Problem Analysis #2

36.67s
9,845 toks
Problem Statement

In the following, zz is a complex number and ii is the imaginary unit. Consider the complex functionf(z)=cotzz2,f(z) = \frac{\cot z}{z^2},where cotz=1tanz\cot z = \dfrac{1}{\tan z}. For a positive integer mm, defineDm=limz0dmdzm(zcotz).D_m = \lim_{z\to 0} \frac{d^m}{dz^m}(z \cot z).If necessary, you may use D2=23D_2 = -\dfrac{2}{3} andlimznπznπsinz=(1)nfor any integer n.\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)f(z). Also, find the order of each pole.
  • [II.] Find the residue of each pole found in I.
  • [III.] Let MM be a positive integer and set R=π(2M+1)R = \pi(2M+1). For each real parameter tt with R2tR2-\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 Ik=Ckf(z)dzI_k = \displaystyle\int_{C_k} f(z)\,dz along CkC_k (k=1,2,3,4k = 1,2,3,4), find limMIk\displaystyle\lim_{M\to\infty} I_k.
  • [IV.] Let CC be the closed loop composed of the four line segments C1,C2,C3C_1, C_2, C_3, and C4C_4 in III. By applying the residue theorem to the complex integral I=Cf(z)dz, I = \oint_C f(z)\,dz, find the value of the infinite series n=11n2. \sum_{n=1}^{\infty} \frac{1}{n^2}.
  • [V.] Now replace f(z)f(z) by the complex function g(z)=cotzz2N, g(z) = \frac{\cot z}{z^{2N}}, where NN is a positive integer. By following the same contour method as in I–IV, express the infinite series n=11n2N \sum_{n=1}^{\infty} \frac{1}{n^{2N}} in terms of DmD_m. Finally, let pp be the order of the pole of f(z)f(z) at z=0z=0 (from I), and let rr be the residue of f(z)f(z) at z=0z=0 (from II). Let S2S_2 denote the value of n=11n2\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} obtained in IV. From the expression in V, let TT be the value of n=11n2\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} obtained by setting N=1N=1 in your general formula.Compute the integerK=999(p21)(3r)(D2)S2T.\mathcal{K}= 999\,(p^2 - 1)\,(-3r)\,(-D_2)\,\frac{S_2}{T}.\
Ground Truth
5328
Execution Trace
Reasoning Process

\boxed{5328}

Reasoning Process

[THOUGHT] We are to compute K = 999*(p^2 - 1)(-3r)(-D_2)*(S_2/T) with given values: p=3, r=-1/3, D_2=-2/3, S_2=π²/6, T=S_2 (since T is the same sum as S_2 when N=1). We use Python to verify the calculation. [/THOUGHT] [PYTHON] p = 3 r = -1/3 D2 = -2/3 import math S2 = math.pi2 / 6 T = S2 K = 999 * (p2 - 1) * (-3 * r) * (-D2) * (S2 / T) print(K) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 5328.0 [/PYTHON OUTPUT]
Reasoning Process

\boxed{5328}

Final Model Output
5328

Comparative Analysis (P2) (Average across runs)

Avg Time Taken

K-EXAONE-236B-A23B
448.4
Solar-Open-100B
223.1
Llama-VARCO-8B-Instruct
149.6
Gemini-3-Pro-Preview
144.7
Kanana-2-30B-Thinking
97.9
Grok-4.1-fast
89.6
K-EXAONE-236B-A23B
72.4
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
Kanana-2-30B-Thinking-2601
33.5
HCX-007(high)
32.2
axk1
18.0
EXAONE-4.0.1-32B (high)
15.1
GPT-5.2 (high)
11.9

Avg Token Usage

K-EXAONE-236B-A23B
57849.7
Solar-Open-100B
36387.0
Kanana-2-30B-Thinking
18183.0
Grok-4.1-fast
14348.0
Solar-Pro-2 (31B)(high)
13266.0
Llama-VARCO-8B-Instruct
12669.0
K-EXAONE-236B-A23B
11254.0
Gemini-3-Pro-Preview
10743.0
Kanana-2-30B-Thinking-2601
9313.5
Deepseek-V3.2
8345.0
Claude-Opus-4.5
8331.0
EXAONE-4.0.1-32B (high)
8303.0
GPT-5.1 (high)
5813.0
axk1
5525.0
HCX-007(high)
5438.0
GPT-5.2 (high)
4500.0
A.X-4.0 (72B)
3825.0