From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: "Composite Root" of a function
Date: 10 Feb 1999 14:02:14 -0500
Newsgroups: sci.math
Keywords: Solve f o f = exp
In article <36C1B5F1.72DB8AE2@flash.net>,
Matt McLelland wrote:
>By composite root of a function g, I mean a function f such that
>f(f(x))=g(x).
>
>Is there a composite root of e^x over the reals?
>Is there any easy way to characterize those functions which have
>composite roots?
This question has come up a few times before, so I keep a canned
quotation. Good luck, ZVK(Slavek).
Marek Kuczma: Functional Equations in a Single Variable
Monografie Matematyczne 46, Warsaw 1968
In Ch. XV, Sec. 6, he writes [modified for ASCII format]:
"For the equation
(*) f^2(x) = e^x,
a real analytic solution has been found by H. Kneser.
This solution, however, is not single-valued (Baker)
and, as pointed out by G. Szekeres, there is no
uniqueness attached to the solution. It seems reasonable
to admit f(x)=F^(1/2)(x), where F^u is the regular
iteration group of g(x)=e^x, as the "best" solution of
the equation (*) (best behaved at infinity). However,
we do not know whether this solution is analytic for
x>0.
[Kuczma defines and discusses regular iterations at
infinity in Chapter IX, Sec 5.]
References:
Baker, I.N.: The iteration of entire transcendental
functions and the solution of the functional equation
f(f(z))=F(z), Math. Ann. 120(1955), pp. 174-180
Kneser, H.: Reele analytische Loesungen der Gleichung
f(f(x))=e^x und verwandten Funktionalgleichungen,
J. reine angew. Math. 187(1950), pp. 56-67
Szekeres, G.: Fractional iterations of exponentially
growing functions,
J. Australian Math. Soc. 2(1961/62), pp. 301-320
==============================================================================
From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: Hard function to plot?
Date: 12 Mar 1999 16:36:02 GMT
Newsgroups: sci.math.num-analysis
In article <36E93472.504A429E@ic.ac.uk>,
Raphael Clifford writes:
|> If f(f(x)) = e^x then how would I plot f(x) given that I don't know how
|> to solve for f explicitly? Is there a numerical method I could apply?
|> Thanks in advance,
|> Raphael
|>
this function has been discussed previously in this group.
for a list of references see
H. Kneser, Reele analytische L\"osungen der Gleichung $\phi(\phi(x))=e^x$.
J. f. reine angew. Math. 187, (1950) 56--67.
G. Szekeres, Fractional iteration of exponentially growing functions. J.
Australian Math. Soc. 2, (1962) 301-320.
M. Kuczma, Functional Equations in a Single Variable,
PWN-Polish Scientific Publishers, Warsaw, 1968.
C. Horowitz, Iterated logarithms of entire functions. Israel J. Math. 29
(1978) 31--42.
M. Kuczma and B. Choczewski and R. Ger, Iterative Functional Equations,
Cambridge University Press, Cambridge, 1990.
P. Walker, Infinitely differentiable generalized logarithmic and exponential
functions, Mathematics of Computation 57 (1991) 723-733.
K. Iga, Continuous half-iterates of functions, Manuscript.
URL
you cannot directly solve this equation, hence even such nice commands
as ImplicitSolve from Mathematice would not help. An idea which
comes into mind is series expansion. you assume a series expansion
of f, substitute it into the x-position of itself, and expand.
this gives a sequence of (nonlinear eqautions) for the coefficients
which hopefully can be solved uniquely using the information from the
functional equation. I have not tried that. good luck
peter
==============================================================================
From: ikastan@sol.uucp (ilias kastanas 08-14-90)
Subject: Re: solutions to composite equations
Date: 6 May 1999 22:04:13 GMT
Newsgroups: sci.math
In article <7gltcn$161$1@gannett.math.niu.edu>,
Dave Rusin wrote:
@For the umpteenth time the discussion of solutions to g o g = exp
@has arisen.
@
@theodore hwa wrote that g cannot be entire:
@>An entire function must always grow at least as fast as the function
@>(a constant times) |z|, or else it is constant. Obviously h is not
@>constant, so h(e^h(z)) must grow like c|e^(d|z|)|, c,d constants, so it
@>cannot be equal to the right side of (*) which has linear growth.
@
@Polya, ca 1926, one of the London journals. If anyone has a precise
@reference handy I'd appreciate it.
@
If f, g are entire and f(g()) has finite order then either
g is a polynomial and f has finite order, or else g is not a polynomial,
g has finite order, and f has order 0. "On an integral function of
an integral function", Journal London Math. Soc. 1 (1926), p. 12-15.
