| Author |
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Rob Johnson
Guest
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 Posted:
Tue Nov 29, 2005 12:21 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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In article <nYidnRlaTNcguRbeRVn-tA@rcn.net>,
Jerry Avins <jya@ieee.org> wrote:
| Quote: | Rob Johnson wrote:
In article <7jatgdv7.fsf@ieee.org>,
Randy Yates <yates@ieee.org> wrote:
David C. Ullrich <ullrich@math.okstate.edu> writes:
On Mon, 28 Nov 2005 03:48:08 GMT, Randy Yates <yates@ieee.org> wrote:
israel@math.ubc.ca (Robert Israel) writes:
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.
Should be a point in the deleted neighborhood of the expansion point, not
ANY point, right?
Not sure exactly what you mean by that, but in any case no, what
he said is precisely correct. We have a segment on which we're
assuming that our entire function vanishes; if p is ANY point
of that segment then all the derivatives of the function
vanish at p.
That is a claim. What I was trying to do was see how you prove the
claim. To do that, you must choose a point z1 on the segment for which
f(z1) = 0 and in which z1 != z0, where z0 is the expansion point of
the Taylor series. That is because if you let z1 = z0, then the Taylor
series expansion becomes zero no matter what the derivatives are,
since 0 = (z1 - z0) = (z1 - z0)^2 = ..., and thus we cannot conclude
that the derivatives are zero from this case.
If z1 != z0, then
f(z1) = 0
= f(z0) + (f'(z0)/1!)(z1 - z0) + (f''(z0)/2!)(z1 - z0)^2 + ...,
and since f(z0) = 0,
(f'(z0)/1!)(z1 - z0) + (f''(z0)/2!)(z1 - z0)^2 + ... = 0
I'm still not quite sure how to show that this implies all the
derivatives are zero.
Suppose f is identically 0 on some open interval. Pick ANY point, z0,
in this interval. For any other z also in that interval,
f(z) - f(z0) 0 - 0
------------ = ------ = 0
z - z0 z - z0
Therefore, taking the limit as z -> z0 also gives 0. Thus, f' is 0 on
the same open interval, so f'' is identically 0, f''' is identically 0,
etc. So all derivatives of f are identically 0 on that open interval.
It seems to me that the function vanishing is just a special case. If a
continuous function with continuous derivatives of all orders is
perfectly planar in a finite-size patch, then the function describes
that plane everywhere. Even that can be generalized.
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The generalization that comes to mind is that if two entire functions
are identical on a non-empty open set, then they are identical on the
whole real line (or complex plane if working there). This is an easy
corollary of the statement about vanishing functions. Simply consider
the difference of the two functions; the difference, being an entire
function vanishing on a non-empty open set, must be identically zero
everywhere. Thus, the two functions are identical.
In your case, the other entire function is a constant function, which
in our case, was the zero function.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying |
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Jerry Avins
Guest
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| Posted:
Tue Nov 29, 2005 1:16 am Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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Rob Johnson wrote:
...
| Quote: | The generalization that comes to mind is that if two entire functions
are identical on a non-empty open set, then they are identical on the
whole real line (or complex plane if working there). This is an easy
corollary of the statement about vanishing functions. Simply consider
the difference of the two functions; the difference, being an entire
function vanishing on a non-empty open set, must be identically zero
everywhere. Thus, the two functions are identical.
In your case, the other entire function is a constant function, which
in our case, was the zero function.
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Exactly so.
jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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David C. Ullrich
Guest
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| Posted:
Wed Nov 30, 2005 5:16 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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On 29 Nov 2005 11:15:28 -0800, "Randy Yates" <yates@ieee.org> wrote:
| Quote: | As I said, didn't believe that anyone would misunderstand.
Yes, that's quite clear. What is also clear is your opinion of your
writing abilities.
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Um, here's a hint: What he wrote originally was perfectly
clear. Your continuing to insist that it was not makes
you look like a fool.
That's maybe not a very polite thing to say, but it's
a fact that maybe you'd want to know. I mean really.
I didn't say anything at the start of all this, even
though when you express doubt over the fact that
if f vanishes on an interval then f' also vanishes
on that interval it becomes clear that even though
you don't know that the derivative of the zero function
is zero(!) you're arguing with people about calculus.
