I would like to know if I'm doing well the following exercise. I think that I'm doing something wrong (specially in part (3)) but I don't know what...
Having $F(x)=x^2+\frac{3}{16}$ I have to answer the following quetions:
That function has got two fixed points. Which are those points?
What I have done is:
$x^2+\frac{3}{16}=x\space ; \space x^2-x+\frac{3}{16}=0 \space $ so $x_1 =\frac{3}{4}$ and $x_2=\frac{1}{4} \space $.$\space $ So the two fixed points are $\frac{1}{4}$ and $\frac{3}{4}$.
Being $\space x_{k+1}=F(x_k)$ the iteration of the fixed point for that $F$, for which of the points obtained in part (1) is going to converge? (Let's suppose that the starting point is near of the fixed point).
I have done the following:
$F'(x)=2x \space \space;$ and $|F'(x)|$ has to be smaller that $1$ to converge. So $x=\frac{1}{4}$
For the fixed point/points obtained in part (2), how many iteration are needed more or less, so that the absolute accuracy is 10 times less.
What I have done is:
$|p_n-p|\le 10^{-1}|p-p_0|\le K^n|p-p_0|;\space 10^{-1}\le K^n; \space log_{10}10^{-1}\le log_{10}K^n; \space -1\le nlog_{10}K; \space \frac{-1}{log_{10}K}\le n; \space 3.321928\le n$
where $K=F'(\frac{1}{4})=\frac{1}{2}\space$
So we need 4 iterations at least.
Is this ok? Can anyone help me?