# Expanding a function in Gaussian-Hermites

1. Feb 24, 2008

### alfredska

Expanding a function in "Gaussian-Hermites"

This isn't homework or coursework, but seeing as it's most like a homework problem, I figured this would be the best place to ask.

Note that I'm using the physics

1. The problem statement, all variables and given/known data
I would like to expand a function (let's take a gaussian for example) in terms of this series (very similar to the harmonic oscillator, except for a factor of 2 in the exponential):
$$s\left(x\right)=\sum_n\alpha_nH_n\left(x\right)\exp\left(-x^2\right)$$

2. Relevant equations
To find the coefficients:
$$\alpha_n=\int_{-\infty}^{\infty}F\left(x\right)H_n\left(x\right)\exp\left(-x^2\right)dx/Normalization$$

where $$F\left(x\right)=\exp\left(-\frac{x^2}{\sigma^2}\right)$$ in my example

3. The attempt at a solution
I have figured out the normalization of $$H_n\left(x\right)\exp\left(-x^2\right)$$
$$\alpha_n=\int_{-\infty}^{\infty}\left[H_n\left(x\right)\exp\left(-x^2\right)\right]^2dx=\sqrt{\frac{\pi}{2}}\left(2 n-1\right)!!$$

but apparently I'm doing something wrong when I write:
$$\alpha_n=\frac{\int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{\sigma^2}\right)H_n\left(x\right)\exp\left(-x^2\right)dx}{\sqrt{\sqrt{\frac{\pi}{2}}\left(2 n-1\right)!!}}$$

Can you tell me where I'm making my mistake?

Last edited: Feb 24, 2008
2. Feb 24, 2008

### coomast

I looked for the Hermite series expansion in one of my books and this is what I came up with:

The function f(x) can be expanded as:

$$f(x)=\sum_{n=0}^{\infty}A_n\cdot H_n(x)$$

With:

$$A_n=\frac{1}{2^n n! \sqrt{\pi}} \int_{-\infty}^{+\infty} e^{-x^2}f(x) H_n(x)dx$$

And:

$$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}\left(e^{-x^2}\right)$$

It's getting late over here, I will try to check it tomorrow. I think you made a calculation error somewhere. However because the function f(x)=sin(x) is not a polynome, you have to calculate the general integral for n, otherwise it would have been a lot easier. We'll see after a nights sleep.

3. Feb 24, 2008

### alfredska

The $$A_n$$ you've listed is correct for expanding a function in a series of Hermites alone. Since I have a Gaussian tacked on my Hermites, my case is slightly different.

One place I've found out that I'm going wrong, is that the functions which I'm summing are not mutually orthonormal, they're not even orthogonal. Can this be resolved by including a weight function?

Last edited: Feb 24, 2008
4. Feb 24, 2008

### siddharth

I'm guessing $$w(x) = e^{x^2}$$ might work? Since it seems to reduce it to the orthogonality condition of the regular Hermite polynomials

5. Feb 24, 2008

### alfredska

You're right in that it reduces it down to the orthogonality case, but then s(x) would need the weight function thrown in as well, and I'm back to the harmonic oscillator.

6. Feb 25, 2008

### coomast

alfredska, I didn't have time yet to look into it. I will try to do it this week, however I'm not giving any guarantee for a solution. Anyway, I'll come back to it.

7. Feb 25, 2008

### alfredska

Perhaps Solved

Ok, I believe I've figured out how to approach this problem. This is likely just another way of implementing weights, and siddharth was probably correct, but I still had to feel my way through it.

My approach:

Take
$$\exp\left(-\frac{x^2}{2}\right)$$
out of the summation and meld it into $$s\left(x\right)$$, defining a new function
$$g\left(x\right)=s\left(x\right)\exp\left(\frac{x^2}{2}\right)$$.

Now the summation is clearly identifiable as the harmonic oscillator, and
$$\alpha_n=\frac{\int_{-\infty}^{\infty}g\left(x\right)H_n\left(x\right)\exp\left(-\frac{x^2}{2}\right)dx}{\sqrt{2^nn!\sqrt{\pi}}}$$

After I find my coefficients, I only have to revert g(x) back to s(x)

8. Feb 27, 2008

### coomast

This seems to be the right way to solve it. I will try to evaluate the integral for determining the coefficients. However what is the exact function, a sin or a Gauss function? There seems to be something changed if I'm not mistaken.

9. Feb 28, 2008

### alfredska

I did change the function

You're right coomast, I did change the function which is being expanded. The method should apply to either case though. For the sake of anyone who may stumble upon this thread, here are the results:

$$s\left(x\right)=\sum_n \alpha_n \frac{H_n\left(x\right)}{\sqrt{2^nn!\sqrt{\pi}}} \exp\left(-x^2\right)$$

$$\frac{s\left(x\right)}{\exp\left(-\frac{x^2}{2}\right)}=\sum_n \alpha_n \frac{H_n\left(x\right)}{\sqrt{2^nn!\sqrt{\pi}}} \exp\left(-\frac{x^2}{2}\right)$$

$$\alpha_n=\int_{-\infty}^{\infty} \frac{H_n\left(x\right)}{\sqrt{2^nn!\sqrt{\pi}}} \exp\left(-\frac{x^2}{2}\right) \frac{s\left(x\right)}{\exp\left(-\frac{x^2}{2}\right)} dx$$

$$\alpha_n=\int_{-\infty}^{\infty} \frac{H_n\left(x\right)}{\sqrt{2^nn!\sqrt{\pi}}} s\left(x\right) dx$$

Thanks for you interest.
~Matt

Last edited: Feb 28, 2008
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