syms x m t L1 L2 L3 L4 L5 f=exp(-m^2*x^2); g1=taylor(f,x,-2*t,'order',4); g2=taylor(f,x,-t,'order',4); g3=taylor(f,x,0,'order',4); g4=taylor(f,x,t,'order',4); g5=taylor(f,x,2*t,'order',4); A11=(1/t)*int(g1,x,-2.5*t,-1.5*t); A12=(1/t)*int(g2,x,-2.5*t,-1.5*t); A13=(1/t)*int(g3,x,-2.5*t,-1.5*t); A14=(1/t)*int(g4,x,-2.5*t,-1.5*t); A15=(1/t)*int(g5,x,-2.5*t,-1.5*t); A21=(1/t)*int(g1,x,-1.5*t,-0.5*t); A22=(1/t)*int(g2,x,-1.5*t,-0.5*t); A23=(1/t)*int(g3,x,-1.5*t,-0.5*t); A24=(1/t)*int(g4,x,-1.5*t,-0.5*t); A25=(1/t)*int(g5,x,-1.5*t,-0.5*t); A31=(1/t)*int(g1,x,-0.5*t,0.5*t); A32=(1/t)*int(g2,x,-0.5*t,0.5*t); A33=(1/t)*int(g3,x,-0.5*t,0.5*t); A34=(1/t)*int(g4,x,-0.5*t,0.5*t); A35=(1/t)*int(g5,x,-0.5*t,0.5*t); A41=(1/t)*int(g1,x,0.5*t,1.5*t); A42=(1/t)*int(g2,x,0.5*t,1.5*t); A43=(1/t)*int(g3,x,0.5*t,1.5*t); A44=(1/t)*int(g4,x,0.5*t,1.5*t); A45=(1/t)*int(g5,x,0.5*t,1.5*t); A51=(1/t)*int(g1,x,1.5*t,2.5*t); A52=(1/t)*int(g2,x,1.5*t,2.5*t); A53=(1/t)*int(g3,x,1.5*t,2.5*t); A54=(1/t)*int(g4,x,1.5*t,2.5*t); A55=(1/t)*int(g5,x,1.5*t,2.5*t); A=[A11,A12,A13,A14,A15; A21,A22,A23,A24,A25; A31,A32,A33,A34,A35; A41,A42,A43,A44,A45; A51,A52,A53,A54,A55]; P=[1,-2*t;1,-t;1,0;1,t;1,2*t]; L=[L1,L2,L3,L4,L5,0,0]'; G=[g1,g2,g3,g4,g5,1,x]'; B=[A11,A12,A13,A14,A15,1,-2*t; A21,A22,A23,A24,A25,1,-t; A31,A32,A33,A34,A35,1,0; A41,A42,A43,A44,A45,1,t; A51,A52,A53,A54,A55,1,2*t; 1,1,1,1,1,0,0; -2*t,-t,0,t,2*t,0,0]; BB=pinv(B)*G
时间: 2024-04-19 12:28:32 浏览: 175
这段代码是用MATLAB编写的,它计算了一些函数的泰勒展开式,并使用这些展开式计算了一些积分和矩阵运算。最后,它使用广义逆函数计算了一个向量。具体来说,这段代码定义了一些符号变量和函数,然后计算了这些函数在不同区间上的积分,并将结果存储在矩阵A中。接下来,它定义了一些其他矩阵P、L、G和B,并使用广义逆函数计算了矩阵B的伪逆乘以向量G的结果。最后,它将结果存储在矩阵BB中。
相关问题
syms k y m x x0 y0; eqn1 = (45*x)^2+90*m*k+m^m == 16*k^2+16; eqn2 = (45^2-16)*k^2+90*m*k+m^2-16 == 0; eqn3 = k == -15^2/(40^2)*x0/y0; eqn4 = m == 15^2/y0; [solk,soly0,solm,solx0] = solve(eqn1,eqn2,eqn3,eqn4,k,y0,m,x0); disp(solk); disp(soly0); disp(solm); disp(solx0);
根据你提供的方程,使用MATLAB的符号计算工具箱可以求解。以下是求解的MATLAB代码:
```matlab
syms k y m x x0 y0;
eqn1 = (45*x)^2+90*m*k+m^m == 16*k^2+16;
eqn2 = (45^2-16)*k^2+90*m*k+m^2-16 == 0;
eqn3 = k == -15^2/(40^2)*x0/y0;
eqn4 = m == 15^2/y0;
[solk,soly0,solm,solx0] = solve(eqn1,eqn2,eqn3,eqn4,k,y0,m,x0);
disp(solk);
disp(soly0);
disp(solm);
disp(solx0);
```
运行这段代码后,你会得到方程的解。其中`solk`是k的解,`soly0`是y0的解,`solm`是m的解,`solx0`是x0的解。
注意,这里我使用了MATLAB的符号计算工具箱来求解方程。如果你没有这个工具箱,可以尝试使用其他数学计算软件或在线计算器来求解方程。
syms l1 l2 M L=[l1,l2]' R=[1,1]' e1=y-c e2=y-z Y=l1c+l2z J=(y-Y)^2 B=[e1,e2] E=B'*B J=L'EL min J M=R'*L M=1,求解l1,l2,代码
根据所给的目标函数和约束条件,这是一个典型的最小二乘问题。可以使用最小二乘法进行求解。
以下是 MATLAB 代码实现:
```matlab
syms l1 l2 M;
L = [l1,l2]';
R = [1,1]';
e1 = y - c;
e2 = y - z;
Y = l1*c + l2*z;
J = (y - Y)^2;
B = [e1,e2];
E = B'*B;
J = L'*E*L;
M = R'*L;
M = 1;
l = inv(J+M)*R;
l1 = l(1);
l2 = l(2);
```
其中,`syms` 用于声明符号变量,`inv` 用于求矩阵的逆,`'*'` 表示矩阵乘法。
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