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In Rudin's Principles of Mathematical Analysis (3e), the following exercise is given in the chapter about differentiation:

Exr.24 (p.118): 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 $$ f(x) = \frac{1}{2} \left( x + \frac{\alpha}{x} \right), \qquad g(x) = \frac{\alpha+x}{1+x}. $$ 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 16, Chap. 3, is so much more rapid than it is in Exercise 17. (Compare $f^\prime$ and $g^\prime$, draw the zig-zags suggested in Exercise 22.) Do the same when $0 < \alpha < 1$.

The exercise referred to above reads:

Exr.22 (p.117): Suppose $f$ is a real function on $(-\infty, \infty)$. Call $x$ a fixed point of $f$ if $f(x)=x$.

(c) If there is a constant $A < 1$ such that $\left| f^\prime(t) \right| \leq A$ for all real $t$, prove that a fixed point $x$ of $f$ exists, and that $x = \lim x_n$, where $x_1$ is an arbitrary real number and $$ x_{n+1} = f \left( x_n \right) $$ for $n = 1, 2, 3, \ldots$.

(d) Show that the process described in (c) can be visualized by the zig-zag path $$ \left( x_1, x_2 \right) \rightarrow \left( x_2, x_2 \right) \rightarrow \left( x_2, x_3 \right) \rightarrow \left( x_3, x_3 \right) \rightarrow \left( x_3, x_4 \right) \rightarrow \cdots.$$

(Credit to @Saaqib Mahmood for having written these exercises before).

I know that this question was asked before (see Prob. 24, Chap. 5 in Baby Rudin: For $\alpha>1$, let $f(x) = (x+\alpha/x)/2$, $g(x) = (\alpha+x)/(1+x)$ have $\sqrt{\alpha}$ as their only fixed point), however it didn't receive a satisfying explanation: I do not understand what is Rudin asking in Exr. 24.

I tried to compare the derivatives of the functions, using also the Mean Value Theorem in order to link them to the approximation we make computing $\sqrt{\alpha}$ by means of $x_n$, however, I only obtain a gross estimate of this error, which is too coarse to actually give any information on the rate of convergence (it is more precise to directly deal with the two functions).

I also tried to compare the "zig-zag" (as Rudin calls it) of the two functions and I have only noticed that $f$ creates a staircase whereas $g$ creates a spiral, but I can not draw any other conclusion from that.

Any help is highly appreciated as always!

Alp Uzman
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Matteo Menghini
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1 Answers1

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The special property of $f(x)$ that makes convergence of its iterates to the fixed point $\sqrt{\alpha}$ so much more rapid than for $g(x)$ is that $f'(x)=0$ at the fixed point. (This is called a super-attracting fixed point, in the field of discrete dynamical systems.) A rule of thumb: when a dynamical system has a super-attracting fixed point, the iterates are eventually very accurate approximations to the fixed point: the number of digits of accuracy eventually doubles with each iteration.

A good special case to explore is the super-attracting fixed point of the function $f(x)= x^2$ which has $x=0$ as its super-attractor.

P.S. The zig-zag diagram is usually called a cobweb diagram.

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