使用mathdotnet求解线性方程组？

Question

我想解这样的方程，

(4-x)*2 = (y-1)*10 + 2
x = y*2 + 1

方程式以字符串形式提供。有没有办法在 mathdotnet 中表达方程式？只能想办法写表达式了

Answer 1

Math.NET Numerics can solve any linear system numerically，但我想这不是您要找的。

Math.NET Symbolics可以处理符号表达式，虽然这个项目处于早期阶段，还不了解方程的概念。然而，我们仍然可以用它来解决像这样的简单系统，只需做一些工作——就像我们手动做的那样。

首先，让我们定义一个小函数来求解阶数为 1 的单个线性方程：

using Expr = MathNet.Symbolics.Expression;

Expr SolveSimpleRoot(Expr variable, Expr expr)
{
    // try to bring expression into polynomial form
    Expr simple = Algebraic.Expand(Rational.Numerator(Rational.Simplify(variable,expr)));

    // extract coefficients, solve known forms of order up to 1
    Expr[] coeff = Polynomial.Coefficients(variable,simple);
    switch(coeff.Length)
    {
        case 1: return Expr.Zero.Equals(coeff[0]) ? variable : Expr.Undefined;
        case 2: return Rational.Simplify(variable,Algebraic.Expand(-coeff[0]/coeff[1]));
        default: return Expr.Undefined;
    }
}

那么我们可以利用这个来解决系统如下：

// declare variables
var x = Expr.Symbol("x");
var y = Expr.Symbol("y");

// Parse left and right side of both equations
Expr aleft = Infix.ParseOrThrow("(4-x)*2");
Expr aright = Infix.ParseOrThrow("(y-1)*10+2");
Expr bleft = Infix.ParseOrThrow("x");
Expr bright = Infix.ParseOrThrow("y*2+1");

// Solve both equations to x
Expr ax = SolveSimpleRoot(x,aleft-aright); // "8 - 5*y"
Expr bx = SolveSimpleRoot(x,bleft-bright); // "1 + 2*y"

// Equate both terms of x, solve to y
Expr cy = SolveSimpleRoot(y,ax-bx); // "1"

// Substitute term of y into one of the terms of x
Expr cx = Algebraic.Expand(Structure.Substitute(y,cy,ax)); // "3"

// Print expression in Infix notation
Console.WriteLine(Infix.Print(cx)); // x=3
Console.WriteLine(Infix.Print(cy)); // y=1

使用mathdotnet求解线性方程组？

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