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If $f(t)$ is periodic and $f(t) + C \cdot f(t + t_1)$ has the same shape of $f(t)$ for each value of $C$ and $t_1$, then $f(t)$ has the shape of a sine wave.

Is there a simple proof?

Is there an intuitive explanation? I mean without using, for example, the Fourier Transform.

Thanks in advance!

MrMazgari
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Andrew
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  • Hint: the addition property $\cos a\sin b + \cos b\sin a= sin(a+b)$. – paul garrett Nov 19 '15 at 22:56
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    @paulgarrett that accounts for the easier direction, but not the other. – Ben Grossmann Nov 19 '15 at 23:14
  • @Omnomnomnom, well, the hard direction is quasi-obvious from taking Fourier transforms, and I would not know how else to "quantify" over "functions" and address "shape" in "elementary" terms. Making the question precise in elementary terms would be necessary to know what the harder converse is, I think. In the end, I think using the fact that the Fourier transform of $\sin x$ is $(\delta_1-\delta_{-1})/2i$ is the most explanatory and arguably simple. – paul garrett Nov 19 '15 at 23:26
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    With the fourier transform is simple to prove but I wonder if there is a simpler prove for such simple and very special property of periodic functions. Is it possible to view the problem as the solution of a functional equation? (I am not a mathematician, sorry for the imprecisions) – Andrew Nov 19 '15 at 23:45

1 Answers1

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One way to describe what makes translating sine waves special is the following: sine waves $f(t)$ have the property that the subspace of the space of functions spanned by their translates $f(t + t_0), t_0 \in \mathbb{R}$ is $2$-dimensional, spanned by $\cos t, \sin t$. What other functions have this property? (I'm not going to require periodicity yet; this condition is already very restrictive.)

It turns out that the answer is the following. Let $V$ be the subspace spanned by translates, and let me assume that $f(t)$ is smooth. Because $V$ is finite-dimensional, it is in particular closed, so if $g(t) \in V$ is differentiable then $g'(t)$, which is a limit of functions in $V$, is also in $V$. Hence the functions $f(t), f'(t), f''(t) \in V$ must be linearly dependent, which means $f(t)$ satisfies a differential equation of the form

$$a f''(t) + b f'(t) + c f(t) = 0.$$

Solutions to differential equations of this form are very restricted. They are linear combinations of functions of the following forms:

  • $e^{rt}, r \in \mathbb{R}$
  • $t e^{rt}, r \in \mathbb{R}$
  • $e^{at} \cos bt, a, b \in \mathbb{R}$
  • $e^{at} \sin bt, a, b \in \mathbb{R}$

In particular, the only ones that are also periodic are sine waves.

Qiaochu Yuan
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  • Thanks for your answer. If I have understood, if (a) f(t) is any linear combination of two base functions and (b) f(t) is periodic then f(t) has a sine wave shape. Intuitively, what (a) means for f(t)? Why we should impose this constraint? What happen if we consider 3 base functions? – Andrew Nov 20 '15 at 15:20
  • @Andrew: it's the next best thing to the constraint that translates of $f(t)$ are scalar multiples of it, which means that $f(t) = e^{rt}$ for some $r$. That is, a condition of this form is a natural weakening of being invariant or invariant-up-to-scale under some symmetry. In the $3$-dimensional case we get a new class of functions, namely $t^2 e^{rt}$. – Qiaochu Yuan Nov 20 '15 at 17:19