The following is taken from Fourier Series and Integral Transforms by A. Pinkus and S. Zafrany, chapter 1 question 2:

Let define $\forall f,g \in C[-1,1],\quad \left<f,g\right> \overset{\Delta}{=} \int_{-1}^1\frac{f(x)\overline{g(x)}}{\sqrt{1-x^2}}dx. $ Prove the follows:

  1. This is an inner product over $C[-1,1]$
  2. The set $\{T_n \overset{\Delta}{=} \cos(n\arccos{x}) \vert n\in \mathbb{N}\}$ is orthogonal.
  3. By substituting $\theta = \arccos x$ prove that for all $n$, $T_n$ is an n degree polynom of $x$.

I proved (1) and (2), the latter by using the substituion mentioned in (3).
It's straightforward that $\left<T_n,T_n\right> >0$, using the fact that the integrand is odd around 0. For $m\neq n$ we get:
$\left<T_n,T_m\right>=\int_{-1}^1 \frac{\cos(n\arccos x) \cos(m\arccos x)}{\sqrt{1-x^2}} dx=\\ \int_{-1}^1 \frac{\cos((m+n)\arccos x) + \cos((m-n)\arccos x)}{2\sqrt{1-x^2}}dx=\\ \int_{0}^\pi \left(\cos((m+n)\theta) + \cos((m-n)\theta)\right) d\theta=0$.

I'm not sure though how is (3) related to all this, or how to derive it using the above.



For question $(3) $, you can use a strong induction proof.

$$T_0 (x)=\cos (0)=1$$ $$T_1 (x)=\cos (\arccos x)=x $$ $$T_2 (x)=\cos (2\arccos x)=2x^2-1$$

$$T_{n+1}(x)+T_{n-1}(x)=$$ $$\cos ((n+1)\theta) +\cos ((n-1)\theta)=$$ $$2\cos(\theta)T_n (x)=2xT_n (x) $$

  • $\begingroup$ That's elegant! Shouldn't have been stuck on the inner-product I guess... Thank you! $\endgroup$ – galra Jul 19 '17 at 22:25

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