How it is possible to get the denominator pade close form of the function $$\frac{1}{2} \left(\sqrt{2 \pi } e^{x/2} \sqrt{x} \text{erf}\left(\frac{\sqrt{x}}{\sqrt{2}}\right)+2\right)$$ as $$2^{-n-1} \binom{4 n}{2 n} \, _1F_1\left(-n;\frac{1}{2}-2 n;-\frac{x}{2}\right)$$ it possible from Gauss Continue fraction?
Asked
Active
Viewed 113 times
1
-
I have a problem trying to understand the question. A Pade approximant is the ratio of two polynomials of given degrees. Could you clarify for me ? – Claude Leibovici May 09 '15 at 04:25
-
Claude I asking for denominator of Pade approximant salute – May 12 '15 at 09:40
-
May I confess that I never saw a Pade approximant with such a denominator. To me, numerator and denominator are not independent from each other. May be, you could clatify for me. Cheers :-) – Claude Leibovici May 12 '15 at 09:53