What are the solutions for $a(n)$ and $b(n)$ when $a(n+1)=a(n)b(n)$ and $b(n+1)=a(n)+b(n)$?

If you have the following recurrence relations :

$a(n+1)= a(n) b(n)$

$b(n+1)= a(n) + b(n) $

How do you find the form of $a(n)$ and $b(n)$ ? I suspect there isn't a closed form , but a infinit sum is good enough . Also what should the values $a(0)$ and $b(0)$, be so that the limit of $b(n)$ as n goes to infinity is 1 ?

Solution 1:

A continuous version of these equations would be $$\frac{d}{dt}\ln a(t)=\ln b(t)\\ \frac{d}{dt}b(t)=a(t)\\ \ddot{b}=\dot{b}\ln b\\ \dot{b}=b\ln b-b+C\\ t=\int\frac{db}{b\ln b-b+C}$$ Since $t\to\infty$ when $b\to1$, we need $C=1$. The leading terms near $b=1$ are $$t=\int \frac2{(b-1)^2}+\frac2{3(b-1)}-\frac19+\frac{7(b-1)}{135}+...db$$ so to leading order $t=2/(1-b)$, or $b\approx1-2/t$. Then $a=db/dt\approx2/t^2$.
I don't know how much of that carries over to the discrete case.
EDIT:
PRELIM $$\frac1{(n+1)^2}=\frac1{n^2}(1+\frac1n)^{-2}$$ By geometric series,
$$(1+\frac1n)^{-1}=1-\frac1n+\frac1{n^2}-\frac1{n^3}+\cdots$$ If you multiply that series by itself, term by term, you get $$(1+\frac1n)^{-2}=1-\frac2n+\frac3{n^2}-\frac4{n^3}+\cdots\\ \frac1{(n+1)^2}=\frac1{n^2}-\frac2{n^3}+\frac3{n^4}-\cdots$$ Also, a similar thing for logs base e, because $1+1/n$ is near $1$, is $$\ln(1+\frac1n)=\frac1n-\frac1{2n^2}+\frac1{3n^3}-\cdots$$ I can't explain that without calculus, but you can check $\ln1.01$ and $\ln0.99$.
NOW It seems that the following leading-order works. The idea is $$a(n)=f_0(n)+f_1(n)+...\\b(n)=g_0(n)+g_1(n)+...$$ where $f_0$ and $g_0$ are known, and $f_1$ and $g_1$ are small corrections. Since $f_0(n)=2/n^2$, we have $f_0(n+1)\approx 2/n^2-4/n^3+...$. Find a similar expression for $g_0(n+1)$. Put them into $a(n+1)=a(n)b(n)$ and $b(n+1)=a(n)+b(n)$, and suppose $f_1(n+1)\approx f_1(n)$ and $g_1(n+1)\approx g_1(n)$. Solve for $f_1$ and $g_1$. If you assume $f_1$ is $O(n^{-3})$ and $g_1$ is $O(n^{-2})$ you get cancellation, with terms left over from $f_0$ and $g_0$. So you need $f_1=O(n^{-3}\ln n)$ and $f_2=O(n^{-3})$. Now $f_1(n+1)$ has a term $O(n^{-4})$ which you can match with the one from $f_0$.

: $$a(n)=\frac2{n^2}+\frac2{3n^3}(1-8\ln n)+...\\ b(n)=1-\frac2n+\frac{8\ln n}{3n^2}- \frac{32}{9n^3}\left(\ln^2n-\ln n+1/2\right)+...$$