Expansion of Cos(x) in Hermite polynomials

[Dansuer]

[/itex]

Homework Statement

Find the first three coeficents c_n of the expansion of Cos(x) in Hermite Polynomials.
The first three Hermite Polinomials are:
$H_0(x) = 1$
$H_1(x) = 2x$
$H_0(x) = 4x^2-2$

The Attempt at a Solution

I know how to solve a similar problem where the function is a polynomial of finite degree, say x^3. Using the fact that H_n(x) is a polinomial of degree n, i set all the coeficent after c_3 equal to zero and equate the terms with equal degree. I find a system of linear equations and i solve it.
In this case however the taylor series of Cos(x) is a polinomial of infinite degree. I can't apply this method.

I then try to use the orthogonality of the Hermite polynomials

$\int^{∞}_{-∞}e^{-x^2}H_n(x)H_m(x)dx = \sqrt{\pi}2^nn!\delta_{nm}$

From the orthogonality i find the coeficents to be

$c_0 = \frac{1}{\sqrt{\pi}2^nn!}\int^{∞}_{-∞}e^{-x^2} Cos(x) dx$

$c_1 = \frac{1}{\sqrt{\pi}2^nn!}\int^{∞}_{-∞}e^{-x^2} 2xCos(x) dx$

$c_2 = \frac{1}{\sqrt{\pi}2^nn!}\int^{∞}_{-∞}e^{-x^2} (4x^2-2)Cos(x) dx$

Which are three hard integrals i haven't been able to solve. I can't use computer methods as I'm suppose to solve this in an exam with pen and paper. I'm stuck.

Thanks to anyone who takes a look at this.:tongue2:

 

[Dansuer]

I don't think you can do that as the higher hermite polinomials are not zero.

I've figured the first and second integrals and they are

$c_0 = \frac{1}{\sqrt{e}^{4}}$

$c_1 =0$

c_0 differs from your method.

i still need to find the last integral though.

 

[Saitama]

I don't think you can do that as the higher hermite polinomials are not zero.

I've figured the first and second integrals and they are

$c_0 = \frac{1}{\sqrt{e}^{4}}$

$c_1 =0$

c_0 differs from your method.

i still need to find the last integral though.

For $c_2$, you have two integrals.

$$4\int_{-\infty}^{\infty} x^2e^{-x^2}\cos(x)\,dx-2\int_{-\infty}^{\infty} e^{-x^2}\cos(x)\,dx$$

Since you already evaluated $c_0$, you know the value of second integral. To evaluate the first integral, consider the following definite integral:
$$I(a)=\int_{-\infty}^{\infty} e^{-a^2x^2}\cos(x)\,dx=\frac{\sqrt{\pi}}{a}e^{-1/(4a^2)}\,\,\,\,\,\,\,\,(*)$$
You should be able to prove the above result in the same way you evaluated $c_0$.
Differentiate both the sides of $(*)$ with respect to $a$ to obtain:
$$\int_{-\infty}^{\infty}-2ax^2e^{-a^2x^2}\cos(x)\,dx=\sqrt{\pi}\frac{e^{-1/(4a^2)}(1-2a^2)}{2a^4} $$
Substitute $a=1$ and you should be able to obtain the answer after some rearrangement of the above expression.

 

[Dansuer]

Thanks a lot!

 

[Saitama]

Thanks a lot!

Glad to help!