Fourier Transform 16 | Calculating Sums with Fourier Series

TL;DR

Extending f(x)=x^2−π^2 periodically with period 2π produces a Fourier series that converges uniformly on [−π,π], enabling pointwise substitution.

Briefing Cornell Notes

Briefing

A carefully chosen 2π-periodic parabola lets Fourier series turn hard-looking infinite sums into clean, closed-form identities involving powers of π. By extending the function f(x)=x^2−π^2 (defined on −π≤x≤π) periodically, the resulting Fourier series converges uniformly (hence pointwise everywhere), so the series can be equated directly to the function at every x in [−π,π]. That pointwise equality becomes a machine for generating sum formulas.

The construction starts by defining f on −π to π as the parabola with zeros at x=±π and a minimum at x=0. With that definition in hand, the Fourier coefficients are computed using the complex form c_k=(1/2π)∫_{−π}^{π} e^{−ikx}f(x)dx. The constant term c_0 comes from integrating x^2−π^2, yielding c_0=−(2/3)π^2. For k≠0, integration by parts repeatedly reduces the coefficient integrals to expressions involving the derivative f′(x)=2x and boundary terms. The periodicity forces one integral term to vanish, leaving a boundary contribution that simplifies using e^{ikπ}=±1. The final result is c_k=2(−1)^k/k^2 for k≠0.

Building the Fourier series and using Euler’s formula e^{ikx}=cos(kx)+i sin(kx), the real-valued identity collapses to a cosine-only series. On −π≤x≤π, the parabola satisfies x^2−(1/3)π^2 = Σ_{k=1}^{∞} (4/k^2)(−1)^k cos(kx), with uniform convergence ensuring the equality holds pointwise.

Plugging in x=0 removes the cosine factor (cos(0)=1), producing an alternating-sum identity: −(1/3)π^2 = Σ_{k=1}^{∞} (4/k^2)(−1)^k. Rearranging gives Σ_{k=1}^{∞} (−1)^k/k^2 = −π^2/12.

A second payoff comes from Parseval’s identity, which relates the L^2 norm of f to the sum of squared Fourier coefficients. After accounting for the exceptional c_0 term, the remaining coefficients contribute Σ_{k=1}^{∞} |c_k|^2 = Σ_{k=1}^{∞} 4/k^4 (since (−1)^k disappears under absolute value). The right-hand side becomes (1/2π)∫_{−π}^{π} (x^2−π^2)^2 dx, which evaluates to (8/15)π^4. Solving for the remaining series yields Σ_{k=1}^{∞} 1/k^4 = π^4/90.

The overall message is practical: once a function meets the Fourier-series regularity conditions, the resulting expansion can be engineered to extract classic constants from infinite sums. The final segment hints at a next step—whether these conclusions survive under weaker convergence assumptions—setting up the next installment.

Cornell Notes

A periodic parabola f(x)=x^2−π^2 (extended with period 2π) has Fourier coefficients c_0=−(2/3)π^2 and, for k≠0, c_k=2(−1)^k/k^2. The Fourier series converges uniformly on [−π,π], so the series equals the function pointwise there. Converting the complex series to a cosine series gives x^2−(1/3)π^2 = Σ_{k=1}^{∞} (4/k^2)(−1)^k cos(kx). Setting x=0 yields the alternating sum Σ_{k=1}^{∞} (−1)^k/k^2 = −π^2/12, and Parseval’s identity then gives the classic result Σ_{k=1}^{∞} 1/k^4 = π^4/90. These identities show how Fourier series can compute nontrivial infinite sums via boundary terms and orthogonality.

How does choosing f(x)=x^2−π^2 on [−π,π] lead to sum identities?

Extending f periodically makes it a 2π-periodic function whose Fourier series converges uniformly on [−π,π]. Uniform (hence pointwise) convergence allows direct substitution of specific x-values into the Fourier series. Because the Fourier coefficients end up proportional to 1/k^2 with a factor (−1)^k, evaluating the cosine series at x=0 turns the series into an alternating sum of 1/k^2.

