---
title: Fourier Series and Convergence
module: PDEs, Fourier Series, and Boundary Value Problems
moduleNumber: 9
lessonNumber: 1
order: 901
summary: >
  A two-point boundary value problem has nontrivial solutions only at a discrete
  set of eigenvalues, the same trichotomy that governs a singular linear system.
  For y'' + lambda y = 0 with zero endpoints the eigenfunctions are sines and
  cosines, and their orthogonality gives the Euler-Fourier coefficient formulas.
  The convergence theorem fixes when the series returns the function, the Gibbs
  phenomenon measures the overshoot at a jump, and even/odd symmetry produces
  half-range sine and cosine series.
topics: [PDEs, Fourier Series, and Boundary Value Problems]
sources:
  - book: Boyce
    ref: "Ch. 10 — Partial Differential Equations and Fourier Series; §10.1 Two-Point Boundary Value Problems; §10.2 Fourier Series; §10.3 The Fourier Convergence Theorem; §10.4 Even and Odd Functions"
  - book: Simmons
    ref: "Ch. 6 §33 The Fourier Coefficients; §34 The Problem of Convergence; §35 Even and Odd Functions; §37 Orthogonal Functions"
draft: false
---

A function defined on an interval can be written as an infinite sum of sines
and cosines. That expansion turns a partial differential equation into a family
of ordinary ones, one per term. The sines and cosines themselves come from a
boundary value problem posed on the interval, and an algebraic structure on
that problem selects which of them appear.

## Two-point boundary value problems

Until now the data for a second-order equation has been given at a single
point: the [initial value problem](/differential-equations/second-order-linear/homogeneous-constant-coefficients)
prescribes $y(t_0)$ and $y'(t_0)$ and asks for the solution afterward.
Physical problems in space instead prescribe the solution at two different
points, the ends of an interval.

> **Definition (Two-point boundary value problem).** A differential equation
> $y'' + p(x)y' + q(x)y = g(x)$ on $\alpha < x < \beta$ together with
> **boundary conditions** $y(\alpha) = y_0$, $y(\beta) = y_1$ specifying the
> value of the solution (or a derivative) at each endpoint. The problem is
> **homogeneous** when $g \equiv 0$ and both boundary values are zero;
> otherwise it is **nonhomogeneous**.

Moving from one point to two changes the question of existence. An
initial value problem with continuous coefficients has exactly one solution.
A boundary value problem may have one solution, none, or infinitely many.
That trichotomy is the same one that governs the linear algebraic system
$A\mathbf{x} = \mathbf{b}$:[^boyce-101]

- **Unique solution.** If $A$ is nonsingular, $A\mathbf{x} = \mathbf{b}$ has one
  solution for every $\mathbf{b}$, and the homogeneous system $A\mathbf{x} = 0$
  has only $\mathbf{x} = 0$.
- **None or infinitely many.** If $A$ is singular, $A\mathbf{x} = \mathbf{b}$
  has no solution unless $\mathbf{b}$ meets a compatibility condition, in which
  case it has infinitely many; the homogeneous system then has nontrivial
  solutions.

The nonhomogeneous boundary value problem inherits this dichotomy from its
associated homogeneous problem, exactly as a matrix system does.

For example, $y'' + 2y = 0$ with $y(0) = 1$, $y(\pi) = 0$ has the single
solution $y = \cos(\sqrt2\,x) - \cot(\sqrt2\,\pi)\sin(\sqrt2\,x)$: a unique
answer, matching a nonsingular $A$. But $y'' + y = 0$ with $y(0) = 1$,
$y(\pi) = a$ demands $c_1 = 1$ from the left condition and $-c_1 = a$ from the
right; these are compatible only when $a = -1$, and then $y = \cos x + c_2\sin
x$ leaves $c_2$ free. The same equation has no solution for $a \ne -1$ and
infinitely many for $a = -1$.

## The eigenvalue problem

The homogeneous problem that drives this module attaches a parameter to the
equation. Consider

$$
y'' + \lambda y = 0, \qquad y(0) = 0, \quad y(\pi) = 0.
$$

For every $\lambda$ this has the trivial solution $y \equiv 0$. The question
is for which $\lambda$ it has others. This is the differential analogue of
$A\mathbf{x} = \lambda\mathbf{x}$, and the vocabulary carries over.

