Problem Analysis #2

The problem asks us to analyze the function $f(z) = \frac{\cot z}{z^2}$ and compute a value $\mathcal{K}$ based on its properties and related series.

1. Analysis of $f(z)$ and Parameters

• Poles and Order ($p$): The function is $f(z) = \frac{\cos z}{z^2 \sin z}$. The pole at $z=0$ comes from the $z^2$ term and the zero of $\sin z$ at $z=0$. Since $\sin z = z - \frac{z^3}{6} + \dots$, the denominator behaves like $z^2(z) = z^3$. The numerator $\cos 0 = 1 \neq 0$. Thus, the pole at $z=0$ has order $p=3$.

• Residue at $z=0$ ($r$): We need the coefficient of $z^{-1}$ in the Laurent series of $f(z)$ at $z=0$. Using the series expansion of $\cot z$: [ \cot z = \frac{1}{z} - \frac{z}{3} - \frac{z^3}{45} - \dots ] [ f(z) = \frac{1}{z^2} \left( \frac{1}{z} - \frac{z}{3} - \frac{z^3}{45} - \dots \right) = \frac{1}{z^3} - \frac{1}{3z} - \frac{z}{45} - \dots ] The coefficient of $z^{-1}$ is $r = -\frac{1}{3}$.

• Derivative Value ($D_2$): $D_m = \lim_{z\to 0} \frac{d^m}{dz^m}(z \cot z)$. We need $D_2$. The series for $z \cot z$ is: [ z \cot z = 1 - \frac{z^2}{3} - \frac{z^4}{45} - \dots ] This is the Taylor series $\sum_{k=0}^\infty \frac{D_k}{k!} z^k$. Comparing the coefficient of $z^2$: [ \frac{D_2}{2!} = -\frac{1}{3} \implies D_2 = -\frac{2}{3}. ]

• Sums $S_2$ and $T$: $S_2$ is the sum $\sum_{n=1}^\infty \frac{1}{n^2}$. From the residue theorem on the square contour (which vanishes at infinity), we have: [ \text{Res}(f, 0) + \sum_{n \neq 0} \text{Res}(f, n\pi) = 0 ] [ -\frac{1}{3} + \sum_{n \neq 0} \frac{1}{(n\pi)^2} = 0 ] [ \frac{2}{\pi^2} \sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{3} \implies \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}. ] So $S_2 = \frac{\pi^2}{6}$. $T$ is the value obtained from the general formula in Part V for $N=1$. Since the derivation is consistent, $T = S_2$. Thus, $\frac{S_2}{T} = 1$.

2. Calculation of $\mathcal{K}$

The formula is: [ \mathcal{K} = 999,(p^2 - 1),(-3r),(-D_2),\frac{S_2}{T} ]

Substituting the values:

• $p = 3 \implies p^2 - 1 = 8$
• $r = -\frac{1}{3} \implies -3r = 1$
• $D_2 = -\frac{2}{3} \implies -D_2 = \frac{2}{3}$
• $\frac{S_2}{T} = 1$

[ \mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 ] [ \mathcal{K} = 999 \cdot \frac{16}{3} ] [ \mathcal{K} = 333 \cdot 16 ]

[PYTHON] print(333 * 16) [/PYTHON] 5328

[THOUGHT] The calculation is verified. $p=3$, $r=-1/3$, $D_2=-2/3$. $\mathcal{K} = 999 * (9-1) * (-3 * -1/3) * (-(-2/3)) * 1$ $= 999 * 8 * 1 * 2/3$ $= 333 * 16 = 5328$. [/THOUGHT]

\boxed{5328}

The order of the pole at $z=0$ is $p=3$. The residue at $z=0$ is $r = -\frac{1}{3}$. The value $D_2 = -\frac{2}{3}$. The sum $S_2 = \frac{\pi^2}{6}$. The value $T$ obtained from the general formula for $N=1$ is also $\frac{\pi^2}{6}$, so $\frac{S_2}{T} = 1$.

Substituting these values into the expression for $\mathcal{K}$: [ \mathcal{K} = 999 , (3^2 - 1) , \left(-3 \left(-\frac{1}{3}\right)\right) , \left(-\left(-\frac{2}{3}\right)\right) , (1) ] [ \mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} ] [ \mathcal{K} = 333 \cdot 16 = 5328 ]

\boxed{5328}