Groups > sci.math > #603111 > unrolled thread

.999 repeating is almost an infinitely large sequence

Started by	"mitchr...@gmail.com" <mitchrae3323@gmail.com>
First post	2023-06-23 19:37 -0700
Last post	2023-06-26 10:14 -0700
Articles	20 on this page of 151 — 36 participants

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  .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-23 19:37 -0700
    Re: .999 repeating is almost an infinitely large sequence Brennan Otten <nnna@nebnrnea.ea> - 2023-06-24 19:20 +0000
      Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-24 14:59 -0700
        Re: .999 repeating is almost an infinitely large sequence Hoang Kouman <gago@hhngnaoa.om> - 2023-06-24 22:22 +0000
          Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 15:28 -0700
            Re: .999 repeating is almost an infinitely large sequence Bennett Sniders <rtne@etnnneis.bn> - 2023-06-24 23:28 +0000
              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 16:53 -0700
                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 16:58 -0700
                Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 09:30 -0700
                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 12:35 -0700
                    Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 12:59 -0700
                      Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:07 -0700
                        Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:41 -0400
                      Re: .999 repeating is almost an infinitely large sequence Deandre Romijnsen <dsrm@nnrderee.ds> - 2023-06-26 21:36 +0000
                    Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 16:31 -0400
                      Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:34 -0700
                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:35 -0700
              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 16:57 -0700
                Re: .999 repeating is almost an infinitely large sequence Bernard Bueren <buur@enbeebdn.ru> - 2023-06-25 08:00 +0000
                  Re: .999 repeating is almost an infinitely large sequence Vlademire Ramaker <arde@rekeadem.ae> - 2023-06-25 19:18 +0000
                    Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-25 12:29 -0700
                    Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-25 17:54 -0400
                      Re: .999 repeating is almost an infinitely large sequence Nikki Cann <nika@kinianka.ka> - 2023-06-25 22:06 +0000
                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 15:15 -0700
                        Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-25 18:30 -0400
                    Re: .999 repeating is almost an infinitely large sequence Shan Kloeten <ntne@ntnnstke.sn> - 2023-06-26 06:45 +0000
                    Re: .999 repeating is almost an infinitely Barton Houtem <otub@ahutnaee.eb> - 2023-06-26 06:58 +0000
                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 12:47 -0700
                    Re: .999 repeating is almost an infinitely large sequence Buddy Oorschot <ocuo@codhudoh.to> - 2023-06-25 20:22 +0000
                      Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 14:20 -0700
                        Re: .999 repeating is almost an infinitely large sequence Geosvani Peter <ogtv@asvsegtg.vp> - 2023-06-25 21:30 +0000
                          Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 14:51 -0700
                            Re: .999 repeating is almost an infinitely large sequence Dustan Rademaker <mdrn@eurttrrn.de> - 2023-06-25 22:16 +0000
                              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 15:18 -0700
                                Re: .999 repeating is almost an infinitely large sequence Lemuel Marqueringh <argr@hqqgugqi.lr> - 2023-06-25 22:57 +0000
                                Re: .999 repeating is almost an infinitely large sequence Volney <volney@invalid.invalid> - 2023-06-26 00:19 -0400
                                  Re: .999 repeating is almost an infinitely large sequence Markeen Schenck <ecme@kckhhkkc.kn> - 2023-06-26 07:09 +0000
                                    Re: .999 repeating is almost an infinitely large sequence Germaine Breda <eraa@gmeenmad.dr> - 2023-06-26 22:37 +0000
                                  Re: .999 repeating is almost an infinitely large sequence Morris Offermans <oeoe@ssoaanmf.sr> - 2023-06-26 07:22 +0000
                                  A Little Insight (was Re: .999 repeating is almost an infinitely large sequence) whodat <whodaat@void.nowgre.com> - 2023-06-26 14:23 -0500
                                    Re: A Little Insight (was Re: .999 repeating is almost an infinitely large sequence) Jim Pennino <jimp@gonzo.specsol.net> - 2023-06-26 12:57 -0700
                                    Re: A Little Insight (was Re: .999 repeating Foster Smeets <fste@sstffsem.ee> - 2023-06-26 21:14 +0000
                                    Re: A Little Insight (was Re: .999 repeating is almost an infinitely Connor Maas <naoa@nnsamoss.co> - 2023-06-26 21:27 +0000
                                    Re: A Little Insight (was Re: .999 repeating) Valery Schuyler <evrl@leerrael.hc> - 2023-06-27 11:48 +0000
                                      Re: A Little Insight (was Re: .999 repeating) Houston Beek <nnoo@bbookeon.ub> - 2023-06-27 22:23 +0000
                              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 15:20 -0700
                        Re: .999 repeating is almost an infinitely large sequence Volney <volney@invalid.invalid> - 2023-06-25 18:26 -0400
                          Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 15:29 -0700
                          Re: .999 repeating is almost an infinitely large sequence Ezra Venn <rene@nvanarvz.vn> - 2023-06-25 22:49 +0000
                        Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 00:14 +0100
                          Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 16:19 -0700
                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 16:23 -0700
                              Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 00:42 +0100
                                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 16:59 -0700
                                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 17:03 -0700
                                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 17:03 -0700
                                    Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 02:20 +0100
                                      Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 19:31 -0700
                                        Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 14:57 +0100
                                          Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 12:39 -0700
                                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-27 12:20 -0700
                                              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-27 15:07 -0700
                                              Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-28 02:38 +0100
                                                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-27 20:21 -0700
                                                  Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-28 14:51 +0100
                                                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-28 12:55 -0700
                                                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-28 12:56 -0700
                                                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-28 15:00 -0700
                                                      Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-29 10:39 -0700
                                                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-29 14:41 -0700
                                                          Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-29 21:18 -0700
                                                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-30 09:57 -0700
                                          Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-27 15:08 -0700
                                  Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 02:13 +0100
                                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 19:33 -0700
                                      Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 05:53 -0400
                                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 12:30 -0700
                                      Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 14:55 +0100
                                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 12:27 -0700
                                          Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 13:09 -0700
                                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:13 -0700
                                              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:15 -0700
                                              Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 13:21 -0700
                                          Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 23:04 +0100
                                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:09 -0700
                                              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:12 -0700
                                                Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 23:20 +0100
                                                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:33 -0700
                                                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:36 -0700
                                                    Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 19:32 -0400
                                                      Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 16:49 -0700
                                                        Re: .999 repeating is almost an infinitely large sequence Ross Finlayson <ross.a.finlayson@gmail.com> - 2023-06-26 20:31 -0700
                                                          Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-27 16:15 +0100
                                                            Re: .999 repeating is almost an infinitely large sequence Ross Finlayson <ross.a.finlayson@gmail.com> - 2023-06-27 14:32 -0700
                                                              Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-28 02:32 +0100
                                                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-27 12:03 -0700
                                                          Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-27 13:11 -0700
                                                            Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-27 13:27 -0700
                                                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:37 -0700
                                      Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 10:11 -0700
                                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 12:28 -0700
                                          Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 12:57 -0700
                                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:05 -0700
                                              Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 13:17 -0700
                                                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 13:29 -0700
                                                  Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 14:32 -0700
                                                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 14:47 -0700
                                                      Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 14:56 -0700
                                                      Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 15:01 -0700
                                                        Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:06 -0700
                                                          Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 15:15 -0700
                                                            Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:19 -0700
                                                              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:20 -0700
                                                              Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 15:24 -0700
                                                                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:32 -0700
                                                              Re: .999 repeating is almost an infinitely large sequence Maron Andel <nado@eredlemn.rl> - 2023-06-26 22:25 +0000
                                                                Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 15:33 -0700
                                                                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 15:59 -0700
                                                                    Re: .999 repeating is almost an infinitely large sequence Volney <volney@invalid.invalid> - 2023-06-26 22:36 -0400
                                                                      Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-27 12:07 -0700
                                              Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 17:15 -0400
                                                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 14:27 -0700
                                                  Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 17:30 -0400
                                                  Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 14:39 -0700
                                                    Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 15:13 -0700
                              Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 10:02 -0700
                                Re: .999 repeating is almost an infinitely large sequence Duncan Antwerp <anec@cnancrwp.ca> - 2023-06-26 22:20 +0000
                                  Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 15:28 -0700
                                  Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-26 19:43 -0400
                    Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-26 09:18 -0700
                  Re: .999 repeating is almost an infinitely large sequence Devoise Kool <ovve@sdodevlv.el> - 2023-06-28 21:56 +0000
                Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 09:38 -0700
                  Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-26 12:33 -0700
                Re: .999 repeating is almost an infinitely large sequence Fritz Feldhase <franz.fritschee.ff@gmail.com> - 2023-06-26 09:42 -0700
            Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-24 20:05 -0400
              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 17:08 -0700
                Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 17:10 -0700
                  Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-24 17:20 -0700
                    Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-24 17:25 -0700
                  Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-24 20:32 -0400
                Re: .999 repeating is almost an infinitely large sequence FromTheRafters <FTR@nomail.afraid.org> - 2023-06-24 20:20 -0400
              Re: .999 repeating is almost an infinitely large sequence Maxie Snider <eida@esxnnrxd.is> - 2023-06-25 08:06 +0000
            Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-25 23:53 +0100
              Re: .999 repeating is almost an infinitely large sequence "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> - 2023-06-25 15:59 -0700
                Re: .999 repeating is almost an infinitely large sequence Ben Bacarisse <ben.usenet@bsb.me.uk> - 2023-06-26 00:32 +0100
    Re: .999 repeating is almost an infinitely large sequence Ross Finlayson <ross.a.finlayson@gmail.com> - 2023-06-25 01:39 -0700
      Re: .999 repeating is almost an infinitely large sequence "mitchr...@gmail.com" <mitchrae3323@gmail.com> - 2023-06-25 10:05 -0700
        Re: .999 repeating is almost an infinitely large sequence Ross Finlayson <ross.a.finlayson@gmail.com> - 2023-06-25 14:05 -0700
          Re: .999 repeating is almost an infinitely large sequence Ross Finlayson <ross.a.finlayson@gmail.com> - 2023-07-02 10:30 -0700
    Re: .999 repeating is almost an infinitely large sequence "zelos...@gmail.com" <zelos.malum@gmail.com> - 2023-06-26 00:25 -0700
    Re: .999 repeating is almost an infinitely large sequence W <wwwwwwwwwwwwwwwwaaaaaa@hotmail.com> - 2023-06-26 10:14 -0700

