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Homework Help: Help with integral

  1. Oct 12, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the elements of the seqeunce [tex]h_k\left(x\right)=H_k\left(x\right)e^{-\frac{1}{2}x^2} [/tex] and [tex]H_k\left(x\right)=\left(-1\right)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex] have the norm [tex]||h_k||=\sqrt{2^nn!\sqrt{\pi}}[/tex]

    2. Relevant equations

    [tex]||h_k||=\sqrt{\left(h_k^*,h_k\right)}[/tex]
    [tex]\left(h_k^*,h_k\right)=\int{h_k^*h_kdx}[/tex]

    3. The attempt at a solution

    [tex]h_k\left(x\right)=\left(-1\right)^n e^{x^2}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}=\left(-1\right)^n e^{\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex]


    [tex]||h_k||=\sqrt{\int{\left(\left(-1\right)^n e^{\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}\right)^*\left(\left(-1\right)^n e^{\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}\right)dx}}[/tex]

    [tex]=\sqrt{\int{\left(e^{x^2}\frac{d^{2n}}{dx^{2n}}e^{-2x^2}\right)dx}}[/tex]

    I tried using [tex]\int{udv}=uv-\int{vdu}[/tex]

    [tex]u=e^{x^2}[/tex]
    [tex]du=2xe^{x^2}dx[/tex]
    [tex]dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx[/tex]
    [tex]v=\frac{d^{2n}}{dx^{2n-1}}e^{-2x^2}[/tex] I don't think that's right.

    Any help would be appreciated.
     
  2. jcsd
  3. Oct 12, 2009 #2
    The interval that seems to work goes from 0 to infinity.
     
  4. Oct 12, 2009 #3

    gabbagabbahey

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    First, surely you mean :

    [tex]H_k\left(x\right)=\left(-1\right)^k e^{x^2}\frac{d^k}{dx^k}e^{-x^2}[/tex]

    and

    [tex]||h_k||=\sqrt{2^k k!\sqrt{\pi}}[/tex]

    right? :confused:

    Second, [itex]\frac{d^k}{dx^k}[/itex] is a differential operator, and when it operates on a product, you will (not surprisingly!) need to use the product rule.:wink:
     
  5. Oct 12, 2009 #4
    Here is an image of the problem statement -- part (d):
     

    Attached Files:

  6. Oct 12, 2009 #5

    gabbagabbahey

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    Can you post your image to imageshack.us instead? (Attachments cannot be viewed by other posters until they are approved by admin----which often takes a while)
     
  7. Oct 12, 2009 #6
    I guess that is a typo on the instructor's part.

    Here is the product rule:

    [tex]\frac{d}{dx}f\left(x\right)g\left(x\right)=f\left(x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}[/tex]

    What I have in this problem is:

    [tex]\frac{d}{dx}f\left(x\right)[/tex]

    where

    [tex]f\left(x\right)=e^{-x^2}[/tex]

    I'm not seeing how the product rule applies here.
     
  8. Oct 12, 2009 #7
  9. Oct 12, 2009 #8
    If you're talking about taking the derivative n times, then yes, the product rule is used.

    I tried that and I don't see how it helps.

    [tex]\left(\frac{d}{dx}\right)^ne^{-2x^2}[/tex]

    [tex]n=1:[/tex]
    [tex]\left(-4x\right)e^{-2x^2}[/tex]

    [tex]n=2:[/tex]
    [tex]\left(-4x\right)^2e^{-2x^2}+\left(-4\right)e^{-2x^2}[/tex]

    [tex]n=3:[/tex]
    [tex]\left(-4x\right)^3e^{-2x^2}+\left(-12\right)\left(-4x\right)^2e^{-2x^2}+\left(-4\right)^2xe^{-2x^2}[/tex]

    etc...
     
  10. Oct 12, 2009 #9

    gabbagabbahey

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    I'd have to agree!

