# Derivation Of The Spectrum Due To f(t).d(t - T)?????

• posted

What qualifies a failed software engineer (remember Westinghouse) to make such a statement?

Did you ever actually get a degree?

• posted

let's ask crystal, shall we bean?

dr. x

• posted

Yes.

Dwayne

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Yes.

Thanks, but actually I realized that a few hundred posts up. Was actually trying to help the guy with the math in my first one or two replies, but since then it's all been just for the fun of watching the way the guy's technique.

He's pretty good, has things like being insulting and then whining when people reply insultingly, ignoring refutations and then claiming they don't exist, claiming that the fact that there have been a lot of replies proves he must be right, etc down pat. We've got better here on sci.math but he's pretty good. I can't decide whether he's just trolling or really believes he's right...

What fun would that be?

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David C. Ullrich

• posted

the more of this thread I read the more I pity those poor souls

• posted

One of the posts you "_DELIBERATELY_" deleted is the one I quoted earlier, where David C. Ullrich, started out quite politely showing you where you made your mistakes. And in his more recent post, he states:

"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. " Not much 'unnecessary gatuitous and insulting remarks' in that unless you consider the last sentence such. And if you do, then you obviously have a much thinner 'skin' than the way you treat others.

daestrom

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No, the antiderivative ( F(t) ) is not a constant. It is a discontinuous function of t. Where t T it has a value of f(T) (note carefully the case of each statement). It has two possible values depending on whether you are evaluating it at tT. That is not the same thing as what you're saying, "... is a constant..."

daestrom

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Evaluating F(t) where tT you have a value of f(t). If evaluated in the reverse order (i.e. over the interval +inf to zero), you get -f(t) instead of f(t). This is exactly as it should be.

daestrom

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Not true. F(0) = zero, not f(0). This is because F(0) = delta(0 - T) f(0). And the Dirac delta function evaluated at (0-T) is zero, regardless of the value of f(0). (assuming that T > 0)

daestrom

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The 'other errors' are on your part, not David's. (see other replies)

daestrom

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Yes, I was not trying to point the finger at you for the original error, just pointing out that at that particular point you corrected your explanation (for those that may have wondered how the 's' term suddenly moved from denominator to numerator). And that, as I said, "I suspect this error came from ARB's original..." posting, not you per se.

Interesting bit of calculus, I may have to go visit sci.math to see what other interesting things are bandy'd about.

daestrom

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in article snipped-for-privacy@4ax.com, David C. Ullrich at snipped-for-privacy@math.okstate.edu wrote on 12/02/2004 07:16:

Beanie doesn't have pupils (except maybe two beneath his corneas). Beanie (real name Gareth) is not competent enough in the subject matter to be teaching it to anyone.

despite

Even kids would be able to figure out quickly that they cannot learn (except by rote, decidedly a bad way to learn math) from Beanie because what he says makes no sense because it is false. even more so with adults. you cannot systemically learn underlying concepts from misconceptions. and you cannot learn math without learning conceptually.

i imagine that Beanie is unemployed and (due to his character) unemployable.

r b-j

• posted

Oops, where t> T you have F(t) = f(T), not f(t). My mistake.

That should be -f(T) instead of f(T), again, my mistake.

daestrom

• posted

Not too sure about that logic.

When I was doing A-level maths (A-levels are the UK exams at 18, before entering university) we had a really bad maths teacher. Everyone knew she was really bad, so we largely ignored her. We worked extra hard on our own from books, and did lots of exercising with the published past papers. The final result was we all achieved really excellent grades in the exams. From this our teacher probably looked like a star performer, which is a bit sad, but the bottom line is her incompetance actually achieved good results. :-)

Regards, Steve

• posted

in article coopl8\$89g\$ snipped-for-privacy@home.itg.ti.com, Steve Underwood at snipped-for-privacy@dis.org wrote on 12/02/2004 23:21:

...

...

makes sense. i was saying that "they cannot learn ... *from* Beanie...".

just because of two outcomes that might be correlated, it does not mean that one was the cause (shitty teacher) and the other was the effect (good test results). you do not know how well you'ld have done with a really good teacher. perhaps it is that you did so well despite the poor teacher not because of the poor teacher.

i am an advocate of good teaching. good, effective, teachers have to be, at the very least, competent in the subject and in related subjects that they are teaching. my \$0.02 .

r b-j

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I'm don't think that's the same Gareth Evans (way too young). Maybe he can enlighten us. It was evident to me yesterday that that's apparently not an uncommon name in the UK (or Australia).

Eric Jacobsen Minister of Algorithms, Intel Corp. My opinions may not be Intel's opinions.

• posted

The Gareth that is giving you the run around, and that is exactly what he is doing, is in his 50s.

He attended Essex University, although it appears there little evidence he graduated, and has an "varied" employment history. He has admitted being dismissed from on job. All the evidence is he screwed up big time. Other posts by him suggest that this was not an isolated case.

He trolls varies newsgroups with his wind-ups on a range of topics, DSP is one his favorite topics- his theories on sampling first came to light several years ago. They have been refuted many times.

Other topics have included a justification for the Soham murders (for US readers, two teenage girls were murdered in Soham a few years ago).

He appears to be unemployable and becomes fixated on those who have won out arguements against him.

Do some googling and you will find evidence of all the above.

If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS. STARVE THE TROLL.

• posted

Gareth is a pretty common Welsh name. Evans is very common Welsh name. Put them together and you get a fairly common, and very Welsh name :-)

Regards, Steve

• posted

• posted

Hang on a moment... you say that I cannot use the properties of a definite integral.....and then you go on and do so yourself!

If I cannot rely on the property of a definite integral that int -oo^+oo f(t).d(t-T) is f(T) then neither can you!

Well, let us suppose that our f(t) is sin(wt)...how does your claimed anti-derivative then differentiate to become sin(wt).d(t-T)?

Or, let us suppose that our f(t) is e^t...how does your claimed anti-derivative then differentiate to become e^t.d(t-T)?

So, how does your anti-derivative differentiate to become two different products of functions?

Remember, until we actually evaluate the definite integral that we're dealing in whole functions over all t in symbolic maths.

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