> **Definition (Eigenvalue, eigenfunction).** A value of $\lambda$ for which
> the homogeneous boundary value problem has a nontrivial solution is an
> **eigenvalue**; the corresponding nonzero solutions are **eigenfunctions**.
> Like matrix eigenvectors, eigenfunctions are determined only up to a
> multiplicative constant.

The form of the general solution differs for positive, zero, and negative
$\lambda$, so the three cases are treated separately.

| Case | Equation | General solution | Boundary conditions force |
| --- | --- | --- | --- |
| $\lambda = \mu^2 > 0$ | $y'' + \mu^2 y = 0$ | $c_1\cos(\mu x) + c_2\sin(\mu x)$ | $c_1 = 0$, then $\sin(\mu\pi) = 0$ |
| $\lambda = 0$ | $y'' = 0$ | $c_1 x + c_2$ | $c_1 = c_2 = 0$: trivial only |
| $\lambda = -\mu^2 < 0$ | $y'' - \mu^2 y = 0$ | $c_1\cosh(\mu x) + c_2\sinh(\mu x)$ | $c_1 = c_2 = 0$: trivial only |

Only the positive case gives nontrivial solutions. There $\sin(\mu\pi) = 0$ requires $\mu$ to be
a positive integer, so the eigenvalues and eigenfunctions are

$$
\lambda_n = n^2, \qquad y_n(x) = \sin(nx), \qquad n = 1, 2, 3, \dots
$$

Replacing the endpoint $\pi$ by an arbitrary length $L$ shifts the condition
to $\sin(\mu L) = 0$, giving $\mu = n\pi/L$ and

$$
\lambda_n = \frac{n^2\pi^2}{L^2}, \qquad y_n(x) = \sin\!\left(\frac{n\pi x}{L}\right).
$$

These eigenfunctions are the standing shapes that a fixed-end string or a
zero-temperature rod can hold. Each is a half-integer number of arches fitted
into the interval.

$$
% caption: The first three eigenfunctions of $y'' + \lambda y = 0$ with
% zero endpoints: mode $n$ fits $n$ half-waves into $[0,L]$ and vanishes at
% both ends.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axis
\draw[black, ->] (-0.3,0) -- (4.6,0) node[right, black!70] {$x$};
\draw[black, dashed] (4,-1.4) -- (4,1.4);
\node[black!70, font=\footnotesize] at (4,-1.65) {$x=L$};
\node[black!70, font=\footnotesize] at (0,-0.28) {$0$};
% mode 1
\draw[acc, thick, domain=0:4, samples=80, smooth] plot (\x, {1.2*sin(45*\x)});
\node[acc, font=\footnotesize] at (2,1.45) {$n=1$};
% mode 2
\draw[black, thick, domain=0:4, samples=90, smooth] plot (\x, {1.0*sin(90*\x)});
\node[black, font=\footnotesize] at (0.7,-1.2) {$n=2$};
% mode 3
\draw[black, thick, dashed, domain=0:4, samples=120, smooth] plot (\x, {0.85*sin(135*\x)});
\node[black, font=\footnotesize] at (3.5,1.15) {$n=3$};
\end{tikzpicture}
$$

## Orthogonality

The sines and cosines from these problems are mutually orthogonal, and that
orthogonality supports the whole theory. Generalize the dot product of vectors
to functions.

> **Definition (Inner product, orthogonality).** The **inner product** of two
> real functions $u, v$ on $\alpha \le x \le \beta$ is $(u, v) = \int_\alpha^\beta
> u(x)v(x)\,\d x$. The functions are **orthogonal** when $(u, v) = 0$. A set is
> **mutually orthogonal** when every distinct pair is orthogonal.

