Fourier Sine Series — Question 6

Question 6 – Solution

Strategy. Test the operation itself: differentiating weakens coefficient decay, while integration improves it.

Step 1: Reject the formal derivative at a specific point. Formal differentiation produces $$2\sum_{n\geq 1}(-1)^{n+1}\cos(nx).$$ At $x=\pi/2$, the even-indexed terms have magnitude two and do not tend to zero. Thus this series diverges there and cannot be a valid ordinary termwise derivative representation of $f'=1$.

Step 2: Pass from finite integrals to the limit. For each finite sum $S_N$, direct integration gives $$\int_0^x S_N(t)dt=2\sum_{n=1}^N\frac{(-1)^{n+1}}{n^2}(1-\cos nx).$$ Cauchy–Schwarz yields $|\int_0^x(S_N-f)dt|\leq\sqrt{x}\,\|S_N-f\|_2 \leq\sqrt\pi\,\|S_N-f\|_2\to 0$. This bound is independent of $x$, proving uniform convergence of the integrals.

Step 3: State the integrated identity and its error. Consequently $$\boxed{\frac{x^2}{2}=2\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}(1-\cos nx),\qquad 0\leq x\leq\pi.}$$ The tail is bounded uniformly by $4\sum_{n>N}n^{-2}\leq 4/N$. This also proves absolute uniform convergence of the integrated series. At zero both sides vanish; the right endpoint is included by the uniform integral argument, despite the original sine mismatch.

Step 4: Evaluate at the right endpoint. At $x=\pi$ the even terms vanish and each odd term contributes $4/n^2$. Hence $\pi^2/2=4\sum_{j\geq 0}(2j+1)^{-2}$, giving $\boxed{\sum_{j\geq 0}(2j+1)^{-2}=\pi^2/8}$. The integration has a proved limiting argument and summable coefficients; the differentiated series fails even a necessary condition for convergence.