Task
Block scheme of the solution:
С++ implementation:
#include<iostream>
#include<math.h>
#include<iomanip>
using namespace std;
float error;
float factorial(int n)
{
float factorial = 1;
for (n; n>0; n--)
factorial *= n;
return factorial;
}
float sum = 0, item;
void main()
{
cout << "(-1^k*x^(2k-1))/(!(k+x)*!k)" << endl;
cout << "Input the error:"; cin >> error;
for (int x = 1; x <= 5; x++)
{
for (int k = 0; k <= 18; k++)//k - float overflowing border
{
item = (pow(-1, k)*pow(x, 2 * k - 1)) / (factorial(k + x)*factorial(k));//calculate item
if (error <= fabs(item))
{
sum += item;//summing items
cout << setw(5) << x << setw(5) << k << setw(25) << item << setw(25) << sum << endl;//intervals & output
}
}
cout << "______________________________________________________________overflow" << endl;
}
cout << endl << "sum=" << sum << endl;//output sum
system("pause");//pause console
}
Screenshot of the working program:
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