5

$$2 + \sin(2\pi\cdot t) + 3\cos(3\pi\cdot t) - 5\sin(7t-\frac{\pi}{4})$$

Is there any manual, easy, way of knowing such a function is not periodic? I'd love to know if there's any method which doesn't require usage of computer.

I'm thinking since the period is defined as $T = \frac{2\pi}{\omega}$, where $\omega = \gcd{(2\pi,3\pi,7)} = \text{Undefined}$, this is the reason why the function isn't periodic? But is this always the case?

B. Lee
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  • i think it is because of 2. your reason is wrong see this link http://math.stackexchange.com/questions/681750/sum-of-two-periodic-functions-is-periodic – R.N Sep 01 '15 at 17:58
  • See http://math.stackexchange.com/questions/897987/principal-period-of-sin-frac3x4-cos-frac2x5 OR http://math.stackexchange.com/questions/873723/how-to-find-the-period-of-the-sum-of-two-trigonometric-functions – lab bhattacharjee Sep 01 '15 at 18:09

1 Answers1

3

Off the top of my head, I would say that the intersection of all of the sets of periods (I include trivial periods here) would have to contain a nonzero value. This would mean that each of the functions in the sum would share that value as a period, so the sum would as well (it may or may not be the minimal period).

For the first function, the set of periods is $\mathbb R$ (the function is constant).

For the second function, the set of periods is $\mathbb Z$.

For the third function, the set of periods is $\frac23\mathbb Z$.

For the fourth function, the set of periods is $\frac{2\pi}{7}\mathbb Z$.

The fourth function "ruins" it for the rest, since there is no rational multiple of $\pi$ that is also a nonzero integer. Note that $\mathbb Z\subset \frac23 \mathbb Z\subset\mathbb R$, so it comes down to the fact that $\mathbb Z\cap \frac{2\pi}{7}\mathbb Z = \{0\}$.

MPW
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