You may well have a point here, if my error is in not evaluating the definite integral over the whole term.

However, other errors in what you propose nullify your argument, (see other posts)

- posted
18 years ago

You may well have a point here, if my error is in not evaluating the definite integral over the whole term.

However, other errors in what you propose nullify your argument, (see other posts)

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- posted
18 years ago

Yes - I corrected that in a later post. However, the error cancelled out because the effect of consecutive differentiation and integration was -s/-s giving unity.

- posted
18 years ago

I pointed out explicitly, in lines you snipped, that Airy had written e^{-st}/s where it should have been e^{-st} s. I thought that I'd fixed the error at that point, it's possible that some erroneous cut&pasted lines were not edited properly.

(Also pointed out at least twice so far that this is somewhat amusing. Before looking at it I knew that his integration by parts must be wrong, since it gave the wrong answer, but I assumed that the problem was with something subtle about delta functions. Turns out that in addition to not getting the antiderivative right his amazing conclusion is based in part on simply differentiating an exponential incorrectly!

It's easy to see why this was the point at which he suddenly decided he needed to announce that he was not reading any more replies... Although announcing that he'd deleted the replies containing math and then also complaining that the replies he was reading didn't contain any math was a very bold step.)

- posted
18 years ago

So

C.L. Phillips, J.M. Parr, Signals, Systems & Transforms, 2nd ed., Prentice Hall, Copyright 1999, p. 46-50, 210-211

really defines the Laplace Transform as an integral from -infinity to infinity, instead of the more usual 0 to infinity?

David C. Ullrich

- posted
18 years ago

The bold step that I took was to make a stand against infantile ad hominem attacks as is your wont.

You continue with your motivation to insult in your posting below.

If you have anything of value to offer, then offer it in a mature post.

Shame on you. You should know better.

(If you adopt the same arrogant stance to your pupils as you do here, and if any of those pupils are reading this NG, you can be sure that you are the laughing stock in your school (if not so already))

- posted
18 years ago

The limits of the bilateral LT are from -oo to +oo. The limits of the unilateral LT are from 0 to +oo. Both are usual.

I thought you__ _BOASTED_ __of being a mathematician of 20 years' standing?

- posted
18 years ago

The irony of it, the irony.

- posted
18 years ago

Indeed. But you left out the best bit, irony-wise:

David C. Ullrich

- posted
18 years ago

I don't think I BOASTED of that, but yes, I mentioned it. I have enough experience to know that something I've never seen before is not necessarily wrong.

David C. Ullrich

- posted
18 years ago

If you take f(T) exp(-st), (assume s > 0), evaluate at t = infinity, t = 0 and subtract you get -f(t). If F is an antiderivative of f(t) delta(t-T), so F(t) = 0 for t < T, F(t) = f(T) for t > T, and you do the same evaluate-subtract with F(t)e^{-st} you get 0.

This was clear from the difference in your calculation and my correct version of the same calculuation, btw.

David C. Ullrich

- posted
18 years ago

Hilarious. I think you might want to take Calc 101 again before lecturing us on all this stuff.

If int means int_0^infinty then yes it is, that's exactly whu it's wrong.

Making less and less sense - if you'd done that evaluation already then there would be no "t" appearing at this point.

It's really good of you to warn me about what a laughingstock I'm making of myself, btw.

David C. Ullrich

- posted
18 years ago

No you don't.

You get -f(T).

- posted
18 years ago

I see that you cannot resist the temptation to lower the tone.

- posted
18 years ago

If you had any sense you'd stop making an idiot of yourself over the math, stop being so utterly hilarious complaining about the tone of people's comments, and just take a calculus class.

David C. Ullrich

- posted
18 years ago

I see that you cannot resist the temptation to lower the tone.

- posted
18 years ago

In an earlier post David pointed out legitimate problems with the math behind your argument. Rather than read, and learn from, that post, you "accidentally" deleted it. I suggest you find the post on Google and read it. You may learn something.

Charles Perry P.E.

- posted
18 years ago

I suggest that you read what is written before you open your mouth and come out with untruths.

You may learn something.

I did not "accidentally" delete anything. In response to a barrage of quite unnecessary gratuitous and insulting remarks by Mr.Ullrich, I__ _DELIBERATELY_ __deleted things.

If Mr.Ullrich had anything valuable to offer, then it was lost in the fog of his childish outbursts.

There is no concept of legitimacy. Law does not come into it.

- posted
18 years ago

in article snipped-for-privacy@4ax.com, David C. Ullrich at snipped-for-privacy@math.okstate.edu wrote on 12/02/2004 07:54:

i just realized now how far this is cross-posted. i suspect that David is hanging out on sci.math (but i dunno). David, he really doesn't get it. and he doesn't want to get it. if it were up to Beanie, we would rewrite all the calculus textbooks.

let's not feed the troll. maybe if we stop feeding it, it will shrivel up and blow away.

r b-j

- posted
18 years ago

But the gives a different result if the integration is done in the opposite direction, and therefore cannot be valid.

Well, here is what David C. Ullrich posted....

- posted
18 years ago

Sounds a bit far fetched to me. Was that before or after the Westinghouse affair?

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