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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.

BFI01
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  • Do you know how to compute the wavelength of a sinusoid? – Brian61354270 Feb 10 '20 at 20:59
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    $\sin\left(\frac{\pi\ln x}{\ln2}\right)$? It has roots at $2^n$ for all $n$, behaves badly near zero though. https://www.desmos.com/calculator/sflp0xbtr9 – krc Feb 10 '20 at 20:59

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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.

Neal
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