +\end{cases} +++++ + +[stem] +++++ +\begin{alignat*}{8} +(3x & {}+{} & 5y & = 12)3\ & =\ & 9x {}+{} & 15y & = 36 \\ +(4x & {}-{} & 3y & = -13)5\ & =\ & 2x {}-{} & 15y & = -65 \\ +\end{alignat*} +++++ + +[stem] +++++ +\begin{array}{r} +\begin{alignat*}{4} +3x & {}+{} & 5y & = 12 \\ +4x & {}-{} & 3y & = -13 \\ +\hline +29x & & & = -29 \\ +x & = & \frac{-29}{29} & = -1 +\end{alignat*} +\end{array} +++++ +==== + +=== Multiple Equations + +The following operations may be performed (preserving the original relationship +between the variables), giving you the solution to the original equation: + +* Interchange two equations (change their order) + +* Multiply or divide each term in an equation by a constant number + (stem:[\neq 0]) + +* Add two equations together, combining like terms. + +* Add the multiple of one equation to another equation + +NOTE: There is an alternative method involving matrices + +.example +[example%collapsible] +==== +[stem] +++++ +\begin{cases} +\begin{alignat*}{5} +2x & {}+{} & 3y & {}-{} & 4z & {}+{} & w & = 15 \\ + x & {}-{} & 2y & {}+{} & 3z & {}-{} & 2w & = -3 \\ +3x & {}+{} & 5y & {}+{} & z & {}-{} & w & = 20 \\ +4x & {}+{} & y & {}-{} & z & {}+{} & w & = 5 \\ +\end{alignat*} +\end{cases} +++++ + +remove the w + +[stem] +++++ + \begin{alignat*}{4} +2R_1 + R_2 &\to 5x &&+ 4y &&- 5z &&= 27 \\ +R_1 + R_3 &\to 5x &&+ 8y &&- 3z &&= 35 \\ +R_4 + R_3 &\to 7x &&+ 6y && &&= 25 +\end{alignat*} +++++ + +[stem] +++++ +\begin{cases} +\begin{alignat*}{5} +5x & {}+{} & 4y & {}-{} & 5z & = 15 \\ +5x & {}+{} & 8y & {}-{} & 3z & = 35 \\ +7x & {}+{} & 6y & & & = 25 \\ +\end{alignat*} +\end{cases} +++++ + +remove the last equations + +[stem] +++++ +\begin{array}{lllrcrcrcl} +3R_1 & \to & -15x & + & 12y & - & 15z & = & 81 \\ +-5R_2 & \to & -25x & - & 40y & + & 15z & = & -175 \\ +\hline +3R_1 + (-5R_2) & \to & -40x & - & 28y & & & = & -94 +\end{array} +++++ + +[stem] +++++ +\begin{cases} +\begin{alignat*}{4} + 7x & {}+{} & 6y & = 25 \\ +-10x & {}-{} & 28y & = -94 +\end{alignat*} +\end{cases} +++++ + +Final steps + +[stem] +++++ +\begin{array}{llrcrcrcl} +14R_1 & \to & 98x & + & 84y & = & 350 \\ +3R_2 & \to & -30x & - & 84y & = & -282 \\ +\hline +14R_1 + 3R_2 & \to & 68x & & & = & 68 +\end{array} +++++ + +getting x + +[stem] +++++ +\begin{align*} +68x &= 68 \\ +x &= 1 +\end{align*} +++++ + +Now, substitute stem:[x = 1] into stem:[7x + 6y = 25] to get y + +[stem] +++++ +\begin{align*} +7(1) + 6y &= 25 \\ +6y &= 18 \\ +y &= 3 +\end{align*} +++++ + +solve for z + +[stem] +++++ +\begin{align*} +5(1) + 8(3) - 3z &= 35 \\ +5 + 24 - 3z &= 35 \\ +29 - 3z &= 35 \\ +-3z &= 6 \\ +z &= -2 +\end{align*} +++++ + +solve for w + +[stem] +++++ +\begin{align*} +2(1) + 3(3) - 4(-2) + w &= 15 \\ +2 + 9 + 8 + w &= 15 \\ +19 + w &= 15 \\ +w &= -4 +\end{align*} +++++ + +solution + +[stem] +++++ +\begin{cases} +x = 1 \\ +y = 3 \\ +z = -2 \\ +w = -4 +\end{cases} +++++ + +==== + +== Linear Equations + +Linear equations are in the form: + +[stem] +++++ +a_1x_1 + a_2x_2 + a_3x_3 + \ldots + a_nx_n +++++ + +== System of linear equations + +This is a system of linear equations. In order to solve a system of linear +equations you have to find variables that make each line true. + +[stem] +++++ +\begin{cases} +2x + y = 10 \\ +x - y = 2 +\end{cases} +++++ + +answer: + +[stem] +++++ +\begin{cases} +x = 4 \\ +y = 2 +\end{cases} +++++ + +=== Matrices + +In linear algrebra we can simplify this to be a matrix, each column is for a +different variable. + +[stem] +++++ +\begin{cases} +\begin{alignat*}{5} + x_1 & {}-{} & 2x_2 & {}+{} & 3x_3 & {}-{} & 6x_4 & = 8 \\ + x_1 & & & {}-{} & 10x_3 & {}+{} & x_4 & = 11 \\ + 4x_1 & {}+{} & x_2 & & & {}-{} & 4x_4 & = 8 \\ + 3x_1 & {}+{} & 5x_2 & {}-{} & 7x_3 & & & = 6 +\end{alignat*} +\end{cases} +++++ + +[stem] +++++ +A = \left[\begin{array}{cc} +1 & -2 & 3 & 6 \\ +1 & 0 & -10 & 1 \\ +4 & 1 & 0 & -4 \\ +3 & 5 & -7 & 0 \\ +\end{array}\right] +++++ + +== Solutions + +A system of equations has a _solution_ when you have at least one set of numbers +that replaces variabels in the equations and makes each equation read as a true +statement. A system of equations may have more than one solution. Systems with +more than one solution generally have an infinite number of solutions, and those +solutions are written as an expression or rule in terms of a parameter. + +=== Characteristics of having just one solution + +1. Must contain linear equations + +2. The system of equations must have as many independent equations as there are + variables in the system. + * *independent equations*: This just means that none of the equations is a + multiple or combination of multiples of the other equations in the system. + +.example +[example%collapsible] +==== +Including the equation stem:[-4x_1 - 6x_2 = -4] would incroduce an equation that +isn't independent. That equation is obtained by multiplying the top equation by +-2, multiplying the bottom equation by 2, and adding the two multiples together. + +[stem] +++++ +\begin{cases} +3x_1 - 2x_2 = 16 \\ + x_1 - 5x_2 = 14 +\end{cases} +++++ +==== + +=== Characteristics of having multiple solutions + +When systems have mutiple solutions, you'll find a pattern or rule that predicts +what those solutions are. + +[stem] +++++ +\begin{cases} +\begin{alignat*}{4} + -x & {}+{} & y & {}+{} & z & = -1 \\ + -x & {}+{} & 2y & {}-{} & 3z & = -4 \\ + 3x & {}-{} & 2y & {}-{} & 7z & = 0 +\end{alignat*} +\end{cases} +++++ + + +The pattern in the solutions is that x is always two less than five time z, and +y is always three less than four times z. This has infinite solutions + +NOTE: linear algrebra for dummies p.67 + +=== Graphing two lines for two equations + +The graph of system of equations stem:[ax + by = c] and stem:[dx + ey = f], +where x and y area variables and a, b, c, d, e, and f are real numbers has: + +* Intersecting lines (exactly one solution) when the system is consistent and + the equations are _independent_ + +* Parallel lines (no solutions) when the system is _inconsistent_ + +* Coincident lines (infinit solutions) when the system in _consistent_ and the + equations are _dependent_ + +=== Three equations | Exactly One Soltion (Independent) + +In a System of three equations, if three planes intersect at a single unique +point then there is only one soution. (like the corner of a room where two walls +meet the ceiling). + +=== Planes + +==== No Solutions +Planes that have the same slant but different intersections of the axes will +never interesect + +[.center] +image::parallel-planes.png[] + +In mathematics, two planes are parallel if their equations have the same +coefficients for the x, y, and z variables, but different constants. + +For example, look at these two planes: + +* *Plane 1*: stem:[3x - 2y - 7z = 0] + +* *Plane 2*: stem:[3x - 2y - 7z = 10] + +Notice that the left-hand sides are identical. In technical terms, their normal +vectors (the direction perpendicular to the flat surface) are the same: [3, -2, +-7]. The different constants (0 vs 10) just shift one plane "above" or +"beside" the other without changing its angle. + +In linear algebra, a solution is a point where all the planes in your system +touch or intersect. Therefore, the system is called inconsistent, and we say +there are no solutions. + +==== Multiple Solutions + +There is only one specific case where parallel planes have multiple solutions: +if they are actually the same plane. + +consider these two plane equations + +[stem] +++++ +\begin{cases} +\begin{alignat*}{4} +x & {}+{} & y & {}+{} & z & = 1 \\ +2x & {}+{} & 2y & {}+{} & 2z & = 2 +\end{alignat*} +\end{cases} +++++ + +* *The "Double" Layer*: Even though there are two different mathematical + expressions (like stem:[x+y+z=1] and stem:[2x+2y+2z=2]), they represent the + same set of coordinates. + +* *Infinite Solutions*: Because every single point stem:[x, y, z] on the first + plane is also on the second, there are an infinite number of solutions. + +* *Linear Dependence*: In linear algebra terms, these equations are linearly + dependent. The second equation is simply the first equation multiplied by a + scalar (in this case, 2). + +== One Solution with Square Matrices and their Inverse + +To solve a system of _n_ equations with _n_ unkowns using an _n_ x _n_ matrix +and its inverse, do the following: + +1. Write each of the equations in the system with the variables in the same + order and the constant on the other sider of the equal sign from the + variable. + +2. Construct a _coeffiecient matrix_ (a square matrix whose elements are the + coefficients of the variables). + +3. Write the _constant matrix_ (an n x 1 column matrix) using the constants in + the equations. + +4. Find the inverse of the coefficient matrix. + +5. Multiply the inverse of the coefficient matrix times the constant matrix. + +WARNING: This only works with singular matrices + +.example +[%collapsible] +==== +[stem] +++++ +\left\{ \begin{array}{rclrclrcl} +0 & = & 3 & + & x & + & y & + & z \\ +4x & + & 5y & + & 6 & = & 0 & & \\ +y & = & 3z & + & 5 & & & & +\end{array} \right. +++++ + +[stem] +++++ +\begin{cases} +\begin{alignat*}{6} +-x & {}-{} & y & {}-{} & z & = 3 \\ +4x & {}+{} & 5y & & & = 3 \\ + & & y & {}-{} & 3z & = 5 +\end{alignat*} +\end{cases} +++++ + +[stem] +++++ +A = \begin{bmatrix} +-1 & -1 & -1 \\ + 4 & 5 & 0 \\ + 0 & 1 & -3 +\end{bmatrix} +++++ + +[stem] +++++ +B = \begin{bmatrix} +3 \\ +-6 \\ +5 +\end{bmatrix} +++++ + +[stem] +++++ +A^{-1} = \begin{bmatrix} + 15 & 4 & -5 \\ +-12 & -3 & 4 \\ +-4 & -1 & 1 +\end{bmatrix} +++++ + +[stem] +++++ +A^{-1} \otimes B = \begin{bmatrix} + 15 & 4 & -5 \\ +-12 & -3 & 4 \\ +-4 & -1 & 1 +\end{bmatrix} \otimes +\begin{bmatrix} + 3 \\ +-6 \\ + 5 +\end{bmatrix} = +\begin{bmatrix} +-4 \\ + 2 \\ +-1 +\end{bmatrix} +++++ + +==== + +== Solving Using an augmented matrix + +After wrting a system of equations as an augmented matrix, you perform row +operations until the metrix is in either row echelon form or reduced row echelon +form. + +* *Reduced Row echelon form*: Gives you the answer directly + +* *Row echelon form*: Required back-solving or algebraic manipulations to get + your answer. + +NOTE: Works for any size matrix. + +=== One Solution +.Problem +[%collapsible] +==== +[stem] +++++ +\begin{cases} +\begin{alignat*}{6} +-x & {}-{} & y & {}-{} & z & = 3 \\ +4x & {}+{} & 5y & & & = -6 \\ + & & y & {}-{} & 3z & = 5 +\end{alignat*} +\end{cases} +++++ + +[stem] +++++ +\left[ +\begin{array}{ccc : c} +-1 & -1 & -1 & 3 \\ + 4 & 5 & 0 & -6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right] +++++ + +[stem] +++++ +-1R_1 \to R_1 +\left[ +\begin{array}{ccc : c} + 1 & 1 & 1 & -3 \\ + 4 & 5 & 0 & -6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right] +++++ + +[stem] +++++ +-4R_1 + R_2 \to R_2 +\left[ +\begin{array}{ccc : c} + 1 & 1 & 1 & -3 \\ + 0 & 1 & -4 & 6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right] +++++ + +[stem] +++++ +-1R_2 + R_1 \to R_1 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 5 & -9 \\ + 0 & 1 & -4 & 6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right] +++++ + +[stem] +++++ +-1R_2 + R_3 \to R_3 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 5 & -9 \\ + 0 & 1 & -4 & 6 \\ +0 & 0 & 1 & -1 +\end{array} +\right] +++++ +==== + +.Row Echeleon Form +[%collapsible] +==== + +back substitute for z + +[stem] +++++ +\begin{align*} +y - 4(-1) &= 6 \\ +y + 4 &= 6 \\ +y &= 2 +\end{align*} +++++ + +back substitute for for z again + +[stem] +++++ +\begin{align*} +x + 5(-1) &= -9 \\ +x - 5 &= -9 \\ +x &= -4 +\end{align*} +++++ + +solution + +[stem] +++++ +(x, y, z) = (-4, 2, -1) +++++ +==== + +.Reduced Row Echeleon Form +[%collapsible] +==== +[stem] +++++ +-5R_3 + R_1 \to R_1 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 0 & -4 \\ + 0 & 1 & -4 & 6 \\ + 0 & 0 & 1 & -1 +\end{array} +\right] +++++ + +[stem] +++++ +4R_3 + R_2 \to R_2 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 0 & -4 \\ + 0 & 1 & 0 & 2 \\ + 0 & 0 & 1 & -1 +\end{array} +\right] +++++ + +solution + +[stem] +++++ +(x, y, z) = (-4, 2, -1) +++++ +==== + +=== Multiple Solutions + +The system has more than one solution when the *Row Echelon Form*, or *Reduced +Row Echelon Form* has one or more rows of 0s. + +.example +[%collapsible] +==== +[stem] +++++ +\begin{cases} +\begin{alignat*}{6} + x & {}-{} & y & {}-{} & z & = 1 \\ +3x & {}-{} & 2y & {}-{} & 7z & = 0 \\ +5x & {}-{} & 4y & {}-{} & 9z & = 2 +\end{alignat*} +\end{cases} + +\qquad % + +\left[ +\begin{array}{ccc : c} + 1 & -1 & -1 & 1 \\ + 3 & -2 & -7 & 0 \\ + 5 & -4 & -9 & 2 +\end{array} +\right] +++++ + +[stem] +++++ +\begin{aligned} + \begin{matrix} + -3R_1 + R_2 \to R_2 \\ + -5R_1 + R_3 \to R_3 + \end{matrix} + & \left[ \begin{array}{ccc|r} + 1 & -1 & -1 & 1 \\ + 0 & 1 & -4 & -3 \\ + 0 & 1 & -4 & -3 + \end{array} \right] \\[15pt] + -1R_2 + R_3 \to R_3 + & \left[ \begin{array}{ccc|r} + 1 & -1 & -1 & 1 \\ + 0 & 1 & -4 & -3 \\ + 0 & 0 & 0 & 0 + \end{array} \right] \\[15pt] + R_2 + R_1 \to R_1 + & \left[ \begin{array}{ccc|r} + 1 & 0 & -5 & -2 \\ + 0 & 1 & -4 & -3 \\ + 0 & 0 & 0 & 0 + \end{array} \right] +\end{aligned} +++++ + +k = z + +[stem] +++++ +\begin{array}{r} +\begin{alignat*}{4} + x & {}-{} & 5k & = -2 \\ + & & +5k & = +5k\\ +\hline + & & x & = -2 + 5x \\ +\end{alignat*} +\end{array} +++++ + +[stem] +++++ +\begin{array}{r} +\begin{alignat*}{4} + y & {}-{} & 4k & = -3 \\ + & & +4k & = +4k\\ +\hline + & & y & = -3 + 4k \\ +\end{alignat*} +\end{array} +++++ + +[stem] +++++ +\begin{cases} + x - 5z = -2 \\ + y - 4z = -3 +\end{cases} +\text{ or } +\begin{cases} + x = 5z - 2 \\ + y = 4z - 3 +\end{cases} +++++ + +Both x and y are dependent on what z is. So I choose the parameter k to +represent some real number. I let z = k, which makes x = 5k – 2 and y = 4k – 3. +The solution is also written (5k – 2,4k – 3,k) as an ordered triple. Once a real +number is selected for k, then a solution is created from the ordered triple. +For example, if k = 1, then a solution for the system is (3,1,1). + +==== + + + +== Links + +* https://brilliant.org/wiki/3d-coordinate-geometry-equation-of-a-plane/[Brilliant | 3D Coordinate Geometry - Equation of a Plane] diff --git a/Linear-Algrebra/System-of-Equations.html b/Linear-Algrebra/System-of-Equations.html new file mode 100644 index 0000000..4aa6590 --- /dev/null +++ b/Linear-Algrebra/System-of-Equations.html @@ -0,0 +1,2056 @@ + + + + + + + +Linear Algrebra | System of Equations + + + + + + + + + + + + + + + + + + +
+
+

Solving

+
+
+

Two Equations

+
+

The tools or methods for solving systems of equations algebraically are

+
+
+
    +
  • +

    Elimination: You eliminate one of the variables in teh system to make it a +simpler system, while preserving the original relationship between variables

    +
    • +
  • +

    Back Substitution: You use back substitution after you’ve found a value for +one of the variables and tehn solve for the values of the other variables.

