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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?

Dr_Be
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AxiomaticApproach
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  • The interesting point is that Chebyshev polynomial $T_n$ is a polynomial of exactly the degree $n$ – gammatester Oct 24 '16 at 12:45
  • Induction seems to be a good starting point. Where does the "circular logic" come into play? Starting with $n=0, 1$ is trivial. Assume the induction hypotheses to be true for $1, 2, \ldots, n$ then show that $T_{n+1}$ is a polynomial of degree $n+1$. That should be straightforward since multiplying a polynomial by $x$ results in a polynomial of degree plus one. As for the second part, set $x=\cos(\theta)$. – Dr_Be Oct 24 '16 at 12:52
  • @BerndH ... I guess I did not use circular logic then in the first place. To me, my way of proving it just felt too simple :) – AxiomaticApproach Oct 24 '16 at 12:55

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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: $\cos(a+b)+\cos(a-b)=2\cos(a)\cos(b)$.

C. Falcon
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