Fourier Series

Introduction Orthogonality of Trigonometric Functions Fourier Coefficients Complex Exponential Form Notation / Sign Convention Parseval's Identity Convergence Properties

Introduction

The central idea of Fourier series is that a wide class of periodic functions can be represented as an infinite sum of sine and cosine functions. This remarkable result, developed by Joseph Fourier in his 1822 work "Théorie analytique de la chaleur" (The Analytical Theory of Heat) while studying heat conduction, has profound implications across mathematics, physics, engineering, and modern computer science.

Definition: Fourier Series Consider a function \(f: \mathbb{R} \to \mathbb{R}\) that is periodic with period \(2L\), meaning \(f(x + 2L) = f(x)\) for all \(x \in \mathbb{R}\). The Fourier series of \(f\), for \(f\) integrable over one period, is the series on the right of \[ f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right) \] where the Fourier coefficients are given by: \[ \begin{align*} a_0 &= \frac{1}{L} \int_{-L}^{L} f(x) \, dx \\\\ a_n &= \frac{1}{L} \int_{-L}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right) \, dx, \quad n \geq 1 \\\\ b_n &= \frac{1}{L} \int_{-L}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right) \, dx, \quad n \geq 1 \end{align*} \] The symbol \(\sim\) attaches the series to \(f\) without asserting convergence. The sense in which the series represents \(f\) is made precise in the convergence section. When \(f\) is piecewise smooth or of bounded variation, the series converges to \(f(x)\) at every point of continuity and to the midpoint of the one-sided limits at a jump discontinuity.

The constant term \(\frac{a_0}{2}\) represents the average value of the function over one period. Each term in the series represents a harmonic component. The \(n\)-th term oscillates \(n\) times over one period of \(f\). This decomposition reveals the frequency content of the signal, a perspective that proves invaluable in both theoretical analysis and practical computation.

Insight: Fourier Decomposition in CS

While Fourier series may seem purely theoretical, the discrete computational methods explored in the Fourier transform and FFT are actively used throughout modern technology.

  • Audio and image compression:
    MP3, AAC, and JPEG use discrete cosine transforms (closely related to Fourier series) to compress data by identifying which frequency components humans are least sensitive to.
  • Speech recognition and synthesis:
    Fourier-based spectrograms are commonly used to convert audio waveforms into features that neural networks can process.
  • Music information retrieval:
    Applications like Shazam identify songs by analyzing their frequency signatures using Fast Fourier Transforms (FFT).
  • Time series forecasting:
    Machine learning models for financial data, weather prediction, and sensor data often use Fourier features to capture periodic patterns (daily, weekly, seasonal cycles).

The fundamental idea of decomposing complex signals into simple periodic components remains at the heart of how many modern systems process audio, images, and temporal data. Understanding Fourier series provides the mathematical foundation for these ubiquitous technologies.

Orthogonality of Trigonometric Functions

The Fourier series formula claims that a wide class of periodic functions can be decomposed into sines and cosines. But how do we extract the individual coefficients? The answer lies in the same principle that makes orthogonal projections work in finite-dimensional spaces. Inner products with orthogonal basis functions isolate each component. We define an inner product on the space of functions over \([-L, L]\) by: \[ \langle f, g \rangle = \int_{-L}^{L} f(x)g(x) \, dx \]

The trigonometric functions satisfy the following orthogonality relations:

Theorem: Orthogonality relations of Trigonometric Functions

\[ \begin{align*} \int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right)\cos\left(\frac{n\pi x}{L}\right) \, dx &= \begin{cases} 0 & \text{if } m \neq n \\\\ L & \text{if } m = n \neq 0 \\\\ 2L & \text{if } m = n = 0 \end{cases} \\\\ \int_{-L}^{L} \sin\left(\frac{m\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right) \, dx &= \begin{cases} 0 & \text{if } m \neq n \\\\ L & \text{if } m = n \neq 0 \\\\ 0 & \text{if } m = n = 0 \end{cases} \\\\ \int_{-L}^{L} \cos\left(\frac{m\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right) \, dx &= 0 \end{align*} \] where \(m\) and \(n\) are any nonnegative integers.

Note that when \(m = n = 0\), \(\cos(0) = 1\), giving us the constant function whose self-inner product is \(2L\).

Proof:

All three identities reduce to integrating \(\cos\frac{k\pi x}{L}\) or \(\sin\frac{k\pi x}{L}\) for an integer \(k\). A direct computation gives, for any integer \(k \neq 0\), \[ \int_{-L}^{L} \cos\frac{k\pi x}{L} \, dx = \frac{L}{k\pi}\sin\frac{k\pi x}{L}\bigg|_{-L}^{L} = 0, \quad \int_{-L}^{L} \sin\frac{k\pi x}{L} \, dx = 0, \] the second integral vanishing by odd symmetry. For \(k = 0\) the cosine integral equals \(2L\) and the sine integral equals \(0\).

Cosine-cosine.
Using the product-to-sum identity \(\cos A \cos B = \tfrac{1}{2}[\cos(A-B) + \cos(A+B)]\), \[ \int_{-L}^{L} \cos\frac{m\pi x}{L}\cos\frac{n\pi x}{L} \, dx = \frac{1}{2}\int_{-L}^{L}\!\!\cos\frac{(m-n)\pi x}{L}\,dx + \frac{1}{2}\int_{-L}^{L}\!\!\cos\frac{(m+n)\pi x}{L}\,dx. \] For \(m \neq n\) both integers \(m-n\) and \(m+n\) are nonzero, so both integrals vanish. For \(m = n \geq 1\) the first integral equals \(2L\) and the second vanishes (since \(m+n \geq 2\)), yielding \(L\). For \(m = n = 0\) both integrands are constant \(1\), giving \(\tfrac{1}{2}(2L + 2L) = 2L\).

