Gmail Calendar Documents Reader Web more »
Recently Visited Groups | Help | Sign in
Google Groups Home
Message from discussion y(f(x))=y(x)+x

View parsed - Show only message text

Path: g2news1.google.com!news3.google.com!news.glorb.com!cyclone.bc.net!nntp.itservices.ubc.ca!not-for-mail
From: Daniel Geisler <dan...@tetration.org>
Newsgroups: sci.math.research
Subject: Re: y(f(x))=y(x)+x
Date: Mon, 10 Jan 2005 23:33:02 -0800
Organization: Posted via Supernews, http://www.supernews.com
Lines: 28
Approved: Robert Israel <isr...@math.ubc.ca>, moderator for sci.math.research
Message-ID: <10u6vv7959fjbe1@corp.supernews.com>
References: <vz6xjmdri49p@legacy> <crqpof$oh$1@news.math.niu.edu>
Reply-To: dan...@tetration.org
Mime-Version: 1.0
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
Content-Transfer-Encoding: 7bit
X-Trace: nntp.itservices.ubc.ca 1105458992 852 137.82.36.63 (11 Jan 2005 15:56:32 GMT)
X-Complaints-To: abuse@interchange.ubc.ca
User-Agent: Mozilla Thunderbird 1.0 (Windows/20041206)
Originator: isr...@math.ubc.ca (Robert Israel)

Dave Rusin wrote:
> In article <vz6xjmdri49p@legacy>, Maxim  <MOsadc...@nes.ru> wrote:
> 
>>Is it possible to solve equation y(f(x))=y(x)+x if f(.) is a known
>>function? (e.g., f(x)=exp(x)-1  ?)
> 
> 
> What do you mean by "solve"?
> 
> Exactly this example is treated as a prototype for a broad family of
> similar problems in a paper by Szekeres; he argues that there is one
> particularly optimal solution  y and in an accompanying paper provides
> tables of numerical values. (He uses the analysis to describe a
> one-parameter family of functions  f_s  with  f_s o f_t = f_{s+t} and
> in particular describes a functional "square root"  f_{1/2}  of  f = f_1.)
> 
>    MR0141905 (25 #5302)
>    Szekeres, G.
>    Fractional iteration of exponentially growing functions. 
>    J. Austral. Math. Soc. 2 1961/1962 301--320.        MSC section 39.99
>  
> dave
> 
See 
http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A052122
for f(x) such that f(f(x)) = exp(x)-1.
Daniel


Create a group - Google Groups - Google Home - Terms of Service - Privacy Policy
©2009 Google