# “Fill in the Gaps” Trig function with integer zeros another trig function doesn't have

I have an interesting challenge involving roots of trig functions. I'm wondering if there is a method of creating a function that hits the integer roots that $$\sin(\frac{\pi}{5}x)*\sin(\frac{\pi}{3}x)*\sin(\frac{\pi}{4}x)$$ doesn't (i.e. instead of 3, 4, 5, 6, 8, 9, 10, 12..., it has roots at 2, 7, 11...). A generalized method for doing so is more of the thing I am asking for rather than the specific answer to this problem. Thanks for all help and creative answers!

• rollback to previous version - vandalizing one's own question is a no-no on math.SE. If there is a real need, you can convince the mods to disassociate this question from you. – achille hui Nov 22 '18 at 4:29

The function $$\sin\left(\frac{\pi}{b}(x-a)\right)$$ has roots $$...,\ a - b,\ a,\ a + b,\ a + 2b,\ ...$$ You can combine the roots of functions together by taking their product. So if you can break the roots you want into finitely many arithmetic sequences, you can a find pretty simple function that has those numbers as its roots. In your example, $$\sin\left(\frac{\pi}{3} (x - 1)\right) \sin\left(\frac{\pi}{3} (x - 2)\right)$$ fills in the gaps.
• Very nice. When formatting, use "\sin" in mathjas to get $\sin$ instead of $sin$. – Ethan Bolker Nov 22 '18 at 1:03
• Thank you for the answer. The function I had in mind was a little more complicated, like $$\sin(\frac{\pi}{2}x)*\sin(\frac{\pi}{3}x)*\sin(\frac{\pi}{4}x)$$ – Ryan Shesler Nov 22 '18 at 2:07