Sine-sine.
We use \(\sin A \sin B = \tfrac{1}{2}[\cos(A-B) - \cos(A+B)]\). If either \(m = 0\) or \(n = 0\), the integrand is identically zero. Otherwise \(m, n \geq 1\), and the same case analysis as above gives \(0\) when \(m \neq n\) and \(L\) when \(m = n\).

Sine-cosine.
The identity \(\sin A \cos B = \tfrac{1}{2}[\sin(A+B) + \sin(A-B)]\) reduces the integral to a sum of sine integrals, each of which vanishes on the symmetric interval \([-L, L]\) by odd symmetry.

These relations show that the set \[ \left\{1, \cos\left(\frac{\pi x}{L}\right), \sin\left(\frac{\pi x}{L}\right), \cos\left(\frac{2\pi x}{L}\right), \sin\left(\frac{2\pi x}{L}\right), \ldots\right\} \] forms an orthogonal system for the space of square-integrable functions on \([-L, L]\).

This system is moreover complete. Every square-integrable function can be approximated arbitrarily well in the \(L^2\) norm by finite linear combinations of these functions, so the system serves as an orthogonal basis of \(L^2[-L, L]\). Completeness combines the Riesz-Fischer theorem, proved on the page devoted to \(L^p\) completeness and convergence, with a density argument. We expand on this in the convergence section below.

Fourier Coefficients

With the orthogonality relations established, we can now derive the coefficient formulas rigorously. The strategy is the same as in finite-dimensional linear algebra. To find the component along a basis vector, we take the inner product of the function with that vector and divide by its squared norm.

A note on term-by-term integration.
The derivations below interchange the infinite sum with the integral, which is not automatic for all convergent series. This step is justified, for instance, when the Fourier series converges uniformly on \([-L, L]\), as is the case when \(f\) is continuously differentiable with \(f(-L) = f(L)\). The same step is justified more generally under the \(L^2\)-convergence framework developed in the convergence section below. There the inner products \(\langle f, \cos(n\pi x/L)\rangle\) and \(\langle f, \sin(n\pi x/L)\rangle\) are taken directly without series manipulations, and yield the same coefficient formulas. We proceed formally here and defer the analytic justification.

To find \(a_0\), integrate both sides of the Fourier series: \[ \begin{align*} \int_{-L}^{L} f(x) \, dx &= \int_{-L}^{L} \frac{a_0}{2} \, dx + \sum_{n=1}^{\infty} \left( a_n \int_{-L}^{L} \cos\left(\frac{n\pi x}{L}\right) \, dx + b_n \int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right) \, dx \right) \\\\ &= L a_0 \end{align*} \] Therefore: \[ a_0 = \frac{1}{L} \int_{-L}^{L} f(x) \, dx \]

To find \(a_n\) for \(n \geq 1\), multiply both sides by \(\cos\left(\frac{m\pi x}{L}\right)\) and integrate: \[ \begin{align*} \int_{-L}^{L} f(x)\cos\left(\frac{m\pi x}{L}\right) \, dx &= \int_{-L}^{L} \frac{a_0}{2}\cos\left(\frac{m\pi x}{L}\right) \, dx \\\\ &\quad + \sum_{n=1}^{\infty} \left( a_n \int_{-L}^{L} \cos\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi x}{L}\right) \, dx + b_n \int_{-L}^{L} \sin\left(\frac{n\pi x}{L}\right)\cos\left(\frac{m\pi x}{L}\right) \, dx \right) \\\\ &= L a_m \end{align*} \] Therefore: \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right) \, dx, \quad n \geq 1 \]

Similarly, multiplying by \(\sin\left(\frac{m\pi x}{L}\right)\) and integrating gives: \[ b_n = \frac{1}{L} \int_{-L}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right) \, dx, \quad n \geq 1 \]

Note: The two cosine formulas can be unified as: \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right) \, dx, \quad n \geq 0 \] where for \(n = 0\), we have \(\cos(0) = 1\), giving back \(a_0 = \frac{1}{L}\int_{-L}^{L} f(x) \, dx\).

The appearance of \(\frac{a_0}{2}\) in the series (rather than \(a_0\)) is a convention that makes the formula symmetric. These formulas allow us to compute the Fourier series representation of any suitable periodic function.

Example: Square Wave

Consider the square wave function with period \(2L\): \[ f(x) = \begin{cases} 1 & \text{if } 0 \lt x \lt L \\\\ -1 & \text{if } -L \lt x \lt 0 \end{cases} \]

First, compute \(a_0\): \[ a_0 = \frac{1}{L} \int_{-L}^{L} f(x) \, dx = \frac{1}{L}\left(\int_{-L}^{0} (-1) \, dx + \int_{0}^{L} 1 \, dx\right) = \frac{1}{L}(-L + L) = 0 \]

For \(n \geq 1\), since \(f\) is an odd function and \(\cos\left(\frac{n\pi x}{L}\right)\) is even, their product is odd: \[ a_n = \frac{1}{L} \int_{-L}^{L} f(x)\cos\left(\frac{n\pi x}{L}\right) \, dx = 0. \]

