Fourier Transform 11 | Sum Formulas for Sine and Cosine

Q: How does the proof turn a finite sum of cosines into a closed-form expression without a summation sign?

Cosine is rewritten as an average of complex exponentials: $\cos(kx)=\tfrac12(e^{ikx}+e^{-ikx})$. The finite sum over $k$ becomes a sum of exponentials that can be shifted to start at index 0, making it a geometric series in $Q=e^{ix}$. The geometric-sum formula $\sum_{k=0}^{m}Q^k=(1-Q^{m+1})/(1-Q)$ applies for $Q\neq 1$, i.e., for $x\notin 2\pi\mathbb{Z}$. After substituting back and simplifying, the result is expressed using sine functions (through the conversion of $e^{i\theta}$ differences into $\sin(\theta)$).

Q: Why does the finite-sum formula fail at $x=0$ and other multiples of $2\pi$?

In the geometric-sum step, the denominator contains $1-e^{ix}$. When $x$ is a multiple of $2\pi$, $e^{ix}=1$, so the denominator becomes zero. That’s exactly the obstruction: the geometric-series closed form is undefined because the original exponential sum’s simplification divides by $1-Q$.

Q: How are the endpoint values at $x=0$ and $x=2\pi$ justified?

The lemma’s uniform convergence initially holds only on $[\varepsilon,2\pi-\varepsilon]$. To extend to the full closed interval, a Weierstrass M-test is invoked: $\sum 1/k^2$ converges, and $|\cos(kx)|\le 1$ provides a uniform majorant. This gives uniform convergence on $[0,2\pi]$. With uniform convergence, the limit function is continuous, and since the identity already matches on the open interval, continuity forces it to hold at the endpoints as well.

TL;DR

A finite sum of cosines is converted into a closed form by rewriting cosine with complex exponentials and evaluating a geometric series in $e^{i x}$ .

Briefing Cornell Notes

Briefing

A precise closed-form expression for a cosine Dirichlet-type series is derived and then used to extend convergence results all the way to the endpoints of the interval. The payoff is an identity that holds for every real x between 0 and 2π (including 0 and 2π):

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

That formula matters because it was a crucial ingredient in the earlier step toward proving Parseval’s identity for a particular step function; without it, the chain of Fourier-series arguments can’t be completed.

The derivation starts by proving a finite “sum of cosines” identity. Using the exponential representation of cosine, cos(kx) is rewritten as an average of complex exponentials. The resulting finite sum becomes a geometric series in $e^{i x}$ , which can be evaluated in closed form. After algebraic simplification, the expression is written using the sine function (via $sin (\cdot)$ terms that appear after converting complex exponentials back to real quantities). A key restriction emerges: the geometric-sum denominator vanishes when $x$ is a multiple of 2π, so the finite-sum formula is valid for $x \in / 2 π Z$ .

Next comes a lemma that controls a related series involving $sin (k x /2)$ -type terms divided by $k$ . The lemma shows that a sign-function series converges uniformly to a simple limit on any closed subinterval that stays a positive distance away from 0 and 2π. The proof uses the earlier cosine-sum formula by expressing the sign function as an integral of cosine, then applying integration by parts to produce a factor of $1/ (n + \frac{1}{2})$ -type decay. Uniform convergence follows by bounding the supremum norm: the denominator’s sine term stays away from zero on $[ε, 2 π - ε]$ , so the whole expression shrinks uniformly as $n \to \infty$ .

With that lemma in hand, the main theorem is obtained by integrating the sign-series identity. Integration turns the $1/ k$ -type structure into the desired $1/ k^{2}$ cosine series, and uniform convergence justifies exchanging the limit with the integral—an operation that would be risky without uniform control. The remaining challenge is the endpoints x=0 and x=2π, where the earlier lemma didn’t apply directly. To bridge that gap, a Weierstrass M-test argument provides uniform convergence on the full interval, and continuity of the limit function forces the identity to extend to the boundary.

Finally, the constant term is pinned down by integrating both sides over a full period. The cosine terms average to zero over $[0, 2 π]$ , leaving a simple polynomial integral that determines the constant $π^{2} /6$ . The result is a complete, endpoint-valid formula for the cosine series that underpins the Fourier-series proof strategy.

Cornell Notes

The series $\sum_{k = 1}^{\infty} cos (k x) / k^{2}$ is evaluated in closed form on $[0, 2 π]$ . The work begins with a finite cosine-sum identity derived by rewriting cosine using complex exponentials and summing a geometric series, yielding an expression in terms of sine functions (with a restriction excluding multiples of 2π). A key lemma then proves uniform convergence for a related sign-function series on $[ε, 2 π - ε]$ , using an integral representation of the sign function, integration by parts, and supremum-norm bounds. Integrating the lemma produces the $1/ k^{2}$ cosine series; uniform convergence plus continuity extends the identity to the endpoints. The remaining constant is determined by integrating over a full period, using that cosine averages to zero.

