6

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.

Rhys Hughes
  • 12,842

3 Answers3

5

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.

Mohammad Riazi-Kermani
  • 68,728
  • 1
    Attractors are cases where the absolute value of the derivative of the function being iterated is less than $1$. When you are close to a fixed point the distance to the fixed point is multiplied by about the derivative at each step. If that is less than $1$ in absolute value the iteration will converge. – Ross Millikan May 13 '18 at 21:00
  • Thanks for a very informative comment. I should have mentioned this point in my answer. – Mohammad Riazi-Kermani May 13 '18 at 21:13
  • I thought it would be helpful. You are welcome to incorporate it. – Ross Millikan May 13 '18 at 21:29
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 $x \approx .30990693238\cdots$.

Leucippus
  • 26,329
  • By $W_0$, do you mean the inverse Lambert W-function ? – Our Jun 26 '18 at 03:46
  • @onurcanbektas $W_{0}(x)$ is the Lambert W function valid for $x \geq -1$ and is the principle branch. – Leucippus Jun 26 '18 at 03:50
  • What I meant is that the LHS is the image of $W(-x \ln(2))$, so I assumed you have applied the inverse function to both sides. – Our Jun 26 '18 at 05:35
0

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.

user
  • 154,566