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The motivation of this question is pure curiosity.

Working on this problem, I tried to find the zero of function $$f(x)=m\,(m-1)^{\frac{1}{m}-1}\, x^{1-\frac{1}{m}}+x-1 \quad \quad \text{where} \quad m>1$$

I arrived to the conclusion that $$x_{(m)}=\frac 12+\sum_{n=1}^p \frac {P_n(L)}{m^n}+O\left(\frac 1{m^{p+1}}\right)\qquad\quad \text{where} \quad L=-\log(2m)$$ where the polynomials are of odd degree (repeated).

My question is :

Is there a way to fully automate the process to find the coefficients of these polynomials up to a selected order $p$ ?

Your answer could be Yes, No or provide more. Thanks in advance.

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  • $\begingroup$ This seems to be relevant: mathematica.stackexchange.com/a/202201/9469 $\endgroup$ Commented May 3 at 8:52
  • $\begingroup$ Yet another post on the same topic: mathematica.stackexchange.com/q/98467/9469 $\endgroup$ Commented May 3 at 8:54
  • $\begingroup$ See also series solutions here: mathoverflow.net/a/249098/41145 $\endgroup$ Commented May 3 at 8:55
  • $\begingroup$ @yarchik. Thanks for the links but they do not answer the question : generate the coefficients of the $P_n(L)$. I have a problem with the results in the accepted answer since the coefficients do not match the ones I did produced in the linked question. Cheers :-) $\endgroup$ Commented May 3 at 13:55

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I completely revised my previous answer which contained some error. Now, the table of the cited problem is reproduced.

The new strategy is to put the asymptotic expansion ($a_k = P_{k}(L)$) $$ x = \frac{1}{2} + \sum_{k=1}^{\infty} \frac{a_k}{m^{k}} \qquad (m\to \infty) $$ directly into $f(x)=0$ and solve for the first $p$ coefficients $\{a_k\}_{k=1}^{p}$. The MMA codes go something like this:

ClearAll[f,m,a]; f = m (m - 1)^(1/m - 1) x^(1 - 1/m) + x - 1; p = 10; f1 = f /. x -> 1/2 + Sum[a[k] m^(-k), {k, 1, p}]

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eq = LogicalExpand[Series[f1, {m, Infinity, p}] == 0]; ans = Solve[eq, Table[a[k], {k, 1, p}]][[1]]; Short[ans, 6]

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Grid[Table[{k, Collect[PowerExpand[a[k] /. ans /. m -> Exp[-L]/2], L]}, {k, 1, p}], Dividers -> All]

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  • $\begingroup$ Thank you ! This is the perfect answer to m'y question $\endgroup$ Commented May 3 at 10:32
  • $\begingroup$ I think there is a typo in the recursive identity. $\endgroup$ Commented May 3 at 12:05
  • $\begingroup$ @yarchik Sorry, I cannot find a typo. Please edit for me. $\endgroup$ Commented May 3 at 12:09
  • $\begingroup$ I guess the intention was $x_{m+1}=x_m+f(x_m)$ $\endgroup$ Commented May 3 at 12:16
  • $\begingroup$ @yarchik Well, I thought of using the recursion like that, too. But I found a solution without using recursion --- just a series expansion around $m=\infty$ was enough. The equation $x_{(m)}=x_{(m)} - f(x_{(m)})$ holds simply because $f(x_{(m)})=0$ by definition. $\endgroup$ Commented May 3 at 13:34

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