Find the limit using a calculator

Question

We have $u_0 = 6$ and $u_{n+1} = \dfrac{1}{2} u_n + \dfrac{1}{u_n}$. We can use our graphing calculator to make a 'web diagram' (no idea what it is called in English, and I can't find it. It sometimes resembles a spider's web).

When I use my calculator for very high values of n I get the same answer, $12.164$.

Is this the limit?
How would I be able to obtain this limit without the graphing calculator? Is it just the intersection with the line $y=x$?

Perhaps you have transcribed the function incorrectly. The one given converges to nothing near $12.164$ (it actually approaches $\sqrt{2}$). — George V. Williams, Jun 02 '13 at 23:48

Ma Ming · Accepted Answer · 2013-06-03T00:09:35.683

6

When $x>\sqrt{2}$, one has $\sqrt{2}<\frac{x}{2}+\frac{1}{x}<x$, then $u_0>u_1>\dots>\sqrt{2}$. So the series converges, say $u_n\to u$, then $u=\sqrt{2}$ by solving $\frac{u}{2}+\frac{1}{u}=u$.

(For completeness: if $0<u_0<\sqrt{2}$, then $u_1>\sqrt{2}$, the result will be the same; If $u_0<0$, then the limit is $-\sqrt{2}$ by symmetry.)

edited Jun 03 '13 at 00:09

answered Jun 02 '13 at 23:55

Ma Ming

7,482

1

+1 I like that you showed convergence (that the sequence does in fact converge to some limit $u$), and then revealed how to solve for the limit $u$. – amWhy Jun 03 '13 at 00:02

DonAntonio · Answer 2 · 2013-06-03T00:32:15.637

5

Assuming the sequence $\,\{u_n\}\,$ converges to a limit $\,u\,$ ,we get from arithmetic of limits:

$$u=\frac12u+\frac1 u\implies\frac12u=\frac1u\implies u^2=2\implies u=\sqrt 2$$

edited Jun 03 '13 at 00:32

answered Jun 02 '13 at 23:45

DonAntonio

211,718
17
136
287

Does that take into account that $u_0 = 6$? – StighurtAndersson Jun 02 '13 at 23:47
@StighurtAndersson Yep, $u_n>0$. – Ma Ming Jun 02 '13 at 23:48
Oh, it doesn't matter: it may as well be $,u_0=1029399^{10292};$ . The limit, if it exists, still is $,\sqrt 2,$ .Carefully calculate the first $;5-6;$ elements, the sequence very rapidly converges to that value (at least beginning with $,6,$) – DonAntonio Jun 02 '13 at 23:49
Yes, than I spotted the mistake, when I put it in my calculator I added a term $+6$. When I removed it, it gave me $\sqrt{2}$ too. – StighurtAndersson Jun 02 '13 at 23:50
The $\ge$ under Ind. Hyp. is not correct. – Ma Ming Jun 03 '13 at 00:17
I think it is, @MaMing...or else can you point at the mistake? Thanks. – DonAntonio Jun 03 '13 at 00:18
By induction hypothesis, $\frac{1}{4}u_n^2 \ge \frac{1}{2}$. – Ma Ming Jun 03 '13 at 00:22
Yes, I think I see your point @MaMing. Thanks, I shall try to edit that part. – DonAntonio Jun 03 '13 at 00:26
Indeed, it was hopeless... – DonAntonio Jun 03 '13 at 00:32
@DonAntonio $\frac{1}{2} u+\frac{1}{u}\ge 2\sqrt{\frac{1}{2} u\frac{1}{u}}=\sqrt{2}$. – Ma Ming Jun 03 '13 at 00:42
1

Yes, I know @MaMing, but that's how you did it, not how I tried...Thanks. – DonAntonio Jun 03 '13 at 00:45
let us continue this discussion in chat – Ma Ming Jun 03 '13 at 08:40
1

@DonAntonio: I completely agree, dear friend, what you say in your presentation. You're right, what you say it is true. – Piquito Jan 03 '16 at 10:53

score 3 · Answer 3 · answered Jun 03 '13 at 00:03

Just for fun, the OP was actually iterating $$u_n=\frac{1}{2} u_{n-1}+\frac{1}{u_{n-1}}+6$$ If this has a limit, it is found by $u=0.5u+\frac{1}{u}+6$ or $\frac{u}{2}-\frac{1}{u}-6=0$ or $\frac{u^2}{2}-6u-1=0$.

This has exactly one positive solution, namely $6+\sqrt{38}\approx 12.1644$.

score 0 · Answer 4 · answered Jan 03 '16 at 10:12

0

If you want to search, the common English term is Cobweb Diagram

answered Jan 03 '16 at 10:12

murkle

411

Find the limit using a calculator

4 Answers4