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The derivative of $\sqrt{x}$ doesn't have a defined value at x = 0. How then do I find its maclaurin series expansion? Or can it only be approximated with a Taylor series at some value x != 0?

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See Puiseux series. –  Bill Dubuque Mar 26 '12 at 23:32
    
@dragon This might be a crazy thought but take the series around $\epsilon>0$ and then let it tend to $0$, maybe you get something interesting, maybe you don't. –  Pedro Tamaroff Mar 27 '12 at 0:08
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up vote 5 down vote accepted

It doesn't have a MacLaurin series. It can be expressed as a Taylor series around values of $x\gt0$.

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If it cannot be approximated at x=0, this means that $\sqrt{x}$ doesn't possess a tangent at x=0. But if you look at its graph, it sure does look like there could be a tangent at x=0. –  hollow7 Mar 26 '12 at 23:27
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@dragon: The tangent line is the $y$-axis itself. The correspondence between "good" local linear approximations and differentiability breaks down for vertical asymptotes. (The derivative of $\sqrt{x}$, which is $x^{-1/2}$, is undefined at $x=0$ and blows up from the right side.) –  anon Mar 26 '12 at 23:41
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Since $f'(0)$ does not exist, there is no Maclaurin series at $x=0$, but it does have a Taylor series about any $x>0$. One can use the general binomial theorem to derive the series. For example, $$ \begin{align} \sqrt{1+x} &=1+\frac12x-\frac18x^2+\dots+\binom{1/2}{k}x^k+\dots\\ &=1+\sum_{n=1}^\infty\left(-\frac14\right)^{n-1}\binom{2n-2}{n-1}\frac{x^n}{2n} \end{align} $$

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