Problem Analysis #2

K-EXAONE-236B-A23B · Run 1

538.47s

74,144 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

We first determine the required quantities from the function $f(z)=\frac{\cot z}{z^2}$ .

Pole at $z=0$ :
$\cot z = \frac{1}{z} - \frac{z}{3} - \frac{z^3}{45} - \cdots$
$\Rightarrow f(z) = \frac{1}{z^3} - \frac{1}{3z} + \cdots$
Hence the order of the pole at $z=0$ is $p = 3$ , and the residue is $r = -\frac{1}{3}$ .

Given: $D_2 = -\dfrac{2}{3}$ .

Sum $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2} = S_2$ :
From the residue theorem applied to the square contour in part IV, one obtains $S_2 = \frac{\pi^2}{6}$ .

General sum for $g(z)=\frac{\cot z}{z^{2N}}$ :
By the same contour method,

\sum_{n=1}^{\infty}\frac{1}{n^{2N}} = -\frac{\pi^{2N} D_{2N}}{2\,(2N)!}.

For $N=1$ this gives

T = -\frac{\pi^{2} D_2}{2\cdot 2!} = -\frac{\pi^{2} D_2}{4} = -\frac{\pi^{2}}{4}\left(-\frac{2}{3}\right) = \frac{\pi^{2}}{6}.

Thus $S_2 = T$ , so $\displaystyle\frac{S_2}{T}=1$ .

Compute $\mathcal{K}$ :

\begin{aligned} p^2-1 &= 3^2 - 1 = 8,\\ -3r &= -3\left(-\frac{1}{3}\right) = 1,\\ -D_2 &= -\left(-\frac{2}{3}\right) = \frac{2}{3},\\ \frac{S_2}{T} &= 1. \end{aligned}

\mathcal{K} = 999 \times 8 \times 1 \times \frac{2}{3} \times 1 = 999 \times \frac{16}{3} = 999 \times \frac{16}{3} = 333 \times 16 = 5328.

\boxed{5328}

Reasoning Process

[THOUGHT]
We compute K using the formula:
K = 999 * (p2 - 1) * (-3 * r) * (-D_2) * (S_2 / T)
with p = 3, r = -1/3, D_2 = -2/3, S_2 = pi2/6, T = pi**2/6.
We'll compute in Python for verification.

[/THOUGHT] [PYTHON] import math

p = 3 r = -1/3 D2 = -2/3 S2 = math.pi2 / 6 T = math.pi2 / 6

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

Tool Output

[PYTHON OUTPUT] 5328.0 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT]
The problem requires computing the integer $\mathcal{K} = 999\,(p^2 - 1)\,(-3r)\,(-D_2)\,\frac{S_2}{T}$ , where:

$p = 3$ (order of the pole of $f(z)$ at $z=0$ ),
$r = -\frac{1}{3}$ (residue of $f(z)$ at $z=0$ ),
$D_2 = -\frac{2}{3}$ (given),
$S_2 = \frac{\pi^2}{6}$ (value of $\sum_{n=1}^\infty \frac{1}{n^2}$ from part IV),
$T = \frac{\pi^2}{6}$ (value of $\sum_{n=1}^\infty \frac{1}{n^2}$ from part V with $N=1$ ).

Since $S_2 = T$ , their ratio is $1$ . Substituting values:

\mathcal{K} = 999 \cdot (3^2 - 1) \cdot \left(-3 \cdot \left(-\frac{1}{3}\right)\right) \cdot \left(-(-\frac{2}{3}\right) \cdot 1

Simplifying step-by-step:

$p^2 - 1 = 8$ ,
$-3r = -3 \cdot (-\frac{1}{3}) = 1$ ,
$-D_2 = -(-\frac{2}{3}) = \frac{2}{3}$ .

