Apparently equilivant series.

[the tree]

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

 

With the series [math]u_{n+1}=u_{n} - \frac{{u_n}^2 - 2}{2 u_n }[/math]

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 [math]u_n = \frac{p}{q}[/math] where p and q are integers. [math]u_{n+1}=\frac{(2p)p - 1}{(2pq)}[/math].

 

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 [math]u_n= \frac{p}{q}[/math] in your first formula:

[math]\frac{p}{q}-\frac{\frac{p^2}{q^2}-2}{\frac{2p}{q}}[/math]

Multiply numerator and denominator of the second fraction by q

[math]\frac{p}{q}- \frac{\frac{p^2}{q}-2q}{2p}[/math]

Get common denominator 2pq

[math]\frac{2p^2}{2pq}- \frac{p^2- 2q^2}{2pq}[/math]

and combine fractions

[math]\frac{p^2+2q^2}{2pq}[/math]

 

The denominator of that is exactly the same the other form

[math]\frac{2p^2-1}{2pq}[/math]

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

[math]p^2+ 2q^2= 2p^2- 1[/math]

That is the same as [math]2q^2= p^2-1[/math] 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.