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?
2026-03-25 02:57:49.1774407469
close form denominator pade approximation
114 Views Asked by user167276 https://math.techqa.club/user/user167276/detail AtRelated Questions in PADE-APPROXIMATION
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