How would I determine a fundamental period in general for a trigonometric polynomial?
e.g.
$$f(x) = 3 - \cos x + \cos 2x + 4 \sin 2x - 3 \sin 5x$$
has fundamental period $2 \pi$. I think it's because each trigonometric term's lowest common period is $2 \pi $? How would I find out for general trig polynomials with different terms with non-integer "degrees"?
e.g.
$$f(x) = 3 - \cos x + \sin 0.29 x - \tan 9.5x$$
?
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Twenty-six colours
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One can still try to find the "greatest common divisor" of a set of real numbers $\{r_1,r_2,\dots\}$: if there exist any numbers $d$ at all such that $r_1/d,r_2/d,\dots$ are all integers, there will be a greatest one $g$. If this $g$ exists, then the fundamental period of $\sum a_j \cos(r_j x)$ (say) will be $2\pi/g$. In your example, for instance, we have $g=0.01$ and so the fundamental period is $400\pi$.
It is quite possible for this greatest common divisor not to exist: take for example $r_1=1$ and $r_2=\sqrt2$. In such circumstances, the corresponding function (like $\sin x+\sin(\sqrt2x)$) is not actually periodic.
Greg Martin
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Thanks. What conditions do we require for that a sum of periodic functions to be periodic? Their division must be rational (or something like that)? – Twenty-six colours Jun 02 '17 at 11:24
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It's true that a sufficient condition is that the individual periods are all rational multiples of some fixed nonzero real number (equivalently, that the $\Bbb Q$-vector space spanned by those periods has $\Bbb Q$-dimension $1$). There are silly examples, like $\sin x + \sin(\sqrt2x) - \sin x$, that show that such a condition isn't literally necessary; but it might be morally necessary. – Greg Martin Jun 02 '17 at 16:55