Fourier Sine Series — Question 9

Question 9

For $0<\varepsilon<\pi/2$, define a unit-area narrow pulse centered at $\pi/2$: $$f_\varepsilon(x)=\begin{cases}(2\varepsilon)^{-1},&|x-\pi/2|<\varepsilon,\\ 0,&\text{otherwise},\end{cases}\qquad 0\leq x\leq\pi.$$ Use $b_n(\varepsilon)=(2/\pi)\int_0^\pi f_\varepsilon\sin nx\,dx$. You may use Parseval and Bessel’s inequality for square-integrable functions.

Tasks

1. Verify the area and derive all sine coefficients. Interpret the value of $\sin z/z$ at $z=0$ by its limit.

2. Find the limit of each fixed coefficient as $\varepsilon\downarrow 0$. Prove that the limiting coefficient sequence cannot be the sine coefficients of any $L^2(0,\pi)$ function.

3. Compute $\|f_\varepsilon\|_2^2$ and use Parseval to find $\sum_{n\geq 1}b_n(\varepsilon)^2$. Explain why a point-mass limit is not an ordinary square-integrable function.

4. Use $|\sin z/z-1|\leq z^2/6$ to bound the largest coefficient error among $1\leq n\leq N$. Derive this inequality from $\sin z/z=\int_0^1\cos(zt)dt$ and explain why the resulting estimate is not uniform over all mode numbers.

Question 9 – Solution

Strategy. Separate a limit at each fixed frequency from a limit of the complete coefficient sequence in squared norm.

Step 1: Integrate the narrow pulse. The height times the width is one. Writing $a=\pi/2$, integration gives $$b_n(\varepsilon)=\frac{\cos(n(a-\varepsilon))-\cos(n(a+\varepsilon))} {\pi\varepsilon n} =\boxed{\frac 2\pi\sin(n\pi/2)\frac{\sin(n\varepsilon)}{n\varepsilon}}.$$ The factor $\sin z/z$ has limiting value one at zero. Even modes vanish.

Step 2: Examine the fixed-mode limit. For each fixed $n$, $b_n(\varepsilon)\to\beta_n=(2/\pi)\sin(n\pi/2)$. Every odd-indexed $\beta_n$ has magnitude $2/\pi$, so $\sum\beta_n^2$ diverges. Bessel’s inequality would bound this sum for any $L^2$ function. Hence no square-integrable function has all these limiting coefficients.

Step 3: Track the diverging energy. Direct integration gives $\boxed{\|f_\varepsilon\|_2^2=1/(2\varepsilon)}$. Parseval therefore implies $$\boxed{\sum_{n\geq 1}b_n(\varepsilon)^2=\frac 1{\pi\varepsilon}.}$$ The energies grow without bound while area stays one. Indeed, integration against a continuous test function tends to its value at $\pi/2$, by continuity and averaging over the shrinking interval. This is the point-mass limit, not an $L^2$ function or an ordinary pointwise sine representation of one.

Step 4: Quantify finite-band convergence. From $|1-\cos s|\leq s^2/2$ and the supplied integral representation, $|\sin z/z-1|\leq\int_0^1z^2t^2/2\,dt=z^2/6$. Thus $$\boxed{\max_{1\leq n\leq N}|b_n(\varepsilon)-\beta_n| \leq\frac{N^2\varepsilon^2}{3\pi}.}$$ For each fixed $N$ this tends to zero. It grows with $N$, so it does not justify uniform convergence across all frequencies or interchange with the infinite energy sum.