Merge branch 'develop' of github.com:srele96/sk-experiments into feat/algorithms

Author: srele96

math/13.02.2024.md

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+# 13.02.2024
+
+Prove the following identities:
+
+```math
+\sin^4 \alpha + \cos^2 \alpha + \sin^2 \alpha \cos^2 \alpha = 1
+```
+
+As I was looking at the expression I realized if I split out the $\sin^4 \alpha$ to $\sin^2 \alpha \sin^2 \alpha$ and then looked at the remainder of an expression and realized:
+
+```math
+\begin{align*}
+
+\sin^2 \alpha \sin^2 \alpha + \sin^2 \alpha \cos^2 \alpha + \cos^2 \alpha = 1\\
+
+\sin^2 (\sin^2 \alpha + \cos^2 \alpha) + \cos^2 \alpha = 1\\
+
+\sin^2 \alpha + \cos^2 \alpha = 1\\
+
+1 = 1
+
+\end{align*}
+```
+
+Pure algebraic manipulation!
+
+Aside from that...
+
+```math
+\begin{align*}
+\frac{\sin \alpha}{1 - \cos \alpha} = \frac{1 + \cos \alpha}{\sin \alpha}\\
+\frac{\sin \alpha}{1 - \cos \alpha} - \frac{1 + \cos \alpha}{\sin \alpha} = 0
+\\
+\frac{\sin^2 \alpha - (1 + \cos \alpha)(1 - \cos \alpha)}{(1 - \cos \alpha) \sin \alpha} = 0
+\\
+\frac{\sin^2 \alpha - 1 + \cos^2 \alpha}{(1 - \cos \alpha) \sin \alpha} = 0
+\\
+\frac{1 - 1}{(1 - \cos \alpha) \sin \alpha} = 0
+\\
+0 = 0
+\end{align*}
+```
+
+I started only a month ago, yet I feel quite bad for inability to solve these problems in a matter of seconds. It took me like half an hour for each of these problems looking at formulas and displaying a problem in different states to notice what algebraic transformation to apply.

math/README.md

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+# Math
+
+The start of my journey in math.
+
+[https://pripremna.matf.bg.ac.rs/](https://pripremna.matf.bg.ac.rs/)
+
+## Schedule
+
+- Saturday 10:15 _(in person)_ in the morning, 3 classes
+- Saturday 10:40 _(online)_ in the morning, 3 classes
+- Monday, pay and send the photo of the payment to the teacher
+
+[The newline doesn't work issue.](https://github.com/mathjax/MathJax/issues/2312)
+
+## Update
+
+Through reading of [paul's online notes](https://tutorial.math.lamar.edu/Classes/Alg/DividingPolynomials.aspx) I realized that through my learning of mathematics 7 years ago wasn't fruitful. I could learn the problems and how to solve them, but I could solve only problems I solved with my math teacher. Two years ago when I started learning and reflecting on my learning experience, solving various algorithmic problems, reading about programming, learning various topics that feel difficult and cause confusion, I realized I was learning incorrectly. Now I want to create as much context when learning new stuff. Do not get bound by any limitations of a certain topic and how I have to learn the topic at all. I can start and quit any time I want. I realized that through contextualizing various difficult concepts I can learn them. Make them my own, think about them, etc...

