Fourier Transform 7 | Complex Fourier Series [dark version]

Q: How do Euler’s formulas convert cosine and sine terms into exponentials without changing the function space?

Euler’s formula gives cos(x)=\tfrac12(e^{ix}+e^{-ix}). A matching identity for sine is sin(x)=\tfrac{1}{2i}(e^{ix}-e^{-ix}). Substituting these into any trigonometric polynomial lets like exponential terms be collected. The coefficients for e^{ikx} become complex numbers built from the original cosine/sine coefficients. If a sine (or cosine) component was absent, the corresponding exponential coefficient simply becomes 0, so the conversion is exact in both directions.

Q: Why does the span of {1, cos(kx), sin(kx)} match the span of {e^{ikx}} for k between −n and n?

Both sets contain 2n+1 functions and generate the same subspace. The cosine/sine span uses {1, cos(x)…cos(nx), sin(x)…sin(nx)}. The exponential span uses {e^{-inx},…,e^{-ix},1,e^{ix},…,e^{inx}}. Because each cosine and sine can be rewritten as a linear combination of exponentials (and vice versa), each basis set can be expressed in terms of the other, so the spans coincide.

Q: What role does the inner product normalization (1/(2π)) play in the complex Fourier series?

The L2 inner product over one period is defined using an integral over [−π,π], with complex conjugation on one factor. Choosing the factor 1/(2π) normalizes the exponential basis so that the functions e^{ikx} become orthonormal. That orthonormality is what makes the orthogonal projection—and therefore the Fourier coefficient formula—take a simple, single-integral form.

Q: How does orthogonal projection lead to the one-line coefficient formula for c_k?

Fourier series coefficients come from projecting f onto the 2n+1-dimensional subspace spanned by the basis functions. With an orthonormal basis {e^{ikx}}, the projection coefficient for each basis vector is just the inner product of f with that vector. This yields c_k = (1/(2π))∫_{−π}^{π} f(x) e^{-ikx} dx, valid for every integer k in the range, including k=0.

Q: If f is real-valued, why can the complex Fourier series still produce a real function?

The coefficients c_k and basis functions e^{ikx} are complex in general, so the partial sum can look complex. But for real f, the coefficients satisfy conjugate symmetry (so terms for k and −k pair up). As a result, the imaginary parts cancel in the full sum, leaving a real-valued Fourier series approximation.

TL;DR

Complex Fourier series rewrite cosine and sine bases exactly using Euler’s formulas, turning two real basis types into one exponential basis.

Briefing Cornell Notes

Briefing

Complex Fourier series turn the usual cosine–sine Fourier series into a cleaner, one-formula framework by switching to complex exponentials. The payoff is practical: every trigonometric polynomial built from cos(kx) and sin(kx) can be rewritten exactly as a linear combination of exponentials e^{ikx}, with no loss of information—only simpler algebra.

The key mechanism is Euler’s formula and its companion for sine. Cosine splits as cos(x)=\tfrac12(e^{ix}+e^{-ix}), while sine becomes sin(x)=\tfrac{1}{2i}(e^{ix}-e^{-ix}). Using these identities, any expression made from cosine and sine terms—such as a·cos(2x)+b·cos(2x)+c·sin(2x)—can be regrouped into exponentials with complex coefficients. The coefficients for e^{i2x} and e^{-i2x} emerge by collecting like exponential terms. Even the “missing” sine or cosine pieces are handled automatically: if the original combination had no sine component, the corresponding exponential coefficient simply becomes zero. That exact equivalence means complex-valued trigonometric polynomials don’t need separate cosine/sine bookkeeping.

This equivalence also reshapes the underlying subspaces. For a fixed n, the real-style span built from {1, cos(x),…,cos(nx), sin(x),…,sin(nx)} has 2n+1 basis functions. In the complex formulation, the same subspace can be spanned more compactly by exponentials {e^{-inx},…,e^{-ix},1,e^{ix},…,e^{inx}}—again 2n+1 functions. The mapping between the two descriptions is explicit: for k>0, the exponential coefficients combine the cosine and sine coefficients via factors of 1/2 and 1/(2i); for k<0, the same pattern holds with index adjustments; and for k=0, the constant term matches directly.

The normalization and inner product structure make the complex version especially tidy. With the L2 inner product over a 2π-period, the factor 1/(2π) is chosen so the exponential functions form an orthonormal system. The basis functions are e_k(x)=e^{ikx} for k from −n to n, and orthogonality comes from integrating products of exponentials over [−π,π] (with complex conjugation). Because the Fourier series is an orthogonal projection onto the 2n+1-dimensional subspace, the projection formula collapses into a single sum of exponentials.

As a result, the complex Fourier series of a square-integrable 2π-periodic function f is written as a sum over k=−n,…,n of c_k e^{ikx}. The coefficients are given by one integral formula for every k:

c_k = (1/(2π)) ∫_{−π}^{π} f(x) e^{-ikx} dx.

This holds for k=0 as well, so there’s no special-case handling for the constant term. Even when f is real-valued, the series is expressed using complex coefficients, but the final sum remains real because the imaginary parts cancel. The complex formulation keeps the same approximation power while making the coefficient computation and basis notation substantially simpler—setting up a smoother path for further Fourier approximation work.