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#603495

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-27 12:20 -0700
Message-ID	<u7fcq7$1ffv8$4@dont-email.me>
In reply to	#603370

On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>
>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>
>>>>> I can write the [Julia] set the same way:
>>>> (your correction applied)
>>>>
>>>>> i[n] = i[n - 1]^2 + c
>>>>>
>>>>> c = -.75+0i
>>>>>
>>>>> i[0] = -1+0i
>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>> know what you are getting at, but it's interesting that you think you
>>>> have defined any set at all.
>>>>
>>>
>>> I can also write the recursion as:
>>>
>>> i[n + 1] = i[n]^2 + c
>>>
>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>
>> It's impossible to forget!  A mathematician defining a set would define
>> a set, not a recurrence relation.
>>
> 
> I feed of off recurrence relations! Like to try to turn recursive 
> algorithms into iterative algorithms. Too many fractals on the brain. 
> Then, I also like to try to get a direct formula to a specific iterate 
> in a recurrence relation. A direct formula instead of an iterative 
> formula. Think of getting the result of the 42'nd iteration without 
> having to iterate up to it. A direct formula.

Ben, if you can create a direct formula for an escape time fractal, say, 
a Julia set, lets start there. That would be amazing, and you would need 
to praised!

If I gave you a point say, .75+.1i, and you can give me a formula that 
would produce the result of the 42 iterate without having to manually 
iterate up to 42, well, that would be amazing. I don't think its 
possible, but, who knows! :^)

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#603507

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-27 15:07 -0700
Message-ID	<u7fmir$1glcn$1@dont-email.me>
In reply to	#603495

On 6/27/2023 12:20 PM, Chris M. Thomasson wrote:
> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> I can write the [Julia] set the same way:
>>>>> (your correction applied)
>>>>>
>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>
>>>>>> c = -.75+0i
>>>>>>
>>>>>> i[0] = -1+0i
>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>> know what you are getting at, but it's interesting that you think you
>>>>> have defined any set at all.
>>>>>
>>>>
>>>> I can also write the recursion as:
>>>>
>>>> i[n + 1] = i[n]^2 + c
>>>>
>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>
>>> It's impossible to forget!  A mathematician defining a set would define
>>> a set, not a recurrence relation.
>>>
>>
>> I feed of off recurrence relations! Like to try to turn recursive 
>> algorithms into iterative algorithms. Too many fractals on the brain. 
>> Then, I also like to try to get a direct formula to a specific iterate 
>> in a recurrence relation. A direct formula instead of an iterative 
>> formula. Think of getting the result of the 42'nd iteration without 
>> having to iterate up to it. A direct formula.
> 
> Ben, if you can create a direct formula for an escape time fractal, say, 
> a Julia set, lets start there. That would be amazing, and you would need 
> to praised!
> 
> If I gave you a point say, .75+.1i, and you can give me a formula that 
> would produce the result of the 42 iterate without having to manually 
> iterate up to 42, well, that would be amazing. I don't think its 
> possible, but, who knows! :^)

If you can give me the direct formula for, say:

c = -.75+.1i  // A julia set

i_0 = .4+.03i // A random point in the 2-ary plane

i_n+1 = i_n^2 + c // The recursion relationship
...

