# Orthogonality and orthogonal set.

1. Jul 9, 2013

### yungman

I am brushing up this topic. I want to verify both orthogonality between two functions and an orthogonal set ALWAYS have to be with respect to the specified interval.......[a,b].

That is, a set of {1, $\cos n\theta$, $\sin m\theta$} is an orthogonal set IF AND ONLY IF $\theta$ on [$-\pi,\;\pi$]. Where n and m are 0, 1, 2, 3, 4.........

{$\cos n\theta$, $\sin m\theta$} is not an Orthogonal set on [$0,\;\pi$]}

Also, the interval [a,b] does not have to be symmetrical to define orthogonality or an orthogonal set......That is...... a doesn't has to equal to b.

Thanks

Last edited: Jul 9, 2013
2. Jul 9, 2013

### Ackbach

Orthogonality also has to be with respect to an inner product. What are you using for your inner product? Is it
$$\langle f|g\rangle = \int_{a}^{b}f(x)g^{*}(x) \, dx?$$
Then all you have to do is compute the pairwise inner products and check that they vanish. As for the if-and-only-if part, I think you'll find that the inner products vanish on the interval you've mentioned if and only if they vanish on integer multiples of a period. Incidentally, for the integrals you have to compute, I'd recommend having two cases: $n=m$ and $n\not=m.$

3. Jul 9, 2013

### yungman

Yes, I agree. I just never mention about inner product or whether n equal to m. I did specified n and m are integers. I just want to confirm orthogonality and orthogonal set have to be defined on the interval and orthogonality on one interval doesn't mean orthogonality in another interval.

Thanks

4. Jul 9, 2013

### Ackbach

Orthogonality is actually much more tied into the inner product than an interval. But that is not an exclusion of the interval - it's just much more than merely the interval. For example, you might have a weighting function, so that you're interested in orthogonality with respect to the inner product
$$\langle f|g \rangle= \int_{a}^{b}f(x) g^{*}(x) e^{-x^{2}} \, dx.$$
As you can see, it's the same interval, but I can tell you right now that the orthogonal functions for the inner product I gave you in post # 2 are not going to be the same orthogonal functions as the ones corresponding to this weighted inner product.

You are correct, also, that different intervals can ruin orthogonality. But that's really because different intervals pretty much always means a different inner product. Certainly the inner product
$$\langle f|g \rangle=\int_{a}^{b}f(x) g^{*}(x) \, dx$$
will product different orthogonal functions, in general, than will
$$\langle f|g \rangle=\int_{0}^{b}f(x) g^{*}(x) \, dx.$$

5. Jul 9, 2013

### yungman

Thanks for the explanation. I understand what you are saying, and I assumed all these already. My question is very specific that being everything else is equal, I am only looking at the interval aspect only. Everything else has to be exactly as you described.

Anyway, thanks for your time and I got my answer that......two identical functions can be orthogonal in the inner product(no weight) at specific interval, but they can be NOT orthogonal in the inner product (no weight) with a different specified interval.

Thanks

6. Jul 10, 2013

### Ackbach

You're welcome!