# Write sin in terms of Hermite polynomials

1. May 17, 2014

### skrat

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. May 17, 2014

### pasmith

Hint: Consider $z = ik$ for some strictly positive $k$.

3. May 17, 2014

### skrat

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. May 17, 2014

### vela

Staff Emeritus
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. May 17, 2014

### skrat

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. May 17, 2014

### vela

Staff Emeritus
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. May 17, 2014

### skrat

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. May 17, 2014

### skrat

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. May 17, 2014

### vela

Staff Emeritus
Looks good so far. You can still simplify it a bit further.

10. May 17, 2014

### skrat

Yes I thought about that too. Since only odd numbers survive I think this should be ok:

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

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