advanced34 min studyLesson 371 in Mathematics

Trigonometric systems and Fourier coefficients

Treat periodic frequencies as an orthonormal system, compute coefficients with a fixed normalization, and distinguish energy from pointwise recovery.

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What you’ll understand

  • Compute complex Fourier coefficients with normalized circle measure.
  • Use orthogonality to calculate projection error and energy.
  • Separate coefficient decay from different modes of series convergence.

A periodic signal can be compared with oscillations of every integer frequency. Fourier coefficients measure those comparisons, rather than merely listing the peaks visible in a plot. The underlying geometry becomes precise once the domain and integration measure are fixed. We use the circle of period two pi with normalized measure, so the constant function has norm one. This choice keeps the complex exponentials orthonormal and makes energy calculations transparent. Later lessons will distinguish the geometric meaning of these coefficients from the more delicate question of reconstructing values at individual points.

Fix the circle and its measure

Let T=R/(2πZ)\mathbb T=\mathbb R/(2\pi\mathbb Z) and dμ=dx/(2π)d\mu=dx/(2\pi). For an integrable complex-valued function, define its coefficient at integer nn by pairing with einxe^{inx}. The negative sign in the coefficient integral is the complex conjugate required by the inner product. Translating a function changes coefficient phases, while multiplying by a character shifts frequency indices. A different period or an unnormalized measure changes formulas, so normalization must be recorded before applying Parseval or comparing numerical values.

Orthogonality follows by integrating ei(n−m)xe^{i(n-m)x} over a full period: the result is one when the integers agree and zero otherwise. Real functions satisfy c−n=cn‾c_{-n}=\overline{c_n}, because conjugating the coefficient integral reverses the sign of the frequency. This symmetry does not require the coefficients themselves to be real. A sine contributes purely imaginary coefficients with opposite signs. Its complex coefficients therefore encode both amplitude and phase; replacing them by magnitudes alone loses information about the original signal.

cn=∫Tf(x)e−inx dμ(x),∫Teinxeimx‾ dμ(x)=δnmc_n=\int_{\mathbb T}f(x)e^{-inx}\,d\mu(x),\qquad\int_{\mathbb T}e^{inx}\overline{e^{imx}}\,d\mu(x)=\delta_{nm}
All circle integrals in this lesson use normalized measure dx/(2π). The coefficient is defined for f in L¹.↔ Scroll wide equations horizontally.

Partial sums are orthogonal projections

For f∈L2(T)f\in L^2(\mathbb T), the symmetric partial sum SNf=∑∣n∣≤NcneinxS_Nf=\sum_{|n|\le N}c_ne^{inx} is the orthogonal projection onto those frequencies. Its error is perpendicular to every retained mode. Pythagoras shows that it minimizes squared error among trigonometric polynomials with that frequency range, and Bessel's inequality bounds the retained coefficient energy by the input energy. This is an exact Hilbert-space statement; it does not require the graph to look smooth or each pointwise error to shrink monotonically.

The exponentials are complete in this Hilbert space. One route to completeness first approximates continuous periodic functions by suitable trigonometric averages and then uses density of continuous functions in the square-integrable class. Completeness upgrades Bessel's inequality to Parseval and makes the symmetric projections converge in squared norm. The functions in this space are equivalence classes modulo null sets, so values changed at a single point do not alter any coefficient, energy, or norm convergence statement.

∥f∥L2(μ)2=∑n∈Z∣cn∣2,∥f−SNf∥22=∑∣n∣>N∣cn∣2(f∈L2(T))\|f\|_{L^2(\mu)}^2=\sum_{n\in\mathbb Z}|c_n|^2,\qquad\|f-S_Nf\|_2^2=\sum_{|n|>N}|c_n|^2\quad(f\in L^2(\mathbb T))
Parseval and the tail identity express completeness and orthogonal projection in the normalized L² space.↔ Scroll wide equations horizontally.

What coefficients alone do not promise

An integrable periodic function has coefficients tending to zero as frequency grows, by the Riemann–Lebesgue lemma. This necessary decay does not imply that their absolute sum is finite, and it does not by itself establish pointwise reconstruction. In the square-integrable setting the coefficient sequence is square summable, but square summability alone still leaves pointwise questions separate. Always name the target topology: convergence of energy error, convergence at a point, and uniform convergence make different claims about the same sequence of polynomials.

Finite sampling adds another distinction. A discrete Fourier transform of sampled values gives coefficients of a finite data vector, and unresolved frequencies can alias into lower ones. It is not automatically the same as the exact integral coefficients of a continuous signal. An analytic coefficient calculation provides a reference against which sampling and quadrature can be checked. When a model is band limited and the sampling conditions are supplied, the bridge can be made precise; without those conditions an attractive sampled spectrum can conceal frequencies that were never resolved.

A real signal with imaginary coefficients

For f(x)=2+3cos(x)−4sin(2x), find all nonzero complex Fourier coefficients and its normalized squared L² norm.

  1. Expand cosine into half the sum of the positive and negative first-frequency exponentials. This gives c0=2 and c1=c−1=3/2.

  2. Use sin(2x)=(exp(2ix)−exp(−2ix))/(2i). Multiplying by −4 gives c2=2i and c−2=−2i. All other coefficients vanish, and the negative-frequency coefficients are conjugates of their positive-frequency partners.

  3. Sum their squared magnitudes: 4+2 times 9/4+2 times 4=33/2. Direct trigonometric orthogonality gives the same result: the constant contributes four, the cosine contributes 9/2, and the sine contributes eight.

The result

The squared norm is 33/2, with imaginary second-frequency coefficients despite f being real. There is no additional 2π factor under normalized circle measure.

Explore the model

Interactive lab

Interactive lab · spectral

Adding one orthogonal frequency

Pause and Reset, then set both controls to N=1 and decay p=1. At time zero predict value sin(π/4)=√2/2 at x=π/4. Reset and set N=2 and p=1; the new term there is sin(π/2)/2=0.5. Inspect the additional spatial oscillation. The instrument draws this specified finite sine sum; it does not calculate coefficients of an arbitrary input.

Adding one orthogonal frequencyA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA new orthogonal frequency has its own energy, even when its pointwise contribution adds to or cancels existing modes.

Check your understanding

For a real integrable periodic function, what relation must its complex Fourier coefficients satisfy?

Try it yourself

For f(x)=1+2cos(3x), calculate its normalized energy and the squared error after retaining only frequencies with |n|≤2.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The nonconstant mode lies outside the retained range. Apply Parseval separately to the retained part and the omitted part.

Compare with the solution

The nonzero coefficients are c0=1 and c3=c−3=1, so total squared norm is three. The projection S2f is just the constant one; its omitted coefficient energy is 1²+1²=2. Thus the squared L² error is two, while the actual L² error is √2. The projection is still optimal among polynomials of degree at most two, although increasing its constant or adding unrelated low frequencies cannot reconstruct the missing third frequency.

Take it with you

  • Fourier coefficients depend on the domain and measure convention.
  • Square-integrable Fourier sums are orthogonal projections with exact energy identities.
  • Norm reconstruction and pointwise reconstruction are separate assertions.

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