Are we able to compute the probability of the given function to be positive?

Question

I have the following function $$ \cos \frac{n x}{2}\;\cos \frac{(m-n)x}{2} \;\bigg( 2\cos \frac{m x}{2} +\cos \frac{(2n-m)x}{2}\bigg) $$ with three parameters $m,n,x>0$ where $m>n$. These parameters are continuous and distributed uniformly.

Then, is it possible to compute the probability of which this function is negative or positive, at least for fixed values of $m$ and $n$, say $n=1$ and $m=3$?

In other words, we know that $\sin mx$ for a random value of $x$ is $50$ percent positive and $50$ percent negative, so the probability of being positive and negative is $\frac{1}{2}$. I mean can we say something like this for the given function?

Answer 1

You cannot have a uniform distribution over an infinite interval, so you seem to be looking for some limit of the proportion. It seems $m$ and $n$ (or at least their ratio) do make a difference: if $m=n$ then the expression is never negative, while otherwise it seems to be negative for a relatively small proportion of the possible $x$.

If $\frac mn$ is rational the expression is periodic and you can do calculations over a cycle. You might start by looking at when the expression is $0$, which are in effect double roots, and the gaps between the roots, noting that the expression is $3$ when $x=0$ and at the end of the cycle.

So for example if $m=3$ and $n=1$ then the cycle is of length $2\pi$, and the roots in $[0, \frac2\pi)$ are:

• $x=\cos^{-1}\left(\frac14\right)$
• $x=\frac12 \pi$
• $x=\pi$ (a double root)
• $x=\frac32 \pi$
• $x=2\pi-\cos^{-1}\left(\frac14\right)$

so the proportion of the cycle for which this expression is positive is $\frac12 + \frac1{\pi}\cos^{-1}\left(\frac14\right) \approx 0.91957$ and is negative is $\frac12 - \frac1{\pi}\cos^{-1}\left(\frac14\right) \approx 0.08043$