Problem Analysis #2

[THOUGHT] To verify D_2 and the residue, I will use sympy to compute the second derivative of z * cot(z) at z=0 or the series expansion. [/THOUGHT] [PYTHON] import sympy as sp z = sp.symbols('z') f = z * sp.cot(z) series = sp.series(f, z, 0, 5) print(series)

deriv2 = sp.diff(f, z, 2).limit(z, 0) print(deriv2) [/PYTHON]

[THOUGHT] The poles of $f(z)$ are at $z = 0$ (order 3) and $z = n\pi$ for nonzero integers $n$ (simple poles). The residue at $z = 0$ is $r = -\frac{1}{3}$. Thus, $p = 3$. Also, $D_2 = -\frac{2}{3}$. In IV, $S_2 = \frac{\pi^2}{6}$. In V, for $N=1$, $T = -\frac{\pi^2}{2} \cdot \frac{D_2}{2!} = \frac{\pi^2}{6}$. Then $\mathcal{K} = 999 \cdot (3^2 - 1) \cdot (-3 \cdot -\frac{1}{3}) \cdot (-\frac{2}{3}) \cdot \frac{S_2}{T} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 = 5328$. [/THOUGHT] $\boxed{5328}$

Explanation

The reasoning proceeds through the problem's parts to compute $\mathcal{K}$.

For part I, $f(z) = \frac{\cot z}{z^2}$ has poles where $\sin z = 0$, so at $z = n\pi$ for integer $n$. At $z = n\pi$ with $n \neq 0$, $\cot z$ has a simple pole with residue 1, and $z^2 \neq 0$, yielding simple poles. At $z = 0$, the Laurent series of $\cot z \approx \frac{1}{z} - \frac{z}{3}$ gives $f(z) \approx \frac{1}{z^3} - \frac{1}{3z}$, confirming a pole of order 3. Thus, $p = 3$.

For part II, the residue at $z = 0$ is the coefficient of $z^{-1}$ in the Laurent series of $f(z)$, which is $-\frac{1}{3}$, so $r = -\frac{1}{3}$. At $z = n\pi$ ($n \neq 0$), the residue is $\frac{1}{(n\pi)^2}$.

Parts III and IV use the square contour $C$ with side length $R = \pi(2M+1)$. As $M \to \infty$, the integrals over each $C_k$ vanish due to the decay of $|f(z)|$ on the boundaries. By the residue theorem, $\oint_C f(z) \, dz = 2\pi i \sum \text{residues inside } C$. The poles inside are at $z = 0$ and $z = n\pi$ for $|n| \leq M$, $n \neq 0$. Thus, $0 = 2\pi i \left[ -\frac{1}{3} + 2 \sum_{n=1}^M \frac{1}{(n\pi)^2} \right]$ in the limit, yielding $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$, so $S_2 = \frac{\pi^2}{6}$.

For part V, $g(z) = \frac{\cot z}{z^{2N}}$ has residue at $z = 0$ given by the coefficient of $z^{-1}$ in its Laurent series. Let $h(z) = z \cot z = \sum_{m=0}^\infty \frac{D_m}{m!} z^m$. Then $g(z) = \frac{h(z)}{z^{2N+1}}$, and the $z^{-1}$ term occurs when $m - 2N - 1 = -1$, so $m = 2N$, giving $\operatorname{Res}(g, 0) = \frac{D_{2N}}{(2N)!}$. The contour argument yields $\sum_{n=1}^\infty \frac{1}{n^{2N}} = -\frac{\pi^{2N}}{2} \cdot \frac{D_{2N}}{(2N)!}$. For $N=1$, $T = -\frac{\pi^2}{2} \cdot \frac{D_2}{2!} = -\frac{\pi^2}{2} \cdot \frac{-2/3}{2} = \frac{\pi^2}{6}$.

Finally, substitute into $\mathcal{K} = 999 (p^2 - 1) (-3r) (-D_2) \frac{S_2}{T}$: $p^2 - 1 = 9 - 1 = 8$, $-3r = -3 \cdot (-\frac{1}{3}) = 1$, $-D_2 = - (-\frac{2}{3}) = \frac{2}{3}$, $\frac{S_2}{T} = 1$. Thus, $\mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 = 999 \cdot \frac{16}{3} = 333 \cdot 16 = 5328$.