Sum of a series: Difference between revisions
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</pre> |
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=={{header|Java}}== |
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public class Sum{ |
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public static double f(double x){ |
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return 1/(x*x); |
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} |
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public static void main(String[] args){ |
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double start = 1; |
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double end = 1000; |
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double sum = 0; |
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for(double x = start;x <= end;x++) sum += f(x); |
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System.out.println("Sum of f(x) from " + start + " to " + end +" is " + sum); |
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} |
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} |
Revision as of 04:49, 22 February 2008
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
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.
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 ) ); }
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); } }