What is the inverse of the digamma function? Specifically, how can I solve for $x$ in:$$ ψ(x)=1$$$[x ≈ 3.20317146837693106929448152]$ without the digamma?
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1What does the displayed line mean? Just that $\psi(x) = 1$? And also then impose that $1<x$ ? Seems an unclear way to write what you mean... – coffeemath Mar 27 '19 at 15:59
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https://www.wolframalpha.com/input/?i=digamma(x)%3D1 – cgiovanardi Mar 28 '19 at 02:28
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http://bariskurt.com/calculating-the-inverse-of-digamma-function/ – cgiovanardi Mar 28 '19 at 02:34
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Addressing the more general problem of equation $$\psi (x)=k\tag 1$$ there is an interesting very recent paper (see here) which proves the following inequalities for the inverse of the digamma function $$ \frac{1}{\log \left(1+e^{-x}\right)} \lt \psi^{-1} (x) \lt e^{x}+\frac{1}{2}$$ and the left bound seems to be a very good approximation of the considered function.
So, for a high accuracy, to find the zero of equation $(1)$, the starting value $$x_0=\frac{1}{\log \left(1+e^{-k}\right)}$$ should be quite good.
Trying with the case $k=1$ given in the post, the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 3.1922192845297391106277924019427161296056687330974 \\ 1 & 3.2031497228160267807035282180503621244715140401157 \\ 2 & 3.2031714682913142526893626274911788166676509691070 \\ 3 & 3.2031714683769310692931543294021138774273629843645 \\ 4 & 3.2031714683769310692944815249115036749619303929984 \\ 5 & 3.2031714683769310692944815249115036749619307119220 \end{array} \right).$$
Claude Leibovici
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