Exponential Taylor Series

Question

This is the code I have so far, which is a little messy since I am still trying to figure out how to set it up, but I cannot figure out how to get the output. This code is supposed to take a Taylor Series polynomial of an exponential, and check the amount of iterations it takes to get the approximation.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
/*Prototype for functions used*/
double factorial (int);

int main()
{
   double input = 0;
   double exp_val;
   double delta = 1;
   int f =0;
   int n = 0;
   double taylor;
   int total;
   printf("Plese enter the exponent to check for convergence:\n");
   scanf("%lf", &input);
   exp_val = exp(input);
   printf("  #     Iter      e^X      Sum     Diff\n");
   printf("----   ------   -------   -----  --------");

   while(delta > 0.00001)
   {
      f = factorial(n);
      taylor = ((pow(input,n))/ f);
      delta = (exp_val - taylor);
      printf("%d %f %f %f/n", (n+1), exp_val, taylor, delta);
      n++;
   }
   system("pause");


}

double factorial (int n)
{
  int r = 0;
  int sum = 1;
  int total = 0;
  if (n == 0)
    return total =1;
  else
  {
     for(r; r<n; r++)
     {
        sum = sum * r;
        total = sum + 1;

     }

     return total;
  }

}

Answer 1

Here, I have fixed it, without changing your approach, except for the parts I really had to. One thing we have to clarify before the code is how Taylor Polynomials are made. It is not the first term plus the nth term, rather the sum of all terms from the first term till the nth term. So you definitely have to increase the taylor variable by the current nth term instead of the other way.

Here's the code, with brief comments in it as the explanation:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

/*Prototype for functions used*/
unsigned long long factorial( int );    // <-- made it return unsigned long long

int main( )
{
    double input = 0;
    double exp_val;
    double delta = 1;
    unsigned long long f = 0;   // <-- changed its type
    int n = 0;
    double taylor = 0;  // <-- initialized with 0
    printf( "Plese enter the exponent to check for convergence:\n" );
    scanf( "%lf", &input );
    exp_val = exp( input );
    printf( " #          e^X            Sum           Diff\n" );        // <-- made some cosmetic changes
    printf( "---      ---------      ---------      ---------\n" );     // <-- added \n

    while ( delta > 0.00001 )
    {
        f = factorial( n );
        taylor += ( ( pow( input, n ) ) / f );  // += instead of =
        delta = ( exp_val - taylor );
        printf( "%2d    %12f   %12f   %12f\n", ( n + 1 ), exp_val, taylor, delta ); // <-- replaced / with \ before the n
        n++;                                                                        // and made some edits to make it look better
    }
    system( "pause" );
    return 0;           // <-- better add this
}

unsigned long long factorial( int n )   // <-- made it return unsigned long long
{
    int r = 0;
    unsigned long long sum = 1; // <-- changed its type
    if ( n == 0 )
        return sum; // <-- this
    else
    {
        for ( r; r<n; r++ )
        {
            sum *= r + 1;   // <-- changed this
        }

        return sum; // <-- and this
    }
}

You have to keep in mind that you may not input too high values to it. Anything higher than input == 4 kind of breaks it, because, you see, even with 4, it can reduce the error delta beneath the threshold first only with the 19th cycle. The programme seemingly fails with n == 5 due to inaccurate calculation of pow( 5, 21 ) / factorial( 21 ) when n reaches 21:

0.000034    // the result this programme finds
0.0000093331055943447405008542892329719 // the result Calculator finds

So, yeah... If you want this programme to work with bigger input values, you'll need a better approach. Not calculating the nth term from scratch and calculating it from the (n - 1)th term instead could help until somewhat bigger input values, as the others had said.

Answer 2

A couple issue:

1. Change int r = 0; ... for(r; r<n; r++) to int r; ... for(r=1; r<=n; r++) or int r = 1; ... for(; r<=n; r++)

2. Change printf("%d %f %f %f/n" to printf("%d %f %f %f\n" Add \n

3. Change "... --------" to "... --------\n"

4. Change delta = (exp_val - taylor); to delta = fabs(exp_val - taylor);

5. Change to double taylor = 0.0; Initialize it.

6. Change to taylor += ((pow(input,n))/ f); Note: +=

7. Minor: "Please" not "Plese".

8. Minor: Drop int total;