    It applies when you calculate [itex]h_k^*h_k[/itex],

    [tex]h_k^*h_k=\left(\left(-1\right)^k e^{\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}\right)\left(\left(-1\right)^k e^{\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}\right)=e^{\frac{1}{2}x^2}\frac{d^k}{dx^k}\left(e^{-\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}\right)[/tex]

    The leftmost [tex]\frac{d^k}{dx^k}[/itex] acts on the product [tex]e^{-\frac{x^2}{2}}\frac{d^k}{dx^k}e^{-x^2}[/itex]
     
  11. Oct 12, 2009 #10
    [tex]
    \frac{d}{dx}f\left(x\right)g\left(x\right)=f\left( x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}
    [/tex]

    [tex]f\left(x\right)=e^{-\frac{1}{2}x^2}[/tex]
    [tex]g\left(x\right)=\frac{d^n}{dx^n}e^{-x^2}[/tex]

    [tex]\frac{df}{dx}=\left(-x\right)e^{-\frac{1}{2}x^2}[/tex]
    [tex]\frac{dg}{dx}=\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}[/tex]

    [tex]f\left( x\right)\frac{dg}{dx}+g\left(x\right)\frac{df}{dx}=e^{-\frac{1}{2}x^2}\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}+\frac{d^n}{dx^n}e^{-x^2}\left(-x\right)e^{-\frac{1}{2}x^2}[/tex]

    [tex]=e^{-\frac{1}{2}x^2}\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}-\frac{d^n}{dx^n}xe^{-\frac{3}{2}x^2}[/tex]

    Looks like it's getting more complicated.
     
  12. Oct 12, 2009 #11

    gabbagabbahey

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    EDIT: Nvm. There's no need to use the product rule the way I suggested in my previous post, [itex]h_k(x)[/itex] is a polynomial, not an operator...I should get some sleep.:redface:

    Just continue on with the expression you had in your first post (but use either [itex]n[/itex] or [itex]k[/itex], don't mix and match) and integrate from [itex]-\infty[/itex] to [itex]\infty[/itex].

    Also,

    [tex]dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx=\frac{d}{dx}\left(\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}\right)dx[/tex]

    [tex] \implies v=\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}[/tex]

    which is what I think you meant in your 1st post.

    Further Edit:

    [tex]h_k^*h_k=e^{x^2}\left(\frac{d^k}{dx^k}e^{-x^2}\right)^2\neq e^{x^2}\left(\frac{d^{2k}}{dx^{2k}}e^{-2x^2}\right)[/tex]
     
    Last edited: Oct 12, 2009
  13. Oct 12, 2009 #12
    [tex]f\left(x\right)=\left(-1\right)^ne^{-\frac{1}{2}x^2}[/tex]
    [tex]g\left(x\right)=\frac{d^n}{dx^n}e^{-x^2}[/tex]

    [tex]f\left(x\right)g\left(x\right)f\left(x\right)g\left(x\right)=\left(-1\right)^ne^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}\left(-1\right)^ne^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex]

    [tex]=\left(-1\right)^{2n}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex]

    [tex]=e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex]

    [tex]=e^{-\frac{1}{2}x^2}\frac{d^n}{dx^n}e^{-\frac{3}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}[/tex]

    Leibniz Rule:

    [tex]\frac{d^n}{dx^n}fg=\sum_{k=0}^n{\frac{n!}{k!\left(n-k\right)!}\frac{d^k}{dx^k}f\frac{d^{n-k}}{dx^{n-k}}g}[/tex]

    [tex]\frac{d^n}{dx^n}e^{-\frac{3}{2}x^2}\frac{d^n}{dx^n}e^{-x^2}=\sum_{k=0}^n{\frac{n!}{k!\left(n-k\right)!}\frac{d^k}{dx^k}e^{-\frac{3}{2}x^2}\frac{d^{n-k}}{dx^{n-k}}\frac{d^n}{dx^n}e^{-x^2}}[/tex]

    [tex]=\sum_{k=0}^n{\frac{n!}{k!\left(n-k\right)!}\frac{d^k}{dx^k}e^{-\frac{3}{2}x^2}\frac{d^{2n-k}}{dx^{2n-k}}e^{-x^2}}[/tex]

    :(
     
  14. Oct 12, 2009 #13
    [tex]u=e^{x^2}[/tex]
    [tex]du=2xe^{x^2}dx[/tex]
    [tex]dv=\frac{d^{2n}}{dx^{2n}}e^{-2x^2}dx[/tex]
    [tex]v=\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}[/tex]

    Substitute back into:

    [tex]\int{udv}=uv-\int{vdu}[/tex]

    [tex]=e^{x^2}\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}-\int{\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}2xe^{x^2}dx}[/tex]

    [tex]=e^{x^2}\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}-\int{2xe^{x^2}\frac{d^{2n-1}}{dx^{2n-1}}e^{-2x^2}dx}[/tex]

    I don't think this is getting me anywhere.
     
