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Christian S.
Guest
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 Posted:
Mon Nov 21, 2005 1:14 am Post subject:
Re: "Trivial" |
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On Sun, 20 Nov 2005 13:35:39 -0500, Jerry Avins <jya@ieee.org> wrote:
| Quote: | David C. Ullrich wrote:
...
No, in mathematics "trivial" does not mean commonplace
or ordinary. It doesn't even mean "obvious" or "easy",
quite.
Interesting! What does it mean? Even in common parlance, the word has
strayed far from its Latin roots.
Isolated farms were what one typically found along roads, (vias).
Services for travelers concentrated at crossroads where the
opportunities for business were naturally greater. Settlements often
developed there. From the road configuration, such towns were described
as "crucial".
Forks and T intersections tended to spawn what we would call a hick
town, rather than a town. After the intersection of three roads, such
towns were described as "trivial".
News tended to be classified according the layout of where it came from.
...
Jerry
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See here
http://mathworld.wolfram.com/Trivial.html
Chris |
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Randy Yates
Guest
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| Posted:
Mon Nov 21, 2005 1:15 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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David C. Ullrich <ullrich@math.okstate.edu> writes:
| Quote: | On Sat, 19 Nov 2005 16:35:06 GMT, Randy Yates <yates@ieee.org> wrote:
David C. Ullrich <ullrich@math.okstate.edu> writes:
[...]
I suppose that the term "compact support" might not
be familiar, sorry. But that doesn't make the argument
itself non-trivial,
[...]
David, I disagree. The term "trivial" means "commonplace, ordinary":
No, in mathematics "trivial" does not mean commonplace
or ordinary. It doesn't even mean "obvious" or "easy",
quite.
|
So say you.
--
% Randy Yates % "Though you ride on the wheels of tomorrow,
%% Fuquay-Varina, NC % you still wander the fields of your
%%% 919-577-9882 % sorrow."
%%%% <yates@ieee.org> % '21st Century Man', *Time*, ELO
http://home.earthlink.net/~yatescr |
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Jaap Spies
Guest
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| Posted:
Mon Nov 21, 2005 5:15 pm Post subject:
Re: "Trivial" |
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José Carlos Santos wrote:
| Quote: | On 21-11-2005 15:02, Jaap Spies wrote:
There is that old story about (my) professor Freundenthal, in a lecture
he once said: "this is trivial". Hesistating, starting to think,
leaving the lecture room. After a while, maybe 10 minutes later, he
came back
with the words: "yes it is trivial". Continuing the lecture.
I think this story is told around the world with different names, but
it is one of my favourites when it comes to trivia(l).
I knew this story with Hardy instead of Freundenthal.
Best regards,
Jose Carlos Santos
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Hardy is certainly the more famous! Obfuscating by misspelling the
name of Hans Freudenthal.
Jaap Spies |
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David C. Ullrich
Guest
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| Posted:
Mon Nov 21, 2005 5:16 pm Post subject:
Re: "Trivial" |
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On Sun, 20 Nov 2005 13:35:39 -0500, Jerry Avins <jya@ieee.org> wrote:
| Quote: | David C. Ullrich wrote:
...
No, in mathematics "trivial" does not mean commonplace
or ordinary. It doesn't even mean "obvious" or "easy",
quite.
Interesting!
|
Also true. For example trivial results in some field that
almost nobody knows anything about are certainly not
commonplace.
| Quote: | What does it mean?
|
Not the sort of thing where one could give a mathematically
precise definition. Something like "easy, once you see it,
if you have the right background"...
| Quote: | Even in common parlance, the word has
strayed far from its Latin roots.
Isolated farms were what one typically found along roads, (vias).
Services for travelers concentrated at crossroads where the
opportunities for business were naturally greater. Settlements often
developed there. From the road configuration, such towns were described
as "crucial".
Forks and T intersections tended to spawn what we would call a hick
town, rather than a town. After the intersection of three roads, such
towns were described as "trivial".
News tended to be classified according the layout of where it came from.
...
Jerry
|
************************
David C. Ullrich |
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David C. Ullrich
Guest
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| Posted:
Mon Nov 21, 2005 5:16 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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On Sun, 20 Nov 2005 20:13:02 GMT, Randy Yates <yates@ieee.org> wrote:
| Quote: | David C. Ullrich <ullrich@math.okstate.edu> writes:
On Sat, 19 Nov 2005 16:35:06 GMT, Randy Yates <yates@ieee.org> wrote:
David C. Ullrich <ullrich@math.okstate.edu> writes:
[...]