Say I want to know the 42'nd iterate aka i_42 without having to iterate 
42 times to find out the damn answer.

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#603522

From	Ben Bacarisse <ben.usenet@bsb.me.uk>
Date	2023-06-28 02:38 +0100
Message-ID	<87leg4nz79.fsf@bsb.me.uk>
In reply to	#603495

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> I can write the [Julia] set the same way:
>>>>> (your correction applied)
>>>>>
>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>
>>>>>> c = -.75+0i
>>>>>>
>>>>>> i[0] = -1+0i
>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>> know what you are getting at, but it's interesting that you think you
>>>>> have defined any set at all.
>>>>>
>>>>
>>>> I can also write the recursion as:
>>>>
>>>> i[n + 1] = i[n]^2 + c
>>>>
>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>
>>> It's impossible to forget!  A mathematician defining a set would define
>>> a set, not a recurrence relation.
>>>
>> I feed of off recurrence relations! Like to try to turn recursive
>> algorithms into iterative algorithms. Too many fractals on the
>> brain. Then, I also like to try to get a direct formula to a specific
>> iterate in a recurrence relation. A direct formula instead of an
>> iterative formula. Think of getting the result of the 42'nd iteration
>> without having to iterate up to it. A direct formula.
>
> Ben, if you can create a direct formula for an escape time fractal, say, a
> Julia set, lets start there.

Where do you see this exchange going?

-- 
Ben.

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#603527

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-27 20:21 -0700
Message-ID	<u7g8vi$1m2dk$1@dont-email.me>
In reply to	#603522

On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>
>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>
>>>>>>> I can write the [Julia] set the same way:
>>>>>> (your correction applied)
>>>>>>
>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>
>>>>>>> c = -.75+0i
>>>>>>>
>>>>>>> i[0] = -1+0i
>>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>> have defined any set at all.
>>>>>>
>>>>>
>>>>> I can also write the recursion as:
>>>>>
>>>>> i[n + 1] = i[n]^2 + c
>>>>>
>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>
>>>> It's impossible to forget!  A mathematician defining a set would define
>>>> a set, not a recurrence relation.
>>>>
>>> I feed of off recurrence relations! Like to try to turn recursive
>>> algorithms into iterative algorithms. Too many fractals on the
>>> brain. Then, I also like to try to get a direct formula to a specific
>>> iterate in a recurrence relation. A direct formula instead of an
>>> iterative formula. Think of getting the result of the 42'nd iteration
>>> without having to iterate up to it. A direct formula.
>>
>> Ben, if you can create a direct formula for an escape time fractal, say, a
>> Julia set, lets start there.
> 
> Where do you see this exchange going?
> 

Not sure... Maybe you might find some free time to try to develop an IFS 
that can generate the Mandelbrot set? I think it is most likely 
impossible. ;^)

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#603556

From	Ben Bacarisse <ben.usenet@bsb.me.uk>
Date	2023-06-28 14:51 +0100
Message-ID	<87a5wjoft2.fsf@bsb.me.uk>
In reply to	#603527

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>> 
>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>>
>>>>>>>> I can write the [Julia] set the same way:
>>>>>>> (your correction applied)
>>>>>>>
>>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>>
>>>>>>>> c = -.75+0i
>>>>>>>>
>>>>>>>> i[0] = -1+0i
>>>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>>> have defined any set at all.
>>>>>>>
>>>>>>
>>>>>> I can also write the recursion as:
>>>>>>
>>>>>> i[n + 1] = i[n]^2 + c
>>>>>>
>>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>>
>>>>> It's impossible to forget!  A mathematician defining a set would define
>>>>> a set, not a recurrence relation.
>>>>>
>>>> I feed of off recurrence relations! Like to try to turn recursive
>>>> algorithms into iterative algorithms. Too many fractals on the
>>>> brain. Then, I also like to try to get a direct formula to a specific
>>>> iterate in a recurrence relation. A direct formula instead of an
>>>> iterative formula. Think of getting the result of the 42'nd iteration
>>>> without having to iterate up to it. A direct formula.
>>>
>>> Ben, if you can create a direct formula for an escape time fractal, say, a
>>> Julia set, lets start there.
>> Where do you see this exchange going?
>
> Not sure... Maybe you might find some free time to try to develop an IFS
> that can generate the Mandelbrot set? I think it is most likely
> impossible. ;^)

I am glad to see you are not being serious.

-- 
Ben.

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#603589

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-28 12:55 -0700
Message-ID	<u7i36n$1ro1f$2@dont-email.me>
In reply to	#603556

On 6/28/2023 6:51 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>
>>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>>>
>>>>>>>>> I can write the [Julia] set the same way:
>>>>>>>> (your correction applied)
>>>>>>>>
>>>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>>>
>>>>>>>>> c = -.75+0i
>>>>>>>>>
>>>>>>>>> i[0] = -1+0i
>>>>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>>>> have defined any set at all.
>>>>>>>>
>>>>>>>
>>>>>>> I can also write the recursion as:
>>>>>>>
>>>>>>> i[n + 1] = i[n]^2 + c
>>>>>>>
>>>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>>>
>>>>>> It's impossible to forget!  A mathematician defining a set would define
>>>>>> a set, not a recurrence relation.
>>>>>>
>>>>> I feed of off recurrence relations! Like to try to turn recursive
>>>>> algorithms into iterative algorithms. Too many fractals on the
>>>>> brain. Then, I also like to try to get a direct formula to a specific
>>>>> iterate in a recurrence relation. A direct formula instead of an
>>>>> iterative formula. Think of getting the result of the 42'nd iteration
>>>>> without having to iterate up to it. A direct formula.
>>>>
>>>> Ben, if you can create a direct formula for an escape time fractal, say, a
>>>> Julia set, lets start there.
>>> Where do you see this exchange going?
>>
>> Not sure... Maybe you might find some free time to try to develop an IFS
>> that can generate the Mandelbrot set? I think it is most likely
>> impossible. ;^)
> 
> I am glad to see you are not being serious.
> 

:^) Actually, one of my friends has been looking for the holy grail for 
a while now. He has got kind of sorta close, but no cigar.

