Fourier Cosine Series — Question 2

Question 2 – Solution

Strategy. Retain the interior slope jump when integrating a continuous function with a corner.

Step 1: Integrate the linear pieces. The triangular areas give $\boxed{m=[a^2+(\pi-a)^2]/(2\pi)}$. Twice integrating the pieces separately gives $$\boxed{a_n=\frac 2{\pi n^2}\left[1+(-1)^n-2\cos(na)\right],\quad n\geq 1.}$$ The endpoint slopes $-1,1$ contribute $1+(-1)^n$, and the interior slope jump two contributes $-2\cos(na)$. None of these terms can be dropped.

Step 2: Specialize to the midpoint cusp. At $a=\pi/2$ the mean is $\pi/4$. Odd modes vanish, and among the even modes exactly $n=4j+2$ survive, with coefficient $8/(\pi n^2)$. The extension has period $\pi$; its maxima occur precisely at multiples of $\pi$, so every positive period must be a positive multiple of $\pi$. Its fundamental period is therefore $\boxed{\pi}$. Absolute coefficient summability gives uniform convergence, and the continuous piecewise linear even extension identifies the sum at all points, including cusps.

Step 3: Invert the first coefficient. The general formula gives $a_1=-4\cos a/\pi$. Strict monotonicity of cosine on $(0,\pi)$ yields $$\boxed{a=\arccos(-\pi a_1/4),\qquad -4/\pi<a_1<4/\pi.}$$ The open measurement range corresponds exactly to interior cusp locations.

Step 4: Describe what magnitudes lose. Reflection gives $f_{\pi-a}(x)=f_a(\pi-x)$ and $a_n^{\mathrm{reflected}}=(-1)^na_n$, with unchanged mean. Thus all magnitudes coincide for $a$ and $\pi-a$. In fact $m=\pi/4+(a-\pi/2)^2/\pi$, so the mean leaves precisely those two possibilities unless $a=\pi/2$, when they coincide. Signed odd coefficients resolve the nonsymmetric ambiguity.