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wsowens | 5 years ago
pq^(k-1)log(q)
be (k-1)pq^(k-1)log(q)
Apologies if I'm off base here, I'm pretty rusty on infinite series.Great piece, this is really making this concept more intuitive for me.
wsowens | 5 years ago
pq^(k-1)log(q)
be (k-1)pq^(k-1)log(q)
Apologies if I'm off base here, I'm pretty rusty on infinite series.Great piece, this is really making this concept more intuitive for me.
ethanweinberger|5 years ago