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I have an iterative process that starts with

$$x_1 = \log_{10}(a)$$

Following iterations are as follows:

$$x_2 = \log_{10}(a-b\cdot x_1)$$

$$x_3 = \log_{10}(a-b\cdot x_2)$$

$$x_4 = \log_{10}(a-b\cdot x_3)$$


and so on indefinitely. (Assume $a$ and $b$ are known and positive)

Once these are nested, it reads:

$$x = \log_{10}(a-b\cdot \log_{10}(a-b\cdot \log_{10}(a-b \cdot \ldots \log_{10}(a-b\cdot \log_{10}(a))\ldots)))$$

This seems straight forward enough to have a simpler (non-iterative) solution, but in all my searching I can't find anything.

Any help would be appreciated.


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have you any additional knowledge about $a$ and $b$? – V-X Mar 13 '13 at 7:11

Take a look at the Banach fixed-point theorem. Under certain conditions this leads to the solution being: $$x=\log_{10}(a-bx)$$ Of course, this cannot be solved analytically, but with use of the Lambert W function: $$x = \frac{a}{b} - \frac{1}{\log 10}\cdot W\left(\frac{10^{a/b} \log 10}{b}\right)$$

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If this serie has solution, it converges very fast:

$\lim_{n\rightarrow\infty} x_n = \log_{10}(a-bx_n)$

You have to find exact region where the solution exists($a-bx_n>0$ for each $n$) and after that you need to solve $x_n = \log_{10}(a-bx_n)$ which doesn't seem to have simple analytical solution.

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Well V-X beat me to it. But a bit more detail, consider the function $f(x)=a-b\log_{10}x$. This map is a contraction. There is a certain domain (too lazy to work out the details) but within that domain, this map has an attractor meaning no matter what you starting guess $x_1$ is (within your domain of convergence), the iteration $$x_{n+1}=f(x_n)$$ will converge to the same value which is its unique fixed point. This fixed point is the solution to $$x=f(x)$$ which has no nice closed-form analytic solution. But if you allow the Lambert W function where $W$ is the principal solution to $$z=W(z)e^{W(z)}$$ then the above fixed point can be written as $$x=\frac{b\cdot W(\frac{10^{a/b}\cdot \log(10)}{b})}{\log(10)}.$$

Addendum: playing with it numerically, it looks like in the domain, you get a unique real fixed point. Outside the domain, you still have a fixed point but it can be a complex number.

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