Fourier series

The top plot draws a function $f$ of period $2\pi$ in ink and its partial sum $S_N$ in yellow: the constant term and every cosine and sine up to frequency $N$. The strip below multiplies $f$ by one probe wave and shades the signed area between the product and the axis. The bottom strip lays out the coefficients found this way, cosines above and sines below.

Why does one area pick out one coefficient when $f$ holds all the waves at once? Over a full period, any two different waves from $1, \cos x, \sin x, \cos 2x, \sin 2x, \dots$ multiply to a product with zero area, while a sine or cosine times itself has area $\pi$. Multiplying a sum of these waves by $\sin kx$ and integrating therefore removes every term except $b_k \sin kx$ and leaves $\pi b_k$. In the two-sine highlight the strip splits the product in two, and the part with the other frequency cancels itself. The waves behave like perpendicular axes and each coefficient is the projection onto one of them, so adding a term never changes the coefficients already found.

With enough terms, the partial sums overshoot beside a jump. As $N$ grows, the peak narrows and moves toward the jump, but the overshoot does not vanish: for the square wave the excess tends to about 8.949% of the jump. Each $S_N$ is continuous, so it has to cross the jump over a short stretch, and on any interval containing the jump its largest error cannot shrink to 0. Continuity alone does not fix the size: the 8.949% comes from computing the partial sums, and every jump of a piecewise smooth function ends up with the same fraction.

Smoother functions need fewer terms. Here $f$ means the function repeated forever, so ends at $\pm\pi$ that do not meet are a jump too. Integrating by parts moves the derivative from the wave onto $f$ and brings out a factor $1/k$. Across a jump the first step leaves a boundary term with a factor $1/k$: the square wave has $b_k = 4/(k\pi)$ for odd $k$ and 0 for even $k$. If $f$ is continuous but its slope jumps, as at the triangle wave's corners, the first step goes through and the second leaves the term, so the coefficients shrink like $1/k^2$. On the log scale, compare the bars with the dashed $1/k$ and $1/k^2$ lines.

The energy readout applies the same perpendicularity to squares. The integral of $f^2$ splits into the squares of the coefficients, with no cross terms:$$\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)^2\,dx = \frac{a_0^2}{2} + \sum_{k \ge 1} \left(a_k^2 + b_k^2\right).$$The readout gives the share of this total that $S_N$ already holds; the rest is its squared error, measured the same way. For the square wave $S_1$ holds about 81% and $S_9$ about 96%, while the overshoot is still there: a narrow peak adds little area. A partial sum can be close to $f$ on average and still miss it near a jump.