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Summation proof. 
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#1
Sep1911, 11:06 PM

P: 292

1. The problem statement, all variables and given/known data
Show that the summation notation satisfies the following property: [itex]\displaystyle\sum\limits_{i=1}^n(\displaystyle\sum\limits_{j=1}^m aij) = \displaystyle\sum\limits_{j=1}^m(\displaystyle\sum\limits_{i=1}^n aij) [/itex] 2. Relevant equations N/A 3. The attempt at a solution [itex]\displaystyle\sum\limits_{i=1}^n(\displaystyle\sum\limits_{j=1}^m aij) = \displaystyle\sum\limits_{i=1}^n ai_{1} + \displaystyle\sum\limits_{i=1}^n ai_{2} + ... +\displaystyle\sum\limits_{i=1}^n ai_{n} = \displaystyle\sum\limits_{j=1}^m(\displaystyle\sum\limits_{i=1}^n aij) [/itex] Have I proven this sufficiently or have I skipped a step? If I skipped a step, which one was it? Thanks in advance. 


#2
Sep1911, 11:25 PM

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Sci Advisor
PF Gold
P: 9,249

1. The problem statement, all variables and given/known data
Show that the summation notation satisfies the following property: [tex]\sum_{i=1}^n\bigg(\sum_{j=1}^m a_{ij}\bigg) = \sum_{j=1}^m\bigg(\sum_{i=1}^n a_{ij}\bigg) [/tex] 2. Relevant equations N/A 3. The attempt at a solution [tex]\sum_{i=1}^n\bigg(\sum_{j=1}^m a_{ij}\bigg) = \sum_{i=1}^n a_{i1} + \sum\limits_{i=1}^n a_{i2} + \cdots +\sum_{i=1}^n a_{im} = \sum_{j=1}^m\bigg(\sum_{i=1}^n a_{ij}\bigg) [/tex] I would at least have written out the step [tex]\sum_{i=1}^n(a_{i1}+\cdots+a_{im})=\sum_{i=1}^n a_{i1} + \sum\limits_{i=1}^n a_{i2} + \cdots +\sum_{i=1}^n a_{im}.[/tex] If you want to do these things rigorously, you need to avoid the ... notation and use induction. If you use tex tags instead of itex, you don't need to type "displaystyle" all the time. (Use tex tags only when you want the math to appear on a separate line). Hit the quote button to see how I prefer to type the math above. 


#3
Sep2011, 06:13 PM

P: 292




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