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Given a finite summation of sines and cosines of the form $$\sum\limits_{i=1}^{n} \left(A_i\sin(\omega_ix)\right)+\sum_{j=1}^p \left(B_j\cos(\sigma_jx)\right)$$ where $A_i,\omega_i\in\mathbb{Z}$ $\forall i\in[1,n]\cap\mathbb{N}$ and $B_j,\sigma_j\in\mathbb{Z}$ $\forall j\in[1,p]\cap\mathbb{N}$. How do we prove that the least period of the summation is the LCM of the periods of the constituent sines and cosines?

The challenge I am having is in proving that the LCM of individual periods is in fact the smallest period of the resulting periodic summation.

user5402
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Sidd
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  • What are the "$\ldots$"? The first two terms you write has a sine and then a cosine; what comes next? – hmakholm left over Monica Sep 10 '15 at 23:57
  • It's not a true statement (at least without some restrictions on the parameters). The least period of $\sin x + \cos 99 x + (-1)\cos 99x$ is not $99$, but $1$. – Greg Martin Sep 11 '15 at 00:33
  • @HenningMakholm: The dots mean that there are finitely more terms, each either a sine or a cosine, with the coefficients of $x$ as described. – Sidd Sep 11 '15 at 00:36
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    @GregMartin: Actually it is $2\pi$. And that is, in that particular case, the least common multiple of the periods of the three terms, which are $2\pi$, $\frac{2\pi}{99}$, and $\frac{2\pi}{99}$. – hmakholm left over Monica Sep 11 '15 at 00:46
  • The periods are of the form $\frac{2\pi}{\omega_i}$ or $\frac{2\pi}{\sigma_j}$. Let $T$ be the LCM of these periods and $f(x)$ your function. It's straightforward to prove that $f(x+T)=f(x)$. – user5402 Sep 11 '15 at 00:59
  • @whatever: But the question asks for a proof of the fact that this is the 'least' period of the summation. – Sidd Sep 11 '15 at 01:00
  • Prove that $f\left(x+\frac{T}{n}\right)\not=f(x)$ for any natural number $n>1$. I'm too lasy to write a proof now. – user5402 Sep 11 '15 at 01:02
  • @whatever: Why does n have to be a natural number only? – Sidd Sep 11 '15 at 01:08
  • Fair enough, I should have written $\sin 2\pi x + \cos \frac{2\pi}{99} x + (-1) \cos \frac{2\pi}{99} x$; the LCM of the periods is $99$ but the least period is $1$. My point is that some extra assumptions are needed to make the statement true, so trying to prove it as is isn't going to work. – Greg Martin Sep 12 '15 at 06:54

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