even and odd functions
Some shapes look the same in a mirror placed on the vertical axis; others flip to their negative. A function is even if its graph is symmetric across the y-axis, so f(-x) = f(x) — like a parabola or a cosine. It is odd if rotating the graph 180 degrees about the origin leaves it unchanged, so f(-x) = -f(x) — like a line through the origin or a sine. Most functions are neither, but every function splits uniquely into an even part and an odd part, and that split is exactly what a Fourier series cares about.
The payoff is a labor-saving rule for the coefficients. Because cosine is even and sine is odd, an even f has a Fourier series made only of cosines (all b_n = 0), and an odd f has one made only of sines (all a_n = 0). The reason is symmetry of integrals: the integral over a symmetric interval -L to L of an odd integrand is automatically 0, while an even integrand can be computed as twice the integral from 0 to L. So when f is even, f times sin is odd and its integral vanishes, killing every b_n; when f is odd, f times cos is odd and kills every a_n. Recognizing the parity of f before integrating can halve the work and is one of the first things to check.
Beyond saving algebra, parity is how you steer a Fourier series toward a desired form. If you have a function defined only on 0 to L (say a temperature profile along half a rod) you get to choose how to extend it to -L to L: extend it as an even function to get a pure cosine series, or as an odd function to get a pure sine series. That deliberate choice is the half-range expansion, and it is the standard move for matching the boundary conditions of a heat or wave problem.
f(x) = x^2 on (-pi, pi) is even, so its series is all cosines: x^2 = pi^2/3 + 4 sum (-1)^n cos(n x)/n^2. f(x) = x on (-pi, pi) is odd, so its series is all sines.
Spotting the symmetry tells you half the coefficients are zero before you integrate a single thing.
Parity is judged relative to the chosen center of the interval; a function that looks 'odd-ish' is only truly odd if f(-x) = -f(x) about that exact center, and a nonzero average (like x^2) forces a nonzero a_0 even though the symmetry is even.