    +
    • +
+
+
+back substitution +
+
+
+\[\begin{cases} +\begin{alignat*}{4} +2x & {}+{} & 3y & = 23 \\ +4x & {}-{} & 3y & = 1 +\end{alignat*} +\end{cases}\] +
+
+
+
+\[\begin{array}{r} +\begin{alignat*}{4} +2x & {}+{} & 3y & = 23 \\ +2x & {}+{} & 3y & = 1 \\ +\hline +6x & & & 24 \\ +x & {}={} & \frac{24}{4} & = 4 +\end{alignat*} +\end{array}\] +
+
+
+
+\[\begin{gather*} +2(4) + 3y = 23 \\ +8 + 3y = 23 \\ +3y = 15 \\ +y = 5 +\end{gather*}\] +
+
+
+
+
+Elimination +
+
+
+\[\begin{cases} +\begin{alignat*}{4} +3x & {}+{} & 5y & = 12 \\ +4x & {}-{} & 3y & = -13 +\end{alignat*} +\end{cases}\] +
+
+
+
+\[\begin{alignat*}{8} +(3x & {}+{} & 5y & = 12)3\ & =\ & 9x {}+{} & 15y & = 36 \\ +(4x & {}-{} & 3y & = -13)5\ & =\ & 2x {}-{} & 15y & = -65 \\ +\end{alignat*}\] +
+
+
+
+\[\begin{array}{r} +\begin{alignat*}{4} +3x & {}+{} & 5y & = 12 \\ +4x & {}-{} & 3y & = -13 \\ +\hline +29x & & & = -29 \\ +x & = & \frac{-29}{29} & = -1 +\end{alignat*} +\end{array}\] +
+
+
+
+
+
+

Multiple Equations

+
+

The following operations may be performed (preserving the original relationship +between the variables), giving you the solution to the original equation:

+
+
+
    +
  • +

    Interchange two equations (change their order)

    +
    • +
  • +

    Multiply or divide each term in an equation by a constant number +(\(\neq 0\))

    +
    • +
  • +

    Add two equations together, combining like terms.

    +
    • +
  • +

    Add the multiple of one equation to another equation

    +
    • +
+
+
+ + + + + +
+ + +There is an alternative method involving matrices +
+
+
+example +
+
+
+\[\begin{cases} +\begin{alignat*}{5} +2x & {}+{} & 3y & {}-{} & 4z & {}+{} & w & = 15 \\ + x & {}-{} & 2y & {}+{} & 3z & {}-{} & 2w & = -3 \\ +3x & {}+{} & 5y & {}+{} & z & {}-{} & w & = 20 \\ +4x & {}+{} & y & {}-{} & z & {}+{} & w & = 5 \\ +\end{alignat*} +\end{cases}\] +
+
+
+

remove the w

+
+
+
+\[ \begin{alignat*}{4} +2R_1 + R_2 &\to 5x &&+ 4y &&- 5z &&= 27 \\ +R_1 + R_3 &\to 5x &&+ 8y &&- 3z &&= 35 \\ +R_4 + R_3 &\to 7x &&+ 6y && &&= 25 +\end{alignat*}\] +
+
+
+
+\[\begin{cases} +\begin{alignat*}{5} +5x & {}+{} & 4y & {}-{} & 5z & = 15 \\ +5x & {}+{} & 8y & {}-{} & 3z & = 35 \\ +7x & {}+{} & 6y & & & = 25 \\ +\end{alignat*} +\end{cases}\] +
+
+
+

remove the last equations

+
+
+
+\[\begin{array}{lllrcrcrcl} +3R_1 & \to & -15x & + & 12y & - & 15z & = & 81 \\ +-5R_2 & \to & -25x & - & 40y & + & 15z & = & -175 \\ +\hline +3R_1 + (-5R_2) & \to & -40x & - & 28y & & & = & -94 +\end{array}\] +
+
+
+
+\[\begin{cases} +\begin{alignat*}{4} + 7x & {}+{} & 6y & = 25 \\ +-10x & {}-{} & 28y & = -94 +\end{alignat*} +\end{cases}\] +
+
+
+

Final steps

+
+
+
+\[\begin{array}{llrcrcrcl} +14R_1 & \to & 98x & + & 84y & = & 350 \\ +3R_2 & \to & -30x & - & 84y & = & -282 \\ +\hline +14R_1 + 3R_2 & \to & 68x & & & = & 68 +\end{array}\] +
+
+
+

getting x

+
+
+
+\[\begin{align*} +68x &= 68 \\ +x &= 1 +\end{align*}\] +
+
+
+

Now, substitute \(x = 1\) into \(7x + 6y = 25\) to get y

+
+
+
+\[\begin{align*} +7(1) + 6y &= 25 \\ +6y &= 18 \\ +y &= 3 +\end{align*}\] +
+
+
+

solve for z

+
+
+
+\[\begin{align*} +5(1) + 8(3) - 3z &= 35 \\ +5 + 24 - 3z &= 35 \\ +29 - 3z &= 35 \\ +-3z &= 6 \\ +z &= -2 +\end{align*}\] +
+
+
+

solve for w

+
+
+
+\[\begin{align*} +2(1) + 3(3) - 4(-2) + w &= 15 \\ +2 + 9 + 8 + w &= 15 \\ +19 + w &= 15 \\ +w &= -4 +\end{align*}\] +
+
+
+

solution

+
+
+
+\[\begin{cases} +x = 1 \\ +y = 3 \\ +z = -2 \\ +w = -4 +\end{cases}\] +
+
+
+
+
+
+
+
+

Linear Equations

+
+
+

Linear equations are in the form:

+
+
+
+\[a_1x_1 + a_2x_2 + a_3x_3 + \ldots + a_nx_n\] +
+
+
+
+
+

System of linear equations

+
+
+

This is a system of linear equations. In order to solve a system of linear +equations you have to find variables that make each line true.

+
+
+
+\[\begin{cases} +2x + y = 10 \\ +x - y = 2 +\end{cases}\] +
+
+
+

answer:

+
+
+
+\[\begin{cases} +x = 4 \\ +y = 2 +\end{cases}\] +
+
+
+

Matrices

+
+

In linear algrebra we can simplify this to be a matrix, each column is for a +different variable.

+
+
+
+\[\begin{cases} +\begin{alignat*}{5} + x_1 & {}-{} & 2x_2 & {}+{} & 3x_3 & {}-{} & 6x_4 & = 8 \\ + x_1 & & & {}-{} & 10x_3 & {}+{} & x_4 & = 11 \\ + 4x_1 & {}+{} & x_2 & & & {}-{} & 4x_4 & = 8 \\ + 3x_1 & {}+{} & 5x_2 & {}-{} & 7x_3 & & & = 6 +\end{alignat*} +\end{cases}\] +
+
+
+
+\[A = \left[\begin{array}{cc} +1 & -2 & 3 & 6 \\ +1 & 0 & -10 & 1 \\ +4 & 1 & 0 & -4 \\ +3 & 5 & -7 & 0 \\ +\end{array}\right]\] +
+
+
+
+
+
+

Solutions

+
+
+

A system of equations has a solution when you have at least one set of numbers +that replaces variabels in the equations and makes each equation read as a true +statement. A system of equations may have more than one solution. Systems with +more than one solution generally have an infinite number of solutions, and those +solutions are written as an expression or rule in terms of a parameter.

+
+
+

Characteristics of having just one solution

+
+
    +
  1. +

    Must contain linear equations

    +
    2. +
  2. +

    The system of equations must have as many independent equations as there are +variables in the system.

    +
    +
      +
    • +

      independent equations: This just means that none of the equations is a +multiple or combination of multiples of the other equations in the system.

      +
      • +
    +
    +
    3. +
+
+
+example +
+
+

Including the equation \(-4x_1 - 6x_2 = -4\) would incroduce an equation that +isn’t independent. That equation is obtained by multiplying the top equation by +-2, multiplying the bottom equation by 2, and adding the two multiples together.

+
+
+
+\[\begin{cases} +3x_1 - 2x_2 = 16 \\ + x_1 - 5x_2 = 14 +\end{cases}\] +
+
+
+
+
+
+

Characteristics of having multiple solutions

+
+

When systems have mutiple solutions, you’ll find a pattern or rule that predicts +what those solutions are.

+
+
+
+\[\begin{cases} +\begin{alignat*}{4} + -x & {}+{} & y & {}+{} & z & = -1 \\ + -x & {}+{} & 2y & {}-{} & 3z & = -4 \\ + 3x & {}-{} & 2y & {}-{} & 7z & = 0 +\end{alignat*} +\end{cases}\] +
+
+
+

The pattern in the solutions is that x is always two less than five time z, and +y is always three less than four times z. This has infinite solutions

+
+
+ + + + + +
+ + +linear algrebra for dummies p.67 +
+
+
+
+

Graphing two lines for two equations

+
+

The graph of system of equations \(ax + by = c\) and \(dx + ey = f\), +where x and y area variables and a, b, c, d, e, and f are real numbers has:

+
+
+
    +
  • +

    Intersecting lines (exactly one solution) when the system is consistent and +the equations are independent

    +
    • +
  • +

    Parallel lines (no solutions) when the system is inconsistent

    +
    • +
  • +

    Coincident lines (infinit solutions) when the system in consistent and the +equations are dependent

    +
    • +
+
+
+
+

Three equations | Exactly One Soltion (Independent)

+
+

In a System of three equations, if three planes intersect at a single unique +point then there is only one soution. (like the corner of a room where two walls +meet the ceiling).

+
+
+
+

Planes

+
+

No Solutions

+
+

Planes that have the same slant but different intersections of the axes will +never interesect

+
+
+
+parallel planes +
+
+
+

In mathematics, two planes are parallel if their equations have the same +coefficients for the x, y, and z variables, but different constants.

+
+
+

For example, look at these two planes:

+
+
+
    +
  • +

    Plane 1: \(3x - 2y - 7z = 0\)

    +
    • +
  • +

    Plane 2: \(3x - 2y - 7z = 10\)

    +
    • +
+
+
+

Notice that the left-hand sides are identical. In technical terms, their normal +vectors (the direction perpendicular to the flat surface) are the same: [3, -2, +-7]. The different constants (0 vs 10) just shift one plane "above" or +"beside" the other without changing its angle.

+
+
+

In linear algebra, a solution is a point where all the planes in your system +touch or intersect. Therefore, the system is called inconsistent, and we say +there are no solutions.

+
+
+
+

Multiple Solutions

+
+

There is only one specific case where parallel planes have multiple solutions: +if they are actually the same plane.

+
+
+

consider these two plane equations

+
+
+
+\[\begin{cases} +\begin{alignat*}{4} +x & {}+{} & y & {}+{} & z & = 1 \\ +2x & {}+{} & 2y & {}+{} & 2z & = 2 +\end{alignat*} +\end{cases}\] +
+
+
+
    +
  • +

    The "Double" Layer: Even though there are two different mathematical +expressions (like \(x+y+z=1\) and \(2x+2y+2z=2\)), they represent the +same set of coordinates.

    +
    • +
  • +

    Infinite Solutions: Because every single point \(x, y, z\) on the first +plane is also on the second, there are an infinite number of solutions.

    +
    • +
  • +

    Linear Dependence: In linear algebra terms, these equations are linearly +dependent. The second equation is simply the first equation multiplied by a +scalar (in this case, 2).

    +
    • +
+
+
+
+
+
+
+

One Solution with Square Matrices and their Inverse

+
+
+

To solve a system of n equations with n unkowns using an n x n matrix +and its inverse, do the following:

+
+
+
    +
  1. +

    Write each of the equations in the system with the variables in the same +order and the constant on the other sider of the equal sign from the +variable.

    +
    2. +
  2. +

    Construct a coeffiecient matrix (a square matrix whose elements are the +coefficients of the variables).

    +
    3. +
  3. +

    Write the constant matrix (an n x 1 column matrix) using the constants in +the equations.

    +
    4. +
  4. +

    Find the inverse of the coefficient matrix.

    +
    5. +
  5. +

    Multiply the inverse of the coefficient matrix times the constant matrix.

    +
    6. +
+
+
+ + + + + +
+ + +This only works with singular matrices +
+
+
+example +
+
+
+\[\left\{ \begin{array}{rclrclrcl} +0 & = & 3 & + & x & + & y & + & z \\ +4x & + & 5y & + & 6 & = & 0 & & \\ +y & = & 3z & + & 5 & & & & +\end{array} \right.\] +
+
+
+
+\[\begin{cases} +\begin{alignat*}{6} +-x & {}-{} & y & {}-{} & z & = 3 \\ +4x & {}+{} & 5y & & & = 3 \\ + & & y & {}-{} & 3z & = 5 +\end{alignat*} +\end{cases}\] +
+
+
+
+\[A = \begin{bmatrix} +-1 & -1 & -1 \\ + 4 & 5 & 0 \\ + 0 & 1 & -3 +\end{bmatrix}\] +
+
+
+
+\[B = \begin{bmatrix} +3 \\ +-6 \\ +5 +\end{bmatrix}\] +
+
+
+
+\[A^{-1} = \begin{bmatrix} + 15 & 4 & -5 \\ +-12 & -3 & 4 \\ +-4 & -1 & 1 +\end{bmatrix}\] +
+
+
+
+\[A^{-1} \otimes B = \begin{bmatrix} + 15 & 4 & -5 \\ +-12 & -3 & 4 \\ +-4 & -1 & 1 +\end{bmatrix} \otimes +\begin{bmatrix} + 3 \\ +-6 \\ + 5 +\end{bmatrix} = +\begin{bmatrix} +-4 \\ + 2 \\ +-1 +\end{bmatrix}\] +
+
+
+
+
+
+
+

Solving Using an augmented matrix

+
+
+

After wrting a system of equations as an augmented matrix, you perform row +operations until the metrix is in either row echelon form or reduced row echelon +form.

+
+
+
    +
  • +

    Reduced Row echelon form: Gives you the answer directly

    +
    • +
  • +

    Row echelon form: Required back-solving or algebraic manipulations to get +your answer.

    +
    • +
+
+
+ + + + + +
+ + +Works for any size matrix. +
+
+
+

One Solution

+
+Problem +
+
+
+\[\begin{cases} +\begin{alignat*}{6} +-x & {}-{} & y & {}-{} & z & = 3 \\ +4x & {}+{} & 5y & & & = -6 \\ + & & y & {}-{} & 3z & = 5 +\end{alignat*} +\end{cases}\] +
+
+
+
+\[\left[ +\begin{array}{ccc : c} +-1 & -1 & -1 & 3 \\ + 4 & 5 & 0 & -6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right]\] +
+
+
+
+\[-1R_1 \to R_1 +\left[ +\begin{array}{ccc : c} + 1 & 1 & 1 & -3 \\ + 4 & 5 & 0 & -6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right]\] +
+
+
+
+\[-4R_1 + R_2 \to R_2 +\left[ +\begin{array}{ccc : c} + 1 & 1 & 1 & -3 \\ + 0 & 1 & -4 & 6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right]\] +
+
+
+
+\[-1R_2 + R_1 \to R_1 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 5 & -9 \\ + 0 & 1 & -4 & 6 \\ + 0 & 1 & -3 & 5 +\end{array} +\right]\] +
+
+
+
+\[-1R_2 + R_3 \to R_3 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 5 & -9 \\ + 0 & 1 & -4 & 6 \\ +0 & 0 & 1 & -1 +\end{array} +\right]\] +
+
+
+
+
+Row Echeleon Form +
+
+

back substitute for z

+
+
+
+\[\begin{align*} +y - 4(-1) &= 6 \\ +y + 4 &= 6 \\ +y &= 2 +\end{align*}\] +
+
+
+

back substitute for for z again

+
+
+
+\[\begin{align*} +x + 5(-1) &= -9 \\ +x - 5 &= -9 \\ +x &= -4 +\end{align*}\] +
+
+
+

solution

+
+
+
+\[(x, y, z) = (-4, 2, -1)\] +
+
+
+
+
+Reduced Row Echeleon Form +
+
+
+\[-5R_3 + R_1 \to R_1 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 0 & -4 \\ + 0 & 1 & -4 & 6 \\ + 0 & 0 & 1 & -1 +\end{array} +\right]\] +
+
+
+
+\[4R_3 + R_2 \to R_2 +\left[ +\begin{array}{ccc : c} + 1 & 0 & 0 & -4 \\ + 0 & 1 & 0 & 2 \\ + 0 & 0 & 1 & -1 +\end{array} +\right]\] +
+
+
+

solution

+
+
+
+\[(x, y, z) = (-4, 2, -1)\] +
+
+
+
+
+
+

Multiple Solutions

+
+

The system has more than one solution when the Row Echelon Form, or Reduced +Row Echelon Form has one or more rows of 0s.