I suppose that the term "compact support" might not
be familiar, sorry. But that doesn't make the argument
itself non-trivial,
[...]
David, I disagree. The term "trivial" means "commonplace, ordinary":
No, in mathematics "trivial" does not mean commonplace
or ordinary. It doesn't even mean "obvious" or "easy",
quite.
So say you.
|
Lemme check... yes, you're right: so say I.
You never saw the cartoon / heard the joke? Two
mathematicians talking. One says something is trivial.
The other doesn't see it right away. After a half hour
of intense scribbling at the blackboard he finally
says "yeah, you're right, it _is_ trivial".
Whether you approve of that use of the word or not,
the reason it's a joke is that it _happens_. Which
says something about the meaning of the term "trivial"
in mathematics.
************************
David C. Ullrich |
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Robert Low
Guest
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| Posted:
Mon Nov 21, 2005 5:16 pm Post subject:
Re: "Trivial" |
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David C. Ullrich wrote:
| Quote: | Not the sort of thing where one could give a mathematically
precise definition. Something like "easy, once you see it,
if you have the right background"...
|
It's the sort of thing that's obvious to the sort of
person who finds that sort of thing obvious.
Or something like that... |
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Jaap Spies
Guest
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| Posted:
Mon Nov 21, 2005 5:16 pm Post subject:
Re: "Trivial" |
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David C. Ullrich wrote:
| Quote: | On Sun, 20 Nov 2005 13:35:39 -0500, Jerry Avins <jya@ieee.org> wrote:
David C. Ullrich wrote:
...
No, in mathematics "trivial" does not mean commonplace
or ordinary. It doesn't even mean "obvious" or "easy",
quite.
Interesting!
Also true. For example trivial results in some field that
almost nobody knows anything about are certainly not
commonplace.
What does it mean?
Not the sort of thing where one could give a mathematically
precise definition. Something like "easy, once you see it,
if you have the right background"...
|
There is that old story about (my) professor Freundenthal, in a lecture
he once said: "this is trivial". Hesistating, starting to think, leaving
the lecture room. After a while, maybe 10 minutes later, he came back
with the words: "yes it is trivial". Continuing the lecture.
I think this story is told around the world with different names, but
it is one of my favourites when it comes to trivia(l).
Jaap Spies |
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José Carlos Santos
Guest
|
| Posted:
Mon Nov 21, 2005 5:16 pm Post subject:
Re: "Trivial" |
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On 21-11-2005 15:02, Jaap Spies wrote:
| Quote: | There is that old story about (my) professor Freundenthal, in a lecture
he once said: "this is trivial". Hesistating, starting to think, leaving
the lecture room. After a while, maybe 10 minutes later, he came back
with the words: "yes it is trivial". Continuing the lecture.
I think this story is told around the world with different names, but
it is one of my favourites when it comes to trivia(l).
|
I knew this story with Hardy instead of Freundenthal.
Best regards,
Jose Carlos Santos |
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mebden
Guest
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| Posted:
Thu Nov 24, 2005 5:16 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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I can't believe we don't have a complete answer after 6 pages of text.
The closest is David's, but there is one thing missing. Apparently "it'
quite easy to show that if the signal is time-limited then the Fourie
transform extends to an entire function in the plane". Well, once we ar
shown why this is easy, we are done.
So close.... |
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robert bristow-johnson
Guest
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| Posted:
Fri Nov 25, 2005 11:04 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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in article P9idnemUg6NaeRjeRVn-jA@giganews.com, mebden at mebden@yahoo.com
wrote on 11/24/2005 11:05:
| Quote: | I can't believe we don't have a complete answer after 6 pages of text.
The closest is David's, but there is one thing missing. Apparently "it's
quite easy to show that if the signal is time-limited then the Fourier
transform extends to an entire function in the plane". Well, once we are
shown why this is easy, we are done.
So close....
|
mebden, you didn't cross post to sci.math where the real anal guyz are.
if someone can prove to me that for an entire function (one that is
analytic, or has all derivatives existing, everywhere in the complex plane)
that if it is zero for any segment longer than a single point, then it is
zero everywhere, i think i can make the case because it isn't too hard to
show that a timelimited function of time (with nonzero and finite energy)
has, for its Fourier Transform, analytic spectrum. if the spectrum was
bandlimited, then it would be exactly zero for a segment of frequency and,
being analytic, that would mean it was identically zero.
