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Write sin in terms of Hermite polynomials

  1. May 17, 2014 #1
    1. The problem statement, all variables and given/known data
    Write ##sin(ax)## for ##a \in \mathbb{R}##. (Use generating function for appropriate ##z##)


    2. Relevant equations

    ##e^{2xz-z^2}=\sum _{n=0}^{\infty }\frac{H_n(x)}{n!}z^n##

    3. The attempt at a solution

    No idea what to do.

    My idea was that since ##sin(ax)=\frac{1}{2i}(e^{iax}-e^{-iax})## I tried to calculate ##z## from ##e^{iax}=e^{2xz-z^2}## but I doubt this method really works.

    Could somebody give me a hint?
     
  2. jcsd
  3. May 17, 2014 #2

    pasmith

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    Hint: Consider [itex]z = ik[/itex] for some strictly positive [itex]k[/itex].
     
  4. May 17, 2014 #3
    I don't see how this would be any different to my idea?

    ##e^{iax}=e^{2xz-z^2}##

    ##iax=2xz-z^2## Now this part is already confusing since this should be true for ALL x or in other words ##z\neq z(x)##. The same if I set ##z=ik##.

    If I ignore that, than this is of course quadratic equation for z:

    ##z_{1,2}=x\pm \sqrt{x^2-iax}## and also a very similar solutions for ##k## if ##z=ik##.

    I don't think that ##z## is meant to be a function of anything? Or am I wrong here?
     
  5. May 17, 2014 #4

    vela

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    Why don't you actually try what pasmith suggested? Don't make assumptions based on how you think it's supposed to work out. Just set z=ik and work out what the lefthand side equals.
     
  6. May 17, 2014 #5
    I try to not make a fool out of myself so don't be offended but I did consider his hint before replying.

    Anyhow, ##2xz-z^2=2xik+k^2=iax## which, like I already mentioned, is quadratic equation for ##k##.

    ##k_{1,2}=-ix\pm \sqrt{iax-x^2}##
     
  7. May 17, 2014 #6

    vela

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    Oh, okay, now I see what you were trying.

    Why are you solving for ##k##? What would that buy you? Again, I'll suggest you work out what the lefthand side works out to, not just the exponent.
     
  8. May 17, 2014 #7
    Because I was looking for ##z##.

    Hmm, let's try it ##e^{2xik+k^2}=e^{2xik}e^{k^2}=e^{k^2}(cos(2kx)+isin(2kx))##

    Does this mean that ##k=\frac{a}{2}##? How do I continue from here?
     
  9. May 17, 2014 #8
    I think I got it:

    for ##z=ik## the exponent ##e^{2xik+k^2}=e^{k^2}(cos(2kx)+isin(2kx)##

    for ##z=-ik## the exponent ##e^{-2xik+k^2}=e^{k^2}(cos(2kx)-isin(2kx)##

    No simply the difference between the equations above:

    ##e^{2xik+k^2}-e^{-2xik+k^2}=2ie^{k^2}sin(2kx)##

    ##sin(2kx)=\frac{1}{2ie^{k^2}}(e^{2xik+k^2}-e^{-2xik+k^2})## now note that ##z_1=ik## and ##z_2=-ik##

    ##sin(2kx)=\frac{1}{2ie^{k^2}}(e^{2xz_1-z_1^2}-e^{2xz_2-z_2^2})##

    ##sin(2kx)=\frac{1}{2ie^{k^2}}\sum _{n=0}^{\infty }\frac{H_n(x)}{n!}(ik)^n[1-(-1)^n)##

    What do you think?
     
  10. May 17, 2014 #9

    vela

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    Looks good so far. You can still simplify it a bit further.
     
  11. May 17, 2014 #10
    Yes I thought about that too. Since only odd numbers survive I think this should be ok:


    ##sin(2kx)=\frac{1}{2ie^{k^2}}\sum _{n=0}^{\infty }\frac{H_n(x)}{n!}(ik)^n[1-(-1)^n]##

    ##sin(2kx)=\frac{1}{e^{k^2}}\sum _{n=0}^{\infty }(-1)^n\frac{H_{2n+1}(x)}{(2n+1)!}k^{2n+1}##
     
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