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Let $(u_n(x))$ be a sequence of functions in $(0,\infty)$ such that: $u_1(x)=x$, $u_{n+1}=\frac12\left(u_n(x)+\frac1{u_n(x)}\right)$ for $n\in\mathbb{N}$. Check if $u_n(x)$ converge unfiormly in $[a,b]$ when $0<a<b$.

I tried to show that $u_n(x)\to{x}$ by showing that $(u_n(x))$ is monotone, but didn't succeed. I would like to get a hint.

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Why do you need two different domains: $(0, \infty)$ and $[a, b]$? –  Tunococ Jan 30 '13 at 12:26
    
I don't think it converges to $x$, by the way. Consider what happens for $x=1/2$ and $x=2$. After one iteration, they are both equal, hence they'll converge to the same thing, which couldn't be the case if $u_n(x)\to x$. –  Clayton Jan 30 '13 at 12:30
    
I need to show uniform convergence only in the closed interval $[a,b]$ –  Gyt Jan 30 '13 at 12:31
    
I believe it converges to $1$. –  Tunococ Jan 30 '13 at 12:31
    
If it converges to a positive function, it converges to $y$ such that $y=(y+1/y)/2$, thus $y=1$. –  1015 Jan 30 '13 at 12:41

3 Answers 3

up vote 1 down vote accepted

Assuming $x > 0$. For each $x$, we can show that $u_n(x) \to 1$ as $n \to \infty$ as follows:

If $u_0(x) < 1$, then $u_1(x) > 1$.

If $u_n(x) \ge 1$ for any $n$, then $u_{n+1} - u_n = \frac 12\left(\frac 1{u_n(x)} - u_n(x)\right) < 0$ and $u_{n+1}(x) = \frac 12\frac{(1-u_n(x))^2}{u_n(x)} + 1$. Therefore, the sequence $u_n(x)$ is monotonically decreasing and bounded below by $1$ when $n > 0$, hence convergent.

We can now take the limit $n \to \infty$ in the relation $u_{n+1}(x) = \frac 12\left(u_n(x) + \frac 1{u_n(x)}\right)$ to get $u(x) = \frac 12\left(u(x) + \frac{1}{u(x)}\right)$ where $u(x)$ is the pointwise limit of $u_n(x)$. The equation is quadratic: $$ u(x)^2 - 1 = 0. $$ However, we already know that $u_n(x) \ge 1$ for all $x > 0$, so $u(x) = 1$.

Next, we show that the convergence is uniform in any closed interval $[a, b]$ inside $(0, \infty)$. Let $s_n = \sup_{x\in[a,b]}u_n(x)$ for $n > 0$. (We use the fact that $x \in [a, b]$ here to show that $s_1 < \infty$.) We see that for $n > 0$, $s_n \ge 1$ and \begin{align*} s_{n+1} & = \frac 12\sup_{x\in[a,b]}\left(u_n(x) + \frac{1}{u_n(x)}\right)\\ & \le \frac 12 s_n + \frac 12 \\ s_{n+1} - 1 & \le \frac 12 (s_n - 1). \end{align*} Hence $s_n \to 1$ as $n \to \infty$. We can now conclude that $u_n \to 1$ uniformly.

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Study the function $$ f(t)=\frac{1}{2}\left(t+\frac{1}{t} \right) $$ on $(0,+\infty)$ first. It decreases on $(0,1]$ and it increases on $[1,+\infty)$. There is a unique golbal minimum $f(1)=1$.

Then observe that $$ g(t)=f(t)-t=\frac{1-t^2}{2t} $$ is positive on $(0,1)$ and negative on $(1,+\infty)$.

For each fixed $x>0$, the sequence $u_n(x)$ is defined recursively by $u_1(x)=x$ and $u_{n+1}(x)=f(u_n(x))$. The behavior of this sequence depends on the position of $x$ with respect to $1$.

If $x=1$, then it is a fixed point of $f$ so $U_n(1)=1$ for all $n$ and it converges to $u(1)=1$.

Now fix $x>1$. First prove by induction that $U_n(x)>1$ for all $n$, using the study of $f$ above. Next deduce from the study of $g$ that $u_{n+1}(x)=f(u_n(x))<u_n(x)$ for all $n$. It follows that the sequence $(u_n(x))$ is decreasing and bounded below by $1$, so it converges to $u(x)\geq 1$. Now $u(x)$ has to fulfill the condition $u(x)=f(u(x))$ (by passing to the limit in $u_{n+1}(x)=f(u_n(x))$). This yields the quadratic equation $u(x)^2-1=0$, so $u(x)=1$.

Now fix $x<1$. We have $u_1(x)=x<1$. But $u_2(x)=f(u_1(x))=f(x)>1$. Now apply the previous paragraph to $f(x)$ instead of $x$. This shows that $U_n(x)$ starts decreasing from rank $n=2$ and it converges to $1$.

So we have shown that $u_n$ converges pointwise to the constant function $u(x)=1$.

Uniform convergence, now.

Let us first consider that case $[a,b]=[1,b]$ with $b\geq 1$. Fix $1\leq x\leq y$. Recall that $u_n(x)$ and $u_n(y)$ both belong to $[1,+\infty)$ for all $n$, from the convergence study above. Also, $f$ is increasing on $[1,+\infty)$. At the initial step, we have $1\leq U_1(x)=x\leq y=u_1(y)$. So $1=f(1)\leq u_2(x)=f(u_1(x))\leq u_2(y)=f(u_2(y))$. Therefore, a proof by induction shows that $1\leq u_n(x)\leq u_n(y)$ for all $n$ In particular, we have $$ 1\leq u_n(x)\leq u_n(b) $$ for all $x\in[ 1,b]$, and for all $n$. Uniform convergence on $[1,b]$ follows by the Squeeze Theorem.

Now on $[a,1]$ for $a\leq 1$. Take $a\leq x\leq 1$, i.e. $u_1(a)\leq u_1(x)\leq 1$. Apply f, which is decreasing there. This yields $1\leq U_2(x)\leq u_2(a)$. From now on, everything behaves like in the previous paragraph. So $$ 1\leq U_n(x)\leq u_n(a) $$ for all $x\in[a,1]$, for all $n\geq 2$. By the Squeeze Theorem, we have uniform convergence on $[a,1]$.

The general case $[a,b]$ follows easily from the cases $[a,1]$ and $[1,b]$.

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It's obvious that $u_n(x)$ converges pointwise to $1$ in $[a,b]$.
Consider each $x_0\in[a,b]$, as $u_n(x_0)\geq1$ and ${u_n(x_0)}$ decreasing when $n\geq 2$ ($u_{n+1}(x_0)-u_n(x_0)=\frac{1}{2}(\frac{1}{u_n(x_0)}-u_n(x_0))\leq0$), we have $u_n(x_0)$ converge. Easily, we can get it converges to $1$. What's more, it shows $u_n(x)\ge u_{n+1}(x)\ (\forall x\in[a,b]\ \text{and}\ n\ge2)$.
Applying Dini's Theorem, we have $u_n(x)$ converge unfiormly to $1$ in $[a,b]$.
Q.E.D.

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