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Find all functions $f:\mathbb R \to \mathbb R$ that satisfy: $$ f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x.$$

Some progress: I plugged-in $\dfrac{x-1}{x}$ and $\dfrac{1}{1-x}$, got a system of equations and solving I got $f(x) = \dfrac{x^3-4x^2-3x-3}{4x^2-4x}$. But after testing it back looks like it's not sufficient.

Pikachuu.
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  • I don't have any non-trivial progress. – Pikachuu. Jul 22 '14 at 18:15
  • Now, I plugged-in $\dfrac{x-1}{x}$ and $\dfrac{1}{1-x}$. I got a system of equations and solving I got $f(x) = \dfrac{x^3-4x^2-3x-3}{4x^2-4x}$. But after testing it back looks like it's not sufficient. Is this correct? – Pikachuu. Jul 22 '14 at 18:25
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    The issue here is that you want a function from $\mathbb{R} \to \mathbb{R}$, your function does not exist at zero. – Joel Jul 22 '14 at 18:45
  • Ok... so what can we do? – Pikachuu. Jul 22 '14 at 19:01

1 Answers1

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Set $T(x) = \frac{x}{x-1}$. Then the composition $T^3(x) = x$. We have $f(Tx) = a x + b f(x)$ for $ a= \frac{7}{3}$ and $b = -\frac{1}{3}$, so \begin{align*} f(x) &= f(T^3(x)) \\ &= a T^2(x) + b f(T^2 x) \\ &= a T^2(x) + a b T(x) + b^2 f(Tx) \\ &= a T^2(x) + a b T(x) +a b^2 x + b^3 f(x). \end{align*} That relation gives $f$ as a rational function of $x$.

anomaly
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  • Wait so what is f(x)? – Pikachuu. Jul 22 '14 at 19:59
  • I like this solution a lot. I believe any function satisfying the functional equation must also satisfy your list of equalities. However, I wonder if your rational function will have a pole at $x=1$, since we have $T(x)$ isolated here. In which case, there is no solution since the question calls for a function defined on all of $\mathbb{R}$. – Joel Jul 22 '14 at 20:38
  • @Pikachuu: Just solve for $f$ above. – anomaly Jul 22 '14 at 20:50
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    @Joel: The functional equation for $f$ isn't defined at $0$, so presumably it's only required to hold outside the orbit ${0, 1, \infty}$ of $0$ under $T$. Choosing arbitrary $f(0), f(1)$ with $f(1) + 3f(0) = 7$ should be enough. (I don't know, not having worked out the explicit form for $f$, where the resulting function would be continuous.) Alternatively, we could just only require that $f$ be well-defined away from poles. It seems like the question should really about a function on the Riemann sphere, to neatly avoid issues of the functional equation or $f$ itself not being defined. – anomaly Jul 22 '14 at 20:55
  • You must show it is sufficient, I believe there are no solutions – Pikachuu. Jul 22 '14 at 20:55
  • @anomaly Is it chance/trials that led you to note that $T^3(x)=x$, or is there some kind of technique/rationale behind it ? – Gabriel Romon Jul 23 '14 at 06:33
  • @G.T.R: I figured some iterate would be trivial for three reasons: (a) the problem probably wouldn't have a nice solution otherwise (you'd just break $\mathbb{R}$ into separate orbits, pick an arbitrary basepoint, and wind up with a recurrence relation that probably wouldn't have a nice closed-form solution); (b) I remembered that $PSL(2, \mathbb{Z})$ is an amalgamated free product of finite groups with generators $\sigma(z) = z + 1$ and $\tau(z) = -1/z$, and $f =\sigma \tau$; and (c) plugging $T^3(x)$ into Wolfram Alpha showed that it was just $x$. :) – anomaly Jul 23 '14 at 08:04