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?

How is the second recurrence well defined tree?

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

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.

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