# How to construct and oscillation with exponentially growing period times?

I'm searching for the (maybe even smooth) "oscillating" function

$$f(t)=A\sin{\left(g(t)\right)},$$

such that there are zeroes at times $t_n=T^n$ for some fixed number $T$. So this will not really be periodic, it will be a motion which makes one full turn at exponentially growing gaps, like for example

$$t_1=2\ \ \text{sec},\ \ t_2=4\ \text{sec},\ \ t_3=8\ \text{sec},\ \ t_4=16\ \text{sec},\ ...$$

Which function does that? Is there a corresponding Newtonian equation of motion?

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So you want $g(T^n) = n\pi$, $g(x) = \log_T x \cdot \pi$ should do ... – martini May 2 '12 at 12:34
Of course that one is not analytic at $x=0$. It is possible (but maybe a bit complicated) to get an analytic function with $g(T^n) = n \pi$ for all nonnegative integers $n$. – Robert Israel May 6 '12 at 7:38
@RobertIsrael: Isn't it more natural to start at $x=1$? – user26872 May 6 '12 at 18:40

$$y''(x)+y(x) = 0, \hspace{5ex} y(0) = 0, \hspace{5ex} y'(0) = 1$$ with solution $y(x) = \sin x.$
Change coordinates, let $x(t) = \log t$ and $f(t) = y( x(t)) = \sin(\log t)$. The zeros of $f(t)$ occur when $\log t_n = n \pi$, that is, when $t_n = (e^\pi)^n$. Notice that $$\frac{d}{d x} = \frac{d t}{d x} \frac{d}{dt} = t \frac{d}{dt} \hspace{5ex} \textrm{and so}\hspace{5ex} \frac{d^2}{dx^2} = \left(t \frac{d}{dt}\right)\left(t \frac{d}{dt}\right) = t^2 \frac{d^2}{dt^2} + t \frac{d}{dt}.$$ Therefore, the function $f(t)$ satisfies the differential equation $$t^2 f''(t) + t f'(t) + f(t) = 0, \hspace{5ex} f(1) = 0,\hspace{5ex} f'(1) = 1.$$
As indicated in the comments, if instead we choose $x(t) = \pi \log_T t$, the zeros will be at $t_n = T^n$. The differential equation satisfied by $f(t) = \sin(\pi \log_T t)$ is then $$t^2 f''(t) + t f'(t) + \left(\frac{\pi}{\log T}\right)^2 f(t) = 0,$$ We can think of this roughly as a harmonic oscillator with time-dependent restoring force.
Addendum: Find below a plot of $f(t) = \sin(\pi \log_2 t)$.