〈 Berger | Dillon 〉

Let S(x) denote the number of trials until the sum exceeds x. In this case we want S(1). Suppose the first trial equals f < x, then: S(x) = 1 + integral of S(x - f) between 0 and x => dS(x)/dx = S(x) and we know that S(0) = 1 Hence, S(x) = e^x and S(1) = e

I feel like this would be super easy to understand with proper notation and if I wasn't drinking

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