Sum of a series
Sum of a series
You are encouraged to solve this task according to the task description, using any language you may know.
You are encouraged to solve this task according to the task description, using any language you may know.
Display the sum of a finite series for a given range.
For this task, use S(x) = 1/x^2, from 1 to 1000. (This approximates the Riemann zeta function. The Basel problem solved this: zeta(2) = π2/6.)
C++
#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 / ( x * x ) ); }
D
module series ; import std.stdio ; T series(T)(T function(int) t, int end, int start = 1 /* 0 if zero base*/ ) { T sum = 0 ; for(int i = start ; i <= end ; i++) sum += t(i) ; return sum ; } real term(int n){ return 1.0L/(n*n) ; } void main(){ writef("sum@[1..1000] = ", series(&term, 1000)) ; }
Forth
: sum ( fn start count -- fsum ) 0e bounds do i s>d d>f dup execute f+ loop drop ; :noname ( x -- 1/x^2 ) fdup f* 1/f ; ( xt ) 1 1000 sum f. \ 1.64393456668156 pi pi f* 6e f/ f. \ 1.64493406684823
J
S=: % @ *: NB. inverse of square of input +/ S >: i. 1000 1.64393 (*:o.1)%6 NB. pi squared over six, for comparison 1.64493
Java
public class Sum{ public static double f(double x){ return 1/(x*x); } public static void main(String[] args){ double start = 1; double end = 1000; double sum = 0; for(double x = start;x <= end;x++) sum += f(x); System.out.println("Sum of f(x) from " + start + " to " + end +" is " + sum); } }