CODEWITHSUNDEEP
cmath.h
rh939
rh939

Reputation: 55

Why tanh function from math.h library give wrong result?

#include<stdio.h>
#include<math.h>
#define PI 2*acos(0.0)
int main(void)
{
    double theta;
    theta=tanh(1/(sqrt(3.0)));
    printf("With tanh function = %lf\n",theta);
    printf("Actual value = %lf\n",PI/6.0);
    return 0;
}

Output:

With tanh function = 0.520737

Actual value = 0.523599

Why are these two values different? It should me same as my understanding.

Upvotes: 2

Views: 346

Answers (2)

rici
rici

Reputation: 241881

You've got that identity completely wrong.

The actual identity is

tanh-1(i ⁄ √3) = πi ⁄ 6 (where i is the imaginary unit, √-1)

C11 can easily validate that:

#define _XOPEN_SOURCE 700
#include<stdio.h>
#include<math.h>
#include<complex.h>

int main(void)
{
    complex double theta=catanh(I/sqrt(3.0));
    printf("With atanh function = %lf\n",cimag(theta));
    printf("Actual value = %lf\n",M_PI/6);
    return 0;
}

(Live on coliru: http://coliru.stacked-crooked.com/a/f3df5358a2be67cd):

With atanh function = 0.523599
Actual value = 0.523599

M_PI will be in math.h in any Posix compliant system. Apparently, on Windows you use

#define _USE_MATH_DEFINES

but I have no idea whether Visual Studio supports complex.h.

Upvotes: 6

Keith Thompson
Keith Thompson

Reputation: 263507

Your program has a couple of minor flaws, but none that cause it to misbehave.

Your PI macro should be parenthesized:

#define PI (2*acos(0.0))

but you happen to get away without the parentheses because of the way you use it.

The correct format for printing a double value is actually %f, but %lf is accepted as well. (%Lf is for long double. %f also works for float, because float arguments to variadic functions are promoted to double). This also doesn't affect your program's behavior.

In fact, your program is working correctly. I've confirmed using an HP 42S emulator that tanh(1/(sqrt(3.0))) is approximately 0.520737 (I get 0.520736883716).

The problem is your assumption that the result should be π/6.0. It isn't.

Upvotes: 3

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