FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups 
 ProfileProfile   PreferencesPreferences   Log in to check your private messagesLog in to check your private messages   Log inLog in 
Forum index » Science and Technology » Math
Fourier Transform / Decay / Canot set
Post new topic   Reply to topic Page 1 of 1 [12 Posts] View previous topic :: View next topic
Author Message
David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Thu Jul 20, 2006 11:55 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

On 20 Jul 2006 13:51:42 -0400, hrubin@odds.stat.purdue.edu (Herman
Rubin) wrote:

Quote:
In article <e9mvt4$46l$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.

The question is what kind of decay can I expect in f as x -> infinity ?

Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (-1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.

But more can be said.

ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(n-k) pi + t/3^k)/cos(3^(n-k) pi))
= sum_{k=1}^infty ln(cos(t/3^k) - sin(t/3^k) tan(3^(n-k) pi))

The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) -> sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))

i.e. (-1)^n f(3^n pi + t) -> f(t) f(pi) as n -> infty
(and this should be uniform on, say, t in [-1,1]).
Therefore...

Therefore not much. The numbers of the form 3^n*pi+t,
t in a bounded set, have asymptotic density 0.

I believe that there is a theorem which states that
if, in the definition of f, the "3" is replaced by
2 or smaller, the resulting characteristic function
corresponds to a density.

By "correspond to a density" mean that the measure
we're taking the Fourier transform of is absolutely
continuous? If so I don't _think_ it's quite that
simple (not sure).

Quote:
So f(x)*f(sqrt(3)*x) is
at least in L^2.

What I am really interested is what L^p(R) spaces is f in ?

... None except L^infty.

John Schutkeker's shoe is safe.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


************************

David C. Ullrich
Back to top
David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Thu Jul 20, 2006 11:53 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

On Thu, 20 Jul 2006 01:59:27 EDT, don <ctcowan@sfu.ca> wrote:

Quote:
In article
13869322.1153345299269.JavaMail.jakarta@nitrogen.math
forum.org>,
don <ctcowan@sfu.ca> wrote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for
real x.

This is the Fourier transform of a measure supported
on the usual Cantor set
(middle third) but translated so symmetric about 0.


The question is what kind of decay can I expect in f
as x -> infinity ?

[...]




thanks for the help.


New question:

How about if I use a fat Cantor set instead ?

Since the measure is "smoother " I would expect the transform to have better decay.

I had a comment on this this morning, decided to look up
something I might have been misremembering.

Even for Cantor sets of measure zero, obtained by using a
constant ratio of dissection, the decay is not a monotone
function of the "smoothness", ie of the dimension!

Bizarre true fact: Say you construct a Cantor set by repeatedly
removing the middle (1-2a)-th part of the intervals, keeping
the left a-th and the right a-th. Then the Fourier transform
of that measure tends to 0 at infinity if and only if 1/a is
not a Pisot number. Pisot numbers are algebraic, so there's
only countably many of them.

In particular you can have a Cantor set that's definitely
thinner than the middle-thirds set, so the measure is in
some sense less smooth, such that the Fourier transform
does tend to zero.

(And since this is a perfectly ordinary Cantor set
with a constant ratio and measure zero and everything,
if anything appearing worse to the naked eye than the
middle thirds set, I think that someone really _should_
eat his shoe...))

Quote:
Can I expect this to be in some L^p space ? (p < inf.)



thanks


craig


************************

David C. Ullrich
Back to top
Robert B. Israel
science forum Guru


Joined: 24 Mar 2005
Posts: 2151

PostPosted: Thu Jul 20, 2006 8:25 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

In article <e9ofre$32s0@odds.stat.purdue.edu>,
Herman Rubin <hrubin@odds.stat.purdue.edu> wrote:
Quote:
In article <e9mvt4$46l$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.

The question is what kind of decay can I expect in f as x -> infinity ?

Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (-1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.

But more can be said.

ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(n-k) pi + t/3^k)/cos(3^(n-k) pi))
= sum_{k=1}^infty ln(cos(t/3^k) - sin(t/3^k) tan(3^(n-k) pi))

The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) -> sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))

i.e. (-1)^n f(3^n pi + t) -> f(t) f(pi) as n -> infty
(and this should be uniform on, say, t in [-1,1]).
Therefore...

Therefore not much. The numbers of the form 3^n*pi+t,
t in a bounded set, have asymptotic density 0.

