Expansion of Cos(x) in Hermite polynomials

  • Thread starter Dansuer
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  • #1
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[/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:
[itex]H_0(x) = 1[/itex]
[itex]H_1(x) = 2x[/itex]
[itex]H_0(x) = 4x^2-2[/itex]


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

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

From the orthogonality i find the coeficents to be

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

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

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

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:
 
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Answers and Replies

  • #2
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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

[itex] c_0 = \frac{1}{\sqrt{e}^{4}} [/itex]

[itex] c_1 =0 [/itex]

c_0 differs from your method.

i still need to find the last integral though.
 
  • #3
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92
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

[itex] c_0 = \frac{1}{\sqrt{e}^{4}} [/itex]

[itex] c_1 =0 [/itex]

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.
 
  • #5
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