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

On Fri, 3 Dec 2004 07:27:45 -0000, "Nimrod" , paused briefly between bottles, to write:

Which let's be honest, is pretty much everybody.

Nick.

I didn't say any such thing. I said that you need an antiderivative in the integration by parts formula. That formula you quote is _why_ the antiderivative is what I say it is.

By definition an antiderivative for f(t).d(t-T) would be F, where

F(x) = int_0^x f(t).d(t-T).

If x < T then that gives F(x) = 0, while if x > T that gives F(x) = f(T), precisely because of that formula above.

Impossible to answer this question since you insist generalized functions have nothing to do with it - these antiderivatives are not differentiable in the classical sense, they have derivatives in the sense of distributions.

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David C. Ullrich

Just explained that in a different post.

True, and of no relevance whatever to the question of what an antiderivative is.

Hint: You're _really_ showing over and over that you have no understanding of _basic_ calculus. You keep kindly informing me that I'm making a laughingstock of myself, presumably you'd want to know how utterly ignorant you're revealing yourself to be.

PS: Look up the definition of "ad hominem". I'm not saying you must be wrong because you're stupid, I'm saying you're looking very stupid because the things you're saying are wrong, at such a basic level.

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David C. Ullrich

By what definition? - You have presented here a definite integral and not an anti-derivative which contradicts earlier assertions by you.

Indeed, in the only related definition of which I am aware, the limits of integration should be +/- oo and not 0 to x. To which definition in your "By definition" do you refer?

Where does that assertion come from?

Remember - you proscribe the definite integral int -oo^+oo f(t).d(t-T)

Apart from throwing in the term "generalised functions", how does the presented mathematics for your derivatives differ from those of the classical sense?

Remember that any one function in your class of generalised functions represents the properties of the whole class.

I tangled with him in another newsgroup. He is untrainable and is proud of that fact.

/BAH

/BAH

Subtract a hundred and four for e-mail.

An ad hominem attack from you of no value to the discussion.

Shame on you.

You should no better.

I do not "keep kindly informing you that you're making a laughingstock of myself". I have suggested that once and once only.

Please get your facts correct.

You say that the things quoted below from me are true and then you say that I'm "utterly ignorant".

You are inconsistent.

That does not agree with the definition with which I work, which is int -oo^+oo f(t).d(t-T) = f(T). In that evaluation, all dependence upon t disappears.

What definition are you using to claim discontinuity?

If you evaluate the anti-derivative given by Mr.Ullrich (without accepting it as a correct formula, BTW) then you would get the properties where F(t) = 0 for t > T and f(T) for t < T.

f(t).

if F(t) = f(t).d(t-T), then with T equal to zero, this comes down to f(t).d(t).

This might be undefined because it is not expressed under an integral sign, but it is (probably) not 0.

If it were to be expressed under an integral sign, then it would be f(0) as I stated.

get lost bean

get lost bean

get lost bean

get lost bean

That "evaluation" of the _antiderivative_ is simply _wrong_.

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David C. Ullrich

You really need to find a calculus book somewhere. If F(x) = int_-infinty^x f(t) then F' = f; this is a thing called the Fundamental Theorem of Calculus.

(In the present context int_0^x is the same, since we're assuming that f(t) = 0 for t < 0).

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David C. Ullrich

The antiderivative is not even continuous, much less differentiable in the classical sense.

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David C. Ullrich

I suppose it should be no surprise that you can't follow the definition of a term like "ad hominem" any better than you follow definitions in mathematics.

Look it up. Google "ad hominem" - the first hit is

That's a correct definition, and nothing I've said here fits. (If you don't understand why not, I gave an explanation in the "PS" below.)

Uh, no.

No. What you say about the value of the definite integral is correct - your idea that the integral from -infinity to infinity can be used in place of the antiderivative in an integration by parts shows amazing ignorance.

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David C. Ullrich