# Scott Garrabrant comments on Concise Open Problem in Logical Uncertainty

• You are correct that you use the fact that 1+eps is at approximately e^(eps).

The concrete way this is used in this proof is replacing the ln(1+3eps) you subtract from bad1 when the environment is a 1 with 3eps=(bound—bad1) /​ (next—this), and replacing the ln(1-3eps/​2) you subtract from bad1 when the environment is a 0 with −3eps/​2=-(bound—bad1) /​ (next—this)/​2

I believe the inequality is the wrong direction to just use e^(eps) as a bound for 1+eps, but when next-this gets big, the approximation gets close enough.

• In case anyone shared my confusion:

The while loop where we ensure that eps is small enough so that

bound > bad1() + (next—this) * log((1 - p1) /​ (1 - p1 - eps))

is technically necessary to ensure that bad1() doesn’t surpass bound, but it is immaterial in the limit. Solving

bound = bad1() + (next—this) * log((1 - p1) /​ (1 - p1 - eps))

gives

eps >= (1/​3) (1 - e^{ -[bound—bad1()] /​ [next—this]] })

which, using the log(1+x) = x approximation, is about