4

I was searching for information about the power series representation for $\sqrt x$ valid for complex numbers so I found and 'hacked' this equation here: http://www.maeckes.nl/Reeksen/kwadraatwortel%20GB.html to make it work with complex numbers. It is now published on the above site on page 2. $$\sqrt{x} = x\sum_{n=0}^{\infty}\frac{(-1)^{n-1}(2n)!}{(n!)^2(2n-1)}\left ( \frac{1}{4x}-\frac{1}{4} \right )^n,\ \ \ \ \Re(x)>1$$

Now I am looking for the corresponding equation for $\Re(x) < 0$ (or $\Re(x) < -1$), with two prerequisites:
1: No fractions in the exponents.
2: It must hold for complex $x$.

This is the closest I have been able to get so far:
$$\frac{1}{\sqrt{x+2}\sqrt{x-2}}=\sum_{n=1}^{\infty}\frac{(2 n-2)!}{((n-1)!)^2}(x+2)^{-n}$$ Any help is appreciated.

Bob the Builder
  • 55

2 Answers2

5

For $|-1/z-1|< 1$ ie. for $\Re(z) < -1/2$

$$i\sum_{n=0}^\infty {-1/2\choose n} (-1/z-1)^n=i(1-1/z-1)^{-1/2}= z^{1/2}$$ The RHS is the branch analytic for $\Re(z) <-1/2$ and such that $(-1)^{1/2}=i$.

reuns
  • 77,999
2

The exact series expansion of $\sqrt{x}$ for $x\geq1$ is :

$x-\sum_{k=1}^\infty \frac{(2k)!(x-1)^k}{(k!)^22^{2k}(2k-1)x^{k-1}}$

And the exact series expansion of $\sqrt{x}$ for $0\leq x\leq1$ is :

$1-\sum_{k=1}^\infty \frac{(2k)!(1-x)^k}{(k!)^22^{2k}(2k-1)}$

The explanations is included in my researches at:

https://www.researchgate.net/profile/Mones_Jaafar I hope you find it useful.

mones
  • 21
  • 1
    As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Jan 06 '22 at 07:20
  • it was addressed to Claude Leibovic comment and "question", the first one is not explained but my researches were explained – mones Jan 06 '22 at 14:17
  • Thank you for your contribution! As I tend to collect these formulas, I am always fascinated to find new ones. – Bob the Builder Jan 07 '22 at 18:59