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In J, you can square a vector and get a vector of the squared entries. So "sum under square" really is doing "sqrt(sum(square(vector)))", and more generally "(a
by Rangi42 11y ago
In J, you can square a vector and get a vector of the squared entries. So "sum under square" really is doing "sqrt(sum(square(vector)))", and more generally "(a under b) x" does "b^-1(a(b(x)))".
- matchu 11y agoMm, I thought that might be what's up, but had no clue how J actually works xD Thanks! Even without the mapping wrinkle, though, this is definitely a deviation from the examples the article provides: instead of unsquaring the vector we initially squared, we unsquare the sum.