I am interested in a series expansion of $\sqrt{x}$ in powers of $x$ for real $x$ around some arbitrary point $0<a<1$ (with the respective radius of convergence for $x$). I start with the well known Taylor series for $$ \sqrt{1+x'} = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n}x'^n,$$ for $x'\in(-1,1)$. Then I do the transformation $x'\rightarrow x - a $, for $a\in(0,1)$ to arrive at a double sum $$ \sum_{0\le i \le n}^{\infty} \binom{\frac{1}{2}}{n}\binom{n}{i}x^i(-a)^{n-i}.$$ Now I like to rearrange the thing for powers of $x$, after some gymnastics I arrive at $$\sum_{m=0}^{\infty} x^m \sum_{0\le j \le m} \binom{\frac{1}{2}}{m+j}\binom{m+j}{j}(-a)^j.$$ At this point I feel that the second sum could be simplified further, but I could not succeed in doing so, yet. Can this expression be further simplified?

$\mathbf{Note:}$ An interesting special case would be the algebraic expression for the limit $a\rightarrow 1^-$, as it would represent a series expansion of $\sqrt{x}$ around $x=0$ (which is not possible because $\sqrt{x}$ is not differentiable at 0).


For the case of $a=1$:

Just as Tom Cooney, it is not very clear to me.

Anyway, using a CAS, what I got is $$a_m=\sum_{j=0}^m(-1)^j \binom{\frac{1}{2}}{m+j}\binom{m+j}{j}=-\frac{2 (-1)^m (m+1) \binom{\frac{1}{2}}{2 m+1} \binom{2 m+1}{m+1}}{2 m-1}$$ The problem is that $$\Phi(x)=\sum_{m=0}^\infty a_mx^m=\, _3F_2\left(-\frac{1}{2},\frac{1}{4},\frac{3}{4};\frac{1}{2},2;-4 x\right)+\frac{3}{8} x \, _3F_2\left(\frac{1}{2},\frac{5}{4},\frac{7}{4};\frac{3}{2},3;-4 x\right)$$ where appear the generalized hypergeometric function.


There is something interesting to notice : noting $f_k$ the value of the last given expression for $x_k=10^{2k}-1$, the following values are obtained $$f_1=10.1250000$$ $$f_2=100.0506048$$ $$f_3=1000.017094$$ $$f_4=10000.00551$$ $$f_5=100000.0018$$ which are closer and closer to $\sqrt{x_k+1}$.

For infinite values of $x$, the asymptotics of $\Phi(x)$ is given by $$\Phi(x)=\sqrt{x}-\frac{3 \,\Gamma \left(-\frac{3}{4}\right)}{16\,\sqrt{\pi }\ \Gamma \left(\frac{7}{4}\right)}\frac{1}{\sqrt[4]{x}}+O\left(\frac{1}{x^{3/4}}\right)$$ If, as discussed on chat, you really want to see $\sqrt{x+1}$, the asymptotic would be $$\Phi(x)=\sqrt{x+1}+\frac{\Gamma \left(\frac{5}{4}\right)}{\sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}\frac{1}{\sqrt[4]{x}}+O\left(\frac{1}{x^{1/2}}\right)$$

  • $\begingroup$ Sorry for the typo ! $\endgroup$ Apr 2 '16 at 15:06
  • $\begingroup$ The problem is that the function "looks" like $\sqrt{1+x}$ but it is not. $\endgroup$ Apr 2 '16 at 15:08
  • $\begingroup$ I did not mention the motivation for that, since I thought it might be too distracting, but I wanted to use an algebraic equivalent of $\sqrt$ for comming up with the "square root" of an operator (namely the generator operator in second quantization). So maybe its not too bad. Thank you for your help with the CAS! $\endgroup$ Apr 2 '16 at 15:15
  • $\begingroup$ @Franky_GTH. There are "funny" things ! Look at my edit. $\endgroup$ Apr 3 '16 at 4:26
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Apr 3 '16 at 6:29

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