If the attacker managed to obtain a nonce that gives the double‐ spending transaction a weight of at least 3n0 during the time interval of length t0, then the attack succeeds. The probability of this event is

formula from the whitepaper

My question: How does the above probability expression end with only this term (t0μ)/w1 without the exponential term?

1 Answer 1


It's Taylor series of the exp function to the 1st order:


exp(-x) = 1 + (-x)^1/1! + (-x)^2/2! + ....

Since white paper assumes x to be small, you forget about higher order terms and only take 1st order. What remains is an approximation for exp(-x), since x^2 is really tiny, once x is already small: (so is x^3 and so on)

Resulting in:

exp(-x) = 1 - x which is same as 1 - exp(-x) = x

  • Thanks! Your answer is correct. Yes, the Maclaurin series ((-x)^2)/2! will be even tinier. To verify 1 - exp(-x) = x, I tried with actual number and confirmed it is true when x is small and false when x is big.
    – Casey Yeow
    Apr 7, 2018 at 0:54

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