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Use the Banach fixed point theorem to show that the following sequence converges. What is the limit of this sequence? $$\left(\frac{1}{3}, \frac{1}{3+\frac{1}{3}}, \frac{1}{3+\frac{1}{3+\frac{1}{3}}}, \dots\right)$$

I inferred the closed form of the sequence $$\begin{align} x_0 &= \frac{1}{3} \\ x_n &= \frac{1}{3 + x_{n-1}} \end{align}$$

and then the function that would, I assume, be that which would be applied repeatedly to converge the sequence (per the Banach fixed point theorem) $$f(x)=\frac{1}{3+x}$$


That's about as far as I know to go for sure, but I assumed then that I needed to prove $f(x)$ was a contraction mapping (as the hypothesis of the Banach fixed point theorem dictates), that is

$$d(f(x),f(y)) \leq r d(x,y)$$ where $r \in [0, 1)$, and so

$$ \begin{align} \frac{1}{3+x} - \frac{1}{3+y} &\leq r(x - y) \\ \frac{1}{(x - y)(3 + x)} - \frac{1}{(x - y)(3 + y)} &\leq r \\ \frac{-1}{(x + 3)(y + 3)} &\leq r \end{align}$$

Although I don't think the above proves anything useful at all. (Unless $(x+3)(y+3) < -1$, where it is shown that $\exists r \in [0,1) : \dots$.)

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You're on the right track (though you forgot the rather crucial absolute value signs in your inequality). Try showing that $f$ is a contraction map on some subinterval of $\Bbb R$. (It isn't even defined at $-3$, so we have to consider a proper subinterval, anyway.) Since we're taking our initial point to be $\frac13$, we might as well consider $[0,\infty)$. We could instead use $[\alpha,\infty)$ for some $-3<\alpha<0$, if we liked, but $0$ works just fine.

Show that if $x\in[0,\infty)$ then $f(x)\in[0,\infty).$ Then for all $x,y\in[0,\infty)$, you can see that $$\left|\frac1{3+x}-\frac1{3+y}\right|=\left|\frac{y-x}{(3+x)(3+y)}\right|=\frac1{(3+x)(3+y)}|x-y|\le r|x-y|,$$ where $r=???$

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  • $\begingroup$ Thanks for noticing the absolute value stuff - I define a new command \abs#1 for that, and MathJax doesn't know how to define new commands ;) But since we know that $|x-y|\geq0$, we can safely simplify out the absolute value (except in the trivial case where $x=y$, where the inequality holds from the get-go). Then, since $x,y\geq0$, the fraction is strictly positive, satisfying $r\in [0,1)$. But what about actually finding its limit? $\endgroup$ – Sean Allred Apr 28 '13 at 11:12
  • $\begingroup$ But wait a minute ... there's no upper bound on $r$ here... $\endgroup$ – Sean Allred Apr 28 '13 at 11:15
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    $\begingroup$ Since $x,y\ge 0$, then $3+x,3+y\ge 3,$ so $(3+x)(3+y)\ge9$, and so $$\frac1{(3+x)(3+y)}\le\frac19.$$ That's how we get our $r$. As for finding the limit, remember that it will be the fixed point of $f$ in $[0,\infty).$ In particular, it will be the positive solution to $$\frac1{3+x}=x.$$ $\endgroup$ – Cameron Buie Apr 28 '13 at 15:11
  • $\begingroup$ Thanks for the clarification - it seems I was thinking about it all wrong. :) $\endgroup$ – Sean Allred Apr 28 '13 at 15:16
  • $\begingroup$ Glad to help out. $\endgroup$ – Cameron Buie Apr 28 '13 at 16:05
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Hint: Use the mean value theorem to prove your map $f(x)=\frac{1}{3+x}$ is contraction on the interval $[0,\infty)$. See here for detailed techniques.

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