Fourier Sine Series — Question 1

Question 1 – Solution

Strategy. Determine the odd extension before applying the convergence theorem at its endpoints.

Step 1: Compute the coefficients and extension. Integration by parts yields $$\int_0^\pi x\sin(nx)dx=[-x\cos(nx)/n]_0^\pi +[\sin(nx)/n^2]_0^\pi=-\pi(-1)^n/n.$$ Hence $\boxed{b_n=2(-1)^{n+1}/n}$. The odd extension equals $x$ on $(-\pi,\pi)$ and repeats every $2\pi$. At an odd multiple of $\pi$ its left and right limits are $\pi$ and $-\pi$. The join value may be chosen zero for an odd periodic representative.

Step 2: Apply the convergence theorem. The series equals $x$ for $0<x<\pi$. At zero it equals zero by continuity of the extension, and at $\pi$ it equals the jump average zero: $$\boxed{\sum_{n\geq 1}\frac{2(-1)^{n+1}}n\sin(nx) =x\ (0\leq x<\pi),\qquad \text{sum at }\pi=0.}$$ This does not reproduce the separately assigned value $f(\pi)=\pi$.

Step 3: Rule out uniform convergence near the join. On the closed interval, $|f(\pi)-S_N(\pi)|=\pi$ for every $N$. On the open interval, continuity of each finite sum gives $\lim_{x\uparrow\pi}|x-S_N(x)|=\pi$, so the supremum error is at least $\pi$. Neither asserted uniform convergence holds, despite pointwise convergence inside.

Step 4: Evaluate an interior numerical series. At $x=\pi/2$, only odd modes remain and their sine values alternate: $\pi/2=2(1-1/3+1/5-\cdots)$. Therefore the convergent series equals $\boxed{\pi/4}$. The graph displays a finite approximation; the convergence theorem, not endpoint agreement of the original data, justifies the interior sum.