Problem Analysis #2

33.55s

8,331 toks

Problem Statement

In the following, $z$ is a complex number and $i$ is the imaginary unit. Consider the complex function $f(z) = \frac{\cot z}{z^2},$ where $\cot z = \dfrac{1}{\tan z}$ . For a positive integer $m$ , define $D_m = \lim_{z\to 0} \frac{d^m}{dz^m}(z \cot z).$ If necessary, you may use $D_2 = -\dfrac{2}{3}$ and $\lim_{z\to n\pi} \frac{z - n\pi}{\sin z} = (-1)^n\quad\text{for any integer } n.$

[I.] Find all poles of $f(z)$ . Also, find the order of each pole.
[II.] Find the residue of each pole found in I.
[III.] Let $M$ be a positive integer and set $R = \pi(2M+1)$ . For each real parameter $t$ with $-\dfrac{R}{2} \le t \le \dfrac{R}{2}$ , consider the four line segments $\begin{align*} C_1:\ & z(t) = \frac{R}{2} + it, \\ C_2:\ & z(t) = -t + i\frac{R}{2}, \\ C_3:\ & z(t) = -\frac{R}{2} - it, \\ C_4:\ & z(t) = t - i\frac{R}{2}. \end{align*}$ These four oriented segments form the boundary of a square centered at the origin, traversed counterclockwise. For each complex integral $I_k = \displaystyle\int_{C_k} f(z)\,dz$ along $C_k$ ( $k = 1,2,3,4$ ), find $\displaystyle\lim_{M\to\infty} I_k$ .
[IV.] Let $C$ be the closed loop composed of the four line segments $C_1, C_2, C_3$ , and $C_4$ in III. By applying the residue theorem to the complex integral $I = \oint_C f(z)\,dz,$ find the value of the infinite series $\sum_{n=1}^{\infty} \frac{1}{n^2}.$
[V.] Now replace $f(z)$ by the complex function $g(z) = \frac{\cot z}{z^{2N}},$ where $N$ is a positive integer. By following the same contour method as in I–IV, express the infinite series $\sum_{n=1}^{\infty} \frac{1}{n^{2N}}$ in terms of $D_m$ . Finally, let $p$ be the order of the pole of $f(z)$ at $z=0$ (from I), and let $r$ be the residue of $f(z)$ at $z=0$ (from II). Let $S_2$ denote the value of $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ obtained in IV. From the expression in V, let $T$ be the value of $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ obtained by setting $N=1$ in your general formula.Compute the integer $\mathcal{K}= 999\,(p^2 - 1)\,(-3r)\,(-D_2)\,\frac{S_2}{T}.$ \

Ground Truth

5328

Execution Trace

Reasoning Process

[THOUGHT] Let me work through this problem step by step.

Part I: Find all poles of f(z) = cot(z)/z²

cot(z) = cos(z)/sin(z), so f(z) = cos(z)/(z² sin(z))

Poles occur where sin(z) = 0, i.e., at z = nπ for integer n.

At z = 0: We need to analyze the order. Near z = 0:

sin(z) ≈ z - z³/6 + ...
cos(z) ≈ 1 - z²/2 + ...
cot(z) = cos(z)/sin(z) ≈ (1/z)(1 - z²/3 + ...)
f(z) = cot(z)/z² ≈ 1/z³ - 1/(3z) + ...

So z = 0 is a pole of order 3.

At z = nπ (n ≠ 0): sin(z) has a simple zero, cos(nπ) = (-1)^n ≠ 0, so cot(z) has a simple pole. Since z² is non-zero at z = nπ, f(z) has a simple pole at z = nπ for n ≠ 0.

So p = 3 (order of pole at z = 0).

Part II: Find residues

At z = nπ (n ≠ 0): Res[f(z), nπ] = lim_{z→nπ} (z - nπ) · cot(z)/z² = lim_{z→nπ} (z - nπ)/sin(z) · cos(z)/z² = (-1)^n · (-1)^n / (nπ)² = 1/(n²π²)

At z = 0: We need the coefficient of 1/z in the Laurent expansion. z·cot(z) = z·cos(z)/sin(z) Near z = 0: z·cot(z) = 1 - z²/3 - z⁴/45 - ... So cot(z)/z² = 1/z³ - 1/(3z) - z/45 - ...

The residue at z = 0 is r = -1/3.

Part III: Limits of contour integrals

As M → ∞, the integrals along the sides of the square go to 0 because |cot(z)| is bounded on these contours and 1/z² → 0.

Part IV: Using residue theorem

The contour encloses poles at z = 0, ±π, ±2π, ..., ±Mπ.

∮C f(z) dz = 2πi × (sum of residues) = 2πi × [(-1/3) + 2∑{n=1}^{M} 1/(n²π²)]

As M → ∞, the contour integral → 0, so: 0 = 2πi × [(-1/3) + 2∑{n=1}^{∞} 1/(n²π²)] 1/3 = 2∑{n=1}^{∞} 1/(n²π²) ∑_{n=1}^{∞} 1/n² = π²/6

So S₂ = π²/6.

