How Do L and U Sums Differ in Double Summation?

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SUMMARY

The discussion focuses on the differences between L and U sums in double summation, specifically for the nxn array defined by bij = aiaj. The sums are defined as L = Ʃ(i=1 to n)Ʃ(j=1 to i) bij and U = Ʃ(j=1 to n)Ʃ(i=1 to j) bij. The key conclusion is that by symmetry, L equals U, demonstrating that both sums ultimately yield the same result despite their different formulations.

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  • Understanding of double summation notation
  • Familiarity with sequences and arrays
  • Basic knowledge of symmetry in mathematical expressions
  • Ability to manipulate algebraic expressions involving indices
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  • Practice with specific examples of double summation using small arrays
  • Explore the properties of symmetric sums in mathematics
  • Learn about the implications of L and U sums in combinatorial contexts
  • Investigate advanced topics in series convergence and divergence
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Homework Statement


Consider any sequence a1, a2,..., an and the nxn array of values bij = aiaj. Which terms in the array are involved in the sums L = Ʃ(between i=1 and n)Ʃ(between j=1 and i) bij
and U = Ʃ(between j=1 and n)Ʃ(between i=1 and j) bij?

Also, by symmetry, show that L=U.

Homework Equations



n/a

The Attempt at a Solution



I've just started this topic and the double summation is really confusing me. Any guidance or suggestions on how to approach this would be really useful thanks.
 
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dbatten said:

Homework Statement


Consider any sequence a1, a2,..., an and the nxn array of values bij = aiaj. Which terms in the array are involved in the sums L = Ʃ(between i=1 and n)Ʃ(between j=1 and i) bij
and U = Ʃ(between j=1 and n)Ʃ(between i=1 and j) bij?

Also, by symmetry, show that L=U.

Homework Equations



The Attempt at a Solution



I've just started this topic and the double summation is really confusing me. Any guidance or suggestions on how to approach this would be really useful thanks.

Just do a small example by hand. Use maybe a 5 by 5 array of bi,j .

a1a1 a1a2 a1a3 a1a4 a1a5

a2a1 a2a2 a2a3 a2a4 a2a5

a3a1 a3a2 a3a3 a3a4 a3a5

a4a1 a4a2 a4a3 a4a4 a4a5

a5a1 a5a2 a5a3 a5a4 a5a5


 

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