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I'm stuck on this seemingly easy question, hoping someone can explain to me how to prove it.

Let $$x_1 = \frac{a^2+b^2+2}{2(a+b)}$$

and for $n > 1$ define

$$x_{n+1} = \frac{x_n^2+1}{2x_n}=\frac{x_n}{2}+\frac{1}{2x_n}$$

Then the question asks to show that $x_n \to 1 \text{ or } -1$ depending on whether $a+b$ is positive or negative respectively.

I started by rewriting $x_1$ as

$$x_1 = \text{sign}(a+b)\left( \frac{a^2+b^2+2}{2|a+b|}\right) $$

Inserting this into the above formula for $x_n$, $n>1$, gives us:

$$x_2 = \text{sgn}(a+b)\left( \frac{(a^2+b^2+2)^2+|a+b|^2}{4|a+b|(a^2+b^2+2)} \right)$$

And it seems like putting this back in to get $x_3$ will just become very complex, let along obtaining a general formula for $x_n$.

Am I missing an easier way to go about this?

potato12
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2 Answers2

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Hint: If it converges, call the limit $x$. Then $x=x/2+1/(2x)\implies (1/2)x^2=1/2\implies x=\pm 1$.

To show it converges, try to show it's monotone and bounded. Look at $f(x)=x/2+1/(2x)$. Check the derivative and initial values.

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Yes: the recurrence relation is defined by a continuous function its domain $\mathbf R^*$. So, if the sequence is well-defined and converges, its limit $\ell$ is a fixed point of of the function, i.e. it satisfies the equation $$\ell==\frac12\Bigl(\ell+\frac1\ell\Bigr)\iff \ell=\frac 1\ell\iff\ell^2=1$$ Note this sequence is indeed well-defined, which means no term is $0$, because $x_{n+1}=0$ if and only if $x_n^2+1=0$, and we're in $\mathbf R$.

To determine whether the limit is $1$ or $-1$, it suffices to observe that all terms of the sequence has the same sign, which therefore will be the sign of ˆ$x_1$, i.e. the sign of $a+b$. Indeed the recurrence relation can rewritten as $$x_nx_{n+1}=\frac12(x_n^2+1)>0.$$ Note: this relation comes from the relation in Newton's algorithm to obtain the square root of a positive number $a$: $$x_{n+1}=\frac12\Bigl(x_n+\frac a{x_n}\Bigr).$$

Bernard
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  • The point about sign determining $\pm 1$ makes perfect sense to me (thanks a lot for your explanation!), but I still don't understand how to establish it indeed converges to $1$ or $-1$. Is this a consequence of them being fixed points? And if so, what conditions should I need to verify on $f(x)$? Apologies for my lack of understanding here. For example, is it a result of continuity of $f(x)$? – potato12 Sep 22 '19 at 12:00
  • Actually, I took for granted it converges (because of Newton's algorithm). Continuity only implies the limit is a fixed point, provided it converges. To prove convergence, you can use the monotone convergence theorem:if the sequence is positive, it is decreasing for $ x_1>1$, increasing for $0<x_1<1$. Symmetric result if $x_1<0$. – Bernard Sep 22 '19 at 12:13
  • Many thanks! I will try that – potato12 Sep 22 '19 at 12:18
  • B;t.w. it's easier to see it graphically: a sequence defined by $x_{n+1}=f(x_n)$ is increasing if the terms of the sequence are all above the straight line $y=x$, decreasing if they're all below. So you have to study and plot the defining function. – Bernard Sep 22 '19 at 12:26