Derivation of Snell's law

Fermat's principle says that the path taken between two points by a ray of light is the path that can be traversed in the least time.

How long does it take for light to pass from $A_1$ to $A_2$? If $c_1, c_2$ are the speeds of light passing through media $1,2$ respectively, then the time needed will be $$t(x)=\dfrac{\sqrt{h_1^2+x^2}}{c_1}+\dfrac{\sqrt{h_2^2+(a-x)^2}}{c_2}.$$ Then at the critical point, $$t'(x)=\dfrac{1}{c_1}\dfrac{x}{\sqrt{h_1^2+x^2}}-\dfrac{1}{c_2}\dfrac{a-x}{\sqrt{h_2^2+(a-x)^2}}=0.$$ Now comes the surprising moment: $$\dfrac{x}{\sqrt{h_1^2+x^2}}=\sin \alpha_1,\quad \dfrac{x}{\sqrt{h_1^2+x^2}}=\sin \alpha_2.$$ This actually gives us $$\dfrac{\sin \alpha_1}{c_1}=\dfrac{\sin \alpha_2}{c_2}$$ or $$\dfrac{\sin \alpha_1}{\sin \alpha_2}=\dfrac{c_1}{c_2},$$ which is Snell's law. In words, it says that when the sines of the angles in the different media are in the same proportion as the propagation velocities, the time of travel from $A_1$ to $A_2$ is minimised.

Reference:
Mathematical Analysis I by Vladimir A. Zorich

Chebyshev polynomials

Recall $$\cos(\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \alpha.$$ We then have $$\cos((n+1)\theta)=\cos \theta \cos n\theta-\sin \theta \sin n\theta\\ \cos((n-1)\theta)=\cos \theta \cos n\theta+\sin \theta \sin n\theta\\ \cos((n+1)\theta)=2\cos \theta \cos n\theta-\cos ((n-1)\theta)\quad (*)$$We claim that for each non-negative integer $n$, there exist integers $c_i$ such that $$\cos n\theta=\sum_{i=0}^n c_i\cos^i\theta$$ (can be proved by induction). Namely $$\cos(n\theta)=T_n(\cos \theta).$$ Let $x=\cos \theta$. Then we have Chebyshev polynomials $$T_n(x)=\cos(n\arccos x)$$ for $x\in [-1,1]$ (since $\arccos x$ is only defined in $[-1,1]$).

The equation $(*)$ also leads to Chebyschev recurrence relation $$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x).$$ In fact, each $T_n$ is actually an algebraic polynomial of degree n: $$T_0(x)=\cos(0)=1\\ T_1(x)=\cos(\arccos x)=x\\ T_2(x)=\cos (2\arccos x)=\cos^2 y-\sin^2 y=\cos^2 y-(1-\cos^2 y)=2x^2-1,\quad \cos y=x\\ T_3(x)=\cos (3\arccos x)=4x^3-3x$$ They can also be derived using the recurrence relation.

Application
Errors in Polynomial Interpolation

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Automorphism group of the complex plane

The automorphism group of the complex plane is $$\text{Aut}(\Bbb C)=\{\text{analytic bijections}\;f:\Bbb C\to \Bbb C\}.$$Such automorphisms have the form $$f(z)=az+b,\quad a,b\in \Bbb C, a\neq 0.$$ (To be proved later) If $f(z)=az+b,g(z)=a'z+b'$, then $$f\circ g(z)=aa'z+(ab'+b)\\f^{-1}(z)=a^{-1}z-a^{-1}b.$$ Expressed in matrix form, we have $$\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}a'&b'\\0&1\end{pmatrix}=\begin{pmatrix}aa'&ab'+b\\0&1\end{pmatrix}\\ \begin{pmatrix}a&b\\0&1\end{pmatrix}^{-1}=\begin{pmatrix}a^{-1}&-a^{-1}b\\0&1\end{pmatrix},$$ which encode the formulas for composing and inverting automophisms of the plane.

The correspondence $$f(z)=az+b \longleftrightarrow \begin{pmatrix}a&b\\0&1\end{pmatrix}$$ is a natural group isomorphism. We now define the parabolic group as $$P=\bigg\{\begin{pmatrix}a&b\\0&1\end{pmatrix}\bigg|\; a,b\in \Bbb C, a\neq 0\bigg\}.$$ Thus $$\text{Aut}(C)\cong P.$$ Two subgroups of $P$ are its Levi component $$M=\bigg\{\begin{pmatrix}a&0\\0&1\end{pmatrix}\bigg|\; a\in \Bbb C, a\neq 0\bigg\},$$ describing the dilations $f(z)=az$, and its unipotent radical $$N=\bigg\{\begin{pmatrix}1&b\\0&1\end{pmatrix}\bigg|\; b\in \Bbb C\bigg\},$$ describing the translations $f(z)=z+b$. The parabolic group takes the form $$P=MN=NM$$ since $$\begin{pmatrix}a&b\\0&1\end{pmatrix}=\begin{pmatrix}a&0\\0&1\end{pmatrix}\begin{pmatrix}1&\color{blue}{a^{-1}b}\\0&1\end{pmatrix}=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}.$$ Geometrically, any affine map is the composition of a translation and a dilation, in which the order of translation and dilation does not matter. Also, $M$ normalizes $N$, meaning that $$m^{-1}nm\in N\quad \forall m\in M, n\in N$$ since $$\begin{pmatrix}a^{-1}&0\\0&1\end{pmatrix}\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}=\begin{pmatrix}1&a^{-1}b\\0&1\end{pmatrix}.$$ Geometrically, it says that a dilation followed by a translation followed by the reciprocal dilation is again a translation. On the other hand, $N$ does not normalize $M$.

Motivation for the definition of a group

We know that $x+1=2$ implies $x=1$. When solving such an equation, we are actually working with the properties of a group. $$\begin{align}x+1&=2\quad &\text{integers under +}\\x+1+(-1)&=2+(-1)\quad &\text{inverses}\\x+1+(-1)&=1\quad &\text{closure under +}\\x+[1+(-1)]&=1\quad &\text{associativity}\\x+0&=1\quad &\text{identity}\\x&=1\end{align}$$ As it turns out, the definition of a group is the simplest definition that will let you solve a basic equation. How amazing is that!