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Exercise 2168Linear algebra › Inner product spacesL1Added 3 Nov 2026

Fourier coefficients and orthogonal projection

#Fourier coefficients#trigonometric basis#orthogonal projection

Statement

On the space of continuous 2π2\pi-periodic functions with inner product ⟨f,g⟩=1π∫−ππf(t)g(t) dt\langle f,g\rangle=\frac1\pi\int_{-\pi}^\pi f(t)g(t)\,dt, show that the functions 1,cos⁡t,sin⁡t,cos⁡2t,sin⁡2t,…1,\cos t,\sin t,\cos 2t,\sin 2t,\ldots are orthogonal.

Hint

Compute ⟨cos⁡mt,cos⁡nt⟩\langle \cos mt,\cos nt\rangle, ⟨sin⁡mt,sin⁡nt⟩\langle\sin mt,\sin nt\rangle, ⟨cos⁡mt,sin⁡nt⟩\langle\cos mt,\sin nt\rangle using product-to-sum formulas.

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Solution

Orthogonality of cosines. For m≠nm\neq n:
⟨cos⁡mt,cos⁡nt⟩=1π∫−ππcos⁡mtcos⁡nt dt=12π∫−ππ[cos⁡(m−n)t+cos⁡(m+n)t] dt=0.\langle\cos mt,\cos nt\rangle=\frac1\pi\int_{-\pi}^\pi\cos mt\cos nt\,dt=\frac1{2\pi}\int_{-\pi}^\pi[\cos(m-n)t+\cos(m+n)t]\,dt=0.
For m=nm=n, it equals 11.

Orthogonality of sines. Similarly, ⟨sin⁡mt,sin⁡nt⟩=δmn\langle\sin mt,\sin nt\rangle=\delta_{mn}.

Orthogonality between cosine and sine. ⟨cos⁡mt,sin⁡nt⟩=0\langle\cos mt,\sin nt\rangle=0 because the integrand is odd.

Orthogonality with constant. ⟨1,cos⁡mt⟩=0\langle 1,\cos mt\rangle=0 and ⟨1,sin⁡nt⟩=0\langle 1,\sin nt\rangle=0 (integral of cosine/sine over a period).

Conclusion. These functions form an orthogonal family. The Fourier coefficients an=⟨f,cos⁡nt⟩a_n=\langle f,\cos nt\rangle, bn=⟨f,sin⁡nt⟩b_n=\langle f,\sin nt\rangle are the projections of ff onto these orthogonal functions.