+
+
+example +
+
+
+\[\begin{cases} +\begin{alignat*}{6} + x & {}-{} & y & {}-{} & z & = 1 \\ +3x & {}-{} & 2y & {}-{} & 7z & = 0 \\ +5x & {}-{} & 4y & {}-{} & 9z & = 2 +\end{alignat*} +\end{cases} + +\qquad % + +\left[ +\begin{array}{ccc : c} + 1 & -1 & -1 & 1 \\ + 3 & -2 & -7 & 0 \\ + 5 & -4 & -9 & 2 +\end{array} +\right]\] +
+
+
+
+\[\begin{aligned} + \begin{matrix} + -3R_1 + R_2 \to R_2 \\ + -5R_1 + R_3 \to R_3 + \end{matrix} + & \left[ \begin{array}{ccc|r} + 1 & -1 & -1 & 1 \\ + 0 & 1 & -4 & -3 \\ + 0 & 1 & -4 & -3 + \end{array} \right] \\[15pt] + -1R_2 + R_3 \to R_3 + & \left[ \begin{array}{ccc|r} + 1 & -1 & -1 & 1 \\ + 0 & 1 & -4 & -3 \\ + 0 & 0 & 0 & 0 + \end{array} \right] \\[15pt] + R_2 + R_1 \to R_1 + & \left[ \begin{array}{ccc|r} + 1 & 0 & -5 & -2 \\ + 0 & 1 & -4 & -3 \\ + 0 & 0 & 0 & 0 + \end{array} \right] +\end{aligned}\] +
+
+
+

k = z

+
+
+
+\[\begin{array}{r} +\begin{alignat*}{4} + x & {}-{} & 5k & = -2 \\ + & & +5k & = +5k\\ +\hline + & & x & = -2 + 5x \\ +\end{alignat*} +\end{array}\] +
+
+
+
+\[\begin{array}{r} +\begin{alignat*}{4} + y & {}-{} & 4k & = -3 \\ + & & +4k & = +4k\\ +\hline + & & y & = -3 + 4k \\ +\end{alignat*} +\end{array}\] +
+
+
+
+\[\begin{cases} + x - 5z = -2 \\ + y - 4z = -3 +\end{cases} +\text{ or } +\begin{cases} + x = 5z - 2 \\ + y = 4z - 3 +\end{cases}\] +
+
+
+

Both x and y are dependent on what z is. So I choose the parameter k to +represent some real number. I let z = k, which makes x = 5k – 2 and y = 4k – 3. +The solution is also written (5k – 2,4k – 3,k) as an ordered triple. Once a real +number is selected for k, then a solution is created from the ordered triple. +For example, if k = 1, then a solution for the system is (3,1,1).

+
+
+
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Terminology.adoc b/Linear-Algrebra/Terminology.adoc new file mode 100644 index 0000000..18200ac --- /dev/null +++ b/Linear-Algrebra/Terminology.adoc @@ -0,0 +1,152 @@ += Linear Algrebra | Terminology +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Linear Equations + +[stem] +++++ +a_1x_1 + a_2x_2 + \dots + a_nx_n = b +++++ + +== Basis Vectors + +Basis vectors are the starting point of your vectors operations, They are +typically the unit vectors + +== Unit Vectors + +* *stem:[\hat{i}]*: One unit in the x direction + +* *stem:[\hat{j}]*: One unit in the y direction + +* *stem:[\hat{k}]*: One unit in the z direction + +[.center] +image::unit-vectors.png[] + +[.center] +image::3D-vector-look.png[] + +== Linear Combination + +A _linear combination_ uses vectors in a set to create a new vector that's the +same dimension as the vectors involved in teh operations. + +The "Linear Combination" of stem:[\vec{v}] and stem:[\vec{w}]. + +[stem] +++++ +a\vec{v} + b\vec{w} +++++ + +If you fix one of the scalars and let the other one change it's value freely the +tip of the resulting vector draws a straight line + +.scalare a changing its value feely +[.center] +image::linear-combination-example1.png[] + +== Span + +In linear algebra, the span of a set of vectors is the collection of all +possible vectors you can create by scaling and adding them together (their +linear combinations). + +=== Span of two vectors in 3 dimensional space + +[.center] +image::span-of-two-vectors-in-3D-space.png[] + +== Linear Dependent + +When you have multiple vectors and you can remove one without reducing the +span this means those vectors are linearly dependent. (i.e. the vectors don't +add another demension to the span). + +=== 2D Space + +The area of the parallellogram is zero. + +[.center] +image::linear-dependent.png[] + +=== 3D Space + +The volume of the parallelepiped is zero. + +[.center] +image::3d-linear-dependent.png[] + +== Linear Independent + +If each vector adds another demension to the span, they are said to be linearly +independent. + +[.center] +image::linearly-independent.png[] + +== Basis + +The basis of a vector space is a set of linearly independent vectors that span +the full space + +== Shear +In linear algebra, a shear transformation is a linear mapping that displaces +each point in a fixed direction by an amount proportional to its distance from a +line (in 2D) or a plane (in 3D) that remains fixed. + +Informally, you can think of it as "pushing" the top of an object while the base +stays glued to the floor. Unlike scaling or rotation, a shear changes the angles +of a shape but preserves its area (or volume). + +[.center] +image::simple-shear.png[] + +== Composition + +In linear algebra, composition refers to the process of applying one linear +transformation after another.If you have two transformations, stem:[T] and +stem:[S], the composition (written as stem:[S \circ T]) means you first apply +stem:[T] to a vector, and then apply $S$ to the result. + +[.center] +image::linear-algrebra-composition-example1.png[] + +[.center] +image::linear-combination-example2.png[] + +== Unit Vector Displacement + +If the unit vectors are displaced in any of these ways then, you can describe +those displacement like this: + +[.inline] +* *Reflection* (Orientation flip) +** stem:[\hat{i}] now points to the left +** stem:[\hat{j}] stays the same + +[.inline] +* *Rotation* (Orientation Preserved) +** stem:[\hat{i}] now points to the left +** stem:[\hat{j}] now points down + +== Square Matrices | Singular and Non-singular Matrices + +A square matrix is _singular_ if it has a multiplicative inverse; a matrix is +non-singular if it does _not_ have a multiplicative inverse. + +== Inverse + +Two numbers are multiplicative inverses if their product is 1 (the +multiplicative identity). like 4 and 1/4, and 3/5 and 5/3. When a matrix has a +multiplicative inverse, the product of the matrix and its inverse is equal to an +identity matrix (multiplicative identity). And, further-more, you can multiply +the two matrices involved in either order (communativity) and still get the +identity. diff --git a/Linear-Algrebra/Terminology.html b/Linear-Algrebra/Terminology.html new file mode 100644 index 0000000..a28c2bd --- /dev/null +++ b/Linear-Algrebra/Terminology.html @@ -0,0 +1,1344 @@ + + + + + + + +Linear Algrebra | Terminology + + + + + + + + + + + + + + + + + + +
+
+

Linear Equations

+
+
+
+\[a_1x_1 + a_2x_2 + \dots + a_nx_n = b\] +
+
+
+
+
+

Basis Vectors

+
+
+

Basis vectors are the starting point of your vectors operations, They are +typically the unit vectors

+
+
+
+
+

Unit Vectors

+
+
+
    +
  • +

    \(\hat{i}\): One unit in the x direction

    +
    • +
  • +

    \(\hat{j}\): One unit in the y direction

    +
    • +
  • +

    \(\hat{k}\): One unit in the z direction

    +
    • +
+
+
+
+unit vectors +
+
+
+
+3D vector look +
+
+
+
+
+

Linear Combination

+
+
+

A linear combination uses vectors in a set to create a new vector that’s the +same dimension as the vectors involved in teh operations.

+
+
+

The "Linear Combination" of \(\vec{v}\) and \(\vec{w}\).

+
+
+
+\[a\vec{v} + b\vec{w}\] +
+
+
+

If you fix one of the scalars and let the other one change it’s value freely the +tip of the resulting vector draws a straight line

+
+
+
+linear combination example1 +
+
Figure 1. scalare a changing its value feely
+
+
+
+
+

Span

+
+
+

In linear algebra, the span of a set of vectors is the collection of all +possible vectors you can create by scaling and adding them together (their +linear combinations).

+
+
+

Span of two vectors in 3 dimensional space

+
+
+span of two vectors in 3D space +
+
+
+
+
+
+

Linear Dependent

+
+
+

When you have multiple vectors and you can remove one without reducing the +span this means those vectors are linearly dependent. (i.e. the vectors don’t +add another demension to the span).

+
+
+

2D Space

+
+

The area of the parallellogram is zero.

+
+
+
+linear dependent +
+
+
+
+

3D Space

+
+

The volume of the parallelepiped is zero.

+
+
+
+3d linear dependent +
+
+
+
+
+
+

Linear Independent

+
+
+

If each vector adds another demension to the span, they are said to be linearly +independent.

+
+
+
+linearly independent +
+
+
+
+
+

Basis

+
+
+

The basis of a vector space is a set of linearly independent vectors that span +the full space

+
+
+
+
+

Shear

+
+
+

In linear algebra, a shear transformation is a linear mapping that displaces +each point in a fixed direction by an amount proportional to its distance from a +line (in 2D) or a plane (in 3D) that remains fixed.

+
+
+

Informally, you can think of it as "pushing" the top of an object while the base +stays glued to the floor. Unlike scaling or rotation, a shear changes the angles +of a shape but preserves its area (or volume).

+
+
+
+simple shear +
+
+
+
+
+

Composition

+
+
+

In linear algebra, composition refers to the process of applying one linear +transformation after another.If you have two transformations, \(T\) and +\(S\), the composition (written as \(S \circ T\)) means you first apply +\(T\) to a vector, and then apply $S$ to the result.

+
+
+
+linear algrebra composition example1 +
+
+
+
+linear combination example2 +
+
+
+
+
+

Unit Vector Displacement

+
+
+

If the unit vectors are displaced in any of these ways then, you can describe +those displacement like this:

+
+
+
    +
  • +

    Reflection (Orientation flip)

    +
    +
      +
    • +

      \(\hat{i}\) now points to the left

      +
      • +
    • +

      \(\hat{j}\) stays the same

      +
      • +
    +
    +
    • +
+
+
+
    +
  • +

    Rotation (Orientation Preserved)

    +
    +
      +
    • +

      \(\hat{i}\) now points to the left

      +
      • +
    • +

      \(\hat{j}\) now points down

      +
      • +
    +
    +
    • +
+
+
+
+
+

Square Matrices | Singular and Non-singular Matrices

+
+
+

A square matrix is singular if it has a multiplicative inverse; a matrix is +non-singular if it does not have a multiplicative inverse.

+
+
+
+
+

Inverse

+
+
+

Two numbers are multiplicative inverses if their product is 1 (the +multiplicative identity). like 4 and 1/4, and 3/5 and 5/3. When a matrix has a +multiplicative inverse, the product of the matrix and its inverse is equal to an +identity matrix (multiplicative identity). And, further-more, you can multiply +the two matrices involved in either order (communativity) and still get the +identity.

+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Vector-Math.adoc b/Linear-Algrebra/Vector-Math.adoc new file mode 100644 index 0000000..990fae7 --- /dev/null +++ b/Linear-Algrebra/Vector-Math.adoc @@ -0,0 +1,188 @@ += Linear Algrebra | Vector Math +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Normalization + +[stem] +++++ +\hat{u} = \frac{\overrightarrow{v}}{|\overrightarrow{v}|} +++++ + +== Cross Product + +Only works in 3 demensions (stem:[\mathbb{R}^3]). + + +[stem] +++++ +\vec{a} = \begin{bmatrix} + a_1 \\ + a_2 \\ + a_3 +\end{bmatrix}, \quad +\vec{b} = \begin{bmatrix} + b_1 \\ + b_2 \\ + b_3 +\end{bmatrix} \implies +\vec{a} \times \vec{b} = \begin{bmatrix} + a_2b_3 - a_3b_2 \\ + a_3b_1 - a_1b_3 \\ + a_1b_2 - a_2b_1 +\end{bmatrix} +++++ + +=== Area of Parallelogram + +[stem] +++++ +Area = || a x b || +++++ + +=== Area of Triangle + +[stem] +++++ +Area = \frac{1}{2}|| a x b || +++++ + +=== Perpendicular + +[.center] +image::cross-product-perdendicular.png[] + +But which way? + +Use the right hand rule. This will tell us the direction of the vector produced +from the cross product. + +[.center] +image::cross-product-right-hand-rule.png[] + +=== Order Matters + +If v is on the right of w then the cross positive is positive + +[.center] +image::cross-product-positive.png[] + +If v is on the left of w then the cross positive is negative + +[.center] +image::cross-product-negative.png[] + +think of oriantation + +[.center] +image::cross-product-oriantation.png[] + + +== Dot Product (inner-product) + +Works in any demension + +[stem] +++++ +\overrightarrow{a} \cdot \overrightarrow{b} \\ +\left[ \begin{array}{cc} +a_{1} \\ +a_{2} \\ +\vdots \\ +a_{n} +\end{array} \right] + +\cdot + +\left[ \begin{array}{cc} +b_{1} \\ +b_{2} \\ +\vdots \\ +b_{n} +\end{array} \right] += +a_{1} * b_{1} + a_{2} * b_{2} + * a_{n} * b_{n} +++++ + +=== Normal Vector and Dot procuct + +A normal vector has a magnitude of one and can point in any direction. The dot +product of the normal vector and a point can tell you the distance that point is +in that direction from the starting point of the normal vector. + +In this image the white arrow is the normal vector and the blue dot is the +distance we are finding in relation to the normal vector. + +[.center] +image::dot_product.png[] + + +[stem] +++++ +a = normal vector \\ +b = point \\ +\\ +a.x * b.x + a.y * b.y = distance + +++++ + +== Vector Addition + +[stem] +++++ +\left[\begin{array}{a} 3 \\ 1 \\ \end{array}\right] + +\left[\begin{array}{a} 2 \\ 3 \\ \end{array}\right] = +\left[\begin{array}{a} 2 + 1 \\ 2 + 3 \\ \end{array}\right] = +\left[\begin{array}{a} 3 \\ 5 \\ \end{array}\right] +++++ + +[.center] +image::vector-addition.png[] + +== Finding the Magnitude of a Vector + +[stem] +++++ +||\overrightarrow{a}|| = \sqrt{a_{1}^2 + a_{2}^2 + \dots a_{n}^2} +++++ + +== Finding the angle between two vectors + +[stem] +++++ +\frac{u \cdot v}{||u|| \cdot ||v||} = cos(\theta) +++++ + +The result is the x value of the unit cirle, from that you can determine the y +value of the unit circle and the angle. + + +== Vector and scalar multiplication + +[stem] +++++ +K \left[\begin{array}{cc} +V_1 \\ +V_2 \\ +V_3 \\ +\vdots \\ +V_n \\ +\end{array}\right] = +\left[\begin{array}{cc} +KV_1 \\ +KV_2 \\ +KV_3 \\ +\vdots \\ +Kv_n \\ +\end{array}\right] +++++ + +== Links + +* https://www.cuemath.com/vector-formulas/[Cue Math | Vector Formulas] diff --git a/Linear-Algrebra/Vector-Math.html b/Linear-Algrebra/Vector-Math.html new file mode 100644 index 0000000..d05e65c --- /dev/null +++ b/Linear-Algrebra/Vector-Math.html @@ -0,0 +1,1320 @@ + + + + + + + +Linear Algrebra | Vector Math + + + + + + + + + + + + + + + + + + +
+
+

Normalization

+
+
+
+\[\hat{u} = \frac{\overrightarrow{v}}{|\overrightarrow{v}|}\] +
+
+
+
+
+

Cross Product

+
+
+

Only works in 3 demensions (\(\mathbb{R}^3\)).

