Looking for a specific recursion relation

Question

I don't know how to do a search for information on a specific equation. It's \(\displaystyle f(n + 1) = 2 - \dfrac{d(n)}{f(n)}\), where d(n) is more or less arbitrary. It came up in some work I've been doing and I can't seem to get anywhere with it. Being non-linear it may not even have a closed form solution. There are two other ways to look at it. It's a non-linear difference equation: \(\displaystyle f \Delta f + f(f - 2) = d\) and it can also be considered as a continued fraction. (I'm going to be looking up that idea tonight.) Any thoughts? -Dan

Answer 1

topsquark said: Good thought but the problem is that most searches only turn up the linear equations. -Dan Click to expand... My suggestion seems to predominantly turn up fixed point recursion.

Answer 2

CaptainBlack said: I don't know if this will help, but I would search for something like "recursion x(n+1)=f(x(n))" Click to expand... Good thought but the problem is that most searches only turn up the linear equations. -Dan

Answer 3

topsquark said: I don't know how to do a search for information on a specific equation. It's \(\displaystyle f(n + 1) = 2 - \dfrac{d(n)}{f(n)}\), where d(n) is more or less arbitrary. It came up in some work I've been doing and I can't seem to get anywhere with it. Being non-linear it may not even have a closed form solution. There are two other ways to look at it. It's a non-linear difference equation: \(\displaystyle f \Delta f + f(f - 2) = d\) and it can also be considered as a continued fraction. (I'm going to be looking up that idea tonight.) Any thoughts? -Dan Click to expand... I don't know if this will help, but I would search for something like "recursion x(n+1)=f(x(n))"