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In my experience, math.stackexchange likes mathematical rigour and presupposes some mathematical maturity. There are a fair number of professional mathematician

by zodiac 13y ago

In my experience, math.stackexchange likes mathematical rigour and presupposes some mathematical maturity. There are a fair number of professional mathematicians using the site. Sure, if you just want a "for dummies" understanding, a hand-wavy example like the second highest voted answer suffices. But if you want to learn it properly (prove theorems, etc), you need to know concepts like complex numbers and integration.

mturmon 13y ago
If you wanted something more deep than the square wave example, you could explain why the complex exponential is used. This would provide important insight, because a smart reader might say, OK fine, square wave approximation, but why don't we use step functions or gaussian-shaped bumps or anything else as basis functions. By the Stone-Weierstrass theorem (http://en.wikipedia.org/wiki/Stone–Weierstrass_theorem http://en.wikipedia.org/wiki/Stone–Weierstrass_theorem) we know that lots (really, lots!) of sets of functions would work, not just complex exponentials. The story I tell myself about that is, "complex exponentials are the (unique) eigenfunctions of linear time-invariant systems". Simply put, this means that if S(.) is a system that has derivatives and integrals, and scaling and shifting in time, then S(exp(i omega t)) = const * exp(i omega t) I.e., you put in a complex exponential, you get out a (scaled) complex exponential. That's not going to happen, in general, for other basis functions. This property is (under some not too restrictive conditions) basically a defining property of the complex exponential. (Basically, notice that d/dt [exp (a t)] = a exp(a t) -- which turns out to have a converse as well.) Then, because S is linear, a linear combo of complex exponentials (on the input of S) transforms into a weighted combo of the same exponentials on the output. To me, that's the deeper reason that Fourier analysis is important. Those sines and cosines are mathematically inevitable, right from the moment in high school when you learned d/dx e^x = e^x. Noted on that SE page (the OP) is a related, but subsidiary property, that the Fourier bases are also the eigenfunctions of the wave equation. But the wave equation is linear, so this follows from the above reasoning. I imagine, but I don't know, that the usefulness of the Fourier basis for quantum-mechanical analysis also follows from linearity.

by zodiac 13y ago

mturmon 13y ago

If you wanted something more deep than the square wave example, you could explain why the complex exponential is used. This would provide important insight, because a smart reader might say, OK fine, square wave approximation, but why don't we use step functions or gaussian-shaped bumps or anything else as basis functions. By the Stone-Weierstrass theorem (http://en.wikipedia.org/wiki/Stone–Weierstrass_theorem http://en.wikipedia.org/wiki/Stone–Weierstrass_theorem) we know that lots (really, lots!) of sets of functions would work, not just complex exponentials. The story I tell myself about that is, "complex exponentials are the (unique) eigenfunctions of linear time-invariant systems". Simply put, this means that if S(.) is a system that has derivatives and integrals, and scaling and shifting in time, then S(exp(i omega t)) = const * exp(i omega t) I.e., you put in a complex exponential, you get out a (scaled) complex exponential. That's not going to happen, in general, for other basis functions. This property is (under some not too restrictive conditions) basically a defining property of the complex exponential. (Basically, notice that d/dt [exp (a t)] = a exp(a t) -- which turns out to have a converse as well.) Then, because S is linear, a linear combo of complex exponentials (on the input of S) transforms into a weighted combo of the same exponentials on the output. To me, that's the deeper reason that Fourier analysis is important. Those sines and cosines are mathematically inevitable, right from the moment in high school when you learned d/dx e^x = e^x. Noted on that SE page (the OP) is a related, but subsidiary property, that the Fourier bases are also the eigenfunctions of the wave equation. But the wave equation is linear, so this follows from the above reasoning. I imagine, but I don't know, that the usefulness of the Fourier basis for quantum-mechanical analysis also follows from linearity.