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Let $\varphi:\mathbb{R}\backslash\{3\}\to \mathbb{R}$ a periodic function so that forall $x\in \mathbb{R}$ $$\varphi(x+4)=\frac{\varphi(x)-5}{\varphi(x)-3}$$ Find the period the $\varphi$.

Ivan Neretin
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Roiner Segura Cubero
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  • Did you meant $\mathbb{R}\backslash{3}$? All real numbers excluding 3? – Ian Miller Oct 20 '15 at 02:33
  • Yes @Ian Miller – Roiner Segura Cubero Oct 20 '15 at 02:34
  • @IanMiller it is perfectly acceptable notation. I don't understand the downvote, this seems like it isn't a homework question. Unless it is trivial? – Almentoe Oct 20 '15 at 02:37
  • What downvote? I just was commented about the notation. – Ian Miller Oct 20 '15 at 02:41
  • @IanMiller Bad grammar, I did not mean that it was you. – Almentoe Oct 20 '15 at 02:45
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    You can't have a periodic function defined on all of $\Bbb R$ except at 3, other than artificially restricting it. The periodicity condition defines a value at $3$ as well. (I didn't downvote it, either.) – Paul Sinclair Oct 20 '15 at 02:46
  • @ Paul Sinclair: How about tan (x)? It is undefined at at all $\pi/2 + n\pi$, but otherwise periodic. – P Vanchinathan Oct 20 '15 at 02:50
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    @Paul my thought is due to the exclusion of 3 then we must exclude 7 and 11 and so on so that function must have a periodicity of 4 (or at least $\frac{4}{n}$). Not sure how to prove it. – Ian Miller Oct 20 '15 at 02:52
  • Perhaps a trivial observation might help: $f: \mathbb{R} \backslash { 3 } \to \mathbb{R} \backslash { 1 }, f(y) := \frac{y-5}{y-3}$ is invertible, with inverse $g(z) = \frac{3z-1}{z-1}.$

    I will continue with my investigation.

    I would invite someone to check this, however. I am up late with insomnia.

     – Almentoe Oct 20 '15 at 02:59
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    @PVanchinathan - The problem states that $\varphi$ is defined everywhere in $\Bbb R$ except at $3$, The statement makes no exceptions for periodic offsets of $3$. Either Ian is correct in his interpretation, or, as I suspect is more likely, Roiner was trying to exclude $\varphi(x) = 3$, since that would make the ratio undefined, but put the restriction on the wrong end. In either case, the problem would be incorrectly stated which is why I would like to see a clarification. – Paul Sinclair Oct 20 '15 at 03:04
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    Hint: if $f(x) = \frac{x-5}{x-3}$ then $f\circ f \circ f \circ f = x$ – Winther Oct 20 '15 at 03:21
  • @Winther How did you think of that?! – Almentoe Oct 20 '15 at 03:49
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    @Almentoe It's just a natural thing to try (from solving similar problems in the past). This approach is also outlined in the answer below. It does not solve the problem completely though as one must still do some work to find the smallest period. – Winther Oct 20 '15 at 03:57

2 Answers2

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With $f(x) = \frac{x-5}{x-3}$ the functional equation can be written

$$\varphi(x+4) = f\circ \varphi(x)$$

A direct calculation shows that $$f\circ f\circ f\circ f = \text{Id} \implies \varphi(x+16) = \varphi(x)$$ which implies that the period satisfy $T = \frac{16}{m}$ for some $m\in\mathbb{N}$. We can rule out $4\mid m$ as a possibillity since

$$f\circ \varphi(x) = \varphi(x+4) = \varphi\left(x+ T \cdot \frac{m}{4}\right) = \varphi(x)\implies \varphi(x) = 2\pm i \not\in\mathbb{R}$$

and we can also rule out $m \equiv 2\pmod 4$ by the same type of argument.

The remaining values $m\equiv \pm 1\pmod{4}$ are all possible and we can prove this by explcitly constructing solutions with the desired period. The form of the equation looks similar to the addition formula for $\tan(x+y)$ so taking the ansatz $\varphi(x) = A\tan(kx) + B$ in the functional equation and solving for $A,B,k$ we find the following family of solutions

$$\varphi(x) = \tan\left(\frac{\pi x}{T}\right)+2$$

where $T = \frac{16}{4n\pm 1}$ with $n\in\mathbb{N}$.

Winther
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  • Can you please elaborate how you obtained $φ(x+16)=φ(x)$ in second step? I understood till the last step, but not this one. Thank you! – Gaurang Tandon Jan 05 '18 at 02:02
  • @GaurangTandon First show that $f\circ f\circ f\circ f(t) = t$ then take $t = \varphi(x)$ and use $\varphi(x+4) = f\circ \varphi(x)$ over and over again to get: $\varphi(x) = f\circ f\circ f\circ f(\varphi(x)) = f\circ f\circ f\circ \varphi(x+4) = f\circ f\circ \varphi(x+8) = f\circ \varphi(x+12) = \varphi(x+16)$. – Winther Jan 05 '18 at 02:07
  • Understood. Thanks! Also, "which implies that the period satisfy" I think you are talking about fundamental period here? – Gaurang Tandon Jan 05 '18 at 02:29
  • And can you please detail how you arrived at the value $2\pm i$ near the middle? I have understood everything till that step, but where did you get that $2\pm i$ from? Thanks! – Gaurang Tandon Jan 05 '18 at 02:54
  • @GaurangTandon You get it by solving $f\circ \varphi(x) = \varphi(x)$ for $\varphi(x)$ (it becomes a quadratic equation). – Winther Jan 05 '18 at 10:01
  • Ok thanks for your help! – Gaurang Tandon Jan 05 '18 at 11:38
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Defining $f(x) = {x - 5\over x-3}$, we have $\varphi(x + 4) = f(\varphi(x))$. More generally, $\varphi(x + 4k) = f^{(k)}(\varphi(x))$. Assuming that the period is a multiple of $4$, for some $k$, $\varphi(x) = y = f^{(k)}(y)$.

I suspect the trick is to find a value of $k$ such that $y = f^{(k)}(y)$ has real solutions.

Paul Sinclair
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