Sequence solutions of $ax=e^x$

Question

This question comes from my answer to: Solving $4x = e^x$ without graphing and looking for intersection

Here I've used a sequence of nested exponentials constructed from $$ x=\frac{1}{a}e^x $$ and a sequence of nested logarithms constructed from $$ x=\log a+\log x $$ that converge to the two solutions of the equation $4x=e^x$.

But I'm not able to proof why this works and why one sequence converges to one solution and the other sequence converges to the other solution.

A brief surf on the web doesn't give references to this "method". Does someone know some references or can show where and when this method converges to the solutions?

Answer 1

As you may know by now, this is called Fixed-Point Iteration (read the properties section). It allows you to easily calculate things in a rather easy way sometimes.

For some input values, however, Fixed-Point Iteration will fail to converge.

For the second method with logarithms, I readily have where it converges, and it may be found on another question.

For the first method, with $x=\frac1ae^x$, it will most likely converge everywhere the logarithmic method fails, however, that may not always be true for other functions.

It also works best if you stay away from values close to the boundaries of convergence (or whatever technical name those boundaries have).

Answer 2

This method is called the fixed point method. For a differentiable function $h$, we have that the sequence defined by $x_{n+1}=h(x_n)$ for a given $x_0$ converges if $h'(x) <1$ for all $x$ between $x_0$ and the fixed point $x^*=h(x^*)$