What does the iteration of this function approach as $n\rightarrow\infty$?

Question

Suppose I have the function $$f(x)=\frac{4x}{1+\frac{x}{M}}.$$ Denote the $n$th iteration as $f^n(x)$. What is $\lim_{n\rightarrow\infty}{f^n(x)}$?

This is from a physics problem and the book claims that the limit is $3M$, though it says this without justification. It says it becomes clear once you draw a graph and observe that $x=3M$ is a fixed point.

But this is hardly rigorous. Is there any rigorous way to prove this limit?

Answer 1

We have (after slightly re-arranging the RHSide): $f^{n+1}(x) = \frac{4}{1/M + 1/f^n(x)}$. Now, assume we have a limit of $f^n(x)$ and $n \to \infty$ denoted by $l$. Replacing it in the above equation, one has: $l = 4 / (1/M + 1/l)$, which gives $l=3M$.

Answer 2

Let's assume that there is a fixed point, $x_0$. In that case $f(x_0)=x_0$. $$f(x_0)=\frac{4Mx_0}{M+x_0}=x_0$$ Then $$x_0^2+Mx_0-4Mx_0=0$$ This can be written as $$x_0(x_0-3M)=0$$ The possible solutions are $x_0=0$ and $x_0=3M$. You would need to check the stability of these points.

Answer 3

Indeed, taking Chip's answer one step farther: if we define $g(x) = 3M/f(x)$, then $g$ satisfies the iteration $$ g^{n+1}(x) = \frac{3M}{f^{n+1}(x)} = \frac{3M(1/M+1/f^n(x))}{4} = \frac{3+g^n(x)}4. $$ This recursion is pretty easy to solve—we get $$ g^n(x) = 1-\frac{1-x}{4^n}, \quad\text{so } f^n(x) = \frac{3M}{1-(1-x)/4^n}. $$ This converges exponentially fast to $3M$ as long as the denominator never equals $0$, that is, for all $x$ not of the form $1-4^k$ for $k\ge0$.

Answer 4

$$f(x)=\frac{4x}{1+\frac{x}{M}}=x$$

has the solution $$x=3M$$ Thus $x=3M$ is a fixed point.

In order to show that it is an attractor you may find the derivative of $f(x)$ at $x=3M$ and show that its magnitude is less than $1$

The derivative at $x=3M$ is $$f'(3M)= 1/4$$

Thus the fixed point is indeed an attrctor.