Subject: f(x+1) = exp(f(x))
From: Michael Leumas
Date: Mon, 20 Jan 1997 12:36:13 -0500
Define a function f(x) by
f(0) = 0; f(1) = 1; f(x+1) = exp(f(x)), x real, f continuous,
differentiable.
Is anyone familiar with work on such a function? If so, I'd appreciate
any references you could provide.
My former colleague, Enrico Federighi (now retired), asserts the
following (without proof):
Let b(x) = f(x) - ln(x+1). Then:
(1) b(-1) = ln(1 + b1), where b1 is the second term in the Taylor series
for b(x);
(2) b(x) has a radius of convergence of 2.
Subject: postdoctoral position
From: renardyy@corona.math.vt.edu (Yuriko Renardy)
Date: 20 Jan 1997 20:44:17 GMT
Position vacant: Postdoctoral Research Associate
I have funding from the National Science Foundation for a
postdoctoral research associate to work with me for approximately two
years. The project concerns interfacial dynamics in bicomponent
liquids, viscous, thermal and viscoelastic effects. I am looking for
a candidate in dynamical systems (the project may involve
bifurcations in thermal convection problems), and
numerical methods related to pdes (the project will involve modifying a
Navier-Stokes code), programming in Fortran, familiarity with
workstation environments (we have Suns, Decs and Ibm/riscs in the
department, Pcs and Macs), visualization software,
and modeling. For more on the type of work I do, my web page is
http://www.math.vt.edu/people/renardyy
The applicant should send vita and letters of reference, the fax number is 540
231 5960. Position will begin in the fall 1997.
(Applications from graduate students wishing to work with me are always
welcome. The department provides employment for graduate students as
teaching assistants. More information on this can be found on the
departmental web page http://www.math.vt.edu)
Send correspondence to:
Prof. Yuriko Renardy, Dept of Mathematics, 460 McBryde Hall,Virginia Tech,
Blacksburg, VA 24061-0123
Email: renardyy@math.vt.edu
Subject: Re: extensions of measures
From: hardy@umnstat.stat.umn.edu (Michael Hardy)
Date: 20 Jan 1997 23:37:59 GMT
In article <5b68ds$2q0@epx.cis.umn.edu>, modulo
correction of a typo, I wrote:
> Suppose B is a subalgebra of a Boolean algebra A,
> and m : B ---> [0,1] is a normalized measure. ("Normalized" means
> m(1)=1, "measure" is intended to imply that the values of m are
> always non-negative and m is countably additive.)
>
> Suppose
>
> (1) x is a member of the interval [0,1], and
>
> (2) p is a member of A but not of B, and
>
> (3) for every q in B such that q
> (4) for every q in B such that q >or= p we have m(q) >or= x.
>
> Let C be the smallest subalgebra of A that includes B and contains p.
>
> Can we conclude that m can be extended to a measure on C satisfying the
> constraint m(p)=x?
Ilias Kastansas answered
this affirmatively. I continued
> Can we conclude that m can be extended to a measure on all of A
> satisfying the constraint m(p)=x?
Then in response to this Ilias wrote:
> No. Surely we can handle finitely many p... But we cannot
> expect to extend Lebesgue measure to A = Power([0,1]), even without
> requiring translation-invariance (e.g. it is consistent that the
> continuum is aleph_1).
But I'm not sure this is a legitimate counterexample, because
Lebesgue measure on the sigma-algebra of Lebesgue measurable sets is
not quite a "measure on a Boolean algebra". This term is usually
defined so as to require measures to be strictly increasing, i.e. if
a < b then m(a) < m(b). Treat two Lebesgue measurable sets as
equivalent if they differ by a set of measure zero, and look at the
measure you get on the algebra of equivalence classes, and that's a
"measure on a Boolean algebra".
Mike Hardy
Michael Hardy
hardy@stat.umn.edu
Subject: Re: f(x+1) = exp(f(x))
From: robin@fireant.ma.utexas.edu (Robin Michaels)
Date: 21 Jan 1997 14:44:09 GMT
Michael Leumas (michael.leumas@jhuapl.edu) wrote:
: Define a function f(x) by
:
: f(0) = 0; f(1) = 1; f(x+1) = exp(f(x)), x real, f continuous,
: differentiable.
:
: Is anyone familiar with work on such a function? If so, I'd appreciate
: any references you could provide.
:
: My former colleague, Enrico Federighi (now retired), asserts the
: following (without proof):
:
: Let b(x) = f(x) - ln(x+1). Then:
:
: (1) b(-1) = ln(1 + b1), where b1 is the second term in the Taylor series
: for b(x);
:
: (2) b(x) has a radius of convergence of 2.
This is rather unexpected, because f is highly non-unique.
If we write F for the operation x-> f(x)
and E for x-> exp(x)
and S for x-> x+1
We want to find F related to a C1 function satisfying:
FS = EF
Since f(x) will be monotonic, we can find an inverse, so
FS(F^-1) = E
Now, let k(x) be a smooth function with k(x)+1=k(x+1) and
|k'(x)|<1 For example k(x)=x+w sin(2\pi x) with |w|<1.
Then if K denotes x -> k(x), we have
(FK) S ((K^-1)(F^-1))= F (KS(K^-1)) (F^-1) = FS(F^-1) = E
since K commutes with S.
Hence if f(x) is a suitable function, so is f(x+w sin( 2\pi x) )
which is somewhat annoying.
