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Expansion of Cos(x) in Hermite polynomials

  1. Jun 12, 2014 #1
    [/itex]1. The problem statement, all variables and given/known data
    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]


    3. 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:
     
    Last edited: Jun 12, 2014
  2. jcsd
  3. Jun 12, 2014 #2
    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.
     
  4. Jun 12, 2014 #3
    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. Jun 12, 2014 #4
    Thanks a lot!:biggrin:
     
  6. Jun 12, 2014 #5
    Glad to help! :smile:
     
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