Talk:Functional square root

I'm having trouble finding relevant material for this article, with the exception of a few Usenet posts and the like. In particular, I can't find anything informative on Kneser's result. Maybe I'm using the wrong term? Help would be appreciated. Fredrik Johansson 15:59, 29 September 2006 (UTC)[reply]

Abuse of notation?

"One notation that expresses that f is a functional square root of g is f = g^1/2." f is a functional square root. Not the functional square root. If there's more than one, they can't all be equal to the same.

I'm not sure if it's more an abuse of = or an abuse of the superscript notation. With numbers, there's the concept of principal value, so that x² = y doesn't necessarily imply x = y^1/2. But has anybody come up with a definition of a principal functional square root? -- Smjg (talk) 12:26, 12 June 2010 (UTC)[reply]

Not that I know of and there may be many of them even when constrained to be continuous and monotoic. For a monotonic increasing function they'd normally want a monotonically increasing square root though. Normally you can't ensure they are holomorphic so if you can you're lucky. Dmcq (talk) 12:42, 12 June 2010 (UTC)[reply]

Graph slightly wrong

"The green triangle is lim k->0 sin[k](x)" isn't quite right, it should read sin[1/k](x) 94.245.127.12 (talk) 11:23, 8 August 2013 (UTC)[reply]

What do you mean? The green triangle is the 0th iterate, so, basically, x, except with the proper periodicity. It is your sin[1/m](x) for enormous, not infinitesimal m. Track the orange and back curves. The opposite limit you may be contemplating without realizing it, is the k →∞ limit, m=0, where the figure basically falls flat onto the x-axis. Have you read up on iterated functions? Cuzkatzimhut (talk) 11:46, 8 August 2013 (UTC)[reply]

Analytic functions with a fixed point have analytic functional square roots

According to my understanding of a discussion in Scott Aaronson's blog, the class of analytic functions f such that f(0) = 0 is closed under functional square root, and the i+1st coefficient can easily be calculated from the first i coefficients (see Richard Stanley, Enumerative Combinatorics 2, Exercise 5.52). So, for example, e^x - 1 has an analytic functional square root. It follows that if f is an analytic function with a real fixed point, then f has an analytic functional square root. So, for example, sqrt(2)^x has an analytic functional square root. The question for 2^x and for e^x is open as far as I know.

71.224.195.6 (talk) 22:36, 20 July 2014 (UTC)Richard Beigel[reply]

It's true that exp(x)-1 has a functional square root which can be described using a (rational coefficient) power series, which allows us to calculate its values to arbitrary precision. But I'm pretty sure that it's not analytic -- the series is evidently divergent. Joule36e5 (talk) 05:11, 5 August 2015 (UTC)[reply]