Question

Write a program that approximates the sum of this geometric series using a user- specified number of terms. For example, if you named your program approx, calling approx(3) should use the first 3 terms of the series to calculate the approximation:

Approximate result = 1/2 + 1/4 + 1/8 = 0.875...
Note: the values are all powers of 1⁄2: (1/2)1, (1/2)2, (1/2)3, ...

Next, have your program calculate the difference (rounded to 3 decimal places) between the value “1” and the approximation calculated by your program:

Difference = 1 – approximate result = 0.125

HINT: You will probably want to define some variables before your for loop, such as a variable to keep track of the value of the current term and a variable to keep track of the sign of the current term. For example, initially, you can have approx = 0. As the program iterates through the loop, approx can be modified using approx = approx + current.

(NOTE: Make sure your program is not doing integer division (see p. 59). If you are using a version of Python earlier than Python 3, the division operator “/” may not give you the result you are expecting.)

You may have noticed that the approximation, at least using just 3 terms, is not very accurate. Modify your program from part a to look at all how the approximation improves as more terms are used. Your modified program will NOT do the comparison to 1 – it will just print out the value of the current approximation. For example, if you name your modified program approx2, calling approx2(3) will print out the following:

0.5 0.75 0.875

Now have your modified program round responses to 2 decimal places. How many terms are needed for the approximation to return a result of 0.99 or greater?

Answer 1

#include <iostream>
 
double f(double x);
 
int main()
{
    unsigned int start = 1;
    unsigned int end = 1000;
    double sum = 0;
 
    for( unsigned int x = start; x <= end; ++x )
    {
        sum += f(x);
    }
    std::cout << "Sum of f(x) from " << start << " to " << end << " is " << sum << std::endl;
    return 0;
}
 
 
double f(double x)
{
    return ( 1.0 / ( x * x ) );
}