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METH2 · 1

Fourier series

11 concepts · 24 questions

Fourier coefficients, orthogonality, convergence (Dirichlet), sine and cosine series, Parseval's identity and the Gibbs phenomenon.

Overview 1The plan of this section

Periodic functions are sums of sines and cosines; this is linear algebra in a space of functions.

  1. Orthogonality of the trigonometric system (Prop. 1.1) and Fourier series (Def. 1.2), with a worked example (Example 1.3).
  2. Dirichlet's convergence theorem (Thm 1.4).
  3. Sine and cosine series via odd and even extensions (Prop. 1.5); complex form (Def. 1.6).
  4. Parseval's identity (Thm 1.7) and the Gibbs phenomenon (Example 1.8); differentiation and decay of coefficients (Prop. 1.9).

Orthogonal sines/cosines ⇒ coefficients by integrals; Dirichlet convergence; odd/even extensions; Parseval; Gibbs.

Proposition 1.1Orthogonality of the trigonometric system

On [−L,L][-L, L], for integers m,n≥0m, n\ge0:

∫−LLcos⁡mπxLcos⁡nπxL dx=∫−LLsin⁡mπxLsin⁡nπxL dx=Lδmn (m,n≥1),∫−LLsin⁡mπxLcos⁡nπxL dx=0,\int_{-L}^L\cos\frac{m\pi x}L\cos\frac{n\pi x}L\,\mathrm{d}x = \int_{-L}^L\sin\frac{m\pi x}L\sin\frac{n\pi x}L\,\mathrm{d}x = L\delta_{mn}\ (m, n\ge1),\qquad\int_{-L}^L\sin\frac{m\pi x}L\cos\frac{n\pi x}L\,\mathrm{d}x = 0,

and ∫−LL1 dx=2L\int_{-L}^L1\,\mathrm{d}x = 2L.

{1,cos⁡nπxL,sin⁡nπxL}\{1, \cos\frac{n\pi x}L, \sin\frac{n\pi x}L\} are orthogonal on [−L,L][-L, L]; each has squared norm LL (the constant: 2L2L).

Proof

Definition 1.2Fourier series

For an integrable 2L2L-periodic ff, the Fourier series is

a02+∑n≥1(ancos⁡nπxL+bnsin⁡nπxL),an=1L∫−LLf(x)cos⁡nπxL dx,bn=1L∫−LLf(x)sin⁡nπxL dx.\frac{a_0}2 + \sum_{n\ge1}\Big(a_n\cos\frac{n\pi x}L + b_n\sin\frac{n\pi x}L\Big),\quad a_n = \frac1L\int_{-L}^Lf(x)\cos\frac{n\pi x}L\,\mathrm{d}x,\quad b_n = \frac1L\int_{-L}^Lf(x)\sin\frac{n\pi x}L\,\mathrm{d}x.

The formulas are forced: if ff equals such a series (with termwise integration allowed), multiplying by a basis function and integrating isolates one coefficient.

an=1L∫fcos⁡a_n = \frac1L\int f\cos, bn=1L∫fsin⁡b_n = \frac1L\int f\sin; the constant term is a02\frac{a_0}2 (the mean of ff).

Example 1.3The square wave and the sawtooth

  • Square wave f=−1f = -1 on (−π,0)(-\pi, 0), 11 on (0,π)(0, \pi): odd, bn=2π∫0πsin⁡nx=2(1−(−1)n)nπb_n = \frac2\pi\int_0^\pi\sin nx = \frac{2(1 - (-1)^n)}{n\pi}, so f∼4π(sin⁡x+sin⁡3x3+sin⁡5x5+⋯)f\sim\frac4\pi\big(\sin x + \frac{\sin3x}3 + \frac{\sin5x}5 + \cdots\big).
  • Sawtooth f(x)=xf(x) = x on (−π,π)(-\pi, \pi): odd, bn=2(−1)n+1nb_n = \frac{2(-1)^{n+1}}n, so x∼2∑(−1)n+1nsin⁡nxx\sim2\sum\frac{(-1)^{n+1}}n\sin nx. Putting x=π2x = \frac\pi2 in the square wave gives 1−13+15−⋯=π41 - \frac13 + \frac15 - \cdots = \frac\pi4.

Square wave 4π∑n oddsin⁡nxn\frac4\pi\sum_{n\text{ odd}}\frac{\sin nx}n; sawtooth 2∑(−1)n+1nsin⁡nx2\sum\frac{(-1)^{n+1}}n\sin nx.

