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- Newsgroups: sci.math
- Subject: Re: Linearity question
- Message-ID: <a_rubin.718923697@dn66>
- From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
- Date: 12 Oct 92 21:01:37 GMT
- References: <105268@siemens.siemens.com>
- Organization: Beckman Instruments, Inc.
- Nntp-Posting-Host: dn66.dse.beckman.com
- Lines: 62
-
- In <105268@siemens.siemens.com> kostas@siemens.com. (Konstantinos Diamantaras) writes:
-
- >Hi netters,
-
- >in the course of my research I came up with the following,
- >I think iteresting question
-
- >Suppose I have a continuous real function f(x) of a single
- >real parameter x. I want to "design" two functions which
- >I will call "addition" (+) and "multiplication" (*)
- >(clearly not the usual ones) such that f is linear in them
-
- >f(x+y)=f(x)+f(y), f(a*x)=a*f(x)
-
- >In addition, I would like these functions to have some of the
- >standard addition and multiplication properties. In particular,
-
- >(a) I want both functions to be smooth
- > (first derivative exists is enough for the moment)
- >(b) I want both to be abelian groups (with respect to the real
- > numbers)
- >(c) I want them to have the property
- > (x+y)*z = (x*z)+(y*z)
-
- Are you trying to say, in (b) that:
-
- (R,+) forms a group with identity z.
- (R - {z},*) forms a group with identity e.
-
- ?
-
- Or equivalently, that (b) and (c) state that (R,'+','*') forms a field?
-
- If so, I have a partial solution:
-
- f(x) = f(e '*' x) = x '*' f(e)
-
- Conversely, if we assume we have "+" and "*", then writing f(x) = x '*' q,
-
- f(x '+' y) = (x '+' y) '*' q = (x '*' q) '+' (y '* 'q) = f(x) '+' f(y), and
- f(a '*' x) = (a '*' x) '*' q = a '*' (x '*' q) = a '*' f(x).
-
- If we assume, as I believe must be the case, that (R,'+','*') must be
- isomorphic to (R,+,*), there is a 1-1 function k such that:
-
- k(x '+' y) = k(x) + k(y), and
- k(x '*' y) = k(x) * k(y), and, finally,
-
- k(f(x)) = k(x '*' q) = k(x) k(q).
-
- I don't want to work out the necessary conditions for k to exist here, but
- it is clear that if f has a first derivative, f(0) = 0 < f'(0) ~= 1,
- 0 < f'(x), f has no other fixed points, and f maps R onto R, then we can
- choose a smooth k with k'(0) = 1 such that k(f(x)) = k(x) f'(0).
-
-
-
- --
- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
- 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
- My opinions are my own, and do not represent those of my employer.
- My interaction with our news system is unstable; please mail anything important.
-