C# Matemáticas cuestión de clase

Question

necesito para calcular Tanh-1 en C#
(y Sinh-1 y Cosh-1)

No me encontré en la biblioteca de matemáticas .. ¿Alguna sugerencia?

EDITAR: Tanh not Tan !!

Answer 1

Necesitas derivarlos a sí mismo usando las funciones existentes, por ejemplo, Math.sin

Usted puede encontrar esto útil:

Secant Sec(X) = 1/Cos(X) 
Cosecant Cosec(X) = 1/Sin(X) 
Cotangent Cotan(X) = 1/Tan(X) 
Inverse Sine Arcsin(X) = Atn(X/Sqr(-X * X + 1)) 
Inverse Cosine Arccos(X) = Atn(-X/Sqr(-X * X + 1)) + 2 * Atn(1) 
Inverse Secant Arcsec(X) = 2 * Atn(1) - Atn(Sgn(X)/Sqr(X * X - 1)) 
Inverse Cosecant Arccosec(X) = Atn(Sgn(X)/Sqr(X * X - 1)) 
Inverse Cotangent Arccotan(X) = 2 * Atn(1) - Atn(X) 
Hyperbolic Sine HSin(X) = (Exp(X) - Exp(-X))/2 
Hyperbolic Cosine HCos(X) = (Exp(X) + Exp(-X))/2 
Hyperbolic Tangent HTan(X) = (Exp(X) - Exp(-X))/(Exp(X) + Exp(-X)) 
Hyperbolic Secant HSec(X) = 2/(Exp(X) + Exp(-X)) 
Hyperbolic Cosecant HCosec(X) = 2/(Exp(X) - Exp(-X)) 
Hyperbolic Cotangent HCotan(X) = (Exp(X) + Exp(-X))/(Exp(X) - Exp(-X)) 
Inverse Hyperbolic Sine HArcsin(X) = Log(X + Sqr(X * X + 1)) 
Inverse Hyperbolic Cosine HArccos(X) = Log(X + Sqr(X * X - 1)) 
Inverse Hyperbolic Tangent HArctan(X) = Log((1 + X)/(1 - X))/2 
Inverse Hyperbolic Secant HArcsec(X) = Log((Sqr(-X * X + 1) + 1)/X) 
Inverse Hyperbolic Cosecant HArccosec(X) = Log((Sgn(X) * Sqr(X * X + 1) + 1)/X) 
Inverse Hyperbolic Cotangent HArccotan(X) = Log((X + 1)/(X - 1))/2 
Logarithm to base N LogN(X) = Log(X)/Log(N) 

Answer 2

Debe definirlos usted mismo.

http://en.wikipedia.org/wiki/Hyperbolic_function#Inverse_functions_as_logarithms

-1  1 1 + x 
tanh x = — ln ————— 
      2 1 - x 

    -1    _______ 
sinh x = ln (x + √ x² + 1) 

    -1    _______ 
cosh x = ln (x + √ x² - 1) 

Answer 3

Para fórmulas de NET-ify David Relihan:

public static class MathHelper 
{ 
    // Secant 
    public static double Sec(double x) 
    { 
     return 1/Math.Cos(x); 
    } 

    // Cosecant 
    public static double Cosec(double x) 
    { 
     return 1/Math.Sin(x); 
    } 

    // Cotangent 
    public static double Cotan(double x) 
    { 
     return 1/Math.Tan(x); 
    } 

    // Inverse Sine 
    public static double Arcsin(double x) 
    { 
     return Math.Atan(x/Math.Sqrt(-x * x + 1)); 
    } 

    // Inverse Cosine 
    public static double Arccos(double x) 
    { 
     return Math.Atan(-x/Math.Sqrt(-x * x + 1)) + 2 * Math.Atan(1); 
    } 


    // Inverse Secant 
    public static double Arcsec(double x) 
    { 
     return 2 * Math.Atan(1) - Math.Atan(Math.Sign(x)/Math.Sqrt(x * x - 1)); 
    } 

    // Inverse Cosecant 
    public static double Arccosec(double x) 
    { 
     return Math.Atan(Math.Sign(x)/Math.Sqrt(x * x - 1)); 
    } 

    // Inverse Cotangent 
    public static double Arccotan(double x) 
    { 
     return 2 * Math.Atan(1) - Math.Atan(x); 
    } 

    // Hyperbolic Sine 
    public static double HSin(double x) 
    { 
     return (Math.Exp(x) - Math.Exp(-x))/2 ; 
    } 

    // Hyperbolic Cosine 
    public static double HCos(double x) 
    { 
     return (Math.Exp(x) + Math.Exp(-x))/2 ; 
    } 

    // Hyperbolic Tangent 
    public static double HTan(double x) 
    { 
     return (Math.Exp(x) - Math.Exp(-x))/(Math.Exp(x) + Math.Exp(-x)); 
    } 

    // Hyperbolic Secant 
    public static double HSec(double x) 
    { 
     return 2/(Math.Exp(x) + Math.Exp(-x)); 
    } 

    // Hyperbolic Cosecant 
    public static double HCosec(double x) 
    { 
     return 2/(Math.Exp(x) - Math.Exp(-x)); 
    } 

    // Hyperbolic Cotangent 
    public static double HCotan(double x) 
    { 
     return (Math.Exp(x) + Math.Exp(-x))/(Math.Exp(x) - Math.Exp(-x)); 
    } 

    // Inverse Hyperbolic Sine 
    public static double HArcsin(double x) 
    { 
     return Math.Log(x + Math.Sqrt(x * x + 1)) ; 
    } 

    // Inverse Hyperbolic Cosine 
    public static double HArccos(double x) 
    { 
     return Math.Log(x + Math.Sqrt(x * x - 1)); 
    } 

    // Inverse Hyperbolic Tangent 
    public static double HArctan(double x) 
    { 
     return Math.Log((1 + x)/(1 - x))/2 ; 
    } 

    // Inverse Hyperbolic Secant 
    public static double HArcsec(double x) 
    { 
     return Math.Log((Math.Sqrt(-x * x + 1) + 1)/x); 
    } 

    // Inverse Hyperbolic Cosecant 
    public static double HArccosec(double x) 
    { 
     return Math.Log((Math.Sign(x) * Math.Sqrt(x * x + 1) + 1)/x) ; 
    } 

    // Inverse Hyperbolic Cotangent 
    public static double HArccotan(double x) 
    { 
     return Math.Log((x + 1)/(x - 1))/2; 
    } 

    // Logarithm to base N 
    public static double LogN(double x, double n) 
    { 
     return Math.Log(x)/Math.Log(n); 
    } 
} 

Answer 4

También hay fórmula más rápida para tanh la computación, que requieren sólo una exp (), porque tanh se relaciona con función logística:

tanh (x) = 2/(1 + exp (-2 * x)) - 1
también
tanh (x) = 1 - 2/(1 + exp (2 * x))

Ver: http://en.wikipedia.org/wiki/Logistic_function