Theorem 1.4Dirichlet's convergence theorem

If ff is 2L2L-periodic and piecewise C1C^1 (finitely many jumps, with one-sided derivatives), its Fourier series converges at every xx to 12(f(x−)+f(x+))\frac12\big(f(x^-) + f(x^+)\big): to f(x)f(x) where ff is continuous, and to the midpoint of the jump otherwise. If in addition ff is continuous everywhere, the convergence is uniform. (Proof via the Dirichlet kernel; omitted.)

Piecewise C1C^1 ⇒ Fourier series → average of left and right limits; continuous + piecewise C1C^1 ⇒ uniform convergence.

Proposition 1.5Sine and cosine series

A function on (0,L)(0, L) can be expanded as a sine series ∑bnsin⁡nπxL\sum b_n\sin\frac{n\pi x}L with bn=2L∫0Lfsin⁡nπxLb_n = \frac2L\int_0^Lf\sin\frac{n\pi x}L (the series of its odd 2L2L-periodic extension), or as a cosine series a02+∑ancos⁡nπxL\frac{a_0}2 + \sum a_n\cos\frac{n\pi x}L with an=2L∫0Lfcos⁡nπxLa_n = \frac2L\int_0^Lf\cos\frac{n\pi x}L (even extension). The sine series converges to 00 at x=0,Lx = 0, L; the cosine series has derivative 00 there formally.

On (0,L)(0, L): sine series (odd extension, bn=2L∫0Lb_n = \frac2L\int_0^L) or cosine series (even extension). Choose by the boundary conditions.

Proof

Definition 1.6Complex Fourier series

For 2π2\pi-periodic ff, f∼∑n∈Zcneinxf\sim\sum_{n\in\mathbb{Z}}c_ne^{inx} with cn=12π∫−ππf(x)e−inx dxc_n = \frac1{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}\,\mathrm{d}x. Then cn=an−ibn2c_n = \frac{a_n - ib_n}2 and c−n=an+ibn2c_{-n} = \frac{a_n + ib_n}2 for n≥1n\ge1, c0=a02c_0 = \frac{a_0}2; for real ff, c−n=cn‾c_{-n} = \overline{c_n}.

cn=12π∫fe−inxc_n = \frac1{2\pi}\int fe^{-inx}; einxe^{inx} are orthogonal for ⟨f,g⟩=∫fgˉ\langle f, g\rangle = \int f\bar g.

Theorem 1.7Parseval's identity

For square-integrable ff on (−L,L)(-L, L),

1L∫−LLf2 dx=a022+∑n≥1(an2+bn2),\frac1L\int_{-L}^Lf^2\,\mathrm{d}x = \frac{a_0^2}2 + \sum_{n\ge1}(a_n^2 + b_n^2),

and the partial sums converge to ff in mean square. The partial sum SNS_N is the best approximation to ff in mean square by trigonometric polynomials of degree ≤N\le N (Bessel's inequality gives ≤\le; equality is completeness, quoted).

Parseval: energy of ff = sum of squared coefficients. Evaluates sums like ∑1n2\sum\frac1{n^2}.

Proof

Example 1.8The Gibbs phenomenon

Near a jump, partial sums overshoot: for the square wave, max⁡SN→2π∫0πsin⁡tt dt≈1.179\max S_N\to\frac2\pi\int_0^\pi\frac{\sin t}t\,\mathrm{d}t\approx1.179 as N→∞N\to\infty, an overshoot of about 9%9\% of the jump size (22), located at distance ≈πN\approx\frac\pi{N} from the jump. The overshoot never disappears; it only narrows, consistent with convergence being pointwise but not uniform.

Gibbs: ≈ 9% overshoot of the jump that narrows but never vanishes.

Proposition 1.9Smoothness and decay of coefficients

If ff is 2π2\pi-periodic, continuous and piecewise C1C^1, then f′f' has Fourier coefficients nbnnb_n (for cos⁡\cos) and −nan-na_n (for sin⁡\sin); i.e. termwise differentiation gives the series of f′f'. Consequently smoother ff has faster-decaying coefficients: jumps give O(1n)O(\frac1n), kinks O(1n2)O(\frac1{n^2}), and f∈Ckf\in C^k gives o(n−k)o(n^{-k}).

Differentiate termwise if ff is continuous and periodic; smoothness ⇔ fast coefficient decay.

Proof

Check 1.10Check: computing Fourier series

  • Use symmetry first: odd ⇒ sines only, even ⇒ cosines only.
  • Integrate by parts; cos⁡nπ=(−1)n\cos n\pi = (-1)^n, sin⁡nπ=0\sin n\pi = 0.
  • Check decay: jumps ∼1n\sim\frac1n, kinks ∼1n2\sim\frac1{n^2}.
  • Check a value: the series at a continuity point must give f(x)f(x); at a jump, the midpoint.

Symmetry, parts, (−1)n(-1)^n, decay rate, check one value.