Problem Analysis #2

Gemini-3-Pro-Preview · Run 1

144.71s
10,743 toks
Problem Statement

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

f(z)=cot⁡zz2,f(z) = \frac{\cot z}{z^2},

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

Dm=lim⁡z→0dmdzm(zcot⁡z).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

lim⁡z→nπz−nπsin⁡z=(−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 −R2≤t≤R2-\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 lim⁡M→∞Ik\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=1∞1n2.\sum_{n=1}^{\infty} \frac{1}{n^2}.
  • [V.] Now replace f(z)f(z) by the complex function
g(z)=cot⁡zz2N,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=1∞1n2N\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=1∞1n2\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} obtained in IV. From the expression in V, let TT be the value of ∑n=1∞1n2\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2} obtained by setting N=1N=1 in your general formula.Compute the integer

K=999 (p2−1) (−3r) (−D2) S2T.\mathcal{K}= 999\,(p^2 - 1)\,(-3r)\,(-D_2)\,\frac{S_2}{T}.

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Ground Truth
5328
Execution Trace
Reasoning Process

The problem asks us to analyze the function f(z)=cot⁡zz2f(z) = \frac{\cot z}{z^2} and compute a value K\mathcal{K} based on its properties and related series.

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

  • Poles and Order (pp): The function is f(z)=cos⁡zz2sin⁡zf(z) = \frac{\cos z}{z^2 \sin z}. The pole at z=0z=0 comes from the z2z^2 term and the zero of sin⁡z\sin z at z=0z=0. Since sin⁡z=z−z36+…\sin z = z - \frac{z^3}{6} + \dots, the denominator behaves like z2(z)=z3z^2(z) = z^3. The numerator cos⁡0=1≠0\cos 0 = 1 \neq 0. Thus, the pole at z=0z=0 has order p=3p=3.

  • Residue at z=0z=0 (rr): We need the coefficient of z−1z^{-1} in the Laurent series of f(z)f(z) at z=0z=0. Using the series expansion of cot⁡z\cot z:

cot⁡z=1z−z3−z345−…\cot z = \frac{1}{z} - \frac{z}{3} - \frac{z^3}{45} - \dots f(z)=1z2(1z−z3−z345−… )=1z3−13z−z45−…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 (D2D_2): Dm=lim⁡z→0dmdzm(zcot⁡z)D_m = \lim_{z\to 0} \frac{d^m}{dz^m}(z \cot z). We need D2D_2. The series for zcot⁡zz \cot z is:
zcot⁡z=1−z23−z445−…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 $z^2$:
D22!=−13  ⟹  D2=−23.\frac{D_2}{2!} = -\frac{1}{3} \implies D_2 = -\frac{2}{3}.
  • Sums S2S_2 and TT: S2S_2 is the sum ∑n=1∞1n2\sum_{n=1}^\infty \frac{1}{n^2}. From the residue theorem on the square contour (which vanishes at infinity), we have:
Res(f,0)+∑n≠0Res(f,nπ)=0\text{Res}(f, 0) + \sum_{n \neq 0} \text{Res}(f, n\pi) = 0 −13+∑n≠01(nπ)2=0-\frac{1}{3} + \sum_{n \neq 0} \frac{1}{(n\pi)^2} = 0 2π2∑n=1∞1n2=13  ⟹  ∑n=1∞1n2=π26.\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 K\mathcal{K}

The formula is:

K=999 (p2−1) (−3r) (−D2) S2T\mathcal{K} = 999\,(p^2 - 1)\,(-3r)\,(-D_2)\,\frac{S_2}{T}

Substituting the values:

  • p=3  ⟹  p2−1=8p = 3 \implies p^2 - 1 = 8
  • r=−13  ⟹  −3r=1r = -\frac{1}{3} \implies -3r = 1
  • D2=−23  ⟹  −D2=23D_2 = -\frac{2}{3} \implies -D_2 = \frac{2}{3}
  • S2T=1\frac{S_2}{T} = 1
K=999⋅8⋅1⋅23⋅1\mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 K=999⋅163\mathcal{K} = 999 \cdot \frac{16}{3} K=333⋅16\mathcal{K} = 333 \cdot 16

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

[THOUGHT] The calculation is verified. p=3p=3, r=−1/3r=-1/3, D2=−2/3D_2=-2/3. K=999∗(9−1)∗(−3∗−1/3)∗(−(−2/3))∗1\mathcal{K} = 999 * (9-1) * (-3 * -1/3) * (-(-2/3)) * 1 =999∗8∗1∗2/3= 999 * 8 * 1 * 2/3 =333∗16=5328= 333 * 16 = 5328. [/THOUGHT]

5328\boxed{5328}

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

The order of the pole at z=0z=0 is p=3p=3. The residue at z=0z=0 is r=−13r = -\frac{1}{3}. The value D2=−23D_2 = -\frac{2}{3}. The sum S2=π26S_2 = \frac{\pi^2}{6}. The value TT obtained from the general formula for N=1N=1 is also π26\frac{\pi^2}{6}, so S2T=1\frac{S_2}{T} = 1.

Substituting these values into the expression for K\mathcal{K}:

K=999 (32−1) (−3(−13)) (−(−23)) (1)\mathcal{K} = 999 \, (3^2 - 1) \, \left(-3 \left(-\frac{1}{3}\right)\right) \, \left(-\left(-\frac{2}{3}\right)\right) \, (1) K=999⋅8⋅1⋅23\mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} K=333⋅16=5328\mathcal{K} = 333 \cdot 16 = 5328

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