# Thread: linear algebra Gram-Schmidt in Function Spaces problem

1. ## linear algebra Gram-Schmidt in Function Spaces problem

I attached the problem in the attachments, its easier to read this way. I know it seems long but the problems themselves aren't too long. We never really went over this stuff so I'm lost. If someone could guide me through these step by step that would really help for when I have my exam. Thank you so much.

2. So, for problem 6.1, what ideas do you have? (This guy's got hilarious footnotes, by the way. If it weren't for the typos, I'd have thought the author was Griffiths.)

3. According to the footnotes I think for 6.1 i'm supposed to show the derivative of an even function is an odd function and the same for the opposite. I'm not sure how to do this with the actual function given. Yeah, he likes to go crazy with the footnotes.

4. Right. Now, in the Rodrigues representation for the Hermite polynomials, you're taking the nth derivative of a function. Do you notice anything about that function?

5. There's already a derivative in it

6. Hmm. Not exactly what I was looking for. Ok. You've got your Hermite polynomial, and it's got a Rodrigues formula representation:

$\displaystyle \displaystyle{H_{n}(x)=(-1)^{n}e^{x^{2}}\frac{d^{n}}{dx^{n}}\,e^{-x^{2}}}.$

You're taking the nth derivative of the function $\displaystyle e^{-x^{2}}.$ What do you notice about the function $\displaystyle e^{-x^{2}},$ all by itself?

7. Umm... well the only thing I notice is that it goes from negative to positive with each derivative you take

8. Actually, with each derivative you take you're going to start getting additional terms from the product rule, right?

I'm thinking more along these lines: suppose, in that exponential function (also known as the Gaussian bell curve), that I replace x with -x. What happens?

9. I didn't think of that. But that makes sense, the function keeps getting longer and longer.
Even with the -x wouldn't it basically do the same as with x except with the alternating negative? Or am I just not getting where your going?

10. You were doing so great:

Even with the -x wouldn't it basically do the same as with x...
and then you wrote

...except with the alternating negative?
The answer to the question, "What changes when I replace x with -x in the function $\displaystyle e^{-x^{2}}$?" is... nothing! Sorry about the trick question there, as well as asking you to read my mind. But the fact that nothing changes is precisely the point. That means the function $\displaystyle e^{-x^{2}}$ is even. Now, if you examine the rest of the expression for the zeroth Hermite polynomial, you can see that it is what: even or odd?

I'm keying off of Footnote 10, in case you're wondering.

11. Oh, so replacing it with -x is the same because of the squared correct? And I think the rest of it would also be even then since the only other x is in $\displaystyle e^{x^{2}}$

12. Precisely. There's a convention that the zeroth-order derivative of a function is just the function right back at you. So you can see that the exponentials cancel, giving you, indeed, the zeroth Hermite polynomial.

Now, start taking derivatives of the exponential function formally. That is, don't actually compute the closed-form expressions of them. You take a derivative of an even function. That gives you a what kind of function? The derivative is in turn multiplied by an even function. The result is what kind of function? (Hint: it's either even or odd, but you have to justify the result.)

13. I took the first derivative and got
$\displaystyle H'(x) = -2xe^{x^{2}}(-2xe^{x^{2}}) -4x^{2}e^{-x^{2}}(-2xe^{x^{2}}}$ which is hopefully right. Now if i plug in -x the function isn't the same so its odd. Then i take it again and I should get an even function which i would prove the same way. Is this all I need for 6.1 if I did it correctly?

14. Oh, I know you can do better than that. Derivative operators only look to the right. Don't take any derivatives of the $\displaystyle e^{x^{2}}$ factor. Also clean things up a bit and use equals signs where you mean equals.

A function is not odd merely because substituting -x for x changes something. It has to change in a particular way: f(x)=-f(-x) for all x is what an odd function is like.

15. wait, don't i need to take the derivative of the whole function? So wouldn't there be a product rule and I would be taking the derivative of $\displaystyle e^{x^{2}}$ if i'm taking the derivative of the whole function.

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