Continued function?

batman · #1 $Old$ September 23rd, 2009, 05:23 AM

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)

halbard · #2 $Old$ September 25th, 2009, 03:59 PM

Quote:

Originally Posted by CaptainBlack $View Post$

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.

CaptainBlack · #3 $Old$ September 26th, 2009, 03:01 AM

Quote:

Originally Posted by halbard $View Post$

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

CB