math/derivative-first-principle-arcsin.md

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+# Derivative of \(\frac{d}{dx}\arcsin{x}\) by the first principle
+
+Start by using the limit definition:
+
+\[
+\frac{d}{dx} \arcsin{x} =
+\lim\_{h \to 0} \frac{\arcsin(x+h) - \arcsin{x}}{h}
+\]
+
+Derive the identity for \(\arcsin{x}-\arcsin{y}\):
+
+\[
+\sin(\alpha-\beta)=\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}
+\\
+\arcsin(\sin(\alpha-\beta))=\arcsin(\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta})
+\\
+\alpha-\beta=\arcsin(\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta})
+\]
+
+By substitution:
+
+\[
+\alpha=\arcsin{x} \implies \sin{\alpha}=x \quad
+-\pi/2\leq\alpha\leq\pi/2 \quad
+\text{and} \quad -1 \leq x \leq 1
+\\
+\beta=\arcsin{y} \implies \sin{\beta}=y \quad
+-\pi/2\leq\beta\leq\pi/2 \quad
+\text{and} \quad -1 \leq y \leq 1
+\\
+\text{Bounds on } \alpha \text{ result in choice of the positive square root.}
+\\
+\cos{\alpha} = \sqrt{1 - \sin^2{\alpha}} = \sqrt{1 - x^2}
+\\
+\text{Bounds on } \beta \text{ result in choice of the positive square root.}
+\\
+\cos{\beta} = \sqrt{1 - sin^2{\beta}} = \sqrt{1 - y^2}
+\]
+
+Therefore:
+
+\[
+\arcsin{x} - \arcsin{y} = \arcsin(x\sqrt{1 - y^2} - y\sqrt{1 - x^2})
+\]
+
+Back to the limit and apply the derived identity:
+
+\[
+\frac{d}{dx} \arcsin{x} =
+\lim_{h \to 0} \frac{\arcsin(x+h) - \arcsin{x}}{h} =
+\\
+\lim_{h \to 0} \frac{\arcsin((x+h)\sqrt{1 - x^2} - x\sqrt{1 - (x+h)^2})}{h}
+\]
+
+Apply substitution to get rid of the function in the numerator.
+
+\[
+\arcsin((x+h)\sqrt{1 - x^2} - x\sqrt{1 - (x+h)^2}) = t
+\\
+\sin(\arcsin((x+h)\sqrt{1 - x^2} - x\sqrt{1 - (x+h)^2})) = \sin{t}
+\\
+(x+h)\sqrt{1 - x^2} - x\sqrt{1 - (x+h)^2} = \sin{t}
+\]
+
+Solve for h:
+
+\[
+(x+h)\sqrt{1 - x^2} - \sin{t} = x\sqrt{1 - (x+h)^2}
+\\
+[(x+h)\sqrt{1 - x^2} - \sin{t}]^2 = [x\sqrt{1 - (x+h)^2}]^2
+\\
+(x+h)^2\lvert1 - x^2\rvert - 2(x+h)\sqrt{1 - x^2}\sin{t} + \sin^2{t} = x^2\lvert1 - (x+h)^2\rvert
+\]
+
+_(I am possibly wrong here. This is another substitution and I might not be able to rely on restrictions of x and y.)_ Because of restrictions \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1\) in identity \(\arcsin{x} - \arcsin{y} = \arcsin(x\sqrt{1-y^2} - y\sqrt{1-x^2})\) we know that the two expressions \(1-x^2 > 0\) and \(1 - y^2 > 0\) are positive. Therefore we can use the positive definition of the absolute value.
+
+Therefore:
+
+\[
+(x+h)^2(1 - x^2) - 2(x+h)\sqrt{1 - x^2}\sin{t} + \sin^2{t} = x^2(1 - (x+h)^2)
+\\
+(x+h)^2(1 - x^2) - 2(x+h)\sqrt{1 - x^2}\sin{t} + \sin^2{t} = x^2 - x^2(x+h)^2
+\\
+(x+h)^2(1 - x^2) + x^2(x+h)^2 - 2(x+h)\sqrt{1 - x^2}\sin{t} + \sin^2{t} - x^2 = 0
+\\
+(x+h)^2(1 - x^2 + x^2) - 2(x+h)\sqrt{1 - x^2}\sin{t} + \sin^2{t} - x^2 = 0
+\\
+(x+h)^2 - 2(x+h)\sqrt{1 - x^2}\sin{t} + \sin^2{t} - x^2 = 0
+\\
+[1](x+h)^2 - [2\sqrt{1 - x^2}\sin{t}](x+h) + [\sin^2{t} - x^2] = 0
+\]
+
+We notice that it is a quadratic equation, where:
+
+\[
+a = 1 \\
+b = -2\sqrt{1 - x^2}\sin{t} \\
+c = \sin^2{t} - x^2 \\
+\]
+
+Solve for \((x + h)\) using quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
+
+\[
+(x+h) = \frac{2\sqrt{1 - x^2}\sin{t} \pm \sqrt{[2\sqrt{1 - x^2}\sin{t}]^2 - 4[\sin^2{t} - x^2]}}{2}
+\\
+(x+h) = \frac{2\sqrt{1 - x^2}\sin{t} \pm \sqrt{4(1 - x^2)\sin^2{t} - 4\sin^2{t} + 4x^2}}{2}
