# functionnal equation

• Aug 14th 2006, 10:22 AM
tize
functionnal equation
Hello !
I have a problem to solve this question :
Find all real continuous functions f verifing : f(x+1)=f(x)+f(1/x)
Have you ever seen this, could you help me please ?
• Aug 14th 2006, 11:49 AM
ThePerfectHacker
There is something that bothers me. You say that [tex]f[tex] is countinous on the number line. Yet, it is undefined for any negative integer!
• Aug 19th 2006, 03:16 AM
CaptainBlack
Quote:

Originally Posted by tize
Hello !
I have a problem to solve this question :
Find all real continuous functions f verifing : f(x+1)=f(x)+f(1/x)
Have you ever seen this, could you help me please ?

Can you tell us some of background to this equation,
its context etc?

RonL
• Aug 19th 2006, 11:12 AM
tize
I would like but I don't know anything about this function except that it is continuous. Somebody ask me about such a function and I don't know where does it come from ...
• Aug 19th 2006, 04:46 PM
ThePerfectHacker
Quote:

Originally Posted by tize
I would like but I don't know anything about this function except that it is continuous. Somebody ask me about such a function and I don't know where does it come from ...

What makes it unusual that is a composition of,
$\displaystyle f(1/x)$ which usually are not solved for in functional equations.
• Aug 19th 2006, 08:25 PM
Jameson
Like PH said, this has a cyclical property to it and try replacing all x's with 1/x. The right side of the equation yields the same thing, showing that $\displaystyle f(\frac{1}{x}+1)=f(x+1)$.
• Aug 20th 2006, 11:57 PM
tize
Yes we can already proove that :
$\displaystyle f(0)=f(\Phi^{-1})=f(-\Phi)=0$ where $\displaystyle \Phi$ is the Gold number : $\displaystyle \frac{1+\sqrt 5}{2}$
and
$\displaystyle f$ has a limit when $\displaystyle x\rightarrow\infty$, $\displaystyle f\rightarrow f(1)$