BELOW IS A PROBLEM STATEMENT THAT I CANNOT FIGURE OUT. IT HAS TO BE IN C PROGRAM NOT C++. I ALSO TYPED THE PROFESSORS EXAMPLE BELOW FOR REFERENCE, BUT HOWEVER, WE ARE REQUIRED TO COME UP WITH OUR OWN CODE THAT HAS THE SAME OUTCOME.

Problem Statement: Develop a C Program for Exponential Value Determination using Numerical Approximation for e as given by the Taylor series expansion; This series approximation for e is given by the series written below:

e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! .... we stop at 6! division in this program.

Hence, compute the relative error and absolute error after computing the true value of e using the math function exp().

(PROFESSORS )C-Program example:

/* Exponential Value Approximation ('e' value in Math Computations) */

/* Approximation for 'e' is given by the series written below: */

/* e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! .... we stop at 6! division in this program */

/* Hence compute the relative and absolute error */

#include

#include

#include /* Need this include to use math library functions */

int main()

{

printf(" The computed approximation for exponential 'e': (using 7 terms of series)");

printf(" The series for e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6!");

double eapprox=2; /* intialize eapprox value as 2*/

double etruevalue = exp(1.0);

int n; /* term number in the approx series */

double sum = 0; /* assign sum variable to zero */

printf(" The eapprox value = %lf for n=1, first two terms",eapprox);

/* add each term in the series and add each term to eapprox value */

n=2; /* do the sum for n =2, third term 1/2! term */

sum = sum + 1.0/n;

eapprox = eapprox + sum;

printf(" The eapprox value = %lf for n=2, for three terms",eapprox);

/* do the sum for n =3, fourth term 1/3! term */

n++; /* this is the same as n = n+1, shortened as n++, now n = 3 */

sum = sum*(1.0/n);

eapprox = eapprox + sum;

printf(" The eapprox value = %lf for n=3, for four terms",eapprox);

/* do the sum for n =4, fifth term 1/4! term */

n++; /* this is the same as n = n+1, shortened as n++, now n = 4 */

sum = sum*(1.0/n);

eapprox = eapprox + sum;

printf(" The eapprox value = %lf for n=4, for five terms",eapprox);

/* do the sum for n =5, sixth term 1/5! term */

n++; /* this is the same as n = n+1, shortened as n++, n = 5 */

sum = sum*(1.0/n);

eapprox = eapprox + sum;

printf(" The eapprox value = %lf for n=5, for six terms",eapprox);

/* do the sum for n =6, seventh term 1/6! term */

n++; /* this is the same as n = n+1, shortened as n++, n = 6 */

sum = sum*(1.0/n);

eapprox = eapprox + sum;

printf(" The eapprox value = %lf for n=6, for seven terms",eapprox);

/* Print e true value */

printf(" The e true value (math function computed) is = %lf ",etruevalue);

/* Print rel error and abs error on computing e approx value*/

double relerror=0, abserror=0;

abserror = etruevalue - eapprox;

relerror = abserror/etruevalue*100;

printf(" The absolute error on computing approx e = %lf", abserror);

printf(" The relative error in percent on computing approx e = %lf ", relerror);

return 0;

}

Output of this Program:

The computed approximation for exponential 'e': (using 7 terms of series)

The series for e = 1 + 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6!

The eapprox value = 2.000000 for n=1, first two terms

The eapprox value = 2.500000 for n=2, for three terms

The eapprox value = 2.666667 for n=3, for four terms

The eapprox value = 2.708333 for n=4, for five terms

The eapprox value = 2.716667 for n=5, for six terms

The eapprox value = 2.718056 for n=6, for seven terms

The e true value (math function computed) is = 2.718282

The absolute error on computing approx e = 0.000226

The relative error in percent on computing approx e = 0.008324