Barrett Osborn

2022-11-22

The Taylor Series of Gamma Function exists or not.

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Expert

If you want the Taylor series, you basically need the ${n}^{th}$ derivative of $\mathrm{\Gamma }\left(x\right)$. These express in terms of the polygamma function. Considering
${d}_{n}=\frac{{\left[\mathrm{\Gamma }\left(x\right)\right]}^{\left(n\right)}}{\mathrm{\Gamma }\left(x\right)}$
the first terms are
${d}_{1}={\psi }^{\left(0\right)}\left(x\right)$
${d}_{2}={\psi }^{\left(0\right)}\left(x{\right)}^{2}+{\psi }^{\left(1\right)}\left(x\right)$
${d}_{3}={\psi }^{\left(0\right)}\left(x{\right)}^{3}+3{\psi }^{\left(1\right)}\left(x\right){\psi }^{\left(0\right)}\left(x\right)+{\psi }^{\left(2\right)}\left(x\right)$
${d}_{4}={\psi }^{\left(0\right)}\left(x{\right)}^{4}+6{\psi }^{\left(1\right)}\left(x\right){\psi }^{\left(0\right)}\left(x{\right)}^{2}+4{\psi }^{\left(2\right)}\left(x\right){\psi }^{\left(0\right)}\left(x\right)+3{\psi }^{\left(1\right)}\left(x{\right)}^{2}+{\psi }^{\left(3\right)}\left(x\right)$
which "simplify" (a little !) when you perform the expansion around $x=a$, $a$ being a positive integer.

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