Why do integration by parts and periodicity simplify the coefficient calculation for c_k (k≠0)?

For k≠0, c_k involves ∫_{−π}^{π} e^{−ikx}f(x)dx. Integration by parts shifts derivatives onto f, producing terms with f′(x)=2x. A second integration by parts step introduces an integral of e^{−ikx} over [−π,π], which vanishes because e^{−ikx} is 2π-periodic and completes an integer number of oscillations. That cancellation leaves only boundary terms, which simplify using e^{ikπ}=±1, giving the factor (−1)^k.

What is the key Fourier-series identity obtained for x in [−π,π]?

After converting the complex exponential series to a real cosine series, the result is x^2−(1/3)π^2 = Σ_{k=1}^{∞} (4/k^2)(−1)^k cos(kx). The cosine-only form comes from taking the real part (the original function is real-valued) and from cosine being even, so cos(−kx)=cos(kx).

How does x=0 produce the alternating sum Σ (−1)^k/k^2?

At x=0, cos(k·0)=1 for all k. The identity becomes −(1/3)π^2 = Σ_{k=1}^{∞} (4/k^2)(−1)^k. Dividing both sides by 4 gives Σ_{k=1}^{∞} (−1)^k/k^2 = −π^2/12.

How does Parseval’s identity produce Σ 1/k^4 = π^4/90?

Parseval relates (1/2π)∫_{−π}^{π} |f(x)|^2 dx to Σ_{k=−∞}^{∞} |c_k|^2. The constant term c_0 must be handled separately, while for k≠0 the coefficients satisfy c_k=2(−1)^k/k^2, so |c_k|^2=4/k^4 (the sign disappears). The integral of (x^2−π^2)^2 over [−π,π] evaluates to (8/15)π^4, and solving for the remaining series yields Σ_{k=1}^{∞} 1/k^4 = π^4/90.

Review Questions

What are the explicit Fourier coefficients c_0 and c_k (k≠0) for the periodic extension of f(x)=x^2−π^2, and how does periodicity determine the (−1)^k factor?
Why is it legitimate to plug x=0 into the Fourier series, and what convergence property makes that step reliable?
Using the cosine-series identity, derive the alternating sum for 1/k^2 step by step, then check it against the final closed form −π^2/12.

Key Points

1
Extending f(x)=x^2−π^2 periodically with period 2π produces a Fourier series that converges uniformly on [−π,π], enabling pointwise substitution.
2
The constant Fourier coefficient is c_0=−(2/3)π^2, obtained by integrating x^2−π^2 over [−π,π].
3
For k≠0, repeated integration by parts plus the vanishing of ∫_{−π}^{π} e^{−ikx}dx yields c_k=2(−1)^k/k^2.
4
Converting the complex series to a real cosine series gives x^2−(1/3)π^2 = Σ_{k=1}^{∞} (4/k^2)(−1)^k cos(kx).
5
Setting x=0 turns the cosine series into an alternating sum, giving Σ_{k=1}^{∞} (−1)^k/k^2 = −π^2/12.
6
Parseval’s identity converts the L^2 norm of f into Σ |c_k|^2, leading to Σ_{k=1}^{∞} 1/k^4 = π^4/90 after evaluating (1/2π)∫_{−π}^{π} (x^2−π^2)^2 dx.

Highlights

A single Fourier-series identity for a periodic parabola produces two famous constants: Σ_{k=1}^{∞} (−1)^k/k^2 = −π^2/12 and Σ_{k=1}^{∞} 1/k^4 = π^4/90.

Boundary terms from integration by parts, combined with e^{ikπ}=±1, force the coefficient pattern c_k=2(−1)^k/k^2.

Uniform convergence makes it safe to equate the Fourier series with the original function at specific points like x=0.

Parseval’s identity turns squared Fourier coefficients into a computable integral of (x^2−π^2)^2, yielding the π^4/90 result.