Finding an explicit formula from a recursive formula.

Question

I have the recurrence relation:

$$g(k, 0, x) = k,$$

$$g(k, n, x) = \dfrac{1}{2} \log_{k}{\left(\dfrac{k^{g(k, n - 1, x)}x}{g(k, n - 1, x)}\right)},$$

and I would like to solve it, if it is possible.

By the way, $\lim_{n \to \infty}{g(k, n, x)} = f^{-1}(k, x), f(k, x) = k^{x}x$.

Answer 1

I understand simplify the $n$-th term not solve the recurrence relation.

Noting $u_n=g(k,n,x)$ one has $$2u_n = \log_{k}{\left(\dfrac{k^{g(k, n - 1, x)}x}{g(k, n - 1, x)}\right)}$$ i.e.$$2u_n=\log_{k}\frac {k^{u_{n-1}}x}{u_{n-1}}\iff k^{2u_n}=\frac {k^{u_{n-1}}x}{u_{n-1}}$$ Hence $$u_{n-1}k^{(2u_n-u_{n-1})}=x$$

►"Answer" (Bis):Please allow me a note about this question I can not entirely understood before. A recurrence relation is well defined when initial conditions are given. In this case, it appears $u_0=k$ to be sufficient. Anyway, one has $$ 2u_n=\log_{k}\frac {k^{u_{n-1}}x}{u_{n-1}} ; u_0=k \Rightarrow x=k^{(u_1+1-k)}$$ So, the variable $x$ must be constant? The answer to this is, simply, here the recurrence relation is not of numbers but of functions, only the first term $u_0$ is constant. (I did not see this and I was looking for numbers). What follows is my "answer" as I found it, however I beg you read the REMARK below. We have$$ u_{n-1}k^{(2u_n-u_{n-1})}=x \Rightarrow \frac {u_n}{u_{n-1}}=k^{(3u_n-u_{n-1}-2u_{n+1})} $$ Note both telescopic (1) and (2):

(1) $$\frac {u_1}{u_0}\cdot\frac {u_2}{u_1}\cdot\frac {u_3}{u_2}\cdot.......\cdot\frac{u_n}{u_{n-1}}=\frac{u_n}{u_0}$$ (2)$$\sum_{k=0}^{k=n}(3u_k-u_{k-1}-2u_{k+1})=-2u_{n+1} +u_n+2u_1-u_0$$ Hence $$\frac {u_n}{u_0}=k^{(-2u_{n+1} +u_n+2u_1-u_0)}$$ But $$2u_1=\log_{k}\frac{k^{u_0}x}{u_0}\Rightarrow 2u_1-u_0=\log_{k}\frac {x}{u_n}$$ i.e. $$u_n=k^{(-2u_{n+1}+u_n+\log_{k}x)}$$ Thus $$\boxed{ 2u_{n+1}=u_n+\log_{k}\frac {x}{u_n}}$$ Let us calculate three terms (the first one is the given initial condition). $$u_0=k$$ $$u_1=\frac {k+\log_{k}\frac {x}{k}}{2}$$ $$2u_2=\frac {k+\log_{k}\frac {x}{k}}{2}+\log_{k}\frac {2x}{k+\log_{k}\frac {x}{k}}$$ The terms become more and more complicated as n increases.

REMARK.- See this and related, please. All the above can be shortened considerably from the given by @Taylor formulation and the "simplification" given before. If the terms of the recurrence are regarded as what they are, functions and not numbers, then the "answer" is easy and short. Actually this "answer" (Bis) is exactly the formulation given by @Taylor!!! which is easy to verify.