"fsolve" 不适用于系统非线性方程
"fsolve" doesn't work for system nonlinear equation
我有这些等式：
syms pm pr teta s
A1 = -2 * b1 * pm + 2 * b2 * pr + b * teta + (1-t) * s + (1-p) * a + c * (b1 - b2);
A2 = 2 * b2 * pm + 2 * b1 * pr + (1-b) * teta + t * s + p * a + c * (b1 - b2);
A3 = b * pm + (1-b) * pr - n * teta - c;
A4 = (1-t) * pm + t * pr - k * s - c;
eqns = [A1,A2,A3,A4];
F=@(pm, pr, teta, s) [A1
A2
A3
A4];
x0 = [10, 10, 10, 10];
fsolve(F, x0)
我该如何解决？
（当我使用fsolve时，报错：FSOLVE requires all values returned by functions to be of data type double）
自从您标记了 Mathematica
A1 = -2*b1*pm + 2*b2*pr + b*teta + (1 - t)*s + (1 - p)*a + c*(b1 - b2);
A2 = 2*b2*pm + 2*b1*pr + (1 - b)*teta + t*s + p*a + c*(b1 - b2);
A3 = b*pm + (1 - b)*pr - n*teta - c;
A4 = (1 - t)*pm + t*pr - k*s - c;
FullSimplify[Solve[{A1 == 10, A2 == 10, A3 == 10, A4 == 10}, {pm, pr, teta, s}]]
pm -> ((k ((-1 + b)^2 + 2 b1 n) + n t^2) (b (10 + c) k +
n (10 + c + 10 k - a k - b1 c k + b2 c k +
a k p - (10 + c) t)) + ((-1 + b) (10 + c) k +
k n (-10 + b1 c - b2 c + a p) - (10 + c) n t) (b k - b^2 k +
n (2 b2 k + t - t^2)))/(k n (1 - 2 b1 k + b^2 (1 + 4 b2 k) +
2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t -
4 (b1 + b2) n t + (1 + 4 b2 n) t^2 +
2 b (-1 + 2 b1 k - 2 b2 k + t)))
pr -> (c +
b^2 (10 + c - (-20 + a + 2 b1 c - 2 b2 c) k) + 2 b1^2 c k n -
b2 (20 + c - 2 (-10 + a) k + 2 b2 c k) n - a (1 + 2 b2 k) n p -
2 b1 k (10 + c + n (10 - a p)) + 10 (1 + n - 2 t) -
c (2 + b2 n) t +
n (-30 + a + 20 b2 + a p) t + (10 + c - (-20 + a) n +
2 b2 c n) t^2 - b1 n (c + 20 t + c t (-1 + 2 t)) +
b (k (-10 + a + 20 b1 - 20 b2 - a p) + 20 (-1 + t) +
c (-2 + 3 b1 k - 3 b2 k + 2 t)))/(1 - 2 b1 k +
b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t -
4 (b1 + b2) n t + (1 + 4 b2 n) t^2 +
2 b (-1 + 2 b1 k - 2 b2 k + t))
teta -> (10 - 20 b2 - b2 c -
20 b2 k + 2 a b2 k + 40 b2^2 k + 2 b2^2 c k +
2 b1^2 (20 + 3 c) k - a p -
2 a b2 k p + (-30 + 3 b2 (20 + c) + a (1 + p)) t - (-20 + a +
2 b2 (20 + c)) t^2 +
b (-10 - 4 b1^2 c k + a p +
b1 (20 + 40 k - 2 a k + c (3 + 4 b2 k - 2 t)) + 20 t - a t +
b2 (20 + c - 2 a k + 4 a k p - 2 (20 + c) t)) +
b1 (2 k (-10 + a p) + 20 (-1 + t) + c (-3 + (5 - 2 t) t)))/(1 -
2 b1 k + b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t -
4 (b1 + b2) n t + (1 + 4 b2 n) t^2 +
2 b (-1 + 2 b1 k - 2 b2 k + t))
s -> (20 b1 + b1 c - b2 c -
2 b^2 (-10 + b1 c + b2 (20 + c)) + 20 b1 n + 40 b1^2 n -
20 b2 n + 40 b2^2 n + 2 b1^2 c n + 4 b1 b2 c n +
2 b2^2 c n - (b1 - b2) (20 + c) t +
4 b1 (-10 + b1 c - b2 c) n t - 10 (-1 + 2 b2 + t) +
a (-1 - b^2 - 2 b1 n + p + 2 (b1 + b2) n p + t - p t +
2 n (b1 + b2 - 2 b2 p) t - b (-2 + p + t)) +
b (-(-10 + b2 (20 + c)) (-3 + 2 t) + b1 (-20 + c - 2 c t)))/(1 -
2 b1 k + b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t -
4 (b1 + b2) n t + (1 + 4 b2 n) t^2 +
