Exercise 5.24

Question

Exercise 24:
    The process described in part (c) of Exercise 22 can of course also be applied to functions that map $(0,\infty)$ to $(0,\infty)$.

Fix some $\alpha>1$, and put $$f(x)=\frac{1}{2}\biggl(x+\frac{\alpha}{x}\biggr),\quad\quad g(x)=\frac{\alpha+x}{1+x}.$$ Both $f$ and $g$ have $\sqrt{\alpha}$ as their only
    fixed point in $(0,\infty)$. Try to explain, on the basis of properties of $f$ and $g$, why the convergence in Exercise 3.16 is so much more rapid than it is in Exercise 3.17.
    Do the same when $0<\alpha<1$.

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

The distance between the successive elements of the sequence $x_n=f(x_{n-1})$ is given by the function $F(x)=f(x)-x$, while the distance between the successive elements of
    the sequence $y_n=g(y_{n-1})$ is given by the function $G(y)=g(y)-y$. We have
    $$\displaylines{
    F(x)=-\frac{1}{2}\biggl(x-\frac{\alpha}{x}\biggr) \cr
    F\bigl(\sqrt{\alpha})=0,\quad F'\bigl(\sqrt{\alpha}\bigr)=-1 \cr
    G(y)=\frac{\alpha-y^2}{1+\alpha}\cr
     G\bigl(\sqrt{\alpha})=0,\quad G'\bigl(\sqrt{\alpha}\bigr)=-\frac{2}{1+\sqrt{\alpha}}.
    }$$
    By Taylor's theorem we have near $\sqrt{\alpha}$
    \begin{align*}
        F(x) &= -(x-\sqrt{\alpha})+K_1(x-\sqrt{\alpha})^2 \
        G(y) &= -\frac{2}{1+\sqrt{\alpha}}(y-\sqrt{\alpha})+K_2(y-\sqrt{\alpha})^2
    \end{align*}
    Now $\sqrt{\alpha}-x$ is the distance from $x$ to $\sqrt{\alpha}$, so we see that as $x_n$ approaches $\sqrt{\alpha}$ the differences between the successive elements of the
    first sequence become very close to the distance to $\sqrt{\alpha}$, but not so much in the case of the second sequence, which has the additional factor of
    $2/\bigl(1+\sqrt{\alpha})$. In other words, the difference between the successive steps of the $x_n$ and the distance to $\sqrt{\alpha}$ is quadratic in $x-\sqrt{\alpha}$,
    while in the case of the second sequece the difference is only linear in $y-\sqrt{\alpha}$.

Exercise 5.24

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Exercise 5.24

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