For the sine coefficients (\(f\) is odd, \(\sin\left(\frac{n\pi x}{L}\right)\) is odd, so their product is even): \[ \begin{align*} b_n &= \frac{1}{L} \int_{-L}^{L} f(x)\sin\left(\frac{n\pi x}{L}\right) \, dx \\\\ &= \frac{1}{L}\left(\int_{-L}^{0} (-1)\sin\left(\frac{n\pi x}{L}\right) \, dx + \int_{0}^{L} \sin\left(\frac{n\pi x}{L}\right) \, dx\right) \\\\ &= \frac{1}{L}\left[\frac{L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\bigg|_{-L}^{0} - \frac{L}{n\pi}\cos\left(\frac{n\pi x}{L}\right)\bigg|_{0}^{L}\right] \\\\ &= \frac{1}{n\pi}\left[(\cos(0) - \cos(-n\pi)) - (\cos(n\pi) - \cos(0))\right] \\\\ &= \frac{1}{n\pi}\left[(1 - \cos(n\pi)) - (\cos(n\pi) - 1)\right] \\\\ &= \frac{2}{n\pi}(1 - \cos(n\pi)). \end{align*} \] Since \(\cos(n\pi) = (-1)^n\): \[ b_n = \begin{cases} \frac{4}{n\pi} & \text{if } n \text{ is odd} \\\\ 0 & \text{if } n \text{ is even} \end{cases} \]

Therefore, the Fourier series is: \[ \begin{align*} f(x) &\sim \frac{4}{\pi}\sum_{k=0}^{\infty} \frac{\sin\left(\frac{(2k+1)\pi x}{L}\right)}{2k+1} \\\\ &= \frac{4}{\pi}\left(\sin\left(\frac{\pi x}{L}\right) + \frac{\sin\left(\frac{3\pi x}{L}\right)}{3} + \frac{\sin\left(\frac{5\pi x}{L}\right)}{5} + \cdots\right). \end{align*} \]

Complex Exponential Form

The real trigonometric form of the Fourier series involves separate cosine and sine coefficients, which can be notationally cumbersome. By combining them into complex exponentials via Euler's formula, we obtain a mathematically powerful representation that treats positive and negative frequencies symmetrically. This complex form is the standard starting point for the Fourier transform.

Assume \(f\) is periodic with \(f(-L) = f(L)\). The derivation below is a formal manipulation of the series. The analytic hypotheses needed for convergence are the same as in the real form, and are discussed in the convergence section. Using Euler's formula: \[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \] we can express trigonometric functions as: \[ \cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} \]

We can rewrite the Fourier series: \[ \begin{align*} f(x) &= \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right) \\\\ &= \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \left( \frac{e^{in\pi x/L} + e^{-in\pi x/L}}{2} \right) + \sum_{n=1}^{\infty} b_n \left( \frac{e^{in\pi x/L} - e^{-in\pi x/L}}{2i} \right) \\\\ &= \frac{a_0}{2} + \frac{1}{2}\sum_{n=1}^{\infty} (a_n - ib_n) e^{in\pi x/L} + \frac{1}{2}\sum_{n=1}^{\infty} (a_n + ib_n) e^{-in\pi x/L} \end{align*} \]

We now reindex the first summation by the dummy-variable substitution \(n \mapsto -n\). Under this substitution, \(e^{in\pi x/L}\) becomes \(e^{-i n\pi x/L}\), the summation range \(n = 1, 2, 3, \ldots\) becomes \(n = -1, -2, -3, \ldots\), and the coefficient \((a_n - i b_n)\) becomes \((a_{-n} - i b_{-n})\) (here \(-n\) is positive, so \(a_{-n}, b_{-n}\) are already-defined real Fourier coefficients): \[ \begin{align*} f(x) &= \frac{a_0}{2} + \frac{1}{2}\sum_{n=-\infty}^{-1} (a_{-n} - i b_{-n})\, e^{-i n\pi x/L} + \frac{1}{2}\sum_{n=1}^{\infty} (a_n + i b_n)\, e^{-i n\pi x/L}. \end{align*} \]

The two sums now share the same exponential \(e^{-in\pi x/L}\) but have different coefficient formulas depending on the sign of \(n\). To unify them into a single summation \(\sum_{n=-\infty}^{\infty}\), we define the complex Fourier coefficients \(c_n\) by cases:

\[ c_n = \begin{cases} \frac{1}{2}(a_n + ib_n) & \text{if } n \gt 0, \\\\ \frac{a_0}{2} & \text{if } n = 0, \\\\ \frac{1}{2}(a_{-n} - ib_{-n}) & \text{if } n \lt 0. \end{cases} \]

The two sums then collapse into a single summation over all integers. Substituting these definitions, we obtain the complex form of the Fourier series:

Definition: Complex Form of the Fourier Series

\[ f(x) \sim \sum_{n= -\infty}^{\infty} c_n e^{\frac{-i n \pi x}{L}} \] where, for \(n \gt 0\), the complex Fourier coefficients reduce to a single closed formula: \[ \begin{align*} c_n &= \frac{1}{2}(a_n + ib_n) \\\\ &= \frac{1}{2L} \int_{-L}^{L} f(x) \left( \cos \left(\frac{n \pi x}{L}\right) + i \sin \left(\frac{n \pi x}{L}\right) \right) \, dx \\\\ &= \frac{1}{2L} \int_{-L}^{L} f(x)e^{\frac{i n\pi x}{L}} \, dx \end{align*} \] A direct check shows that the cases \(n = 0\) and \(n \lt 0\) lead to the same final expression, so the closed integral formula holds for every integer \(n\). Note that if \(f\) is real-valued, the coefficients satisfy conjugate symmetry: \(c_{-n} = \overline{c_n}\).

Derivation using Orthogonality:

A complex function \(\phi(x)\) is orthogonal to another complex function \(\psi(x)\) over an interval \(a \leq x \leq b\) if \[ \int_a^b \overline{\phi}\psi \, dx = 0 \] where \(\overline{\phi}\) is the complex conjugate of \(\phi\).

