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I am thinking either it does not have a Taylor series since there does not seem a way to define the $n^{\rm th}$ derivative of $x^2$ or its Taylor series is just $x^2$ and radius of covergence is infinity. Don't know which argument is correct? Any help is appreciated. Thanks!

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    The $n$th derivative of $x^2$ is $0$ for $n \geq 3$.2012-11-26
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    yes of course,but since the taylor series is defined using the n-th derivative, are you suggesting that the Taylor series is 0?!?2012-11-26
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    Are you finding Taylor series at the point $x=0$?2012-11-26
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    A consequence of Antonio's point is that the terms of order greater than $2$ are $0$, not that the whole Taylor series is $0$.2012-11-26

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