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In a problem I'm working on, I develop an expression: $$\sum_{i=1}^NB_i =K_4 \frac{\sum_{i=1}^NM_i}{K_2+\sum_{i=1}^NM_i}$$ What I really want is an expression for an individual $B_i$. Through various means, I can demonstrate that $$B_i =K_4 \frac{M_i}{K_2+\sum_{i=1}^NM_i}$$ Arriving at this result took a considerable effort because I was avoiding the "naive" idea to simply remove the summation signs on $B_i$ and $M_i$. However, that naive approach does produce the correct result in this instance.

My Question: Is there a way to know if the "naive" approach is valid when equating finite summations with the same number of objects?

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  • $\begingroup$ It's completely coincidental, you were right to expend considerable effort. $\endgroup$
    – Kwame Gonzales
    Jan 26, 2016 at 23:46
  • $\begingroup$ I'm never happy with coincidence as an explanation in mathematics. There must be some underlying reason that it's correct in this instance - whether or not this reason is exploitable is another thing entirely $\endgroup$
    – Bamboo
    Jan 27, 2016 at 0:21

1 Answer 1

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Set $$ Y=K_4\frac{\sum_{i=1}^NM_i}{K_2+\sum_{i=1}^NM_i} $$ Now you can choose $$ B_1=Y,\quad B_i=0, 1<i\le N $$ or $$ B_n=Y,\quad B_i=0, 1\le i<N $$ or, of course, infinitely many other ways to have the same sum on the left-hand side as on the right-hand side (unless $N=1$, of course).

Without other relations between $B_i$ and $M_i$ there's no way to remove the summation from both sides.

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  • $\begingroup$ This was my reasoning for avoiding the naive approach. I guess what I'm really asking is what kind of restrictions are implied by the naive approach... I'll update my question if you don't mind $\endgroup$
    – Bamboo
    Jan 27, 2016 at 0:46
  • $\begingroup$ Actually, now that I've thought about it a bit more, this has a trivial answer... $\endgroup$
    – Bamboo
    Jan 27, 2016 at 0:51

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