Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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?

share|cite|improve this question
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
up vote 5 down vote accepted

Consider the equation of motion for a simple harmonic oscillator,
$$y''(x)+y(x) = 0, \hspace{5ex} y(0) = 0, \hspace{5ex} y'(0) = 1$$ 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 $f(t) = \sin(\pi \log_2 t)$.

Plot of $f(t) = \sin(\pi \log_2 t)$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.