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Math Help - Wilsons Theorem?

  1. #16
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    If you have a power series and the terms a_n are sufficiently bounded, say, |a_n| \leq n! then the series converges absolutely on (-1,1) and so it is okay to differenciate term-by-term.
    Why is this?
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  2. #17
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    Quote Originally Posted by chiph588@ View Post
    Why is this?
    If \sum_{n=0}^{\infty}a_nx^n converges on (-\varepsilon,\varepsilon) for some \varepsilon >0 then this function is \mathcal{C}^{\infty}
    on this open interval i.e. infinitely differenciable and the derivative is given by term-by-term differenciation. This is just a result from analysis.


    If |a_n| are too huge then \sum_{n=0}^{\infty}a_nx^n will only converge at x=0 but if they are bounded big enough then there will be an an interval centered at 0 on which it converges.
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