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A question about Taylor series expansion

  • Thread starter hefnrh
  • Start date
  • #1
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Homework Statement


Find the Taylor series expansion for f(x)=x*e^(-x^2) about x = -1


Homework Equations





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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #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]
 
  • #3
D H
Staff Emeritus
Science Advisor
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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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