OOM Summation

2020-11-28

How do you sum a set of order-of-magnitude values? For example, if my personal library contains E0 dieting books, E0 classical literature volumes, E1 textbooks, and E2 fantasy books, then how many books do I have total? Clearly the fantasy books dominate and I have E2 total books. So to get the sum total, I take the maximum E-value of the set?

Sometimes, but this is not true when there are many of the lower terms. For example let's look at the world population by country. To get a picture of it, we take the values from wikipedia and group them by their order-of-magnitude:

OOM Count
E1 0
E2 1
E3 6
E4 13
E5 41
E6 44
E7 92
E8 41
E9 3

To get the contribution of each order-of-magnitude, we multiply with their count (i.e. add with their count's E-value):

OOM Count Contribution
E1 0 0
E2 1 E2
E3 6 E4
E4 13 E5
E5 41 E7
E6 44 E8
E7 92 E9
E8 41 E10
E9 3 E9

So now we take the maximum of the contributions and we get E10 from the E8 countries (take that, China, India, and U.S.A!), which is in fact ~ the world population at 7.8E9.

I wouldn't expect to use this technique all that much since it's a bit laborious for what I want, but it's nice to keep in mind and serves as a conceptual sanity check.

Follow-up: What are some pernicious examples? What is the process of binning by E-value? This probably has a common name.

NOTE: v ~ \(E(round(log_{10}(v)))\)