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#603590

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-28 12:56 -0700
Message-ID	<u7i3a1$1ro1f$3@dont-email.me>
In reply to	#603556

On 6/28/2023 6:51 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>
>>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>>>
>>>>>>>>> I can write the [Julia] set the same way:
>>>>>>>> (your correction applied)
>>>>>>>>
>>>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>>>
>>>>>>>>> c = -.75+0i
>>>>>>>>>
>>>>>>>>> i[0] = -1+0i
>>>>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>>>> have defined any set at all.
>>>>>>>>
>>>>>>>
>>>>>>> I can also write the recursion as:
>>>>>>>
>>>>>>> i[n + 1] = i[n]^2 + c
>>>>>>>
>>>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>>>
>>>>>> It's impossible to forget!  A mathematician defining a set would define
>>>>>> a set, not a recurrence relation.
>>>>>>
>>>>> I feed of off recurrence relations! Like to try to turn recursive
>>>>> algorithms into iterative algorithms. Too many fractals on the
>>>>> brain. Then, I also like to try to get a direct formula to a specific
>>>>> iterate in a recurrence relation. A direct formula instead of an
>>>>> iterative formula. Think of getting the result of the 42'nd iteration
>>>>> without having to iterate up to it. A direct formula.
>>>>
>>>> Ben, if you can create a direct formula for an escape time fractal, say, a
>>>> Julia set, lets start there.
>>> Where do you see this exchange going?
>>
>> Not sure... Maybe you might find some free time to try to develop an IFS
>> that can generate the Mandelbrot set? I think it is most likely
>> impossible. ;^)
> 
> I am glad to see you are not being serious.
> 

Fwiw, here is one of my tries wrt trying to get an overall Mandelbrot 
shape from a Julia set:

http://paulbourke.net/fractals/cubicjulia/

Try plotting that one. It take a lot of iterations to fill in the spiral 
formations.

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#603597

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-28 15:00 -0700
Message-ID	<u7iahb$1sg74$1@dont-email.me>
In reply to	#603556

On 6/28/2023 6:51 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>
>>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>>>
>>>>>>>>> I can write the [Julia] set the same way:
>>>>>>>> (your correction applied)
>>>>>>>>
>>>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>>>
>>>>>>>>> c = -.75+0i
>>>>>>>>>
>>>>>>>>> i[0] = -1+0i
>>>>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>>>> have defined any set at all.
>>>>>>>>
>>>>>>>
>>>>>>> I can also write the recursion as:
>>>>>>>
>>>>>>> i[n + 1] = i[n]^2 + c
>>>>>>>
>>>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>>>
>>>>>> It's impossible to forget!  A mathematician defining a set would define
>>>>>> a set, not a recurrence relation.
>>>>>>
>>>>> I feed of off recurrence relations! Like to try to turn recursive
>>>>> algorithms into iterative algorithms. Too many fractals on the
>>>>> brain. Then, I also like to try to get a direct formula to a specific
>>>>> iterate in a recurrence relation. A direct formula instead of an
>>>>> iterative formula. Think of getting the result of the 42'nd iteration
>>>>> without having to iterate up to it. A direct formula.
>>>>
>>>> Ben, if you can create a direct formula for an escape time fractal, say, a
>>>> Julia set, lets start there.
>>> Where do you see this exchange going?
>>
>> Not sure... Maybe you might find some free time to try to develop an IFS
>> that can generate the Mandelbrot set? I think it is most likely
>> impossible. ;^)
> 
> I am glad to see you are not being serious.
> 

Here is a strange formation I noticed in some of my work that kind of 
resembles the essence of a low iteration Mandelbrot set:

https://i.ibb.co/dgHxtCf/image.png

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#603634

From	"mitchr...@gmail.com" <mitchrae3323@gmail.com>
Date	2023-06-29 10:39 -0700
Message-ID	<4e493cfb-ae96-438f-8385-140b9b3c3b7cn@googlegroups.com>
In reply to	#603597

On Wednesday, June 28, 2023 at 3:00:22 PM UTC-7, Chris M. Thomasson wrote:
> On 6/28/2023 6:51 AM, Ben Bacarisse wrote: 
> > "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> > 
> >> On 6/27/2023 6:38 PM, Ben Bacarisse wrote: 
> >>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>> 
> >>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote: 
> >>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote: 
> >>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>>>>> 
> >>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote: 
> >>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>>>>>>> 
> >>>>>>>>> I can write the [Julia] set the same way: 
> >>>>>>>> (your correction applied) 
> >>>>>>>> 
> >>>>>>>>> i[n] = i[n - 1]^2 + c 
> >>>>>>>>> 
> >>>>>>>>> c = -.75+0i 
> >>>>>>>>> 
> >>>>>>>>> i[0] = -1+0i 
> >>>>>>>> No, this is not the definition of any set, much less a Julia set.  I 
> >>>>>>>> know what you are getting at, but it's interesting that you think you 
> >>>>>>>> have defined any set at all. 
> >>>>>>>> 
> >>>>>>> 
> >>>>>>> I can also write the recursion as: 
> >>>>>>> 
> >>>>>>> i[n + 1] = i[n]^2 + c 
> >>>>>>> 
> >>>>>>> I find this easy to read. Keep in mind that I am not a mathematician. 
> >>>>>> 
> >>>>>> It's impossible to forget!  A mathematician defining a set would define 
> >>>>>> a set, not a recurrence relation. 
> >>>>>> 
> >>>>> I feed of off recurrence relations! Like to try to turn recursive 
> >>>>> algorithms into iterative algorithms. Too many fractals on the 
> >>>>> brain. Then, I also like to try to get a direct formula to a specific 
> >>>>> iterate in a recurrence relation. A direct formula instead of an 
> >>>>> iterative formula. Think of getting the result of the 42'nd iteration 
> >>>>> without having to iterate up to it. A direct formula. 
> >>>> 
> >>>> Ben, if you can create a direct formula for an escape time fractal, say, a 
> >>>> Julia set, lets start there. 
> >>> Where do you see this exchange going? 
> >> 
> >> Not sure... Maybe you might find some free time to try to develop an IFS 
> >> that can generate the Mandelbrot set? I think it is most likely 
> >> impossible. ;^) 
> > 
> > I am glad to see you are not being serious. 
> >
> Here is a strange formation I noticed in some of my work that kind of 
> resembles the essence of a low iteration Mandelbrot set: 
> 
> https://i.ibb.co/dgHxtCf/image.png

If the Mandelbrot set is considered perfect why does some have to be left out?