Asymptotic density doesn't matter here. The L^p norm requires the
integral of |f|^p, not the asymptotic average of |f|^p, to be
finite. So if f is bounded away from 0 on, say, the intervals
(3^n pi - 1, 3^n pi + 1), then it can't be in L^p for any
p < infinity.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Back to top
Herman Rubin
science forum Guru


Joined: 25 Mar 2005
Posts: 730

PostPosted: Thu Jul 20, 2006 5:51 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

In article <e9mvt4$46l$1@nntp.itservices.ubc.ca>,
Robert Israel <israel@math.ubc.ca> wrote:
Quote:
In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.

The question is what kind of decay can I expect in f as x -> infinity ?

Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (-1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.

But more can be said.

ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(n-k) pi + t/3^k)/cos(3^(n-k) pi))
= sum_{k=1}^infty ln(cos(t/3^k) - sin(t/3^k) tan(3^(n-k) pi))

The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) -> sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))

i.e. (-1)^n f(3^n pi + t) -> f(t) f(pi) as n -> infty
(and this should be uniform on, say, t in [-1,1]).
Therefore...

Therefore not much. The numbers of the form 3^n*pi+t,
t in a bounded set, have asymptotic density 0.

I believe that there is a theorem which states that
if, in the definition of f, the "3" is replaced by
2 or smaller, the resulting characteristic function
corresponds to a density. So f(x)*f(sqrt(3)*x) is
at least in L^2.

Quote:
What I am really interested is what L^p(R) spaces is f in ?

... None except L^infty.

John Schutkeker's shoe is safe.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada


--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Back to top
Herman Rubin
science forum Guru


Joined: 25 Mar 2005
Posts: 730

PostPosted: Thu Jul 20, 2006 5:41 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

In article <Xns9805CD7BEA1A7lkajehoriuasldfjknak@207.115.17.102>,
John Schutkeker <jschutkeker@sbcglobal.net.nospam> wrote:
Quote:
don <ctcowan@sfu.ca> wrote in
news:13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org:

let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual
Cantor set (middle third) but translated so symmetric about 0.

The question is what kind of decay can I expect in f as x -> infinity
?

Any comments or suggestion would be greatly appreciated

If you get any kind of decay in f, as x -> infinity, I'll eat my shoe.
Taking the Fourier transform of the Cantor set is asking for trouble.

I have already commented about it not approaching 0.

However, and I have not checked this, I believe that the
measure of the set of reals between 0 and T for which
|f(x)| > c is o(T) for all c > 0.

One reason for believing this is if one takes

g(x) = product_{k=1}^infinity cos(x 4^{-k} ),

then g(x)*g(2x) is the characteristic function of a
uniform distribution.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Back to top
Herman Rubin
science forum Guru


Joined: 25 Mar 2005
Posts: 730

PostPosted: Thu Jul 20, 2006 5:19 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
Quote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set (middle third) but translated so symmetric about 0.


Quote:
The question is what kind of decay can I expect in f as x -> infinity ?

If one considers f(3^k * 2pi), this is the same for all
k and is not zero.


Quote:
What I am really interested is what L^p(R) spaces is f in ?

Certainly not for p <= 2. As f is bounded, if it is in
any L^p, it is in all higher ones, and if it were in L^2,
the measure of which it is a transform would have to have
a density.



Quote:
Any comments or suggestion would be greatly appreciated



Quote:
craig


--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
Back to top
Robert B. Israel
science forum Guru


Joined: 24 Mar 2005
Posts: 2151

PostPosted: Thu Jul 20, 2006 5:04 pm    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

don wrote:

Quote:
New question:

How about if I use a fat Cantor set instead ?

Since the measure is "smoother " I would expect the transform to have better decay.


Can I expect this to be in some L^p space ? (p < inf.)

