Recursive formula for e

Question

Looking for a finite recursive formula for the constant e preferably using standard operators (ones a calculator could carry out)

i.e. a formula of the form $ x_{n+1}=f(x_n)$ where $\lim_{n\rightarrow\infty}(x_n)=e$ for some $x_1$

Bonus points if it works for any $x_1$

lim as n approaches infinity of $(1+1/n)^n)$ is equal to $e$. — Юрій Ярош, Apr 24 '18 at 18:51
@ЮрійЯрош That's not recursive though, also most calculators can't do limits. — andrew, Apr 24 '18 at 18:53
@andrew But you asked for the calculator to calculate $x_n$ and to have $$e=\lim x_n$$ A calculator can definitely calculate $\left(1+ \frac{1}{n} \right)^n$. — N. S., Apr 24 '18 at 18:55
@andrew By recursive formula you mean formula for finding better and better approximations of $e$ ? Because, I know recursive formula as the way of defining sequence but not defining a number. — Юрій Ярош, Apr 24 '18 at 18:56
https://math.stackexchange.com/questions/1171572/a-recursive-formula-to-approximate-e-prove-or-disprove — , Apr 24 '18 at 18:57
@ЮрійЯрош Yes, is a formula that finds better and better approximations of e not a recursive function? — andrew, Apr 24 '18 at 19:02
@andrew You mean definition of e ? Definition of e is a limit of a sequence. For approximating $e$ you can take limit definition of it and substitute big value of $n$. — Юрій Ярош, Apr 24 '18 at 19:04
In addition to my answer, I'd also like to point out that this question has been looked at before. — Brian Tung, Apr 24 '18 at 19:13
You could be interested to my question https://math.stackexchange.com/q/2622195 — Fabio Lucchini, Apr 24 '18 at 19:16
@Brain this was linked in a comment above and looks quite interesting but I'd really like a sequence which only has the $x_n$ and $x_{n+1}$ terms. — andrew, Apr 24 '18 at 19:16
@Fabio This first relation given in the second question you linked looks interesting but didn't seem to work for the couple numbers I tried for $a_1$, have I made some kind of mistake or is there a set range of values $a_1$ can take? — andrew, Apr 24 '18 at 19:26
@andrew: You are right there is a mistake in the formula; I edited it now. — Fabio Lucchini, Apr 24 '18 at 19:30

Brian Tung · Accepted Answer · 2018-04-24T19:15:22.160

One approach is to use Newton's method on a function known to have a zero at $e$—for instance, $f(x) = \ln x - 1$. Then $f'(x) = 1/x$, and we can iterate as

$$ x_{n+1} = x_n (2-\ln x_n) $$

Of course, this isn't very satisfying, since the natural log is sitting right there in the expression. (I also haven't looked at the interval of convergence.) But it is available on most scientific calculators.

I'll have to give this more thought to see if something clever can be done to avoid the natural log (or exp, for that matter).

Also, as I mentioned in the comments, this question has been looked at before on Math.SE, with the nice result

$$ x_{n+1} = x_n - \frac1n x_{n-1} $$

with $x_1 = x_2 = 1$. This is not exactly in the form desired by the OP, but it's pretty nice all the same.

Newton's method is the reason I started thinking about this but I couldn't think of any functions with roots at e. Thanks for the help! — andrew, Apr 24 '18 at 19:13

score 0 · Answer 2 · answered Apr 24 '18 at 18:56

0

$$x_{n+1}= \left( 1+ \frac{\max\{ |x_n|,1 \} }{n}\right)^{\frac{\max\{ |x_n|,1 \}}{n}}$$

answered Apr 24 '18 at 18:56

N. S.

132,525

Recursive formula for e

2 Answers2