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How does this poor man's log10 calculator trick work?
- lonelappde 7y agolog (x) is approximated by √x iterated n times x^(0.5^n), multiplied by a constant equal to (2^n + [fudge factor for √'s finite precision]) The best answer is a link to https://m.imgur.com/s9eBx5Z https://m.imgur.com/s9eBx5Z which solves the problem in math notation. There's another comment that describes that in words, and a bunch of reddit chitchat.
- SamReidHughes 7y agoThe best explanation was Vietoris’s: https://www.reddit.com/r/explainlikeimfive/comments/fyh9qr/eli5_how_does_the_poor_mans_log₁₀_calculator/fn207gm/ https://www.reddit.com/r/explainlikeimfive/comments/fyh9qr/e...
- jackhalford 7y agoThe same trick (approximating with a first order Taylor series), is also the basis of quake's famous fast inverse square root [1]. I love these methods because they always involve a "magic constant". [1] https://en.m.wikipedia.org/wiki/Fast_inverse_square_root https://en.m.wikipedia.org/wiki/Fast_inverse_square_root