The functions $\sin(m\pi x/L)$ and $\cos(m\pi x/L)$ for $m = 1, 2, \dots$ are
mutually orthogonal on $-L \le x \le L$. Direct integration, using the
product-to-sum identities, gives the orthogonality relations

$$
\int_{-L}^{L} \cos\!\left(\frac{m\pi x}{L}\right)\cos\!\left(\frac{n\pi x}{L}\right)\d x =
\begin{cases} 0, & m \ne n,\\ L, & m = n,\end{cases}
$$

$$
\int_{-L}^{L} \sin\!\left(\frac{m\pi x}{L}\right)\sin\!\left(\frac{n\pi x}{L}\right)\d x =
\begin{cases} 0, & m \ne n,\\ L, & m = n,\end{cases}
$$

$$
\int_{-L}^{L} \cos\!\left(\frac{m\pi x}{L}\right)\sin\!\left(\frac{n\pi x}{L}\right)\d x = 0 \quad \text{for all } m, n.
$$

Sines are orthogonal to cosines; distinct sines are orthogonal to each other,
as are distinct cosines; and a function integrated against itself returns its
squared length $L$. Integration plays the role of the dot product, and the
trigonometric functions form an orthogonal basis in the space of functions on
$[-L, L]$.

$$
% caption: Orthogonality as a Gram matrix: the inner product of any two
% distinct basis functions over a full period is zero, so only the diagonal
% survives.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\foreach \i in {0,1,2,3}{
  \foreach \j in {0,1,2,3}{
    \ifnum\i=\j
      \fill[acc!16] (\j,-\i) rectangle ++(1,1);
      \node at (\j+0.5,-\i+0.5) {$L$};
    \else
      \node[black] at (\j+0.5,-\i+0.5) {$0$};
    \fi
  }
}
\draw[black] (0,1) grid[step=1] (4,-3);
\node[black, font=\footnotesize] at (0.5,1.3) {$c_1$};
\node[black, font=\footnotesize] at (1.5,1.3) {$s_1$};
\node[black, font=\footnotesize] at (2.5,1.3) {$c_2$};
\node[black, font=\footnotesize] at (3.5,1.3) {$s_2$};
\node[black, font=\footnotesize] at (-0.4,0.5) {$c_1$};
\node[black, font=\footnotesize] at (-0.4,-0.5) {$s_1$};
\node[black, font=\footnotesize] at (-0.4,-1.5) {$c_2$};
\node[black, font=\footnotesize] at (-0.4,-2.5) {$s_2$};
\node[black!70, font=\footnotesize] at (2,-3.5) {$c_k$ cosine basis, $s_k$ sine basis};
\end{tikzpicture}
$$

## The Euler-Fourier formulas

Suppose a function $f$ on $-L \le x \le L$ can be written as the trigonometric
series

$$
f(x) = \frac{a_0}{2} + \sum_{m=1}^{\infty}\left[a_m\cos\!\left(\frac{m\pi x}{L}\right) + b_m\sin\!\left(\frac{m\pi x}{L}\right)\right].
$$

Orthogonality recovers every coefficient. Multiply both sides by $\cos(n\pi
x/L)$ and integrate from $-L$ to $L$. On the right, every product integrates to
zero except the one term with $m = n$ in the cosine sum, which contributes
$a_n L$. Hence $\int_{-L}^L f(x)\cos(n\pi x/L)\,\d x = a_n L$. The same maneuver
with $\sin(n\pi x/L)$ isolates $b_n$.

> **Definition (Fourier series).** The **Fourier series** of an integrable
> function $f$ of period $2L$ is the series above with coefficients given by
> the **Euler-Fourier formulas**
> $$
> a_n = \frac{1}{L}\int_{-L}^{L} f(x)\cos\!\left(\frac{n\pi x}{L}\right)\d x, \quad n = 0, 1, 2, \dots
> $$
> $$
> b_n = \frac{1}{L}\int_{-L}^{L} f(x)\sin\!\left(\frac{n\pi x}{L}\right)\d x, \quad n = 1, 2, 3, \dots
> $$

Writing the constant term as $a_0/2$ lets the single formula for $a_n$ cover
$n = 0$ as well: $a_0 = \frac{1}{L}\int_{-L}^L f\,\d x$ is twice the average of
$f$. Each coefficient is computed independently of the others, and each depends
only on the values of $f$ on one period. The coefficient $a_n$ is the
projection of $f$ onto the $n$-th cosine, divided by that cosine's squared
length, precisely the vector formula $a_i = (\mathbf{u}\cdot\mathbf{v}_i)/
(\mathbf{v}_i\cdot\mathbf{v}_i)$ read one dimension at a time.[^simmons-37]