  15. Oct 12, 2009 #14

    gabbagabbahey

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    See my "further edit" of my last post.
     
  16. Oct 12, 2009 #15
    Well then, if you don't mind me asking, what is it?
     
  17. Oct 12, 2009 #16

    gabbagabbahey

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    Try using

    [tex]u=e^{x^2}\left(\frac{d^k}{dx^k}e^{-x^2}\right)[/tex]

    and

    [tex]dv=\left(\frac{d^k}{dx^k}e^{-x^2}\right)dx=\frac{d}{dx}\left(\frac{d^{k-1}}{dx^{k-1}}e^{-x^2}\right)dx[/tex]

    to integrate by parts.
     
  18. Oct 12, 2009 #17

    [tex]u=e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)[/tex]
    [tex]du=\left(2xe^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)\right)dx[/tex]

    [tex]dv=\left(\frac{d^n}{dx^n}e^{-x^2}\right)dx[/tex]
    [tex]v=\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)[/tex]

    [tex]uv-\int{vdu}[/tex]
    [tex]=e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e^{-x^2}\right)\right)dx}[/tex]

    :cry:
     
  19. Oct 12, 2009 #18

    gabbagabbahey

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    Okay, now use the fact that

    [tex]\left(\frac{d^n}{dx^n}e^ {-x^2}\right)=\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d}{dx}e^ {-x^2}\right)=\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)[/tex]

    and

    [tex]\left(\frac{d^{n+1}}{dx^{n+1}}e^ {-x^2}\right)=\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d^2}{dx^2}e^ {-x^2}\right)[/tex]

    to simplify

    [tex]\left[2xe^{x^2}\left(\frac{d^n}{dx^n}e^ {-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e ^{-x^2}\right)\right][/tex]

    ...
     
  20. Oct 12, 2009 #19
    [tex]
    e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^n}{dx^n}e^ {-x^2}\right)+e^{x^2}\left(\frac{d^{n+1}}{dx^{n+1}}e ^{-x^2}\right)\right)dx}
    [/tex]

    [tex]
    =e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)\right)+e^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d^2}{dx^2}e^ {-x^2}\right)\right)\right)dx}
    [/tex]

    Not seeing how this can be simplified much.
     
  21. Oct 12, 2009 #20

    gabbagabbahey

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    Calculate [itex]\frac{d^2}{dx^2}e^{-x^2}[/itex] explicitly and then simplify...

    Also, don't forget your integration limits!

    [tex]||h_n(x)||=\left.e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\right|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)\right)+e^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(\frac{d^2}{dx^2}e^ {-x^2}\right)\right)\right)dx}[/tex]

    The terms

    [tex]\left(\frac{d^n}{dx^n}e^{-x^2}\right)[/tex]

    and

    [tex]\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)[/tex]

    are both going to be some polynomials multiplied by [itex]e^{-x^2}[/itex], so

    [tex]e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)[/tex]

    is also going to be some polynomial times [itex]e^{-x^2}[/itex].

    So when you integrate from [itex]-\infty[/itex] to [itex]\infty[/itex], what does this term become? (Remember, a negative exponential decreases faster than any polynomial can increase:wink: )

    Basically, you want to end up with something along the lines of [itex]||h_n(x)||\propto ||h_{n-1}(x)||[/itex]; giving you a recurrence relation. After, you need only calculate [itex]||h_0(x)||[/itex] explicitly, and then you can determine [itex]||h_n(x)||[/itex] for all [itex]n[/itex] from this recurrence relation.
     
