Suppose, for example, using Fourier series techniques to solve a differential equation leads to the following:

$a_0 + \sum_{n=1}^{\infty}a_n\sin(nx)+b_n\cos(nx)=4\sin x$

At this point, why can you equate the coefficients of $\cos(nx)$ and $\sin(nx)$ on the two sides the equation. I don't understand why this step works.

With very many thanks,


  • $\begingroup$ Do you mean i. $a_n=b_n$ or ii. $a_1=4, a_{\neq 1}=b_n=0$? $\endgroup$ – draks ... Jun 6 '12 at 10:59
  • $\begingroup$ If $f$ and $g$ are equal as functions, then by definition their Fourier coefficients must be equal. $\endgroup$ – Willie Wong Jun 6 '12 at 11:14
  • $\begingroup$ Perhaps it follows from this statement of Plancherel's theorem? $\endgroup$ – bgins Jun 6 '12 at 11:22

Informally you can recover the coefficient by averaging both sides against $\sin mx$:

$$\frac{1}{\pi}\int_{-\pi}^{\pi}\left( a_0 + \sum_{n=1}^\infty a_n\sin(nx) + b_n\cos(nx)\right)\sin(mx)\, dx =\frac{1}{\pi}\int_{-\pi}^\pi 4\sin x \sin (mx)\, dx, $$ because the left hand side becomes then

$$a_0 \frac{1}{\pi}\int_{-\pi}^\pi\sin(mx)\, dx+ \sum_{n=1}^\infty a_n\frac{1}{\pi}\int_{-\pi}^\pi\sin(nx)\sin(mx)\, dx+b_n\frac{1}{\pi}\int_{-\pi}^\pi\cos(nx)\sin(mx)\, dx =a_m $$

while the left hand side is $4\delta_{1m}$. This procedure, similar to the original reasoning by Fourier, requires the interchange of integral and sum in left hand side, which can be done if the series converges uniformly. However, that request is excessively restrictive: turns out that convergence in $L^2$ sense will suffice. This can be proven by means of Hilbert space methods.

  • $\begingroup$ Thanks! That makes it really clear. $\endgroup$ – Froskoy Jun 7 '12 at 10:29
  • $\begingroup$ @Froskoy: You can find a better explanation in the same spirit in the introductory chapters of every book on harmonic analysis. A very good choice is Stein-Shakarchi's "Fourier Analysis", volume 1. $\endgroup$ – Giuseppe Negro Jun 7 '12 at 12:08

What you want is exactly saying that the functions $x \mapsto \cos( nx)$ and $x \mapsto \sin(nx)$ are linearly indepedent. You can actually prove quite more very simply : they form an orthogonal family in $L^2$ (you prove it by checking that $\int \cos(nx) \cos(mx) = 1$ if $n=m$, 0 otherwise) (with suitable bounds of integration and renormalization constants).

  • 3
    $\begingroup$ Not quite, I'm afraid. Linear independence deals with finite sets of functions. Here we have an infinite series. $\endgroup$ – Giuseppe Negro Jun 6 '12 at 16:49
  • $\begingroup$ true. then let's say they form a linearly independant orthogonal family... they can therefore be completed in an orthogonal basis. the point is, you don't need the "generating" aspect of the (hilbert) basis, only the "linearly independent" one. $\endgroup$ – Glougloubarbaki Jun 7 '12 at 12:56
  • $\begingroup$ I understand what you want to say and I agree with it; but I still find it is poorly written. The generalization of the concept of basis to infinite dimensional Banach spaces is indeed rather complicated: see Schauder basis. The standard nice case is that of orthogonal systems in Hilbert space: therefore, you do need orthogonality to prove what the OP asks for. $\endgroup$ – Giuseppe Negro Jun 7 '12 at 14:40

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