> **Worked example.** Let $f(x) = 0$ on $-L < x < 0$ and $f(x) = L$ on
> $0 < x < L$, extended with period $2L$. The constant term is
> $a_0 = \frac{1}{L}\int_0^L L\,\d x = L$, and for $m \ge 1$,
>
> $$
> a_m = \frac{1}{L}\int_0^L L\cos\!\left(\frac{m\pi x}{L}\right)\d x = \frac{L}{m\pi}\sin(m\pi) = 0.
> $$
>
> Every cosine coefficient vanishes. The sine coefficients are
>
> $$
> b_m = \frac{1}{L}\int_0^L L\sin\!\left(\frac{m\pi x}{L}\right)\d x = \frac{L}{m\pi}\bigl(1 - \cos(m\pi)\bigr) =
> \begin{cases} 0, & m \text{ even},\\[2pt] \dfrac{2L}{m\pi}, & m \text{ odd},\end{cases}
> $$
>
> so the square wave is a sum of odd sines with amplitudes falling off as $1/m$:
>
> $$
> f(x) = \frac{L}{2} + \frac{2L}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n-1}\sin\!\left(\frac{(2n-1)\pi x}{L}\right).
> $$

## The convergence theorem

The Euler-Fourier formulas manufacture coefficients from any integrable $f$,
but the resulting series need not converge to $f$, and constructed examples
show it can fail at isolated points. Convergence requires a mild smoothness
hypothesis, the same **piecewise continuity** used to define the
[Laplace transform](/differential-equations/laplace/laplace-definition-ivps):
$f$ is piecewise continuous on an interval if the interval splits into finitely
many pieces on each of which $f$ is continuous and has finite one-sided limits.
Write $f(c^+)$ and $f(c^-)$ for the right- and left-hand limits at $c$.

> **Theorem (Fourier convergence).** Let $f$ and $f'$ be piecewise continuous
> on $-L \le x < L$, and let $f$ be extended periodically with period $2L$.
> Then the Fourier series of $f$ converges to $f(x)$ at every point where $f$
> is continuous, and to the average $\tfrac12\bigl(f(x^+) + f(x^-)\bigr)$ of
> the one-sided limits at every jump.[^boyce-103]

The hypotheses are sufficient, not necessary, and the proof is beyond scope,
but the statement gives the applications everything they need. At a jump the series
splits the difference: for the square wave, the endpoints $x = 0, \pm L$ carry
a jump between $0$ and $L$, and every sine term vanishes there, leaving the sum
$L/2$, the midpoint.

A Fourier series of infinitely differentiable terms can converge to a
discontinuous sum, and the convergence is not uniform.

> **Definition (Gibbs phenomenon).** Near a jump discontinuity, the partial
> sums of a Fourier series overshoot the function on each side of the jump. As
> more terms are added the overshoot narrows toward the jump but does not
> shrink in height; its limiting size is about $9\%$ of the jump. The error's
> least upper bound therefore does not fall to zero as $n \to \infty$.

$$
% caption: A five-term partial sum of the square wave. The approximation
% tracks the flat sections but overshoots at each jump, the Gibbs phenomenon,
% and the overshoot height persists as terms are added.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% axes
\draw[black, ->] (-0.3,0) -- (4.7,0) node[right, black!70] {$x$};
\draw[black, dashed] (0,-1.3) -- (0,1.3);
\node[black!70, font=\footnotesize] at (2,-1.55) {two periods};
% ideal square wave (period 2): +1 on (0,2), -1 on (2,4)
\draw[black, thick] (0,1) -- (2,1);
\draw[black, thick] (2,-1) -- (4,-1);
\draw[black, dashed] (2,-1) -- (2,1);
\draw[black, dashed] (0,-1) -- (0,1);
\draw[black, dashed] (4,-1) -- (4,1);
% 5-term partial sum, amplitude 4/pi ~ 1.273, L=2 so k-th odd harmonic = 90*(2k-1)*x deg
\draw[acc, thick, domain=0.02:3.98, samples=240, smooth]
  plot (\x, {1.273*(sin(90*\x) + sin(270*\x)/3 + sin(450*\x)/5 + sin(630*\x)/7 + sin(810*\x)/9)});
\node[acc, font=\footnotesize] at (1,1.55) {overshoot};
\draw[acc!70, ->] (1,1.4) -- (0.3,1.2);
\end{tikzpicture}
$$

The rate of decay of the coefficients governs the speed of convergence.
Smoother functions decay faster: the triangular wave, whose coefficients drop
like $1/m^2$, needs about $21$ terms for uniform error below $0.01$, while the
square wave's $1/m$ coefficients converge far more slowly. In general, the more
continuous derivatives $f$ has, the faster its series converges.

