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Math Help - power series and analitic continuation

  1. #1
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    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
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  2. #2
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    Re: power series and analitic continuation

    Quote Originally Posted by capea View Post
    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.
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  3. #3
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    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
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