equations 2.0.1

An equation solving library written in Dart. It also works with complex numbers and fractions.

An equation solving library written purely in Dart

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Thanks to equations you're able to solve polynomial and nonlinear equations with ease. It's been written in "pure" Dart, meaning that it has no dependency on any framework. It can be used with Flutter for web, desktop and mobile. Here's a summary of the contents of the package:

• Algebraic and all of its subtypes, which can be used to solve algebraic equations (also known as polynomial equations);
• Nonlinear and all of its subtypes, which can be used to solve nonlinear equations;
• SystemSolver and all of its subtypes, which can be used to solve systems of linear equations;
• Complex, which is used to easily handle complex numbers;
• Fraction, from the fraction package which helps you working with fractions.

This package is meant to be used with Dart 2.12 or higher because the code is entirely null safe. Please don't hesitate to open a PR or file an issue if you wish to contribute to the growth of this package!

Algebraic equations

Use one of the following classes to find the roots of a polynomial. You can use both complex numbers and fractions as coefficients.

Solver name	Equation	Params field
Constant	f(x) = a	a ∈ C
Linear	f(x) = ax + b	a, b ∈ C
Quadratic	f(x) = ax2 + bx + c	a, b, c ∈ C
Cubic	f(x) = ax3 + bx2 + cx + d	a, b, c, d ∈ C
Quartic	f(x) = ax4 + bx3 + cx2 + dx + e	a, b, c, d, e ∈ C
Laguerre	Any polynomial P(xi) where xi are coefficients	xi ∈ C

There's a formula for polynomials up to the fourth degree, as explained by Galois Theory. Roots of polynomials whose degree is 5 or higher, must be seeked using Laguerre's method or any other root-finding algorithm. For this reason, we suggest to go for the following approach:

• Use Linear to find the roots of a polynomial whose degree is 1.
• Use Quadratic to find the roots of a polynomial whose degree is 2.
• Use Cubic to find the roots of a polynomial whose degree is 3.
• Use Quartic to find the roots of a polynomial whose degree is 4.
• Use Laguerre to find the roots of a polynomial whose degree is 5 or higher.

Note that Laguerre can be used with any polynomials, so you could use it (for example) to solve a cubic equation as well. Laguerre internally uses loops, derivatives and other mechanics that are much slower than Quartic, Cubic, Quadratic and Linear so use it only when really needed. Here's how you can solve a cubic:

// f(x) = (2-3i)x^3 + 6/5ix^2 - (-5+i)x - (9+6i)
final equation = Cubic(
  a: Complex(2, -3),
  b: Complex.fromImaginaryFraction(Fraction(6, 5)),
  c: Complex(5, -1),
  d: Complex(-9, -6)
);

final degree = equation.degree; // 3
final isReal = equation.isRealEquation; // false
final discr = equation.discriminant(); // -31299.688 + 27460.192i

// f(x) = (2 - 3i)x^3 + 1.2ix^2 + (5 - 1i)x + (-9 - 6i)
print("$equation");
// f(x) = (2 - 3i)x^3 + 6/5ix^2 + (5 - 1i)x + (-9 - 6i)
print(equation.toStringWithFractions());

/*
 * Prints the roots of the equation:
 *
 *  x1 = 0.348906207844 - 1.734303423032i
 *  x2 = -1.083892638909 + 0.961044482775
 *  x3 = 1.011909507988 + 0.588643555642
 * */
for (final root in equation.solutions()) {
  print(root);
}

Alternatively, you could have used Laguerre to solve the same equation:

// f(x) = (2-3i)x^3 + 6/5ix^2 - (-5+i)x - (9+6i)
final equation = Laguerre(
  coefficients: [
    Complex(2, -3),
    Complex.fromImaginaryFraction(Fraction(6, 5)),
    Complex(5, -1),
    Complex(-9, -6),
  ]
);

/*
 * Prints the roots of the equation:
 *
 *  x1 = 1.0119095 + 0.5886435
 *  x2 = 0.3489062 - 1.7343034i
 *  x3 = -1.0838926 + 0.9610444
 * */
for (final root in equation.solutions()) {
  print(root);
}

As we've already pointed out, both ways are equivalent but Laguerre internally performs operations on matrices and calculates determinants many times so it's computationally slower than Cubic. Use Laguerre only when the degree of your polynomial is greater or equal than 5.

final quadratic = Algebraic.from(const [
  Complex(2, -3),
  Complex.i(),
  Complex(1, 6)
]);

final quartic = Algebraic.fromReal(const [
  1, -2, 3, -4, 5
]);

The factory constructor Algebraic.from() automatically returns the best type of Algebraic according with the number of parameters you've passed.