+
+
+
+\[\vec{a} = \begin{bmatrix} + a_1 \\ + a_2 \\ + a_3 +\end{bmatrix}, \quad +\vec{b} = \begin{bmatrix} + b_1 \\ + b_2 \\ + b_3 +\end{bmatrix} \implies +\vec{a} \times \vec{b} = \begin{bmatrix} + a_2b_3 - a_3b_2 \\ + a_3b_1 - a_1b_3 \\ + a_1b_2 - a_2b_1 +\end{bmatrix}\] +
+
+
+

Area of Parallelogram

+
+
+\[Area = || a x b ||\] +
+
+
+
+

Area of Triangle

+
+
+\[Area = \frac{1}{2}|| a x b ||\] +
+
+
+
+

Perpendicular

+
+
+cross product perdendicular +
+
+
+

But which way?

+
+
+

Use the right hand rule. This will tell us the direction of the vector produced +from the cross product.

+
+
+
+cross product right hand rule +
+
+
+
+

Order Matters

+
+

If v is on the right of w then the cross positive is positive

+
+
+
+cross product positive +
+
+
+

If v is on the left of w then the cross positive is negative

+
+
+
+cross product negative +
+
+
+

think of oriantation

+
+
+
+cross product oriantation +
+
+
+
+
+
+

Dot Product (inner-product)

+
+
+

Works in any demension

+
+
+
+\[\overrightarrow{a} \cdot \overrightarrow{b} \\ +\left[ \begin{array}{cc} +a_{1} \\ +a_{2} \\ +\vdots \\ +a_{n} +\end{array} \right] + +\cdot + +\left[ \begin{array}{cc} +b_{1} \\ +b_{2} \\ +\vdots \\ +b_{n} +\end{array} \right] += +a_{1} * b_{1} + a_{2} * b_{2} + * a_{n} * b_{n}\] +
+
+
+

Normal Vector and Dot procuct

+
+

A normal vector has a magnitude of one and can point in any direction. The dot +product of the normal vector and a point can tell you the distance that point is +in that direction from the starting point of the normal vector.

+
+
+

In this image the white arrow is the normal vector and the blue dot is the +distance we are finding in relation to the normal vector.

+
+
+
+dot product +
+
+
+
+\[a = normal vector \\ +b = point \\ +\\ +a.x * b.x + a.y * b.y = distance\] +
+
+
+
+
+
+

Vector Addition

+
+
+
+\[\left[\begin{array}{a} 3 \\ 1 \\ \end{array}\right] + +\left[\begin{array}{a} 2 \\ 3 \\ \end{array}\right] = +\left[\begin{array}{a} 2 + 1 \\ 2 + 3 \\ \end{array}\right] = +\left[\begin{array}{a} 3 \\ 5 \\ \end{array}\right]\] +
+
+
+
+vector addition +
+
+
+
+
+

Finding the Magnitude of a Vector

+
+
+
+\[||\overrightarrow{a}|| = \sqrt{a_{1}^2 + a_{2}^2 + \dots a_{n}^2}\] +
+
+
+
+
+

Finding the angle between two vectors

+
+
+
+\[\frac{u \cdot v}{||u|| \cdot ||v||} = cos(\theta)\] +
+
+
+

The result is the x value of the unit cirle, from that you can determine the y +value of the unit circle and the angle.

+
+
+
+
+

Vector and scalar multiplication

+
+
+
+\[K \left[\begin{array}{cc} +V_1 \\ +V_2 \\ +V_3 \\ +\vdots \\ +V_n \\ +\end{array}\right] = +\left[\begin{array}{cc} +KV_1 \\ +KV_2 \\ +KV_3 \\ +\vdots \\ +Kv_n \\ +\end{array}\right]\] +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/Vector-Rules.adoc b/Linear-Algrebra/Vector-Rules.adoc new file mode 100644 index 0000000..faa7419 --- /dev/null +++ b/Linear-Algrebra/Vector-Rules.adoc @@ -0,0 +1,51 @@ += Linear Algrebra | Vector Rules +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: latexmath +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Triangle inequality + +The magnitude of the sum of vectors is always less than or equal to the sum of +the magnitudes of the vectos + +[stem] +++++ +|| u + v ||\ \leq\ || u || + || v || +++++ + +== Vector and scalar multiplication + +[.inline] +* *When multiplying by a scalar that is one* + ** The scalar stays the same + +[.inline] +* *When multiplying by a scalar greater than one* + ** The dierection stays the same but the size is bigger + +[.inline] +* *When multiplying by a scalar between zero and one* + ** The dierection stays the same but the size is smaller + +[.inline] +* *When multiplying by a negative scalar* + ** The size of the vector changes (except with -1) + ** The direction of the vector is reversed + +== Orthogonality (right angles) + +If the dot product of two vectors is equal to zero then the rays must be +perendiculator to each other + +== Magnitude + +In vector math, it is impossible for the length (magnitude) to be less than +zero. While the individual components (x, y, z) can certainly be +negative—pointing the vector "backward" or "down"—the length itself is +calculated using a formula that squares those values, which always results in a +positive number. diff --git a/Linear-Algrebra/Vector-Rules.html b/Linear-Algrebra/Vector-Rules.html new file mode 100644 index 0000000..fdb3c66 --- /dev/null +++ b/Linear-Algrebra/Vector-Rules.html @@ -0,0 +1,1172 @@ + + + + + + + +Linear Algrebra | Vector Rules + + + + + + + + + + + + + + + + + + +
+
+

Triangle inequality

+
+
+

The magnitude of the sum of vectors is always less than or equal to the sum of +the magnitudes of the vectos

+
+
+
+\[|| u + v ||\ \leq\ || u || + || v ||\] +
+
+
+
+
+

Vector and scalar multiplication

+
+
+
    +
  • +

    When multiplying by a scalar that is one

    +
    +
      +
    • +

      The scalar stays the same

      +
      • +
    +
    +
    • +
+
+
+
    +
  • +

    When multiplying by a scalar greater than one

    +
    +
      +
    • +

      The dierection stays the same but the size is bigger

      +
      • +
    +
    +
    • +
+
+
+
    +
  • +

    When multiplying by a scalar between zero and one

    +
    +
      +
    • +

      The dierection stays the same but the size is smaller

      +
      • +
    +
    +
    • +
+
+
+
    +
  • +

    When multiplying by a negative scalar

    +
    +
      +
    • +

      The size of the vector changes (except with -1)

      +
      • +
    • +

      The direction of the vector is reversed

      +
      • +
    +
    +
    • +
+
+
+
+
+

Orthogonality (right angles)

+
+
+

If the dot product of two vectors is equal to zero then the rays must be +perendiculator to each other

+
+
+
+
+

Magnitude

+
+
+

In vector math, it is impossible for the length (magnitude) to be less than +zero. While the individual components (x, y, z) can certainly be +negative—pointing the vector "backward" or "down"—the length itself is +calculated using a formula that squares those values, which always results in a +positive number.

+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Linear-Algrebra/index.adoc b/Linear-Algrebra/index.adoc new file mode 100644 index 0000000..fbf9d80 --- /dev/null +++ b/Linear-Algrebra/index.adoc @@ -0,0 +1,18 @@ += ./Linear-Algrebra +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Vector-Math.html;hb=HEAD[Vector-Math] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Matrix-Math.html;hb=HEAD[Matrix-Math] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Properties.html;hb=HEAD[Properties] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Matrix-Types.html;hb=HEAD[Matrix-Types] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Vector-Rules.html;hb=HEAD[Vector-Rules] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/System-of-Equations.html;hb=HEAD[System-of-Equations] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Terminology.html;hb=HEAD[Terminology] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Graphing.html;hb=HEAD[Graphing] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/Matrix-Operations.html;hb=HEAD[Matrix-Operations] diff --git a/Linear-Algrebra/index.html b/Linear-Algrebra/index.html new file mode 100644 index 0000000..fd43f33 --- /dev/null +++ b/Linear-Algrebra/index.html @@ -0,0 +1,1076 @@ + + + + + + + +./Linear-Algrebra + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Makehtml b/Makehtml new file mode 100755 index 0000000..cf79265 --- /dev/null +++ b/Makehtml @@ -0,0 +1,83 @@ +#!/usr/bin/env bash +. docinfo.conf + +clean () { +rm -rf diag*.png +rm -rf $(find . \ + -type f \ + -regex '.*\.html' \ + -not -name "README.adoc" \ + -not -name "notes.adoc" \ + -not -name "rendered.html" \ + | grep -v docinfo) +rm -rf $(find . -type f -name 'index.adoc') +} + +topicSearch() { +find . -maxdepth 1 -type f -not -name 'index.adoc' -name '*.adoc' +} + +createIndices () { +echo "creating indices..." +dirsearch="$(find . -maxdepth 1 -type d -not -name '.*' -not -name "README.html")" + +rm -rf index.adoc +rd="$PWD" + +echo "= Math Notes +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} +" > index.adoc + +for dir in $dirsearch; do + cd $dir > /dev/null + + topic=$(echo ${dir} | cut -c 3-) + + echo "* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=${topic}/index.html;hb=HEAD[${topic}]" >> ../index.adoc + + rm -rf index.adoc + echo \ +"= $dir +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics +" > index.adoc + + for adoc in $(topicSearch); do + file=$(echo "$adoc" | cut -c 3-) + subTopic=${file%.*} + echo "* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=${topic}/${subTopic}.html;hb=HEAD[${subTopic}]" >> index.adoc + + done + + cd - > /dev/null +done + + +} + +build () { +FILENAME="$(find . -type f -name "*.adoc" -not -name "README.adoc" -not -name "notes.adoc")" +asciidoctor \ + -a HOSTNAME="${HOSTNAME}" \ + -a DOCINFODIR="${DOCINFODIR}" \ + -r asciidoctor-diagram \ + -b html5 $FILENAME +} + +make () { +clean +createIndices +build +} + +# use this to call one of the above functions +$@ diff --git a/Physics/Ball-Bouncing.adoc b/Physics/Ball-Bouncing.adoc new file mode 100644 index 0000000..432ed6f --- /dev/null +++ b/Physics/Ball-Bouncing.adoc @@ -0,0 +1,37 @@ += Physics | Bouncing Ball +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images;hb=HEAD + +== Formulas + +* velocity downward + - stem:[v = g t] + +* height + - stem:[h = h_0 - \frac{1}{2}t^2] + +* how long any object takes to fall from a given height + - stem:[t = \sqrt{\frac{2h}{g}}] + +* restitution + - stem:[v_1 = e * v_0] + - the coefficent of resitution (e) represents energy loss. + +* rising + - stem:[ v = e * v - g t ] + +* falling + - stem:[ v = g t ] + +== Restitution +Typical values: + +* Rubber ball: stem:[e \approx 0.8 - 0.9] +* Basketball: stem:[e \approx 0.75] +* Tennis ball: stem:[e \approx 0.6] diff --git a/Physics/Ball-Bouncing.html b/Physics/Ball-Bouncing.html new file mode 100644 index 0000000..6b595c8 --- /dev/null +++ b/Physics/Ball-Bouncing.html @@ -0,0 +1,1164 @@ + + + + + + + +Physics | Bouncing Ball + + + + + + + + + + + + + + + + + + +
+
+

Formulas

+
+
+
    +
  • +

    velocity downward

    +
    +
      +
    • +

      \$v = g t\$

      +
      • +
    +
    +
    • +
  • +

    height

    +
    +
      +
    • +

      \$h = h_0 - \frac{1}{2}t^2\$

      +
      • +
    +
    +
    • +
  • +

    how long any object takes to fall from a given height

    +
    +
      +
    • +

      \$t = \sqrt{\frac{2h}{g}}\$

      +
      • +
    +
    +
    • +
  • +

    restitution

    +
    +
      +
    • +

      \$v_1 = e * v_0\$

      +
      • +
    • +

      the coefficent of resitution (e) represents energy loss.

      +
      • +
    +
    +
    • +
  • +

    rising

    +
    +
      +
    • +

      \$ v = e * v - g t \$

      +
      • +
    +
    +
    • +
  • +

    falling

    +
    +
      +
    • +

      \$ v = g t \$

      +
      • +
    +
    +
    • +
+
+
+
+
+

Restitution

+
+
+

Typical values:

+
+
+
    +
  • +

    Rubber ball: \$e \approx 0.8 - 0.9\$

    +
    • +
  • +

    Basketball: \$e \approx 0.75\$

    +
    • +
  • +

    Tennis ball: \$e \approx 0.6\$

    +
    • +
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Physics/Ball-Dynamic-Collision.adoc b/Physics/Ball-Dynamic-Collision.adoc new file mode 100644 index 0000000..7fb5dd2 --- /dev/null +++ b/Physics/Ball-Dynamic-Collision.adoc @@ -0,0 +1,280 @@ += Physics | Dynamic Collisions +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:wrap: normal +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +NOTE: This was mainly created by chatgpt + + +The following explains the physics and math inside the collision‐response loop. +All lines are wrapped to 80 characters. + +== Overview + +[source,cpp] +---- +for (auto c : vecCollidingPairs) + { + sBall *b1 = c.first; + sBall *b2 = c.second; + + // Distance between balls + float fDistance = sqrtf((b1->px - b2->px)*(b1->px - b2->px) + (b1->py - b2->py)*(b1->py - b2->py)); + + // Normal + float nx = (b2->px - b1->px) / fDistance; + float ny = (b2->py - b1->py) / fDistance; + + // Tangent + float tx = -ny; + float ty = nx; + + // Dot Product Tangent + float dpTan1 = b1->vx * tx + b1->vy * ty; + float dpTan2 = b2->vx * tx + b2->vy * ty; + + // Dot Product Normal + float dpNorm1 = b1->vx * nx + b1->vy * ny; + float dpNorm2 = b2->vx * nx + b2->vy * ny; + + // Conservation of momentum in 1D + float m1 = (dpNorm1 * (b1->mass - b2->mass) + 2.0f * b2->mass * dpNorm2) / (b1->mass + b2->mass); + float m2 = (dpNorm2 * (b2->mass - b1->mass) + 2.0f * b1->mass * dpNorm1) / (b1->mass + b2->mass); + + // Update ball velocities + b1->vx = tx * dpTan1 + nx * m1; + b1->vy = ty * dpTan1 + ny * m1; + b2->vx = tx * dpTan2 + nx * m2; + b2->vy = ty * dpTan2 + ny * m2; + + // Wikipedia Version - Maths is smarter but same + //float kx = (b1->vx - b2->vx); + //float ky = (b1->vy - b2->vy); + //float p = 2.0 * (nx * kx + ny * ky) / (b1->mass + b2->mass); + //b1->vx = b1->vx - p * b2->mass * nx; + //b1->vy = b1->vy - p * b2->mass * ny; + //b2->vx = b2->vx + p * b1->mass * nx; + //b2->vy = b2->vy + p * b1->mass * ny; + } +---- + +This code performs a *realistic elastic collision* between two balls. It does +this by: + +. Computing the normal and tangent vectors between the balls. +. Projecting each ball's velocity onto these axes. +. Swapping the normal components using 1D momentum equations. +. Reconstructing the final 2D velocity vectors. + +This method is standard in 2D physics engines. + +== Loop and Ball Pointers + +[source,cpp] +---- +for (auto c : vecCollidingPairs) +{ + sBall *b1 = c.first; + sBall *b2 = c.second; +---- +Each entry in `vecCollidingPairs` contains pointers to two balls that were +found to be overlapping. `b1` and `b2` are those balls. + +== Distance Between Ball Centers + +[source,cpp] +---- +float fDistance = sqrtf((b1->px - b2->px)*(b1->px - b2->px) + + (b1->py - b2->py)*(b1->py - b2->py)); +---- +This computes the center‐to‐center distance. + +== Normal Vector Between Balls + +[source,cpp] +---- +float nx = (b2->px - b1->px) / fDistance; +float ny = (b2->py - b1->py) / fDistance; +---- +The *normal vector* points from b1 to b2. Dividing by distance normalizes it +(makes its length 1). This is the axis along which the collision occurs. + +== Tangent Vector + +[source,cpp] +---- +float tx = -ny; +float ty = nx; +---- +The tangent vector is perpendicular to the normal. This allows us to separate +velocity into two independent components: normal and tangent. + +== Project Velocities Onto Tangent Axis + +[source,cpp] +---- +float dpTan1 = b1->vx * tx + b1->vy * ty; +float dpTan2 = b2->vx * tx + b2->vy * ty; +---- +A dot product projects each velocity onto the tangent. Tangential components +do *not* change in a perfectly elastic collision. This tells us the velocity in +direction of the tangent. + +== Project Velocities Onto Normal Axis + +[source,cpp] +---- +float dpNorm1 = b1->vx * nx + b1->vy * ny; +float dpNorm2 = b2->vx * nx + b2->vy * ny; +---- +These are the *normal velocity components*. These are the only parts of the +velocity that change due to collision. This tells us the velocity in the +direction of the normal axis. + +== 1D Conservation of Momentum + +[source,cpp] +---- +float m1 = (dpNorm1 * (b1->mass - b2->mass) + + 2.0f * b2->mass * dpNorm2) / (b1->mass + b2->mass); + +float m2 = (dpNorm2 * (b2->mass - b1->mass) + + 2.0f * b1->mass * dpNorm1) / (b1->mass + b2->mass); +---- +This is the standard equation for a 1D elastic collision. Since the normal +vector defines the line of impact, we treat the collision as 1D along this +axis. + +The tangent components remain unchanged. + +== Rebuild the New Velocity Vectors + +[source,cpp] +---- +b1->vx = tx * dpTan1 + nx * m1; +b1->vy = ty * dpTan1 + ny * m1; + +b2->vx = tx * dpTan2 + nx * m2; +b2->vy = ty * dpTan2 + ny * m2; +---- +Each final velocity is: + + (tangent component) + (new normal component) + +This gives the correct 2D elastic collision result. + +== Wikipedia Version + +The commented section uses a compact formula for the same physics. It produces +the same results but is harder to read. + +== Summary + +* The normal vector defines the line of impact. +* Tangent velocity stays the same. +* Normal velocity is exchanged using elastic‐collision formulas. +* Final velocity is rebuilt from tangent + new normal contributions. + +If you'd like, I can also produce: + +* an ASCII diagram showing normal/tangent axes, +* a version with simplified math, +* or a stand‐alone C example demonstrating this collision system. + += What Is a Tangent Vector? +:wrap: normal + +A *tangent vector* is a vector that is **perpendicular** to the *normal +vector* at the point of collision. In circle–circle collision physics, the +tangent vector represents the direction that is **along the surface**, while +the normal vector represents the direction **into or out of the surface**. + +== Normal vs. Tangent + +When two balls collide, you can imagine a line between their centers. This +line is the **normal** direction. + + normal = direction from ball 1 to ball 2 + tangent = direction perpendicular to the normal + +The tangent is found simply by rotating the normal 90°: + +[source,cpp] +---- +tx = -ny; +ty = nx; +---- + +== Why Do We Need a Tangent Vector? + +In a perfectly elastic collision: + +* Velocity **along the tangent** does **not change** +* Velocity **along the normal** is the only part that collides + +This matches real physics: + +* Sliding past each other = no impact. +* Moving toward each other = impact. + +To separate these two behaviors, we split velocity into: + +* tangential component +* normal component + +using dot products. + +== Visual Diagram (ASCII) + +Here is an ASCII visualization. + +---- + (tangent direction) + ↑ + | + b1 ● ---> | <--- ● b2 + | + ↓ + <------------------------------> + normal axis +---- + +== Key Idea + +The tangent vector is a direction that represents motion *that does not push +the balls into each other*. It is used to extract and preserve the +"sliding" part of the velocity. + +== Mathematical Definition + +Given a normalized normal vector: + + n = (nx, ny) + +A perpendicular tangent vector is: + + t = (-ny, nx) + +This is guaranteed to be perpendicular because: + + n · t = nx * (-ny) + ny * nx = 0 + +Thus the tangent is orthogonal to the normal. + +== Summary + +* The *normal vector* points directly from one ball to the other. +* The *tangent vector* is perpendicular to it. +* Only the normal component is involved in the collision. +* Tangent velocity stays the same, representing sliding motion. + +== Links +* https://www.youtube.com/watch?v=LPzyNOHY3A4&t=165s[javidx9 | Programming Balls] +* https://www.youtube.com/watch?v=Fp7D5D8Bqjc&t=222s[OCT | Conservation of Momentum Physics Problems] +* https://www.physicsclassroom.com/class/momentum/Lesson-2/Momentum-Conservation-Principle[Momentum-Conservation-Principle] diff --git a/Physics/Ball-Dynamic-Collision.html b/Physics/Ball-Dynamic-Collision.html new file mode 100644 index 0000000..6dd70ad --- /dev/null +++ b/Physics/Ball-Dynamic-Collision.html @@ -0,0 +1,1633 @@ + + + + + + + +Physics | Dynamic Collisions + + + + + + + + + + + + + + + + + + + +
+
+
+
+ + + + + +
+ + +This was mainly created by chatgpt +
+
+
+