## Even and odd symmetry

Symmetry cuts the work in half and dictates which functions appear. Recall
$f$ is **even** when $f(-x) = f(x)$ and **odd** when $f(-x) = -f(x)$. Products
follow sign rules: even times even is even, odd times odd is even, even times
odd is odd. The integral consequences are what matter:

$$
\int_{-L}^{L} f\,\d x = 2\int_0^L f\,\d x \;\; (f \text{ even}), \qquad
\int_{-L}^{L} f\,\d x = 0 \;\; (f \text{ odd}).
$$

Apply these to the Euler-Fourier formulas. If $f$ is even, then $f(x)\sin(n\pi
x/L)$ is odd and integrates to zero, so every $b_n$ vanishes. If $f$ is odd,
every $a_n$ vanishes.

> **Definition (Cosine and sine series).** An even periodic function has a
> **Fourier cosine series** $f(x) = \tfrac{a_0}{2} + \sum a_n\cos(n\pi x/L)$
> with $a_n = \tfrac{2}{L}\int_0^L f\cos(n\pi x/L)\,\d x$ and all $b_n = 0$. An
> odd periodic function has a **Fourier sine series** $f(x) = \sum b_n\sin(n\pi
> x/L)$ with $b_n = \tfrac{2}{L}\int_0^L f\sin(n\pi x/L)\,\d x$ and all $a_n = 0$.

| Symmetry of $f$ | Series | Nonzero coefficients | Integral range |
| --- | --- | --- | --- |
| Even | cosine only | $a_n$ | $\int_0^L$, doubled |
| Odd | sine only | $b_n$ | $\int_0^L$, doubled |
| Neither | both | $a_n$ and $b_n$ | $\int_{-L}^{L}$ |

### Half-range expansions

A function given only on $0 \le x \le L$ can be expanded as either a sine or a
cosine series by first extending it to $[-L, L]$ with the symmetry you want,
then extending periodically.[^simmons-35]

- **Even (cosine) extension.** Reflect across the $y$-axis: $g(x) = f(-x)$ for
  $-L < x < 0$. Its cosine series represents $f$ on $[0, L]$.
- **Odd (sine) extension.** Reflect through the origin: $h(x) = -f(-x)$ for
  $-L < x < 0$, with $h(0) = 0$. Its sine series represents $f$ on $(0, L)$.

The same $f(x) = x$ on $[0, L]$ becomes a triangular wave under even extension
and a sawtooth under odd extension, with different series but the same values
on the original interval. The choice is dictated by the boundary conditions of
the problem being solved: a rod held at zero temperature at both ends calls for
a sine series, because sines vanish at the ends.

$$
% caption: A function on the half-interval $[0,L]$ (solid) extended two ways:
% the even reflection (left, dashed) builds a cosine series, the odd
% reflection (right, dashed) builds a sine series.
\begin{tikzpicture}[scale=0.95, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% ---- even extension (left panel) ----
\begin{scope}
\draw[black, ->] (-2.2,0) -- (2.4,0) node[right, black!70, font=\footnotesize] {$x$};
\draw[black, ->] (0,-0.3) -- (0,1.7);
% original on [0,2]: a rising line f=x/2
\draw[acc, thick] (0,0) -- (2,1.4);
% even reflection on [-2,0]
\draw[thick, dashed] (0,0) -- (-2,1.4);
\node[acc, font=\footnotesize] at (1.3,1.55) {$f$};
\node[black, font=\footnotesize] at (0,-0.55) {even extension};
\end{scope}
% ---- odd extension (right panel) ----
\begin{scope}[xshift=6.2cm]
\draw[black, ->] (-2.2,0) -- (2.4,0) node[right, black!70, font=\footnotesize] {$x$};
\draw[black, ->] (0,-1.7) -- (0,1.7);
\draw[acc, thick] (0,0) -- (2,1.4);
% odd reflection on [-2,0]
\draw[thick, dashed] (0,0) -- (-2,-1.4);
\node[acc, font=\footnotesize] at (1.3,1.55) {$f$};
\node[black, font=\footnotesize] at (0,-1.95) {odd extension};
\end{scope}
\end{tikzpicture}
$$

The same $f$ expanded both ways gives two series with the same values on
$[0, L]$ but different smoothness off it.