If I understand you correctly, you're now taking the Fourier transform
of a
constant times the indicator function I_C of your "fat Cantor set".
Since I_C
is in L^1 and L^2, by the Hausdorff-Young theorem its Fourier transform
is
in L^p for 2 <= p <= infinity. But actually more can be said. Suppose
your
"fat Cantor set" is obtained by starting with an interval C_0 of length
s_0 = 1
and removing open subintervals; at the n'th stage you remove 2^{n-1}
middle
subintervals, each of length r_n < s_{n-1}, from C_{n-1}, and what
remains,
C_n, is the union of 2^n subintervals, each of length s_n, where
2 s_n + r_n = s_{n-1}.
Thus the Lebesgue measure of C_n is
m(C_n) = 2^n s_n = (1 - r_n/s_{n-1}) m(C_{n-1}).
The intersection of all C_n is C, with
m(C) = product_{n=1}^infty (1 - r_n/s_{n-1})
and for it to be "fat" you want this infinite product to be nonzero,
which is
the case iff sum_{n=1}^infty r_n/s_{n-1} < infty. The simplest way is
to take r_n/s_{n-1} = w^n for some constant w, 0 < w < 1.

Now the Fourier transform of the indicator function of an interval, say
(a,b) is
g_(a,b)(x) = (exp(-iax) - exp(-ibx)) i/x
which is in all L^p for p > 1. Noting that |g_(a,b){(x)| <= min(b-a,
2/|x|),
||g_(a,b)||_p <= c (b-a)^(1-1/p) where c is a constant (depending only
on p).
Now since C_n \ C_{n-1} consists of 2^{n-1} intervals of length r_n,
the Fourier
transform of its indicator function has L^p norm at most
2^{n-1} c r_n^(1-1/p). The L^p norm of the Fourier transform of I_C is
then
bounded by
c + sum_{n=1}^infty 2^(n-1) c r_n^(1-1/p).
Now if r_n/s_{n-1} = w^n we have r_n < w^n/2^(n-1), so this L^p norm is
bounded by c (1 + sum_{n=1}^infty 2^((n-1)/p) w^(n(1-1/p)))
which converges if w < 2^(1/(1-p)). So in this case, the Fourier
transform
is in L^p at least for infinity >= p > (ln(w) - ln(2))/ln(w) > 1.

However, if instead of making w constant we make r_n decrease rapidly
enough that (r_n/s_{n-1})^(1/n) -> 0+, we should get the Fourier
transform
in L^p for all p in (1,infty].

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Back to top
David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Thu Jul 20, 2006 10:53 am    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

On Wed, 19 Jul 2006 17:41:09 EDT, don <ctcowan@sfu.ca> wrote:

Quote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set (middle third) but translated so symmetric about 0.


The question is what kind of decay can I expect in f as x -> infinity ?

None. For example if N is a positive integer then

f(3^N 2 pi) = f(2 pi) <> 0.

Quote:
What I am really interested is what L^p(R) spaces is f in ?

None except for p = infinity. This follows from the above, plus
the fact that f' is bounded (either differentiate the formula
above or note that f' is also the Fourier transform of a
certain (complex) measure.)

Quote:
Any comments or suggestion would be greatly appreciated



craig


************************

David C. Ullrich
Back to top
don11154
science forum beginner


Joined: 05 Jul 2005
Posts: 39

PostPosted: Thu Jul 20, 2006 5:59 am    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

Quote:
In article
13869322.1153345299269.JavaMail.jakarta@nitrogen.math
forum.org>,
don <ctcowan@sfu.ca> wrote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for
real x.

This is the Fourier transform of a measure supported
on the usual Cantor set
(middle third) but translated so symmetric about 0.


The question is what kind of decay can I expect in f
as x -> infinity ?

Note that f(3x) = cos(x) f(x). In particular, f(3^n
pi) = (-1)^n f(pi)
(and that is nonzero) for every positive integer n,
so there is no
decay.

But more can be said.

ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(n-k) pi +
+ t/3^k)/cos(3^(n-k) pi))
= sum_{k=1}^infty ln(cos(t/3^k) - sin(t/3^k)
k) tan(3^(n-k) pi))

The terms for k <= n are of course ln(cos(t/3^k)),
while all terms
are dominated by c/3^k for some c (depending on t),
so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) -> sum_{k=1}^infty
ln(cos(t/3^k)) = ln(f(t))

i.e. (-1)^n f(3^n pi + t) -> f(t) f(pi) as n -> infty
(and this should be uniform on, say, t in [-1,1]).
Therefore...

What I am really interested is what L^p(R) spaces
is f in ?

... None except L^infty.

John Schutkeker's shoe is safe.

Robert Israel
israel@math.ubc.ca
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia Vancouver,
BC, Canada




thanks for the help.


New question:

How about if I use a fat Cantor set instead ?

