I'm quite a novice at this so any help would be greatly appreciated. You can view what I've written so far at the bottom of this post. Thanks!

Scan a double variable x and evaluate y=exp(x) by

using the math.h library of functions.

Your objective in this homework is to compare the so calculated

value of y with the approximate value Y obtained by

using 1, 2, 3, 4, and 5 leading terms of the Taylor series:

exp(x) = x^0/0! + x^1/1! + x^2/2! + x^3/3! + x^4/4! + ....

Recall that k!=k*(k-1)*(k-2)* ... *3*2*1 and 0!=1.

Use a do/while loop in conjunction with a switch statement

switch(). Enter a character 1 for one term approximation,

2 for two terms, etc., by using getchar() function. Use

0 to exit the program. Within switch, use default:

if (char != '0') => unrecognized operator.

Let your opening case calculate the fourth term x^4/4!, let the

subsequent case calculate the x^3/3! term, etc. (In this way,

you wan't need to use repeatedly a break statement within your switch).

Evaluate the corresponding Y, print the result in one line,

and the relative error (Y-y)/y in another line.

Execute your program for all five cases (with corresponding

printouts).

............................................

Your output should be like this:

Enter x: 1.25

True value of exp(1.2500) = 3.490343

Enter a character 1-5 (0 to exit):

1

1 term(s) approximation

Approximate exp(1.2500) = 1.000000

Relative error = -71.349520 percent

Enter a character 1-5 (0 to exit):

2

2 term(s) approximation

Approximate exp(1.2500) = 2.250000

Relative error = -35.536421 percent

Enter a character 1-5 (0 to exit):

3

3 term(s) approximation

Approximate exp(1.2500) = 3.031250

Relative error = -13.153233 percent

Enter a character 1-5 (0 to exit):

4

4 term(s) approximation

Approximate exp(1.2500) = 3.356771

Relative error = -3.826905 percent

Enter a character 1-5 (0 to exit):

5

5 term(s) approximation

Approximate exp(1.2500) = 3.458496

Relative error = -0.912428 percent

Enter a character 1-5 (0 to exit):

6

unrecognized operator

Enter a character 1-5 (0 to exit):

0

Code:

#include<stdio.h> #include<math.h> /* Scanning a double variable x and evaluating y=exp(x) by using the math.h library of functions */ int main() { /* Defining variables */ int k, N; char ch; double x, y, Y; /* Two empty lines and data input + true value of exp(x) at beginning */ printf("\n\nEnter x:\n\n"); scanf("%lf", &x); y=exp(x); printf("True value of exp(%lf) = %9.6f\n", x, y); /* Data input: do/while statement with a switch statement within */ do { printf("Enter x:\n"); scanf("%d", &N); y=0; switch(N) { case 5; y-= } } }