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!

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    $\begingroup$ 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. $\endgroup$ – 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.

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    $\begingroup$ Very nice. When formatting, use "\sin" in mathjas to get $\sin$ instead of $sin$. $\endgroup$ – Ethan Bolker Nov 22 '18 at 1:03
  • $\begingroup$ @EthanBolker Thank you. $\endgroup$ – WhatToDo Nov 22 '18 at 1:04
  • $\begingroup$ Thank you for the answer. The function I had in mind was a little more complicated, like $$\sin(\frac{\pi}{2}x)*\sin(\frac{\pi}{3}x)*\sin(\frac{\pi}{4}x)$$ $\endgroup$ – Ryan Shesler Nov 22 '18 at 2:07
  • $\begingroup$ @RyanShesler That function doesn't do what you said you wanted. It doesn't have a root at x=5, for example. $\endgroup$ – WhatToDo Nov 22 '18 at 2:10
  • $\begingroup$ Sorry I meant like this function has roots at 2, 3, 4, 6, 8, 9, 10. What's a way of finding another function with roots at 1, 5, 7, 11... Almost like a generalization of your answer to functions with more than one 'a' $\endgroup$ – Ryan Shesler Nov 22 '18 at 2:14

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