Problem Analysis #2

K-EXAONE-236B-A23B · Run 1

30.34s
7,628 toks
Problem Statement

Evaluate the integral

I=P(x)eix(x3i)8dx,I = \int_{-\infty}^{\infty} \frac{P(x)\, e^{ix}}{(x - 3i)^{8}}\,dx,

where

P(x)=x714ix6+(8560i)x5(240+210i)x4+(420840i)x3(560+1680i)x2+(13442688i)x+3840.P(x) = x^{7} - 14ix^{6} + (85 - 60i)x^{5} - (240 + 210i)x^{4} + (420 - 840i)x^{3} - (560 + 1680i)x^{2} + (1344 - 2688i)x + 3840.

If the value of the integral is I=a+biI = a + bi, ii is the imaginary unit, compute

10a+b.\lfloor 10a \rfloor + \lfloor b \rfloor.
Ground Truth
7
Execution Trace
Reasoning Process

[THOUGHT] We need to evaluate the integral:

I=P(x)eix(x3i)8dxI = \int_{-\infty}^{\infty} \frac{P(x) e^{ix}}{(x - 3i)^8} dx

where P(x)P(x) is a degree-7 polynomial with complex coefficients.

We can use contour integration in the complex plane. The integrand is f(z)=P(z)eiz(z3i)8f(z) = \frac{P(z) e^{iz}}{(z - 3i)^8}. The pole is at z=3iz = 3i, which is in the upper half-plane. Since we have eize^{iz}, for z=x+iyz = x + iy, eiz=ei(x+iy)=eixeye^{iz} = e^{i(x+iy)} = e^{ix} e^{-y}. For y>0y > 0, eye^{-y} decays as y+y \to +\infty. So we can close the contour in the upper half-plane (since eize^{iz} decays in the upper half-plane? Actually, check: eiz=ei(x+iy)=eixeye^{iz} = e^{i(x+iy)} = e^{ix} e^{-y}. For y>0y > 0, eye^{-y} decays, yes. So we can close in the upper half-plane.

The contour is the real axis plus a semicircle at infinity in the upper half-plane. The integral over the semicircle vanishes as radius RR \to \infty because eize^{iz} decays exponentially and P(z)/(z3i8)z7/z8=1/z|P(z)|/(|z-3i|^8) \sim |z|^7/|z|^8 = 1/|z| for large z|z|, so the integrand decays like 1/R1/R, and the length is πR\pi R, so integral tends to 0? Actually, need to check: eiz=ey|e^{iz}| = e^{-y}. On the semicircle, z=Reiθz = R e^{i\theta}, y=Rsinθy = R \sin\theta, so eiz=eRsinθ|e^{iz}| = e^{-R \sin\theta}. For θ[0,π]\theta \in [0,\pi], sinθ0\sin\theta \ge 0, so the maximum is at θ=π/2\theta = \pi/2, where sinθ=1\sin\theta = 1, giving decay eRe^{-R}. So the integrand decays exponentially in the upper half-plane. So the semicircle integral vanishes as RR \to \infty.

Thus, by the residue theorem, the integral equals 2πi2\pi i times the residue at z=3iz = 3i.

So we need to compute:

I=2πiResz=3iP(z)eiz(z3i)8.I = 2\pi i \cdot \operatorname{Res}_{z=3i} \frac{P(z) e^{iz}}{(z - 3i)^8}.

Since the pole is of order 8, the residue is given by:

Resz=3if(z)=17!limz3id7dz7[(z3i)8f(z)].\operatorname{Res}_{z=3i} f(z) = \frac{1}{7!} \lim_{z \to 3i} \frac{d^7}{dz^7} \left[ (z - 3i)^8 f(z) \right].

But (z3i)8f(z)=P(z)eiz(z - 3i)^8 f(z) = P(z) e^{iz}.

Thus:

Resz=3if(z)=17!d7dz7(P(z)eiz)z=3i.\operatorname{Res}_{z=3i} f(z) = \frac{1}{7!} \left. \frac{d^7}{dz^7} \left( P(z) e^{iz} \right) \right|_{z=3i}.

So we need to compute the 7th derivative of P(z)eizP(z) e^{iz} evaluated at z=3iz = 3i.

We can compute this using Leibniz's rule for the 7th derivative of a product:

d7dz7[P(z)eiz]=k=07(7k)P(k)(z)d7kdz7keiz.\frac{d^7}{dz^7} [P(z) e^{iz}] = \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(z) \frac{d^{7-k}}{dz^{7-k}} e^{iz}.

