StevenXClontz’s avatarStevenXClontz’s Twitter Archive—№ 6,639

    1. …in reply to @sbagley
      @sbagley Also the step that seems to assume log(1+n/100) \approx n/100 is doing a lot of heavy lifting here.
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  1. …in reply to @StevenXClontz
    @sbagley (I guess it's because log(1+0/100)=0/100 and d/dx(log(1+x/100))=1/(100+x) and d/dx(x/100)=1/100.)
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