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(I had asked in this question how should I begin to add together these two trigonometric functions: $z(t) = 3\cos(3t + \pi/5) + 4\cos(4t+\pi/8)$, in order to obtain the period of their sum). Now, I wonder how I might do so, should the angular frequency of the first be altered from $3$ to $\sqrt{10}$: $$z(t) = 3\cos\left(\sqrt{10}t + \frac\pi5\right) + 4\cos\left(4t+\frac\pi8\right)$$

which I think to be considerably more difficult, for obeys no longer the relation $z(t+2π) = z(t)$.

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V.E.
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  • The answer is the same, just with $\sqrt{10}$ instead of $3$. One difference is that the function is no longer periodic, but periodicity was nowhere used or assumed in that answer. – dxiv Jul 17 '21 at 05:15
  • I do still think the resultant function to be periodic; and I have plotted it on Desmos to confirm it. – V.E. Jul 17 '21 at 05:23
  • What do you think the period could be? It's not, see Period of sum of sinusoids for example. – dxiv Jul 17 '21 at 05:29
  • I think it periodic for no better reason than that is is formed of the sum of two periodic functions; and that the graph of it shows a regular pattern, which repeats itself again every $~7.754+0.13 = 7.88$s. (But as to how I should have obtained this number algebraically, I know not...) – V.E. Jul 17 '21 at 05:38
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    Trust the math more than your eyes ;-) A linear combination $\cos(at)+\cos(bt)$ is periodic iff $a/b \in \mathbb Q$, and the same applies to your function $-$ but $\sqrt{10}/4 \not\in\mathbb Q$. You cannot reliably tell periodicity from a graph, any more than you can tell whether a Lissajous curve is mathematically closed on the oscilloscope. – dxiv Jul 17 '21 at 05:42

2 Answers2

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The given function $z(t) = 3\cos\left(\sqrt{10}t + \frac\pi5\right) + 4\cos\left(4t+\frac\pi8\right)$ is not a periodic function!

If $L > 0$ is a period of $z$, then $L$ is a period of $z'' + \alpha^2 z$ for any $\alpha$. In particular, if one set $\alpha$ to $\sqrt{10}$ and $4$, one find $L$ is a period of

$$z'' + 10z = -24\cos(4t + \frac{\pi}{8}) \quad\text{ and }\quad z'' + 16z = 18\cos(\sqrt{10}t+ \frac{\pi}{5})$$ This forces $L$ to be integer multiples of $\frac{2\pi}{4}$ and $\frac{2\pi}{\sqrt{10}}$ and as a corollary, $\frac{\sqrt{10}}{4} = \frac{2\pi}{4} / \frac{2\pi}{\sqrt{10}}$ is rational. Since this isn't the case, $L$ doesn't exist and $z$ is not periodic.

achille hui
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Let $ T$ be the period then $z(t+T)=z(t)$ $$3\cos(\sqrt{10}t+\frac{\pi}{5}+\sqrt{10}T)+4\cos(4t+\frac{\pi}{8}+4T)=3\cos(\sqrt{10}t+\frac{\pi}{5})+4\cos(4t+\frac{\pi}{8})$$
$\exists p,m $ all integers such that
$\sqrt{10}T=2\pi p,4T=2\pi m$

$$ \sqrt{10}T=\frac{4T}{m}p$$ $$\sqrt{10}=\frac{4}{m}p$$ so $\sqrt{10}$ is rational number impossible
the giving function has no period

IDRISS
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