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Infinite Series - Finding the 10th derivative

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Let T(x) = [tex]\sum^{\infty}_{k=0}[/tex] [tex]\frac{1}{2^k}[/tex] [tex]\frac{(x-3)}{k!}[/tex]k be the Taylor series for a function f. What is the value of f10(3), the tenth derivative of f at x = 3?

    3. The attempt at a solution

    I have a very small idea of actually starting this problem. Can I just derive the infinite series by taking derivatives of [tex]\frac{1}{2^k}[/tex] [tex]\frac{(x-3)}{k!}[/tex]k 10 times?

    Basically, what is the first step in this problem?
  2. jcsd
  3. Apr 11, 2009 #2


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    I'm sure you mean (1/2^k)*(x-3)^k/k!. Yes, think about taking the derivative 10 times and then putting x=3. Only one term will survive. What is it? Practice by thinking about x^k. If you take 10 derivatives of that and then put x=0 you will almost always get 0 unless k has a particular value. What is it?
  4. Apr 11, 2009 #3
    Man, Dick, you're very helpful!

    Thanks I got (1/2)^10 or 9.765 * 10^-4.

    You're right about that one term. When k = 10, then its 0^0, or 1. Thanks for the tips! Much appreciated!
  5. Apr 11, 2009 #4


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    I just want to emphasize the significance of the problem.

    It's too easy to let your thinking run in only one direction -- you compute derivatives so that you may find the Taylor series to your function.

    But the other direction is also important -- sometimes we want to find the derivative, and it turns out that we can compute the Taylor series easily. So, we compute the Taylor series so that we may find a derivative!

    For example, suppose that we have

    [tex]f(x) = \frac{1}{1-x}[/tex]

    and we want to know the eighteenth derivative:

    [tex]f^{(18)}(0) = \, ?[/tex]
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