Part V: General formula

For g(z) = cot(z)/z^{2N}, the residue at z = 0 involves D_{2N-1}/(2N-1)!

Following similar analysis: ∑{n=1}^{∞} 1/n^{2N} = (-1)^{N+1} × D{2N}/(2(2N)!) × (2π)^{2N}

Wait, let me recalculate this more carefully using Python.

[/THOUGHT] [PYTHON] from sympy import *

z = symbols('z')

Part I: Order of pole at z=0

f(z) = cot(z)/z^2 = cos(z)/(z^2 * sin(z))

Let's expand z*cot(z) around z=0

z_cot_z = z * cot(z) expansion = series(z_cot_z, z, 0, 10) print("z*cot(z) expansion:", expansion)

cot(z)/z^2 = (z*cot(z))/z^3

So we need to find the Laurent expansion of cot(z)/z^2

cot_z_over_z2 = cot(z)/z**2 laurent = series(cot_z_over_z2, z, 0, 4) print("cot(z)/z^2 expansion:", laurent)

The leading term is 1/z^3, so order of pole at z=0 is 3

p = 3 print(f"Order of pole at z=0: p = {p}")

Part II: Residue at z=0

Residue is the coefficient of 1/z in the Laurent expansion

From z*cot(z) = 1 - z^2/3 - z^4/45 - ...

cot(z)/z^2 = 1/z^3 - 1/(3z) - z/45 - ...

So residue at z=0 is -1/3

r = Rational(-1, 3) print(f"Residue at z=0: r = {r}")

D_2 = lim_{z->0} d^2/dz^2 (z*cot(z))

Let's compute this

d2_z_cot_z = diff(z_cot_z, z, 2) D2 = limit(d2_z_cot_z, z, 0) print(f"D_2 = {D2}")

Part IV: Sum of 1/n^2

From residue theorem: 0 = -1/3 + 2sum_{n=1}^infty 1/(n^2pi^2)

So sum = pi^2/6

S2 = pi**2/6 print(f"S_2 = {S2}")

Part V: For N=1, we should get the same result

The general formula involves D_{2N}

For N=1: sum = -D_2 * pi^2 / (2 * 2!) (need to verify sign and factor)

Let's verify the formula

zcot(z) = sum_{k=0}^infty D_k z^k / k!

where D_0 = 1, D_1 = 0, D_2 = -2/3, ...

For g(z) = cot(z)/z^{2N}, residue at z=0 is D_{2N}/(2N)!

Residue at z=npi (n != 0) is 1/(npi)^{2N}

From residue theorem:

0 = D_{2N}/(2N)! + 2sum_{n=1}^infty 1/(npi)^{2N}

sum_{n=1}^infty 1/n^{2N} = -D_{2N}/(2N)! * pi^{2N} / 2

For N=1:

T = -D_2/(2!) * pi^2 / 2 = -(-2/3)/2 * pi^2 / 2 = (1/3) * pi^2 / 2 = pi^2/6

D_2 = Rational(-2, 3) T = -D_2 / factorial(2) * pi**2 / 2 print(f"T (from formula with N=1) = {T}") print(f"T simplified = {simplify(T)}")

Now compute K

K = 999 * (p^2 - 1) * (-3r) * (-D_2) * S_2/T

K = 999 * (p**2 - 1) * (-3*r) * (-D_2) * S2/T print(f"K = {K}") print(f"K simplified = {simplify(K)}") [/PYTHON]

Tool Output

[PYTHON OUTPUT] z*cot(z) expansion: 1 - z**2/3 - z**4/45 - 2*z**6/945 - z**8/4725 + O(z**10) cot(z)/z^2 expansion: z**(-3) - 1/(3*z) - z/45 - 2*z**3/945 + O(z**4) Order of pole at z=0: p = 3 Residue at z=0: r = -1/3 D_2 = -2/3 S_2 = pi**2/6 T (from formula with N=1) = pi**2/6 T simplified = pi**2/6 K = 5328 K simplified = 5328 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The Python calculation confirms all the values:

p = 3 (order of pole at z=0)
r = -1/3 (residue at z=0)
D_2 = -2/3
S_2 = π²/6
T = π²/6 (from the general formula with N=1)

Computing K: K = 999 × (p² - 1) × (-3r) × (-D_2) × (S_2/T) K = 999 × (9 - 1) × (-3 × (-1/3)) × (-(-2/3)) × (π²/6)/(π²/6) K = 999 × 8 × 1 × (2/3) × 1 K = 999 × 8 × 2/3 K = 5328 [/THOUGHT]

\boxed{5328}

Final Model Output

5328

Comparative Analysis (P2) (Average across runs)

Avg Time Taken

Llama-VARCO-8B-Instruct

149.6

Gemini-3-Pro-Preview

144.7

Grok-4.1-fast

89.6

Solar-Pro-2 (31B)(high)

71.1

GPT-5.1 (high)

55.4

Deepseek-V3.2

48.1

A.X-4.0 (72B)

43.9

Claude-Opus-4.5

33.5

HCX-007(high)

32.2

EXAONE-4.0.1-32B (high)

15.1

Avg Token Usage