Fourier Transform 11 | Sum Formulas for Sine and Cosine [dark version]

Q: Why rewrite $\cos(kx)$ using complex exponentials, and what does it buy in the proof?

Writing $\cos(kx)=\tfrac12(e^{ikx}+e^{-ikx})$ turns a cosine sum into a sum of exponentials. After an index shift, the exponential sum becomes a geometric series in $q=e^{ix}$. That geometric-series formula removes the summation symbol and yields a closed expression involving $e^{ix}$, which can then be converted back into real sine functions.

Q: What is the key restriction on $x$ when using the geometric-series formula?

The geometric sum requires $q\ne 1$. Here $q=e^{ix}$, so $q=1$ happens exactly when $x$ is a multiple of $2\pi$. That’s why the derived sine-ratio expression for the finite cosine sum is valid for $x\notin 2\pi\mathbb{Z}$, and boundary handling later becomes a separate step.

Q: How does the proof justify exchanging an infinite series with an integral?

It proves uniform convergence on $[\varepsilon,2\pi-\varepsilon]$. With uniform convergence, the limit can be interchanged with integration safely (avoiding the danger of pulling limits out of integrals without control). The proof then extends convergence to the full interval using a Weierstrass M-test.

Q: How does integrating the sine series produce the cosine $1/k^2$ series?

The sine identity has terms $\sin(kx)/k$. Integrating $\sin(kx)/k$ with respect to $x$ gives $-\cos(kx)/k^2$ (up to constants from the integration bounds). Because uniform convergence allows term-by-term integration, the resulting expression matches $\sum_{k=1}^{\infty}\cos(kx)/k^2$ as a quadratic polynomial plus a constant.

Q: How is the constant $C$ determined once the series is known to equal a quadratic plus $C$?

Integrate both sides over one full period $[0,2\pi]$. Each $\cos(kx)$ term averages to zero over a full period, so the left side vanishes. The right side then reduces to the integral of the quadratic polynomial plus $2\pi C$, letting the proof solve for $C=\pi^2/6$.

TL;DR

The cosine series $\sum_{k = 1}^{\infty} cos (k x) / k^{2}$ has an explicit closed form $\frac{x ^{2}}{4} - \frac{π x}{2} + \frac{π ^{2}}{6}$ for all $x \in [0, 2 π]$ .

Briefing Cornell Notes

Briefing

A key payoff of the proof is an explicit closed-form for the cosine-weighted Dirichlet-type series

$k = 1 \sum \infty \frac{cos ( k x )}{k ^{2}} = \frac{x ^{2}}{4} - \frac{π x}{2} + \frac{π ^{2}}{6} (0 \leq x \leq 2 π),$

including the boundary points. Getting the exact value matters because it plugs directly into the earlier step used to establish Parseval’s identity for a particular step function—so the whole Fourier-transform chain depends on this cosine sum being nailed down precisely, not just shown to converge.

The argument starts by proving a finite-sum identity for cosine terms using complex exponentials. The cosine is rewritten as $cos (k x) = \frac{1}{2} (e^{ik x} + e^{- ik x}),$ which turns a cosine sum into a sum of exponentials. After an index shift, the exponential sum becomes a geometric series in $q = e^{i x}$ . Applying the geometric-sum formula eliminates the summation symbol and yields a closed expression involving complex exponentials; rewriting that fraction in real terms produces a sine-ratio form. A crucial restriction appears: the geometric-series step fails when $q = 1$ , which corresponds to $x$ being a multiple of $2 π$ . So the cosine-sum formula holds for $x \in / 2 π Z$ .

With that sine-ratio tool in hand, the proof then targets a related sine series: $k = 1 \sum \infty \frac{sin ( k x )}{k} = \frac{π - x}{2} (0 < x < 2 π) .$ To justify exchanging limits and integrals later, the proof establishes uniform convergence on any closed subinterval $[ε, 2 π - ε]$ . It does this by expressing the sine series through an integral of cosine terms, then bounding the resulting remainder using supremum norms. The denominator in the sine-ratio expression stays away from zero on $[ε, 2 π - ε]$ , producing a $1/ n$ -type factor that forces the remainder to vanish uniformly.

The final leap to the cosine $1/ k^{2}$ series uses integration. Since integrating $sin (k x) / k$ term-by-term produces $cos (k x) / k^{2}$ , the proof integrates the sine-series identity from $x_{0}$ to $x$ . Uniform convergence on $[ε, 2 π - ε]$ legitimizes swapping the infinite sum with the integral. The result is that the cosine series must equal a quadratic polynomial in $x$ plus a constant $C$ . Determining $C$ requires extending the identity to the boundary points $0$ and $2 π$ . The proof uses a Weierstrass M-test to upgrade uniform convergence across the full interval, then relies on continuity: two continuous functions that match on $(0, 2 π)$ must match everywhere.

To compute $C$ , the proof integrates both sides over one full period $[0, 2 π]$ . The cosine terms integrate to zero over a period, leaving only the polynomial contribution, which yields $C = π^{2} /6$ . Substituting back gives the final closed form for $\sum_{k = 1}^{\infty} cos (k x) / k^{2}$ on $[0, 2 π]$ , completing the technical step needed for the broader Fourier-series framework.