Since the measure is "smoother " I would expect the transform to have better decay.


Can I expect this to be in some L^p space ? (p < inf.)



thanks


craig
Back to top
Robert B. Israel
science forum Guru


Joined: 24 Mar 2005
Posts: 2151

PostPosted: Thu Jul 20, 2006 4:13 am    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

In article <13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org>,
don <ctcowan@sfu.ca> wrote:
Quote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set
(middle third) but translated so symmetric about 0.

The question is what kind of decay can I expect in f as x -> infinity ?

Note that f(3x) = cos(x) f(x). In particular, f(3^n pi) = (-1)^n f(pi)
(and that is nonzero) for every positive integer n, so there is no
decay.

But more can be said.

ln(f(3^n pi + t)/f(3^n pi))
= sum_{k=1}^infty ln(cos(3^(n-k) pi + t/3^k)/cos(3^(n-k) pi))
= sum_{k=1}^infty ln(cos(t/3^k) - sin(t/3^k) tan(3^(n-k) pi))

The terms for k <= n are of course ln(cos(t/3^k)), while all terms
are dominated by c/3^k for some c (depending on t), so by the
Dominated Convergence Theorem for sums,
ln(f(3^n pi + t)/f(3^n pi)) -> sum_{k=1}^infty ln(cos(t/3^k)) = ln(f(t))

i.e. (-1)^n f(3^n pi + t) -> f(t) f(pi) as n -> infty
(and this should be uniform on, say, t in [-1,1]).
Therefore...

Quote:
What I am really interested is what L^p(R) spaces is f in ?

.... None except L^infty.

John Schutkeker's shoe is safe.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Back to top
John Schutkeker
science forum Guru Wannabe


Joined: 30 May 2005
Posts: 172

PostPosted: Thu Jul 20, 2006 12:11 am    Post subject: Re: Fourier Transform / Decay / Canot set Reply with quote

don <ctcowan@sfu.ca> wrote in
news:13869322.1153345299269.JavaMail.jakarta@nitrogen.mathforum.org:

Quote:
let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual
Cantor set (middle third) but translated so symmetric about 0.

The question is what kind of decay can I expect in f as x -> infinity
?

Any comments or suggestion would be greatly appreciated

If you get any kind of decay in f, as x -> infinity, I'll eat my shoe.
Taking the Fourier transform of the Cantor set is asking for trouble.
Back to top
don11154
science forum beginner


Joined: 05 Jul 2005
Posts: 39

PostPosted: Wed Jul 19, 2006 9:41 pm    Post subject: Fourier Transform / Decay / Canot set Reply with quote

let

f(x) = product_{k=1}^infinity cos(x 3^{-k} ) for real x.

This is the Fourier transform of a measure supported on the usual Cantor set (middle third) but translated so symmetric about 0.


The question is what kind of decay can I expect in f as x -> infinity ?


What I am really interested is what L^p(R) spaces is f in ?



Any comments or suggestion would be greatly appreciated



craig
Back to top
Google

Back to top
Display posts from previous:   
Post new topic   Reply to topic Page 1 of 1 [12 Posts] View previous topic :: View next topic
The time now is Sun Apr 21, 2019 12:58 am | All times are GMT
Forum index » Science and Technology » Math
Jump to:  

Similar Topics
Topic Author Forum Replies Last Post
No new posts A series related to the Hilbert transform larryhammick@telus.net Math 0 Fri Jul 21, 2006 3:28 am
No new posts Exponential Growth/Decay Computers Guru Math 3 Mon Jul 17, 2006 4:28 pm
No new posts non-periodic boundaries in split-step fourier for NLS kemmy num-analysis 0 Sat Jul 15, 2006 2:16 pm
No new posts split step fourier for perturbed NLS kemmy num-analysis 0 Fri Jul 14, 2006 9:20 pm
No new posts Fourier theory help Abstract Dissonance Math 1 Wed Jul 12, 2006 11:22 am

Copyright © 2004-2005 DeniX Solutions SRL
Other DeniX Solutions sites: Electronics forum |  Medicine forum |  Unix/Linux blog |  Unix/Linux documentation |  Unix/Linux forums  |  send newsletters
 


Powered by phpBB © 2001, 2005 phpBB Group
[ Time: 0.0260s ][ Queries: 20 (0.0052s) ][ GZIP on - Debug on ]