# Continued function?

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• September 23rd 2009, 04:23 AM
batman
Continued function?
I want to solve y=f(y+f(y+f(y+f(y+...)))), where f(x) is some bounded function increasing in x.

First, what is the name of this kind of problem? I know of continued fractions, but here f(x) isn't of the form f(x)=1/(a+x).

Second, how can I find y? (numerically is fine)
• September 25th 2009, 02:59 PM
halbard
Quote:

Originally Posted by CaptainBlack
If this has a solution for a given function f, then:

$y=f(y+f(y))$

Hang on, if $y=f(y+f(y+f(y+f(y+\dots))))$ isn't $y=f(2y)$?

Then all you need to find are the fixed points of the function $g(y)=f(2y)$. This can sometimes be realised by a simple iteration $y_{n+1}=g(y_n)$ for a suitable starting value $y_0$, but you shouldn't rule out the need for more advanced methods.
• September 26th 2009, 02:01 AM
CaptainBlack
Quote:

Originally Posted by halbard
Hang on, if $y=f(y+f(y+f(y+f(y+\dots))))$ isn't $y=f(2y)$?

Then all you need to find are the fixed points of the function $g(y)=f(2y)$. This can sometimes be realised by a simple iteration $y_{n+1}=g(y_n)$ for a suitable starting value $y_0$, but you shouldn't rule out the need for more advanced methods.

Yes, now you mention it. That is what I started with but for some reason changed it to what I had posted (Shake)

CB
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