Cornell Notes

The proof derives an exact formula for the cosine series $\sum_{k = 1}^{\infty} cos (k x) / k^{2}$ on $[0, 2 π]$ , not just its convergence. It first converts finite cosine sums into geometric sums using $cos (k x) = \frac{1}{2} (e^{ik x} + e^{- ik x})$ , producing a sine-ratio expression (valid away from $x \in 2 π Z$ ). Next, it establishes a companion sine-series identity $\sum_{k = 1}^{\infty} sin (k x) / k = (π - x) /2$ on $(0, 2 π)$ with uniform convergence on $[ε, 2 π - ε]$ . Integrating that sine identity term-by-term yields the cosine $1/ k^{2}$ series as a quadratic polynomial plus a constant $C$ . Uniform convergence on the full interval (via a Weierstrass M-test) and periodic integration determine $C = π^{2} /6$ , giving the closed form for all $x \in [0, 2 π]$ .

Why rewrite $cos (k x)$ using complex exponentials, and what does it buy in the proof?

Writing

cos (k x) = \frac{1}{2} (e^{ik x} + e^{- ik x})

turns a cosine sum into a sum of exponentials. After an index shift, the exponential sum becomes a geometric series in

q = e^{i x}

. That geometric-series formula removes the summation symbol and yields a closed expression involving

e^{i x}

, which can then be converted back into real sine functions.

What is the key restriction on $x$ when using the geometric-series formula?

The geometric sum requires

q \neq = 1

. Here

q = e^{i x}

, so

q = 1

happens exactly when

x

is a multiple of

2 π

. That’s why the derived sine-ratio expression for the finite cosine sum is valid for

x \in / 2 π Z

, and boundary handling later becomes a separate step.

How does the proof justify exchanging an infinite series with an integral?

It proves uniform convergence on

[ε, 2 π - ε]

. With uniform convergence, the limit can be interchanged with integration safely (avoiding the danger of pulling limits out of integrals without control). The proof then extends convergence to the full interval using a Weierstrass M-test.

How does integrating the sine series produce the cosine $1/ k^{2}$ series?

The sine identity has terms

sin (k x) / k

. Integrating

sin (k x) / k

with respect to

x

gives

- cos (k x) / k^{2}

(up to constants from the integration bounds). Because uniform convergence allows term-by-term integration, the resulting expression matches

\sum_{k = 1}^{\infty} cos (k x) / k^{2}

as a quadratic polynomial plus a constant.

How is the constant $C$ determined once the series is known to equal a quadratic plus $C$ ?

Integrate both sides over one full period

[0, 2 π]

. Each

cos (k x)

term averages to zero over a full period, so the left side vanishes. The right side then reduces to the integral of the quadratic polynomial plus

2 π C

, letting the proof solve for

C = π^{2} /6

Review Questions

Where exactly does the condition $x \in / 2 π Z$ enter, and why is it not automatically resolved by taking limits?
What role does uniform convergence on $[ε, 2 π - ε]$ play in the step where the series is integrated term-by-term?
Why does matching two continuous functions on $(0, 2 π)$ force equality on $[0, 2 π]$ ?

Key Points

1
The cosine series $\sum_{k = 1}^{\infty} cos (k x) / k^{2}$ has an explicit closed form $\frac{x ^{2}}{4} - \frac{π x}{2} + \frac{π ^{2}}{6}$ for all $x \in [0, 2 π]$ .
2
Finite cosine sums are converted into geometric sums by rewriting $cos (k x)$ with complex exponentials, enabling a closed-form sine-ratio expression.
3
The geometric-sum step requires $e^{i x} \neq = 1$ , so intermediate formulas exclude $x \in 2 π Z$ ; boundary points are handled later by convergence and continuity arguments.
4
A companion identity $\sum_{k = 1}^{\infty} sin (k x) / k = (π - x) /2$ on $(0, 2 π)$ is proved with uniform convergence on $[ε, 2 π - ε]$ .
5
Uniform convergence is used to justify exchanging an infinite series with an integral when deriving the $1/ k^{2}$ cosine series.
6
The constant term is found by integrating over $[0, 2 π]$ ; periodicity makes the cosine terms integrate to zero, leaving only the polynomial contribution.
7
Weierstrass M-test plus continuity extends the identity to the boundary points $0$ and $2 π$ .

Highlights

Turning cosine sums into geometric series via cos(kx)=21​(eikx+e−ikx) is the engine that removes the summation symbol.

Uniform convergence on [ε,2π−ε] is what makes term-by-term integration legitimate.

The boundary values x=0 and x=2π are recovered using uniform convergence on the whole interval and continuity, not by re-running the geometric-sum restriction.

Integrating over a full period kills every cos(kx) term, making the constant π2/6 fall out cleanly.

Topics

Fourier Series
Geometric Sum
Uniform Convergence
Term-by-Term Integration
Sine and Cosine Series

Mentioned

M-test