int sum (int n)
{
if (n<=1) return n;
else return n*n + sum (n-1);
}
#include <iostream> using namespace std; int main() { int i,sum; // variables sum = 0; // initialize sum /* recursive addition of squares */ for (i = 1; i <= 30; i++) sum = sum + (i * i); cout << sum <<" is the sum of the first 30 squares." << endl; return 0; }
Yes, but a recursive function running for a long time would eventually cause your program to crash.
For some algorithms recursive functions are faster, and there are some problems that can only be solved through recursive means as iterative approaches are computationally infeasible.
Guess you meant: can a recursive function call predefined functions? Answer: sure, why not.
int Nodes (Tree *t) { int sum= 0; if (t) { sum+=1; if (t->left) sum += Nodes (t->left); if (t->right) sum += Nodes (t->right); } return sum; }
#include <iostream> using namespace std; int main() { int i,sum; // variables sum = 0; // initialize sum /* recursive addition of squares */ for (i = 1; i <= 30; i++) sum = sum + (i * i); cout << sum <<" is the sum of the first 30 squares." << endl; return 0; }
cos2x + sin2x = 1; cosh2x + sinh2x = 1.
The sum of their squares is 10.
If the regression sum of squares is the explained sum of squares. That is, the sum of squares generated by the regression line. Then you would want the regression sum of squares to be as big as possible since, then the regression line would explain the dispersion of the data well. Alternatively, use the R^2 ratio, which is the ratio of the explained sum of squares to the total sum of squares. (which ranges from 0 to 1) and hence a large number (0.9) would be preferred to (0.2).
Yes, but a recursive function running for a long time would eventually cause your program to crash.
There is a calculation error.
i love u darling
J. C. E. Dekker has written: 'Recursive equivalence types' -- subject(s): Recursive functions
For some algorithms recursive functions are faster, and there are some problems that can only be solved through recursive means as iterative approaches are computationally infeasible.
split 10 in two parts such that sum of their squares is 52. answer in full formula
8081 can be the sum of two perfect squares because its perfect squares are 41 x41+80x80=1681+6400. Answer=1681+6400
Sum of squares? Product?