= Solution
The <square wave> alternates between a low interval of length five and a high interval of length five; its <integral average> is three. In the real <Fourier series> convention
$$
f(x)\sim\frac{a_0}{2}+\sum_{n=1}^\infty\left[a_n\cos\frac{n\pi x}{5}+b_n\sin\frac{n\pi x}{5}\right],
$$
integration over one period gives
$$
a_0=\frac15\int_{-5}^5 f(x)\,dx=6,\qquad a_n=\frac65\int_0^5\cos\frac{n\pi x}{5}\,dx=0,
$$
and
$$
b_n=\frac65\int_0^5\sin\frac{n\pi x}{5}\,dx=\frac6{n\pi}[1-(-1)^n].
$$
Thus only odd <harmonics> occur:
$$
\boxed{f(x)\sim3+\frac{12}{\pi}\sum_{j=0}^\infty\frac{\sin((2j+1)\pi x/5)}{2j+1}.}
$$
The <Dirichlet-Jordan convergence theorem> makes the sum equal to the function at its continuity points and to the mean of its one-sided limits at a jump. Consequently \b[the sum is 3 at each of $x=-5,0,5$], regardless of any separately assigned values of the original function at those endpoints.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2-square-wave-fourier.png]
{title=A periodic square wave and its Fourier approximations}
{description=Original plot of the alternating low and high intervals and two partial Fourier sums. The jump values of the Fourier series are their midpoint; finite sums exhibit Gibbs oscillations.}
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