Suppose $y=f(x)$ is defined implicitly as the solution of the following equation
$$\Gamma \left(\frac{1}{x+1},0\right)=2 \Gamma \left(\frac{1}{x+1},e^y\right)$$
Where $\Gamma$ is the upper incomplete Gamma function.
The plot of $f(x)$ looks linear, how do I solve for the slope/intercept?
Motivation: this gives the range of values for which the loss curve of gradient descent is well modeled by the power law. Equivalently, when generalized exponential integral is well approximated by $x^{n-1}$ term
Inverse gamma regularized $Q^{-1}(a,x)$ gives:
$$y=\ln\left(Q^{-1}\left(\frac1{x+1},\frac12\right)\right)$$
where the $\frac12$ implies a connection with the gamma distribution median. Therefore the $y$ intercept of $f(x)$ is:
$$\left(0, \ln\left(Q^{-1}\left(1,\frac12\right)\right)\right)=(0,\ln(\ln(2)))$$
Using oblique asymptote limits:
$$y\sim mx+a=-\ln(2)(x+1)-\gamma\\m=\lim_{x\to0}\frac y x=\lim_{x\to0}x\ln\left(\sqrt[x]\frac{x!}2\right)=-\ln(2)\\a=\lim_{x\to\infty} \ln\left(Q^{-1}\left(\frac1{x+1},\frac12\right)\right)+\ln(2)x=\lim_{x\to0}\frac{\ln(x!)}x-\ln(2)=-\gamma-\ln(2)$$
with the $a$ limit using and $m$ limit using $Q^{-1}(a,x)\sim \sqrt[a]{(1-x)a!}$ with the Euler Mascheroni constant. Using the OP’s window: