Mathematics

Peter Tamaroff

12:00 AM

Suppose $a>0$.

Let's just assume nothing, but only the initial condition $a_1>a_2$ or $a_1<a_2$

user45170

Ok

Peter Tamaroff

If $a_1=a_2$ then the sequence is just constant (check this)

So if $a_1>a_2$ then we have the following: $$\eqalign{
& {a_1} > {a_2} \cr
& {a_1} + 1 > {a_2} + 1 \cr
& \frac{1}{{{a_1} + 1}} < \frac{1}{{{a_2} + 1}} \cr
& {a_2} < {a_3} \cr} $$

user45170

You mean with a_1 the starting value, right? And a_2 the second value in the sequence?

Peter Tamaroff

@user45170 Yes.

$$\eqalign{
& {a_2} < {a_3} \cr
& {a_2} + 1 < {a_3} + 1 \cr
& \frac{1}{{{a_2} + 1}} > \frac{1}{{{a_3} + 1}} \cr
& {a_3} > {a_4} \cr} $$

You see the inequality sign changes every time.

$$\eqalign{
& {a_1} > {a_2} \cr
& {a_2} < {a_3} \cr
& {a_3} > {a_4} \cr
& {a_4} < {a_5} \cr
& \cdots > \cdots \cr
& \cdots < \cdots \cr} $$

And indeed what we have to prove is that this sequence has two monotone convergent subsequences, of even and odd terms. Then, we have to show they both converge to the asme limit.

In particular, one is monotone increasing and one monotone decreasing.

user45170

The monotonicity is clear for the subsequents

Peter Tamaroff

12:12 AM

@user45170 Proof?

user45170

Since $a_{2n}<=a_{2n+1}$ and $a_{2n-1}>=a_{2n}$

Peter Tamaroff

@user45170 What you want to prove is $a_{2n} > \text{ or }< a_{2n-2}$ and $a_{2n+1}> \text{ or }<a_{2n-1}$

user45170

We showed it for the first values, by induction it must be possible to show it for general n

Peter Tamaroff

@user45170 We didn't show that.

Victor

Hi, the math formula doesn't show up on my computer, can somebody help?

Peter Tamaroff

12:20 AM

@Victor Here

user45170

I am a little bit confused now, you also assume that a_1 can take rational numbers?

Victor

@PeterTamaroff - It seems not working on Google chrome browser...

Peter Tamaroff

@user45170 Why not?

user45170

Yes, you are right, it should not be a problem

Victor

@PeterTamaroff - Thanks, it works now

Hi, anyone could answer my question here: math.stackexchange.com/questions/222364/… ?

Peter Tamaroff

12:28 AM

@Victor It seems it is an application of a fixed point of a rational function.

When he says "quickly" he should say we gain one correct decimal per iteration, and that is basically because each iteration "powers by $1$".

Victor

@PeterTamaroff - Where can i find the reference for how this work?

Peter Tamaroff

@Victor Some research on functions of the type $$f(x)=\frac {ax+b}{cx+d}$$ should do. In the context of complex analysis these are called "Möbius Tranformations"

Note that the inverse of $f$ is also of that type, and composing two functions of these type produces also a function of this type.

user45170

I tried to show which inequality symbol < or > is right, but it is not that simple, may you could help

Peter Tamaroff

@user45170 I must confess I have not thought that out.

Victor

@PeterTamaroff - Thanks for the clues, But could you answer the question for me on the main site?

Peter Tamaroff

12:34 AM

@Victor Did he really get $100,000$ dollars? I mean, it is not even something that big.

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$Mathematics$ Mathematics