expm1() in C++

The expm1(x) function returns e^x – 1 where x is an argument and e is mathematical constant with value equal to 2.71828. Syntax:

double expm1() (double x);
float expm1() (float x);
long double expm1() (long double x);

• The expm1() function takes a single argument and computes e^x -1.
• The expm1() function returns e^x -1 if we pass x in the argument.

1. It is mandatory to give both the arguments otherwise it will give error no matching function for call to ‘expm1()’.
2. If we pass string as argument we will get error no matching function for call to ‘expm1(const char [n]).
3. If we pass std::numeric_limits::max() we will get -2147483648.

Time Complexity: O(1)
Auxiliary Space: O(1)

Examples:

Input : expm1(5.35)
Output : 209.608

Input : expm1(-5)
Output : -0.993262

# CODE 1

CPP

// CPP implementation of the
// above function
#include <cmath>
#include <iostream>
using namespace std;

int main()
{
    double x = 5.35, answer;
    answer = expm1(x);

    cout << "e^" << x << " - 1 = "
        << answer << endl;

    return 0;
}

Output

e^5.35 - 1 = 209.608

# CODE 2

CPP

// CPP implementation of the
// above function
#include <cmath>
#include <iostream>
using namespace std;

int main()
{
    int x = -5;
    double answer;
    answer = expm1(x);

    cout << "e^" << x << " - 1 = "
        << answer << endl;

    return 0;
}

Output

e^-5 - 1 = -0.993262