The roots of the sine wave should start at 1, then: 2, 4, 8, 16, 32, etc.
I have tried playing around with formulas such as:
y = sin(sqrt(x))
And I believe this is the key however I am not sure how to progress from here. Perhaps I am wrong though.
Best Answer
If you want your function to be a sine curve with roots at $2^j\ :\ j = 0, 1, 2, \ldots$ then you need to precompose $\sin(\cdot)$ with some function $g$ such that $g(2^j) = k\pi$ for some $k$.
Let's come up with one. What if we try for $g(2^j) = j\pi$? Then $g$ has to move $j$ from the exponent down to the coefficient, so let's try a logarithm: $\ln(2^j) = j\ln(2)$. Scaling the result to get things to work out yields: $$ g(x) = \frac{\pi}{\ln(2)} \ln(x)$$ which you can check has $g(2^j) = j\pi$.
Since $\sin(j\pi) = 0$ for all integers $j$, we must have that $$ \sin(g(2^j)) = \sin(j\pi) = 0 $$ so the function $x\mapsto \sin(g(x))$ satisfies your stated condition.