+\\
+(x+h) = \frac{2\sqrt{1 - x^2}\sin{t} \pm \sqrt{4\sin^2{t} - 4x^2sin^2{t} - 4\sin^2{t} + 4x^2}}{2}
+\\
+(x+h) = \frac{2\sqrt{1 - x^2}\sin{t} \pm \sqrt{4x^2 - 4x^2sin^2{t}}}{2}
+\\
+(x+h) = \frac{2\sqrt{1 - x^2}\sin{t} \pm \sqrt{4x^2(1 - sin^2{t})}}{2}
+\\
+(x+h) = \frac{2\sqrt{1 - x^2}\sin{t} \pm 2\sqrt{x^2}\sqrt{1 - sin^2{t}}}{2}
+\\
+(x+h) = \sqrt{1 - x^2}\sin{t} \pm \sqrt{x^2}\sqrt{1 - sin^2{t}}
+\\
+(x+h) = \sqrt{1 - x^2}\sin{t} + \lvert{x}\rvert(\pm\sqrt{1 - sin^2{t}})
+\\
+h = \sqrt{1 - x^2}\sin{t} + \lvert{x}\rvert(\pm\sqrt{1 - sin^2{t}}) - x
+\]
+
+Here we notice that the square root is a \(\cos{t}\):
+
+\[
+\cos{t}=\pm\sqrt{1 - sin^2{t}}
+\]
+
+_(I might be wrong here. I don't know how to handle the absolute value of x.)._ And in the context of the limit and substitution:
+
+\[\text{The variables we solve for are } \mathbf{t} \text{ and } \mathbf{h} \text{. Therefore the variable } \mathbf{x} \text{ is treated as a constant.}\]
+
+Since the absolute value of a constant is always positive value, we proceed as follows:
+
+\[
+h = \sqrt{1 - x^2}\sin{t} + x\cos{t} - x
+\\
+h \to 0 \implies t \to 0
+\\
+\sqrt{1 - x^2}\sin{t} + x\cos{t} - x = 0
+\\
+0 + x - x = 0
+\\
+0 = 0
+\]
+
+Therefore, we can apply the following substitution to the limit:
+
+\[
+\arcsin((x+h)\sqrt{1 - x^2} - x\sqrt{1 - (x+h)^2}) = t
+\\
+h = \sqrt{1 - x^2}\sin{t} + x\cos{t} - x
+\\
+h \to 0 \implies t \to 0
+\]
+
+Back to the limit and apply the substitution:
+
+\[
+\frac{d}{dx} \arcsin{x} =
+\lim_{h \to 0} \frac{\arcsin(x+h) - \arcsin{x}}{h} =
+\\
+\lim_{h \to 0} \frac{\arcsin((x+h)\sqrt{1 - x^2} - x\sqrt{1 - (x+h)^2})}{h} =
+\\
+\lim_{t \to 0} \frac{t}{\sqrt{1 - x^2}\sin{t} + x\cos{t} - x}
+\]
+
+Here we notice that:
+
+\[[x\cos{t}] \rightarrow x \text{ as } t \to 0 \]
+
+Therefore:
+
+\[
+\lim_{t \to 0} \frac{t}{\sqrt{1 - x^2}\sin{t} + x\cos{t} - x} =
+\\
+\lim_{t \to 0} \frac{t}{\sqrt{1 - x^2}\sin(t) + x - x}
+\\
+\lim_{t \to 0} \frac{t}{\sqrt{1 - x^2}\sin(t)}
+\]
+
+By the common limit and by the limit law:
+
+\[
+\lim_{t \to 0} \frac{\sin{t}}{t} = 1
+\\
+[\lim_{t \to 0} \frac{\sin{t}}{t}]^{-1} = [1]^{-1}
+\\
+\lim_{t \to 0} [\frac{\sin{t}}{t}]^{-1} = [1]^{-1}
+\\
+\lim_{t \to 0} \frac{t}{\sin{t}} = 1
+\]
+
+Hence, we conclude:
+
+\[
+\lim_{t \to 0} \frac{t}{\sqrt{1 - x^2}\sin(t)} = \frac{1}{\sqrt{1 - x^2}}
+\]

math/resources.txt

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+https://en.wikipedia.org/wiki/Algebra
+https://en.wikipedia.org/wiki/Arithmetic
+https://en.wikipedia.org/wiki/Variable_(mathematics)
+https://en.wikipedia.org/wiki/Equation_solving
+https://en.wikipedia.org/wiki/System_of_linear_equations
+https://en.wikipedia.org/wiki/Set_(mathematics)
+https://en.wikipedia.org/wiki/Mathematical_object
+https://en.wikipedia.org/wiki/Equivalence_class
+https://en.wikipedia.org/wiki/Congruence_relation#Basic_example
+https://en.wikipedia.org/wiki/Modular_multiplicative_inverse
+https://www.educationquizzes.com/knowledge-bank/what-are-bodmas-bidmas-and-pemdas/
+https://en.wikipedia.org/wiki/Indeterminate_(variable)
+https://en.wikipedia.org/wiki/Polynomial
+https://www.toppr.com/guides/maths-formulas/algebra-formula/
+https://www.splashlearn.com/math-vocabulary/coefficient
+https://www.reddit.com/r/math/
+https://www.reddit.com/r/math/comments/1af16z5/uses_of_complex_analysis/