Problem with infinite product using iterating of a function: $ \exp(x) = x \cdot f^{\circ 1}(x)\cdot f^{\circ 2}(x) \cdot \ldots $

Solution 1:

Say that $f$ solves $(\ast)$ for $g$ if $g(x)=x\cdot f(x)\cdot f\circ f(x)\cdot f\circ f\circ f(x)\cdots$, in the sense that the product in the RHS converges and that its value is $g(x)$, for every positive $x$.

The computations in the post and in some comments show that, if $f$ solves $(\ast)$ for $g:x\mapsto c\mathrm e^x$ for some positive $c$ then $f:x\mapsto x-\log x$, and that $f:x\mapsto x-\log x$ indeed solves $(\ast)$ for $g:x\mapsto\mathrm e^{x-1}$ and not for any other $g:x\mapsto c\mathrm e^x$.

More generally, note that if $f$ solves $(\ast)$ for $g$ then $g(f(x))=f(x)\cdot f\circ f(x)\cdot f\circ f\circ f(x)\cdots$ hence $$ x\cdot g(f(x))=g(x). $$ Assuming that $g$ is invertible, for example because $g$ is increasing, we consider the function $Tg$ defined by $$ Tg(x)=g^{-1}(g(x)/x). $$ Then the only possible solution of $(\ast)$ is $f=Tg$, in particular $f$ is unique and such that $f(1)=1$ when it exists. Thus, for $(\ast)$ to have solutions, one must assume that $g(1)=1$.

Assume furthermore that $g'(1)$ exists, thus, $g(1+\varepsilon)=1+g'(1)\varepsilon+o(\varepsilon)$ when $\varepsilon\to0$. Then $Tg(1+\varepsilon)=1+\eta$ with $g'(1)\eta+o(\eta)=(g'(1)-1)\varepsilon+o(\varepsilon)$. If $g'(1)\lt1/2$, the point $1$ is repulsive for $f$ hence the product in the RHS of $(\ast)$ diverges and $(\ast)$ has no solution. If $g'(1)\gt1/2$, the ratio $(g'(1)-1)/g'(1)$ is in $(-1,1)$ hence the infinite product converges and $(\ast)$ has the unique solution $Tg$.

In the specific case when $g(x)=\mathrm e^{x-1}$, then $g'(1)=1$ hence $(\ast)$ has a unique solution $f=Tg$ and one knows that $Tg:x\mapsto x-\log x$.

In the specific case when $g(x)=b^{x-1}$ for some positive $b$, then $g'(1)=\log b$ hence $(\ast)$ has a unique solution $f=Tg$ when $b\gt\sqrt{\mathrm e}$ and one knows that $Tg:x\mapsto x-\log_b x$, and $(\ast)$ has no solution when $b\lt\sqrt{\mathrm e}$.

In the limit regime $b=\sqrt{e}$, a more careful analysis of the sequences $(x_n)$ defined by $1+x_{n+1}=Tg(1+x_n)$ shows that $(x_n)$ is ultimately alternated (decreasing in modulus and alternatively positive and negative for every $n$ large enough), thus the infinite product $\prod\limits_n(1+x_n)$ converges and $(\ast)$ has a unique solution $f=Tg:x\mapsto x-2\log x$ in this case too.

Solution 2:

The constant c from (1a) is canceled out in step (3). The constant c must be calculated after f(x) is obtained. One way is numerically. Other way is algebraically, but I don't know yet how to.