# Continued function?

• Sep 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)
• Sep 25th 2009, 02:59 PM
halbard
Quote:

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

\$\displaystyle y=f(y+f(y))\$

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

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

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

Then all you need to find are the fixed points of the function \$\displaystyle g(y)=f(2y)\$. This can sometimes be realised by a simple iteration \$\displaystyle y_{n+1}=g(y_n)\$ for a suitable starting value \$\displaystyle 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