Ilias
==============================================================================
From: ikastan@sol.uucp (ilias kastanas 08-14-90)
Subject: Re: f(f(x)) = g(x)
Date: 11 May 1999 04:27:46 GMT
Newsgroups: sci.math
In article <3736F3A5.B9C95656@hut.fi>,
Pertti Lounesto wrote:
@If g(x) = exp(x), I presume there is no everywehere analytic solution f
@to f(f(x)) = g(x). The best approximation, with 20 points in the interval
@-0.5 < x < 1.5, I could find, was
@
@ f(x) = 0.387+1.207x+0.0965x^2+0.00109x^3+0.00534x^4,
@
@which falls off considerably.
Right; a power series convergent for all x would yield an f(z)
that would be entire (z in C). But f(f(z) = exp(z) is impossible, by
Polya's theorem (recent post).
If we only ask that f be continuous, there are lots of solutions.
The technique is similar to one used by Hardy (if not by du Bois-Reymond):
Pick an a, 0 < a < 1. We'll have f(0) = a, and hence f(a) = 1. Pick
a continuous, increasing h on [0,a] s.t. h(0) = a, h(a) = 1 and let
f(x) = h(x) there. Extend f to [a,1] by f(x) = exp(f^-1 (x)). Then
extend it likewise to [1,e^a], to [e^a,e]... and so on. (You can similarly
extend it in the other direction, left of 0, by f(x) = f^-1 (exp(x)) ).
E.g. h(x) = x + 1/2.
Can f be differentiable? Sure. Just pick h differentiable, and
s.t. at "break" points left-hand f' = right-hand f'. For this, you need
h'(a) = 1/h'(0) (easy computation). E.g. h(x) = 27/16 x^2 + 1/2 x + 4/9.
Clearly, this can go on; we can make f be C^2, C^3, ...
So... can f actually be real-analytic (C^w)? (c.f. 1/ 1+z^2 has
a pole at z = i... but 1/ 1+x^2 is C^w on R; power series depending
on x). Here is a slightly extravagant argument that the answer is "yes": Rou-
ghly, let E be the equivalence relation induced on R by exp (x equivalent
to exp(x) equivalent to exp(exp(x))...). The quotient R/E is a real-analytic
manifold, diffeomorphic to (the usual real-analytic) S_1. By a theorem due
(I think) to H. Whitney, any two such C^w structures are isomorphic; i.e.
there is a C^w f_1 carrying one into the other. Using f_1 we can lift
the antipodal mapping of S_1 ( x |-> x + 1/2, S_1 being R mod 1) and get
an f as desired.
Ilias
==============================================================================
From: hwatheod@leland.Stanford.EDU (theodore hwa)
Subject: Re: Numerical solutions to composite equations
Date: 2 May 1999 02:21:22 GMT
Newsgroups: sci.math
John Creighton (jwcrght@mta.ca) wrote:
:
: But does this matter when g(x)=e^h(x)=e^(ln|z|) * e^(i*arg(z)). The second term of this product
: will give the same value no matter if arg(z)=c, c+pi, c=2+2pi ect.
But the function doesn't work, because your steps leading to it were
not valid due to the multi valuedness of the logarithm.
Your function above is g(z) = |z| e^(i * arg(z)). If w is real,
then according to this definition, g(w) = w. Then g(g(w)) = w, not e^w.
So this g(z) is not a solution to g(g(z)) = e^z.
e^z is a counterexample, if we require the solution function g(z) to be
entire.
Suppose g(g(z))=e^z. First assume g has no zeroes, then g has an entire
logarithm, so that g(z) = e^h(z). Then we have g(g(z)) = e^(e^(h(z)).
e^(e^(h(z)) = e^z, then e^(h(z)) = z + 2 pi k i, where k must be an
integer, and therefore independent of z since everything else is in the
function is continuous. This is a contradiction since the right side has
a zero at -2pi k i, but the left side does not. (My earlier post about
e^(h(z)) giving a continuous branch of logarithm is incorrect.)
Finally, suppose g does have zeroes. Then since g(g(z)) has no zeroes, if
b is a zero of g(z), then g(g(z)) = e^z implies g must omit the value b.
Since g omits the value b, it has the form g(z) = b + e^(h(z)).
If b = 0 we are in the previous case again. Otherwise we get
g(g(z)) = e^z
b + e^(b + e^(h(z))) = e^z
e^(b + e^(h(z))) = (e^z) - b
which is again a contradiction since the right has zeroes for any value of
z = ln b, but the left side is an exponential and so does not.
Ted