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Here's my question. Let $b_2$, ..., $b_d \in \mathbb{C}$ ($d$ is an integer greater than 2), and consider the functional equation

$$V(z^d)=dz^{d-1} V(z)+(b_2 z^{d-2} + b_3 z^{d-3}+\ldots + b_d)$$

In this article : http://www.math.harvard.edu/~ctm/papers/home/text/papers/wp/wp.pdf , page 33, McMullen claims that it is clear this functional equation admits a unique analytic solution on $\{z$ $s.t. |z|>1\}$, and that the solution is

$$V(z)=\frac{-z}{d} \sum_{k=2}^d \sum_{n=0}^{\infty} \frac{b_k z^{-k d^n}}{d^n}$$

Of course I believe him =) , but I've been staring at the equation and the solution for some time, and it's not clear to me that the solution is unique (or how he found the solution). All I can do is check it is indeed a solution. So :

Question 1 : why is it unique ?

Question 2 : can you see a method on how the formula was obtained ?

PS : I mentionned the article, but I believe there is no need to refer to it. this problem should be self contained.

Thanks in advance !

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1 Answer 1

First, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2308.pdf

Let $\begin{cases}z_1=\ln z\\V_1(z_1)=V(z)\end{cases}$ ,

Then $V_1(dz_1)=de^{(d-1)z_1}V_1(z_1)+\sum\limits_{k=2}^db_ke^{(d-k)z_1}$

Then, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2303.pdf

Let $\begin{cases}z_2=\ln z_1\\V_2(z_2)=V_1(z_1)\end{cases}$ ,

Then $V_2(z_2+\ln d)=de^{(d-1)e^{z_2}}V_2(z_2)+\sum\limits_{k=2}^db_ke^{(d-k)e^{z_2}}$

$z_2\to z_2\ln d$ :

$V_2(z_2\ln d+\ln d)=de^{(d-1)e^{z_2\ln d}}V_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)e^{z_2\ln d}}$

$V_2((z_2+1)\ln d)=de^{(d-1)d^{z_2}}V_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$

$\therefore V_{2,c}((z_2+1)\ln d)=de^{(d-1)d^{z_2}}V_{2,c}(z_2\ln d)$

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)\prod\limits_{z_2}de^{(d-1)d^{z_2}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_product#Rules,

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(\prod\limits_{z_2}e^{d^{z_2}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(e^{\sum\limits_{z_2}d^{z_2}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_sum#Antidifferences_of_exponential_functions,

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}\left(e^{\frac{d^{z_2}}{d-1}}\right)^{d-1}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_{2,c}(z_2\ln d)=\Theta_1(z_2)d^{z_2}e^{d^{z_2}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

Let $V_2(z_2\ln d)=d^{z_2}e^{d^{z_2}}W_2(z_2\ln d)$ ,

Then $d^{z_2+1}e^{d^{z_2+1}}W_2((z_2+1)\ln d)=de^{(d-1)d^{z_2}}d^{z_2}e^{d^{z_2}}W_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$

$d^{z_2+1}e^{d^{z_2+1}}W_2((z_2+1)\ln d)=d^{z_2+1}e^{d^{z_2+1}}W_2(z_2\ln d)+\sum\limits_{k=2}^db_ke^{(d-k)d^{z_2}}$

$W_2((z_2+1)\ln d)=W_2(z_2\ln d)+\sum\limits_{k=2}^d\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$

$W_2(z_2\ln d)=\Theta_1(z_2)+\sum\limits_{z_2}\sum\limits_{k=2}^d\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$W_2(z_2\ln d)=\Theta_1(z_2)+\sum\limits_{k=2}^d\sum\limits_{z_2}\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Antidifference#Mueller.27s_formula,

$\because\lim\limits_{z_2\to+\infty}\dfrac{b_ke^{-kd^{z_2}}}{d^{z_2+1}}=0$

$\therefore W_2(z_2\ln d)=\Theta_1(z_2)-\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^{z_2+n}}}{d^{z_2+n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_2(z_2\ln d)=\Theta_1(z_2)d^{z_2}e^{d^{z_2}}-e^{d^{z_2}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^nd^{z_2}}}{d^{n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_2(z_2\ln d)=\Theta_1(z_2)e^{z_2\ln d}e^{e^{z_2\ln d}}-e^{e^{z_2\ln d}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^ne^{z_2\ln d}}}{d^{n+1}}$ , where $\Theta_1(z_2)$ is an arbitrary periodic function with unit period

$V_2(z_2)=\Theta(z_2)e^{z_2}e^{e^{z_2}}-e^{e^{z_2}}\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^ne^{z_2}}}{d^{n+1}}$ , where $\Theta(z_2)$ is an arbitrary periodic function with period $\ln d$

$V(z)=\Theta(\ln\ln z)z\ln z-z\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_ke^{-kd^n\ln z}}{d^{n+1}}$ , where $\Theta(z)$ is an arbitrary periodic function with period $\ln d$

$V(z)=\Theta(\ln\ln z)z\ln z-z\sum\limits_{k=2}^d\sum\limits_{n=0}^\infty\dfrac{b_kz^{-kd^n}}{d^{n+1}}$ , where $\Theta(z)$ is an arbitrary periodic function with period $\ln d$

The solution should be not unique.

So McMullen claims should be wrong. I think he should first make deep introspection about the form of the general solution of the functional equation $V(z^d)=f(z)V(z)+g(z)$ .

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-1, although the post does have some entertainment value. –  user31373 Sep 14 '12 at 0:17

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