I am running through some numbers and found a mistake. Turns out I had to do $\sum(b-a)$ instead of $\sum(a-b)$. But, I only have the end result (I do not have a source data anymore).

Can just flip the sign of the end result?

Or, is $\sum(b-a)$ equal to $-\sum(a-b)$ ?

  • 2
    $\begingroup$ Note that $$\sum\limits_i (a_i-b_i)=(a_1-b_1)+(a_2-b_2)+\cdots=-(b_1-a_1)-(b_2-a_2)-\cdots=-\sum\limits_i (b_i-a_i)$$ $\endgroup$ – TheSimpliFire Apr 9 at 9:43

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