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Taylor Series

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Write the Taylor series of the function f(x) = ([tex]\pi[/tex] -x)^-2 around a = 0



    2. Relevant equations

    ([tex]\pi[/tex] - x)^-2 = f(a) + f'(a)(x-a) + [f''(a)(x-a)^2]/(2!) +......+ [f^n(a)(x-a)^n]/(n!)


    3. The attempt at a solution

    This is what i have and i am not sure i am showing it correctly or compleatly.

    f(x) = ([tex]\pi[/tex] -x)^-2
    f'(x) = 2([tex]\pi[/tex] - x)^-3
    f''(x) = 6([tex]\pi[/tex] - x)^-4
    f'''(x) = 24([tex]\pi[/tex] - x)^-5

    ([tex]\pi[/tex] - x)^-2 = f(a) + [f'(a)(x-a)^2]/(2!) +......+ [f^n(a)(x-a)^n]/(n!)

    = ([tex]\pi[/tex]-0)^-2 + [(2([tex]\pi[/tex]-0)^-3)(x-0)^2]/(2!) + [(6([tex]\pi[/tex]-0)^-4)(x-0)^3]/(3!) + ......

    = [tex]\pi[/tex]^-2 + 2([tex]\pi[/tex]^-3)x +[6([tex]\pi[/tex]^-4)(x^2)]/(2!) + [24([tex]\pi[/tex]^-5)(x^3)]/(3!) + ........
     
    Last edited: Mar 23, 2010
  2. jcsd
  3. Mar 23, 2010 #2
    You seem to have written the series wrong: where is the f(a) term? And why isn't the f'(a) term multiplied by (x-a)? Your derivatives are correct, now you just need to substitute into the (correct) series that you wrote in the "relevant equations" section...
     
  4. Mar 23, 2010 #3
    I looked over my work again and found where i messed up, thanks.
    the final answer i get is:

    ([tex]\pi[/tex]-x)^-2 = [tex]\pi[/tex]^-2 + 2([tex]\pi[/tex]^-3)(x) + 3([tex]\pi[/tex]^-4)(x^2) +....
     
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