import sympy as sp
from sympy import I, pi, exp, solve, N

# Known Efimov s0 transcendental equation (simplified form from Faddeev kernel)
# The root s0 satisfies (roughly): s0 * cot(s0 * pi/2) or equivalent form derived from recoupling.
# Standard numerical value:
s0_known = 1.00624

# SO(2,1) Casimir in radial QM: effective lambda(lambda-1) with lambda = 1/2 + i s0 (bound regime)
s = sp.symbols('s', real=True, positive=True)
lambda_eff = sp.Rational(1,2) + I * s

casimir = lambda_eff * (lambda_eff - 1)
print("Casimir eigenvalue (symbolic):")
sp.pprint(casimir.simplify())

# Real part after reduction gives - (s**2 + 1/4)
real_part = sp.re(casimir)
print("\nReal part (effective strength): - (s**2 + 1/4)")
sp.pprint(real_part)

# Numerical check
s0_val = N(s0_known)
eff_strength = - (s0_val**2 + 0.25)
print(f"\nFor s0 ≈ {s0_val:.5f}, effective λ_sym ≈ {eff_strength:.5f}")

# Example transcendental solver (placeholder for full kernel equation)
# In full derivation, solve something like: integral kernel = eigenvalue
# For illustration, known root finder:
def efimov_transcendental(s):
    # Approximate form; replace with exact from 3a-math.md
    return sp.sin(s * pi) / (s * (sp.exp(2*s*sp.acosh(2)) - 1))  # placeholder

# Solve numerically near 1.0
s_var = sp.symbols('s_var')
eq = efimov_transcendental(s_var)
# roots = solve(...); use nsolve with guess 1.0
print("\nIn practice: nsolve the exact kernel equation from eigenvalue-flow notes.")