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Let $f\colon \Bbb R\to \Bbb R-\{3\}$ be a function with the property that there exist $T>0$ such that $f(x+T)=\frac{f(x)-5}{f(x)-3}$ for every $x\in \Bbb R$. Prove that $f(x)$ is periodic and find the period of $f(x)$.

For $f(x)$ to be periodic, we need to prove $f(x+\lambda)=f(x)$, where $\lambda$ is the fundamental period of $f(x)$. But there seems no path to prove this in the given definition of the function. Please help me.

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Well I seem to have found something $$ f(x)=\frac{f(x)-5}{f(x)-3} $$ replace $ x \rightarrow x+T $ to get $$ f(x+2T)=\frac{2f(x)-5}{f(x)-2} $$ now once again replace $ x \rightarrow x+2T $ to get $$ f(x+4T)=f(x)$$

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Hint: Find expressions of the form $f(x+kT)=\frac{a_kx+b_k}{c_kx+d_k}$, for $k=0$ (trivial), $k=1$ (given), $k=2$, $k=3$, and so on. You'll be surprised (soon enough).

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  • $\begingroup$ ,i did not get it Sir,it may be easy but not clicking me.I tried your suggestions on paper. $\endgroup$ – Brahmagupta Sep 25 '15 at 15:56

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