For \(-\infty \lt n \lt \infty\), the exponentials \(e^{\frac{-i n \pi x}{L}}\) form an orthogonal set, as the computation below verifies: \[ \int_{-L}^L \left(\overline{e^{\frac{- i m\pi x}{L}}}\right) e^{\frac{ - i n\pi x}{L}} \, dx = \begin{cases} 0 & \text{if } m \neq n \\\\ 2L & \text{if } m = n \end{cases} \] because \(\left(\overline{e^{\frac{- i m\pi x}{L}}}\right) = e^{\frac{i m\pi x}{L}} \).

We now multiply the complex Fourier series by \(e^{\frac{i m \pi x}{L}}\) and integrate from \(-L\) to \(L\): \[ \int_{-L}^L f(x) e^{\frac{i m \pi x}{L}} \, dx = \sum_{n= -\infty}^{\infty} c_n \int_{-L}^L e^{\frac{-i n \pi x}{L}} e^{\frac{i m \pi x}{L}} \, dx. \] Using the complex orthogonality condition, only the \(m = n\) term survives, and thus we obtain the complex Fourier coefficients: \[ \begin{align*} &\int_{-L}^L f(x) e^{\frac{i m \pi x}{L}} \, dx = 2Lc_m \\\\ &\Longrightarrow c_m = \frac{1}{2L} \int_{-L}^{L} f(x)e^{\frac{i m\pi x}{L}} \, dx. \end{align*} \]

Notation / Sign Convention

Fourier conventions differ between fields and between textbooks, and formulas copied from one source into another are a common source of sign errors and stray factors of \(2\pi\). For series, a convention is fixed by two independent choices: the sign of the exponent in the coefficient integral, and the place where the normalizing factor sits. For the transform there is a third choice, described below. Some sources also normalize the basis itself, using \(\frac{1}{\sqrt{2L}}e^{-\frac{in\pi x}{L}}\), whose coefficients are \(\sqrt{2L}\) times ours, and others work on \([0, T]\) with exponents \(\pm\frac{2\pi i n x}{T}\), which is the same system with \(T = 2L\). This section records the choices made on this site and gives a dictionary for translating results from other sources.

Convention used in this text:

We adopt the following complex-exponential convention for the Fourier series: \[ \begin{align*} f(x) &\sim \sum_{n=-\infty}^{\infty} c_n e^{-\frac{i n\pi x}{L}}, \\\\ c_n &= \frac{1}{2L}\int_{-L}^{L} f(x)\,e^{+\frac{i n\pi x}{L}}\,dx. \end{align*} \] The exponent is negative in the series and positive in the coefficient integral. Most analysis texts, most engineering and signal-processing texts, and the standard FFT libraries use the opposite sign.

Why this sign:

Neither sign is more correct. We choose the sign used for the characteristic function \(\mathbb{E}\bigl[e^{i\xi X}\bigr]\), the expectation of \(e^{i\xi X}\), of a random variable \(X\) in probability theory, so that formulas about characteristic functions carry over without translation. For \(X\) with density \(p\), the Fourier transform of this site, defined below, gives exactly \(\mathbb{E}\bigl[e^{i\xi X}\bigr] = \hat{p}(\xi)\), with no change of sign and no constant. The same sign matches time evolution in physics. A stationary quantum state of energy \(E\) evolves by the factor \(e^{-iEt/\hbar}\), where \(\hbar\) is the reduced Planck constant. Writing a function of time as a superposition of the waves \(e^{-i\omega t}\), as the inverse transform below does with \(e^{-ix\xi}\), therefore assigns the positive frequency \(\omega = E/\hbar\) to a positive energy. In space, physics uses the opposite sign, as described below. The heat, wave, and Laplace pages of this site use the convention throughout. The cost is on the computational side, since the standard FFT libraries use the opposite sign, and the discrete part of the Fourier transform page switches to theirs.

Properties of the basis:

The basis functions \(e^{-\frac{i n\pi x}{L}}\) are eigenfunctions of the derivative operator: \[ \frac{d}{dx}\,e^{-\frac{i n\pi x}{L}} = -\frac{i n\pi}{L}\,e^{-\frac{i n\pi x}{L}}, \] so applying the derivative twice multiplies by \(-\frac{n^2\pi^2}{L^2}\), and these nonpositive eigenvalues of the second-derivative operator match the negative semidefinite nature of the Laplacian in PDEs.

These basis functions also satisfy the orthogonality relation computed above. The conjugate relationship \(\overline{e^{-\frac{i n\pi x}{L}}} = e^{+\frac{i n\pi x}{L}}\) is what ensures that multiplying the series by \(e^{+\frac{i m\pi x}{L}}\) and integrating isolates the coefficient \(c_m\) directly. Both properties hold under the opposite sign as well, with \(n\) replaced by \(-n\). The choice of sign is a convention, not a matter of correctness.

Energy and the normalized inner product:

With the normalized inner product \(\langle f, g \rangle = \frac{1}{2L}\int_{-L}^{L} f(x)\,\overline{g(x)}\,dx\), the complex exponential system forms an orthonormal basis of \(L^2[-L, L]\), and the map \(f \mapsto \{c_n\}\) is unitary. The resulting identity between the \(L^2\) norm of \(f\) and the coefficient sum is stated and proved in the next section.

The opposite sign convention:

Under the opposite sign, the series and its coefficients read \[ \begin{align*} f(x) &\sim \sum_{n=-\infty}^{\infty} \tilde{c}_n e^{+\frac{i n\pi x}{L}}, \\\\ \tilde{c}_n &= \frac{1}{2L}\int_{-L}^{L} f(x)\,e^{-\frac{i n\pi x}{L}}\,dx. \end{align*} \] The two conventions are related by \(\tilde{c}_n = c_{-n}\). Since this is just a re-indexing, all mathematical properties (orthogonality, completeness, Parseval's identity) remain valid in both conventions. Analysis texts often take the period to be \(2\pi\), that is, \(L = \pi\), and write \(\hat{f}(n)\) for \(\tilde{c}_n\). That \(\hat{f}(n)\) uses the negative sign, whereas the transform \(\hat{f}(\xi)\) of this site, defined below, uses the positive sign. The same hat can mean opposite signs in different sources.

