Problem Analysis #44

[THOUGHT] To resolve the issue and accurately determine g(m) for different m, I will use Python to numerically find the number of solutions to y = m (h(y) - 1) for various values of m around the critical values. This will show where g(m) jumps and what the values are. I will define h(y) piecewise, compute a = LambertW(2)/2, and use fsolve with multiple starting points to find all distinct roots for selected m. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve from sympy import lambertw

w = lambertw(2).evalf() a = float(w / 2) print(f"a = {a}")

def h(y): y = float(y) if abs(y) <= 1: return 0.5 * y * (5 - y*y) elif y > 1: return np.exp(y - 1) + 1 else: return -np.exp(-y - 1) - 1

def eq(y, m): return float(y) - m * (h(y) - 1)

def find_roots(m, starts): roots = [] for start in starts: root, info, ier, msg = fsolve(eq, start, args=(m,), full_output=True) if ier == 1 and abs(eq(root, m)) < 1e-6: root_r = round(float(root[0]), 6) if all(abs(root_r - round(float(r),6)) > 1e-4 for r in roots): roots.append(root[0]) return sorted(roots)

starts_neg = np.linspace(-20, -0.01, 50) starts_pos = np.linspace(0.01, 20, 50) starts = np.concatenate((starts_neg, starts_pos))

ms = [0.1, 0.3, 0.35, 0.4, a-0.01, a, a+0.01, 0.5, 0.8, 0.99, 1, 1.01, 2, -1, -0.5] for m in ms: roots = find_roots(m, starts) print(f"For m={m:.4f}, number of roots: {len(roots)}, roots: {[f'{r:.4f}' for r in roots]}") </PYTHON>

[THOUGHT] Fixed the code: made equation scalar-to-scalar, used fsolve with wrapper func(z) -> [equation(z[0],m)] to make it vectorized for fsolve. Used np.concatenate for starts. Deduplicated roots using rounding and checks. Computed a = LambertW(2)/2 ≈0.4263, b=1. Selected m values around a and b, including negatives and others to see g(m). Used 100 starts negative, 100 positive up to ±30. This will show the number of roots for each m, revealing where jumps occur and the values of g(a), lim a+ g(m), g(b). [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve from sympy import lambertw

a_sym = lambertw(2).evalf() a = float(a_sym / 2) print("a =", a)

def h(y): if abs(y) <= 1.0: return 0.5 * y * (5.0 - y * y) elif y > 1.0: return np.exp(y - 1.0) + 1.0 else: return -np.exp(-y - 1.0) - 1.0

def equation(y, m): hy = h(y) return y - m * (hy - 1.0)

starts_neg = np.linspace(-30, -0.1, 100) starts_pos = np.linspace(0.1, 30, 100) all_starts = np.concatenate((starts_neg, starts_pos))

def find_roots(m): roots = [] for start in all_starts: def func(z): return [equation(z[0], m)] sol = fsolve(func, [start]) candidate = float(sol[0]) if abs(equation(candidate, m)) < 1e-5: cand_round = round(candidate, 6) if all(abs(cand_round - round(float(r), 6)) >= 1e-4 for r in roots): roots.append(candidate) unique_roots = sorted([r for r in set(round(root, 5) for root in roots)]) return len(unique_roots), unique_roots

m_vals = [0.1, 0.2, 0.3, 0.35, 0.4, a-0.02, a-0.01, a, a+0.01, a+0.02, 0.45, 0.5, 0.7, 0.9, 0.99, 1.0, 1.01, 1.1, 2.0, -0.5, -1.0, 0.0] for m in m_vals: num, rts = find_roots(m) print("m={:.4f}: {} roots: {}".format(m, num, ["{:.4f}".format(r) for r in rts])) </PYTHON]

[THOUGHT] Fixed syntax and logic issues: used a list to collect roots with distance check >1e-4 to deduplicate. Used np.array for func return. Defined func inside with args no, but used lambda-like. Used simple print with round. Increased starts density. Added m_vals around 0, 0.3,0.4,1. Also negative. Computed a. This will print number of roots for each m, revealing g(m) values and where it jumps. Expect jump at a~0.426 from 3 to 1, but to find b. [/THOUGHT] [PYTHON] import numpy as np from scipy.optimize import fsolve from sympy import lambertw

w = lambertw(2).evalf() a = float(w / 2) print('a =', a)

def h(y): y = float(y) if abs(y) <= 1: return 0.5 * y * (5 - y*y) elif y > 1: return np.exp(y

[THOUGHT] The critical values are $a = \frac{W(2)}{2} \approx 0.426$ and $b=1$. Analysis shows $g(m)=3$ for $0<m<a$, $g(a)=2$, $g(m)=1$ for $m>a$. Thus $g(a)=2$, $\lim_{m\to a^+} g(m)=1$. For $b=1$, $g(b)=1$, and $\left( \frac{\ln 1}{1} \right)^2=0$, so the term is $0$. The expression is $2 \times 1 + 1 \times 0 = 2$. \boxed{2}

Explanation

The reasoning