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  1. #1
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    taylor

    when you say that a function which is m times continuously differentiable is analytic, does it mean that in the taylor formula, the remainder tends to 0 as m tends to infinity?
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  2. #2
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    A function is analytic if it has a convergent taylor series in a particular domain.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    when you say that a function which is m times continuously differentiable is analytic, does it mean that in the taylor formula, the remainder tends to 0 as m tends to infinity?
    If the function is C^{\infty} this is what it means.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Prove It View Post
    A function is analytic if it has a convergent taylor series in a particular domain.
    A convergent Taylor series which actually converges to the function! Take for example the classic \displaystyle f(x)=\begin{cases}e^{\frac{-1}{x^2}} & \mbox{if}\quad x>0\\ 0 & \mbox{if}\quad x\leqslant 0\end{cases}. Then, f is infinitely differentiable and its Maclaurin series congerges, but it doesn't converge to f!
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    when you say that a function which is m times continuously differentiable is analytic, does it mean that in the taylor formula, the remainder tends to 0 as m tends to infinity?
    An important detail is that if a complex function is differentiable in z=z_{0}, then it is analytic in z=z_{0} and it is representable as Taylor function 'somewhere around' z=z_{0}... that isn't necessarly true for an 'ordinary' real function...

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    \chi \sigma
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