mathematica如何求解含有三角函数的方程?FindRoot[{3 x1^3 + 2 x2 - 5 + Sin[x1 - x2] Sin[x1 + x2] == 0,-x1*Exp[x1 - x2] + x2 (4 + 3 x2^2) + 2 x3 + Sin[x2 - x3] Sin[x2 + x3] - 8 == 0,-x2*Exp[x2 - x3] + 4 x3 - 3 ==0},{x1,Pi/2},{x2,2},{x3,0}]

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mathematica如何求解含有三角函数的方程?FindRoot[{3 x1^3 + 2 x2 - 5 + Sin[x1 - x2] Sin[x1 + x2] == 0,-x1*Exp[x1 - x2] + x2 (4 + 3 x2^2) + 2 x3 + Sin[x2 - x3] Sin[x2 + x3] - 8 == 0,-x2*Exp[x2 - x3] + 4 x3 - 3 ==0},{x1,Pi/2},{x2,2},{x3,0}]

mathematica如何求解含有三角函数的方程?FindRoot[{3 x1^3 + 2 x2 - 5 + Sin[x1 - x2] Sin[x1 + x2] == 0,-x1*Exp[x1 - x2] + x2 (4 + 3 x2^2) + 2 x3 + Sin[x2 - x3] Sin[x2 + x3] - 8 == 0,-x2*Exp[x2 - x3] + 4 x3 - 3 ==0},{x1,Pi/2},{x2,2},{x3,0}]
mathematica如何求解含有三角函数的方程?
FindRoot[{3 x1^3 + 2 x2 - 5 + Sin[x1 - x2] Sin[x1 + x2] ==
0,-x1*Exp[x1 - x2] + x2 (4 + 3 x2^2) + 2 x3 +
Sin[x2 - x3] Sin[x2 + x3] - 8 == 0,-x2*Exp[x2 - x3] + 4 x3 - 3 ==
0},{x1,Pi/2},{x2,2},{x3,0}]
得出结果是{x1 -> 1.,x2 -> 1.,x3 -> 1.}

mathematica如何求解含有三角函数的方程?FindRoot[{3 x1^3 + 2 x2 - 5 + Sin[x1 - x2] Sin[x1 + x2] == 0,-x1*Exp[x1 - x2] + x2 (4 + 3 x2^2) + 2 x3 + Sin[x2 - x3] Sin[x2 + x3] - 8 == 0,-x2*Exp[x2 - x3] + 4 x3 - 3 ==0},{x1,Pi/2},{x2,2},{x3,0}]
你不是已经解出来了吗?这种超越方程一般都是用FindRoot求数值解的,因为往往没有解析解.