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February 25

Rank of an approximation sequence for a given accuracy

How can the rank of an approximation sequence of an irrational number (like e) be determined for a given accuracy with m decimal digits, for instance in the classic rendering of e as : $e=\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)^{n}$ ? What equation must be solved in order to get the rank involved in the approximation? (Thanks)--82.137.9.121 (talk) 01:08, 25 February 2017 (UTC)[reply]

e-(1+n^{-1})^{n}\approx en^{-1}\leq 10^{-m}

Bo Jacoby (talk) 10:27, 25 February 2017 (UTC).[reply]

I got

e-(1+n^{-1})^{n}\approx {\frac {e}{2n}}

, or more specifically

(1+{\frac {1}{n}})^{n}\approx e(1-{\frac {1}{2n}}+{\frac {11}{24n^{2}}}\dots )

. The OP said it was just an example though so I'm not sure that answers the question. Would a link to Numerical analysis help? --RDBury (talk) 15:11, 25 February 2017 (UTC)[reply]

I made a mistake. You are correct. Bo Jacoby (talk) 17:28, 25 February 2017 (UTC).[reply]

I think you mean Order of approximation rather than rank? Dmcq (talk) 15:45, 25 February 2017 (UTC)[reply]

There are good answers, indeed. The next step is to formulate a general case analysis from this specific examples from list_of_representations_of_e#As_the_limit_of_a_sequence for a specific irrational number, e in this case, to any irrational number approximated as a limit of sequence in terms of series acceleration and rate of convergence, that is to get a better approximation to m digits of an irrational at a lower rank of the approximation sequence or at a lower iteration of the approximation. Taking another representation for the same irrational e, for instance replacing n with n!(factorial) what gain in accuracy can be get in the same iteration for a better converging sequence?--82.137.8.83 (talk) 10:54, 26 February 2017 (UTC)[reply]

Also what function is more easily to calculate values and apply in the equation obtained, the exponential or the natural logarithm?--82.137.8.83 (talk) 11:02, 26 February 2017 (UTC)[reply]