Fourier Sine Series — Question 2

Question 2 – Solution

Strategy. Analyze a moving maximum of a finite sum rather than substitute a moving point into a pointwise convergence claim.

Step 1: Find the nonzero modes. Direct integration gives $b_n=2(1-(-1)^n)/(\pi n)$: zero for even $n$ and $4/(\pi n)$ for odd $n$. Thus $P_M$ is the stated truncation. The odd extension has jump averages zero at both endpoints; its sum is one on $(0,\pi)$.

Step 2: Locate the first maximum. A finite geometric sum, or telescoping $2\sin x\cos((2j+1)x)$, gives $\sum_{j=0}^{M-1}\cos((2j+1)x)=\sin(2Mx)/(2\sin x)$. Differentiating the finite sum is legitimate, so $$\boxed{P_M'(x)=\frac 2\pi\frac{\sin(2Mx)}{\sin x},\qquad x_M=\frac\pi{2M}.}$$ The derivative is positive before $x_M$ and negative just after it, for $M\geq 1$.

Step 3: Pass to the moving-peak limit correctly. Since $P_M(0)=0$, substitution $t=2Mx$ gives $$P_M(x_M)=\frac 2\pi\int_0^\pi \frac{\sin t}{2M\sin(t/(2M))}\,dt \longrightarrow\boxed{\frac 2\pi\int_0^\pi\frac{\sin t}{t}\,dt}.$$ Write the integrand as $(\sin t/t)(z/\sin z)$, where $z=t/(2M)$. Since $z/\sin z$ extends continuously with value one at zero, this factor tends uniformly to one for $0\leq t\leq\pi$. Also $|\sin t/t|\leq 1$. These facts justify the integral limit; its value is approximately $1.179$.

Step 4: Prove a persistent excess. The supplied inequalities imply $\int_0^\pi\sin t/t\,dt\geq\pi/2-\pi^3/144+1/\pi>\pi/2$, since $\pi^4<144$. The limiting peak is strictly above one. The peak location moves toward the jump, so convergence at each fixed interior point is consistent with this persistent overshoot. The square wave jumps by two across zero.