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge." |
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Jerry Avins
Guest
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| Posted:
Fri Nov 25, 2005 11:42 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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robert bristow-johnson wrote:
| Quote: | in article P9idnemUg6NaeRjeRVn-jA@giganews.com, mebden at mebden@yahoo.com
wrote on 11/24/2005 11:05:
I can't believe we don't have a complete answer after 6 pages of text.
The closest is David's, but there is one thing missing. Apparently "it's
quite easy to show that if the signal is time-limited then the Fourier
transform extends to an entire function in the plane". Well, once we are
shown why this is easy, we are done.
So close....
mebden, you didn't cross post to sci.math where the real anal guyz are.
if someone can prove to me that for an entire function (one that is
analytic, or has all derivatives existing, everywhere in the complex plane)
that if it is zero for any segment longer than a single point, then it is
zero everywhere, i think i can make the case because it isn't too hard to
show that a timelimited function of time (with nonzero and finite energy)
has, for its Fourier Transform, analytic spectrum. if the spectrum was
bandlimited, then it would be exactly zero for a segment of frequency and,
being analytic, that would mean it was identically zero.
|
I can't prove it, but maybe I can convince you anyway. If a function is
zero in a finite patch, however small, then all it's derivatives are
also zero there, and it can never get off the ground, so to speak.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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robert bristow-johnson
Guest
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| Posted:
Sat Nov 26, 2005 1:15 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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in article F-GdnfN3p-a40BrenZ2dnUVZ_t-dnZ2d@rcn.net, Jerry Avins at
jya@ieee.org wrote on 11/25/2005 12:42:
| Quote: | robert bristow-johnson wrote:
if someone can prove to me that for an entire function (one that is
analytic, or has all derivatives existing, everywhere in the complex plane)
that if it is zero for any segment longer than a single point, then it is
zero everywhere, i think i can make the case
....
I can't prove it, but maybe I can convince you anyway. If a function is
zero in a finite patch, however small, then all it's derivatives are
also zero there, and it can never get off the ground, so to speak.
|
i've been previously convinced, Jerry, for the same language of reason.
looks like Robert Israel has put it in sufficient language to be a proof, so
i will assemble the full proof tonight. (still might not cut it for the
sci.math guyz.)
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge." |
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Robert Israel
Guest
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| Posted:
Sat Nov 26, 2005 1:16 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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In article <BFACAE36.C6CE%rbj@audioimagination.com>,
robert bristow-johnson <rbj@audioimagination.com> wrote:
| Quote: | mebden, you didn't cross post to sci.math where the real anal guyz are.
|
I'm really not all that anal...
| Quote: | if someone can prove to me that for an entire function (one that is
analytic, or has all derivatives existing, everywhere in the complex plane)
that if it is zero for any segment longer than a single point, then it is
zero everywhere, i think i can make the case because it isn't too hard to
show that a timelimited function of time (with nonzero and finite energy)
has, for its Fourier Transform, analytic spectrum. if the spectrum was
bandlimited, then it would be exactly zero for a segment of frequency and,
being analytic, that would mean it was identically zero.
|
If p is any point on such a segment, then all derivatives at p must be 0.
The function is the sum of its Taylor series, which is 0.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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Jerry Avins
Guest
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| Posted:
Sat Nov 26, 2005 7:48 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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Robert Israel wrote:
...
| Quote: | If p is any point on such a segment, then all derivatives at p must be 0.
The function is the sum of its Taylor series, which is 0.
|
Hey! Just about what I said! I fully expected you'd shred it! Zowie! I'm
not so out of it after all!
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Bob Cain
Guest
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| Posted:
Sat Nov 26, 2005 9:15 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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Robert Israel wrote:
| Quote: | In article <BFACAE36.C6CE%rbj@audioimagination.com>,
robert bristow-johnson <rbj@audioimagination.com> wrote:
mebden, you didn't cross post to sci.math where the real anal guyz are.
I'm really not all that anal...
if someone can prove to me that for an entire function (one that is
analytic, or has all derivatives existing, everywhere in the complex plane)
that if it is zero for any segment longer than a single point, then it is
zero everywhere, i think i can make the case because it isn't too hard to
show that a timelimited function of time (with nonzero and finite energy)
has, for its Fourier Transform, analytic spectrum. if the spectrum was
bandlimited, then it would be exactly zero for a segment of frequency and,
being analytic, that would mean it was identically zero.
If p is any point on such a segment, then all derivatives at p must be 0.
The function is the sum of its Taylor series, which is 0.
|
Aw, c'mon. It can't be _that_ easy. :-)
Bob
--
"Things should be described as simply as possible, but no simpler."
A. Einstein |
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