Thus:

$$\mathcal{K}$$ = 999 $$\cdot$$ 8 $$\cdot$$ 1 $$\cdot$$ $$\frac$${_so I'll compute this in Python to confirm. </THOUGHT> [PYTHON] import math p = 3 r = -1/3 D2 = -2/3 S2 = math.pi**2 / 6 T = math.pi**2 / 6 K = 999 * (p**2 - 1) * (-3 * r) * (-D2) * (S2 / T) K [/PYTHON] 5328.0

Tool Output

[PYTHON OUTPUT] 5328.0 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT]
We need to compute the integer

\mathcal{K} = 999\,(p^2 - 1)\,(-3r)\,(-D_2)\,\frac{S_2}{T}

with

$p = 3$ (order of pole of $f(z) = \frac{\cot z}{z^2}$ at $z=0$ )
$r = -\frac{1}{3}$ (residue of $f(z)$ at $z=0$ )
$D_2 = -\frac{2}{3}$ (given)
$S_2 = \frac{\pi^2}{6}$ (sum from part IV)
$T = \frac{\pi^2}{6}$ (sum from part V with $N=1$ )

Since $S_2 = T$ , $\frac{S_2}{T} = 1$ .

Now compute step-by-step:

p^2 - 1 = 9 - 1 = 8

-3r = -3 \cdot \left(-\frac{1}{3}\right) = 1

-D_2 = -\left(-\frac{2}{3}\right) = \frac{2}{3}

Thus

\mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 = 999 \cdot \frac{16}{3} = 333 \cdot 16 = 5328

We verify with Python:

</THOUGHT> [PYTHON] import math

p = 3 r = -1/3 D2 = -2/3 S2 = math.pi2 / 6 T = math.pi2 / 6

K = 999 * (p**2 - 1) * (-3 * quest: I realize I made a small error in the Python code block in the previous thought. I'll correct it now.

In the previous step, I had a Python block, but I need to ensure it's correct. Let me write the correct code:

[THOUGHT] We have:

p = 3
r = -1/3
D2 = -2/3
S2 = math.pi**2 / 6
T = math.pi**2 / 6

Compute K = 999 * (p**2 - 1) * (-3 * r) * (-D2) * (S2 / T)

Let's compute this in Python. </think>

[THOUGHT] We compute K using the given values. </THOUGHT> [PYTHON] import math

p = 3 r = -1/3 D2 = -2/3 S2 = math.pi2 / 6 T = math.pi2 / 6

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

Tool Output

[PYTHON OUTPUT] SyntaxError: unterminated string literal (detected at line 9) (<string>, line 9) [/PYTHON OUTPUT]

Reasoning Process

5328

Reasoning Process

[THOUGHT] The integer 𝒦 = 5328.

$\boxed{5328}$

Reasoning Process

[THOUGHT] The integer $\mathcal{K} = 999 \times (p^2 - 1) \times (-3r) \times (-D_2) \times \frac{S_2}{T}$ , with $p = 3$ , $r = -\frac{1}{3}$ , $D_2 = -\frac{2}{3}$ , $S_2 = \frac{\pi^2}{6}$ , $T = \frac{\pi^2}{6}$ .
Since $S_2 / T = 1$ , we compute:

$p^2 - 1 = 8$
$-3r = -3 \times (-\frac{1}{3}) = 1$
$-D_2 = -(-\frac{2}{3}) = \frac{2}{3}$ Thus $\mathcal{K} = 999 \times 8 \times 1 \times \frac{2}{3} = 999 \times \frac{16}{3} = 333 \times 16 = 5328$ .

\boxed{5328}

Reasoning Process

[THOUGHT] We'll compute K using Python. [/THOUGHT] [PYTHON] import math

p = 3 r = -1/3 D2 = -2/3 S2 = math.pi2 / 6 T = math.pi2 / 6

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

Tool Output

[PYTHON OUTPUT] 5328.0 [/PYTHON OUTPUT]

Reasoning Process

[THOUGHT] The integer $\mathcal{K} = 5328$ .

\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