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The derivative of a Taylor series?

  1. Jun 27, 2012 #1
    I took my first calculus class over the last two semesters, and my teacher and I privately worked on some harder material together. Toward the end of the school year he gave me a question that I never answered and never found an answer for. It asked me to find the derivative of a Taylor series. I don't remember the specific problem, but if someone could shed some light, that'd be lovely!
     
  2. jcsd
  3. Jun 27, 2012 #2

    mathwonk

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    you are kidding right? i.e. what do you think?
     
  4. Jun 27, 2012 #3
    I'm not kidding at all. At the time I received the problem I'd never heard of a Taylor series, and I haven't seen a problem with one since. It's something I've never had the opportunity to learn at school thus far.
     
  5. Jun 27, 2012 #4
    I'm not sure if this is exactly what you're looking for, but if f(x)=Ʃanxn in some interval, then f'(x)=Ʃnanxn-1 in the interior of that interval; that is, you can differentiate the series termwise. This is because of uniform convergence, which can be shown with the Weierstrass M-test.
     
  6. Jun 27, 2012 #5
    Really, it's a very plausible generalization of what you already know for finite polynomials.
     
  7. Jun 27, 2012 #6
    I'm really just caught up in looking at the sigma notation.. It throws me out of whack. I'm assuming I'm making a much bigger deal out of this than I should be.

    This may be a better question:

    If f(x) = f(a) + [f'(a)/1!](x-a) + [f"(a)/2!](x-a)^2 + ...
    Then f'(x) = ?
     
    Last edited: Jun 27, 2012
  8. Jun 27, 2012 #7

    pwsnafu

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    Exactly what you would get if you differentiated each term individually.
    Provided it converges.
     
  9. Jun 27, 2012 #8
    badamann, just use the power rule.
     
  10. Jun 28, 2012 #9
    Thank you. I'm not sure why I made it out to be quite so difficult... :(
     
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