# Homework Help: Orthonormal functions using Gramm Schmidt

1. May 13, 2010

### scigal89

How do you choose the interval? Does it matter conceptually that the Legendre polynomials, for instance, are orthonormalised over (-1,1)? Does that mean that even though they are not orthonormal over all space even though you can obviously graph them over all space?

Specifically, I want to use Gramm Schmidt on the functions sin(x)/x and sin(2x)/(2x) and looking at their behavior at the origin but also comparing their long range oscillations. Therefore, I'm thinking 0 to 2pi might not be a good choice, but negative pi to pi or negative infinity to infinity are better choices. How do I choose the appropriate interval?

2. May 13, 2010

### Dickfore

Last edited by a moderator: Apr 25, 2017
3. May 13, 2010

### scigal89

Right. But the process involves integrating over some space, which is why I asked about the bounds. I was thinking that given my functions, 0 to 2pi might not be a good choice, but negative pi to pi or negative infinity to infinity are better choices. How do I choose the appropriate interval? Similarly, is it correct to say that for Legendre polynomials generated using Gramm Schmidt on (-1,1) that they are not orthonormal outside that interval?

Last edited by a moderator: Apr 25, 2017
4. May 13, 2010

### Dickfore

The inner product in the vector space you are considering has to be defined. This would mean, for example, that if your vector space is a functional space, say, C2[a, b] and the inner product involves a definite integral, then the bounds are a and b.

Think physically what your "functions" represent. I think that will tell you what the interval of interest is.

5. May 13, 2010

### scigal89

That's what I originally thought (over all space for my "functions"). Yet, that made me wonder, for something like the orthonormal Legendre polynomials, which are generated by Gram Schmidt on (-1,1) they are not used merely within those bounds...

6. May 13, 2010

### Dickfore

Actually, they are. The argument $x$ in Legendre polynomials comes from $x = \cos \theta$.

7. May 13, 2010

### scigal89

I didn't know that! I never thought of it parametrically before. Interesting... To be honest, though, I'm still not sure about the bounds for my case. I figure since "physically" they have to do with finding an electron at some displacement, it is extended over all space from negative infinity to infinity, so those should be my bounds - is that correct reasoning?

8. May 13, 2010

### Dickfore

I would have gone with those.