Convergence radius of $\log(\Gamma(\exp(x)))$?

Question

In the context of iteration of functions I'm working with the power series for $$ \small f(x)=\log(\Gamma(\exp(x))) =\sum_{k=1}^\infty a_k x^k \sim -0.577216 x + 0.533859 x^2 + 0.325579 x^3 + 0.125274 x^4+ O(x^5)$$ and want to write about its radius of convergence $\rho_f$ . What I've done was to scale the coefficients of the power series such that they appear to become roughly constant in their magnitude. If I write $$ c_k = \pi^k a_k $$ then it seems that the coefficients at indexes $ c_{4j},c_{4j+2}$ up to $j=128$ become roughly of magnitude $0.008$ and that at indexes$ c_{4j+1},c_{4j+3}$ about 1/100 of that - but still decreasing. It seems, if I increase the scaling factor only a little bit over $\pi$ I always run into a divergent sequence of $c_k$ - only that they "begin to diverge" later, where it is not visible using truncated power series.

If I rescale $$ d_k = (k+1) \pi^k a_k $$ then I get impressive convergence of the $d_{4j} \to 2$ , $d_{4j+2} \to -2$ while $d_{4j+1}, d_{4j+3}$ still decrease in magnitude; so this looks even much better. However, the introduction of the $ (k+1)$ cofactor does not extend that radius if we have a sequence of geometric decrease (I think so).

So I would tend to say, that the radius of convergence of $ \rho_f \sim \pi $ - is that meaningful?

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[update] It seems, a bit help of wolfram alpha and decoding its output gave a good idea such that I found now a meaningful decomposition in term of zetas, stirling numbers 2nd kind and factorials: $ \displaystyle \quad \small \begin{array} {} a_1 &=& (- 1 \gamma &.&.&.&)/1!\\ a_2 &=& (- 1 \gamma & +1 \zeta(2) &.&.&)/2! \\ a_3 &=& (- 1 \gamma & +3 \zeta(2) & -2 \zeta(3) &.&)/3! \\ a_4 &=& (- 1 \gamma & +7 \zeta(2) & -12 \zeta(3) &+6 \zeta(4)&)/4! \\ \ldots& & \ldots \end{array} $
(I should have seen this before because I already had an expression for that series using the matrix-product $ \small \operatorname{fS2F}\cdot \operatorname{ZETA} \cdot \operatorname{fS1F}$, where the names of the matrices are simply from my Pari/GP-conventions but indicate their contents) However, this does not immediately answer my question about the range of convergence but maybe is a step in the correct direction

Answer 1

Hints:

• the series converges up to the next singularity
• $\Gamma(x)$ is never 0
• $\Gamma(x)$ has a simple pole for $x=0,-1,-2, \dots$
• $\exp(x)$ is never 0
• $\exp(\pm i\pi) = -1$
• the singularities of $\log(x)$ are the branch points at $x=0,\infty$.

These hints should be enough to convince you that the nearest singularity of the function $f(x) = \log (\Gamma(e^{x}))$ to the origin is at $x=\pm i \pi$ which sets the convergence radius to $\pi$.

The next singularities are then at $x= \log(2) \pm i \pi$ and $x=\pm i 3\pi$ and so on.

Answer 2

Here are some suggestions:

1) Try to use Carleman Matrix, since you are using iteration of functions or

2) You can work on the function

$$ g(x)=f'(x) = e^{x}\psi(e^x) $$ and then integrate the resulting series term by term.