What is the code for ( e^x = 1 + x/1! + x^2/2! + x^3/3! + ..... )
anyway can help me, plzzz :cry: :rolleyes:
What is the code for ( e^x = 1 + x/1! + x^2/2! + x^3/3! + ..... )
anyway can help me, plzzz :cry: :rolleyes:
Use a 'for' loop. Show what you think works, and tell us exactly what each line does and why.
In doing this, you'll probably solve your problem, but if that fails, you'll have people here willing to help you.
Are you trying to just get the natural exponential of a number?
There's a function in math library, if I'm not wrong.
Try to see if the math.h of your compiler has a double exp (double) function.
See
http://unixhelp.ed.ac.uk/CGI/man-cgi?exp+3
For microsoft, the function prototype may be different:
thanks for replyiing me,
But i wanna write the code for (e^x = 1 + x/1! + x^2/2! + x^3/3! + ..... ) without using math library.
#include<iostream.H>
int main()
{
//(e^x = 1 + x/1! + x^2/2! + x^3/3! + .....)
int result = 1;
int x = 2;
int z = 1;
int e = 1;
for (int i = 1 ; i < 4 ; i++)
{
// This is the factorial part & its not working well, i need to fix tht, plz help (i dont wanna use any function)
while ( z != 0)
{
result *= z;
z--;
}
//cout << "Result = " << result << endl;
//cout << "Z = " << z << endl;
e = e + x^i/result;
z++;
}
cout << "The Sum = " << e << endl;
return 0;
}
I think you know how to take a factorial of a number, but your variable usage is not good.
while ( z != 0)
{
result *= z;
z--;
}
Now, honestly, that works if you first set z to the value of whatever you want to take the factorial of, and set result to one. You're setting z equal to one before the for loop, and then incrementing it at the end of each loop, but you decrement it to zero before that, so your while loop is *entirely pointless*.
Just multiply result by i, you don't need a while/for loop or whatever.
-Fredric
#include<iostream.H>
int main()
{
//(e^x = 1 + x/1! + x^2/2! + x^3/3! + .....)
int result = 1;
int x = 2;
int e = 1;
for (int i = 1 ; i < 4 ; i++)
{
result *= i;
//cout << "Result = " << result << endl;
e = e + x^i/result;
}
cout << "The Sum = " << e << endl;
return 0;
}
<< moderator edit: added [code][/code] tags >>
You mean like this??? :?:
I didn't see it the first time, but the ^ doesn't do what you think it does in C++, if you want to take m to the n power then you use the pow function like so...
#include <math.h>
...
double m=2.0, n=3.0, result;
result = pow(m, n);
That would return 2 to the 3rd, or 8. Hope that helps a bit.
-Fredric
Hi, I have some issues with your variable names. result is actually not the result, and e is actually the natural log constant in your comments. Made some changes to the var and trying to do away with the pow fn.
Not sure if you want to restrict your input to just integral values, but output will most definitely be double?
int main()
{
//(e^x = 1 + x/1! + x^2/2! + x^3/3! + .....)
double x = 2.0; // input value
int factorial = 1;
double pow_x = 1.0;
double result = 1.0;
cout << "e^" << x << " = 1";
for (int i = 1 ; i < 4 ; i++)
{
factorial *= i;
pow_x *= x;
cout << " + " << pow_x << "/" << factorial;
result = result + pow_x/factorial;
}
cout << endl << "= " << result << endl;
return 0;
}
I've not tried the code, but should be ok ...
We're a friendly, industry-focused community of 1.18 million developers, IT pros, digital marketers, and technology enthusiasts learning and sharing knowledge.