The Fourier transform: a dictionary

For the Fourier transform, the sign and the placement of \(2\pi\) are joined by the third choice: whether the frequency is angular (radians per unit of the variable) or ordinary (cycles per unit). This site uses angular frequency and the convention \[ \begin{align*} \hat{f}(\xi) &= \int_{-\infty}^{\infty} f(x)\,e^{+i x\xi}\,dx, \\\\ f(x) &= \frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(\xi)\,e^{-i x\xi}\,d\xi. \end{align*} \] The same sign appears in the characteristic function \(\mathbb{E}\bigl[e^{i\xi X}\bigr]\) of a random variable \(X\) in probability theory, which equals this transform of the density of \(X\) when \(X\) has one. The first table lists the common alternatives. Its last column expresses each transform \(F\) of a function \(f\) through this site's \(\hat{f}\), so a formula from any of these sources can be translated by substitution.

Convention Forward transform \(F\) \(F\) through \(\hat{f}\)
(a) This site \(\int f(x)\,e^{+ix\xi}\,dx\) \(\hat{f}(\xi)\)
(b) Positive sign, \(\frac{1}{2\pi}\) in front \(\frac{1}{2\pi}\int f(x)\,e^{+i\omega x}\,dx\) \(\frac{1}{2\pi}\hat{f}(\omega)\)
(c) Negative sign, angular \(\int f(t)\,e^{-i\omega t}\,dt\) \(\hat{f}(-\omega)\)
(d) Ordinary frequency \(\int f(x)\,e^{-2\pi i x\nu}\,dx\) \(\hat{f}(-2\pi\nu)\)
(e) Unitary, negative sign \(\frac{1}{\sqrt{2\pi}}\int f(x)\,e^{-ix\xi}\,dx\) \(\frac{1}{\sqrt{2\pi}}\hat{f}(-\xi)\)
(f) Unitary, positive sign \(\frac{1}{\sqrt{2\pi}}\int f(t)\,e^{+i\omega t}\,dt\) \(\frac{1}{\sqrt{2\pi}}\hat{f}(\omega)\)

Where they are used: (a) on this site and, for densities, in probability; (b) in some applied texts on partial differential equations; (c) in engineering texts that use angular frequency, which usually write \(j\) for the imaginary unit; (d) in many analysis texts and in engineering texts that measure frequency in hertz; (e) in some analysis texts and in quantum mechanics in the spatial variable, with \(\xi = p/\hbar\) for the momentum \(p\), and with \(\sqrt{2\pi\hbar}\) in place of \(\sqrt{2\pi}\); (f) as the default of Mathematica's transform.

The second table records how three basic rules change. Write \(G\) for the transform of \(g\) in the same convention. The middle column is Plancherel's identity, and the last column is the transform of the convolution \(f * g\) of the convolution theorem.

Convention Transform of \(f'\) \(\int |f|^2\) Transform of \(f * g\)
(a) \(-i\xi\,\hat{f}\) \(\frac{1}{2\pi}\int |\hat{f}|^2\) \(\hat{f}\,\hat{g}\)
(b) \(-i\omega F\) \(2\pi\int |F|^2\) \(2\pi\, F G\)
(c) \(i\omega F\) \(\frac{1}{2\pi}\int |F|^2\) \(F G\)
(d) \(2\pi i\nu F\) \(\int |F|^2\) \(F G\)
(e) \(i\xi F\) \(\int |F|^2\) \(\sqrt{2\pi}\, F G\)
(f) \(-i\omega F\) \(\int |F|^2\) \(\sqrt{2\pi}\, F G\)

Three patterns are worth remembering. Reversing the sign replaces the frequency by its negative, so the derivative factor changes sign. Placing \(\frac{1}{2\pi}\) on one side only leaves a factor \(2\pi\) or \(\frac{1}{2\pi}\) in Plancherel's identity, while the unitary and ordinary-frequency forms preserve the \(L^2\) norm exactly. The convolution theorem carries no constant exactly when the forward transform has no constant in front of the integral.

The most common physics convention mixes the two signs. A plane wave \(e^{i(kx - \omega t)}\), with wavenumber \(k\) and angular frequency \(\omega\), carries the positive sign in space and the negative sign in time. Expansions in time therefore use the synthesis kernel \(e^{-i\omega t}\), so a transform in time falls under (a), (b), or (f), depending on where it places \(2\pi\). Expansions in space use \(e^{+ikx}\), as in (c), (d) with \(k = 2\pi\nu\), and (e). Some optics and electrical engineering texts write the plane wave as \(e^{i(\omega t - kx)}\), which reverses both signs.

