Sujet : Re: hot to write out this summation:
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.mathDate : 01. Jul 2024, 17:40:12
Autres entêtes
Organisation : Nemoweb
Message-ID : <J1xLmgGlctHdxd-hpcHeDCAuzJY@jntp>
References : 1
User-Agent : Nemo/0.999a
Le 30/06/2024 à 23:38, "Chris M. Thomasson" a écrit :
r[0] = .01
r[1] = .0011
r[2] = .000111
r[3] = .00001111
r[4] = .0000011111
...
Now, think of a possible formula. Something like this shit:
;^) lol.
r[0] = .01
r[1] = .01 * 10^(-1) + .1 * 10^(-3) = .0011
r[2] = .0011 * 10^(-1) + .1 * 10^(-5) = .000111
r[3] = .000111 * 10^(-1) + .1 * 10^(-7) = .00001111
r[4] = .00001111 * 10^(-1) + .1 * 10^(-9) = .0000011111
r[5] = .0000011111 * 10^(-1) + .1 * 10^(-11) = .000000111111
...
Taken to infinity, it would be: .(0)(1) or something? How to properly write it:
.0...1...
?
In the sequence
01 0011 000111
00001111 ...
little by little every index is covered by zero. The limit is an infinite sequence of zeros 000..., if we start to enumerate from the left-hand side as is usual in European literature and in the notation of digits behind the decimal point. Starting from the right-hand side, as is usual in Arabic literature and in denoting the digits of integers, the limit is an infinite sequence of ones 111... . When indexing the digits alternatingly like
..., -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, ...
then the limit of the sequence has infinitely many zeros at the left-hand side and infinitely many ones at the right-hand side. What may happen when writing from top to bottom?
https://www.hs-augsburg.de/~mueckenh/Transfinity/Transfinity/pdf p. 225
Regards, WM