我正在嘗試解決通過比較不同多項式的係數而獲得的多項式方程組。獲取Sage多項式系統的單一解決方案
# Statement of Problem:
# We are attempting to find complex numbers a, b, c, d, e, J, u, v, r, s where
# ((a*x + c)^2)*(x^3 + (3K)*x + 2K) - ((b*x^2 + d*x + e)^2) = a^2*(x - r)^2*(x - s)^3 and
# ((a*x + c)^2)*(x^3 + (3K)*x + 2K)) - ((b*x^2 + d*x + e - 1)^2) = a^2*(x - u)*(x - v)^4
R.<x> = CC['x']
a, b, c, d, e, r, s, u, v, K = var('a, b, c, d, e, r, s, u, v, K')
y2 = x^3 + (3*K)*x + 2*K
q0 = ((a*x + c)^2)*(y2) - ((b*x^2 + d*x + e)^2)
p0 = (a^2)*((x-r)^2)*((x-s)^3)
t = (b^2 - 2*a*c)/a^2
Q0 = q0.expand()
P0 = p0.expand()
P0 = P0.substitute(s = ((t - 2*r)/3))
Relations0 = []
i = 0
while i < 6:
Relations0.append(P0.coefficient(x, n = i) - Q0.coefficient(x, n = i))
i = i+1
q1 = ((a*x + c)^2)*(y2) - ((b*x^2 + d*x + e - 1)^2)
p1 = (a^2)*(x-u)*((x-v)^4)
Q1 = q1.expand()
P1 = p1.expand()
P1 = P1.substitute(u = t - 4*v)
Relations1 = []
i = 0
while i < 6:
Relations1.append(P1.coefficient(x, n = i) - Q1.coefficient(x, n = i))
i = i+1
Relations = Relations0 + Relations1
告訴賢者解決使用solve(Relations, a,b,c,d,e,r,v,K)
多項式的系統似乎非常低效的,並已不僅導致具有其內存限制超過賢者。此外,試圖通過求解一些變量來減少方程和變量的數量也是低效的,並且沒有給出任何有效的結果。由於試圖找到所有解決方案已證明如此困難,是否有任何方法只能提取單個解決方案?