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Why fractional power does not have power series expansions? For example, $f(x)=x^{1/2}$, why the behavior at $0$ disallows a power-series expansion? For what reason?

Thanks in advance.

Ian
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  • There is no problem in expressing $x^{1/2}$ as for example $\sum_0^\infty a_n(x-\pi)^n$. (As is often the case, this expansion will be valid only when $|x-\pi|$ is not too big, and in particular will not be valid at $x=\pi$.) – André Nicolas Jun 25 '13 at 23:02

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If you blindly plug $f(x)=x^{1/2}$ into the Taylor series formula, you get $f(0)+xf'(0)+\dots=0+x(\frac 12 x^{(-1/2)}\mid_{x=0})+\dots$ and plugging $0$ in to the derivative as shown gives a divide by zero. This reflects the vertical tangent at $x=0$.

Ross Millikan
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  • but can it be expressed as other form of power-series, except the Taylor series? – Ian Jun 25 '13 at 22:12
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    The original function is a (single-term) power series. You can also do a Taylor series around any point greater than zero. The fundamental problem is that polynomials can't do vertical tangents. – Ross Millikan Jun 25 '13 at 22:30