Fourier Sine Series — Question 3

Question 3 – Solution

Strategy. Retain the change in slope at the interior corner when integrating by parts.

Step 1: Compute the corner contribution. Put $k=n\pi$. One integration by parts removes the zero endpoint terms; integrating each constant slope then gives $$\int_0^1f\sin(kx)dx =\frac 1{k^2}\left(\frac Ha+\frac H{1-a}\right)\sin(ka).$$ Thus $\boxed{b_n=2H\sin(n\pi a)/[\pi^2a(1-a)n^2]}$. The nonzero slope change, not an endpoint displacement, supplies these terms.

Step 2: Identify missing modes and convergence. For $a=1/3,H=1$, $\boxed{b_n=9\sin(n\pi/3)/(\pi^2n^2)}$. Exactly the multiples of three vanish. The bound $|b_n|\leq 9/(\pi^2n^2)$ gives uniform absolute convergence by the Weierstrass test. The continuous, piecewise linear odd extension satisfies the Fourier convergence theorem, so the sum is the tent at every point, including both endpoints and the corner.

Step 3: Recover the unknown tent. Because $\sin(\pi a)>0$, $b_2/b_1=\sin(2\pi a)/(4\sin\pi a)=\tfrac 12\cos\pi a$. The measured ratio gives $\cos\pi a=1/2$, hence $\boxed{a=1/3}$ uniquely on $(0,1)$. Substituting into $b_1$ gives $$\boxed{H=\frac{2\pi^2b_1}{9\sqrt 3}.}$$

Step 4: Interpret reflection and lost sign information. Under $x\mapsto 1-x$, $\sin(n\pi(1-x))=(-1)^{n+1}\sin(n\pi x)$. A change of variable gives $\boxed{b_n^{\mathrm{reflected}}=(-1)^{n+1}b_n}$. Odd modes are unchanged and even modes reverse sign. All magnitudes therefore coincide, so they cannot distinguish $a$ from $1-a$ when those locations differ.