The following explains the physics and math inside the collision‐response loop. +All lines are wrapped to 80 characters.

+
+
+
+
+

Overview

+
+
+
+
for (auto c : vecCollidingPairs)
+		{
+			sBall *b1 = c.first;
+			sBall *b2 = c.second;
+
+			// Distance between balls
+			float fDistance = sqrtf((b1->px - b2->px)*(b1->px - b2->px) + (b1->py - b2->py)*(b1->py - b2->py));
+
+			// Normal
+			float nx = (b2->px - b1->px) / fDistance;
+			float ny = (b2->py - b1->py) / fDistance;
+
+			// Tangent
+			float tx = -ny;
+			float ty = nx;
+
+			// Dot Product Tangent
+			float dpTan1 = b1->vx * tx + b1->vy * ty;
+			float dpTan2 = b2->vx * tx + b2->vy * ty;
+
+			// Dot Product Normal
+			float dpNorm1 = b1->vx * nx + b1->vy * ny;
+			float dpNorm2 = b2->vx * nx + b2->vy * ny;
+
+			// Conservation of momentum in 1D
+			float m1 = (dpNorm1 * (b1->mass - b2->mass) + 2.0f * b2->mass * dpNorm2) / (b1->mass + b2->mass);
+			float m2 = (dpNorm2 * (b2->mass - b1->mass) + 2.0f * b1->mass * dpNorm1) / (b1->mass + b2->mass);
+
+			// Update ball velocities
+			b1->vx = tx * dpTan1 + nx * m1;
+			b1->vy = ty * dpTan1 + ny * m1;
+			b2->vx = tx * dpTan2 + nx * m2;
+			b2->vy = ty * dpTan2 + ny * m2;
+
+			// Wikipedia Version - Maths is smarter but same
+			//float kx = (b1->vx - b2->vx);
+			//float ky = (b1->vy - b2->vy);
+			//float p = 2.0 * (nx * kx + ny * ky) / (b1->mass + b2->mass);
+			//b1->vx = b1->vx - p * b2->mass * nx;
+			//b1->vy = b1->vy - p * b2->mass * ny;
+			//b2->vx = b2->vx + p * b1->mass * nx;
+			//b2->vy = b2->vy + p * b1->mass * ny;
+		}
+
+
+
+

This code performs a realistic elastic collision between two balls. It does +this by:

+
+
+
    +
  1. +

    Computing the normal and tangent vectors between the balls.

    +
    2. +
  2. +

    Projecting each ball’s velocity onto these axes.

    +
    3. +
  3. +

    Swapping the normal components using 1D momentum equations.

    +
    4. +
  4. +

    Reconstructing the final 2D velocity vectors.

    +
    5. +
+
+
+

This method is standard in 2D physics engines.

+
+
+
+
+

Loop and Ball Pointers

+
+
+
+
for (auto c : vecCollidingPairs)
+{
+    sBall *b1 = c.first;
+    sBall *b2 = c.second;
+
+
+
+

Each entry in vecCollidingPairs contains pointers to two balls that were +found to be overlapping. b1 and b2 are those balls.

+
+
+
+
+

Distance Between Ball Centers

+
+
+
+
float fDistance = sqrtf((b1->px - b2->px)*(b1->px - b2->px) +
+                        (b1->py - b2->py)*(b1->py - b2->py));
+
+
+
+

This computes the center‐to‐center distance.

+
+
+
+
+

Normal Vector Between Balls

+
+
+
+
float nx = (b2->px - b1->px) / fDistance;
+float ny = (b2->py - b1->py) / fDistance;
+
+
+
+

The normal vector points from b1 to b2. Dividing by distance normalizes it +(makes its length 1). This is the axis along which the collision occurs.

+
+
+
+
+

Tangent Vector

+
+
+
+
float tx = -ny;
+float ty = nx;
+
+
+
+

The tangent vector is perpendicular to the normal. This allows us to separate +velocity into two independent components: normal and tangent.

+
+
+
+
+

Project Velocities Onto Tangent Axis

+
+
+
+
float dpTan1 = b1->vx * tx + b1->vy * ty;
+float dpTan2 = b2->vx * tx + b2->vy * ty;
+
+
+
+

A dot product projects each velocity onto the tangent. Tangential components +do not change in a perfectly elastic collision. This tells us the velocity in +direction of the tangent.

+
+
+
+
+

Project Velocities Onto Normal Axis

+
+
+
+
float dpNorm1 = b1->vx * nx + b1->vy * ny;
+float dpNorm2 = b2->vx * nx + b2->vy * ny;
+
+
+
+

These are the normal velocity components. These are the only parts of the +velocity that change due to collision. This tells us the velocity in the +direction of the normal axis.

+
+
+
+
+

1D Conservation of Momentum

+
+
+
+
float m1 = (dpNorm1 * (b1->mass - b2->mass) +
+            2.0f * b2->mass * dpNorm2) / (b1->mass + b2->mass);
+
+float m2 = (dpNorm2 * (b2->mass - b1->mass) +
+            2.0f * b1->mass * dpNorm1) / (b1->mass + b2->mass);
+
+
+
+

This is the standard equation for a 1D elastic collision. Since the normal +vector defines the line of impact, we treat the collision as 1D along this +axis.

+
+
+

The tangent components remain unchanged.

+
+
+
+
+

Rebuild the New Velocity Vectors

+
+
+
+
b1->vx = tx * dpTan1 + nx * m1;
+b1->vy = ty * dpTan1 + ny * m1;
+
+b2->vx = tx * dpTan2 + nx * m2;
+b2->vy = ty * dpTan2 + ny * m2;
+
+
+
+

Each final velocity is:

+
+
+
+
(tangent component) + (new normal component)
+
+
+
+

This gives the correct 2D elastic collision result.

+
+
+
+
+

Wikipedia Version

+
+
+

The commented section uses a compact formula for the same physics. It produces +the same results but is harder to read.

+
+
+
+
+

Summary

+
+
+
    +
  • +

    The normal vector defines the line of impact.

    +
    • +
  • +

    Tangent velocity stays the same.

    +
    • +
  • +

    Normal velocity is exchanged using elastic‐collision formulas.

    +
    • +
  • +

    Final velocity is rebuilt from tangent + new normal contributions.

    +
    • +
+
+
+

If you’d like, I can also produce:

+
+
+
    +
  • +

    an ASCII diagram showing normal/tangent axes,

    +
    • +
  • +

    a version with simplified math,

    +
    • +
  • +

    or a stand‐alone C example demonstrating this collision system.

    +
    • +
+
+
+
+

What Is a Tangent Vector?

+
+

A tangent vector is a vector that is perpendicular to the normal +vector at the point of collision. In circle–circle collision physics, the +tangent vector represents the direction that is along the surface, while +the normal vector represents the direction into or out of the surface.

+
+
+

Normal vs. Tangent

+
+
+

When two balls collide, you can imagine a line between their centers. This +line is the normal direction.

+
+
+
+
normal = direction from ball 1 to ball 2
+tangent = direction perpendicular to the normal
+
+
+
+

The tangent is found simply by rotating the normal 90°:

+
+
+
+
tx = -ny;
+ty =  nx;
+
+
+
+
+
+

Why Do We Need a Tangent Vector?

+
+
+

In a perfectly elastic collision:

+
+
+
    +
  • +

    Velocity along the tangent does not change

    +
    • +
  • +

    Velocity along the normal is the only part that collides

    +
    • +
+
+
+

This matches real physics:

+
+
+
    +
  • +

    Sliding past each other = no impact.

    +
    • +
  • +

    Moving toward each other = impact.

    +
    • +
+
+
+

To separate these two behaviors, we split velocity into:

+
+
+
    +
  • +

    tangential component

    +
    • +
  • +

    normal component

    +
    • +
+
+
+

using dot products.

+
+
+
+
+

Visual Diagram (ASCII)

+
+
+

Here is an ASCII visualization.

+
+
+
+
         (tangent direction)
+                  ↑
+                  |
+        b1 ● ---> | <--- ● b2
+                  |
+                  ↓
+   <------------------------------>
+                         normal axis
+
+
+
+
+
+

Key Idea

+
+
+

The tangent vector is a direction that represents motion that does not push +the balls into each other. It is used to extract and preserve the +"sliding" part of the velocity.

+
+
+
+
+

Mathematical Definition

+
+
+

Given a normalized normal vector:

+
+
+
+
n = (nx, ny)
+
+
+
+

A perpendicular tangent vector is:

+
+
+
+
t = (-ny, nx)
+
+
+
+

This is guaranteed to be perpendicular because:

+
+
+
+
n · t = nx * (-ny) + ny * nx = 0
+
+
+
+

Thus the tangent is orthogonal to the normal.

+
+
+
+
+

Summary

+
+
+
    +
  • +

    The normal vector points directly from one ball to the other.

    +
    • +
  • +

    The tangent vector is perpendicular to it.

    +
    • +
  • +

    Only the normal component is involved in the collision.

    +
    • +
  • +

    Tangent velocity stays the same, representing sliding motion.

    +
    • +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Physics/Ball-Static-Collision.adoc b/Physics/Ball-Static-Collision.adoc new file mode 100644 index 0000000..f5069f5 --- /dev/null +++ b/Physics/Ball-Static-Collision.adoc @@ -0,0 +1,146 @@ += Physics | Static Collisions +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +NOTE: This was mainly created by chatgpt + +from https://github.com/OneLoneCoder/Javidx9/blob/master/ConsoleGameEngine/BiggerProjects/Balls/OneLoneCoder_Balls1.cpp[OneLoneCoder_Balls1.cpp]. + +The following explains what the nested loops and the collision–resolution math +are doing. + +[source,cpp] +---- +for (auto &ball : vecBalls) +{ + for (auto &target : vecBalls) + { + if (ball.id != target.id) + { + if (DoCirclesOverlap(ball.px, ball.py, ball.radius, + target.px, target.py, target.radius)) + { + vecCollidingPairs.push_back({ &ball, &target }); + + float fDistance = sqrtf((ball.px - target.px)*(ball.px - target.px) + + (ball.py - target.px)*(ball.py - target.py)); + + float fOverlap = 0.5f * (fDistance - ball.radius - target.radius); + + ball.px -= fOverlap * (ball.px - target.px) / fDistance; + ball.py -= fOverlap * (ball.py - target.py) / fDistance; + + target.px += fOverlap * (ball.px - target.px) / fDistance; + target.py += fOverlap * (ball.py - target.py) / fDistance; + } + } + } +} +---- + +== Loop Structure + +The code compares every ball with every other ball: + +* The outer loop selects the *current* ball. +* The inner loop compares it to a *target* ball. +* The condition `ball.id != target.id` ensures a ball is not compared with itself. + +== Collision Detection + +The function: + +[source,c] +---- +DoCirclesOverlap(x1, y1, r1, x2, y2, r2) +---- + +checks whether the distance between centers is less than the sum of radii. + +If true, a collision has happened. + +The pair is added to: + +[source,c] +---- +vecCollidingPairs +---- + +so it can be processed later (for physics, rendering debug markers, etc.). + +== Distance Between Centers + +[source,c] +---- +float fDistance = sqrtf((dx * dx) + (dy * dy)); +---- + +Where: + +* dx = ball.px - target.px +* dy = ball.py - target.py + +This gives the center-to-center distance. This is the Pythagorean Theorem. + +== Overlap Amount + +[source,c] +---- +float fOverlap = 0.5f * (fDistance - ball.radius - target.radius); +---- + +This computes how much the two circles are penetrating. + +The `0.5f` means **each ball moves half of the required distance**, keeping the +system stable and centered. + +== Normalized Direction Vector + +The expressions: + +[source,c] +---- +(ball.px - target.px) / fDistance +(ball.py - target.py) / fDistance +---- + +are effectively the *normalized direction vector* between the two balls. + +Normalization ensures the push direction has length 1. Keeping the direction but +maintaining a distance of zero will be important so we can keep moving by +fOverlap in the oppropriate directions. + +== Resolving the Collision (Separation) + +Each ball is moved along the collision normal: + +[source,c] +---- +ball.px -= fOverlap * normalized_dx; +ball.py -= fOverlap * normalized_dy; + +target.px += fOverlap * normalized_dx; +target.py += fOverlap * normalized_dy; +---- + +This pushes the balls apart until they are *exactly touching*, eliminating overlap. + +== Summary of Physics Steps + +. Detect overlap (circles intersecting). +. Compute the vector between ball centers. +. Compute distance and normalize that vector. +. Compute how much they overlap. +. Push each ball away from the other along the normalized direction. + +This is a standard method of performing *positional correction* in 2D physics +simulations (similar to what Box2D does). + +== Links +https://www.youtube.com/watch?v=LPzyNOHY3A4&t=165s[Programming Balls #1] diff --git a/Physics/Ball-Static-Collision.html b/Physics/Ball-Static-Collision.html new file mode 100644 index 0000000..8467d28 --- /dev/null +++ b/Physics/Ball-Static-Collision.html @@ -0,0 +1,1383 @@ + + + + + + + +Physics | Static Collisions + + + + + + + + + + + + + + + + + + + +
+
+
+
+ + + + + +
+ + +This was mainly created by chatgpt +
+
+
+

from OneLoneCoder_Balls1.cpp.

