# Fixed Point Convergence. Finding the interval for which the iteration converges.

I've solved the first part. I think I have something for the second part, but I'm unsure.

A) You are given the fixed point problem $$x=Ax^2$$ where $$A>0$$ is a constant. Compute positive fixed point of the problem.

The fixed point is $$1/A$$, because $$1/A=A\, (1/A)^2$$.

B) For what values of $$A>0$$ will a fixed point iteration $$p_n=A(p_{n-1})^2$$ converge to the positive fixed point?

I think this is from $$0$$ to $$A=1/2$$, $$(0,1/2)$$ but am unsure. This is because the convergence theorem. If we set $$g(x)=Ax^2$$ and compute derivative $$2Ax$$, then $$2Ax\leq K<1$$ according to the convergence theorem, the derivative has to be less than some constant $$K$$ which is less than $$1$$ on the interval of convergence. Therefore $$Ax<1/2$$ $$\implies$$ $$A<1/2$$ for the sequence to converge.

Is this right?

The derivative of $A\,x^2$ evaluated at $x=1/A$ is $2$ for all $A$. The fixed point iteration will never converge to $1/A$ except if $x_0=1/A$. If $0\le x_0<1/A$ it will converge to the aerator $0$. If $x>1/A$ it will converge to $\infty$.