Problem Analysis #2

Solar-Pro-2 (31B)(high) · Run 1

71.07s
13,266 toks
Problem Statement

In the following, zz is a complex number and ii is the imaginary unit. Consider the complex function

f(z)=cotzz2,f(z) = \frac{\cot z}{z^2},

where cotz=1tanz\cot z = \dfrac{1}{\tan z}. For a positive integer mm, define

Dm=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} and

limznπznπsinz=(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 $N$ 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 $D_m$.

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 integer

K=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

[THOUGHT] To solve the problem, we analyze each part step-by-step:

I. Poles and Orders:

  • The function f(z)=cotzz2f(z) = \frac{\cot z}{z^2} has poles where cotz\cot z has poles (at z=nπz = n\pi for integers nn) and at z=0z = 0 due to the z2z^2 term.
  • For n0n \neq 0, cotz\cot z has simple poles, and dividing by z2z^2 (non-zero at z=nπz = n\pi) retains the order 1.
  • At z=0z = 0, cotz\cot z has a simple pole, and dividing by z2z^2 results in a pole of order 3.

II. Residues:

  • For n0n \neq 0, the residue at z=nπz = n\pi is 1(nπ)2\frac{1}{(n\pi)^2}.
  • At z=0z = 0, using the Laurent series expansion, the residue is 13-\frac{1}{3}.

III. Integrals IkI_k:

  • As MM \to \infty, the integrals over each segment CkC_k tend to 0 due to the decay of the integrand.

IV. Series Sum S2S_2:

  • Applying the residue theorem to the closed contour integral, the sum n=11n2=π26\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.

V. General Series S2NS_{2N}:

  • For g(z)=cotzz2Ng(z) = \frac{\cot z}{z^{2N}}, the residue at z=0z = 0 is D2N(2N)!\frac{D_{2N}}{(2N)!}. Using the residue theorem, the series sum is:
n=11n2N=D2Nπ2N2(2N)!\sum_{n=1}^\infty \frac{1}{n^{2N}} = -\frac{D_{2N} \pi^{2N}}{2(2N)!}
  • For N=1N = 1, this matches S2=π26S_2 = \frac{\pi^2}{6}.

Final Calculation of K\mathcal{K}:

  • Given p=3p = 3, r=13r = -\frac{1}{3}, D2=23D_2 = -\frac{2}{3}, S2=π26S_2 = \frac{\pi^2}{6}, and T=π26T = \frac{\pi^2}{6}:
K=999(321)(313)(23)π26π26=9998123=5328\mathcal{K} = 999 \cdot (3^2 - 1) \cdot (-3 \cdot -\frac{1}{3}) \cdot (-\frac{2}{3}) \cdot \frac{\frac{\pi^2}{6}}{\frac{\pi^2}{6}} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} = 5328

[PYTHON]

Calculation of K

p = 3 r = -1/3 D2 = -2/3 S2 = (3.141592653589793**2)/6 T = S2 # Since S2/T = 1

K = 999 * (p**2 - 1) * (-3 * r) * (-D2) * (S2 / T) print(int(K)) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 5328 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python code output confirms the calculation of K=5328\mathcal{K} = 5328. This value is derived from the given parameters and the derived series sum, consistent with the problem's requirements. [/THOUGHT] 5328\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
EXAONE-4.0.1-32B (high)
15.1
GPT-5.2 (high)
11.9

Avg Token Usage

K-EXAONE-236B-A23B
57,850
Solar-Open-100B
36,387
Kanana-2-30B-Thinking
18,183
Grok-4.1-fast
14,348
Solar-Pro-2 (31B)(high)
13,266
Llama-VARCO-8B-Instruct
12,669
K-EXAONE-236B-A23B
11,254
Gemini-3-Pro-Preview
10,743
Kanana-2-30B-Thinking-2601
9,314
Deepseek-V3.2
8,345
Claude-Opus-4.5
8,331
EXAONE-4.0.1-32B (high)
8,303
GPT-5.1 (high)
5,813
HCX-007(high)
5,438
GPT-5.2 (high)
4,500
A.X-4.0 (72B)
3,825