This document contains C++ code that defines several functions to calculate approximations of mathematical expressions and exponential, trigonometric, and factorial functions using Taylor series up to a given order n. It then calls these functions in a main program to calculate and output specific approximations and errors compared to known values.

1. /////////////////////////////////////////////////////////////////////////////////////////
// Brian Goggins
// 2/17/2015
// COSC 281
/////////////////////////////////////////////////////////////////////////////////////////
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
double factorial(double i)
{
double sum = 1;
if (i > 0)
{
while (i >= 1)
{
sum *= i;
i--;
}
return sum;
}
else if (i == 0)
{
return sum = 1;
}
}
double problem1a(double x, double n)
{
double sum = 0;
for (int i = 1; i <= n; i++)
{
sum += pow(x, i);
}
return sum;
}
void problem1b(double x, double n)
{
double sum = 0, sum2=0;
for (int i = 0; i <= n; i++)
{
sum += pow(x, i);
}
sum2 = (1 - pow(x, n + 1)) / (1 - x);
}
double problem2a(double x, double n)
{
double sum = 0, fact = 0;
for (int i = 0; i <= n; i++)
{
fact = factorial(i);
sum += (pow(x, i)) / fact;
}
return sum;
}
double problem2b(double x, double n)

2. {
double sum = 0, fact = 0;
for (int i = 0; i <= n; i++)
{
fact = factorial(2*i);
sum += ((pow(-1, i)) / (fact))*(pow(x, 2 * i));
}
return sum;
}
double problem2c(double x, double n)
{
double sum = 0, fact = 0;
for (int i = 0; i <= n; i++)
{
fact = factorial((2 * i) + 1);
sum += ((pow(-1, i)) / (fact))*(pow(x, (2 * i) + 1));
}
return sum;
}
double problem3(double n)
{
double sum = 0, fact = 0;
for (int i = 0; i <= n; i++)
{
sum += pow(-1, i) / ((2*i)+1);
}
return sum;
}
int main()
{
int exit1 = 0;
double answer = 0, answer2 = 0, pi=4*atan(1);
while (exit1 != 1)
{
//problem 1
answer = problem1a(3, 3);
cout << "(A) " << answer << endl;
answer = problem1a(3, 7);
cout << "(B) " << answer << endl;
answer = problem1a(.2, 3);
cout << "(C) " << answer << endl;
answer = problem1a(.2, 7);
cout << "(D) " << answer << endl;
//problem 2
//a
answer = problem2a(0, 6); answer2 = exp(0) - answer;
cout << "n(A) N1=6 x=0 ttf(x)=" << answer << "ttt" << fabs(answer2) << endl;
answer = problem2a(1, 7); answer2 = exp(1) - answer;
cout << "(A) N1=7 x=1 ttf(x)=" << answer << "tt" << fabs(answer2) << endl;
//b

3. answer = problem2b(0, 6); answer2 = cos(0) - answer; cout <<
"n(B) N1=6 x=0 ttf(x)=" << answer << "ttt" << fabs(answer2) << endl;
answer = problem2b(pi/4, 3); answer2 = cos(pi / 4) - answer;
cout << "(B) N1=3 x=pi/4 tf(x)=" << answer << "tt" << fabs(answer2) << endl;
answer = problem2b(pi/2, 4); answer2 = cos(pi / 2) - answer;
cout << "(B) N1=4 x=pi/2 tf(x)=" << answer << "t" << fabs(answer2) << endl;
//c
answer = problem2c(0, 6); answer2 = sin(0) - answer;
cout << "n(C) N1=6 x=0 ttf(x)=" << answer << "ttt" << fabs(answer2) << endl;
answer = problem2c(pi/4, 2); answer2 = sin(pi / 4) - answer;
cout << "(C) N1=2 x=pi/4 tf(x)=" << answer << "tt" << fabs(answer2) << endl;
answer = problem2c(pi/2, 4); answer2 = sin(pi / 2) - answer;
cout << "(C) N1=4 x=pi/2 tf(x)=" << answer << "ttt" << fabs(answer2) << endl;
//problem 3
answer = 4*(problem3(1000000)); fabs(pi - answer);
cout << "n(A) N=1000000 tt" <<fabs(pi - answer) << endl;
cout << "(B) Pi = " << setprecision(20) << pi << endl;
cout << "close<1> "; cin >> exit1;
}
return 0;}
OUTPUT
(A) 39
(B) 3279
(C) 0.248
(D) 0.249997
(A) N1=6 x=0 f(x)=1 0
(A) N1=7 x=1 f(x)=2.71825 2.78602e-005
(B) N1=6 x=0 f(x)=1 0
(B) N1=3 x=pi/4 f(x)=0.707103 3.56636e-006
(B) N1=4 x=pi/2 f(x)=2.47373e-005 2.47373e-00
(C) N1=6 x=0 f(x)=0 0
(C) N1=2 x=pi/4 f(x)=0.707143 3.62646e-005
(C) N1=4 x=pi/2 f(x)=1 3.54258e-006
(A) N=1000000 9.99999e-007
(B) Pi = 3.1415926535897931
close<1>
Pi approximation 3.14159 26535 89793 23846 26433 83279 50288 41971