1
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.