clangcmp · HackerTrans

5 years agodiscuss

They did. They called it Newton's series

5 years agodiscuss

Disappointing, did not see the extremely easy and quite efficient method of calculating e through continued fractions.

  S =[2,1,2] 
  T =0

  for P in range(200):
    S += [1,1,(4 + 2*P)]

  for x in range(len(S)-1):
    T = 1 / (T + S[-(x+1)])

  print(S[0] + T) # value of e

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