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I am tasked with finding the degree 5 Chebyshev-II polynomial, using the fact that it's orthogonal to those preceding it w.r.t the Chebyshev-II inner product. I am told to use the normalisation that each polynomial be monic (having leading coefficent $1$).

I suppose this mean to use the Gram-Schmidt procedure on the standard basis of $\mathbb{R}[x]$ using the Chebyshev-II i.p:

$$(f,g)=\int_{-1}^1f(x)g(x)(1-x^2)^{1/2}dx$$

so:

$$\begin{align}p_0&=1\\ p_1 &= x - (p_0,x)p_0 = x-(1,x)1=x-\int_{-1}^1x(1-x^2)^{1/2}dx=x\\ p_2 &= x^2 - (p_1,x^2)p_1 - (p_0,x^2)p_0\\ &=x^2-(x,x^2)x-(1,x^2)1\\ &=x^2-x\int_{-1}^1 x^3(1-x^2)^{1/2}dx-\int_{-1}^1 x^2(1-x^2)^{1/2}dx\\ &=x^2-0-\frac\pi8\end{align}$$

But a quick google tells me this is wrong, and I can't for the life of me figure out where the method went wrong - did I evaluate the integrals wrong? Am I using the right inner product? Should I start with a different basis?

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    $\begingroup$ The formula as you apply is valid only for orthonormal bases. You have to divide all the terms $(p_i, x^k) p_i$ by $(p_i, p_i)$. $\endgroup$ – Berci Mar 7 at 22:37
  • $\begingroup$ Ok thanks, I didn't realise. I imagine this will solve the problem. Is there another way of applying the formula that doesn't require normalisation? $\endgroup$ – Gentleman_Narwhal Mar 7 at 22:40

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