Taylor series expansion rearrangement

Question

From the Taylor series expansion of $\sqrt{x}$ provided in Algorithms for approximating $\sqrt{2}$ I got Taylor series expansion of $x^{s}$, $0<s<1$ 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 $x^{i}$.

Answer 1

Unfortunately, for $0<s<1$, 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.