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If i had a situation like, $$\sum^{N}_{i}B_{i}\times \frac{\sum_{i}^{N}|A_{i}|}{\sqrt{\sum^{N}_{i}(A_{i})^{2}}}=0 $$

I am struggling to see how it can be rewritten in a simpler form, I have tried expanding the the sqrt about zero but that gets very messy and confusing, are there any obvious things im missing so that it can be simplified, i no that the answer should be something like, $$\sum^{N}_{i}B_{i}|A_{i}|=0. $$

user3261646
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  • You should fix the subscripts in your sums, the expressions do not make much sense. – TerranDrop Feb 14 '14 at 14:03
  • how do you mean fix, as in put in i=1? – user3261646 Feb 14 '14 at 14:04
  • Yes. Either do just $\sum_i$ or $\sum_{i=1}^N$. $\sum_i^N$ doesn't make sense. As for the question itself, I think you need to share where this came from and how you got here. I'm going to guess you made an error earlier in your derivations. It simply isn't possible to go from your top summation to your bottom summation without more information. – Michael Grant Feb 14 '14 at 14:18
  • Assuming $A_i\neq 0$ for at least one $i$ (because if it didn't, you would have a division by zero), you necessarily have $\sum_i B_i=0$ here. – Michael Grant Feb 14 '14 at 14:22

1 Answers1

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I presume all your subscripts run from $0$. It's the same if they run from $1$ or any other number. If all values $A_i$ equal zero, the formula you wrote does not make sense. If at least one value is larger than $0$, then you can multiply the equation with $$\sqrt{\sum_{i=0}^nA_i^2}$$ and get $$\left(\sum_{i=0}^NB_i\right)\left(\sum_{i=0}^n|A_i|\right)=0$$ This really means that $$(B_1+B_2+\cdots+B_N)\cdot (|A_1|+|A_2|+\cdots +|A_N|)=0.$$

Multiplying the two sums wil yield not only $B_1\cdot |A_1|$, but also $B_1\cdot |A_i|$ for all other values of $i$, meaning the formula is

$$\sum_{i=0}^N\left(\sum_{j=0}^N B_i|A_j|\right)=0.$$

This is sometimes written as

$$\sum_{i,j=0}^NB_i|A_j|,$$ but you must realize, even in this notation, that this means ALL possible combinations of $i,j$ where both $i$ and $j$ are between $0$ and $N$.

5xum
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