The place value of the nth element of @e is 1/n!, so the array of 1's represents e itself. Then we just multiply by 10 in the positional notation and extract the integer part (actually I guess it's the other way around).

Since we calculate the value of the entire number each time it doesn't matter that the digits are not normalized (for example, 0,0,0,20,0 should really be 3,0,0,2,0).

No continued fractions are involved, just the usual series 1+1+1/2+1/6+1/24+... which comes from the power series for exp(x).

In reply to Re^2: Caclulation of e by jdalbec
in thread Caclulation of e by doc

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