Reading another source or code:

  1. Sign. Find the exponent in the forward transform or the coefficient integral. If it is negative, replace \(\xi\) by \(-\xi\), or \(n\) by \(-n\), before comparing with this site.
  2. Placement of \(2\pi\). Check whether \(2\pi\) sits in the exponent (ordinary frequency), in front of the inverse, in front of the forward transform, or as \(\frac{1}{\sqrt{2\pi}}\) on both.
  3. Test pairs. The Gaussian \(e^{-x^2/2}\) fixes the normalization. This site sends it to \(\sqrt{2\pi}\,e^{-\xi^2/2}\), the unitary forms send it to \(e^{-\xi^2/2}\), form (b) sends it to \(\frac{1}{\sqrt{2\pi}}e^{-\omega^2/2}\), and the ordinary-frequency form sends \(e^{-\pi x^2}\) to \(e^{-\pi\nu^2}\). The Gaussian is even, so it cannot detect the sign. The one-sided exponential, equal to \(e^{-x}\) for \(x \gt 0\) and to \(0\) for \(x \lt 0\), detects it. This site sends it to \(\frac{1}{1 - i\xi}\), while (c) sends it to \(\frac{1}{1 + i\omega}\).
  4. Discrete transforms. NumPy, SciPy, MATLAB, and FFTW define the discrete Fourier transform with the negative sign, \(X_k = \sum_{n=0}^{N-1} x_n e^{-2\pi i kn/N}\) for data \(x_0, \ldots, x_{N-1}\). All four compute this unnormalized sum in the forward direction. NumPy, SciPy, and MATLAB divide the inverse by \(N\) by default. FFTW does not, so a forward and backward pass multiplies by \(N\). NumPy and SciPy offer other placements of the factor as an option. For a continuous \(2L\)-periodic \(f\) sampled as \(x_n = f(t_n)\) at the points \(t_n = 2Ln/N\), one period on \([0, 2L)\), the quantity \(\frac{1}{N}X_k\) approximates \(\tilde{c}_k = c_{-k}\) for each fixed \(k\) as \(N\) grows, where a negative \(k\) is read as the index \(k + N\). The discrete part of the Fourier transform page uses the negative-sign convention.

Throughout this text, we consistently use the convention fixed at the top of this section, except on the Fourier transform page from its discrete Fourier transform section onward, which follows the software convention. When consulting other sources or implementing algorithms, run the checks above before combining formulas.

The complex form is often preferred because it simplifies many operations. For a real-valued \(f\), the conjugate symmetry recorded above is what makes the imaginary parts cancel when the series is summed, so that the result is again a real signal.

Parseval's Identity

In finite-dimensional linear algebra, the squared norm of a vector equals the sum of squares of its components in an orthonormal basis, an iterated form of the Pythagorean theorem that appears there as norm preservation under orthonormal coordinates. Parseval's identity is the infinite-dimensional analogue, and it guarantees that no information is lost or created when passing between the time and frequency domains.

Notation. Throughout this section, \(|\cdot|\) denotes the complex modulus. For a complex number \(z = x + iy\), we have \(|z|^2 = x^2 + y^2\). For real numbers, this reduces to the ordinary absolute value.

Theorem: Parseval's Identity

Parseval's identity relates the total energy of a signal in the time domain to its energy in the frequency domain. For a real-valued function \(f\) with Fourier series coefficients \(a_n\) and \(b_n\), it states: \[ \frac{1}{L}\int_{-L}^{L} |f(x)|^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2). \]

In the complex exponential form, which holds for complex-valued \(f\) as well, this becomes: \[ \frac{1}{2L}\int_{-L}^{L} |f(x)|^2 \, dx = \sum_{n=-\infty}^{\infty} |c_n|^2. \]

Insight: Parseval's Identity in Signal Processing and ML

  • Energy Conservation:
    The total "energy" (\(L^2\) norm) of a signal is identical whether measured in the time domain or the frequency domain. This ensures that the transformation itself does not distort the information content of the data.
  • Data Compression (for example, JPEG and MP3):
    Because total energy is preserved, we can truncate small Fourier coefficients that contribute little to the total energy. This allows us to discard "insignificant" data with small energy error, which in practice translates into minimal loss of perceived quality.
  • Feature Selection:
    Parseval's identity allows us to identify which frequency components contain the majority of the signal's energy, providing a rigorous way to reduce dimensionality in machine learning tasks.
  • Noise Filtering:
    Signals typically concentrate energy in a few specific coefficients, whereas white noise tends to spread its energy across all frequencies. This energy distribution analysis is the basis for spectral denoising.
Proof:

We prove the complex form first. A fully rigorous argument proceeds through the partial sums \(S_N f(x) = \sum_{|n|\leq N} c_n e^{-in\pi x/L}\): for each finite \(N\) the double sum below is finite, so interchanging sum and integral is elementary, and the identity \(\|S_N\|_2^2 = \sum_{|n|\leq N} |c_n|^2\) follows. Taking \(N \to \infty\) and using \(L^2\)-convergence \(S_N \to f\) in \(L^2[-L,L]\), established in the convergence section and resting on the Riesz-Fischer theorem of the \(L^p\) completeness page, yields the full identity. We present the calculation in the suggestive infinite-sum form below, with this limiting procedure understood.

Starting with the Fourier series \[ f(x) = \sum_{n=-\infty}^{\infty} c_n e^{-i\frac{n\pi x}{L}}, \] so that \(\overline{f(x)} = \sum_{m=-\infty}^{\infty} \overline{c_m}\, e^{+i\frac{m\pi x}{L}}\), we compute: \[ \begin{align*} \frac{1}{2L}\int_{-L}^{L} |f(x)|^2 \, dx &= \frac{1}{2L}\int_{-L}^{L} f(x) \overline{f(x)} \, dx \\\\ &= \frac{1}{2L}\int_{-L}^{L} \left(\sum_{n=-\infty}^{\infty} c_n e^{-i\frac{n\pi x}{L}}\right) \left(\sum_{m=-\infty}^{\infty} \overline{c_m} e^{+i\frac{m\pi x}{L}}\right) dx \\\\ &= \frac{1}{2L}\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty} c_n \overline{c_m} \int_{-L}^{L} e^{i\frac{(m-n)\pi x}{L}} \, dx. \end{align*} \] By the orthonormality of \(\left\{e^{-i\frac{n\pi x}{L}}\right\}\) with respect to the inner product \(\langle f, g \rangle = \frac{1}{2L}\int_{-L}^{L} f\overline{g} \, dx\): \[ \int_{-L}^{L} e^{i\frac{(m-n)\pi x}{L}} \, dx = \begin{cases} 2L & \text{if } n = m \\\\ 0 & \text{if } n \neq m \end{cases} \] Therefore: \[ \frac{1}{2L}\int_{-L}^{L} |f(x)|^2 \, dx = \sum_{n=-\infty}^{\infty} |c_n|^2. \]