    Last edited: Oct 12, 2009
  22. Oct 12, 2009 #21
    [tex]\frac{d^2}{dx^2}e^{-x^2}[/tex]
    [tex]=\frac{d}{dx}\left(-2x\right)e^{-x^2}[/tex]
    [tex]=-2e^{-x^2}+\left(-2x\right)^2e^{-x^2}[/tex]

    Now I'll put that back into the last equation:

    [tex]

    =e^{x^2}\left(\frac{d^n}{dx^n}e^{-x^2}\right)\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)-\int{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2xe^ {-x^2}\right)\right)+e^{x^2}\left(\frac{d^{n-1}}{dx^{n-1}}\left(-2e^{-x^2}+\left(-2x\right)^2e^{-x^2}\right)\right)\right)dx}

    [/tex]
     
  23. Oct 12, 2009 #22

    gabbagabbahey

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    Actually, instead if using the advice of post #18, just use the Leibniz rule to simplify

    [tex]\left(\frac{d^{n+1}}{dx^{n+1}}e^ {-x^2}\right)=\frac{d^{n}}{dx^{n}}\left(-2xe^ {-x^2}\right)[/tex]

    it's much quicker this way!
     
  24. Oct 12, 2009 #23
    I see that the uv term goes away, as I suspected.

    So I'm left with:

    [tex]
    ||h_n(x)||=-\int_{-\infty}^{\infty}{\left(\frac{d^{n-1}}{dx^{n-1}}e^{-x^2}\right)\left(2xe^{x^2}\frac{d^{n}}{dx^{n}}e^{-x^2}+e^{x^2}\frac{d^{n+1}}{dx^{n+1}}\\e^{-x^2}\right)dx}
    [/tex]

    Now I can see that the first term is a polynomial times [itex]e^{-x^2}[/itex]

    And the second term is too, but the [itex]e^{x^2}[/itex] cancels it out. And the same with the last term.

    So I only have one [itex]e^{-x^2}[/itex] in the integral, which accounts for the [itex]\sqrt{\pi}[/itex] in the answer I'm looking for.

    I can see that every time you differentiate [itex]e^{-x^2}[/itex] you'll get a factor of [itex]-2x[/itex].

    So that's where the [itex]2^n[/itex] comes from.

    But I'm not seeing where the [tex]n![/tex] comes from.

    And what happens to the rest of the terms in the polynomials? Why is [itex]2^n[/itex] the only thing that survives?

    I'm about ready to give up on this. I've spent way too much time on it.
     
  25. Oct 12, 2009 #24
    I can write those derivatives as polynomials:

    [tex]||h_n(x)||=-\int_{-\infty}^{\infty}{\left(a\left(x\right)e^{-x^2}\right)\left(2xe^{x^2}b\left(x\right)e^{-x^2}+e^{x^2}c\left(x\right)e^{-x^2}\right)dx}[/tex]

    Then I can simply:

    [tex]||h_n(x)||=-\int_{-\infty}^{\infty}{\left(a\left(x\right)e^{-x^2}\right)\left(2xb\left(x\right)+c\left(x\right)\right)dx}[/tex]

    I can combine the polynomials:

    [tex]||h_n(x)||=-\int_{-\infty}^{\infty}{f\left(x\right)e^{-x^2}dx}[/tex]

    I can get a factor of [tex]2^n[/tex] out of that polynomial:

    [tex]||h_n(x)||=-2^n\int_{-\infty}^{\infty}{g\left(x\right)e^{-x^2}dx}[/tex]

    Now the integral over an infinite interval of [tex]e^{-x^2}[/tex] results in [tex]\sqrt{\pi}[/tex] multiplied by some factor depending on the power of x integrated along with it. I don't know how to get rid of those factors, or how they somehow are represented by n!.

    Also, do you know of any good integral tables online? Preferably one that has virtually "everything".
     
  26. Oct 12, 2009 #25

    gabbagabbahey

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    Good, now apply the Lebniz rule as suggested in my previous post...

    [tex]\frac{d^{n+1}}{dx^{n+1}}\\e^{-x^2}=\frac{d^{n}}{dx^{n}}\left(-2xe^{-x^2}\right)=???[/tex]
     
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