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Taylor series of an integral function

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    $$ \displaystyle f\left( x \right)=\int\limits_{0}^{x}{\frac{\sin t}{t}dt} $$

    Calculate the Maclaurin series of third order.

    2. Relevant equations



    3. The attempt at a solution

    What I do is:

    $$ \displaystyle f'\left( x \right)=\frac{\sin x}{x} $$

    $$ \displaystyle f''\left( x \right)=\frac{x\cos x-\sin x}{{{x}^{2}}}$$

    and so on...

    but if I try to evaluate theese two on x=0 it doesn't make sense. I've seen I have to transform the integrand to the taylor polynomial but why is my "method" wrong?

    Thanks
     
    Last edited: Sep 10, 2012
  2. jcsd
  3. Sep 10, 2012 #2
    f' and f'' both have specific limiting values at x = 0. For example, in the limit of x approaches zero, f' = 1.
     
  4. Sep 10, 2012 #3

    Dick

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    It's a lot easier if you expand sin(t) in a power series. Then divide by t. You are simply trying to do it the hard way.
     
  5. Sep 10, 2012 #4

    Zondrina

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    Homework Helper

    The beauty of a Taylor series is that it allows you to express functions which are complicated to integrate in terms of an infinite sum of polynomials which are easy to integrate.

    The Maclaurin series is simply the Taylor series centered at 0.

    Recall that the Maclaurin series for [itex]sinx = \sum_{n=0}^{∞}\frac{x^(2n+1)}{(2n+1)!}(-1)^n[/itex]

    Using this fact you should be able to solve your integral.
     
  6. Sep 11, 2012 #5
    I agree. That's how I would have done the problem if it had been posed to me. But I wanted to show how it could also have been done using the more difficult way that the OP had pursued.
     
  7. Sep 11, 2012 #6
    Thank you guys, now I understand ;)
     
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