tag:blogger.com,1999:blog-11295132.post8266076036196212490..comments2019-01-08T20:18:02.133-08:00Comments on A Neighborhood of Infinity: Logarithms and exponentials of functionsDan Piponihttps://plus.google.com/107913314994758123748noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-11295132.post-55107280681434505732017-03-08T11:45:05.155-08:002017-03-08T11:45:05.155-08:00In the Preliminaries section I believe you meant t...In the Preliminaries section I believe you meant the equation before the last to be (f+g) o h = foh + goh<br /><br /><br /><br /> Abdulmonem Al-Dubaisihttps://www.blogger.com/profile/01248413227871568393noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-12877532160966207072017-02-16T07:09:39.682-08:002017-02-16T07:09:39.682-08:00Added some updates to the "small theorems&quo...Added some updates to the "small theorems"Dan Piponihttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-43398731131658480162017-02-15T07:04:41.927-08:002017-02-15T07:04:41.927-08:00Blair S,
You are correct to point out that there ...Blair S,<br /><br />You are correct to point out that there is a problem with law 2. But it might not be the problem you think :-) As it stands, I think it's correct, but for trivial reasons.<br /><br />z is the identity function so z^oa o z^ob = z^oc for any a, b and c.<br /><br />That wasn't my intention so I'll delete it when I have a moment.Dan Piponihttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-41607371761161330942017-02-14T23:14:29.578-08:002017-02-14T23:14:29.578-08:00Hi, I stumbled across your blog and it looked inte...Hi, I stumbled across your blog and it looked interesting. Although, I'm not sure I follow your theorem two, you write z^n o z^m = z^(nm); should the lhs really be (z^n)^m or perhaps the rhs should be z^(n+m)?Blair Shttps://www.blogger.com/profile/10490971825310225119noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-79191456765904821142017-02-09T14:25:43.604-08:002017-02-09T14:25:43.604-08:00I did a search on formal power series on arxiv and...I did a search on formal power series on arxiv and found this: https://arxiv.org/abs/1701.03654<br /><br />It also computes the "functional" logarithm and uses a series like mine, but they ultimately do something different with it and I haven't read it properly yet. They also end by saying they're working on Haskell code.Dan Piponihttps://www.blogger.com/profile/08096190433222340957noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-60331597199087755602017-02-06T23:44:32.290-08:002017-02-06T23:44:32.290-08:00Wonderful post.
I think there's a typo in eq....Wonderful post.<br /><br />I think there's a typo in eq. 2 in the Notation section. Should be<br />z^{\circ m}\circ z^{\circ n} = z^{\circ m + n}Slavomir Kaslevhttps://www.blogger.com/profile/07759469055809975039noreply@blogger.comtag:blogger.com,1999:blog-11295132.post-67975482910444924982017-02-06T13:40:38.729-08:002017-02-06T13:40:38.729-08:00I just spotted this on mathoverflow: http://mathov...I just spotted this on mathoverflow: http://mathoverflow.net/a/215203/1233 which points out some of the connections between combinatorics and some of the things I've been computing.Dan Piponihttps://www.blogger.com/profile/04770325615925828083noreply@blogger.com