26 авг. 2010 г. · The continued fraction for e starts off 2+1/(a+1/(b+⋯)). So 1/(e−2)=a+1/(b+⋯). If you understand one, you understand the other.

12 июл. 2012 г. · The convergents that you normally see listed are those for the standard continued fraction expansion of e, ie, the one with 1 for each numerator.

5 окт. 2016 г. · How do we prove that the simple continued fraction for e2/n=[1;n−12,6n,5n−12,1,1,...]?

17 апр. 2016 г. · I am trying to convert some continued fractions to series by using Euler's continued fraction formula (see the link to Wikipedia).

7 июл. 2017 г. · Note we have Euler's continued fraction formula, which states: a0+a0a1+a0a1a2+⋯+a0a1a2⋯an=a01−a11+a1−a21+a2−⋱⋱an−11+an−1−an1+an.

22 авг. 2022 г. · By truncating e's continued fraction at different points, you obtain the approximation sequence [2,3,8/3,11/4,19/7,87/32,106/39,193/71,1264/465, ...

11 дек. 2015 г. · My metric for "computational efficiency" is achieving the most precise decimal places with a fixed number of terms generated in the continued fraction.

26 сент. 2018 г. · The euler number is a irrational number. And you can have a infinite continued fraction of euler number. But how can you form the coninued ...

24 апр. 2020 г. · Expressing continued fractions through e · 1. i. is almost surely correct. · 1. I established the recursion for i., but I could only numerical ...

19 дек. 2016 г. · Abstract: This note presents an especially short and direct variant of Hermite's proof of the simple continued fraction expansion e=[2,1,2,1,1,4 ...