Write a program that implements the following functions iteratively (no recursio

Question

Write a program that implements the following functions iteratively (no recursion). double factorial (int n) double exponent (double x, int n) The functions implemented should follow below guidelines factorial: Computes n! = n times (n -1) x...x 1 exponent: Computes the sum of first n terms of e^x using the following approximation. F(x, n) = e^x = sigma^n = i = 0 x^i/i! = x^0/0! + x^1/2! +...+ x^n/n! You can use pow() and factorial 0 functions in your exponent function. In main () use argc and argu read the value of n and z from the user and compute and print the approximation of e^x for all values up to n using the function exponent. Print the results as a table as shown below. Also, print the exact value of e^x using the math library function exp(). When you increase the value of i your result should get closer to the result of exp. Name your program assign3.c Sample execution of the program is given below. First parameter is n and second parameter is z. You need to use functions atoi f() and atof() in stdlib.h to convert strings to integer and double respectively.

Explanation / Answer

// C code
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>

double factorial (int n)
{
   double fact = 1;  
   int i;
   for (i = 1; i <= n; ++i)
   {
       fact = fact*i;
   }

   return fact;
}

double exponent(double x, int n)
{
   double result = 0;
  
   int i;
   for (i = 0; i <= n ; ++i)
   {
       result = result + pow(x,i)/factorial(i);
   }

   return result;

}

int main(int argc, char const *argv[])
{
   if(argc < 2)
   {
       printf("Input arguments missing ");
       return 0;
   }

   int n;
   sscanf (argv[1],"%d",&n);

   double x;
   sscanf (argv[2],"%lf",&x);
      
   int i;
   double exponentValue;

   printf(" i Approximation ");
   printf("---------------------------------- ");
   for (i = 0; i <= n; ++i)
   {
       printf(" %d ", i);

       exponentValue = exponent(x,i);

       printf("%0.10lf ",exponentValue);
   }

   printf("Exact Value: %0.10lf ",exponent(x,n));
   return 0;
}

/*
output:

   i   Approximation
----------------------------------
   0   1.0000000000
   1   3.2000000000
   2   5.6200000000
   3   7.3946666667
   4   8.3707333333
   5   8.8002026667
   6   8.9576747556
   7   9.0071659835
   8   9.0207760712
   9   9.0241029815
   10   9.0248349018
Exact Value: 9.0248349018

*/

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