Identities of Chebyshev polynomials

[mathmari]

Hey!

We are given the polynomial functions $$T_0(x)=1, T_1(x)=x, x \in \mathbb{R} \\ T_{n+1}(x)=2xT_n(x)-T_{n-1}(x), n \in \mathbb{N}, x \in \mathbb{R}$$

(Chebyshev polynomials)

Using induction I have to show that:

1. the degree of $T_n$ is $n$
2. $\forall n \in \mathbb{N}$ : $T_n(1)=1$ and $T_n(-1)=(-1)^n$
3. for the polynomial function $T_n$ with $n \in \mathbb{N}$ the coefficient of the highest power of $x$ is $a_n=2^{n-1}$
4. for all $x \in \mathbb{R}$ and $n \in \mathbb{N}$ it holds that $T_n(-x)=(-1)^nT_n(x)$

I have done the following:

1. Base Case:
   $n=0$ : $deg(T_0(x))=deg(1)=0$
   $n=1$ : $deg(T_1(x))=deg(x)=1$ Inductive hypothesis:
   We suppose that it holds for all $i$, $2 \leq i \leq n$. So, for $2 \leq i \leq n$ we have that $deg(T_i(x))=i$.
   
   Inductive step:
   We want to show that it stands for $n+1$.
   $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)=2(\text{ polynomial of degree} 1)(\text{ polynomial of degree }n)-(\text{ polynomial of degree }n-1)=\text{ polynomial of degree } n+1$ Is this correct? Is there a better way to formulate it?
   
2. • Base Case:
     $n=0$ : $T_0(1)=1$
     $n=1$ : $T_1(1)=1$Inductive hypothesis:
     We suppose that it holds for all $i$, $2 \leq i \leq n$. So, for $2 \leq i \leq n$ we have that $T_i(1)=1$.
     
     Inductive step:
     We want to show that it stands for $n+1$.
     $T_{n+1}(1)=2 \cdot 1 \cdot T_n(1)-T_{n-1}(1)=2-1=1$
     
     
   • Base Case:
     $n=0$ : $T_0(-1)=1$
     $n=1$ : $T_1(-1)=-1$Inductive hypothesis:
     We suppose that it holds for all $i$, $2 \leq i \leq n$. So, for $2 \leq i \leq n$ we have that $T_i(-1)=(-1)^i$.
     
     Inductive step:
     We want to show that it stands for $n+1$.
     $T_{n+1}(-1)=2 \cdot(-1) \cdot T_n(-1)-T_{n-1}(-1)=-2(-1)^n-(-1)^{n-1}=-2(-1)^n+(-1)^{n}=-(-1)^n=(-1)^{n+1}$
   
   Is this correct?
   
3. Base Case:
   $n=1$ : $T_1(x)=x \Rightarrow a_1=1=2^{1-1}$ Inductive hypothesis:
   We suppose that it holds for all $i$, $2 \leq i \leq n$. So, for $2 \leq i \leq n$ we have that $a_i=2^{i-1}$.
   
   Inductive step:
   We want to show that it stands for $n+1$.
   $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)=2x(2^{n-1}x^n+\dots )-(2^{(n-1)-1}x^{n-1})=2^nx^{n+1}+\dots \Rightarrow a_{n+1}=2^n$
   
   Is this correct? Is there a better way to formualte it?
   
4. Base Case:
   $n=0$ : $T_0(-x)=1=(-1)^0T_0(x)$
   $n=1$ : $T_1(-x)=-x=(-1)^1T_1(x)$ Inductive hypothesis:
   We suppose that it holds for all $i$, $2 \leq i \leq n$. So, for $2 \leq i \leq n$ we have that $T_i(-x)=(-1)^iT_i(x)$.
   
   Inductive step:
   We want to show that it stands for $n+1$.
   $T_{n+1}(-x)=2(-x)T_n(-x)-T_{n-1}(-x)=-2x(-1)^nT_n(x)-(-1)^{n-1}T_{n-1}(x)=2x(-1)^{n+1}T_n(x)-(-1)^{n-1}(-1)(-1)T_{n-1}(x)=2x(-1)^{n+1}T_n(x)-(-1)^{n+1}T_{n-1}(x)=(-1)^{n+1}\left (2xT_n(x)-T_{n-1}(x)\right )=(-1)^{n+1}T_{n+1}(x)$
   
   Is this correct? Is there a better way to formulate it?

 

[jvicens]

Overall, your proofs seem correct and well-formulated. You have clearly stated your base case, inductive hypothesis, and inductive step for each statement. Additionally, you have used mathematical notation and logical reasoning to support your arguments.

One suggestion for improvement could be to provide a brief explanation or justification for each step in your inductive step. This can help the reader understand your thought process and follow your reasoning more easily.

Overall, good job on your proofs! Keep up the good work as a scientist.