1. Aug 1, 2011

### Char. Limit

All right, so I was wondering... I took a look at generating orthogonal functions (over an interval), and say I have these four:

$$\frac{1}{\sqrt{3}}$$
$$\frac{5}{3} - \frac{2}{3} x$$
$$\frac{11}{3} \sqrt{\frac{5}{3}} - \frac{10}{3} \sqrt{\frac{5}{3}} x + \frac{2}{3} \sqrt{\frac{5}{3}} x^2$$
$$\frac{245}{27} \sqrt{\frac{7}{3}} - \frac{116}{9} \sqrt{\frac{7}{3}} x + \frac{50}{9} \sqrt{\frac{7}{3}} x^2 - \frac{20}{27} \sqrt{\frac{7}{3}} x^3$$

These four polynomials are all orthonormal and orthogonal over the interval [1,4]. Now what I want to know is, is it possible to prove that these are the ONLY polynomials of degree 3 or fewer that satisfy orthonormality and orthogonality?

2. Aug 1, 2011

### Hurkyl

Staff Emeritus
The answer is obviously no -- just remember your linear algebra. (did you notice that you're working in a vector space with inner product?)

But the answer is yes (up to sign) if you include another condition that I think you meant to include. Once you make that condition explicit, I think the proof is straightforward by induction.

3. Aug 1, 2011

### Char. Limit

Umm... sorry, but I don't quite get what you mean. What condition are you referring to?

And let me get out my old linear algebra book... I think I know what you mean though.

4. Aug 1, 2011

### Hurkyl

Staff Emeritus
Well, I was hoping you'd take some time to think about what you mean....

But what I think you are thinking is not simply for an orthonormal basis on the space of degree-3 polynomials, but you actually wanted a basis where each basis vector is a different degree.

5. Aug 1, 2011

### Char. Limit

Yes, that's true. I guess I sort of forgot to mention that... sorry about that.

And after re-reading my linear algebra book, I see what you mean by the earlier comment. Thanks for the help!