Fourier Sine Series — Question 5

Question 5 – Solution

Strategy. Use the zero endpoint values and the constant second derivative to obtain rapidly decaying coefficients.

Step 1: Integrate twice and retain the parity. Because $f(0)=f(\pi)=0$ and $f''=-2$, integration by parts gives $\int_0^\pi f\sin(nx)dx=2(1-(-1)^n)/n^3$. Therefore $$\boxed{b_n=\frac{4(1-(-1)^n)}{\pi n^3} =\begin{cases}8/(\pi n^3),&n\text{ odd},\\0,&n\text{ even}.\end{cases}}$$

Step 2: Bound the entire tail uniformly. The coefficient series is absolutely summable, so it converges uniformly. The continuous odd periodic extension and the convergence theorem identify its sum with $f$ everywhere on $[0,\pi]$. For every $x$, $$|f(x)-S_N(x)|\leq\frac 8\pi\sum_{n>N}\frac 1{n^3} \leq\frac 8\pi\int_N^\infty t^{-3}dt=\boxed{\frac 4{\pi N^2}}.$$ Both endpoint values of the target and all partial sums are zero; there is no endpoint mismatch or jump obstructing uniform convergence.

Step 3: Choose a certified truncation. The stated bound is below $10^{-3}$ when $N>\sqrt{4000/\pi}$. Since this threshold lies between 35 and 36, $\boxed{N=36}$ is the smallest integer certified by this bound. The estimate discarded the missing even modes and possible cancellation; it does not prove that every smaller $N$ fails.

Step 4: Extract an exact numerical sum. Expanding the polynomial square yields $\int_0^\pi x^2(\pi-x)^2dx=\pi^5/30$. Parseval gives $$\frac{\pi^5}{30}=\frac{32}{\pi}\sum_{j=0}^\infty\frac 1{(2j+1)^6}, \qquad \boxed{\sum_{j=0}^\infty\frac 1{(2j+1)^6}=\frac{\pi^6}{960}.}$$ This conclusion uses the complete series and Parseval, not a finite numerical fit.