Fourier Transform 10 | Fundamental Example for Fourier Series [dark version]

TL;DR

Use the 2π-periodic step function h_a(x)=1 on (−π,a] and 0 on (a,π) as the base case for Parseval’s identity.

Briefing Cornell Notes

Briefing

A single, carefully chosen step function is enough to prove Parseval’s identity for all square-integrable functions—because the Fourier-series calculation for that step function can be carried through to the exact norm equality. The key move is to start with a 2π-periodic “Heaviside-type” step function h_a(x) that equals 1 on the interval (−π, a] and 0 on (a, π), with the value at the jump point treated consistently but shown to be irrelevant to the Fourier inner products. Once the Fourier coefficients of this one example are computed explicitly, the remaining work is to sum their squared magnitudes and show the result matches the L2 norm of h_a.

Fourier coefficients c_k are obtained by integrating h_a(x) against the complex exponential basis e_k(x) = e^{ikx}/√(2π) (up to the normalization used in the transcript). Because h_a(x) vanishes on (a, π), every coefficient reduces to an integral over (−π, a]. For k = 0, the coefficient becomes proportional to the interval length a − (−π) = a + π. For k ≠ 0, the coefficient comes from integrating e^{−ikx} over (−π, a], producing an expression involving e^{−ika} and e^{ikπ}. The calculation stays elementary: it’s just exponentials and limits.

To connect Fourier coefficients to Parseval’s identity, the transcript then focuses on the sum of squared magnitudes, Σ_{k} |c_k|^2. After multiplying each coefficient by its complex conjugate, the exponential terms simplify into a real expression containing cos(k(π+a)) divided by k^2. The sum is split into the k = 0 contribution (coming from c_0) and the remaining symmetric k ≠ 0 terms, which can be rewritten as a doubled sum over positive integers. The final step requires evaluating two classical series limits involving 1/k^2 and cos(xk)/k^2.

A general identity for 0 ≤ x ≤ 2π is invoked: the limit of Σ_{k=1}^∞ cos(kx)/k^2 is expressed in closed form using quadratic polynomials in x. Plugging in x = 0 yields Σ 1/k^2 = π^2/6, while choosing x = π + a supplies the cosine-sum value needed for the step function. Substituting both limits into the Parseval sum causes the π^2 terms to cancel cleanly, leaving exactly the L2 norm of h_a. The transcript concludes that Σ_{k∈Z} |c_k|^2 equals ∫_{−π}^{π} |h_a(x)|^2 dx, which for this step function is simply the length of the region where h_a = 1, namely a + π.

That equality for every h_a is treated as the crucial base case. From there, the plan is to extend Parseval’s identity first to general step functions (built from linear combinations of single steps) and then to all L2 functions via approximation, setting up the next stages of the proof.

Cornell Notes

The proof strategy starts with a 2π-periodic step function h_a(x): it equals 1 on (−π, a] and 0 on (a, π). Because h_a is zero on half the interval, each Fourier coefficient c_k reduces to a simple integral over (−π, a]. For k = 0 the coefficient depends on the interval length a + π; for k ≠ 0 it involves exponentials evaluated at a and π. Summing |c_k|^2 requires evaluating series of the form Σ 1/k^2 and Σ cos(kx)/k^2, using a closed-form identity valid for 0 ≤ x ≤ 2π. After substituting x = 0 and x = π + a, the coefficient-sum collapses to the L2 norm of h_a, confirming Parseval’s identity for this base case and enabling extension to all step functions and then all L2 functions.

Why does changing a function at the jump point not affect the Fourier-series/Parseval calculation here?

The step function h_a differs only at a finite set of points (the jump location). Fourier coefficients and L2 norms depend on integrals, and changing values at finitely many points does not change the integral. So the exact value at the boundary point (where the step jumps) does not alter the computed inner products or the L2 norm.

How do the Fourier coefficients c_k simplify for h_a(x)?

Since h_a(x)=0 on (a, π), the integral defining c_k over (−π, π) collapses to an integral over (−π, a]. For k=0, the integrand is constant 1, giving a coefficient proportional to the length a+π. For k≠0, integrating the exponential produces terms like e^{−ika} − e^{ikπ} (with the normalization used in the transcript). The computation is essentially “integrate an exponential over a finite interval.”

What expression appears after computing Σ|c_k|^2, and why does it involve cosines?

Multiplying c_k by its complex conjugate cancels some exponential factors and combines others into real trigonometric terms. The result becomes a sum involving cos(k(π+a)) divided by k^2, plus the separate k=0 contribution. Symmetry in k (the formulas are unchanged under k→−k) lets the sum be rewritten as twice a sum over k=1 to n.

Which two infinite-series values are needed to finish the Parseval sum?

The calculation reduces to two limits: (1) Σ_{k=1}^∞ 1/k^2, obtained by using the cosine-sum formula at x=0, and (2) Σ_{k=1}^∞ cos(kx)/k^2 evaluated at x=π+a. The transcript uses the closed-form identity for 0≤x≤2π to get Σ 1/k^2 = π^2/6 and the corresponding cosine-sum value in terms of (π+a).

How does the final algebra show the coefficient-sum equals the L2 norm of h_a?

After substituting the two series limits, the π^2 terms cancel against the constant contributions coming from c_0 and the cosine-sum part. What remains is proportional to (a+π), which matches ∫_{−π}^{π} |h_a(x)|^2 dx because |h_a(x)|^2=1 exactly on (−π, a] and 0 elsewhere. The transcript summarizes this as the Parseval equality for h_a: Σ_{k∈Z} |c_k|^2 = ||h_a||_2^2.

Review Questions

What are the explicit regions where h_a(x)=1 and where h_a(x)=0, and how does that determine the limits of integration for c_k?
Why does the sum over k≠0 become a doubled sum over positive integers?
Which closed-form series identity is used to evaluate Σ cos(kx)/k^2, and what substitutions for x are made in this proof?

Key Points

1
Use the 2π-periodic step function h_a(x)=1 on (−π,a] and 0 on (a,π) as the base case for Parseval’s identity.
2
Fourier coefficients c_k reduce to integrals over (−π,a] because h_a vanishes on (a,π).
3
For k=0, c_0 depends on the interval length a+π; for k≠0, c_k comes from integrating exponentials over (−π,a].
4
Summing |c_k|^2 turns exponential products into a cosine term of the form cos(k(π+a)) divided by k^2.
5
Evaluating Σ 1/k^2 and Σ cos(kx)/k^2 using a closed-form identity (for 0≤x≤2π) completes the coefficient-sum calculation.
6
After substitution and cancellation, the result equals ∫_{−π}^{π} |h_a(x)|^2 dx = a+π, proving Parseval’s identity for every h_a.
7
The base-case result is then positioned as the foundation for extending Parseval’s identity from step functions to all L2 functions via approximation.

Highlights

A single step function h_a(x) is enough: once Parseval’s identity holds for every h_a, the proof can be extended to all step functions and then all L2 functions.

The Fourier-coefficient integrals collapse to (−π,a] because h_a(x)=0 on (a,π), making the coefficient computation straightforward.

After taking |c_k|^2, exponential terms simplify into real cosine expressions, reducing the Parseval sum to Σ 1/k^2 and Σ cos(kx)/k^2.

Using a closed-form identity for Σ cos(kx)/k^2 at x=0 and x=π+a makes the final algebra cancel down to the L2 norm of h_a.

The final equality matches the geometric fact that ||h_a||_2^2 equals the length of the interval where h_a=1, namely a+π.