Fourier Expansion of cosine

April 10, 2026 · Luciano Muratore

Following our discussion on the Z-transform and system stability, we return to the continuous domain to perform a detailed derivation of a specific Fourier expansion. This analysis demonstrates how parity and trigonometric identities simplify the process of representing periodic functions.

1. Objective

Given $\alpha \in \mathbb{R} \setminus \mathbb{Z}$, we consider the $2\pi$-periodic function:

$f(x) = \cos(\alpha x)$

Our goal is to compute its real Fourier series representation:

$f(x) = \frac{a_{0}}{2} + \sum_{n=1}^{\infty} a_{n} \cos(nx) + \sum_{n=1}^{\infty} b_{n} \sin(nx)$

2. The Fourier Coefficients

The coefficients for a $2\pi$-periodic function are defined by the following integrals:

• Cosine coefficients: $a_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(\alpha x) \cos(nx) dx$
• Sine coefficients: $b_{n} = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(\alpha x) \sin(nx) dx$
• DC Component: $a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(\alpha x) dx$

3. Computation of $b_{n}$

By applying a parity argument, we can immediately simplify our work:

• $\cos(\alpha x)$ is an even function.
• $\sin(nx)$ is an odd function.
• The product of an even and an odd function is odd.

Mathematical Property $\int_{-\pi}^{\pi} g(x) dx = 0$ if $g$ is odd.

Consequently, we conclude that $b_{n} = 0$ for all $n \ge 1$.

4. Computation of $a_{0}$

Evaluating the integral for the constant term:

$a_{0} = \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(\alpha x) dx = \frac{1}{\pi} \left[ \frac{\sin(\alpha x)}{\alpha} \right]_{-\pi}^{\pi} = \frac{\sin(\alpha \pi) - \sin(-\alpha \pi)}{\alpha \pi}$

Since $\sin(-u) = -\sin(u)$, we obtain:

$a_{0} = \frac{2 \sin(\alpha \pi)}{\alpha \pi}$

5. Computation of $a_{n}$

To solve for $a_n$, we utilize the following Key Identity:

Trigonometric Product-to-Sum $\cos A \cos B = \frac{1}{2}(\cos(A-B) + \cos(A+B))$

Applying this with $A = \alpha x$ and $B = nx$:

$\cos(\alpha x) \cos(nx) = \frac{1}{2}(\cos((\alpha - n)x) + \cos((\alpha + n)x))$

The integral becomes:

$a_{n} = \frac{1}{2\pi} \int_{-\pi}^{\pi} (\cos((\alpha - n)x) + \cos((\alpha + n)x)) dx$

Recall that $\int_{-\pi}^{\pi} \cos(kx) dx = \frac{2 \sin(k\pi)}{k}$ for $k \neq 0$. Applying this to both terms:

$a_{n} = \frac{1}{2\pi} \left( \frac{2 \sin((\alpha - n)\pi)}{\alpha - n} + \frac{2 \sin((\alpha + n)\pi)}{\alpha + n} \right)$

6. Simplifying $\sin((\alpha \pm n)\pi)$

Using the identity $\sin(u \pm v) = \sin u \cos v \pm \cos u \sin v$:

$\sin((\alpha \pm n)\pi) = \sin(\alpha\pi) \cos(n\pi) \pm \cos(\alpha\pi) \sin(n\pi)$

Since $\sin(n\pi) = 0$ and $\cos(n\pi) = (-1)^n$, we have:

$\sin((\alpha \pm n)\pi) = (-1)^{n} \sin(\alpha \pi)$

Substituting this back into the expression for $a_n$:

$a_{n} = \frac{(-1)^n \sin(\alpha \pi)}{\pi} \left( \frac{1}{\alpha - n} + \frac{1}{\alpha + n} \right)$

Combining the terms $(\frac{1}{\alpha - n} + \frac{1}{\alpha + n}) = \frac{\alpha + n + \alpha - n}{\alpha^2 - n^2} = \frac{2\alpha}{\alpha^2 - n^2}$, we arrive at the final result:

$a_{n} = \frac{2 \alpha (-1)^{n} \sin(\alpha \pi)}{\pi (\alpha^{2} - n^{2})}$

7. Final Fourier Series Expansion

The full expansion of $\cos(\alpha x)$ for $\alpha \notin \mathbb{Z}$ is:

$\cos(\alpha x) = \frac{\sin(\alpha \pi)}{\alpha \pi} + \sum_{n=1}^{\infty} \frac{2 \alpha (-1)^{n} \sin(\alpha \pi)}{\pi (\alpha^{2} - n^{2})} \cos(nx)$

This series converges to the periodic extension of $\cos(\alpha x)$ and illustrates how non-integer frequencies are “distributed” across the integer harmonics of the Fourier basis.