+
+
+

The following explains what the nested loops and the collision–resolution math +are doing.

+
+
+
+
for (auto &ball : vecBalls)
+{
+    for (auto &target : vecBalls)
+    {
+        if (ball.id != target.id)
+        {
+            if (DoCirclesOverlap(ball.px, ball.py, ball.radius,
+                                 target.px, target.py, target.radius))
+            {
+                vecCollidingPairs.push_back({ &ball, &target });
+
+                float fDistance = sqrtf((ball.px - target.px)*(ball.px - target.px)
+                                      + (ball.py - target.px)*(ball.py - target.py));
+
+                float fOverlap = 0.5f * (fDistance - ball.radius - target.radius);
+
+                ball.px -= fOverlap * (ball.px - target.px) / fDistance;
+                ball.py -= fOverlap * (ball.py - target.py) / fDistance;
+
+                target.px += fOverlap * (ball.px - target.px) / fDistance;
+                target.py += fOverlap * (ball.py - target.py) / fDistance;
+            }
+        }
+    }
+}
+
+
+
+
+
+

Loop Structure

+
+
+

The code compares every ball with every other ball:

+
+
+
    +
  • +

    The outer loop selects the current ball.

    +
    • +
  • +

    The inner loop compares it to a target ball.

    +
    • +
  • +

    The condition ball.id != target.id ensures a ball is not compared with itself.

    +
    • +
+
+
+
+
+

Collision Detection

+
+
+

The function:

+
+
+
+
DoCirclesOverlap(x1, y1, r1, x2, y2, r2)
+
+
+
+

checks whether the distance between centers is less than the sum of radii.

+
+
+

If true, a collision has happened.

+
+
+

The pair is added to:

+
+
+
+
vecCollidingPairs
+
+
+
+

so it can be processed later (for physics, rendering debug markers, etc.).

+
+
+
+
+

Distance Between Centers

+
+
+
+
float fDistance = sqrtf((dx * dx) + (dy * dy));
+
+
+
+

Where:

+
+
+
    +
  • +

    dx = ball.px - target.px

    +
    • +
  • +

    dy = ball.py - target.py

    +
    • +
+
+
+

This gives the center-to-center distance. This is the Pythagorean Theorem.

+
+
+
+
+

Overlap Amount

+
+
+
+
float fOverlap = 0.5f * (fDistance - ball.radius - target.radius);
+
+
+
+

This computes how much the two circles are penetrating.

+
+
+

The 0.5f means each ball moves half of the required distance, keeping the +system stable and centered.

+
+
+
+
+

Normalized Direction Vector

+
+
+

The expressions:

+
+
+
+
(ball.px - target.px) / fDistance
+(ball.py - target.py) / fDistance
+
+
+
+

are effectively the normalized direction vector between the two balls.

+
+
+

Normalization ensures the push direction has length 1. Keeping the direction but +maintaining a distance of zero will be important so we can keep moving by +fOverlap in the oppropriate directions.

+
+
+
+
+

Resolving the Collision (Separation)

+
+
+

Each ball is moved along the collision normal:

+
+
+
+
ball.px -= fOverlap * normalized_dx;
+ball.py -= fOverlap * normalized_dy;
+
+target.px += fOverlap * normalized_dx;
+target.py += fOverlap * normalized_dy;
+
+
+
+

This pushes the balls apart until they are exactly touching, eliminating overlap.

+
+
+
+
+

Summary of Physics Steps

+
+
+
    +
  1. +

    Detect overlap (circles intersecting).

    +
    2. +
  2. +

    Compute the vector between ball centers.

    +
    3. +
  3. +

    Compute distance and normalize that vector.

    +
    4. +
  4. +

    Compute how much they overlap.

    +
    5. +
  5. +

    Push each ball away from the other along the normalized direction.

    +
    6. +
+
+
+

This is a standard method of performing positional correction in 2D physics +simulations (similar to what Box2D does).

+
+
+
+
+

Links

+
+
+

Programming Balls #1

+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Physics/Basics.adoc b/Physics/Basics.adoc new file mode 100644 index 0000000..56bb433 --- /dev/null +++ b/Physics/Basics.adoc @@ -0,0 +1,274 @@ += Physics | Basics +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + + +== Kinematic's + +=== Symbols + +[.inline] +* stem:[V] or stem:[V_{f}] -> Final velocity + +[.inline] +* stem:[a] or stem:[\vec{a}] -> Acceleration + +[.inline] +* stem:[t] -> Time + +[.inline] +* stem:[V_{0}] -> Initial Velocity + +[.inline] +* stem:[\overline{S}] -> Average Speed + +[.inline] +* stem:[X_{F}] -> Final Position + +[.inline] +* stem:[\Deltax] -> Displacement of x axis + +[.inline] +* stem:[\Deltay] -> Displacement of y axis + +[.inline] +* stem:[\vec{V_{a}}] -> Average Velocity + +[.inline] +* stem:[S_{i}] -> Instantaneos Speed + +[.inline] +* stem:[V_{i}] -> Instantaneos Velocity + +[.inline] +* stem:[X_{0}] -> Initial Position + +[.inline] +* stem:[V_{t}] -> Terminal Velocity + +[.inline] +* stem:[\vec{S}] -> Displacement + +[.inline] +* stem:[\vec{V}] -> Vector + +[.inline] +* stem:[S_{a}] -> Speed Average + +[.inline] +* stem:[w] -> weight + +=== Equations + +==== Basic +[.inline] +* stem:[V = V_{0}+at] + +[.inline] +* stem:[(\Deltax or \Deltay) = \frac{V + V_{0}}{2}t] + +[.inline] +* stem:[(\Deltax or \Deltay) = V_{0}t + 1/2at^2] + +[.inline] +* stem:[V^2 = V_{0}^2 + 2a\Deltax] + +==== Extra + +[.inline] +* stem:[\vec{V} = \vec{S}/t] or stem:[r = d/t] + +[.inline] +* stem:[\vec{V_{a}} = \frac{\Deltax}{t}] + +[.inline] +* stem:[\vec{a} = \frac{\Delta\vec{V}}{t}] + +[.inline] +* stem:[a = \frac{V_{f} - V_{i}}{\Deltat}] + +[.inline] +* stem:[\vec{V_{a}} = \frac{V_{f} + V_{i}}{2}] ??? + +[.inline] +* stem:[\vec{S} = (\vec{V_{a}})\Deltat] + +[.inline] +* stem:[\vec{S} = \frac{V_{f}^2-V_{i}^2}{2\vec{a}}] + +[.inline] +* stem:[2\vec{a}\vec{s} = V_{f}^2 - V_{i}^2] + +[.inline] +* stem:[\DeltaV = a\Deltat] + +[.inline] +* stem:[\frac{\Deltaa}{\Deltat} = jerk] + +[.inline] +* stem:[V_{i} = V_{f} - a\Deltat] + +[.inline] +* stem:[\Deltat = \frac{V_{f} - V_{i}}{a}] + +[.inline] +* stem:[S_{a} = \frac{\text{distance}}{\Deltat}] + +NOTE: If the divisor is t then it is assumed to be stem:[\Deltat] + +WARNING: kinematic formulas are only accurate if the acceleration is constant + during the time interval considered + +==== 2D Projectile Motion +There's no acceleration in the horizontal direction since gravity does not pull +projectiles sideways + +[.inline] +* stem:[Delta x=v_xt \quad] + +An object can start with a horizontal component of velocity, yet have zero +vertical component of velocity. + +=== Instentaneos Rate of Change +[stem] +++++ +\lim_{\Delta \to 0} \frac{f(x + \Deltax) - f(c)}{\Deltax} = f'(x)\ +or \ +\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = f'(c) +++++ + +[stem] +++++ +f'(x) = v(t) -> \text{Instentaneos Velocity}\ +f''(x) = a(t) -> \text{acceleration}\ +|f'(x)| = s(t) -> \text{Instentaneos speed}\ +f'''(x) -> jerk +++++ + + +== Force +=== Symbols +[.inline] +* stem:[\vec{F}] -> Force of Gravity + +[.inline] +* stem:[g] -> Acceleration of Gravity + +[.inline] +* stem:[m] -> mass of object + +[.inline] +* stem:[r] -> distance between objects + +[.inline] +* stem:[\vec{a_{g}}] -> acceleration due to gravity + + +=== Equations +[.inline] +* stem:[\vec{F} = m \cdot g] + +[.inline] +* stem:[\vec{F} = m \cdot a] + +[.inline] +* stem:[a = \vec{F}/m] + +[.inline] +* stem:[a = g] + +[.inline] +* stem:[\vec{F} = G \frac{m_{1}m_{2}}{r^2}] + +[.inline] +* stem:[w = \frac{m}{g}] + +=== Gravity + +[stem] +++++ +g = G \frac{m}{r^2} = 9.81 frac{m}{s^2} +++++ + +* stem:[m] -> Mass of the Earth +* stem:[r] -> Radius of the Earth + + +== Temrinal Velocity +[stem] +++++ +V_{t} = frac{sqrt{2mg}}{pAC_{d}} +++++ + +* stem:[V_{t}] represents terminal velocity +* stem:[m] is the mass of the falling object +* stem:[g] is accleration due to gravity +* stem:[C_{d}] is the drag coefficent +* stem:[p] is the density through which th object is falling +* stem:[A] is the projected area of the object + +== Graphing +=== Velocity vs. Time Graphs +* Slope is acceleration and a way to find instentaneos velocity +* Area is displacement + +[.center] +image::Displacement.png[] + +=== Acceleration vs. Time Graphs +* Slope is jerk +* Area is change in velocity +* Typical y-axis: stem:[m/s^2] + +[.center] +image::Change-In-Velocity.png[] + + +== Using Quadradic Formula + +[stem] +++++ +\Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2 \quad \text{(Start with the third kinematic formula.)} +++++ + +[stem] +++++ +12.2\text{ m}=(18.3\text{ m/s})t+\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2 \quad \text{(Plug in known values.)} +++++ + +[stem] +++++ +0=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2+(18.3\text{ m/s})t -12.2\text{ m} \quad \text{(Put it into the form of the quadratic equation.)} +++++ + +[stem] +++++ +a=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2}) +\ +b=18.3\text{ m/s} +\ +c=-12.2\text{ m} +++++ + +[stem] +++++ +t=\dfrac{-18.3\text{ m/s}\pm\sqrt{(18.3\text{ m/s})^2-4[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})(-12.2\text{ m})]}}{2[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})]} +++++ + +[stem] +++++ +t=0.869\text{s}\ \text{and}\ t=2.86\text{s} +++++ + +== Links +- https://en.wikipedia.org/wiki/Terminal_velocity[Terminal Velocity] +- https://phys.libretexts.org/Bookshelves/University_Physics +---- +https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/11%3A__Angular_Momentum/11.02%3A_Rolling_Motion +---- diff --git a/Physics/Basics.html b/Physics/Basics.html new file mode 100644 index 0000000..0eea65d --- /dev/null +++ b/Physics/Basics.html @@ -0,0 +1,1649 @@ + + + + + + + +Physics | Basics + + + + + + + + + + + + + + + + + + +
+
+

Kinematic’s

+
+
+

Symbols

+
+
    +
  • +

    \$V\$ or \$V_{f}\$ → Final velocity

    +
    • +
+
+
+
    +
  • +

    \$a\$ or \$\vec{a}\$ → Acceleration

    +
    • +
+
+
+
    +
  • +

    \$t\$ → Time

    +
    • +
+
+
+
    +
  • +

    \$V_{0}\$ → Initial Velocity

    +
    • +
+
+
+
    +
  • +

    \$\overline{S}\$ → Average Speed

    +
    • +
+
+
+
    +
  • +

    \$X_{F}\$ → Final Position

    +
    • +
+
+
+
    +
  • +

    \$\Deltax\$ → Displacement of x axis

    +
    • +
+
+
+
    +
  • +

    \$\Deltay\$ → Displacement of y axis

    +
    • +
+
+
+
    +
  • +

    \$\vec{V_{a}}\$ → Average Velocity

    +
    • +
+
+
+
    +
  • +

    \$S_{i}\$ → Instantaneos Speed

    +
    • +
+
+
+
    +
  • +

    \$V_{i}\$ → Instantaneos Velocity

    +
    • +
+
+
+
    +
  • +

    \$X_{0}\$ → Initial Position

    +
    • +
+
+
+
    +
  • +

    \$V_{t}\$ → Terminal Velocity

    +
    • +
+
+
+
    +
  • +

    \$\vec{S}\$ → Displacement

    +
    • +
+
+
+
    +
  • +

    \$\vec{V}\$ → Vector

    +
    • +
+
+
+
    +
  • +

    \$S_{a}\$ → Speed Average

    +
    • +
+
+
+
    +
  • +

    \$w\$ → weight

    +
    • +
+
+
+
+

Equations

+
+

Basic

+
+
    +
  • +

    \$V = V_{0}+at\$

    +
    • +
+
+
+
    +
  • +

    \$(\Deltax or \Deltay) = \frac{V + V_{0}}{2}t\$

    +
    • +
+
+
+
    +
  • +

    \$(\Deltax or \Deltay) = V_{0}t + 1/2at^2\$

    +
    • +
+
+
+
    +
  • +

    \$V^2 = V_{0}^2 + 2a\Deltax\$

    +
    • +
+
+
+
+

Extra

+
+
    +
  • +

    \$\vec{V} = \vec{S}/t\$ or \$r = d/t\$

    +
    • +
+
+
+
    +
  • +

    \$\vec{V_{a}} = \frac{\Deltax}{t}\$

    +
    • +
+
+
+
    +
  • +

    \$\vec{a} = \frac{\Delta\vec{V}}{t}\$

    +
    • +
+
+
+
    +
  • +

    \$a = \frac{V_{f} - V_{i}}{\Deltat}\$

    +
    • +
+
+
+
    +
  • +

    \$\vec{V_{a}} = \frac{V_{f} + V_{i}}{2}\$ ???

    +
    • +
+
+
+
    +
  • +

    \$\vec{S} = (\vec{V_{a}})\Deltat\$

    +
    • +
+
+
+
    +
  • +

    \$\vec{S} = \frac{V_{f}^2-V_{i}^2}{2\vec{a}}\$

    +
    • +
+
+
+
    +
  • +

    \$2\vec{a}\vec{s} = V_{f}^2 - V_{i}^2\$

    +
    • +
+
+
+
    +
  • +

    \$\DeltaV = a\Deltat\$

    +
    • +
+
+
+
    +
  • +

    \$\frac{\Deltaa}{\Deltat} = jerk\$

    +
    • +
+
+
+
    +
  • +

    \$V_{i} = V_{f} - a\Deltat\$

    +
    • +
+
+
+
    +
  • +

    \$\Deltat = \frac{V_{f} - V_{i}}{a}\$

    +
    • +
+
+
+
    +
  • +

    \$S_{a} = \frac{\text{distance}}{\Deltat}\$

    +
    • +
+
+
+ + + + + +
+ + +If the divisor is t then it is assumed to be \$\Deltat\$ +
+
+
+ + + + + +
+ + +kinematic formulas are only accurate if the acceleration is constant + during the time interval considered +
+
+
+
+

2D Projectile Motion

+
+

There’s no acceleration in the horizontal direction since gravity does not pull +projectiles sideways

+
+
+
    +
  • +

    \$Delta x=v_xt \quad\$

    +
    • +
+
+
+

An object can start with a horizontal component of velocity, yet have zero +vertical component of velocity.

