Is there any special reason that we use the sines and cosines functions in the Fourier Series, while we know that if we chose any maximal orthonormal system in L2, we would get the same result? Is it something historical or what?
Thanks in advance.
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2$\begingroup$ The sine and cosine are quite natural from the perspective of sound and frequencies. Otherwise, I think it's more natural to take a basis of complex exponentials. $\endgroup$– Qiaochu YuanCommented Jan 14, 2010 at 16:03
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$\begingroup$ Actually by sines and cosines, I had the basis of complex exponentials in my mind. Thank you. $\endgroup$– AxiomCommented Jan 14, 2010 at 16:06
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9$\begingroup$ OH. Well, in that case, the basis of complex exponentials is the one that diagonalizes the derivative. $\endgroup$– Qiaochu YuanCommented Jan 14, 2010 at 16:13
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6$\begingroup$ IMO Terry Tao's article on the Fourier Transform in the Princeton Companion to Mathematics is perhaps the best low-tech response to your question. The high-tech response is that there is a more general theorem called the Peter-Weil Theorem that identifies the relevant basis of L^2 for a compact lie group, apply it to the unit circle and you get $\sin$ and $\cos$ (it also tells you how you can find other basis). $\endgroup$– Ryan BudneyCommented Jan 14, 2010 at 16:21
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9$\begingroup$ @Ryan, I believe that it is weyl, not Weil. $\endgroup$– AnweshiCommented Jan 14, 2010 at 16:29
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4 Answers
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$1$. Mathematical reason.
There is one reason which makes the basis of complex exponentials look very natural, and the reason is from complex analysis. Let $f(z)$ be a complex analytic function in the complex plane, with period $1$.
Then write the substitution $q = e^{2\pi i z}$. This way the analytic function $f$ actually becomes a meromorphic function of $q$ around zero, and $z = i \infty$ corresponds to $q = 0$. The Fourier expansion of $f(z)$ is then nothing but the Laurent expansion of $f(q)$ at $q = 0$.
Thus we have made use of a very natural function in complex analysis, the exponential function, to see the periodic function in another domain. And in that domain, the Fourier expansion is nothing but the Laurent expansion, which is a most natural thing to consider in complex analysis.
Here you can make suitable modifications when $f$ is periodic in some domain which is not the whole complex plane. In that case in the $q$-domain, $f$ will be analytic in some circle around $0$, and you can use that to get a Laurent expansion. The modular forms for instance are defined only in the upper-half plane, and what we get here is called the $q$-expansion.
However from the point of view of Real analysis, $L^p$-spaces etc., any other base would do just as fine as the complex exponentials. The complex exponentials are special because of complex analytic reasons.
$2$. Physical reason.
There are historical reasons also. For instance, in electrical engineering or theory of waves, it is very useful to decompose a function into its frequency components and this is the reason for the great importance of Fourier analysis in electrical engineering or in electrical communication theory. The impedance offered by circuits depends on the frequency of the signal that is being fed in, and a circuit consisting of capacitors, inductors etc. react differently to different frequencies, and thus the sine/cosine wave decomposition is very natural from a physical point of view. And it was from this context, and also the theory of heat conduction, that Fourier analysis developed up.
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8$\begingroup$ The physical reason is a mathematical reason. In a basis of complex exponentials the derivative is diagonalised, and so in this basis the properties of resistors, capacitors and inductors can be described by elementary linear equations rather than by differential equations. $\endgroup$ Commented Jan 14, 2010 at 18:35
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$\begingroup$ Oh that is good to know. I always wondered why. Thanks for pointing out. $\endgroup$– AnweshiCommented Jan 14, 2010 at 19:29
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$\begingroup$ Maybe you or Qiaochu should write that down as an answer? $\endgroup$– AnweshiCommented Jan 14, 2010 at 19:30
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Nobody mentioned that $e^{2\pi inz}$ are the characters of $S^1$.
answered Mar 7, 2010 at 20:17
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1$\begingroup$ That's implicit in mentioning the Peter-Weyl theorem, I guess. $\endgroup$ Commented Mar 8, 2010 at 3:08
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Sines and cosines can also be thought of as the special case of harmonic functions (i.e. Laplacian eigenfunctions) on the real line; harmonic functions can be used to form a natural basis for L2 functions on many spaces.
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because {cos x，cos 2x,...,cos nx,...sin x,sin 2x,...,sin nx,...}form a group of bases of an orthogonal.These bases have a lot of characterization.You can find them easily in any reference book. May it help!
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4$\begingroup$ This reason is already mentioned in the question. What was being asked is why does one choose this basis and not any other one for FT. $\endgroup$ Commented Mar 7, 2010 at 17:28
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$\begingroup$ I think the most basic reason is that cos$^2$s+sin$^2$=1 $\endgroup$ Commented Aug 17, 2010 at 15:35
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