But dmdzmeiz=imeiz\frac{d^{m}}{dz^{m}} e^{iz} = i^m e^{iz}.

So:

d7dz7[P(z)eiz]=eizk=07(7k)P(k)(z)i7k.\frac{d^7}{dz^7} [P(z) e^{iz}] = e^{iz} \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(z) i^{7-k}.

Evaluate at z=3iz = 3i.

Thus:

I=2πi17!eizk=07(7k)P(k)(z)i7kz=3i.I = 2\pi i \cdot \frac{1}{7!} \left. e^{iz} \sum_{k=0}^{7} \binom{7}{k} P^{(k)}(z) i^{7-k} \right|_{z=3i}.

We need to compute P(z)P(z) and its derivatives up to order 7 at z=3iz = 3i. Since PP is a degree-7 polynomial, P(7)(z)P^{(7)}(z) is constant (the leading coefficient times 7!7!). But we can compute everything numerically using Python.

Let's implement in Python.

We'll define the polynomial coefficients from highest degree to constant term. The polynomial is:

P(x)=x714ix6+(8560i)x5(240+210i)x4+(420840i)x3(560+1680i)x2+(13442688i)x+3840.P(x) = x^7 - 14i x^6 + (85 - 60i) x^5 - (240 + 210i) x^4 + (420 - 840i) x^3 - (560 + 1680i) x^2 + (1344 - 2688i) x + 3840.

We'll compute derivatives symbolically or numerically? Since we need derivatives at z=3iz = 3i, we can compute directly by evaluating the polynomial and its derivatives using symbolic differentiation or using a finite difference? Better to use symbolic differentiation with sympy for exactness, but we only need a numeric result for aa and bb. However, we must be careful with rounding. Since the final answer is integer floors, we need sufficient precision.

We can use sympy to compute derivatives exactly (with complex coefficients) and then evaluate at 3i3i, then compute the sum, multiply by factors, and get a complex number. Then compute aa and bb as real and imaginary parts.

But note: The integral result II is a+bia+bi. Then we compute 10a+b\lfloor 10a \rfloor + \lfloor b \rfloor. So we need aa and bb accurately enough to determine floors.

Let's code.

First, import sympy. Define the polynomial with complex coefficients. Compute derivatives up to 7. Then compute the sum. Then multiply by 2πi/7!2\pi i / 7!. Then get real and imaginary parts.

We'll print aa and bb with high precision.

Also, we must ensure that the semicircle contribution vanishes; we already argued yes.

Let's proceed. </think>

[PYTHON] import sympy as sp

Define variable

z = sp.symbols('z', complex=True)

Define polynomial P(z) with complex coefficients

Coefficients from highest degree to constant term

coeffs = [ 1, # x^7 -14sp.I, # x^6 (85 - 60sp.I), # x^5 -(240 + 210sp.I), # x^4 (420 - 840sp.I), # x^3 -(560 + 1680sp.I), # x^2 (1344 - 2688sp.I), # x^1 3840 # constant ]

Build polynomial

P = sum(c * z**(7 - i) for i, c in enumerate(coeffs))

Compute derivatives up to 7

derivatives = [P] for k in range(1, 8): derivatives.append(sp.diff(derivatives[-1], z))

Point of evaluation

z0 = 3*sp.I

Compute the sum S = sum_{k=0}^{7} binom(7,k) * P^{(k)}(z0) * i^{7-k}

S = 0 for k in range(8): Pk = derivatives[k].subs(z, z0) binom = sp.binomial(7, k) power_i = sp.I**(7 - k) S += binom * Pk * power_i

Compute the integral I = 2pii / 7! * S

I = 2sp.pisp.I / sp.factorial(7) * S

Simplify

I_simplified = sp.simplify(I)

Get real and imaginary parts as floats with high precision

I_eval = I_simplified.evalf(30) a = I_eval.real b = I_eval.imag

print("I =", I_eval) print("a =", a) print("b =", b) print("10a =", 10a) print("floor(10a) =", int(10a // 1)) # integer division to get floor for positive? Need careful floor for negative. print("floor(b) =", int(b // 1)) print("floor(10a) + floor(b) =", int(10a // 1) + int(b // 1))

More precise floor: use math.floor

import math print("\nUsing math.floor:") print("floor(10a) =", math.floor(10a)) print("floor(b) =", math.floor(b)) print("Sum =", math.floor(10*a) + math.floor(b)) [/PYTHON]

