linear algebra Gram-Schmidt in Function Spaces problem

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• Aug 17th 2010, 11:27 AM
Ackbeet
For the first-order derivative, you won't actually need the product rule - just the chain rule. But for the second derivative on up, you will need the product rule and the chain rule, and the sum rule, etc. So no, you don't take the derivative of the whole function. Like I said, derivatives are an operator. Operators only operate on what is to their right. Example:

$\displaystyle \displaystyle{x^{2}\frac{d}{dx}\,x^{6}=x^{2}(6x^{5 })=6x^{7}.}$ Another example:

$\displaystyle \displaystyle{\sin(x)\,\frac{d^{2}}{dx^{2}}\,\cos( x)=-\sin(x)\,\frac{d}{dx}\,\sin(x)=-\sin(x)\cos(x).}$

One more example for good measure:

$\displaystyle \displaystyle{\ln(x)\,\frac{d^{2}}{dx^{2}}\,\sinh( x^{2})=2\ln(x)\,\frac{d}{dx}(x\cosh(x^{2}))=2\ln(x )(x(2x)\sinh(x^{2})+\cosh(x^{2})).}$

And you can simplify further, but that's not the point of this example. Get the idea?

However, all of this misses the point. You need to imagine what those derivative operators are doing to the right-most (only!) exponential function. As you keep taking more and more derivatives, what happens to the even-ness of odd-ness of the overall functions?
• Aug 17th 2010, 11:40 AM
SpiffyEh
Oh ok, I was taking the derivative of the whole thing.
When I take the first derivative I get
$\displaystyle -e^{x^{2}}(-2xe^{-x^{2}})$

I'm not sure what happens to the even-ness and odd-ness of functions as you take more derivatives though.
• Aug 17th 2010, 11:46 AM
Ackbeet
Too many minus signs in there, I think, but that's much better.

However, what you should really be doing is taking the derivative of an arbitrary even function. An even function satisfies f(x) = f(-x) for all x. If I take the derivative of both sides of that equation, what do I get?
• Aug 17th 2010, 11:53 AM
SpiffyEh
if you take the derivative of both sides of f(x) = f(-x) you should get f(-x) = f(x)
• Aug 17th 2010, 11:56 AM
Ackbeet
*buzzer, buzzer* You forgot the chain rule. You also forgot to include derivative symbols.

You can do this!
• Aug 17th 2010, 12:07 PM
SpiffyEh
umm... well i took the 2nd derivative of the original function if thats what you mean and got
$\displaystyle -2xe^{x^{2}}(2xe^{-x^{2}})-e^{x^{2}}(2e^{-x^{2}} - 4x^{2}e^{-x^{2}})$
• Aug 17th 2010, 12:10 PM
Ackbeet
I think it'll be more instructive just to consider the equation f(x) = f(-x), and completely ignore the Rodrigues formula for now. What happens when I differentiate both sides of f(x) = f(-x)?
• Aug 17th 2010, 05:00 PM
SpiffyEh
umm you get f'(x) = f'(-x)?
• Aug 17th 2010, 05:12 PM
Defunkt
Quote:

Originally Posted by SpiffyEh
umm you get f'(x) = f'(-x)?

Chain rule on the RHS!
• Aug 17th 2010, 05:13 PM
SpiffyEh
f'(-x)f(-x) + f(-x)f'(-x)?
• Aug 17th 2010, 05:19 PM
Defunkt
Not exactly.

To make it clearer: Let $\displaystyle g(x) = -x$

Then $\displaystyle f(-x) = f(g(x))$ and so $\displaystyle \frac{df(-x)}{dx} = \frac{df(g(x))}{dx} = ?$
• Aug 17th 2010, 05:22 PM
SpiffyEh
$\displaystyle \frac{df(g(x))}{dx} * \frac{dg(x)}{dx}$
• Aug 17th 2010, 05:25 PM
Defunkt
Exactly. Now write that with $\displaystyle -x$ instead of $\displaystyle g(x)$ - what do you get?
• Aug 17th 2010, 05:29 PM
SpiffyEh
$\displaystyle \frac{df(-x)}{dx} * \frac{-dx}{dx}$
• Aug 17th 2010, 05:34 PM
Defunkt
Good.

So to Ackbeet's point - you got that if $\displaystyle f(x)$ is an even function, ie. $\displaystyle f(x) = f(-x) \forall x$ then:
$\displaystyle f'(x) = -f'(-x)$, which means f'(x) is an odd function.
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