Mitchell Raemsch

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#603650

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-29 14:41 -0700
Message-ID	<u7ktq0$28vqb$1@dont-email.me>
In reply to	#603634

On 6/29/2023 10:39 AM, mitchr...@gmail.com wrote:
> On Wednesday, June 28, 2023 at 3:00:22 PM UTC-7, Chris M. Thomasson wrote:
>> On 6/28/2023 6:51 AM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>
>>>> On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>
>>>>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>>>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>>>>
>>>>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>>>>>>
>>>>>>>>>>> I can write the [Julia] set the same way:
>>>>>>>>>> (your correction applied)
>>>>>>>>>>
>>>>>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>>>>>
>>>>>>>>>>> c = -.75+0i
>>>>>>>>>>>
>>>>>>>>>>> i[0] = -1+0i
>>>>>>>>>> No, this is not the definition of any set, much less a Julia set.  I
>>>>>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>>>>>> have defined any set at all.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> I can also write the recursion as:
>>>>>>>>>
>>>>>>>>> i[n + 1] = i[n]^2 + c
>>>>>>>>>
>>>>>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>>>>>
>>>>>>>> It's impossible to forget!  A mathematician defining a set would define
>>>>>>>> a set, not a recurrence relation.
>>>>>>>>
>>>>>>> I feed of off recurrence relations! Like to try to turn recursive
>>>>>>> algorithms into iterative algorithms. Too many fractals on the
>>>>>>> brain. Then, I also like to try to get a direct formula to a specific
>>>>>>> iterate in a recurrence relation. A direct formula instead of an
>>>>>>> iterative formula. Think of getting the result of the 42'nd iteration
>>>>>>> without having to iterate up to it. A direct formula.
>>>>>>
>>>>>> Ben, if you can create a direct formula for an escape time fractal, say, a
>>>>>> Julia set, lets start there.
>>>>> Where do you see this exchange going?
>>>>
>>>> Not sure... Maybe you might find some free time to try to develop an IFS
>>>> that can generate the Mandelbrot set? I think it is most likely
>>>> impossible. ;^)
>>>
>>> I am glad to see you are not being serious.
>>>
>> Here is a strange formation I noticed in some of my work that kind of
>> resembles the essence of a low iteration Mandelbrot set:
>>
>> https://i.ibb.co/dgHxtCf/image.png
> 
> If the Mandelbrot set is considered perfect why does some have to be left out?

If you are referring to the _traditional_ bailout condition, any point 
whose orbits are greater than two from the origin are outside of the 
Mandelbrot set. Humm... You might be interested in a buddhabrot. I did a 
"special" one some years ago that actually plots every point in every 
orbit. It looks like this:

https://i.ibb.co/N2ySbhs/image.png

:^)

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#603675

From	"mitchr...@gmail.com" <mitchrae3323@gmail.com>
Date	2023-06-29 21:18 -0700
Message-ID	<00ea5972-7d7a-48bd-9243-a6572ea8804cn@googlegroups.com>
In reply to	#603650

On Thursday, June 29, 2023 at 2:41:32 PM UTC-7, Chris M. Thomasson wrote:
> On 6/29/2023 10:39 AM, mitchr...@gmail.com wrote: 
> > On Wednesday, June 28, 2023 at 3:00:22 PM UTC-7, Chris M. Thomasson wrote: 
> >> On 6/28/2023 6:51 AM, Ben Bacarisse wrote: 
> >>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>> 
> >>>> On 6/27/2023 6:38 PM, Ben Bacarisse wrote: 
> >>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>>>> 
> >>>>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote: 
> >>>>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote: 
> >>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>>>>>>> 
> >>>>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote: 
> >>>>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes: 
> >>>>>>>>>> 
> >>>>>>>>>>> I can write the [Julia] set the same way: 
> >>>>>>>>>> (your correction applied) 
> >>>>>>>>>> 
> >>>>>>>>>>> i[n] = i[n - 1]^2 + c 
> >>>>>>>>>>> 
> >>>>>>>>>>> c = -.75+0i 
> >>>>>>>>>>> 
> >>>>>>>>>>> i[0] = -1+0i 
> >>>>>>>>>> No, this is not the definition of any set, much less a Julia set. I 
> >>>>>>>>>> know what you are getting at, but it's interesting that you think you 
> >>>>>>>>>> have defined any set at all. 
> >>>>>>>>>> 
> >>>>>>>>> 
> >>>>>>>>> I can also write the recursion as: 
> >>>>>>>>> 
> >>>>>>>>> i[n + 1] = i[n]^2 + c 
> >>>>>>>>> 
> >>>>>>>>> I find this easy to read. Keep in mind that I am not a mathematician. 
> >>>>>>>> 
> >>>>>>>> It's impossible to forget! A mathematician defining a set would define 
> >>>>>>>> a set, not a recurrence relation. 
> >>>>>>>> 
> >>>>>>> I feed of off recurrence relations! Like to try to turn recursive 
> >>>>>>> algorithms into iterative algorithms. Too many fractals on the 
> >>>>>>> brain. Then, I also like to try to get a direct formula to a specific 
> >>>>>>> iterate in a recurrence relation. A direct formula instead of an 
> >>>>>>> iterative formula. Think of getting the result of the 42'nd iteration 
> >>>>>>> without having to iterate up to it. A direct formula. 
> >>>>>> 
> >>>>>> Ben, if you can create a direct formula for an escape time fractal, say, a 
> >>>>>> Julia set, lets start there. 
> >>>>> Where do you see this exchange going? 
> >>>> 
> >>>> Not sure... Maybe you might find some free time to try to develop an IFS 
> >>>> that can generate the Mandelbrot set? I think it is most likely 
> >>>> impossible. ;^) 
> >>> 
> >>> I am glad to see you are not being serious. 
> >>> 
> >> Here is a strange formation I noticed in some of my work that kind of 
> >> resembles the essence of a low iteration Mandelbrot set: 
> >> 
> >> https://i.ibb.co/dgHxtCf/image.png 
> > 
> > If the Mandelbrot set is considered perfect why does some have to be left out?
> If you are referring to the _traditional_ bailout condition, any point 
> whose orbits are greater than two from the origin are outside of the 
> Mandelbrot set. Humm... You might be interested in a buddhabrot. I did a 
> "special" one some years ago that actually plots every point in every 
> orbit. It looks like this: 
> 
> https://i.ibb.co/N2ySbhs/image.png 
> 
> :^)

Don't leave any out.

