3
$\begingroup$

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 $\langle f, g \rangle = \displaystyle\int_a^b f(x) g(x) dx$?

$\endgroup$
2
$\begingroup$

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 $v_n(t) = \sqrt{\frac{2}{b-a}} u_n(\frac{2t - a - b}{b-a})$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Not a simple closed formula but a simple recursive one. See http://en.wikipedia.org/wiki/Orthogonal_polynomials#Example:_Legendre_polynomials and http://en.wikipedia.org/wiki/Legendre_polynomials#Recursive_Definition

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Apr 7 '11 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.