Top-posting makes the whole thing impractical.

- posted
18 years ago

Top-posting makes the whole thing impractical.

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- posted
18 years ago

Nothing wrong with top-posting.

It is the preferred method, because you don't have to page down through much repeated and already-seen history.

Bottom posted articles follow>

- posted
18 years ago

Of course not.

David C. Ullrich

- posted
18 years ago

No problem - I assumed you were talking about someone else, wondered why I didn't see any posts from him... (the spelling with one l is much more common.)

David C. Ullrich

- posted
18 years ago

Of course not. Here the question was about a__ _definition_.__ I didn't claim to have proved I was giving a correct definition.

The definition I gave__ _is_ __correct, and perfectly standard, your ignorance of the matter notwithstanding.

David C. Ullrich

- posted
18 years ago

are you afraid to see your mistake explicitly exposed?

"cluck, cluck"

r b-j

in article snipped-for-privacy@uni-berlin.de, Airy R. Bean at snipped-for-privacy@privacy.net wrote on 11/30/2004 04:17:

- posted
18 years ago

Yes, but the delta "function" is__ _not_ __the limit of a sequence of ordinary functions, at least not in the sense that one speaks of limits in calculus. It's not a function at all.

But it is true that one can define the__ _action_ __of a delta function as a limit - that is, the linear functional defined by the delta "function" is indeed a limit of the linear functionals defined by a sequence of ordinary functions. If you want to look at it that way it's very easy to show from that point of view that the Laplace transform is what it is:

Fix T, and say f_n(t) = n for T < t < T + 1/n, f_n(t) = 0 for other t. Then the f_n__ _do_ __converge to delta(t-T) in a certain sense - in particular the Laplace transform of delta(t-T)__ _is_ __the limit of the Laplace transform of f_n as n -> infinity. Let's say F_n is the LT of f_n. Then

F_n(s) = n int_T^{T+1/n} e^{-st} dt

= n (e^{-s(T+1/n)) - e^{-sT}) / (-s)

= e^{-sT} (1 - e^{-s/n}) / (s/n).

It's an easy calculus exercise to show that (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity. So the limit of F_n(s) is e^{-sT}.

David C. Ullrich

- posted
18 years ago

The integration by parts that you're doing is__ _wrong_.__ The fact that integration by parts works is a theorem. Theorems have hypotheses - whatever delta(t-T) is, it's not a continuous function.

David C. Ullrich

- posted
18 years ago

I explained exactly how to evaluate this Laplace transform correctly in my first post on the topic.

(I also just posted a more complicated proof, using the fact that the delta "function" is in some sense a limit of ordinary functions.)

Guffaw.

Uh, yes you did. In your reply to my first post you said in part

"But. to evaluate the (unilateral) Laplace transform, we either have to be able to present f(t) as a function of exponentials, in which case simple integration applies together with the adding of exponents, or we have to apply Integration by Parts. (or else for a time domain product we have to evaluate an s domain convolution)"

which does say that this and that are the only way to do something.

David C. Ullrich

- posted
18 years ago

Your reputation will not be enhanced by your tendency to resort to ad hominem remarks.

- posted
18 years ago

.....[Pantomime mode ON].....

Oh, yes! It__ _IS_ __a function.

.....[Pantomime mode OFF].....

And I have never defined it as a limit of a sequence of functions any more than I would define a derivative as such a limit.

The limit employed is very much the same limit as used in calculus; t > 0.

- posted
18 years ago

Make your mind up - first you say that it is not a function, then you say that it is the delta function.

- posted
18 years ago

So you have argued a derivation, without using the concept of generalised functions which were in any case irrelevant, to arrive at a result that wasn't in dispute.

I wonder why?

- posted
18 years ago

I am not afraid of anything, be that anything you, or the truth. It is simply that I refuse to deal with you when you indulge yourself in infantile outbursts.

Shame on you.

You should know better, including not indulging in silly yah-boo-sucks tirades as in your latest rant below.

You do nothing for your reputation by so behaving.

- posted
18 years ago

Show where it is wrong, or else resort to infantile sneering.

Oh!!! You're doing that already!

- posted
18 years ago

.....[Pantomime mode ON]..... Oh, no! You didn't! .....[Pantomime mode OFF].....

- posted
18 years ago

So first of all you say that I said, "X is the__ _only_ __way to accomplish Y"

Then you said that I said, "this and that are the only way to do something."

You're not consistent.

In any case, I suggested three ways of proceeding and certainly did not state that "X is the__ _only_ __way to accomplish Y" nor did I state "this and that are the only way to do something.".

- posted
18 years ago

The result I arrived at was that the Laplace transform of delta(t-T) is e^{-sT}. That wasn't in dispute?

Because you're too stupid or stubborn to agree with the much simpler argument proving the same thing.

David C. Ullrich

- posted
18 years ago

Ok. Took me a minute to find the place where you showed us the integration by parts:

" Certainly, if you integrate by parts, you would choose int(f(t).d(t-T)) as the integrated bit to yield f(T), but when I try this, I get 0!......

int(UV) = U.int(V) - int[dU.int(V)] giving.....

int(+/-inf)(f(t).d(t-T).e^(-st)) as .....

with "f(t).d(t-T)" as V and "e^(-sT)" as U .....

f(T).e^(-st) - int(e^(-st)/-s . f(T)).....

f(T).e^(-st) - -s.e^(-st)/-s . f(T).....

f(T).e^(-sT) - f(T).e^(-sT)....

- " Your first step,

f(T).e^(-st) - int(e^(-st)/-s . f(T))

is already wrong. My guess is because of a confusion over definite integrals versus antiderivatives, ie indefinite integrals. The (definite) integral from

0 to infinity of f(t).d(t-T) is indeed f(T). But what we need here is an antiderivative.An antiderivative of f(t).d(t-T) is given by F(t), where F(t) = 0 for t < T and f(T) for t > T. So that line should be

F(t).e^(-st) - int(e^(-st)/-s . F(t)).

Golly, I assumed that you had some idea how to do simple calculus - looking at this I realize you took the derivative wrong. What we really get is this:

F(t).e^(-st) - int(e^(-st)(-s) . F(t)). Now when we put in the limits t = 0 to t = infinity the first term vanishes since F(0) = 0 and e^(-st) tends to 0 as t -> infinty (at least if s > 0; if s < 0 this integration by parts is not going to work.) So out Laplace transform becomes

- int_0^infinity (e^(-st)/-s . F(t))

Since F(t) = 0 for t < T and f(T) for t > T this equals

f(T) int_T^infinity (e^(-st)s ).

Again, it's easy to evaluate the last integral; int_T^infinity (e^(-st)s ) = e^(-sT), so we finally get f(T)e^(-sT) for the Laplace transform.

So now we've found__ _two_ __mistakes in your integration by parts, confusing antiderivatives and definite integrals and taking a very simple derivative incorrectly.

David C. Ullrich

- posted
18 years ago

More properly, "It is preferred by some, because they don't have to page down through much repeated and already-seen history." It is not universally 'preferred' by all use-net users. In fact, there are several use-nettiquite articles on the subject.

... by those that don't want to scroll through the previous context. Others, that don't mind reading the postings in context don't mind at all. In fact, many prefer in-context posting as it is much clearer what the poster is replying to.

In the end, in-context posting, with snippage of unrelated, repeatedly quoted material is 'preferred' by many.

You don't speak for the entire use-net community, so don't pretend that you do.

daestrom

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