expm1() in C++
Last Updated :
02 Feb, 2023
The expm1(x) function returns ex – 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.
- It is mandatory to give both the arguments otherwise it will give error no matching function for call to ‘expm1()’.
- If we pass string as argument we will get error no matching function for call to ‘expm1(const char [n]).
- 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
#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
#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
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