Prove f is periodic

**mathfun** · March 31st 2011, 05:34 AM

Let $f:\mathbb{R} \to \mathbb{R}$ such that:
$\tfrac{{\sqrt 5 + 1}}{2}f\left( x \right) = f\left( {x + 1} \right) + f\left( {x - 1} \right)$ (1)

a) Prove that f is periodic
b) Prove that there are infinitely many functions such that (1) is true

**SpringFan25** · March 31st 2011, 11:18 AM

(a)

Edit major mistake in my working so ive removed it, sorry

(b) let f(x) be any function satisfying the condition. consider g(x)= kf(x)

**Opalg** · March 31st 2011, 12:41 PM

This doesn't explain why such a function has to be periodic, but here is an example of one such function (and sure enough it is periodic, with period 10).

Let $f(x) = \sin(\pi x/5)$ . Then $f(x\pm1) = \sin(\pi x/5 \pm \pi/5)$ , and $f(x+1)+f(x-1) = 2\sin(\pi x/5)\cos(\pi/5) = \frac{\sqrt5+1}2f(x)$ since $\cos(\pi/5) = \frac{\sqrt5+1}4.$

**veileen** · March 31st 2011, 01:19 PM

Well, I didn't make all the calculus, but:

$x\leftarrow x+1 \Rightarrow \frac{\sqrt{5}+1}{2}f(x+1)=f(x+2)+f(x)$
$\frac{\sqrt{5}+1}{2}f(x+1)=\frac{\sqrt{5}+3}{2}f(x )-\frac{\sqrt{5}+1}{2}f(x-1)$
$\frac{\sqrt{5}+3}{2}f(x)-\frac{\sqrt{5}+1}{2}f(x-1)=f(x+2)+f(x) \Rightarrow f(x+2)=\frac{\sqrt{5}+1}{2}(f(x)-f(x-1))$

Now calculate f(x+3), f(x+4)... and find the period. I am pretty sure this is a way it can be solved.

**mathfun** · March 31st 2011, 10:02 PM

Well i have tried many combinations but with no result. With certainty the period is 10 though. If anyone made it, please post a solution

**Opalg** · April 1st 2011, 12:23 AM

In fact, veileen's method works nicely. Let $\tau = \frac{\sqrt5+1}2$ . Then $\tau$ is the golden ratio, and satisfies the quadratic equation $\tau^2-\tau-1=0.$

The function f satisfies $f(x+2) = \tau f(x+1) - f(x)$ . Therefore

$\begin{aligned}f(x+3) &= \tau f(x+2) - f(x+1) \\ &= \tau \bigl(\tau f(x+1) - f(x)\bigr) - f(x+1) \\ &= (\tau^2-1)f(x+1) - \tau f(x),\end{aligned}$

$\begin{aligned}f(x+4) &= \tau f(x+3) - f(x+2) \\ &= \tau \bigl((\tau^2-1)f(x+1) - \tau f(x)\bigr) - \bigl(\tau f(x+1) - f(x)\bigr)) \\ &= (\tau^3-2\tau)f(x+1) - (\tau^2-1)f(x),\end{aligned}$

$\begin{aligned}f(x+5) &= \tau f(x+4) - f(x+3) \\ &= \tau \bigl((\tau^3-2\tau)f(x+1) - (\tau^2-1)f(x)\bigr) - \bigl((\tau^2-1)f(x+1) - \tau f(x)\bigr)) \\ &= (\tau^4-3\tau^2+1)f(x+1) - (\tau^3-2\tau)f(x).\end{aligned}$

Now you can check (using the quadratic equation for $\tau$ ) that $\tau^4-3\tau^2+1 = 0$ and $\tau^3-2\tau = 1$ . Thus $f(x+5) = -f(x)$ and so $f(x+10) = f(x).$

Edit. Looking again at veileen's comment, I notice that my solution would be much simpler if I had used the quadratic equation $\tau^2-\tau-1=0$ at each stage of the calculation, rather than saving it up until the end. In fact,

$f(x+3) = (\tau^2-1)f(x+1) - \tau f(x) = \tau\bigl(f(x+1)-f(x)\bigr)$ (because $\tau^2-1=\tau$ ),

$\begin{aligned}f(x+4) = \tau f(x+3) - f(x+2) &= \tau^2\bigl(f(x+1)-f(x)\bigr) - \bigl(\tau f(x+1) - f(x)\bigr) \\ &= (\tau^2-\tau)f(x+1) - (\tau^2-1)f(x) = f(x+1) - \tau f(x),\end{aligned}$

$f(x+5) = \tau f(x+4) - f(x+3) = \tau\bigl( f(x+1) - \tau f(x)\bigr) - \tau\bigl(f(x+1)-f(x)\bigr) = (\tau-\tau^2)f(x) = -f(x).$

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