Nonlinear equations

Use one of the following classes, representing a root-finding algorithm, to find a root of an equation. Only real numbers are allowed. This package supports the following root finding methods:

Solver name	Params field
Bisection	a, b ∈ R
Chords	a, b ∈ R
Netwon	x0 ∈ R
Secant	a, b ∈ R
Steffensen	x0 ∈ R
Brent	a, b ∈ R
RegulaFalsi	a, b ∈ R

Expressions are parsed using petitparser, a fasts, stable and well tested grammar parser. These algorithms only work with real numbers. Here's a simple example of how you can find the roots of an equation:

final newton = Newton("2*x+cos(x)", -1, maxSteps: 5);

final steps = newton.maxSteps; // 5
final tol = newton.tolerance; // 1.0e-10
final fx = newton.function; // 2*x+cos(x)
final guess = newton.x0; // -1

final solutions = await newton.solve();

final convergence = solutions.convergence.round(); // 2
final solutions = solutions.efficiency.round(); // 1

/*
 * The getter `solutions.guesses` returns the list of values computed by the algorithm
 *
 * -0.4862880170389824
 * -0.45041860473199363
 * -0.45018362150211116
 * -0.4501836112948736
 * -0.45018361129487355
 */
final List<double> guesses = solutions.guesses;

Note that certain algorithms don't guarantee the convergence to a root so read the documentation carefully before choosing the method. You can also calculate the numerical value of a definite integral on an interval:

final bisection = Bisection(
  function: "x^3+2*x-1.2", 
  a: 0, 
  b: 3
);

// Integral from 0 to 3 of x^3+2*x-1.2 dx
final integral = bisection.integrateOn(const SimpsonRule(
  lowerBound: 0,
  upperBound: 3,
  intervals: 40,
));

// The result may vary depending on the 'intervals' you've decided to use
print(integral.result); // 25.65

Of course you can also use any NumericalIntegration subtype outside of the scope of a Nonlinear instance:

final midpointValue = MidpointRule(
  lowerBound: 0,
  upperBound: 3,
).integrate("x^3+2*x-1.2");

final trapezoidValue = TrapezoidalRule(
  lowerBound: 0,
  upperBound: 3,
).integrate("x^3+2*x-1.2");

final simpsonValue = SimpsonRule(
  lowerBound: 0,
  upperBound: 3,
).integrate("x^3+2*x-1.2");

Systems of equations

Use one of the following classes to solve systems of linear equations. Note that only real coefficients are allowed (so double is ok but Complex isn't) and you must define N equations in N variables (so a square matrix is needed).

Solver name	Iterative method
CholeskySolver	❌
GaussianElimination	❌
GaussSeidelSolver	✔️
JacobiSolver	✔️
LUSolver	❌
SORSolver	✔️

In any case, solvers only work with square matrices so N equations in N variables. These solvers are used to find the x in the Ax = b relation. Methods require at least the equation system matrix A and the vector b containing the unknowns. Iterative methods may require additional parameters such as an initial guess or a particular configuration value.

// Solve a system using LU decomposition
final luSolver = LUSolver(
  equations: const [
    [7, -2, 1],
    [14, -7, -3],
    [-7, 11, 18]
  ],
  constants: const [12, 17, 5]
);

final solutions = luSolver.solve(); // [-1, 4, 3]
final determinant = luSolver.determinant(); // -84.0

If you just want to work with matrices (operations, LU decomposition and/or Cholesky decomposition) you can consider the usage of RealMatrix or ComplexMatrix which are a real or complex representation of a matrix.

final matrixA = RealMatrix.fromData(
  columns: 2,
  rows: 2,
  data: const [
    [2, 6],
    [-5, 0]
  ]
);

final matrixB = RealMatrix.fromData(
  columns: 2,
  rows: 2,
  data: const [
    [-4, 1],
    [7, -3],
  ]
);

final sum = matrixA + matrixB;
final sub = matrixA - matrixB;
final mul = matrixA * matrixB;
final div = matrixA / matrixB;

final lu = matrixA.luDecomposition();
final cholesky = matrixA.choleskyDecomposition();

final det = matrixA.determinant();

You can use toString() to print the content of the matrix but there's also the possibility to use toStringAugmented() which prints the augmented matrix (the matrix + one extra column with the known values vector).

final lu = LUSolver(
  equations: const [
    [7, -2, 1],
    [14, -7, -3],
    [-7, 11, 18]
  ],
  constants: const [12, 17, 5]
);

/*
 * Output with 'toString':
 *
 * [7.0, -2.0, 1.0]
 * [14.0, -7.0, -3.0]
 * [-7.0, 11.0, 18.0]
*/
print("$lu");

/*
 * Output with 'toStringAugmented':
 *
 * [7.0, -2.0, 1.0 | 12.0]
 * [14.0, -7.0, -3.0 | 17.0]
 * [-7.0, 11.0, 18.0 | 5.0]
*/
print("${lu.toStringAugmented()}");