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Expanding in a Taylor series

  • Thread starter polygamma
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  • #1
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Homework Statement


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]

Homework Equations




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.
 
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Answers and Replies

  • #2
34,053
9,914
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.
 
  • #3
Stephen Tashi
Science Advisor
7,017
1,237
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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