A function satisfying $f(\frac1{x+1})\cdot x=f(x)-1$ and $f(1)=1$? $f:[0,\infty)\to\mathbb{R}$ is a continuous function which satisfies $f(1)=1$ and:
$$f(\frac1{x+1})\cdot x=f(x)-1$$
Does there exist such a function, if they do, are there infinitely many? And is there any explicit example? I have been able to derive these:


*

*$f(1)\cdot0=f(0)-1$, which means $f(0)=1$.

*$f(\frac1{2})=f(1)-1=0$

*Similarly, $f(\frac23)=-2$ and then $f(\frac35)=\frac{-9}2$ and so on. 


This is going on without any pattern I can see. Is there anything special in the sequence $(1,\frac12,\frac23,...,a_n,...)$ where $a_n=\frac1{a_{n-1}+1}$? I have not been able to use continuity as well. Please help!

Such a function doesn't exist. 
Forget $f(1) = 1$. Just assume $f(x)$ satisfy the functional equation on $[0,\infty)$. We have
$$
\begin{align}
f(x) &= 1 + x f\left(\frac{1}{x+1}\right)\\
&= 1 + x \left[ 1 + \frac{1}{x+1}f\left(\frac{x+1}{x+2}\right)\right]\\
&= 1 + x \left[ 1 + \frac{1}{x+1} + \frac{1}{x+2} f\left(\frac{x+2}{2x+3}\right) \right]\\
&\;\vdots\\
&= 1 + x \left[ 1 + \left( \sum_{k=2}^{p-1} \frac{1}{F_{k-1} x + F_k} \right) + \frac{1}{F_{p-1} x + F_p}f\left(\frac{F_{p-1} x + F_p}{F_p x + F_{p+1}}\right)\right]\\
&\;\vdots
\end{align}
$$
where $F_k$ is the $k^{th}$ Fibonacci number. Notice 
$$\lim_{p\to\infty} \frac{F_{p+1}}{F_p} = \phi = \frac{1+\sqrt{5}}{2}
\quad\implies\quad
\lim_{p\to\infty} \frac{F_{p-1} x + F_p}{F_p x + F_{p+1}} = \frac{1}{\phi}
$$
If we further assume $f(x)$ is continuous at a single point $\displaystyle\;\frac{1}{\phi}$,
the last expression converges as $p \to \infty$ and we get:
$$f(x) = 1 + x + \sum_{k=2}^\infty \frac{x}{F_{k-1} x + F_k}$$
Substitute $x$ by $1$, we get
$$f(1) = 1 + 1 + \sum_{k=2}^\infty\frac{1}{F_{k-1}+F_k} = \frac{1}{F_1} + \frac{1}{F_2} + \sum_{k=3}^\infty \frac{1}{F_k}
= \sum_{k=1}^\infty \frac{1}{F_k}$$
The number at the RHS is known as the 
Reciprocal Fibonacci constant
$\psi$ with a value $$\approx 3.359885666243177553172011302918927179688905133731$$
differs from $1$. 
