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 ?

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- Aug 14th 2006, 10:22 AMtizefunctionnal 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 AMThePerfectHacker
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 AMCaptainBlackQuote:

Originally Posted by**tize**

its context etc?

RonL - Aug 19th 2006, 11:12 AMtize
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 PMThePerfectHackerQuote:

Originally Posted by**tize**

$\displaystyle f(1/x)$ which usually are not solved for in functional equations. - Aug 19th 2006, 08:25 PMJameson
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 PMtize
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 $\displaystyle x\rightarrow\infty$, $\displaystyle f\rightarrow f(1)$