I'm trying to show the following properties for the fixed point iteration $$x_{k+1} = \phi(x_k)$$
used to find the zero $\alpha$ of $f(x)$
if $\phi'(\alpha) \in (0,1)$, the convergence is monotone, i.e. $x^k – \alpha$ has the same sign for every $k$.
if $\phi'(\alpha) \in (-1,0)$, the convergence is oscillating, i.e. the error $x^k – \alpha$ changes sign varying $k$
Here's my attempt:
I know that $$x_{k+1} – \alpha = \phi(x_k)- \phi(\alpha) = \phi'(\xi_k) (x_k – \alpha)$$ for $\xi_k$ between $x_k$ and $\alpha$.
Now, since $|\phi'(\alpha)|<1$, then we know that there exists a $\delta >0$ s.t. for every initial data $x_0$ in $(\alpha – \delta, \alpha+\delta) = I$ we have that $\{ x_k\}_k$ converges to $\alpha$.
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If $\phi'(\alpha) \in (0,1)$, since $$e_{k+1} \approx \phi'(\alpha) e_k$$ (I'm assuming that $\phi'$ does not change a lot in $I$) I have that $\{ e_k \}_k$ decrease but the sign is always the same.
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If $\phi'(\alpha) \in (-1,0)$, then from $$e_{k+1} \approx (\phi'(\alpha))^k e_0$$ I have that for $k$ even the error is positive, while for $k$ odd it is negative (I assumed WLOG $e_0>0$). Of course, as $|\phi'|<1 $ in $I$, I have that the magnitude of $e_k$ decreases. So the plot "$e_k$ vs iteration $k$" will be a sort of zig-zag (where the zig and zag has differen signs (lol)) converging to $0$ as $k$ grows.
Essentially, my argument relies on the fact that $\phi'(\alpha) \approx \phi(x)$ for $x \in I$, i.e. that in the neighbourhood $I$ the derivative does not vary much.
Is my procedure correct?
Best Answer
Some minor adjustments to your argument:
It is likely a typo, but you should have
$$x_{k+1}-\alpha=\phi(x_k)-\phi(\alpha)=\phi\color{red}'(\xi_k)(x_k-\alpha)$$
$$e_{k+1}=\phi'(\xi_k)e_k$$
One should then note that
From this can conclude that as long as $x_k$ is sufficiently close to $\alpha$, then
If $\phi'(\alpha)>0$, then the sign of $e_{k+1}$ is the same $e_k$ (monotone).
If $\phi'(\alpha)<0$, then the sign of $e_{k+1}$ is the opposite of $e_k$ (oscillating).