Cornell Notes

Complex Fourier series replace the cosine–sine basis with complex exponentials, using Euler’s formula to rewrite cos(x) and sin(x) exactly as combinations of e^{ix} and e^{-ix}. For a fixed n, the 2n+1-dimensional subspace spanned by {1, cos(x)…cos(nx), sin(x)…sin(nx)} is identical to the span of {e^{-inx},…,e^{-ix},1,e^{ix},…,e^{inx}}. With the L2 inner product over [−π,π] and the normalization factor 1/(2π), the exponentials e^{ikx} form an orthonormal system, so Fourier series coefficients come from a single orthogonal projection formula. The result is one coefficient rule for all k: c_k = (1/(2π))∫_{−π}^{π} f(x)e^{-ikx}dx, and the Fourier series is ∑_{k=-n}^{n} c_k e^{ikx}.

How do Euler’s formulas convert cosine and sine terms into exponentials without changing the function space?

Euler’s formula gives cos(x)=\tfrac12(e^{ix}+e^{-ix}). A matching identity for sine is sin(x)=\tfrac{1}{2i}(e^{ix}-e^{-ix}). Substituting these into any trigonometric polynomial lets like exponential terms be collected. The coefficients for e^{ikx} become complex numbers built from the original cosine/sine coefficients. If a sine (or cosine) component was absent, the corresponding exponential coefficient simply becomes 0, so the conversion is exact in both directions.

Why does the span of {1, cos(kx), sin(kx)} match the span of {e^{ikx}} for k between −n and n?

Both sets contain 2n+1 functions and generate the same subspace. The cosine/sine span uses {1, cos(x)…cos(nx), sin(x)…sin(nx)}. The exponential span uses {e^{-inx},…,e^{-ix},1,e^{ix},…,e^{inx}}. Because each cosine and sine can be rewritten as a linear combination of exponentials (and vice versa), each basis set can be expressed in terms of the other, so the spans coincide.

What role does the inner product normalization (1/(2π)) play in the complex Fourier series?

The L2 inner product over one period is defined using an integral over [−π,π], with complex conjugation on one factor. Choosing the factor 1/(2π) normalizes the exponential basis so that the functions e^{ikx} become orthonormal. That orthonormality is what makes the orthogonal projection—and therefore the Fourier coefficient formula—take a simple, single-integral form.

How does orthogonal projection lead to the one-line coefficient formula for c_k?

Fourier series coefficients come from projecting f onto the 2n+1-dimensional subspace spanned by the basis functions. With an orthonormal basis {e^{ikx}}, the projection coefficient for each basis vector is just the inner product of f with that vector. This yields c_k = (1/(2π))∫_{−π}^{π} f(x) e^{-ikx} dx, valid for every integer k in the range, including k=0.

If f is real-valued, why can the complex Fourier series still produce a real function?

The coefficients c_k and basis functions e^{ikx} are complex in general, so the partial sum can look complex. But for real f, the coefficients satisfy conjugate symmetry (so terms for k and −k pair up). As a result, the imaginary parts cancel in the full sum, leaving a real-valued Fourier series approximation.

Review Questions

For k>0, how do the cosine and sine coefficients combine to form the exponential coefficient c_k (including the factor involving i)?
Why does orthonormality of {e^{ikx}} make the Fourier coefficient computation reduce to a single inner-product integral?
What changes (and what doesn’t) when switching from the cosine–sine Fourier series to the complex exponential Fourier series?

Key Points

1
Complex Fourier series rewrite cosine and sine bases exactly using Euler’s formulas, turning two real basis types into one exponential basis.
2
Any trigonometric polynomial with complex coefficients can be regrouped into a linear combination of e^{ikx} terms with complex coefficients.
3
For each n, the cosine–sine span of dimension 2n+1 matches the exponential span {e^{ikx} : k=−n,…,n}.
4
Choosing the L2 inner product with normalization 1/(2π) makes the exponentials e^{ikx} an orthonormal system on [−π,π].
5
Fourier series coefficients become orthogonal projection coefficients onto the exponential subspace.
6
The complex Fourier series uses one coefficient formula for all integers k: c_k=(1/(2π))∫_{−π}^{π} f(x)e^{-ikx}dx.
7
Even for real-valued f, the complex series remains real after summing because conjugate-symmetric terms cancel imaginary parts.

Highlights

Euler’s identities convert cos(x) and sin(x) into exact combinations of e^{ix} and e^{-ix}, enabling a one-basis Fourier series.

The 2n+1-dimensional subspace built from {1, cos(kx), sin(kx)} is identical to the span of {e^{ikx}} for k=−n,…,n.

With the right normalization, the exponentials form an orthonormal system, so Fourier coefficients come directly from inner products.

The complex Fourier coefficient formula is uniform: c_k = (1/(2π))∫_{−π}^{π} f(x)e^{-ikx}dx, including k=0.

Real functions can be represented using complex exponentials without producing a complex-valued result after summation.