To obtain the real form, assume \(f\) is real-valued, so that \(a_n, b_n \in \mathbb{R}\), and use the relationships between real and complex coefficients. For \(n \geq 1\): \[ \begin{align*} |c_n|^2 + |c_{-n}|^2 &= \left|\frac{a_n + ib_n}{2}\right|^2 + \left|\frac{a_n - ib_n}{2}\right|^2 \\\\ &= \frac{a_n^2 + b_n^2}{4} + \frac{a_n^2 + b_n^2}{4} = \frac{a_n^2 + b_n^2}{2} \end{align*} \] and \(|c_0|^2 = \left|\frac{a_0}{2}\right|^2 = \frac{a_0^2}{4}\). Thus: \[ \begin{align*} \sum_{n=-\infty}^{\infty} |c_n|^2 &= |c_0|^2 + \sum_{n=1}^{\infty} (|c_n|^2 + |c_{-n}|^2) \\\\ &= \frac{a_0^2}{4} + \sum_{n=1}^{\infty} \frac{a_n^2 + b_n^2}{2}. \end{align*} \] Since the left side equals \(\frac{1}{2L}\int_{-L}^{L} |f(x)|^2 \, dx\), multiplying both sides by 2 gives: \[ \frac{1}{L}\int_{-L}^{L} |f(x)|^2 \, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2). \]

Convergence Properties

Having derived the Fourier coefficients and established Parseval's identity, a fundamental question remains: in what sense does the Fourier series actually converge to the original function \(f\)? The answer is surprisingly nuanced and depends on the regularity of \(f\). Three types of convergence arise naturally, each with different assumptions and implications.

1. Pointwise Convergence:

Theorem: Pointwise Convergence (Dirichlet-Jordan)

If \(f\) is periodic and of bounded variation on \([-L, L]\), then at every point \(x\), the Fourier series converges to the average of the left and right limits: \[ \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos\tfrac{n\pi x}{L} + b_n \sin\tfrac{n\pi x}{L}\right) = \frac{f(x^+) + f(x^-)}{2} \] where \(f(x^+) = \lim_{h \to 0^+} f(x+h)\) and \(f(x^-) = \lim_{h \to 0^-} f(x+h)\). At points of continuity, this equals \(f(x)\).

Functions of bounded variation include most functions encountered in practice, such as piecewise smooth and piecewise monotone functions.

Proof Outline:

A complete proof is beyond our present scope, but the main line of argument is instructive and rests on tools that are for the most part already in place. We outline the four key steps, flagging the two ingredients that are supplied elsewhere.

Step 1: Dirichlet kernel representation.
Substituting the integral formula for \(c_n\) into the partial sum \(S_N f(x) = \sum_{|n|\leq N} c_n e^{-in\pi x/L}\) and interchanging the finite sum with the integral gives \[ S_N f(x) = \frac{1}{2L}\int_{-L}^{L} f(y)\, D_N(y - x)\, dy, \quad D_N(t) = \sum_{n=-N}^{N} e^{in\pi t/L}. \] A geometric-series computation puts \(D_N\) in closed form for \(t\) not a multiple of \(2L\): \[ D_N(t) = \frac{\sin\!\left((N+\tfrac{1}{2})\pi t/L\right)}{\sin(\pi t/(2L))}. \] Note that \(D_N\) is an even function of \(t\) and is periodic with period \(2L\). Both facts will be used below.

Step 2: Kernel mass.
Integrating \(D_N\) term-by-term and keeping only the \(n = 0\) term gives \[ \frac{1}{2L}\int_{-L}^{L} D_N(t)\, dt = 1 \quad \text{for every } N. \] This is the mass condition one would expect of an approximate identity. The obstruction to finishing the argument by a naive approximate-identity estimate is that \(D_N\) is not positive. Its \(L^1\) norm \(\frac{1}{2L}\int_{-L}^{L}|D_N(t)|\,dt\) grows like \(\log N\), so cancellation rather than concentration of mass is what drives convergence.

Step 3: Symmetrization and reduction via bounded variation.
Substituting \(y = x + t\) in the kernel representation and using periodicity together with evenness of \(D_N\), the partial sum takes the symmetrized form \[ S_N f(x) = \frac{1}{2L}\int_0^L \bigl[f(x+t) + f(x-t)\bigr]\, D_N(t)\, dt. \] Combined with the normalization \(\frac{1}{2L}\int_0^L D_N(t)\,dt = \tfrac{1}{2}\) (half of the full-period mass, by evenness and Step 2), this gives \[ S_N f(x) - \tfrac{1}{2}\bigl(f(x^+) + f(x^-)\bigr) = \frac{1}{2L}\int_0^L \Big(\bigl[f(x+t) - f(x^+)\bigr] + \bigl[f(x-t) - f(x^-)\bigr]\Big)\,D_N(t)\,dt. \] The goal is to show this tends to \(0\) as \(N \to \infty\).

