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While learning the Gram-Schmidt orthonormalization process, my text would discuss orthonormalizing the standard basis for a subspace of $C[0,1]$, which is the space of continuous differentiable functions on $[0,1]$ with the inner product $\langle f, g \rangle = \displaystyle\int_0^1 f(x) g(x) dx$.

I know that orthonormalizing the standard basis of the subspace of, say, quadratics (that is, ${1,x,x^2}$) requires setting $w_1 = 1$, then taking $w_2=v_2-\displaystyle\frac{\langle v_2,w_1\rangle}{\langle w_1,w_1 \rangle} w_1$, etc. as per the normal process, and then making a unit vector out of it.

My first question is, is it correct to say that, given $w_1 = v_1$, $w_i = v_i-\displaystyle\sum_{j=2}^i \frac{\langle v_j,w_{j-1}\rangle}{\langle w_{j-1},w_{j-1} \rangle} w_{j-1}$, before normalizing?

Is there a way to find a closed formula for the $i$th vector in the orthonormal basis of $n$th degree polynomials in $C[0,1]$? Even better, is there a way to generalize this to $C[a,b]$, where $\langle f, g \rangle = \displaystyle\int_a^b f(x) g(x) dx$?

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up vote 2 down vote accepted

The orthonormal basis consists of the Legendre polynomials with the appropriate normalization. One formula for them is Rodrigues' formula: the orthonormal polynomials on the interval $[-1,1]$ are $ u_n(x) = \frac{\sqrt{2n+1}}{n!} 2^{-n-1/2} \frac{\partial^n}{\partial x^n} (x^2-1)^n $. For orthonormal polynomials on the interval $[a,b]$ you would take $v_n(t) = \sqrt{\frac{2}{b-a}} u_n(\frac{2t - a - b}{b-a})$.

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The quest for an explicit basis for $C[a,b]$ (in the linear algebra sense) is hopeless, but luckily unnecessary, since infinite sums of nice (orthogonal) functions, most notably $\sin(k\theta)$, $\cos(k\theta)$ are available.

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Not a simple closed formula but a simple recursive one. See and

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so that would make the new orthogonal polynomials be $\tilde{P_n}(x) = P_n((b-a)x+a)$? Or is that the incorrect shift? How would the shift be made? and I'm interested in a recursive formula for the orthonormal form as well, if possible. – Eugene Bulkin Apr 7 '11 at 1:55

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