# Taylor series expansion rearrangement

From the Taylor series expansion of $$\sqrt{x}$$ provided in Algorithms for approximating $\sqrt{2}$ I got Taylor series expansion of $$x^{s}$$, $$0 and then collecting the coefficients of $$x^{i}$$ from the binomial expansions I got the following $$\begin{eqnarray*} x^{\displaystyle s} &=& \sum_{n=0}^{\infty} \frac{(x - 1)^{\displaystyle n}}{n!} \left[ \frac{d^{\displaystyle n}}{dt^{\displaystyle n}} x^{\displaystyle s} \right]_{\displaystyle x=1} \\ &=& 1 + \sum_{n=1}^{\infty}(s)(s-1) ... (s-(n-1)) \frac{1}{n!} (x-1)^{\displaystyle n} \\ &=& \left( 1 + \sum_{n=1}^{\infty} \frac{(-1)^{\displaystyle n}}{n!} (s)(s-1) ... (s-(n-1)) \right) + \\ &&\sum_{i=1}^{\infty} (x)^{\displaystyle i} (\frac{(s)(s-1) ... (s-(i-1))}{i!}) \left( 1+ \sum_{k=1}^{\infty} \frac{(-1)^{\displaystyle k}(s-i) ... (s-(i+k-1))}{k!} \right) \\ \end{eqnarray*}$$ Series in the first big brackets seems to tend to zero but not able to determine for higher powers of $$x$$. Am I on right track. I need help to find solution to this problem. If I am wrong then what is it and what is the actual solution. Is it not possible to get the expansion in terms of $$x^{i}$$.

Unfortunately, for $$0, this is impossible.
The derivative of $$x^s$$ is $$sx^{s-1}$$. Since $$s<1$$, we have that $$s-1<0$$, so $$x$$ is raised to a negative power. We also know that $$0$$ to a negative power is the same as $$1/0$$ to a positive power, which is undefined.
So, the function's derivative is undefined at $$x=0$$. Since you are trying to find the Taylor expansion about $$0$$, the $$x^1$$ term must equal the first derivative at $$0$$, which is undefined. But we cannot have an undefined term in our sum, so there cannot be a solution.