> **Worked example.** Take $f(x) = x$ on $0 \le x \le L$.
>
> _Even (cosine) extension._ The even extension is $|x|$ on $[-L, L]$, a
> triangular wave. All $b_n$ vanish, and
>
> $$
> a_0 = \frac{2}{L}\int_0^L x\,\d x = L, \qquad
> a_m = \frac{2}{L}\int_0^L x\cos\!\left(\frac{m\pi x}{L}\right)\d x = \frac{2L}{m^2\pi^2}\bigl(\cos m\pi - 1\bigr),
> $$
>
> so $a_m = -4L/(m^2\pi^2)$ for $m$ odd and $0$ for $m$ even, giving
>
> $$
> x = \frac{L}{2} - \frac{4L}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}\cos\!\left(\frac{(2n-1)\pi x}{L}\right).
> $$
>
> _Odd (sine) extension._ The odd extension is $x$ on $(-L, L)$, a sawtooth.
> All $a_n$ vanish, and integration by parts gives
>
> $$
> b_m = \frac{2}{L}\int_0^L x\sin\!\left(\frac{m\pi x}{L}\right)\d x = \frac{2L}{m\pi}(-1)^{m+1}, \qquad
> x = \frac{2L}{\pi}\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m}\sin\!\left(\frac{m\pi x}{L}\right).
> $$
>
> The cosine coefficients decay like $1/m^2$ and the series is continuous; the
> sine coefficients decay only like $1/m$, and the sawtooth jumps by $2L$ at
> each odd multiple of $L$, where the series returns the midpoint $0$ and
> overshoots by the Gibbs amount on either side.

$$
% caption: The two periodic extensions of $f(x)=x$ on $[0,L]$: the even
% extension (top) is a continuous triangular wave, the odd extension (bottom)
% a sawtooth with a jump at each end of the base interval.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
% ---- even (triangular) ----
\begin{scope}
\draw[black, ->] (-2.4,0) -- (2.5,0) node[right, black!70, font=\footnotesize] {$x$};
\draw[thick] (-2,0) -- (-1,1) -- (0,0) -- (1,1) -- (2,0);
\node[black, font=\footnotesize] at (1.55,0.95) {triangular};
\node[black, font=\footnotesize] at (-1.7,0.9) {even, cosine series};
\end{scope}
% ---- odd (sawtooth) ----
\begin{scope}[yshift=-3cm]
\draw[black, ->] (-2.4,0) -- (2.5,0) node[right, black!70, font=\footnotesize] {$x$};
\draw[thick] (-2,0) -- (-1,1);
\draw[thick] (-1,-1) -- (1,1);
\draw[thick] (1,-1) -- (2,0);
\draw[black, dashed] (-1,-1) -- (-1,1);
\draw[black, dashed] (1,-1) -- (1,1);
\node[black, font=\footnotesize] at (1.6,0.95) {sawtooth};
\node[black, font=\footnotesize] at (-1.55,0.9) {odd, sine series};
\node[black, font=\footnotesize] at (1,-1.3) {jump at $x=L$};
\end{scope}
\end{tikzpicture}
$$

[^boyce-101]: **Boyce**, _Elementary Differential Equations and Boundary Value Problems_, §10.1 — two-point boundary value problems, the parallel with singular and nonsingular linear systems $A\mathbf{x} = \mathbf{b}$, and the eigenvalue problem $y'' + \lambda y = 0$ with $\lambda_n = n^2\pi^2/L^2$.
[^simmons-37]: **Simmons**, _Differential Equations with Applications and Historical Notes_, §33, §37 — the Fourier coefficients derived from orthogonality, and orthogonal functions as an inner-product geometry with coefficients read as projections.
[^boyce-103]: **Boyce**, §10.3 — the Fourier Convergence Theorem for piecewise continuous $f$ and $f'$, convergence to the mean of one-sided limits at jumps, and the Gibbs phenomenon at discontinuities.
[^simmons-35]: **Simmons**, §35 — even and odd functions, the vanishing of the complementary coefficients, and half-range sine and cosine expansions via symmetric extension.