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Expand the function $z = g(w)$ in a power series of $w$ if $$ w = \frac{e^z - 1}{e^{a z}}, \qquad a \in \mathbb{N} \setminus \{1\}, \quad g(0) = 0. $$ Find the radius $R$ of convergence of the obtained power series.

This task has the Answer: $$ g(w) = \sum_{n=1}^{\infty} \frac{C_{an}^n}{an} w^n, \qquad R = \frac{(a-1)^{a-1}}{a^a}. $$

Solution attempt. It is clear that $f(z) = \frac{e^z - 1}{e^{a z}}$, $g(w)=f^{-1}(w)$, $z_0=w_0=0$ and we must use Lagrange inversion theorem or its modifications. Then $g(w) = \sum_{n=1}^{\infty} c_n w^n$, where $$ n!c_n = \lim\limits_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left( \frac{z-z_0}{f(z)-w_0}\right)^n = \lim\limits_{z\to0}\frac{d^{n-1}}{dz^{n-1}}\left( \frac{z}{f(z)}\right)^n = \lim\limits_{z\to0}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{ze^{a z}}{e^z-1}\right)^n. $$ How to find the (n-1)-th derivative of $\left(\frac{ze^{a z}}{e^z-1}\right)^n$? (Why $c_n=\frac{C_{an}^n}{an}$?)

It is unclear how to obtain the answer even by general Leibniz rule. Moreover, it is unclear how to obtain $R$ using the Cauchy-Hadamard formula even knowing $c_n$ from the answer.

Des
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  • Which step precisely is unclear to you in this answer? https://math.stackexchange.com/questions/tagged/proof-explanation: "For posts seeking explanation or clarification of a specific step in a proof. "Please explain this proof" is off topic (too broad, missing context). Instead, the question must identify precisely which step in the proof requires explanation, and why so." – Anne Bauval Oct 13 '24 at 11:42
  • There is a slightly different expansion using a variant of the theorem. Also see Norlund polynomials which may possibly help – Тyma Gaidash Oct 13 '24 at 16:51
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    Thanks, already thought of variant of Lagrange inversion theorem with composition... – Des Oct 13 '24 at 16:57

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As this question has been tagged complex analysis we present a complex variable proof by way of enrichment.

We seek

$$[w^n] g(w) = \frac{1}{n} [w^{n-1}] g'(w).$$

This is

$$\frac{1}{n} \frac{1}{2\pi i} \int_{|w|=\gamma} \frac{1}{w^n} g'(w) \; dw.$$

Now put $w=(\exp(z)-1)/\exp(az)$ to get

$$\frac{1}{n} \frac{1}{2\pi i} \int_{|z|=\varepsilon} \frac{\exp(anz)}{(\exp(z)-1)^n} \; dz.$$

Next introduce $v=\exp(z)$ to obtain

$$\frac{1}{n} \frac{1}{2\pi i} \int_{|v-1|=\rho} \frac{v^{an-1}}{(v-1)^n} \; dv \\ = \frac{1}{n} \frac{1}{2\pi i} \int_{|v-1|=\rho} \frac{1}{(v-1)^n} \sum_{q=0}^{an-1} {an-1\choose q} (v-1)^q \; dv \\ = \frac{1}{n} {an-1\choose n-1} = \frac{1}{an} {an\choose n}.$$

This is the first part of the task. With the second part for the ratio between consecutive terms we find

$$\frac{an+a}{an} \frac{(an)!}{n! \times (an-n)!} \frac{(n+1)! \times (an+a-n-1)!}{(an+a)!}.$$

Now we have

$$\frac{(an)!}{(an+a)!} = \prod_{q=1}^a \frac{1}{an+q} = \frac{1}{(an)^a} \prod_{q=1}^a \frac{1}{1+q/a/n} \\ = \frac{1}{(an)^a} \exp\left[\sum_{q=1}^a \log \frac{1}{1+q/a/n} \right].$$

With $n$ large we may replace the logarithm with the first term of its Taylor series to get

$$\frac{1}{(an)^a} \exp\left[- \frac{1}{an} \sum_{q=1}^a q \right] = \frac{1}{(an)^a} \exp\left[- \frac{1}{2} \frac{a+1}{n}\right].$$

We also have

$$\frac{(an+a-n-1)!}{(an-n)!} = \prod_{q=1}^{a-1} (an-n+q) = (an-n)^{a-1} \prod_{q=1}^{a-1} (1+q/(a-1)/n) \\ = (an-n)^{a-1} \exp\left[ \sum_{q=1}^{a-1} \log (1+q/(a-1)/n) \right] \sim (an-n)^{a-1} \exp\left[\frac{1}{(a-1)n} \sum_{q=1}^{a-1} q \right] = (an-n)^{a-1} \exp\left[\frac{1}{2} \frac{a}{n} \right].$$

Collecting everything,

$$\frac{(n+1)^2}{n} n^{a-1} (a-1)^{a-1} \frac{1}{n^a a^a} \exp\left[-\frac{1}{2n}\right] \\ = \frac{(n+1)^2}{n^2} \frac{(a-1)^{a-1}}{a^a} \exp\left[-\frac{1}{2n}\right] \sim \frac{(a-1)^{a-1}}{a^a}.$$

This will do it for the radius of convergence.

Remark on the contours. We have $z=w+\cdots$ from the problem statement and the functional equation (computing $c_1$ by substitution) so we may actually take $\varepsilon=\gamma.$ We also have $v=1+z+\cdots$ so we may take $\rho=\varepsilon=\gamma.$ Here $\gamma\lt (a-1)^{a-1}/a^a \lt 1 \lt 2\pi.$ With a non-zero radius of convergence this series about the origin represents an analytic function $g(w)$ there, which justifies the initial application of the CCF to $g'(w).$ Supposing analyticity in a neighborhood of the origin we have obtained just such a branch and the corresponding radius of convergence.

Marko Riedel
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  • Marko Riedel, thank you a lot. You have given a very interesting solution.

    Concerning your remark about contours, I will add that we can also reasoning in the following way. Since $f'(0)\ne0$, there exists a neighborhood in which $f'(z)\ne0$, hence $f$ is locally univalent and its inverse function is also holomorphic and locally univalent in some neighborhood of the point $w=0$.

     – Des Oct 19 '24 at 10:27
  • This task is taken from the topic about local inversion of holomorphic function by means of Lagrange inversion theorem. So the only strange thing is that it is very difficult to find the $(n-1)$-th derivative of $\left(\frac{ze^{a z}}{e^z-1}\right)^n$ at the point $z=0$. – Des Oct 19 '24 at 10:34
  • Re. the edit with $\gamma\lt 2\pi$, the function in $z$ has a pole there. – Marko Riedel Oct 19 '24 at 17:27
  • It turns out that $\gamma<1<2\pi$. What is the purpose of the second inequality? – Des Oct 19 '24 at 18:37
  • Notice that we took $\varepsilon=\gamma$. We don't want to hit the pole at $z=2\pi i.$ – Marko Riedel Oct 19 '24 at 18:39
  • I dont understand: f(z) dosn't have any poles in $\mathbb C$. But $g(w)$ has isolated second-order branch points $w^=\frac{(a-1)^{a-1}}{a^a}e^{2a\pi ki}$ and $g'(w)$ has the poles in $w^$, as $f'(z^)=0$, where $w^=f(z^*)$. – Des Oct 19 '24 at 18:58
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    Well $\exp(anz)/(\exp(z)-1)$ has closest poles at $\pm 2\pi i.$ So we don't want these inside our contour $|z|=\varepsilon.$ – Marko Riedel Oct 19 '24 at 19:03