Whoops. Not sure integration by parts will get you the $\displaystyle H_{1}$ integral. Let me think on that one a bit more.
Correct on the $\displaystyle H_{0}$ normalization!
For the $\displaystyle H_{1}$ normalization, I would do the same dirty trick as for $\displaystyle H_{0}$: square the integral. You're gonna get the following:
Let
$\displaystyle \displaystyle{I=\int_{-\infty}^{\infty}(2x)^{2}e^{-x^{2}}\,dx}.$
Then
$\displaystyle \displaystyle{I^{2}=16\int_{-\infty}^{\infty}x^{2}e^{-x^{2}}\,dx\int_{-\infty}^{\infty}y^{2}e^{-y^{2}}\,dy}$
$\displaystyle \displaystyle{=16\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x^{2}y^{2}e^{-(x^{2}+y^{2})}\,dx\,dy.}$
Switching to polar, we get
$\displaystyle \displaystyle{I^{2}=16\int_{0}^{2\pi}\int_{0}^{\in fty}r^{4}\cos^{2}(\theta)\sin^{2}(\theta)e^{-r^{2}}\,r\,dr\,d\theta}$
$\displaystyle \displaystyle{=16\int_{0}^{2\pi}\cos^{2}(\theta)\s in^{2}(\theta)\,d\theta\int_{0}^{\infty}r^{4}e^{-r^{2}}\,r\,dr.}$
The angle integral succumbs to standard Calc II techniques, perhaps using double-angle formulas. For the integral on the right, substitute $\displaystyle u=r^{2}$ and THEN integrate by parts twice.
Yeah, this is a lot of work! Welcome to special functions.
Wow, I really hope nothing like this is on the exam. I tried it with wolframapha just to see what would happen and got 2sqrt(pi) from that integral. so sqrt(2sqrt(pi)) would be the normalized H_1? Sorry for not going through the steps, i'm trying to understand as much as I can before the exam in a couple hours
WolframAlpha certainly does seem to compute things correctly. That is the correct norm, I believe. So that does it for 6.3, right?
I understand not going through the steps, if you're going for the big picture. You know better than me whether you need the big picture or the details.
So how about 6.4?
Hopefully the big picture is all I need. We only had 3 1 hour long class periods to go over 2 chapters so he really can't make things way too difficult on the exam. Well, at least I hope he thinks that way.
6.4 - I'm a little lost about doing this method with functions rather than vectors. I understand it with vectors.
The procedure is precisely the same as for vectors, except that instead of a dot product, you do integrals, and instead of plugging vectors into the dot product, you're plugging functions into the integrals. It's a fair amount of integration, as you can see. Does that help?
Sorry for not getting back sooner, I tried to get on but it would never load, maybe it was my connection. The test was pretty bad, I messed up somewhere with the arithmetic somewhere but it was too late to turn back which made it even harder. But besides that I got the actual concepts. As long as I got above a 78% I'll keep my A
Yeah, the MHF was down because of problems with the LaTeX Preview feature. It was taking forever to preview LaTeX code output. Now that's fixed (yay!).
Well, here's hoping you did well on your test. If you got the concepts, and it wasn't a multiple choice (which, in my very strong opinion, are an abomination), then surely partial credit would help you out, assuming your teacher does that.
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