monta@cmu-cs-g.ARPA (Peter Monta) (01/27/85)
A nice application of spectral theory is the definition of continuous
functions of linear operators like the square root and the exponential. I
got to wondering whether anything could be said about nonlinear operators.
Can you find, for instance, a square root of the exponential on the reals?
That is, find a function f from the reals to the reals such that
x
f(f(x)) = e for all real numbers x? I sketched for a while on graph paper,
and it seems to me that something like this really should exist: limit at
minus infinity of -1, f(-1)=0, f(0)=1/e, f strictly increasing. Failing
an explicit answer, is there a way to compute f?
Peter Monta
ARPA: monta@cmu-cs-g
UUCP: ...!rochester!cmu-cs-pt!cmu-cs-g!montalambert@boring.UUCP (01/28/85)
A general construction for defining a function f such that
f(f(x)) = g(x) for a given strictly increasing function g
was given in
G.H. Hardy, Orders of Infinity, Cambridge Tracts in
Math. and Math. Physics, No. 12, University Press,
Cambridge, England, 1910 (2nd ed. 1924).
Start with a pair of reals x0 and x1 such that x0 < x1, and
define f on [x0,x1] by taking any strictly increasing
function satisfying
f(x0) = x1, f(x1) = g(x0). (1)
(So we can start by taking the linear function on the
interval [-1, 0] such that f(-1) = 0 and f(0) = 1/e.)
Let h stand for the functional inverse of f. The function h
is now defined on the interval [x1, g(x0)]. By using the
identity
f(x) = g(h(x)) (2)
we know the values of f(x) on the interval [x1, g(x0)]. The
function h is now also defined on [g(x0), g(x1)] and the
process can be repeated indefinitely. For determining f(x)
for values of x < x0, basically the same method is used, but
now by using, instead of (2), the identity
f(x) = h(g(x)). (3)
Since (even for fixed values x0 and x1) there are many
functions satisfying (1), the solution is by no means
unique. However, it turns out that for the square root of
the exponential function there is an analytic solution (i.e.,
a function f such that f is analytic):
H. Kneser, Reelle analytische Loesungen der Gleichung
phi(phi(x)) = e^x und verwandter Funktionalgleichungen
(Real analytic solutions of the equation phi(phi(x)) =
e^x and related functional equations), J. Reine Angew.
Math. 187 (1948), 56-67.
In fact, Kneser solves a more general problem: he constructs
a function PSI satisfying
PSI(exp(z)) = PSI(z) + 1.
The function PHI defined by
PHI(z) = PSI^-1.[PSI(z)+1/2]
is analytic and real over the full real axis, and satisfies
PHI(PHI(z)) = exp(z).
In general, we can now define
(exp^a)(x) = PSI^-1.[PSI(x)+a],
and we have
(exp^1)(x) = exp(x),
(exp^a)((exp^b)(x)) = (exp^(a+b))(x).
--
Lambert Meertens
...!{seismo,philabs,decvax}!lambert@mcvax.UUCP
CWI (Centre for Mathematics and Computer Science), Amsterdambill@ur-cvsvax.UUCP (Bill Vaughn) (01/29/85)
> A nice application of spectral theory is the definition of continuous > functions of linear operators like the square root and the exponential. I > got to wondering whether anything could be said about nonlinear operators. > > Can you find, for instance, a square root of the exponential on the reals? > That is, find a function f from the reals to the reals such that > x > f(f(x)) = e for all real numbers x? I sketched for a while on graph paper, > and it seems to me that something like this really should exist: limit at > minus infinity of -1, f(-1)=0, f(0)=1/e, f strictly increasing. Failing > an explicit answer, is there a way to compute f? > > Peter Monta Z.A. Melzak's book 'Bypasses' (Wiley,1983) contains some interesting ideas concerning this problem (see pp.89-90). The idea is to express the exponential function as a conjugate of a simpler function like multipilcation. Unfortunately he doesn't give an explict solution and the problem of uniqueness is up in the air, but at least it's clear that something like the animal you suggest can be defined i.e. if exp(x) = fbf^-1(x) where f is a function and b a constant then the ath iterate of exp can be expressed as fb^af^-1 where now it makes sense to talk of any real a. Therefore the 1/2 iterate of exp is fb^1/2f^-1. The function f has complex coefficients in its power expansion and the value b is 0.318150 + 1.337236i (this is the smallest magnitude root of the equation exp(x)=x). The rest of the derivation is sketchy and most of it left as an exercise which I haven't the time or inclination to do. By the way, I highly recommend this book. It's got an interesting mathematical/philosphical bent to it. His point is that the concept of conjugacy is a very deep one and occurs in many fields. It is also a technique which helps makes intractable problems tractable. See also 'Companion to Concrete Mathematics', Z.A. Melzak, Wiley, 1983, pp. 51-80.
das@ucla-cs.UUCP (02/07/85)
About a dozen years ago, Bruce Resnick (sp? It's been a while) as a Caltech
undergrad did some investigation of f such that f(f(x)) = sin x. Being an
incurable punster, the subtitle of the paper was "What you have to do twice
in order to sin".
(Bruce is the same fellow who, on the eve of his 20th birthday, was frantic.
When asked why, he explained, "I've got to prove something worthwhile!".
"Why?" "So I can say I did it when I was a teenager.")
On that theme, trivia time: Who killed Galois? What were they fighting over?
What became of the man who killed him? (I don't know the answers.)
-- David Smallberg, das@ucla-cs.ARPA, {ihnp4,ucbvax}!ucla-cs!dasdas@ucla-cs.UUCP (02/07/85)
About a dozen years ago, Bruce Reznick (sp? It's been a while) as a Caltech
undergrad did some investigation of f such that f(f(x)) = sin x. Being an
incurable punster, he subtitled the paper "What you have to do twice in order
to sin".
(Bruce is the same fellow who, on the eve of his 20th birthday, was frantic.
When asked why, he explained, "I've got to prove something worthwhile!".
"Why?" "So I can say I did it when I was a teenager.")
On that theme, trivia time: Who killed Galois? What were they fighting over?
What became of the man who killed him? (I don't know the answers.)
-- David Smallberg, das@ucla-cs.ARPA, {ihnp4,ucbvax}!ucla-cs!das