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Migrated from the MSE.


I am interested in the functional inverse of $$ z=1+w+\cdots+w^{n-1},\quad w\geq0,\ n>1. $$ This function is strictly increasing on $w\geq0$ and thus admits an inverse.


By Lagrange's theorem we may write the inverse: $$ w(z)=a+\sum _{k=1}^\infty g_k\frac{(z-f(a))^k}{k!}, $$ with $f(w)=1+w+\cdots+w^{n-1}$ and $$ g_k=\lim_{w\to a}\partial_w^{k-1}\left({\frac {w-a}{f(w)-f(a)}}\right)^k. $$ Expanding $f(w)$ around $w=1$ gives $f(w)=nF(1,1-n;2;1-w)$, which is a hypergeometric function. Choosing $a=1$ we write $$ \frac{f(w)-f(1)}{w-1}=-n\frac{F(1,1-n,2,1-w)-1}{1-w}=\frac{n(n-1)}{2}F(1,2-n;3;1-w). $$ It follows $$ w(z)=1+\sum _{k=1}^\infty a_k\frac{2^k}{n^k(n-1)^k}\frac{(z-n)^k}{k!}, $$ with $$ a_k=\lim_{w\to 1}\partial_w^{k-1}\left(F(1,2-n;3;1-w)\right)^{-k}. $$ Faà di Bruno's formula then allows us to evaluate the finial limit yielding $$ w(z)=1+\frac{2(z-n)}{n(n-1)}+\sum_{k=2}^\infty\sum_{\ell=1}^{k-1}(-k)^{(\ell)}B_{k-1,\ell}\left(\left\{(-1)^m\tfrac{m!(2-n)_m}{(3)_m}\right\}_{m=1}^{k-\ell}\right)\frac{\left(\frac{2(z-n)}{n(n-1)}\right)^k}{k!}, $$ where $(s)^{(n)}=\Gamma(s+1)/\Gamma(s-n+1)$ is the falling factorial, $(s)_n=\Gamma(s+n)/\Gamma(s)$ is the Pochhammer symbol, and $B_{n,k}$ is the partial Bell polynomial.


Could anyone point me to any research that has been done in studying this function? In particular, I would like to know the radius of convergence for the above series representation. If that is intractable, upper and lower bounds on for the radius of convergence would be helpful too.

Plotting the truncated series in Mathematica, it appears that the series may converge for at least $|z-n|<n-1$.

Also, it is clear that the roots of $f(w)$ are all on the unit circle and thus by the Gauss–Lucas theorem all the roots of $f^\prime$ are inside the unit circle. This may be of help in determining the radius of convergence for the series above.

Edit:

I wrote the following Mathematica code to compute the radius of convergence as a function of $n$:

f[w_, n_] := (1 - w^n)/(1 - w);
fprime[w_, n_] := -((n w^(-1 + n))/(1 - w)) + (1 - w^n)/(1 - w)^2;
wk[n_] := Solve[fprime[w, n] == 0, w][[All, 1, 2]];
ROC[n_] := Min[FullSimplify[Abs[f[wk[n], n] - n]]];

We get $ROC(3):|z-3|<9/4$, $ROC(4):|z-4|<4\sqrt{2/3}$, $ROC(5):|z-5|< 4.24942\dots$, and $ROC(6):|z-6|<5.2226\dots$, all of which are greater than $n-1$.

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  • $\begingroup$ What are your results for $n=3$ and $n=4$? $\endgroup$ Commented Feb 10, 2022 at 17:36
  • $\begingroup$ @GeraldEdgar For $n=3$ we get $$w(z)=1+3\sum_{k=1}^\infty (-k)^{(k-1)}\frac{\left(\frac{z-3}{9}\right)^k}{k!}=\frac{1}{2}(\sqrt{4z-3}-1).$$ The $n=4$ would require a little more work. $\endgroup$ Commented Feb 10, 2022 at 18:05
  • $\begingroup$ @GeraldEdgar So in the $n=3$ case above we can write $w(z)$ in terms of $F(3/2,2;3;-\frac{4}{9}(z-3))$ which converges if $|z-3|<2+\frac{1}{4}$. So maybe in general we have $|z-n|<n-1+\frac{1}{n+1}$? $\endgroup$ Commented Feb 10, 2022 at 18:25
  • $\begingroup$ I think you will have to do $n=4$ before making a conjecture like $n-1+\frac{1}{n+1}$. $\endgroup$ Commented Feb 10, 2022 at 19:02
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    $\begingroup$ In the answer here I wrote the Lagrange inversion for the inverse of $ax+bx^2+\dots +px^n$ mathoverflow.net/questions/249060/… $\endgroup$ Commented Feb 10, 2022 at 22:50

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