# Expansion for $x^{\frac{1}{k}}$

Can we find a expansion for $x^{\frac{1}{k}} \quad (k\in \mathbb{Z})$ using alternative power series?

I can find things like

$(x+1)^{\frac{1}{k}}=1+\frac{x}{k}+\frac{\left(\frac{1}{k}-1\right) x^2}{2 k}+\frac{(k-1) (2 k-1) x^3}{6 k^3}+\frac{\left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^4}{24 k}+\frac{\left(\frac{1}{k}-4\right) \left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^5}{120 k}+\frac{\left(\frac{1}{k}-5\right) \left(\frac{1}{k}-4\right) \left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^6}{720 k}+\frac{\left(\frac{1}{k}-6\right) \left(\frac{1}{k}-5\right) \left(\frac{1}{k}-4\right) \left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^7}{5040 k}+\frac{\left(\frac{1}{k}-7\right) \left(\frac{1}{k}-6\right) \left(\frac{1}{k}-5\right) \left(\frac{1}{k}-4\right) \left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^8}{40320 k}+\frac{\left(\frac{1}{k}-8\right) \left(\frac{1}{k}-7\right) \left(\frac{1}{k}-6\right) \left(\frac{1}{k}-5\right) \left(\frac{1}{k}-4\right) \left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^9}{362880 k}+\frac{\left(\frac{1}{k}-9\right) \left(\frac{1}{k}-8\right) \left(\frac{1}{k}-7\right) \left(\frac{1}{k}-6\right) \left(\frac{1}{k}-5\right) \left(\frac{1}{k}-4\right) \left(\frac{1}{k}-3\right) \left(\frac{1}{k}-2\right) \left(\frac{1}{k}-1\right) x^{10}}{3628800 k}+O\left(x^{11}\right)$

But I would like a expansion of $x^{\frac{1}{k}}$. I know this is will not work to $(x+a)^{\frac{1}{k}}$ when a is equal to 0, but we can find another expansions for $\log(x)$ that looks like power series like these ones.

So, how can I perform this?

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I don't understand the question. What is wrong with the series you get by substituting $u = x+1$ into the above? How is it less desirable than a series in powers of $(x-1)/(x+1)$ like in your link? – Erick Wong Jun 14 '12 at 7:36
GarouDan: Why do you neglect to answer the question asked in the comment above? – Did Jul 14 '12 at 10:01

You can't get this. The function $x\mapsto x^{1/k}$ does not admit a nice extension to the complex plane in a neighborhood of zero.

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Actually, you can only achieve this if and only if 1/k is an integer.

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Hun. We can assume k is integer. Will help me. – GarouDan Jun 14 '12 at 2:34