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A question about Taylor series expansion

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the Taylor series expansion for f(x)=x*e^(-x^2) about x = -1


    2. Relevant equations



    3. The attempt at a solution
    I have tried replacing x with (x-1) and f(x-1) = (x-1)*e^(-(x-1)^2).
    Consider the power series for e^(-(x-1)^2) about x = 0, f(x-1) = (x-1)*(1/e+(2x-x^2)/e+(2x-x^2)^2/2!e+(2x-x^2)^3/3!e+...).
    Let (x-1)*e^(-(x-1)^2) = A + Bx + Cx^2 + Dx^3 + Ex^4 + ...
    So (x-1)*(1/e+(2x-x^2)/e+(2x-x^2)^2/2!e+(2x-x^2)^3/3!e+...) =A + Bx + Cx^2 + Dx^3 + Ex^4 + ...
    Comparing x^n coefficients and I can get A, B,...
    I don't know whether my method is correct and even it is correct ,it is too complicated.
    I want an easier way to solve the problem.
    Can anybody help me?
    Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 12, 2011 #2
    Sounds very complicated to me. I would suggest the following:

    a) multiply you function by the fraction ([itex]\frac{-2}{-2}[/itex]). Now take a look at the numerator. Remind you of anything?

    b) get familar with the Hermite Polynomials, more specifically, their generating function:
    [tex]
    H_r(x)=(-1)^r\,e^{x^2}\, \frac{d^r}{dx^r}(e^{-x^2})
    [/tex]
     
  4. Nov 12, 2011 #3

    D H

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    Staff Emeritus
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    What are the Relevant equations?

    In other words, how would you, in general, go about determining the Taylor expansion for some function f(x) about some point x=x0?
     
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