MathSharp 1.0.11

Use as a simple calculator

var inp = "1 + 2 * log(9, 3)";
var expr = MathS.FromString(inp);
Console.WriteLine(expr.Eval());
>>> 5

Build an expression

var x = MathS.Var("x");
var y = MathS.Var("y");
var c = x * y + x / y;
Console.WriteLine(MathS.Sqr(c));
>>> (x * y + x / y) ^ 2

Substitute variables

var x = MathS.Var("x");
var expr = x * 2 + MathS.Sin(x) / MathS.Sin(MathS.Pow(2, x));
var subs = expr.Substitute(x, 0.3);
Console.WriteLine(subs.Simplify());
>>> 0,9134260185941638

Find derivatives

var x = MathS.Var("x");
var func = MathS.Sqr(x) + MathS.Ln(MathS.Cos(x) + 3) + 4 * x;
var derivative = func.Derive(x);
Console.WriteLine(derivative.Simplify());
>>> 2 * x + -1 * sin(x) / (cos(x) + 3) + 4

Build formulas

var x = MathS.Var("x");
var expr = (x + 3).Pow(x + 4);
Func<NumberEntity, Entity> wow = v => expr.Substitute(x, v).Simplify();
Console.WriteLine(wow(4));
Console.WriteLine(wow(5));
Console.WriteLine(wow(6));
>>> 5764801
>>> 134217728
>>> 3486784401

Render latex

var x = MathS.Var("x");
var y = MathS.Var("y");
var expr = x.Pow(y) + MathS.Sqrt(x + y / 4) * (6 / x);
Console.WriteLine(expr.Latexise());
>>> {x}^{y}+\sqrt{x+\frac{y}{4}}*\frac{6}{x}

Play with complex numbers

var expr = MathS.Pow(MathS.e, MathS.pi * MathS.i);
Console.WriteLine(expr);
Console.WriteLine(expr.Eval());
>>> 2,718281828459045 ^ 3,141592653589793i
>>> -1

Solve eqations

var x = MathS.Var("x");
var equation = (x - 1) * (x - 2) * (MathS.Sqr(x) + 1);
foreach (var re in equation.SolveNt(x))
    Console.Write(re.ToString() + "  ");
>>> 1  2  1i

Full documentation

Function list

MathS.Sin(x) MathS.Cos(x) MathS.Log(num, base), MathS.Pow(base, power) MathS.Sqrt(x) = MathS.Pow(x, 0.5) MathS.Sqr(x) = MathS.Pow(x, 2) MathS.Tan(x) = MathS.Sin(x) / MathS.Cos(x) MathS.Cotan(x) = 1 / MathS.Tan(x) MathS.Sec(x) = 1 / MathS.Cos(x) MathS.Cosec(x) = 1 / MathS.Sin(x) MathS.B(x) = x * MathS.Sin(x) MathS.TB(x) = x * MathS.Cos(x)

MathS.FromString(str) - returns Entity

Entity methods

expr.Derive(x) - derivation for variable x expr.Eval() - evaluation & simplification expr.Latexise() - render to latex expr.SolveNt(expr, from, to, stepCount, precision) - find roots assuming we are solving equation expr=0. The algorithm iterates on [from.Re; to.Re] for real part and on [from.Im; to.Im] for imaginary part. The higher stepCount is, the more roots the function can find Precision - if you get similar roots that you think are equal, you can increase this argument. You can also decrease MathS.EQUALITY_THRESHOLD which is responsible for comparing Numbers.