CODEWITHSUNDEEP
c++math
Al13y
Al13y

Reputation: 1

C++ program to calculate e^x

I know there are many examples of this problem but I tried to write a different one myself. This is using the Taylor series e^x = 1 + x/1! + x^2/2! + x^3/3! + ...... My code compiles and runs but it wont output the correct answer for some imputes and I'm not sure why. is this even usable code or should i scrap it?

    #include <iostream>
    #include <cmath>

    using namespace std;
    int main()
   {
        double final,power,end_n = 1.0,e=1.0,x=2.0, n;
        cout<< "n: ";
       // usually enter 5 for test
        cin>> n;
        while (n>1){
            power = pow(x,n);
            end_n = end_n*n;
            e= (power/end_n)+e;
            n--;
        }
        final =e+x;
        cout<< final;
      return 0;
    }

Upvotes: 0

Views: 7213

Answers (4)

Lutz Lehmann
Lutz Lehmann

Reputation: 26040

You could actually employ a Horner-like scheme that uses the counting down in an essential manner

1 + x/1! + x^2/2! + x^3/3! + … + x^n/n! = (((((x/n+1)*x/(n-1)+1)*x/(n-2)+…)*x/1+1 

    e = 1.0;
    while (n>0){
        e = e*x/n + 1;
        n--;
    }

Compare the approximations of e^x and 1/(e^-x) for positive x for exactness.

Explore (e^(x/4))^4 for better exactness.

Upvotes: 0

Olan
Olan

Reputation: 72

I think you are close. Maybe you want something more like this:

#include <iostream>
#include <cmath>
using namespace std;

double factorial(long n)
{
    double result = n;
    while(--n) result*=n;
}

int main()
{
    long n, power;
    double final, e=1.0, x=2.0;
    cout<< "n: ";
    // usually enter 5 for test
    cin>> n;
    while (n>1)
    {
        power = pow((double)x, (double)n);
        end_n = factorial(n);
        e = (power/end_n)+e;
        n--;
    }
    final = e+x;
    cout<< final;
    return 0;
}

Upvotes: 0

hobbs
hobbs

Reputation: 240531

To be very clear about something that others are hinting at: if your loop counted up from 1 to n then end_n would equal n! at each step. But counting down, it doesn't. Look at the examples from 1 to 5:

Forwards

n | n!
1 | 1
2 | 2
3 | 6
4 | 24
5 | 120

Backwards

n | end_n
5 | 5
4 | 20
3 | 60
2 | 120
1 | 120

Since absolutely none of your denominators are right, it's a surprise if your code is only wrong for some inputs — in fact it's probably only correct for x=0.

Finally, I hope that this is just an exercise for learning. If you really need the value of e^x you should use exp(x).

Upvotes: 1

Blindy
Blindy

Reputation: 67447

I honestly have no idea what your reasoning is, at all. The code for that particular expansion is trivially simple:

double x;
cin >> x;

double oldres, res=1, top=1, bottom=1;
int iter=1;

do {
  oldres=res;                           // to calculate the difference between iterations

  ++iter;                               // next iteration
  top*=x;                               // multiply by one x for each iteration
  bottom*=(iter-1);                     // multiply by the iteration number for each iteration
  res+=top/bottom;                      // and add the fraction to the result
} while(fabs(res-oldres)>.1);           // while the difference is still large

cout << res;                            // done, show the result

Upvotes: 1

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