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I'm trying to use the Gram-Schmidt process to find Chebyshev Polynomials. I don't have any issues obtaining the orthogonal vectors, but I'm having a hard time normalizing them. This should be an extremely simple process but for whatever reason I'm missing something.

Let $u$ represent the orthogonal vectors and $v$ represent the normalized vectors.

I find $u_3 = t^2 - \frac12$. I know $v_3$ should be $2t^2-1$, and that can easily be obtained my multiplying $u$ by the largest denominator. The process for finding $v_k$; however, is defined as $v_k=\frac{u_k}{||u_k||}$. I know that $||u_k||=\sqrt{<u_k,u_k>}$; therefore, $v_k=\frac{u_k}{\sqrt{<u_k,u_k>}}$. For Chebyshev Polynomials, the weighted inner product is $<f,g>=\int{_{-1}^1\frac{f(t)g(t)}{\sqrt{1-t^2}}dt}$. Using that inner product, at $u_3$, the equivalent integral becomes $\int{_{-1}^1\frac{(t^2)^2}{\sqrt{1-t^2}}dt}=\frac{3\pi}{8}$. Leading to $v_3=\frac{t^2 - \frac12}{\sqrt{\frac{3\pi}{8}}}=\frac{\sqrt6(2t^2-1)}{3\sqrt\pi}\ne2t^2-1$

What am I missing here? Can someone please walk me through how to normalize $u_3$?

EDIT
Maybe I am looking at this the wrong way, in that the normalization that I showed above was correct, i.e. $v_3=\frac{\sqrt6(2t^2-1)}{3\sqrt\pi}$? I was assuming that after the normalization step, $v_3$ should equal $T_2(t)=cos(2*cos^{-1}(t))=2t^2-1$, but maybe this is not the case, and I am instead missing some other step? I'm trying to follow the process shown here.

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1 Answer

up vote 0 down vote accepted

The $T_i(x)$s are only orthogonal, not orthonormal. In fact, $\|T_0(x)\|^2=\pi$ and $\|T_i(x)\|^2=\pi/2$ for every $i>0$. So to convert the orthogonal basis obtained from the Gram-Schmidt process into Chebyshev polynomials, you need to manually make the leading coefficient of the $i$-th basis vector equal to $2^{i-1}$ for $i>0$.

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Thanks, that cleared things up. –  Slayer537 Jan 24 '13 at 21:12
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