2 b (-1 + 2 b1 k - 2 b2 k + t))
我有这些等式：
syms pm pr teta s
A1 = -2 * b1 * pm + 2 * b2 * pr + b * teta + (1-t) * s + (1-p) * a + c * (b1 - b2);
A2 = 2 * b2 * pm + 2 * b1 * pr + (1-b) * teta + t * s + p * a + c * (b1 - b2);
A3 = b * pm + (1-b) * pr - n * teta - c;
A4 = (1-t) * pm + t * pr - k * s - c;
eqns = [A1,A2,A3,A4];
F=@(pm, pr, teta, s) [A1
A2
A3
A4];
x0 = [10, 10, 10, 10];
fsolve(F, x0)
我该如何解决？ （当我使用fsolve时，报错：FSOLVE requires all values returned by functions to be of data type double）
自从您标记了 Mathematica
A1 = -2*b1*pm + 2*b2*pr + b*teta + (1 - t)*s + (1 - p)*a + c*(b1 - b2);
A2 = 2*b2*pm + 2*b1*pr + (1 - b)*teta + t*s + p*a + c*(b1 - b2);
A3 = b*pm + (1 - b)*pr - n*teta - c;
A4 = (1 - t)*pm + t*pr - k*s - c;
FullSimplify[Solve[{A1 == 10, A2 == 10, A3 == 10, A4 == 10}, {pm, pr, teta, s}]]
pm -> ((k ((-1 + b)^2 + 2 b1 n) + n t^2) (b (10 + c) k + n (10 + c + 10 k - a k - b1 c k + b2 c k + a k p - (10 + c) t)) + ((-1 + b) (10 + c) k + k n (-10 + b1 c - b2 c + a p) - (10 + c) n t) (b k - b^2 k + n (2 b2 k + t - t^2)))/(k n (1 - 2 b1 k + b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t - 4 (b1 + b2) n t + (1 + 4 b2 n) t^2 + 2 b (-1 + 2 b1 k - 2 b2 k + t)))
pr -> (c + b^2 (10 + c - (-20 + a + 2 b1 c - 2 b2 c) k) + 2 b1^2 c k n - b2 (20 + c - 2 (-10 + a) k + 2 b2 c k) n - a (1 + 2 b2 k) n p - 2 b1 k (10 + c + n (10 - a p)) + 10 (1 + n - 2 t) - c (2 + b2 n) t + n (-30 + a + 20 b2 + a p) t + (10 + c - (-20 + a) n + 2 b2 c n) t^2 - b1 n (c + 20 t + c t (-1 + 2 t)) + b (k (-10 + a + 20 b1 - 20 b2 - a p) + 20 (-1 + t) + c (-2 + 3 b1 k - 3 b2 k + 2 t)))/(1 - 2 b1 k + b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t - 4 (b1 + b2) n t + (1 + 4 b2 n) t^2 + 2 b (-1 + 2 b1 k - 2 b2 k + t))
teta -> (10 - 20 b2 - b2 c - 20 b2 k + 2 a b2 k + 40 b2^2 k + 2 b2^2 c k + 2 b1^2 (20 + 3 c) k - a p - 2 a b2 k p + (-30 + 3 b2 (20 + c) + a (1 + p)) t - (-20 + a + 2 b2 (20 + c)) t^2 + b (-10 - 4 b1^2 c k + a p + b1 (20 + 40 k - 2 a k + c (3 + 4 b2 k - 2 t)) + 20 t - a t + b2 (20 + c - 2 a k + 4 a k p - 2 (20 + c) t)) + b1 (2 k (-10 + a p) + 20 (-1 + t) + c (-3 + (5 - 2 t) t)))/(1 - 2 b1 k + b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t - 4 (b1 + b2) n t + (1 + 4 b2 n) t^2 + 2 b (-1 + 2 b1 k - 2 b2 k + t))
s -> (20 b1 + b1 c - b2 c - 2 b^2 (-10 + b1 c + b2 (20 + c)) + 20 b1 n + 40 b1^2 n - 20 b2 n + 40 b2^2 n + 2 b1^2 c n + 4 b1 b2 c n + 2 b2^2 c n - (b1 - b2) (20 + c) t + 4 b1 (-10 + b1 c - b2 c) n t - 10 (-1 + 2 b2 + t) + a (-1 - b^2 - 2 b1 n + p + 2 (b1 + b2) n p + t - p t + 2 n (b1 + b2 - 2 b2 p) t - b (-2 + p + t)) + b (-(-10 + b2 (20 + c)) (-3 + 2 t) + b1 (-20 + c - 2 c t)))/(1 - 2 b1 k + b^2 (1 + 4 b2 k) + 2 (b1 - 2 (b1^2 + b2^2) k) n - 2 t - 4 (b1 + b2) n t + (1 + 4 b2 n) t^2 + 2 b (-1 + 2 b1 k - 2 b2 k + t))