Write sin in terms of Hermite polynomials

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  • #1
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Homework Statement


Write ##sin(ax)## for ##a \in \mathbb{R}##. (Use generating function for appropriate ##z##)


Homework Equations



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

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?
 

Answers and Replies

  • #2
pasmith
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Homework Statement


Write ##sin(ax)## for ##a \in \mathbb{R}##. (Use generating function for appropriate ##z##)


Homework Equations



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

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?

Hint: Consider [itex]z = ik[/itex] for some strictly positive [itex]k[/itex].
 
  • #3
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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?
 
  • #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.
 
  • #5
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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}##
 
  • #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.
 
  • #7
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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?
 
  • #8
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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?
 
  • #9
vela
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Looks good so far. You can still simplify it a bit further.
 
  • #10
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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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