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

• posted

Where do you get that from?

The definite integral int -oo^+oo f(t).d(t-T) gives us f(T), but there is no information given to us about the indefinite integral nor of the anti-derivative.

Indeed, the definite integral gives us f(T), which is a constant function having no aspect of t in its evaluation and which graphs as a horizontal line from -oo to +oo. This comes from the basic definitions of the Diracian.

Your claimed anti-derivative has a strong dependence on t, and therefore has not come from anything formally defined.

Are you making this up as you go along?

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

• posted

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.

• posted

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

• posted

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

• posted

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?

• posted

Where does that assertion come from?

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

• posted

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.

• posted

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.

• posted

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.

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

You are inconsistent.

• posted

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?

• posted

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).

• posted

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.

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get lost bean

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get lost bean

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get lost bean

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get lost bean

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That "evaluation" of the _antiderivative_ is simply _wrong_.

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David C. Ullrich

• posted

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

• posted

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

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David C. Ullrich

• posted

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.