From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: Smooth nowhere analytic functions
Date: 21 Jan 1999 14:02:25 -0500
Newsgroups: sci.math
In article <785k66$fh4@gap.cco.caltech.edu>,
Douglas J. Zare wrote:
>It is easy to construct functions of a real variable which are smooth and
>nowhere analytic. Most constructions look quite contrived, and I'm looking
>for a natural occurence of such a function to describe to calculus
>students who are not mathematics majors. Perhaps I should rule out
>infinite sums or products, so the following is considered too contrived:
>
>inf / - csc^2 (2^n x) / \
>Sum | e / c_n |
>n=1 \ / /
>
>where c_n is chosen to grow sufficiently rapidly.
>
>Thank you for any suggestions,
>
>Douglas Zare
Occasionally, I give a short lecture within an appropriate course about
the much-abused adjective "natural" (its logical independence from
concepts such as "simple", "applicable", "beneficial", and other
heart-warming categories).
Next: Other than rational functions and their close relatives (the
absolute value function etc.), other interesting functions are (sometimes
locally, piecewise or otherwise restricted) limits of polynomials anyway.
Where should one draw the line between the exponential function (perceived
by many as "natural") and "too contrived functions"?
Having said that, would a Fourier series do the job?
It must have been suggested by others, because it resembles Weierstrass's
example of a continuous, nowhere differentiable function.
Take F(x) = sum[n=0 to infinity] exp(i * 2^n * x) / n!
Both Real(F) and Imag(F) are in this category, and F is easier to handle.
Derivatives: F^(k) (x) = exp(2^n * i * x) * (i * 2^n)^k / n!
uniformly convergent everywhere, hence continuous.
Non-analycity at 0: F^(k)(0) = i^k * exp(2^k)
they grow too fast in magnitude to make coefficients of a function
analytic at 0.
Non-analyticity at other points: The same growth can be observed at all
points pi*b where b is a number with terminating binary representation.
These numbers are dense in R, while analyticity at a point implies
analyticity in an open neighbourhood.
Hope it is not too contrived, ZVK(Slavek).
==============================================================================
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: FAQ error: W _is_ analytic
Date: 29 Sep 1999 11:56:55 -0400
Newsgroups: sci.math
In article <7st45f$1fs$1@nnrp1.deja.com>,
Frédéric van der Plancke wrote:
[...]
>> Try (paying homage to Weierstrass)
>>
>> sum[n=0 to infinity] cos(x*2^n)/n!
>>
>> (a neat exercise on radius of convergence and periodicity)
>
>Thanks for your example - it looks already much more natural
>than mine. To prove it right, I computed the derivatives
>(as sum of the terms derivatives) and proved they grow too
>fast. (Hmm... did not check all details actually)
>But I don't have a clue about your "periodicity" proof. Can
>you give some ?
First, you write the function as the real part of
F(x) = sum[n=0 to infinity] exp(i * x * 2^n) / n!
to make differentiation easier. Then you calculate exactly the derivatives
at x = 0 alone. Check out that
F^(k)(0) = i^k * exp(2^k)
Cauchy-Hadamard formula for the radius of convergence gives, with
conventions about infinities,
1/r = lim sup [k to infinity] (abs(F^(k)(0))^(1/k)
and so r = 0. The same will be true about real parts of the derivatives
(the even-indexed ones have the same magnitude).
This says that the power series at x=0 has radius of convergence 0.
F is 2*pi - periodic, so the same is true for every x which is an integer
multiple of 2*pi.
Throw away the first r terms, and you get a function with period
(2*pi)/2^r, and it has points of non-analyticity at corresponding
fractions of 2*pi. So does F and Real(F).
The points of non-analyticity are now found to be dense in R, and there is
no room for intervals of analyticity.
Cheers, ZVK(Slavek).
==============================================================================
From: jeanfi61
Subject: marginal remark
Date: Thu, 30 Sep 1999 01:56:30 +0200
Newsgroups: sci.math
"Frédéric van der Plancke" a écrit :
> Some time ago I wanted to find a "natural" example of infinitely
> derivable function that is nowhere analytic.
As a marginal remark, it can be proved that, in some sense, "almost all"
infinitely derivable functions [0,1]-->[0,1] are nowhere analytic.
More precisely, writing ||f|| for the uniform norm of f
(||f||=sup{ |f(x)| ; x a real number }), and
d(f,g)=sum[n => 0][min(2^{-n} ; ||f^(n) - g^(n)||)],
then d is a distance on the space of infinitely derivable functions
[0,1]-->[0,1], which makes it complete, and an astute application of the
Baire theorem leads to the fact the subset of functions which are analytic
at one point (or more) is a set of first category, thus nowhere dense.
jeanfi61
==============================================================================
From: jeanfi61
Subject: Re: marginal remark
Date: Thu, 30 Sep 1999 06:58:22 +0200
Newsgroups: sci.math
jeanfi61 (myself) wrote :
> Ö an astute application of the Baire theorem leads to the fact the subset
> of functions which are analytic at one point (or more) is a set of first
> category, thus nowhere dense.
It's obviously a mistake : a first category subset of a Baire space has no
interior (instead of : "is nowhere dense" ó think of Q, subset of R, for
example).
This was pointed out to me by e-mail by an attentive reader, with some
precious complements :
> Ö
>
> First category, yes. Nowhere dense, no. (Everywhere dense, in fact.)
>
> This was first proved by Dietrich Morgenstern [Math. Nachr.
> 12 (1954), 74] and later rediscovered by several others.
>
> The following on-line paper may be of interest:
>
> Kent G. Merryfield, "A nowhere analytic C^infinity function",
> Missouri Journal of Mathematical Sciences, Vol. 4 No. 3
> (Fall 1992), 132-138
> [.ps file]
> [.dvi file]
>
> Merryfield's paper gives a specific construction
> of a nowhere analytic C^infinity function, not the
> stronger Baire category result.
>
> Dave L. Renfro
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