Apparently equilivant series.

[the tree]

BEFORE YOU READ: turns out it was posssibly a false alarm, you may disregard.

 

With the series $u_{n+1}=u_{n} - \frac{{u_n}^2 - 2}{2 u_n }$

Taking u_0 to be 3/2

We get the series

3/2

17/12

577/408

 

I was working them out by hand (my calculator isn't sociable with fractions that require many digits) so only got that far before I got bored or something. I had these numbers in my MSN screen name without explanation so someone tried to guess the pattern.

 

He saw it as:

 

Where $u_n = \frac{p}{q}$ where p and q are integers. $u_{n+1}=\frac{(2p)p - 1}{(2pq)}$.

 

Now working with the idea that in maths, co-incidences are pretty damn rare, why are these two distinctive (looking) algorithms behaving exactly the same?

[river_rat]

How is the second recurrence well defined tree?

[the tree]

Oh, make that positive integers with no common factors then.

[Country Boy]

Let $u_n= \frac{p}{q}$ in your first formula:

$\frac{p}{q}-\frac{\frac{p^2}{q^2}-2}{\frac{2p}{q}}$

Multiply numerator and denominator of the second fraction by q

$\frac{p}{q}- \frac{\frac{p^2}{q}-2q}{2p}$

Get common denominator 2pq

$\frac{2p^2}{2pq}- \frac{p^2- 2q^2}{2pq}$

and combine fractions

$\frac{p^2+2q^2}{2pq}$

 

The denominator of that is exactly the same the other form

$\frac{2p^2-1}{2pq}$

so they will be the same as long as the starting values satisfy

$p^2+ 2q^2= 2p^2- 1$

That is the same as $2q^2= p^2-1$ which happens to be true for p/q= 3/2.

[the tree]

Ah, funky. I couldn't think why they would match generally, but matching only in that case makes a lot more sense, thanks.