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Thread: 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 $\displaystyle a_n$ are sufficiently bounded, say, $\displaystyle |a_n| \leq n!$ then the series converges absolutely on $\displaystyle (-1,1)$ 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 $\displaystyle \sum_{n=0}^{\infty}a_nx^n$ converges on $\displaystyle (-\varepsilon,\varepsilon)$ for some $\displaystyle \varepsilon >0$ then this function is $\displaystyle \mathcal{C}^{\infty}$
    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 $\displaystyle |a_n|$ are too huge then $\displaystyle \sum_{n=0}^{\infty}a_nx^n$ will only converge at $\displaystyle x=0$ but if they are bounded big enough then there will be an an interval centered at $\displaystyle 0$ on which it converges.
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