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26th Derivative of a Function

  1. Dec 16, 2006 #1
    The problem statement, all variables and given/known data

    Given that f(x) = sin(x) for x =/= 0 and f(x) = 1^x for x=0, find the 26th derivative of f at 0. Hint: can you find a power series for f(x)?

    The attempt at a solution

    I have no idea how to solve this problem. Since 1^x is always 1, the first derivative at 0 is 0, so ALL derivatives must be 0, right? I'm confused as to how a power series even comes into play in this problem.
     
  2. jcsd
  3. Dec 16, 2006 #2

    StatusX

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    That's a very strange defintion. Note that f(0) is just a number, so all they had to say was f(x)=1 for x=0, the 1^x bit is superfluous. But moreover, the function is not continuous at x=0, so doesn't have any derivatives, let alone 26. Which leads me to ask, are you sure you copied the question correctly?
     
  4. Dec 16, 2006 #3
    I just noticed that somebody erased a line in my book! It should have been sin(x)/x for x=/=0 and 1 for x=0. That makes a lot more sense.
     
  5. Dec 16, 2006 #4

    arildno

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    Dearly Missed

    I advise you to use the hint given!
     
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