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I've recently started learning about Taylor/Maclauren series and I'm finding it a bit hard to wrap my head around a few things.

So, if $f(x)$ is not infinitely differentiable and we construct a polynomial $p(x)$ such that $p(a) = f(a)$, $f'(a) = p'(a)$, $f''(a) = p''(a)$ etc, for all possible derivatives of $f(x)$, can we say that the functions $f(x)$ and $p(x)$ are equivalent?

Likewise, if $f(x)$ is infinitely differentiable, does the taylor series $p(x)$ become equivalent to $f(x)$ when given an infinite number of terms?

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    You'll have problems with functions like $\exp(-1/x^2)$...2011-11-17
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    For the last question, see [A deceiving Taylor series](http://math.stackexchange.com/questions/79241/a-deceiving-taylor-series).2011-11-17
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    What does it mean to you to say that two functions/expressions are "equivalent"?2011-11-17
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    That `f(x)` and `p(x)` will give the same output for any value of `x`.2011-11-17

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