Fourier Series

A Fourier series of a $2\pi$-periodic function $f$ is

$f(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} \left(a_n \cos(nx) + b_n \sin(nx)\right)$

Where

$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\cos(nx)\, dx, \quad b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx)\, dx$

The Fourier coefficients are derived using the orthogonality relations on $[-\pi, \pi]$:

$\int_{-\pi}^{\pi} \cos(mx)\cos(nx)\, dx = \begin{cases} \pi & m = n \neq 0 \\ 2\pi & m = n = 0 \\ 0 & m \neq n \end{cases}$

$\int_{-\pi}^{\pi} \sin(mx)\sin(nx)\, dx = \begin{cases} \pi & m = n \neq 0 \\ 0 & m \neq n \end{cases}$

$\int_{-\pi}^{\pi} \cos(mx)\sin(nx)\, dx = 0 \quad \mathrm{for}\; all\; m, n$

To find $a_n$Multiply both sides of the Fourier expansion by $\cos(nx)$ and integrate over $[-\pi, \pi]$. By orthogonality, all terms vanish except the $\cos(nx)$ term, yielding $a_n \pi = \int_{-\pi}^{\pi} f(x)\cos(nx)\, dx$. Similarly for $b_n$.

Theorem 7.1 (Dirichlet”s Theorem). If $f$ is $2\pi$-periodic and piecewise smooth, its Fourier Series converges to:

• $f(x)$ at points where $f$ is continuous.
• $\frac{f(x^+) + f(x^-)}{2}$ at jump discontinuities.

$\frac{1}{\pi}\int_{-\pi}^{\pi} |f(x)|^2\, dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2 + b_n^2)$

Intuition. Parseval’s identity is the infinite-dimensional analogue of the Pythagorean theorem: The “energy” of $f$ (its $L^2$ norm squared) equals the sum of the energies of its Fourier Components.

For functions defined on $[0, L]$:

• Cosine series (even extension): $a_n = \frac{2}{L}\int_0^L f(x)\cos\frac{n\pi x}{L}\, dx$ $b_n = 0$.
• Sine series (odd extension): $a_n = 0$ $b_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\, dx$.

Problem. Find the Fourier series of $f(x) = x$ on $(-\pi, \pi)$Extended $2\pi$-periodically.

Solution. $f$ is odd, so $a_n = 0$ for all $n$.

$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} x\sin(nx)\, dx = \frac{2}{\pi}\int_0^{\pi} x\sin(nx)\, dx$.

Integration by parts: $u = x$, $dv = \sin(nx)\, dx$:

$b_n = \frac{2}{\pi}\left[-\frac{x\cos(nx)}{n}\Big|_0^{\pi} + \int_0^{\pi} \frac{\cos(nx)}{n}\, dx\right] = \frac{2}{\pi}\left[-\frac{\pi\cos(n\pi)}{n} + 0\right] = \frac{-2\cos(n\pi)}{n} = \frac{2(-1)^{n+1}}{n}$.

$x \sim 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\sin(nx)$. $\blacksquare$

Problem. Find the Fourier cosine series of $f(x) = x^2$ on $[0, \pi]$.

Using Euler’s formula, the Fourier series can be written in complex form:

$f(x) \sim \sum_{n=-\infty}^{\infty} c_n e^{inx}$

Where $c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}\, dx$.

The relationship with the real coefficients is $c_0 = a_0/2$, $c_n = (a_n - ib_n)/2$ for $n > 0$ And $c_{-n} = \overline{c_n}$ when $f$ is real-valued.

Problem. Using the Fourier series of $f(x) = x$ on $(-\pi, \pi)$Verify Parseval’s identity And deduce $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$.

• Confusing the $a_0/2$ convention with $a_0$; the factor of $1/2$ ensures the formula for $a_n$ works for $n = 0$.
• Forgetting that the Fourier series of a function with jump discontinuities converges to the average of the left and right limits, not to the function value.
• Mixing up sine and cosine series: a cosine series requires an even extension, while a sine series requires an odd extension.
• Neglecting to check piecewise smoothness before applying Dirichlet’s theorem.
• Assuming that term-by-term differentiation of a Fourier series is always valid; it requires the differentiated series to converge.
• Confusing the complex Fourier coefficient $c_n$ with the real coefficients $a_n$ and $b_n$.