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Is there a general function or way in Java to calculate an equation of the form (a+b+…+n)^2 with a,b,n >= 0? [closed]

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-3

I need to find a way to implement an algorithm in Java that can calculate (a+b+...+n)^2 with a,b,n >= 0. The purpose is to use it afterwards in order to calculate Jain's Fairness index for my algorithm in networks. Is there any standard way to do that or any specific library for advanced math that i might have missed?

java math equation

share|improve this question

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

closed as too broad by EJoshuaS, Max Vollmer, meowgoesthedog, thewaywewere, gdlmx Mar 28 at 5:49

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  what do you mean by ... ?
  
  – battlmonstr
  Mar 27 at 21:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You mean like sum = a + b + … + n; return sum * sum?
  
  – Nico Schertler
  Mar 27 at 21:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Your question is unclear. Is the sequence (a, b, ..., n) an arithmetic progression (e.g. 5, 10, 15, 20). If so, it's important that you say so.
  
  – Leo Aso
  Mar 27 at 21:15
  
  

  
  
  
  

add a comment | 

-3

I need to find a way to implement an algorithm in Java that can calculate (a+b+...+n)^2 with a,b,n >= 0. The purpose is to use it afterwards in order to calculate Jain's Fairness index for my algorithm in networks. Is there any standard way to do that or any specific library for advanced math that i might have missed?

java math equation

share|improve this question

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

closed as too broad by EJoshuaS, Max Vollmer, meowgoesthedog, thewaywewere, gdlmx Mar 28 at 5:49

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  what do you mean by ... ?
  
  – battlmonstr
  Mar 27 at 21:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  You mean like sum = a + b + … + n; return sum * sum?
  
  – Nico Schertler
  Mar 27 at 21:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Your question is unclear. Is the sequence (a, b, ..., n) an arithmetic progression (e.g. 5, 10, 15, 20). If so, it's important that you say so.
  
  – Leo Aso
  Mar 27 at 21:15
  
  

  
  
  
  

add a comment | 

I need to find a way to implement an algorithm in Java that can calculate (a+b+...+n)^2 with a,b,n >= 0. The purpose is to use it afterwards in order to calculate Jain's Fairness index for my algorithm in networks. Is there any standard way to do that or any specific library for advanced math that i might have missed?

java math equation

share|improve this question

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

share|improve this question

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

share|improve this question

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

asked Mar 27 at 20:42

Σωκράτης ΚαλαμόπουλοςΣωκράτης Καλαμόπουλος

211 bronze badge

2 Answers
2

active

oldest

votes

1

Just sum those n values in for loop and then multiply it by itself. Or am i missing something?

share|improve this answer

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

add a comment | 

0

The number of possible problems you may encounter is infinite, so you should not be surprised if you often get into a situation where there is no method to help you. Let's suppose that you have an array of numbers, let's suppose it has double elements and the name of the array is input, then:

double sum = 0;
for (int index = 0; index < input.length; index++)
 sum += input[index];
double result = Math.pow(sum, 2);
//output holds the result

If there is a possibility of overflow, then you will need to handle it. Also, you will need to handle the validation that your items are positive.

share|improve this answer

edited Mar 27 at 22:36

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @meowgoesthedog Oups, I thought the op wants the latter, sorry. Please see my edit.
  
  – Lajos Arpad
  Mar 27 at 22:37
  

  
  
  
  

add a comment | 

1

Just sum those n values in for loop and then multiply it by itself. Or am i missing something?

share|improve this answer

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

add a comment | 

1

Just sum those n values in for loop and then multiply it by itself. Or am i missing something?

share|improve this answer

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

add a comment | 

Just sum those n values in for loop and then multiply it by itself. Or am i missing something?

share|improve this answer

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

share|improve this answer

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

answered Mar 27 at 21:15

tomietomie

10188 bronze badges

0

The number of possible problems you may encounter is infinite, so you should not be surprised if you often get into a situation where there is no method to help you. Let's suppose that you have an array of numbers, let's suppose it has double elements and the name of the array is input, then:

double sum = 0;
for (int index = 0; index < input.length; index++)
 sum += input[index];
double result = Math.pow(sum, 2);
//output holds the result

If there is a possibility of overflow, then you will need to handle it. Also, you will need to handle the validation that your items are positive.

share|improve this answer

edited Mar 27 at 22:36

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @meowgoesthedog Oups, I thought the op wants the latter, sorry. Please see my edit.
  
  – Lajos Arpad
  Mar 27 at 22:37
  

  
  
  
  

add a comment | 

0

The number of possible problems you may encounter is infinite, so you should not be surprised if you often get into a situation where there is no method to help you. Let's suppose that you have an array of numbers, let's suppose it has double elements and the name of the array is input, then:

double sum = 0;
for (int index = 0; index < input.length; index++)
 sum += input[index];
double result = Math.pow(sum, 2);
//output holds the result

If there is a possibility of overflow, then you will need to handle it. Also, you will need to handle the validation that your items are positive.

share|improve this answer

edited Mar 27 at 22:36

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @meowgoesthedog Oups, I thought the op wants the latter, sorry. Please see my edit.
  
  – Lajos Arpad
  Mar 27 at 22:37
  

  
  
  
  

add a comment | 

The number of possible problems you may encounter is infinite, so you should not be surprised if you often get into a situation where there is no method to help you. Let's suppose that you have an array of numbers, let's suppose it has double elements and the name of the array is input, then:

double sum = 0;
for (int index = 0; index < input.length; index++)
 sum += input[index];
double result = Math.pow(sum, 2);
//output holds the result

If there is a possibility of overflow, then you will need to handle it. Also, you will need to handle the validation that your items are positive.

share|improve this answer

edited Mar 27 at 22:36

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges

double sum = 0;
for (int index = 0; index < input.length; index++)
 sum += input[index];
double result = Math.pow(sum, 2);
//output holds the result

share|improve this answer

edited Mar 27 at 22:36

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges

answered Mar 27 at 22:15

Lajos ArpadLajos Arpad

30.6k1919 gold badges6565 silver badges123123 bronze badges