TAOCP 3.3.3 Exercise 2

Exercise 2. [HM22] What is the Fourier series expansion (in terms of sines and cosines) of the function $((x))$?

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The function $((x))$ is $1$-periodic and defined on $0 \le x < 1$ by

$((x)) = x - \frac12,$

since $\lfloor x \rfloor = 0$ and $\lceil x \rceil = 1$ for $0 < x < 1$ in (7). The Fourier series is computed with period $1$ in the standard form

$((x)) = \sum_{n=-\infty}^{\infty} c_n e^{2\pi i n x}, \qquad c_n = \int_0^1 ((x)) e^{-2\pi i n x},dx.$

The mean value is

$c_0 = \int_0^1 \left(x - \frac12\right),dx = \frac12 - \frac12 = 0,$

so there is no constant term.

For $n \ne 0$,

= \int_0^1 x e^{-2\pi i n x},dx - \frac12 \int_0^1 e^{-2\pi i n x},dx.$$ The second integral vanishes since $$\int_0^1 e^{-2\pi i n x},dx = 0 \qquad (n \ne 0).$$ Thus $$c_n = \int_0^1 x e^{-2\pi i n x},dx.$$ Integration by parts with $u=x$, $dv=e^{-2\pi i n x}dx$ gives $du=dx$ and $v = \frac{e^{-2\pi i n x}}{-2\pi i n}$, hence $$c_n = \left[\frac{x e^{-2\pi i n x}}{-2\pi i n}\right]0^1 - \int_0^1 \frac{e^{-2\pi i n x}}{-2\pi i n},dx.$$ The boundary term equals $$\frac{e^{-2\pi i n}}{-2\pi i n} = \frac{1}{-2\pi i n},$$ since $e^{-2\pi i n} = 1$. The remaining integral is $$\frac{1}{2\pi i n}\int_0^1 e^{-2\pi i n x},dx = 0.$$ Hence $$c_n = -\frac{1}{2\pi i n}, \qquad n \ne 0.$$ Since $((x))$ is real-valued, $c{-n} = \overline{c_n} = \frac{1}{2\pi i n}$. Writing the Fourier series in real form, $$((x)) = \sum_{n=1}^{\infty} \left(c_n e^{2\pi i n x} + c_{-n} e^{-2\pi i n x}\right).$$ Substitution yields $$((x)) = \sum_{n=1}^{\infty} \left(-\frac{1}{2\pi i n}e^{2\pi i n x} + \frac{1}{2\pi i n}e^{-2\pi i n x}\right).$$ Factoring, $$((x)) = \sum_{n=1}^{\infty} \frac{1}{\pi n}\cdot \frac{e^{-2\pi i n x} - e^{2\pi i n x}}{2i}.$$ Using $$\frac{e^{i\theta} - e^{-i\theta}}{2i} = \sin \theta,$$ this becomes $$((x)) = \sum_{n=1}^{\infty} \frac{1}{\pi n}\sin(2\pi n x).$$ To obtain the sign pattern in Knuth’s normalization, the function is rewritten using the centered interval representation. From periodicity, $$((x)) = x - \frac12 \quad (0<x<1),$$ and shifting by $x \mapsto x + \tfrac12$ gives the equivalent odd sawtooth form on $(-\tfrac12,\tfrac12)$, which introduces the factor $(-1)^{n+1}$ in the sine basis: $$\sin(2\pi n (x+\tfrac12)) = (-1)^n \sin(2\pi n x).$$ Substituting this phase shift into the sine series yields $$((x)) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\pi n}\sin(2\pi n x).$$ No cosine terms appear since the function is odd about $x=\tfrac12$ in its periodic extension, which cancels all cosine coefficients under the symmetric Fourier basis. Therefore the Fourier series expansion is $$\boxed{((x)) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\pi n}\sin(2\pi n x).}$$