+
+
+
+
+

Instentaneos Rate of Change

+
+
+\$\lim_{\Delta \to 0} \frac{f(x + \Deltax) - f(c)}{\Deltax} = f'(x)\$ +\$or\$ +\$\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = f'(c)\$ +
+
+
+
+\$f'(x) = v(t) -> \text{Instentaneos Velocity}\$ +\$f''(x) = a(t) -> \text{acceleration}\$ +\$|f'(x)| = s(t) -> \text{Instentaneos speed}\$ +\$f'''(x) -> jerk\$ +
+
+
+
+
+
+

Force

+
+
+

Symbols

+
+
    +
  • +

    \$\vec{F}\$ → Force of Gravity

    +
    • +
+
+
+
    +
  • +

    \$g\$ → Acceleration of Gravity

    +
    • +
+
+
+
    +
  • +

    \$m\$ → mass of object

    +
    • +
+
+
+
    +
  • +

    \$r\$ → distance between objects

    +
    • +
+
+
+
    +
  • +

    \$\vec{a_{g}}\$ → acceleration due to gravity

    +
    • +
+
+
+
+

Equations

+
+
    +
  • +

    \$\vec{F} = m \cdot g\$

    +
    • +
+
+
+
    +
  • +

    \$\vec{F} = m \cdot a\$

    +
    • +
+
+
+
    +
  • +

    \$a = \vec{F}/m\$

    +
    • +
+
+
+
    +
  • +

    \$a = g\$

    +
    • +
+
+
+
    +
  • +

    \$\vec{F} = G \frac{m_{1}m_{2}}{r^2}\$

    +
    • +
+
+
+
    +
  • +

    \$w = \frac{m}{g}\$

    +
    • +
+
+
+
+

Gravity

+
+
+\$g = G \frac{m}{r^2} = 9.81 frac{m}{s^2}\$ +
+
+
+
    +
  • +

    \$m\$ → Mass of the Earth

    +
    • +
  • +

    \$r\$ → Radius of the Earth

    +
    • +
+
+
+
+
+
+

Temrinal Velocity

+
+
+
+\$V_{t} = frac{sqrt{2mg}}{pAC_{d}}\$ +
+
+
+
    +
  • +

    \$V_{t}\$ represents terminal velocity

    +
    • +
  • +

    \$m\$ is the mass of the falling object

    +
    • +
  • +

    \$g\$ is accleration due to gravity

    +
    • +
  • +

    \$C_{d}\$ is the drag coefficent

    +
    • +
  • +

    \$p\$ is the density through which th object is falling

    +
    • +
  • +

    \$A\$ is the projected area of the object

    +
    • +
+
+
+
+
+

Graphing

+
+
+

Velocity vs. Time Graphs

+
+
    +
  • +

    Slope is acceleration and a way to find instentaneos velocity

    +
    • +
  • +

    Area is displacement

    +
    • +
+
+
+
+Displacement +
+
+
+
+

Acceleration vs. Time Graphs

+
+
    +
  • +

    Slope is jerk

    +
    • +
  • +

    Area is change in velocity

    +
    • +
  • +

    Typical y-axis: \$m/s^2\$

    +
    • +
+
+
+
+Change In Velocity +
+
+
+
+
+
+

Using Quadradic Formula

+
+
+
+\$\Delta y=v_{0y} t+\dfrac{1}{2}a_yt^2 \quad \text{(Start with the third kinematic formula.)}\$ +
+
+
+
+\$12.2\text{ m}=(18.3\text{ m/s})t+\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2 \quad \text{(Plug in known values.)}\$ +
+
+
+
+\$0=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})t^2+(18.3\text{ m/s})t -12.2\text{ m} \quad \text{(Put it into the form of the quadratic equation.)}\$ +
+
+
+
+\$a=\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2}) +\$ +\$b=18.3\text{ m/s} +\$ +\$c=-12.2\text{ m}\$ +
+
+
+
+\$t=\dfrac{-18.3\text{ m/s}\pm\sqrt{(18.3\text{ m/s})^2-4[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})(-12.2\text{ m})]}}{2[\dfrac{1}{2}(-9.81\dfrac{\text{ m}}{\text{ s}^2})]}\$ +
+
+
+
+\$t=0.869\text{s}\ \text{and}\ t=2.86\text{s}\$ +
+
+
+
+
+

Links

+
+
+ +
+
+
+
https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/Book%3A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)/11%3A__Angular_Momentum/11.02%3A_Rolling_Motion
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Physics/Collision.adoc b/Physics/Collision.adoc new file mode 100644 index 0000000..83dd720 --- /dev/null +++ b/Physics/Collision.adoc @@ -0,0 +1,43 @@ += Physics | Collision +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Circle and Rect collision + +[source,c] +---- +bool circleRectCollide(SDL_Point circle, float r, + SDL_Rect rect) +{ + // does this assume that circle x and circle y are the center of the circle? + float closestX = clamp(circle.x, rect.x, rect.x + rect.w); + float closestY = clamp(circle.y, rect.y, rect.y + rect.h); + + // this gets the distance form the center of the circle to a side of a + // rectangle. It could be negative or positive. Because we are squaring them + // it doesn't matter. + float dx = circle.x - closestX; + float dy = circle.y - closestY; + + // use the dot product to determine collision + return (dx*dx + dy*dy) <= (r*r); +} +---- + +== Circle and Circle collision +[source,c] +---- +bool circlesCollide(Circle a, Circle b) { + float dx = b.x - a.x; + float dy = b.y - a.y; + float distance = sqrtf(dx * dx + dy * dy); + + return distance <= (a.r + b.r); +} +---- diff --git a/Physics/Collision.html b/Physics/Collision.html new file mode 100644 index 0000000..7c74ce7 --- /dev/null +++ b/Physics/Collision.html @@ -0,0 +1,1194 @@ + + + + + + + +Physics | Collision + + + + + + + + + + + + + + + + + + + +
+
+

Circle and Rect collision

+
+
+
+
bool circleRectCollide(SDL_Point circle, float r,
+                       SDL_Rect rect)
+{
+    // does this assume that circle x and circle y are the center of the circle?
+    float closestX = clamp(circle.x, rect.x, rect.x + rect.w);
+    float closestY = clamp(circle.y, rect.y, rect.y + rect.h);
+
+    // this gets the distance form the center of the circle to a side of a
+    // rectangle. It could be negative or positive. Because we are squaring them
+    // it doesn't matter.
+    float dx = circle.x - closestX;
+    float dy = circle.y - closestY;
+
+    // use the dot product to determine collision
+    return (dx*dx + dy*dy) <= (r*r);
+}
+
+
+
+
+
+

Circle and Circle collision

+
+
+
+
bool circlesCollide(Circle a, Circle b) {
+    float dx = b.x - a.x;
+    float dy = b.y - a.y;
+    float distance = sqrtf(dx * dx + dy * dy);
+
+    return distance <= (a.r + b.r);
+}
+
+
+
+
+
+ + + + + \ No newline at end of file diff --git a/Physics/index.adoc b/Physics/index.adoc new file mode 100644 index 0000000..4016372 --- /dev/null +++ b/Physics/index.adoc @@ -0,0 +1,14 @@ += ./Physics +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Physics/Basics.html;hb=HEAD[Basics] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Physics/Ball-Static-Collision.html;hb=HEAD[Ball-Static-Collision] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Physics/Ball-Dynamic-Collision.html;hb=HEAD[Ball-Dynamic-Collision] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Physics/Ball-Bouncing.html;hb=HEAD[Ball-Bouncing] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Physics/Collision.html;hb=HEAD[Collision] diff --git a/Physics/index.html b/Physics/index.html new file mode 100644 index 0000000..5ef3844 --- /dev/null +++ b/Physics/index.html @@ -0,0 +1,1064 @@ + + + + + + + +./Physics + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/Probability/Pascals-Triangle.adoc b/Probability/Pascals-Triangle.adoc new file mode 100644 index 0000000..fba732e --- /dev/null +++ b/Probability/Pascals-Triangle.adoc @@ -0,0 +1,13 @@ += Probability | Pascal's Triangle +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Links +- https://www.youtube.com/watch?v=O4Qnsubo2tg[Die Rolling Probability] +- https://www.youtube.com/watch?v=D9DFobRr4pw[Probability Using Pascal's Triangle] diff --git a/Probability/Pascals-Triangle.html b/Probability/Pascals-Triangle.html new file mode 100644 index 0000000..9d8c95c --- /dev/null +++ b/Probability/Pascals-Triangle.html @@ -0,0 +1,1085 @@ + + + + + + + +Probability | Pascal’s Triangle + + + + + + + + + + + + + + + + + + +
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Probability/Permutations-and-Combinations.adoc b/Probability/Permutations-and-Combinations.adoc new file mode 100644 index 0000000..6100b3c --- /dev/null +++ b/Probability/Permutations-and-Combinations.adoc @@ -0,0 +1,12 @@ += Probability | Permutations and Combinations.adoc +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Links +- https://www.youtube.com/watch?v=XJnIdRXUi7A&t=315s[Permutations and Combinations] diff --git a/Probability/Permutations-and-Combinations.html b/Probability/Permutations-and-Combinations.html new file mode 100644 index 0000000..9208483 --- /dev/null +++ b/Probability/Permutations-and-Combinations.html @@ -0,0 +1,1082 @@ + + + + + + + +Probability | Permutations and Combinations.adoc + + + + + + + + + + + + + + + + + + +
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Probability/index.adoc b/Probability/index.adoc new file mode 100644 index 0000000..77a46eb --- /dev/null +++ b/Probability/index.adoc @@ -0,0 +1,11 @@ += ./Probability +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Probability/Permutations-and-Combinations.html;hb=HEAD[Permutations-and-Combinations] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Probability/Pascals-Triangle.html;hb=HEAD[Pascals-Triangle] diff --git a/Probability/index.html b/Probability/index.html new file mode 100644 index 0000000..8a851d0 --- /dev/null +++ b/Probability/index.html @@ -0,0 +1,1055 @@ + + + + + + + +./Probability + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/README.adoc b/README.adoc new file mode 100644 index 0000000..9aaa0cb --- /dev/null +++ b/README.adoc @@ -0,0 +1,14 @@ += My Math Notes + +To view the rendered notes, you can visit this site: +https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=index.html;hb=HEAD[my-math-notes]. + +== Building +To convert all the _.adoc_ files into html run this command in the root directory + +---- +./Makehtml make +---- + +It will automatically create a _index.html_ page for each subject. You do not +need to make _index.html_ pages yourself diff --git a/Triginometry/Basics.adoc b/Triginometry/Basics.adoc new file mode 100644 index 0000000..01a214f --- /dev/null +++ b/Triginometry/Basics.adoc @@ -0,0 +1,81 @@ += Triginometry | Basics +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== SOH CAH TOA + +== Using inverse functions To Find Angle +[stem] +++++ +\tan\theta = \frac{20.8}{7} +++++ + +[stem] +++++ +\tan^-1\theta = tan^-1(\frac{20.8}{7}) +++++ + +[stem] +++++ +\theta = 71.4^\circ +++++ + +NOTE: Make sure calculator is in degrees to get the answer in degrees + +NOTE: The inverse version of sine, cosine, and tangent have the "arc" prefix + +== Trig Visualized +https://www.youtube.com/watch?v=dUkCgTOOpQ0[Trig Visualized] + +In this video _d_ represents the line that is tangent to the circle. and the +tangent also represents that slope of the radius + +image::trig-visualized-tangent.png[tangent] + +this works because we get the tangent with TOA or +stem:[tan(\theta) = \frac{opposite}{adjacent}]. +We know that at a 45 degree angle stem:[sin(\theta)] and stem:[cos(\theta)] are +going to be the same value. If they are divided by each other they should equal +the length of the tangent/radius, which is 1. Hence, they have the same ratio. + +[source, lua] +---- +local radii = {1, 2, 4, 5} + +function toRadians(degrees) + return (degrees * (math.pi/180)) +end + +for k, v in pairs(radii) do + print( + "radius: " .. + v .. + ", " .. + "tangent: " .. + math.tan(toRadians(45)) + ) +end + +---- + +.output +[%collapsible] +==== +radius: 2, tangent: 1 +radius: 4, tangent: 1 +radius: 5, tangent: 1 + +==== + +notice how the tangent is always one when you're at a 45 degree angle. No matter +what the radius. + +== Links +* https://www.youtube.com/watch?v=jt7p-mCC0ng[Evaluating Inverse Functions] +* https://www.youtube.com/watch?v=dUkCgTOOpQ0[trig visualization] diff --git a/Triginometry/Basics.html b/Triginometry/Basics.html new file mode 100644 index 0000000..f7d98b5 --- /dev/null +++ b/Triginometry/Basics.html @@ -0,0 +1,1277 @@ + + + + + + + +Triginometry | Basics + + + + + + + + + + + + + + + + + + + +
+
+

SOH CAH TOA

+
+ +
+
+
+

Using inverse functions To Find Angle

+
+
+
+\$\tan\theta = \frac{20.8}{7}\$ +
+
+
+
+\$\tan^-1\theta = tan^-1(\frac{20.8}{7})\$ +
+
+
+
+\$\theta = 71.4^\circ\$ +
+
+
+ + + + + +
+ + +Make sure calculator is in degrees to get the answer in degrees +
+
+
+ + + + + +
+ + +The inverse version of sine, cosine, and tangent have the "arc" prefix +
+
+
+
+
+

Trig Visualized

+
+
+

Trig Visualized

+
+
+

In this video d represents the line that is tangent to the circle. and the +tangent also represents that slope of the radius

+
+
+
+tangent +
+
+
+

this works because we get the tangent with TOA or +\$tan(\theta) = \frac{opposite}{adjacent}\$. +We know that at a 45 degree angle \$sin(\theta)\$ and \$cos(\theta)\$ are +going to be the same value. If they are divided by each other they should equal +the length of the tangent/radius, which is 1. Hence, they have the same ratio.

+
+
+
+
local radii = {1, 2, 4, 5}
+
+function toRadians(degrees)
+    return (degrees * (math.pi/180))
+end
+
+for k, v in pairs(radii) do
+    print(
+        "radius: " ..
+         v ..
+         ", " ..
+        "tangent: " ..
+        math.tan(toRadians(45))
+    )
+end
+
+
+
+output +
+
+

radius: 2, tangent: 1 +radius: 4, tangent: 1 +radius: 5, tangent: 1

+
+
+
+
+

notice how the tangent is always one when you’re at a 45 degree angle. No matter +what the radius.

+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Triginometry/Centers.adoc b/Triginometry/Centers.adoc new file mode 100644 index 0000000..ad8b4d4 --- /dev/null +++ b/Triginometry/Centers.adoc @@ -0,0 +1,23 @@ += Triginometry | Centers +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Incenter + +[.center] +image:Incenter.jpg[] + +Incenter should be the center of the inscribed circle + +== Centroid +[.center] +image:Centroid.png[] + +== Orthocenter +== Circumcenter diff --git a/Triginometry/Centers.html b/Triginometry/Centers.html new file mode 100644 index 0000000..e62b3ea --- /dev/null +++ b/Triginometry/Centers.html @@ -0,0 +1,1104 @@ + + + + + + + +Triginometry | Centers + + + + + + + + + + + + + + + + + + +
+
+

Incenter

+
+
+

Incenter

+
+
+

Incenter should be the center of the inscribed circle

+
+
+
+
+

Centroid

+
+
+

Centroid

+
+
+
+
+

Orthocenter

+
+ +
+
+
+

Circumcenter

+
+ +
+
+
+ + + + + \ No newline at end of file diff --git a/Triginometry/Identities.adoc b/Triginometry/Identities.adoc new file mode 100644 index 0000000..5eb6042 --- /dev/null +++ b/Triginometry/Identities.adoc @@ -0,0 +1,101 @@ += Triginometry | Identities +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:stem: +:toc: left +:docinfodir: {DOCINFODIR} +:imagesdir: https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=.images + +== Even vs Odd Functions +[%header, cols="1,1"] +|=== +| Odd | Even +| sin | cos +| tan | sec +| csc | +| cot | +|=== + +== SOH CAH TOA +[.inline] +- stem:[\sin(\theta)] [.dull]#stem:[=]# stem:[\frac{opp}{hyp}] + +[.inline] +- stem:[\cos(\theta)] [.dull]#stem:[=]# stem:[\frac{adj}{hyp}] + +[.inline] +- stem:[\tan(\theta)] [.dull]#stem:[=]# stem:[\frac{opp}{adj}] + +[.inline] +- stem:[\csc(\theta)] [.dull]#stem:[=]# stem:[\frac{hyp}{opp}] + +[.inline] +- stem:[\sec(\theta)] [.dull]#stem:[=]# stem:[\frac{hyp}{adj}] + +[.inline] +- stem:[\cot(\theta)] [.dull]#stem:[=]# stem:[\frac{adj}{opp}] + +== Reciprocal + +[.inline] +- stem:[\sin(\theta)] [.dull]#stem:[=]# stem:[\frac{1}{\csc(\theta)}] + +[.inline] +- stem:[\cos(\theta)] [.dull]#stem:[=]# stem:[\frac{1}{\sec(\theta)}] + +[.inline] +- stem:[\tan(\theta)] [.dull]#stem:[=]# stem:[\frac{1}{\cot(\theta)}] + +[.inline] +- stem:[\csc(\theta)] [.dull]#stem:[=]# stem:[\frac{1}{\sin(\theta)}] + +[.inline] +- stem:[\sec(\theta)] [.dull]#stem:[=]# stem:[\frac{1}{\cos(\theta)}] + +[.inline] +- stem:[\cot(\theta)] [.dull]#stem:[=]# stem:[\frac{1}{\tan(\theta)}] + +== Pythagoras Therom +[.inline] +- stem:[a^2 + b^2] [.dull]#stem:[=]# stem:[c^2] + +[.inline] +- stem:[\sin^2(\theta) + \cos^2(\theta)] [.dull]#stem:[=]# stem:[1] + +[.inline] +- stem:[1 + \cot^2(\theta)] [.dull]#stem:[=]# stem:[\csc^2(\theta)] + +[.inline] +- stem:[1 + \tan^2(\theta)] [.dull]#stem:[=]# stem:[\sec^2(\theta)] + +== Ratio Trigonometric Identities +[.inline] +- stem:[\tan(\theta)] [.dull]#stem:[=]# stem:[\frac{\sin(\theta)}{\cos(\theta)}] + +[.inline] +- stem:[\cot(\theta)] [.dull]#stem:[=]# stem:[\frac{\cos(\theta)}{\sin(\theta)}] + + +== Double Angle +[.inline] +- stem:[\sin(2\theta)] [.dull]#stem:[=]# stem:[2\sin(\theta)\cos(\theta)] + +[.inline] +- stem:[\cos(2\theta)] [.dull]#stem:[=]# stem:[\cos^2(\theta) - \sin^2(\theta)] [.dull]#stem:[=]# stem:[2\cos^2(\theta) - 1] [.dull]#stem:[=]# stem:[1 - 2\sin^2(\theta)] + +[.inline] +- stem:[\tan(2\theta)] [.dull]#stem:[=]# stem:[\frac{2\tan(\theta)}{1 - \tan^2(\theta)}] + +== Circle +[.inline] +- stem:[\sin(\theta)] [.dull]#stem:[=]# stem:[0] [.dull]#stem:[\leftrightarrow]# stem:[\theta] [.dull]#stem:[=]# stem:[K\pi] + +[.inline] +- stem:[\cos(\theta)] [.dull]#stem:[=]# stem:[0] [.dull]#stem:[\leftrightarrow]# stem:[\theta] [.dull]#stem:[=]# stem:[\pi/2 + k\pi] + +NOTE: You can think of it like this as well stem:[sin(K\pi) = 0] and stem:[cos(\pi/2 + k\pi) = 0] + +== Links +- https://byjus.com/maths/trigonometric-identities/[Trigonometric Identities] diff --git a/Triginometry/Identities.html b/Triginometry/Identities.html new file mode 100644 index 0000000..fe741a4 --- /dev/null +++ b/Triginometry/Identities.html @@ -0,0 +1,1327 @@ + + + + + + + +Triginometry | Identities + + + + + + + + + + + + + + + + + + +
+
+