How does the proof turn a finite sum of cosines into a closed-form expression without a summation sign?

Cosine is rewritten as an average of complex exponentials:

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

. The finite sum over

k

becomes a sum of exponentials that can be shifted to start at index 0, making it a geometric series in

Q = e^{i x}

. The geometric-sum formula

\sum_{k = 0}^{m} Q^{k} = (1 - Q^{m + 1}) / (1 - Q)

applies for

Q \neq = 1

, i.e., for

x \in / 2 π Z

. After substituting back and simplifying, the result is expressed using sine functions (through the conversion of

e^{i θ}

differences into

sin (θ)

Why does the finite-sum formula fail at $x = 0$ and other multiples of $2 π$ ?

In the geometric-sum step, the denominator contains

1 - e^{i x}

. When

x

is a multiple of

2 π

e^{i x} = 1

, so the denominator becomes zero. That’s exactly the obstruction: the geometric-series closed form is undefined because the original exponential sum’s simplification divides by

1 - Q

What is the role of the lemma about sign-function series, and why is uniform convergence crucial?

The lemma proves that a related series involving sign-type terms converges uniformly to a simple limit on

[ε, 2 π - ε]

. Uniform convergence matters because the main theorem requires integrating a limiting series and exchanging the order of limit and integral. Without uniform convergence, swapping limit and integration could fail. The lemma’s proof produces a decay factor (from integration by parts) and uses the fact that on

[ε, 2 π - ε]

the sine denominator stays bounded away from zero, enabling supremum-norm bounds.

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

The sign function is represented as an integral of cosine: the sign-type term can be written as

\int cos (k t) d t

(up to constants). When the series is integrated term-by-term, the integral of

cos (k t)

introduces a factor of

1/ k

. Since the lemma’s series already has a

1/ k

-structure, the integration effectively upgrades it to

1/ k^{2}

, yielding

\sum cos (k x) / k^{2}

How are the endpoint values at $x = 0$ and $x = 2 π$ justified?

The lemma’s uniform convergence initially holds only on

[ε, 2 π - ε]

. To extend to the full closed interval, a Weierstrass M-test is invoked:

\sum 1/ k^{2}

converges, and

∣ cos (k x) ∣ \leq 1

provides a uniform majorant. This gives uniform convergence on

[0, 2 π]

. With uniform convergence, the limit function is continuous, and since the identity already matches on the open interval, continuity forces it to hold at the endpoints as well.

Review Questions

What geometric-series substitution is used to eliminate the finite cosine sum, and what condition on x prevents division by zero?
Where exactly does uniform convergence get used in the argument, and what operation would be unsafe without it?
How is the constant term in $\sum_{k = 1}^{\infty} cos (k x) / k^{2}$ determined, and why do cosine terms vanish when integrating over a full period?

Key Points

1
A finite sum of cosines is converted into a closed form by rewriting cosine with complex exponentials and evaluating a geometric series in $e^{i x}$ .
2
The geometric-sum denominator forces the finite identity to exclude $x \in 2 π Z$ .
3
A lemma establishes uniform convergence for a related sign-function series on $[ε, 2 π - ε]$ using an integral representation, integration by parts, and supremum-norm bounds.
4
Integrating the sign-series identity upgrades the $1/ k$ structure to the target $1/ k^{2}$ cosine series.
5
Uniform convergence on the full interval $[0, 2 π]$ is extended via a Weierstrass M-test, and continuity then forces the formula to hold at $x = 0$ and $x = 2 π$ .
6
The constant $π^{2} /6$ is determined by integrating both sides over $[0, 2 π]$ , where cosine terms average to zero over a full period.

Highlights

The cosine series has an endpoint-valid closed form: ∑k=1∞​cos(kx)/k2=4x2​−2πx​+6π2​ for all x∈[0,2π].

Uniform convergence is the hinge that allows exchanging a limit with an integral when moving from a sign-function series to the 1/k2 cosine series.

The geometric-series step produces sine-function denominators, explaining why multiples of 2π require special care.

The constant term is pinned down by integrating over a full period, leveraging that cosine averages to zero on [0,2π].

Topics

Fourier Series
Cosine Sum
Uniform Convergence
Geometric Series
Integration by Parts

Mentioned

M-test