Why does iterating in different ways produce different solutions?

Question

Recently, I came across the following equation: $$2^x=4x$$ To solve it, I decided to iterate. Firstly I stated: $$x_{n+1}=\frac{2^{x_n}}{4},x_0=1$$ and found a solution of $x\approx 0.3099069324$.

Then I again rearranged it to: $$x_{n+1}=\log_{2}({4x_n}), x_0=1,$$ and achieved the solution $x=4$.

Despite trying many values, I was unable to get these iterations to find the other solution. What I would like to know is: what is the reasoning behind this?

I have seen a similar thing happen in the case of other iterations, where I iterate for a solution in different ways and get two different (correct) solutions - and it has perplexed me.

Answer 1

You are aiming at fixed points of two different functions.

Some fixed points are attractors and some are repellers.

If you graph your functions you see as you iterate, the points tend to the attractor and stay away from the repeller.

For $f(x) = \frac {2^x}{4}$ the point $x\approx 0.3099069324$ is an attractor and $x=4$ is a repeller.

For the other function it is the other way around.

Answer 2

Sketch a graph for the two functions and you see that $y=\log_2 4x$ is greater than $y=x$ and the iteration tends to the greater solution while $y=2^x/4$ is smaller than $y=x$ and the iteration tends to the smaller solution.

Answer 3

An "exact" solution may be obtained with use of the Lambert W-function which is defined by $W_{0}(x) e^{W_{0}(x)} = x$ for $x \geq -1$. This is seen by the following: \begin{align} 2^{x} &= 4x \\ 4 x &= e^{x \, \ln2} \\ (- x \, \ln2) \, e^{- x \, \ln2} &= - \frac{\ln2}{4} \\ - x \, \ln2 &= W_{0}\left( - \frac{\ln2}{4} \right) \\ x &= - \frac{1}{\ln2} \, W_{0}\left( - \frac{\ln2}{4} \right). \end{align} By calculation this gives a value of $x \approx .30990693238\cdots$.