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Expanding in a Taylor series

  1. Jan 6, 2015 #1
    1. The problem statement, all variables and given/known data
    If [itex] \int_{0}^{1} f(x) g(x) \ dx [/itex] converges, and assuming [itex]g(x)[/itex] can be expanded in a Taylor series at [itex]x=0[/itex] that converges to [itex]g(x)[/itex] for [itex]|x| < 1[/itex] (and perhaps for [itex] x= -1 [/itex] as well), will it always be true that [itex] \int_{0}^{1} f(x) g(x) \ dx = \int_{0}^{1} f(x) \sum_{n=0}^{\infty} a_{n} x^{n} \ dx [/itex]?

    Will the fact that that the series doesn't converge for [itex]x=1[/itex] ever be an issue?

    A couple of examples are [tex] \int_{0}^{1} \frac{f(x)}{1-x} \ dx = \int_{0}^{1} f(x) \sum_{n=0}^{\infty} x^{n} \ dx [/tex] and [tex] \int_{0}^{1} f(x) \ln(1-x) \ dx = -\int_{0}^{1} f(x) \sum_{n=1}^{\infty} \frac{x^{n}}{n} \ dx. [/tex]

    2. Relevant equations


    3. The attempt at a solution
    I want to say that it will always be true since it's just a single point. But I don't know if that's sufficient justification.
     
    Last edited: Jan 6, 2015
  2. jcsd
  3. Jan 6, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Is the sum really inside the integral? If yes: the integral (by definition) does not care about the function value at 1 at all.
    If it is outside, it gets more interesting.
     
  4. Jan 7, 2015 #3

    Stephen Tashi

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    Science Advisor

    What kind of integration has been defined in this analysis course? What definition of a definite integral is used? Are you studying measure theory?
     
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