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#603701

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-30 09:57 -0700
Message-ID	<u7n1im$2jdlf$1@dont-email.me>
In reply to	#603675

On 6/29/2023 9:18 PM, mitchr...@gmail.com wrote:
> On Thursday, June 29, 2023 at 2:41:32 PM UTC-7, Chris M. Thomasson wrote:
>> On 6/29/2023 10:39 AM, mitchr...@gmail.com wrote:
>>> On Wednesday, June 28, 2023 at 3:00:22 PM UTC-7, Chris M. Thomasson wrote:
>>>> On 6/28/2023 6:51 AM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>
>>>>>> On 6/27/2023 6:38 PM, Ben Bacarisse wrote:
>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>>>
>>>>>>>> On 6/26/2023 12:39 PM, Chris M. Thomasson wrote:
>>>>>>>>> On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
>>>>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>>>>>>
>>>>>>>>>>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>>>>>>>>>>> "Chris M. Thomasson" <chris.m.t...@gmail.com> writes:
>>>>>>>>>>>>
>>>>>>>>>>>>> I can write the [Julia] set the same way:
>>>>>>>>>>>> (your correction applied)
>>>>>>>>>>>>
>>>>>>>>>>>>> i[n] = i[n - 1]^2 + c
>>>>>>>>>>>>>
>>>>>>>>>>>>> c = -.75+0i
>>>>>>>>>>>>>
>>>>>>>>>>>>> i[0] = -1+0i
>>>>>>>>>>>> No, this is not the definition of any set, much less a Julia set. I
>>>>>>>>>>>> know what you are getting at, but it's interesting that you think you
>>>>>>>>>>>> have defined any set at all.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> I can also write the recursion as:
>>>>>>>>>>>
>>>>>>>>>>> i[n + 1] = i[n]^2 + c
>>>>>>>>>>>
>>>>>>>>>>> I find this easy to read. Keep in mind that I am not a mathematician.
>>>>>>>>>>
>>>>>>>>>> It's impossible to forget! A mathematician defining a set would define
>>>>>>>>>> a set, not a recurrence relation.
>>>>>>>>>>
>>>>>>>>> I feed of off recurrence relations! Like to try to turn recursive
>>>>>>>>> algorithms into iterative algorithms. Too many fractals on the
>>>>>>>>> brain. Then, I also like to try to get a direct formula to a specific
>>>>>>>>> iterate in a recurrence relation. A direct formula instead of an
>>>>>>>>> iterative formula. Think of getting the result of the 42'nd iteration
>>>>>>>>> without having to iterate up to it. A direct formula.
>>>>>>>>
>>>>>>>> Ben, if you can create a direct formula for an escape time fractal, say, a
>>>>>>>> Julia set, lets start there.
>>>>>>> Where do you see this exchange going?
>>>>>>
>>>>>> Not sure... Maybe you might find some free time to try to develop an IFS
>>>>>> that can generate the Mandelbrot set? I think it is most likely
>>>>>> impossible. ;^)
>>>>>
>>>>> I am glad to see you are not being serious.
>>>>>
>>>> Here is a strange formation I noticed in some of my work that kind of
>>>> resembles the essence of a low iteration Mandelbrot set:
>>>>
>>>> https://i.ibb.co/dgHxtCf/image.png
>>>
>>> If the Mandelbrot set is considered perfect why does some have to be left out?
>> If you are referring to the _traditional_ bailout condition, any point
>> whose orbits are greater than two from the origin are outside of the
>> Mandelbrot set. Humm... You might be interested in a buddhabrot. I did a
>> "special" one some years ago that actually plots every point in every
>> orbit. It looks like this:
>>
>> https://i.ibb.co/N2ySbhs/image.png
>>
>> :^)
> 
> Don't leave any out.

Define any... I plotted all of the orbits for every pixel in the plane 
to get the all plotting buddhabrot.

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#603508

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-27 15:08 -0700
Message-ID	<u7fmln$1glcn$2@dont-email.me>
In reply to	#603334

On 6/26/2023 6:57 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/25/2023 6:20 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> I can write the [Julia] set the same way:
>>> (your correction applied)
>>>
>>>> i[n] = i[n - 1]^2 + c
>>>>
>>>> c = -.75+0i
>>>>
>>>> i[0] = -1+0i
>>> No, this is not the definition of any set, much less a Julia set.  I
>>> know what you are getting at, but it's interesting that you think you
>>> have defined any set at all.
>>>
>>
>> I can also write the recursion as:
>>
>> i[n + 1] = i[n]^2 + c
>>
>> I find this easy to read. Keep in mind that I am not a mathematician.
> 
> It's impossible to forget!  A mathematician defining a set would define
> a set, not a recurrence relation.
> 

Also, if you can find an IFS that creates the Mandelbrot set, you would 
be a god.

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#603294

From	Ben Bacarisse <ben.usenet@bsb.me.uk>
Date	2023-06-26 02:13 +0100
Message-ID	<87o7l33u1j.fsf@bsb.me.uk>
In reply to	#603288

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 6/25/2023 4:42 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>> 
>>> On 6/25/2023 4:19 PM, Chris M. Thomasson wrote:
>>>> On 6/25/2023 4:14 PM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> Check this out:
>>>>>>
>>>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>>>>
>>>>>> Right?
>>>>>
>>>>> No.  There is no n (in ℕ) that makes that equation hold.
>>>>>
>>>>>> Lets check:
>>>>>
>>>>> OK...
>>>>>
>>>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>>>> ...
>>>>>>
>>>>>> See?
>>>>>
>>>>> Do you?  None of i[n] = 0.999...
>>>>>
>>>> When n is taken to infinity, it equals .(9), right?
>>>
>>> n is without limit. Therefore the iterates go on forever and always get
>>> closer and closer to one. .999... = 1
>> Are you trying to say that lim_{n->oo} i[n] = 1?
>
> i[n] just denotes the iterates of:
>
> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>
> the infinite iterates are .(9)

So you are not simply saying that lim_{n->oo} i[n] = 1?  OK.  I wonder
what all these other words mean if not the simple fact that lim_{n->oo}
i[n] = 1.