However, I suspect that if we require that f is analytic this forces a
unique solution.
This subject has been discussed twice in 'Eureka', the journal of the Cambridge
University (Undergraduate) Mathematics Society.
Once was in my article in (I think) issue 53 (on Fractional Iterations of
Functions), and once was in one of the
earliest editions; I can't recall which. Perhaps someone with access to
old Eurekas could have a look ? I seem to recall that one of the members of
the group that squared the square was involved, but I don't recall precisely
who. (either Tutte or Tukey, I think, though)
Cheers,
Robin Michaels
Subject: f(x) * f(-x): please help!
From: rdawson@husky1.stmarys.ca (Robert Dawson)
Date: 21 Jan 1997 13:54:07 GMT
I'm working on a problem that I can reduce to the form: is a function
characterized by its convolution with its 'reflection'? More precisely,
let Tf be the convolution of f with x |--> f(-x). We can also write this as
Tf(y) = \int f(x) f(x+y) dx.
Of course, this is not 'really' invertible; Tf = T(x |--> f(-x)) =
T(x|-->f(x+c)); the transform takes a function, its shifts, and its
reflections to the same transformed function. The question is: is the
transform "otherwise" invertible? In other words, given Tf, can we
determine f up to shift and reflection? The function f is always
real and nonnegative, and we may also assume, if helpful, that its integral
from -infinity to infinity converges.
I suspect that this transform must be well known in the right
places, and probably has been for a century. That's why I thought I should
ask before spending too much time on it. I do know that it is invertible
up to shift & reflection for linear combinations of characteristic
functions.
Attempting to use the Fourier transform to invert T, as we might
with f*f, does not quite work - at least not directly. The Fourier transform
is conjugated by reflection, so that the transform of Tf is the square of the
absolute value of that of f. This loses phase information and so is not 1-1;
the inverse Fourier transform of any function with the right absolute value at
each point is clearly a solution. [For instance, -f is a solution.] However, I
suspect that it can be shown that with the additional assumption of
nonnegativity (or perhaps something a little stronger) the solution is unique
up to reflection and shift.
This is not terribly complicated and I imagine that I could beat
it to death with a stick (perhaps using the near-invertibility on non-negative
simple functions with bounded support, which are dense in the non-negative
functions,to show that the Fourier transform is well-behaved.) However, it
does not look like the sort of thing that would have been ignored for all
these years; can anybody provide me with a name, a reference, etc? Reinventing
the wheel is fine as recreation, but not necessarily a good use of research
time!
AdvTHANKSance!
-Robert Dawson
Subject: Re: f(x) * f(-x): please help!
From: edgar@math.ohio-state.edu (G. A. Edgar)
Date: Tue, 21 Jan 1997 13:49:28 -0600
In article <5c2hpv$88a@News.Dal.Ca>, rdawson@husky1.stmarys.ca (Robert
Dawson) wrote:
> I'm working on a problem that I can reduce to the form: is a function
> characterized by its convolution with its 'reflection'? More precisely,
> let Tf be the convolution of f with x |--> f(-x). We can also write this as
> Tf(y) = \int f(x) f(x+y) dx.
>
> Of course, this is not 'really' invertible; Tf = T(x |--> f(-x)) =
> T(x|-->f(x+c)); the transform takes a function, its shifts, and its
> reflections to the same transformed function. The question is: is the
> transform "otherwise" invertible? In other words, given Tf, can we
> determine f up to shift and reflection? The function f is always
> real and nonnegative, and we may also assume, if helpful, that its integral
> from -infinity to infinity converges.
In general, the answer is "no". The modulus of the Fourier transform is
determined, but the phase is not. The question then becomes, can
two (appropriate--say, integrable) functions have Fourier transforms with
the same modulus? In the literature, this problem may be called "phase
retrieval".
In physics, this includes the problem of reconstructing a crystal from
its x-ray diffraction pattern--Can two different crystals yield
the same diffraction pattern? A. L. Patterson (Physical Review,
vol. 65, 1944) showed that such ambiguity is possible, at least
theoretically.
My former colleague Joe Rosenblatt knows a lot more about this
than I do. See, for example:
Rosenblatt, Joseph, "Phase retrieval"
Comm. Math. Phys. 95 (1984), no. 3, 317--343.
--
Gerald A. Edgar edgar@math.ohio-state.edu
Subject: Eigenvalues of infinite matrix
From: Tom Chou
Date: 21 Jan 1997 13:27:33 -0800
Hello All,
Thanks for the responses on possible things to try for
my original problem of finding eigenvalues of an infinite
unsymmetric matrix.
I rewrote it in terms of a generalized eigenval problem then applied
Cholesky decomposition and reformulated it as a regular eigenvalue
problem with a SYMMETRIC, real, positive definite, infinite matrix.
NOW, does anyone know where to look or knows if convergence
can be proved? Can the lowest eigenvalues (analytically or numerically)
be found exactly? (The spectrum is unbounded above). Can density of
spectrum be found for the inifinite values? In short
what can be said at all, in addition to my numerically finding
some of the lower eigenvals?
The off disgonal elements are still larger within a row or column.
ie, diagonals go as M_ii \sim i^5, Off diagonals as
M_ij \sim i^{5/2}j^{5/2}/(|i-j|)
Thanks folks,
Cheers,
Tom Chou
DAMTP, University of Cambridge
