1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Help with integral
  1. Integration help (Replies: 10)

Loading...