I have the following information: For all $x \in \mathbb{R}$: $$T_0(x) := 1$$ $$T_1(x) := x$$ $$\forall n\in\mathbb{N}: T_{n+1}(x):= 2xT_n(x) - T_{n-1}(x)$$ Now I have to prove that for any $n\in\mathbb{N}$ The function $x\rightarrow T_n(x)$ is a polynomial of degree $n$ .
Furthermore, I have to prove that for $x\in [-1,1]$ $$\forall n\in\mathbb{N}: T_n(x) = \cos(n\ \arccos(x))$$
To prove that the function is a polynomial I tried induction, but I seem to use circular logic instead of proving anything. What confuses me is that the definition for $T_{n +1}$ is an inductive definition in the first place. Any hints?
Induction is definitely the right way to go to proove that $T_n$ is a polynomial of degree $n$. Let $n\geqslant 2$, assume the result holds for $T_{n-2}$ and $T_{n-1}$, then $2XT_{n-1}$ is of degree $n$ and $T_{n-2}$ of degree $n-2<n$, therefore the difference is a polynomial of degree $n$.
Let $\theta\in\mathbb{R}$, one has: $T_0(\cos(\theta))=1=\cos(0\times\theta)$ and $T_1(\cos(\theta))=\cos(\theta)$. Using duplication formula and induction, you can proove that: $$\forall n\in\mathbb{N},T_n(\cos(\theta))=\cos(n\theta).$$ Therefore, since $\cos\circ\arccos=\textrm{id}$ on $[-1,1]$,one has: $$T_n(\theta)=\cos(n\arccos(\theta)).$$
Reminders.
$\deg(P+Q)\leqslant\max(\deg(P),\deg(Q))$ and the equality holds whenever $\deg(P)\neq\deg(Q)$.
Duplication formula: $\cos(a+b)+\cos(a-b)=2\cos(a)\cos(b)$.