# Math Help - power series and analitic continuation

1. ## power series and analitic continuation

hi it will be posible to improve the taylor or laurent series to the funtion
$\frac{1}{1+e^{-x^2}}$
so that the taylor series gives
$\frac{1}{2}+\frac{x^2}{4}-\frac{x^6}{48}...$
and it only convergent for x for -2 to 2 more or less

2. ## Re: power series and analitic continuation

Originally Posted by capea
hi it will be possible to improve the taylor or laurent series to the funtion
$\frac{1}{1+e^{-x^2}}$
so that the taylor series gives
$\frac{1}{2}+\frac{x^2}{4}-\frac{x^6}{48}...$
and it only convergent for x for -2 to 2 more or less
If you think of this as a function of a complex variable, $f(z) = \frac1{1+e^{-z^2}},$ then the radius of convergence of its power series will be the distance from the origin to the nearest singularity. The singularities occur at points where $e^{-z^2} = -1,$ from which you can see that the radius of convergence is $\sqrt\pi\approx1.77.$

3. ## Re: power series and analitic continuation

thank for your quick response i think that taylor expansion is it too poor and only it is value for a few value but if you want a expansion for al real and comples space need a rational expansion i found a way for example the expansion for the function
$\frac{1}{e^{-x^2}+1}$ it is
$-\frac{1}{2}-\frac{x^2}{4}-\frac{x^3}{2 \pi ^2 \left((-1)^{1/4} \sqrt{\pi }-x\right)}-\frac{x^3}{2 \pi ^2 \left((-1)^{3/4} \sqrt{\pi }-x\right)}-\frac{x^3}{18 \pi ^2 \left((-1)^{1/4} \sqrt{3 \pi }-x\right)}-\frac{x^3}{18 \pi ^2 \left((-1)^{3/4} \sqrt{3 \pi }-x\right)}-\frac{x^3}{50 \pi ^2 \left((-1)^{1/4} \sqrt{5 \pi }-x\right)}+\frac{x^3}{2 \pi ^2 \left((-1)^{1/4} \sqrt{\pi }+x\right)}+\frac{x^3}{2 \pi ^2 \left((-1)^{3/4} \sqrt{\pi }+x\right)}+\frac{x^3}{18 \pi ^2 \left((-1)^{1/4} \sqrt{3 \pi }+x\right)}+\frac{x^3}{18 \pi ^2 \left((-1)^{3/4} \sqrt{3 \pi }+x\right)}+\frac{x^3}{50 \pi ^2 \left((-1)^{1/4} \sqrt{5 \pi }+x\right)}$
the taylor series for x=2 gives the value 1.5 while the last expansion gives 0.9834 the correct answer is it $0.982014$