What does the following iteration formula $x_{k+1}=2x_k-cx_{k}^2$ do?

Question

The question is:

What does the following interation formula do?: $$x_{k+1}=2x_k-cx_{k}^2.$$

I tried to identify this with Newtons method. I.e. I tried to bring that into the form $x_{k+1}=x_k-\frac{f(x_0)}{f'(x_0)}$, which leads to: $$(cx_k^2-x_k)f'(x_k)=f(x_k).$$ But then $f(x)$ is something like $e^a$ but these functions doesn't have any roots... Is this still correct and I must note that this iteration formula does not converge or are there any other functions satisfying this equality?

Answer 1

You were on the right track, but stopped a bit early. Instead of

$$f(x) = f'(x) \cdot (cx^2-x)$$

Write that as

$$f'(x) = f(x) \cdot \frac{1}{cx^2-x}$$

So a function that works would be

$$ \begin{align} f(x) &= e^{\int 1/(cx^2-x)\ dx} \\ &= e^{\ln((1-cx)/x)} \\ &= \frac{1-cx}{x} \end{align}$$

Answer 2

You could also see the partial square of a binom on the right side, and complete it as $$ 1-cx_{k+1}=1-2cx_k+(cx_k)^2=(1-cx_k)^2. $$ The sequence $y_k=1-c_k$ is then a subsequence of the geometric sequence, and from $y_k=y_0^{2^k}$ follows the quadratic convergence.

Answer 3

If the iterations converge, they converge to an $x$ such that

$$x=2x-cx^2$$

and $x=0$ or $x=\dfrac1c$.

Now the derivative of $f'$ is $2-2cx$. $f'(0)=2>1$ so the iterations diverge around $0$, and $f'(\frac1c)=0$ and the iterations do converge.