Tool Output
[PYTHON OUTPUT] AttributeError: 'Add' object has no attribute 'real' [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The error occurred because I_eval is a symbolic expression, not a numeric complex number. I need to extract the real and imaginary parts properly using sp.re and sp.im and then evaluate numerically. Let me fix the code to get the correct numerical values for a and b.

import sympy as sp
import math

# Define variable
z = sp.symbols('z', complex=True)

# Define polynomial P(z) with complex coefficients
coeffs = [
    1,                                   # x^7
    -14*sp.I,                            # x^6
    (85 - 60*sp.I),                      # x^5
    -(240 + 210*sp.I),                    # x^4
    (420 - 840*sp.I),                     # x^3
    -(560 + 1680*sp.I),                   # x^2
    (1344 - 2688*sp.I),                   # x^1
    3840                                 # constant
]

# Build polynomial
P = sum(c * z**(7 - i) for i, c in enumerate(coeffs))

# Compute derivatives up to 7
derivatives = [P]
for k in range(1, 8):
    derivatives.append(sp.diff(derivatives[-1], z))

# Point of evaluation
z0 = 3*sp.I

# Compute the sum S = sum_{k=0}^{7} binom(7,k) * P^{(k)}(z0) * i^{7-k}
S = 0
for k in range(8):
    Pk = derivatives[k].subs(z, z0)
    binom = sp.binomial(7, k)
    power_i = sp.I**(7 - k)
    S += binom * Pk * power_i

# Compute the integral I = 2*pi*i / 7! * S
I = 2*sp.pi*sp.I / sp.factorial(7) * S

# Simplify
I_simplified = sp.simplify(I)

# Extract real and imaginary parts symbolically and evaluate numerically
a_exact = sp.re(I_simplified)
b_exact = sp.im(I_simplified)

a = a_exact.evalf(30)
b = b_exact.evalf(30)

print("I =", I_simplified.evalf(30))
print("a =", a)
print("b =", b)
print("10*a =", 10*a)
print("floor(10*a) =", math.floor(10*a))
print("floor(b) =", math.floor(b))
print("Sum =", math.floor(10*a) + math.floor(b))
Tool Output
[PYTHON OUTPUT] I = 12.0278118737437798272569775246 + 40.9828241643297075084210073738*I a = 12.0278118737437798272569775246 b = 40.9828241643297075084210073738 10*a = 120.278118737437798272569775246 floor(10*a) = 120 floor(b) = 40 Sum = 160 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] From the Python output, we have a = 12.0278118737437798272569775246 and b = 40.9828241643297075084210073738. Then 10a = 120.278118737437798272569775246, so floor(10a) = 120. floor(b) = 40. Therefore, floor(10a) + floor(b) = 160.

160\boxed{160}

Final Model Output
160

Comparative Analysis (P2) (Average across runs)

Avg Time Taken

Qwen3-30B-A3B-2507
7431.3
Deepseek-R1-distill-Qwen-32B (high)
859.1
Solar-Open-100B
426.8
Gemini-3-Pro-Preview
199.4
Deepseek-V3.2
116.7
Kanana-2-30B-Thinking-2601
92.2
GPT-oss-20B (high)
86.7
Gemma-3-27B
83.2
Kanana-2-30B-Thinking
72.1
Solar-Pro-2 (31B)(high)
71.2
HCX-007(high)
61.9
A.X-4.0 (72B)
42.6
Llama-VARCO-8B-Instruct
35.1
Grok-4.1-fast
22.6
K-EXAONE-236B-A23B
19.8
Claude-Opus-4.5
15.8
GPT-5.1 (high)
13.1
EXAONE-4.0.1-32B (high)
5.4

Avg Token Usage

Qwen3-30B-A3B-2507
24,110
GPT-oss-20B (high)
19,053
Kanana-2-30B-Thinking-2601
18,048
Gemini-3-Pro-Preview
17,645
Deepseek-R1-distill-Qwen-32B (high)
11,661
Kanana-2-30B-Thinking
11,648
HCX-007(high)
11,315
Solar-Pro-2 (31B)(high)
10,817
Solar-Open-100B
6,687
Deepseek-V3.2
5,996
A.X-4.0 (72B)
5,881
Grok-4.1-fast
5,630
K-EXAONE-236B-A23B
5,505
Llama-VARCO-8B-Instruct
5,237
Claude-Opus-4.5
4,039
EXAONE-4.0.1-32B (high)
3,955
Gemma-3-27B
3,895
GPT-5.1 (high)
3,378