Power series twin

[Angelcat]

Hello there, I couldn't find much data out there regarding the power series for the square root of x so I came up with this equation below.
Since this equation is valid for real x > 0, I want to find the corresponding equation for real x less than 0:

[math]\sqrt{x} = \sum_{n=0}^{\infty}\frac{x(-1)^{n+1}(2n)!}{4^{n}(2n-1)(n!)^2}\left(\frac{1}{x}-1\right)^{n},\ \Re (x)>0[/math]
I would be grateful if anyone would help me with this.

 

[Deleted member 4993]

Hello there, I couldn't find much data out there regarding the power series for the square root of x so I came up with this equation below.
Since this equation is valid for real x > 0, I want to find the corresponding equation for real x less than 0:

[math]\sqrt{x} = \sum_{n=0}^{\infty}\frac{x(-1)^{n+1}(2n)!}{4^{n}(2n-1)(n!)^2}\left(\frac{1}{x}-1\right)^{n},\ \Re (x)>0[/math]
I would be grateful if anyone would help me with this.

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem.

[math]\sqrt{x} = \sum_{n=0}^{\infty}\frac{x(-1)^{n+1}(2n)!}{4^{n}(2n-1)(n!)^2}\left(\frac{1}{x}-1\right)^{n},\ \Re (x)>0[/math]

 

[Angelcat]

I want to find the corresponding equation which covers the negative reals, without fractions in the exponents nor the binomial coefficient.

This is the closest I have been able to come so far:
[math]\frac{1}{\sqrt{x-4} \sqrt{x}}=\sum_{n=0}^{\infty}\frac{(2 n)!}{(n!)^2x^{n+1}},\ \Re (x)<0[/math]

 

[blamocur]

Hello there, I couldn't find much data out there regarding the power series for the square root of x so I came up with this equation below.
Since this equation is valid for real x > 0, I want to find the corresponding equation for real x less than 0:

[math]\sqrt{x} = \sum_{n=0}^{\infty}\frac{x(-1)^{n+1}(2n)!}{4^{n}(2n-1)(n!)^2}\left(\frac{1}{x}-1\right)^{n},\ \Re (x)>0[/math]
I would be grateful if anyone would help me with this.

I must be missing something: since you are only talking about real [imath]x[/imath]'s what is [imath]\Re(x)[/imath] for ?
Also, I have doubts about convergence of your series for small [imath]x>0[/imath], e.g. x = 0.25.

 

[Angelcat]

It was the best series I could find at the moment, hopefully one with x^n for the above equation and one twin equation valid for complex numbers will be available. I agree it would be more correct to set x > 1 instead.

Btw, I never said the equation was invalid for complex x.