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Taylor Series Expansion to Compute Derivatives

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the Taylor series expansion of

    f(x) = (x-1)/(1+(x-1)^2)

    about x=1 and use this to compute f(9)(1) and f(10)(1)

    2. Relevant equations

    The sum from n=0 to infinity of f(k)(c)/(k!) (x-c)k

    3. The attempt at a solution

    I'm not sure how to approach this problem. Using the expansion formula is clearly incorrect as the derivative keeps on expanding. Any help would be greatly appreciated.
     
  2. jcsd
  3. Dec 3, 2011 #2

    HallsofIvy

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    First, simplify by letting u= x- 1. Now the function is [itex]u/(1- u^2)[/itex].

    The sum of the geometric series [itex]\sum_{n=0}^\infty ar^n[/itex] is a/(1- r) so let a= u and [itex]u= r^2[/itex]
     
  4. Dec 7, 2011 #3
    I'm a little confused. I believe the bottom would become 1+u2.

    I tried using the power series for 1/(1-x) (sum from n = 0 to infinity of x^n).

    This left me with the sum from n = 0 to infinity of (x-1)^(2n+1).

    I then wrote out the Taylor Polynomial until the 9th power because I wanted to evaluate f(9)(1).

    When I plug in 1, all terms go to zero. I don't think this is correct.

    Any thoughts?
     
  5. Dec 7, 2011 #4
    Please find my attempt at the solution attached
     

    Attached Files:

  6. Dec 7, 2011 #5

    vela

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    You're fine up to
    [tex]\frac{u}{1+u^2} = u \sum_{n=0}^\infty (-u^2)^n[/tex]You want to pull the sign out first before pulling the factor of u from out front into the series.
    [tex]\frac{u}{1+u^2} = u \sum_{n=0}^\infty (-1)^n(u^2)^n = \sum_{n=0}^\infty (-1)^n u^{2n+1} = \sum_{n=0}^\infty (-1)^n (x-1)^{2n+1}[/tex]
    Now compare this series to the Taylor expansion about x=1:
    [tex]f(x) = \sum_{m=0}^\infty \frac{f^{(m)}(1)}{m!} (x-1)^m[/tex]You should be able to read off what the 9th and 10th derivatives are simply by looking at the series you obtained above.
     
  7. Dec 7, 2011 #6
    I'm sorry, could you please elaborate further on how to find the derivatives?

    I have attached my attempt but I don't think it's correct because I'm getting 0 for both the 9th and 10th derivatives.
     

    Attached Files:

  8. Dec 7, 2011 #7

    vela

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    Like I said, compare the series you got with the generic series for the Taylor expansion.
     
  9. Dec 7, 2011 #8
    I think I am missing something.

    Do I actually have to compute the 9th derivative of the original function?

    Another attempt is attached.
     

    Attached Files:

  10. Dec 7, 2011 #9
    Can someone please help me with this? It is very important.
     
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