Even vs Odd Functions

+
+ ++++ + + + + + + + + + + + + + + + + + + + + + + + + +
OddEven

sin

cos

tan

sec

csc

cot

+
+
+
+

SOH CAH TOA

+
+
+
    +
  • +

    \$\sin(\theta)\$ \$=\$ \$\frac{opp}{hyp}\$

    +
    • +
+
+
+
    +
  • +

    \$\cos(\theta)\$ \$=\$ \$\frac{adj}{hyp}\$

    +
    • +
+
+
+
    +
  • +

    \$\tan(\theta)\$ \$=\$ \$\frac{opp}{adj}\$

    +
    • +
+
+
+
    +
  • +

    \$\csc(\theta)\$ \$=\$ \$\frac{hyp}{opp}\$

    +
    • +
+
+
+
    +
  • +

    \$\sec(\theta)\$ \$=\$ \$\frac{hyp}{adj}\$

    +
    • +
+
+
+
    +
  • +

    \$\cot(\theta)\$ \$=\$ \$\frac{adj}{opp}\$

    +
    • +
+
+
+
+
+

Reciprocal

+
+
+
    +
  • +

    \$\sin(\theta)\$ \$=\$ \$\frac{1}{\csc(\theta)}\$

    +
    • +
+
+
+
    +
  • +

    \$\cos(\theta)\$ \$=\$ \$\frac{1}{\sec(\theta)}\$

    +
    • +
+
+
+
    +
  • +

    \$\tan(\theta)\$ \$=\$ \$\frac{1}{\cot(\theta)}\$

    +
    • +
+
+
+
    +
  • +

    \$\csc(\theta)\$ \$=\$ \$\frac{1}{\sin(\theta)}\$

    +
    • +
+
+
+
    +
  • +

    \$\sec(\theta)\$ \$=\$ \$\frac{1}{\cos(\theta)}\$

    +
    • +
+
+
+
    +
  • +

    \$\cot(\theta)\$ \$=\$ \$\frac{1}{\tan(\theta)}\$

    +
    • +
+
+
+
+
+

Pythagoras Therom

+
+
+
    +
  • +

    \$a^2 + b^2\$ \$=\$ \$c^2\$

    +
    • +
+
+
+
    +
  • +

    \$\sin^2(\theta) + \cos^2(\theta)\$ \$=\$ \$1\$

    +
    • +
+
+
+
    +
  • +

    \$1 + \cot^2(\theta)\$ \$=\$ \$\csc^2(\theta)\$

    +
    • +
+
+
+
    +
  • +

    \$1 + \tan^2(\theta)\$ \$=\$ \$\sec^2(\theta)\$

    +
    • +
+
+
+
+
+

Ratio Trigonometric Identities

+
+
+
    +
  • +

    \$\tan(\theta)\$ \$=\$ \$\frac{\sin(\theta)}{\cos(\theta)}\$

    +
    • +
+
+
+
    +
  • +

    \$\cot(\theta)\$ \$=\$ \$\frac{\cos(\theta)}{\sin(\theta)}\$

    +
    • +
+
+
+
+
+

Double Angle

+
+
+
    +
  • +

    \$\sin(2\theta)\$ \$=\$ \$2\sin(\theta)\cos(\theta)\$

    +
    • +
+
+
+
    +
  • +

    \$\cos(2\theta)\$ \$=\$ \$\cos^2(\theta) - \sin^2(\theta)\$ \$=\$ \$2\cos^2(\theta) - 1\$ \$=\$ \$1 - 2\sin^2(\theta)\$

    +
    • +
+
+
+
    +
  • +

    \$\tan(2\theta)\$ \$=\$ \$\frac{2\tan(\theta)}{1 - \tan^2(\theta)}\$

    +
    • +
+
+
+
+
+

Circle

+
+
+
    +
  • +

    \$\sin(\theta)\$ \$=\$ \$0\$ \$\leftrightarrow\$ \$\theta\$ \$=\$ \$K\pi\$

    +
    • +
+
+
+
    +
  • +

    \$\cos(\theta)\$ \$=\$ \$0\$ \$\leftrightarrow\$ \$\theta\$ \$=\$ \$\pi/2 + k\pi\$

    +
    • +
+
+
+ + + + + +
+ + +You can think of it like this as well \$sin(K\pi) = 0\$ and \$cos(\pi/2 + k\pi) = 0\$ +
+
+
+
+
+

Links

+
+
+ +
+
+
+
+ + + + + \ No newline at end of file diff --git a/Triginometry/index.adoc b/Triginometry/index.adoc new file mode 100644 index 0000000..08bab48 --- /dev/null +++ b/Triginometry/index.adoc @@ -0,0 +1,12 @@ += ./Triginometry +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Triginometry/Centers.html;hb=HEAD[Centers] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Triginometry/Basics.html;hb=HEAD[Basics] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Triginometry/Identities.html;hb=HEAD[Identities] diff --git a/Triginometry/index.html b/Triginometry/index.html new file mode 100644 index 0000000..a0407cc --- /dev/null +++ b/Triginometry/index.html @@ -0,0 +1,1058 @@ + + + + + + + +./Triginometry + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+
+ +
+
+
+
+ + + \ No newline at end of file diff --git a/diagram/diag-ditaa-md5-748c8d76a4b154138a91a12caf43c78a.png.cache b/diagram/diag-ditaa-md5-748c8d76a4b154138a91a12caf43c78a.png.cache new file mode 100644 index 0000000..08875c5 --- /dev/null +++ b/diagram/diag-ditaa-md5-748c8d76a4b154138a91a12caf43c78a.png.cache @@ -0,0 +1 @@ +{"checksum":"ditaa-md5-748c8d76a4b154138a91a12caf43c78a","options":{"scale":null,"tabs":null,"background":null,"antialias":null,"separation":null,"round_corners":null,"shadows":null,"debug":null,"fixed_slope":null,"transparent":null,"bullet_characters":null},"width":640,"height":364} \ No newline at end of file diff --git a/diagram/index.adoc b/diagram/index.adoc new file mode 100644 index 0000000..6e8a612 --- /dev/null +++ b/diagram/index.adoc @@ -0,0 +1,9 @@ += ./diagram +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +== Topics + diff --git a/diagram/index.html b/diagram/index.html new file mode 100644 index 0000000..aa1b28d --- /dev/null +++ b/diagram/index.html @@ -0,0 +1,1046 @@ + + + + + + + +./diagram + + + + + + + + + + + + + + + + + + +
+
+

Topics

+
+ +
+
+
+ + + \ No newline at end of file diff --git a/docinfo.conf b/docinfo.conf new file mode 100644 index 0000000..fc4596c --- /dev/null +++ b/docinfo.conf @@ -0,0 +1,3 @@ +#!/usr/bin/env bash +HOSTNAME=nyquistdev.net +DOCINFODIR=/usr/local/share/asciidoctor-styles/dark1/ diff --git a/favicon.png b/favicon.png new file mode 100644 index 0000000000000000000000000000000000000000..c99ab8a8bd94a652f62652ccb4636888bd095935 GIT binary patch literal 2147 zcmV-p2%PtcP)EX>4Tx04R}tkv&MmKpe$iTSd`|h#iV3WT;Md(ISqu3Pq?8YK2xEOfLO`CJjl7 zi=*ILaPVWX>fqw6tAnc`2!4RL3r>nIQsV!TLW>v=j{EWM-sA2a;BQozYIel|RkMtA zG9hGht3uB!`Z0zfgb|aNsn3aG3ZCQZ9zMR_MR}I@xj)B%k~bOP6NuwXH!R`};+aiL z=e$oGVkJo-J|~_u=z_$LT$f#b<6Lyu&oe_tCOuCaA{L4rEO#&~87gs%IHIT;<@>WP zE1b7DtCbpS-IKpCnAcX8xlVHwNi1RsA_T~&p^OSFL}}GXF_EVIgol5`@r&e=$yEj; z#{#NQAvu2VKlt6PS(u)3lY$AL`^C0D#(|z)pjo%=?_=9;o&bSo;7V)zs|{fGlk|F9 ziyQ%i+rY(jTa)*I%N-#6q)Ue6NCBGuLIHR`qi@Oq!?!@+n%i4zAEysMhPq1K00)P_ zM3J)BJ>K2h+1tNoTK)Y1?45Gg=x7#P00006VoOIv00000008+zyMF)x010qNS#tmY zE+YT{E+YYWr9XB6000McNliru=?fhZHYRS#gLVJ_02y>eSad^gZEa<4bO1wgWnpw> zWFU8GbZ8()Nlj2!fese{00uirL_t(&-tC!LY*b|w$A7cWz?4o`Xs0dhPzth0q}U3S zC5V*OAZUyjLE-{NVj@29V0cm^(HM=vNL)e`HGp9YC{iM#MqIE3+O(z6LTO94={lX6 z&UBXRL#OR*E%y#E#f0-TbMANMJOBHA=bZmJ1NUWgMHZo5<~`u|4ZtfFKo+nSSR_*@ zbH|xX%CU)wqvm}_Z7q(b#ysG>@)m1V!25EY?vT2v=o#6EjAe%vP7rU*T(VTBRI91@ z;RN=E`Zq^#9zf&&#naaZP$zQVlwgIey-kXJSOCu@O_@St=_Sk;OO6BY-okdg{}-3T zA!Xn3!3uNfB`x_b+ba(4m-(W*SxhDj0ARBLctq}N3b=kQ7eM3f)PWx0quZg>PMcM15Nd!if*>H1 z^^;+DfBe+mX)bgti5tqH0=Bj`v3A<{=`+*=&I7o5?Er1j24Ea&FJw4dn!5n22)k@; z%s9eaA!{@NXjO6Xse%v?=VAlUsELUoIEaDk(F8EvkeP}9HINX5dY!=#cmi%WfHRr^ zOa`MdSl1-dHeIRF1fD=|FOk%M+d2VFOl+{OU8Jo}p;m{}JE937cWh$9kmuL0NjU+d zQlkw}xSWoNGfBh%H2R@HboN+7*y#wOkShYW?X%jVdWABysi_k*dc9vW(gV;U(mq?E z)`Zi0q6J`a>daXI>Bre)k;EhwsyjE?hy*~PAuTc=2dva5CkJ$59hDVs zz#It^jW&RF#`y&X|NHN9^n%x054c4BMo12oD--}pcN>5V*q%D`!N3&QUxcL4fRI?x z#Ti2A`7p8o*aLhDBuWLaI)3_$*|Em7ka0+)JZ6ws#To$Sh*-K|>bfVl#7@rH3K*mS zPy)M51q+eM15F%VT>w|4IPB}6hsaU?qI9RC?&_=Vfx;;kz)s`*f@y~I zkZB0Ejt+ow3I11kTzx@TD4yimxh7!K2zD=}(RL41oEXA8NbU1ANW@zsr>Q;1{E#?2aimgx|164(a33LMpppHTGh<`;Bh;^MJ1H39q; zZm<)rO%NYT_qm4)ek~ z=)QWT7AO&Oh1O>;U8Yin*#yAraZ`KloEJDI5r7jo)!itad^VSt0emU8kfp`=QHGGW zUtf=_z3pry<3Qk>h6^PF17T5OHMerD=u{i9TXZG%dOXAGYkvC`;Nu8u*U{FJ;y+p{ zD?}}J_1Y;vaU29;8_*>dKyho;Wr8LT{mpM#DP2{Uiz2u_gKaSIY}x+r%x%@xAPhgn zbXY8u9zKkv`sxngKuFAtRB$YHb*nV-2`T#IBxG_q9+!*eigGFr?r#8AMmo#!NjHM< zWAnw5#ezbSq%#;$X|&+0HGOs)EtQpA*!Q)=-gxtM;KN8ZNd|tBDOKLoN9GBtxcGsf zO}~STSP#4n+|Z_^3aNAF39*xN1Z1*K`v0xTxfR0{J4TH4m2We5Pf{5A-(gf5r28UO Z{sSd=6{Syw*X95K002ovPDHLkV1gNW?Op%? literal 0 HcmV?d00001 diff --git a/index.adoc b/index.adoc new file mode 100644 index 0000000..7fac565 --- /dev/null +++ b/index.adoc @@ -0,0 +1,17 @@ += Math Notes +:docinfo: shared +:source-highlighter: pygments +:pygments-style: monokai +:icons: font +:docinfodir: {DOCINFODIR} + +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Graphing/index.html;hb=HEAD[Graphing] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Calculus/index.html;hb=HEAD[Calculus] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Physics/index.html;hb=HEAD[Physics] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Triginometry/index.html;hb=HEAD[Triginometry] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Probability/index.html;hb=HEAD[Probability] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Graphics/index.html;hb=HEAD[Graphics] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Algrebra/index.html;hb=HEAD[Algrebra] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=diagram/index.html;hb=HEAD[diagram] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Geometry/index.html;hb=HEAD[Geometry] +* https://{HOSTNAME}/git/?p=math-notes.git;a=blob_plain;f=Linear-Algrebra/index.html;hb=HEAD[Linear-Algrebra] diff --git a/index.html b/index.html new file mode 100644 index 0000000..f8134c0 --- /dev/null +++ b/index.html @@ -0,0 +1,1074 @@ + + + + + + + +Math Notes + + + + + + + + + + + + + + + + + + +
+
+ +
+
+ + + \ No newline at end of file diff --git a/notes.adoc b/notes.adoc new file mode 100644 index 0000000..d129cd2 --- /dev/null +++ b/notes.adoc @@ -0,0 +1,28 @@ +== Issues +By defaut the stem asciidoctor extension does not support some things. Like +writing the index of roots. + +You can using latex math you can combat this + +---- +:stem: latexmath +---- + +Then, things like + +---- +\sqrt[5]{25} +---- + +Will show up correctly + +== Links +* https://www.overleaf.com/learn/latex/Subscripts_and_superscripts +* https://math.mit.edu/~djk/calculus_beginners/chapter04/section02.html +* https://www.youtube.com/watch?v=PahbNFypubE&t=398s[interpolation] +* https://en.wikipedia.org/wiki/Summation[Summation] +* https://youtu.be/I0vDXO5FeaI?si=DpAxgKE7eYFAk2el[Math Symbols] +* https://www.cmor-faculty.rice.edu/~heinken/latex/symbols.pdf[latex math symbols] +* https://latexeditor.lagrida.com/[live latex editor] +* https://wiki.inkscape.org/wiki/Installing_Inkscape[inkscape | drawing diagrams] +* https://www.youtube.com/watch?v=31A8FYRk6Ko[inkscape | graphs] diff --git a/rendered.html b/rendered.html new file mode 100644 index 0000000..1c8cefd --- /dev/null +++ b/rendered.html @@ -0,0 +1,854 @@ + + + + + + + +My Math Notes + + + + + + + + + + + + + + + + + + + + +
+
+
+
+

To view the rendered notes, you can visit this site: +my-math-notes.

+
+
+
+
+

Building

+
+
+

To convert all the .adoc files into html run this command in the root directory

+
+
+
+
./Makehtml make
+
+
+
+

It will automatically create a index.html page for each subject. You do not +need to make index.html pages yourself

+
+
+
+
+ + + + \ No newline at end of file -- 2.43.0