>>> I am not exactly sure why:
>>>
>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>>
>>> would not equal .(9) when n is taken to infinity.
>> lim_{n->oo} i[n] = 1.  Is that what you mean?
>> My objection was to this:
>>    i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>> It looks like an equation, but there is no n that satisfies it.
>
> i[n] means use n in the formula,

No.  It denotes a (functional) term in an equation (or other formula).
This, for example is correct:

  i[n] < 1

but this

  i[n] = 1

is not.  (Once defined, you don't have to keep repeating the complex
definition you've come up with for i[n].)

> so lets break it down for the first couple
> of iterates:
>
> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
> ...

All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
other than you can only think in a sort of process/programming mode.

-- 
Ben.

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#603300

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-25 19:33 -0700
Message-ID	<u7atd6$pd1c$2@dont-email.me>
In reply to	#603294

On 6/25/2023 6:13 PM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/25/2023 4:42 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/25/2023 4:19 PM, Chris M. Thomasson wrote:
>>>>> On 6/25/2023 4:14 PM, Ben Bacarisse wrote:
>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>
>>>>>>> Check this out:
>>>>>>>
>>>>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>>>>>
>>>>>>> Right?
>>>>>>
>>>>>> No.  There is no n (in ℕ) that makes that equation hold.
>>>>>>
>>>>>>> Lets check:
>>>>>>
>>>>>> OK...
>>>>>>
>>>>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>>>>> ...
>>>>>>>
>>>>>>> See?
>>>>>>
>>>>>> Do you?  None of i[n] = 0.999...
>>>>>>
>>>>> When n is taken to infinity, it equals .(9), right?
>>>>
>>>> n is without limit. Therefore the iterates go on forever and always get
>>>> closer and closer to one. .999... = 1
>>> Are you trying to say that lim_{n->oo} i[n] = 1?
>>
>> i[n] just denotes the iterates of:
>>
>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>
>> the infinite iterates are .(9)
> 
> So you are not simply saying that lim_{n->oo} i[n] = 1?  OK.  I wonder
> what all these other words mean if not the simple fact that lim_{n->oo}
> i[n] = 1.
> 
>>>> I am not exactly sure why:
>>>>
>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>>>
>>>> would not equal .(9) when n is taken to infinity.
>>> lim_{n->oo} i[n] = 1.  Is that what you mean?
>>> My objection was to this:
>>>     i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>> It looks like an equation, but there is no n that satisfies it.
>>
>> i[n] means use n in the formula,
> 
> No.  It denotes a (functional) term in an equation (or other formula).
> This, for example is correct:
> 
>    i[n] < 1
> 
> but this
> 
>    i[n] = 1
> 
> is not.  (Once defined, you don't have to keep repeating the complex
> definition you've come up with for i[n].)
> 
>> so lets break it down for the first couple
>> of iterates:
>>
>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>> ...
> 
> All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
> other than you can only think in a sort of process/programming mode.
> 

If n is taken to infinity then it should be .9 repeating for infinity, 
right? 0, .9, .99, .999, .9999, ...

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#603326

From	FromTheRafters <FTR@nomail.afraid.org>
Date	2023-06-26 05:53 -0400
Message-ID	<u7bn6o$ti29$1@dont-email.me>
In reply to	#603300

Chris M. Thomasson formulated the question :
> On 6/25/2023 6:13 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>> 
>>> On 6/25/2023 4:42 PM, Ben Bacarisse wrote:
>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>
>>>>> On 6/25/2023 4:19 PM, Chris M. Thomasson wrote:
>>>>>> On 6/25/2023 4:14 PM, Ben Bacarisse wrote:
>>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>>
>>>>>>>> Check this out:
>>>>>>>>
>>>>>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>>>>>>
>>>>>>>> Right?
>>>>>>>
>>>>>>> No.  There is no n (in ℕ) that makes that equation hold.
>>>>>>>
>>>>>>>> Lets check:
>>>>>>>
>>>>>>> OK...
>>>>>>>
>>>>>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>>>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>>>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>>>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>>>>>> ...
>>>>>>>>
>>>>>>>> See?
>>>>>>>
>>>>>>> Do you?  None of i[n] = 0.999...
>>>>>>>
>>>>>> When n is taken to infinity, it equals .(9), right?
>>>>>
>>>>> n is without limit. Therefore the iterates go on forever and always get
>>>>> closer and closer to one. .999... = 1
>>>> Are you trying to say that lim_{n->oo} i[n] = 1?
>>>
>>> i[n] just denotes the iterates of:
>>>
>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>>
>>> the infinite iterates are .(9)
>> 
>> So you are not simply saying that lim_{n->oo} i[n] = 1?  OK.  I wonder
>> what all these other words mean if not the simple fact that lim_{n->oo}
>> i[n] = 1.
>> 
>>>>> I am not exactly sure why:
>>>>>
>>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>>>>
>>>>> would not equal .(9) when n is taken to infinity.
>>>> lim_{n->oo} i[n] = 1.  Is that what you mean?
>>>> My objection was to this:
>>>>     i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>> It looks like an equation, but there is no n that satisfies it.
>>>
>>> i[n] means use n in the formula,
>> 
>> No.  It denotes a (functional) term in an equation (or other formula).
>> This, for example is correct:
>> 
>>    i[n] < 1
>> 
>> but this
>> 
>>    i[n] = 1
>> 
>> is not.  (Once defined, you don't have to keep repeating the complex
>> definition you've come up with for i[n].)
>> 
>>> so lets break it down for the first couple
>>> of iterates:
>>>
>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>> ...
>> 
>> All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
>> other than you can only think in a sort of process/programming mode.
>> 
>
> If n is taken to infinity then it should be .9 repeating for infinity, right? 
> 0, .9, .99, .999, .9999, ...

Each of these are elements of Q not equal to one.

Your n is a natural number which means it is always going to be finite 
because the naturals are a discrete infinite set of finite sets. Using 
limits would mean you have switched to the reals where convergence of 
cauchy sequences is guaranteed but you are no longer dealing with 
isolated points but in an arena where each number can be seen as a 
limit of at least one cauchy sequence.

