Project a profile onto orthogonal modes

Try

A periodic signal contains a constant level and oscillations of different frequencies. To measure one frequency, multiply the signal by a matching sine or cosine and average over a full period. Oscillations at other integer frequencies cancel. This is an inner-product measurement, not a claim that a few sampled points identify an arbitrary function.

Use the interval [−π,π] and the real inner product ∫f(x)g(x)dx. The constant function has squared norm 2π; each nonconstant sine or cosine has squared norm π. That difference explains the familiar half-factor in the constant term.

Predict

For f(x) = 2 + 4cos x − sin(3x), predict which coefficients are nonzero. Before integrating, explain why an odd sine cannot detect the even constant or cosine parts. Then predict whether the conventional coefficient a₀ is 2 or 4.

Experiment

Use aₙ = (1/π)∫f cos(nx) and bₙ = (1/π)∫f sin(nx), with a₀ defined by the same cosine formula at n = 0.

Measurement	Integral	Coefficient
Constant	4π	a₀ = 4
cos x	4π	a₁ = 4
sin(3x)	−π	b₃ = −1

All other coefficients vanish. The expansion is a₀/2 plus the nonconstant modes. Change the signal to 2 + 4cos x + sin(3x): only one measured coefficient changes sign. Its squared contribution to energy does not change.

Understand

Product-to-sum identities establish orthogonality by integrating integer-frequency sines and cosines over a complete period. For example, cos(mx)cos(nx) is half the sum of cos((m−n)x) and cos((m+n)x). When positive integers m and n differ, both integrate to zero; when they agree, the zero-frequency term contributes π.

Let S be the finite span of the constant and modes up to frequency N. The coefficients above make f − P orthogonal to every element of S, where P is the truncated expansion. For any Q in S, write f − Q = (f − P) + (P − Q). Orthogonality eliminates the cross term, giving ∥f − Q∥² = ∥f − P∥² + ∥P − Q∥². Thus P is the unique least-squares approximation in that finite-dimensional space, up to equality almost everywhere.

This finite projection argument does not prove that infinite Fourier series converge pointwise. It proves a best-approximation property for a specified number of modes. Completeness and different convergence senses come next.

On [0,L], boundary conditions suggest different extensions. An odd extension produces sine modes that vanish at endpoints; an even extension produces cosine modes compatible with zero derivative at endpoints when sufficient regularity holds. Extending data also introduces possible endpoint jumps, so the extension is a mathematical choice with consequences.

Apply

Construct a finite signal with two cosine modes and one sine mode, calculate its coefficients by orthogonality and compute the squared error after deleting one mode. Then choose data on [0,L] and explain whether a sine or cosine expansion matches the intended boundary condition. State the interval and normalization every time; changing L changes the frequencies to nπ/L.

Continue

The next lesson separates convergence in average squared error from convergence at individual points. A projection can be optimal in L² while still showing visible oscillation near a jump.