Applying the Jordan decomposition \(f = f_1 - f_2\) of a bounded-variation function into monotone pieces, we reduce the problem to bounding each of four integrals of the form \[ I_{j,\pm}(N) = \frac{1}{2L}\int_0^L \bigl[f_j(x \pm t) - f_j(x^\pm)\bigr]\,D_N(t)\,dt, \] where \(j \in \{1,2\}\) and the sign is chosen accordingly. For each \(I_{j,\pm}\), the integrand \(g(t) := f_j(x \pm t) - f_j(x^\pm)\) is monotone in \(t\) with \(g(0^+) = 0\). One then fixes a small \(\delta \gt 0\) and splits the integral at \(\delta\):

  • Tail \([\delta, L]\). Here \(\sin(\pi t/(2L))\) is bounded away from \(0\), so the integrand \(g(t)/\sin(\pi t/(2L))\) is bounded and integrable. By the Riemann-Lebesgue lemma (Step 4), \(\int_\delta^L g(t) D_N(t)\,dt \to 0\) as \(N \to \infty\).
  • Near-origin \([0, \delta]\). The second mean-value theorem for integrals, which we quote from the classical literature rather than develop here, applies to the monotone function \(g\) on \([0, \delta]\) and yields an expression of the form \(g(\delta^-) \cdot \int_\xi^\delta D_N(t)\,dt\) for some \(\xi \in [0, \delta]\). A classical computation shows that partial integrals of the Dirichlet kernel are bounded uniformly in \(N\) and in the endpoints. There exists a constant \(C\) such that \(\bigl|\int_a^b D_N(t)\,dt\bigr| \leq C\) for all \(0 \le a \lt b \le L\) and all \(N\). Since \(g(\delta^-) \to 0\) as \(\delta \to 0^+\), this piece can be made arbitrarily small by choosing \(\delta\) small, uniformly in \(N\).

Given \(\varepsilon \gt 0\), first fix \(\delta\) small enough to control the near-origin piece, then let \(N \to \infty\) to eliminate the tail. This yields \(I_{j,\pm}(N) \to 0\), and summing the four contributions gives the claimed limit.

Step 4: The Riemann-Lebesgue lemma.
The vanishing of the tail integrals in Step 3 as \(N \to \infty\) rests on the Riemann-Lebesgue lemma: for any function \(g\) integrable on a bounded interval, the oscillatory integrals \(\int g(t)\, e^{\pm iNt}\,dt\) tend to \(0\) as \(N \to \infty\). In our application the relevant tail integrand is bounded, hence square-integrable on \([\delta, L]\), so the \(L^2\) form of the lemma is what we need. That form follows directly from Bessel's inequality and is proved on the forthcoming page on Fourier analysis in Hilbert spaces. The extension to general integrable functions, obtained via density, is also established there.

Putting the four steps together yields the stated pointwise limit.

2. Mean Square (\(L^2\)) Convergence:

For any square-integrable function \(f \in L^2[-L, L]\) (the space of functions with \(\int_{-L}^{L} |f(x)|^2 \, dx \lt \infty\), understood as usual up to equality almost everywhere), the Fourier series converges in the mean square sense: \[ \lim_{N \to \infty} \int_{-L}^{L} \left|f(x) - \left(\frac{a_0}{2} + \sum_{n=1}^{N} \left(a_n\cos\left(\frac{n\pi x}{L}\right) + b_n\sin\left(\frac{n\pi x}{L}\right)\right)\right)\right|^2 dx = 0 \] This mean-square convergence rests on the fact that the trigonometric system is a complete orthogonal basis of \(L^2[-L, L]\). Two ingredients combine to establish this.

First, the Riesz-Fischer theorem, established in the upcoming page on \(L^p\) completeness and convergence, guarantees that \(L^2[-L, L]\) is a complete Hilbert space under the inner product \(\langle f, g\rangle = \frac{1}{2L}\int_{-L}^{L} f(x)\overline{g(x)}\,dx\). Second, the density of trigonometric polynomials in \(L^2[-L, L]\) follows from the Weierstrass approximation theorem combined with the density of continuous periodic functions in \(L^2\). Once these are in hand, partial-sum convergence in \(L^2\) is a standard consequence of general Hilbert-space theory applied to an orthonormal basis, as developed in the page introducing Banach and Hilbert spaces.

The role of Lebesgue integration here is to supply the \(L^2\) framework itself. Without it, the space of "square-integrable functions" would not be complete, and a square-summable sequence of coefficients need not be the coefficient sequence of any function. The result matters for four reasons:


3. The Gibbs Phenomenon:

At jump discontinuities, the partial sums of the Fourier series exhibit persistent oscillations near the discontinuity. If \(f\) is piecewise smooth and has a jump discontinuity of magnitude \(J\), the partial sums overshoot by approximately \(0.0895 \cdot J\) (about 9% of the jump magnitude) on each side of the discontinuity. As \(N \to \infty\), this overshoot does not disappear but becomes increasingly localized near the discontinuity while maintaining its relative amplitude. This behavior is known as the Gibbs phenomenon.

For example, consider the square wave that jumps from \(-1\) to \(+1\) at \(x = 0\). The jump magnitude is \(J = 2\), so the overshoot is approximately \(0.09 \times 2 \approx 0.18\). The partial sums therefore reach approximately \(1.18\) near the positive side of the jump instead of \(+1\), and approximately \(-1.18\) near the negative side instead of \(-1\).

Insight: Gibbs Phenomenon in Signal Processing and ML

The Gibbs phenomenon explains why simply truncating Fourier series introduces ringing artifacts near sharp edges. This is a critical consideration in image compression (JPEG) and audio processing (MP3). In practice, window functions (Hamming, Hanning, Blackman) taper the coefficients to suppress ringing at the cost of frequency resolution. In machine learning, the Gibbs phenomenon appears when Fourier features are used to approximate discontinuous functions. Models like Fourier Neural Operators can struggle near sharp interfaces in PDE solutions, motivating hybrid spectral-spatial architectures that handle discontinuities locally.