level 1
ilove青葱
楼主
Z是一个积分,对上面的Π进行积分,F是数量,对F进行求导,解出F,然后代回Z中,得到的结果化简为什么是一个0呢?代码如下:
Subscript[n, 0] = F
Subscript[n, c][t] = Subscript[n, 0] e^(Subscript[b, c] t)
Subscript[P, g][Subscript[Q, g]] = a - b Subscript[Q, g][F]
Subscript[Q, g][F] = Subscript[Q, 0] + \[Lambda] *Subscript[n, c][t]
Subscript[\[Pi], A][t] =
Subscript[P, g][Subscript[Q, g]]*Subscript[Q, g][F] -
Subscript[C, g] * Subscript[Q, g][F]
Z = Integrate[Subscript[\[Pi], A][t], t]
Solve[D[Z] == 0, F]
F = (e^(-2 t Subscript[b,
c]) (a e^(t Subscript[b, c]) \[Lambda] -
e^(t Subscript[b, c]) \[Lambda] Subscript[C, g] -
2 b e^(t Subscript[b, c]) \[Lambda] Subscript[Q, 0] -
Sqrt[b] e^(t Subscript[b, c]) \[Lambda] Sqrt[Subscript[Q, 0]]
Sqrt[-2 a + 2 a t Log[e] Subscript[b, c] + 2 Subscript[C, g] -
2 t Log[e] Subscript[b, c] Subscript[C, g] +
3 b Subscript[Q, 0] -
2 b t Log[e] Subscript[b, c] Subscript[Q, 0]]))/(b \[Lambda]^2)
Simplify[F]
Simplify[Z]





2023年11月24日 11点11分
1
Subscript[n, 0] = F
Subscript[n, c][t] = Subscript[n, 0] e^(Subscript[b, c] t)
Subscript[P, g][Subscript[Q, g]] = a - b Subscript[Q, g][F]
Subscript[Q, g][F] = Subscript[Q, 0] + \[Lambda] *Subscript[n, c][t]
Subscript[\[Pi], A][t] =
Subscript[P, g][Subscript[Q, g]]*Subscript[Q, g][F] -
Subscript[C, g] * Subscript[Q, g][F]
Z = Integrate[Subscript[\[Pi], A][t], t]
Solve[D[Z] == 0, F]
F = (e^(-2 t Subscript[b,
c]) (a e^(t Subscript[b, c]) \[Lambda] -
e^(t Subscript[b, c]) \[Lambda] Subscript[C, g] -
2 b e^(t Subscript[b, c]) \[Lambda] Subscript[Q, 0] -
Sqrt[b] e^(t Subscript[b, c]) \[Lambda] Sqrt[Subscript[Q, 0]]
Sqrt[-2 a + 2 a t Log[e] Subscript[b, c] + 2 Subscript[C, g] -
2 t Log[e] Subscript[b, c] Subscript[C, g] +
3 b Subscript[Q, 0] -
2 b t Log[e] Subscript[b, c] Subscript[Q, 0]]))/(b \[Lambda]^2)
Simplify[F]
Simplify[Z]