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#603364

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-26 12:30 -0700
Message-ID	<u7cp0o$1215e$3@dont-email.me>
In reply to	#603326

On 6/26/2023 2:53 AM, FromTheRafters wrote:
> Chris M. Thomasson formulated the question :
>> On 6/25/2023 6:13 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>
>>>> On 6/25/2023 4:42 PM, Ben Bacarisse wrote:
>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>
>>>>>> On 6/25/2023 4:19 PM, Chris M. Thomasson wrote:
>>>>>>> On 6/25/2023 4:14 PM, Ben Bacarisse wrote:
>>>>>>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
>>>>>>>>
>>>>>>>>> Check this out:
>>>>>>>>>
>>>>>>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>>>>>>>
>>>>>>>>> Right?
>>>>>>>>
>>>>>>>> No.  There is no n (in ℕ) that makes that equation hold.
>>>>>>>>
>>>>>>>>> Lets check:
>>>>>>>>
>>>>>>>> OK...
>>>>>>>>
>>>>>>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>>>>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>>>>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>>>>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>>>>>>> ...
>>>>>>>>>
>>>>>>>>> See?
>>>>>>>>
>>>>>>>> Do you?  None of i[n] = 0.999...
>>>>>>>>
>>>>>>> When n is taken to infinity, it equals .(9), right?
>>>>>>
>>>>>> n is without limit. Therefore the iterates go on forever and 
>>>>>> always get
>>>>>> closer and closer to one. .999... = 1
>>>>> Are you trying to say that lim_{n->oo} i[n] = 1?
>>>>
>>>> i[n] just denotes the iterates of:
>>>>
>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>>>
>>>> the infinite iterates are .(9)
>>>
>>> So you are not simply saying that lim_{n->oo} i[n] = 1?  OK.  I wonder
>>> what all these other words mean if not the simple fact that lim_{n->oo}
>>> i[n] = 1.
>>>
>>>>>> I am not exactly sure why:
>>>>>>
>>>>>> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
>>>>>>
>>>>>> would not equal .(9) when n is taken to infinity.
>>>>> lim_{n->oo} i[n] = 1.  Is that what you mean?
>>>>> My objection was to this:
>>>>>     i[n] = (floor(1/9 * 10^n) / 10^n) * 9 = .999....
>>>>> It looks like an equation, but there is no n that satisfies it.
>>>>
>>>> i[n] means use n in the formula,
>>>
>>> No.  It denotes a (functional) term in an equation (or other formula).
>>> This, for example is correct:
>>>
>>>    i[n] < 1
>>>
>>> but this
>>>
>>>    i[n] = 1
>>>
>>> is not.  (Once defined, you don't have to keep repeating the complex
>>> definition you've come up with for i[n].)
>>>
>>>> so lets break it down for the first couple
>>>> of iterates:
>>>>
>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>> ...
>>>
>>> All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
>>> other than you can only think in a sort of process/programming mode.
>>>
>>
>> If n is taken to infinity then it should be .9 repeating for infinity, 
>> right? 0, .9, .99, .999, .9999, ...
> 
> Each of these are elements of Q not equal to one.
> 
> Your n is a natural number which means it is always going to be finite 
> because the naturals are a discrete infinite set of finite sets. 

Take n to infinity and it equals .(9)

This is how base 10 represents one.

So, we can say it equals one wrt base 10 representation.

What am I missing here?



> Using 
> limits would mean you have switched to the reals where convergence of 
> cauchy sequences is guaranteed but you are no longer dealing with 
> isolated points but in an arena where each number can be seen as a limit 
> of at least one cauchy sequence.

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#603333

From	Ben Bacarisse <ben.usenet@bsb.me.uk>
Date	2023-06-26 14:55 +0100
Message-ID	<87v8fa2uri.fsf@bsb.me.uk>
In reply to	#603300

"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

> On 6/25/2023 6:13 PM, Ben Bacarisse wrote:
>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:

>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>> ...
>> All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
>> other than you can only think in a sort of process/programming mode.
>
> If n is taken to infinity then it should be .9 repeating for infinity,
> right? 0, .9, .99, .999, .9999, ...

Why do you want someone to validate your strange wording?  If you are
just saying that lim_{i->oo} i[n] = 1 then you should switch to saying
it correctly as a limit.  If you are /not/ saying that, how can anyone
agree that you are right when you won't use other words to explain what
it is that you are trying to say?

The whole exercises is getting more silly by the minute.

-- 
Ben.

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#603362

From	"Chris M. Thomasson" <chris.m.thomasson.1@gmail.com>
Date	2023-06-26 12:27 -0700
Message-ID	<u7coqo$1215e$1@dont-email.me>
In reply to	#603333

On 6/26/2023 6:55 AM, Ben Bacarisse wrote:
> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>> On 6/25/2023 6:13 PM, Ben Bacarisse wrote:
>>> "Chris M. Thomasson" <chris.m.thomasson.1@gmail.com> writes:
> 
>>>> i[0] = (floor(1/9 * 10^0) / 10^0) * 9 = 0
>>>> i[1] = (floor(1/9 * 10^1) / 10^1) * 9 = .9
>>>> i[2] = (floor(1/9 * 10^2) / 10^2) * 9 = .99
>>>> i[3] = (floor(1/9 * 10^3) / 10^3) * 9 = .999
>>>> ...
>>> All correct.  i[n] = .(9) = 1 is not.  I am not sure what the problem is
>>> other than you can only think in a sort of process/programming mode.
>>
>> If n is taken to infinity then it should be .9 repeating for infinity,
>> right? 0, .9, .99, .999, .9999, ...
> 
> Why do you want someone to validate your strange wording?  If you are
> just saying that lim_{i->oo} i[n] = 1 then you should switch to saying
> it correctly as a limit.  

The limit of:

i[n] = (floor(1/9 * 10^n) / 10^n) * 9

when n is taken to infinity is one

Clear enough?


> If you are /not/ saying that, how can anyone
> agree that you are right when you won't use other words to explain what
> it is that you are trying to say?





> 
> The whole exercises is getting more silly by the minute.
>

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#603385

From	Fritz Feldhase <franz.fritschee.ff@gmail.com>
Date	2023-06-26 13:09 -0700
Message-ID	<645b7f2f-146c-4a9a-bd5b-6c36f515a3efn@googlegroups.com>
In reply to	#603362

On Monday, June 26, 2023 at 9:27:29 PM UTC+2, Chris M. Thomasson wrote:

> The limit of:
> i[n] = (floor(1/9 * 10^n) / 10^n) * 9
> when n is taken to infinity is one 
> 
> Clear enough?

*sigh* Why don't you just use standard notation and terminology?

| i[n] = (floor(1/9 * 10^n) / 10^n) * 9 (for all n e IN)
|
| lim_(n-->oo) i[n] = 1.

Hint: The latter can also be written like this

| i[n] --> 1 as n --> oo.
[Read: "i[n] approaches 1 as n approaches oo."]

See (!): https://en.wikipedia.org/wiki/Limit_(mathematics)

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.999 repeating is almost an infinitely large sequence

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