Open sadqiang opened 5 years ago
There is a bug: Linq.formatComplexLambda doesn't handle Constant.I
I think adding a line will fix the bug.
let formatComplexLambda ...
let constant = function
| I -> Some (Expression.Constant (complex 0.0 1.0) :> Expression)
And use j as ImaginaryOne in Parse.
Func<Complex, Complex> f = Parse("z * z + j/2").CompileComplex("z");
Complex c = 1.0 / 2.0 - i / 3.0;
@cdrnet, I have questions: Are the functions at compilation searched in System.Math or System.Numerics.Complex? Is this why Trigonometric.simplify used? So to speak, to avoid missing functions such as Coth? Is it possible to call functions from MathNet.Numerics.Trig or SpecialFunctions? I'd like to use Coth itself and Bessel functions.
coth(sqrt(10000000 j)) = 1, but sinh(sqrt(10000000 j)) = oo - oo j and cosh(sqrt(10000000 j)) = oo - oo j
@diluculo : Has your fixing be committed? I just test but the issue still exist.
I found the following enlightenment while I was sleeping. However, this approach is risky because users can accidentally pass values other than i to the second argument.
using System;
using System.Numerics;
using static System.Console;
using static System.Numerics.Complex;
using static MathNet.Symbolics.SymbolicExpression;
class Program
{
static readonly Complex i = ImaginaryOne;
static void Main()
{
var z = Variable("z");
Func<Complex, Complex, Complex> f = Parse("z * i").CompileComplex(nameof(z), "i");
Complex c = 1 / 2.0 - i / 3;
WriteLine(f(c, i));
}
}
Why didn't you also use nameof(i)
? :-)
using System;
using System.Numerics;
using static System.Console;
using static System.Numerics.Complex;
using static MathNet.Symbolics.SymbolicExpression;
class Program
{
static readonly Complex i = ImaginaryOne;
static void Main()
{
var z = Variable("z");
Func<Complex, Complex, Complex> f = Parse("z * i").CompileComplex(nameof(z), nameof(i));
Complex c = 1 / 2.0 - i / 3;
WriteLine(f(c, i));
}
}
The following code snippet does not compile because there are no suitable operators for Complex and SymbolicExpression operands in
z * z + i/2
.Any comments are welcome!