(Honest. When you took calculus and the teacher said
that f = 0 and asked what f' was what did you say?
You didn't have any idea?)
But then there was the following - this is just too
much:
| Quote: | Is z a variable or a point? If it is a variable, then
what "other" variable is there in your statement,
"For any other z also in that interval, ..."?
If it is a point, then we would be be evaluating a complex
constant f(z) and not a function, and the concept of derivative
would not apply.
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When someone says "Let z be an element of S" and you
insist that he's being unclear it's truly remarkable.
Because you find _exactly_ that language in almost
anything remotely mathematical that you read - if
he's being unclear here then so is everyone who
every wrote a math book. First, there's nothing
unclear about the language. Second, even if there
were something unclear about it, the fact that
that language is _so_ standard means that a reader
should simply assume that it means exactly what
it always means.
************************
David C. Ullrich |
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David C. Ullrich
Guest
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| Posted:
Wed Nov 30, 2005 5:16 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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On Wed, 30 Nov 2005 02:12:03 GMT, Randy Yates <yates@ieee.org> wrote:
| Quote: | rob@trash.whim.org (Rob Johnson) writes:
[...]
z0 is also a variable representing another point, distinct from z (that
is, z <> z0).
That's not what you originally wrote. What you wrote above is indeed clear.
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As was what he originally wrote. Hint: if something is clear to
everyone but you then the problem is not that the writer is
not being clear.
************************
David C. Ullrich |
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Randy Yates
Guest
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| Posted:
Wed Nov 30, 2005 5:17 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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Bullshit.
There are many people who have probably read these posts. Of that
population, some may not understand, but since today's society is all
about "spin," the last thing anyone wants to do in a public newsgroup
is admit that they didn't understand something.
HINT: Get your collective mathematical heads out where the sun shines
and realize that the technological world does not revolve around
academic mathematicians and their set of definitions and
interpretations.
--Randy |
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Robert Low
Guest
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| Posted:
Wed Nov 30, 2005 5:17 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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Randy Yates wrote:
| Quote: | the last thing anyone wants to do in a public newsgroup
is admit that they didn't understand something.
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We must be reading different groups. A large number
of posts made here are people proudly announcing that
they don't understand something, and asking for
help with it. |
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David C. Ullrich
Guest
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| Posted:
Thu Dec 01, 2005 5:16 pm Post subject:
Re: how do you prove a signal that is time-limited cannot be |
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On 30 Nov 2005 08:12:54 -0800, "Randy Yates" <yates@ieee.org> wrote:
| Quote: | Bullshit.
There are many people who have probably read these posts. Of that
population, some may not understand, but since today's society is all
about "spin," the last thing anyone wants to do in a public newsgroup
is admit that they didn't understand something.
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What a strange thing to claim. In this very thread we've seen
several people say they didn't understand several things.
The fact that various things were said to be very simple
didn't stop people from asking for explanations.
But only one person has insisted that he didn't understand
the following three things:
(i) if f(t) = 0 then f'(t) = 0.
(ii) the meaning of the word "another".
(iii) the meaning of the phrase "if z is in S then...".
| Quote: | HINT: Get your collective mathematical heads out where the sun shines
and realize that the technological world does not revolve around
academic mathematicians and their set of definitions and
interpretations.
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This is the sort of frothing at the mouth that keeps people
reading usenet; definite you-can't-make-this-stuff-up stuff.
Math bad. Him mathematician - him bad bad bad.
(In some other context it might be a reasonable thing to
say, for example if some dsp guy said something that's more
or less right but left out some hypotheses and some math
guy butt in complaining about the details of the proof
being incomplete. But as a reply to my claim that the
phrase "if z is another such point" is perfectly clear
it's simply priceless.
Or: in a thread where the original post _asked_ for a
precise mathematical proof of a certain fact it's simply
priceless.)
You're not required to believe that you've been making
a fool of yourself here. In fact I rather hope you don't
believe it, it would be boring if you suddenly realized
that and just slunk away. But it's a fact, whether you
believe it or not.
************************
David C. Ullrich |
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