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Can anyone tell me how can I find the hermite representation of the function $x^2-c$? And it would be very interesting, if anyone could tell me a good source about the Hermite representation of a general function $f$? Thank you very much! Maybe I should add "my" definiton of the Hermite Polynomials: $H_{0}(x)=1$ and $H_{n}(x)=(-1)^{n}e^{\frac{x^2}{2}}\frac{d^{n}}{d^{n}x}\left(e^{-\frac{-x^2}{2}}\right)$ Unfortunately I can't find any useful sources about Hermite Polynomials.

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    You did compute $H_1$ and $H_2$?2012-11-04
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    $H_{1}(x)=x$ and $H_{2}(x)=x^{2}-1$2012-11-04
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    Right, so now you have $1$, $x$ and $x^2-1$ and you're looking for a linear combination of these to form $x^2 - c$. Start at the highest degree monomial and work your way down.2012-11-04

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The following recurrence relation is equivalent to your definition,

$$ H_{n+1}(x) = 2x H_n(x) - 2n H_{n-1}(x) .$$

Using this relation, $H_{-1}=0$, and $H_0 = 1$ we can compute the first few polynomials:

$$ H_0 = 1,$$ $$ H_1 = 2x,$$ $$ H_2 = 4x^2 - 2.$$

Now suppose we want the following polynomial rewritten in terms of Hermite polynomials,

$$ 5x^2-2x+4 \qquad \text{(1)}.$$

We start with the highest power of $x$ and work our way down. In order to get a $5x^2$ I would need $5/4$ of $H_2$ which is $$ \frac{5}{4} H_2 = 5x^2 - \frac{5}{2} \quad \Rightarrow \quad \frac{5}{4} H_2 + \frac{5}{2} = 5x^2 $$

Substituting this expression into equation (1) gives us,

$$ H_2 + \frac{5}{2} - 2x + 4 $$

$$ H_2 - 2x + \frac{13}{2} \qquad \text{(2)} $$

Now we handle the next lowest power of $x$ which is the first power of $x$. We need to rewrite $2x$ in terms of Hermite polynomials and in this case it turns out to be equal to $H_1$. Substituting $H_1$ into (2) gives us,

$$ H_2 - H_1 + \frac{13}{2} \qquad \text{(3)} $$

I believe that the constant term should be obvious considering $H_0=1$.

$$ H_2 - H_1 + \frac{13}{2}H_0 \qquad \text{(4)} $$

So the pattern is:

  1. If $x^n$ is the highest power of $x$ use $H_n$ to eliminate it.
  2. Repeat step 1. until all the powers of $x$ are exhausted.

Hermite expansions of other functions are usually accomplished using their orthogonality property. In the space of square integrable functions on the real line the following holds,

$$ f(x) = \sum_{n=0}^\infty c_n H_n(x) e^{-x^2/2} $$

$$\Rightarrow c_n = \int_{-\infty}^\infty f(x) H_n(x) e^{-x^2/2} dx$$

These integrals can be computed using a method called Gaussian quadrature.


The following are resources that contain information related to orthogonal polynomials.

This has a whole chapter dedicated to orthogonal polynomials,

"Abromowitz and Stegun, Handbook of Mathematical Functions"

http://apps.nrbook.com/abramowitz_and_stegun/index.html

This has more information about orthogonal polynomials. In particular it discusses their use in performing Gaussian quadrature.

"Numerical Recipes in Fortran/C"

http://apps.nrbook.com/fortran/index.html