Question about orthogonal functions

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The discussion revolves around the generation of orthogonal functions over the interval [1,4], specifically four polynomials of degree three or fewer that are orthonormal and orthogonal. It is established that while these polynomials are not the only ones satisfying orthonormality and orthogonality, they can form a unique basis under specific conditions. The conversation highlights the importance of including an additional condition to prove uniqueness, which leads to a straightforward proof by induction. Participants clarify the need for a basis where each vector represents a different degree of polynomial. The dialogue concludes with acknowledgment of the insights gained from revisiting linear algebra concepts.
Char. Limit
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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?
 
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Char. Limit said:
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?
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.
 
Hurkyl said:
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.

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.
 
Char. Limit said:
What condition are you referring to?
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.
 
Hurkyl said:
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.

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!
 
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