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

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David C. Ullrich

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Stupid boy - once again you resort to the use of gratuitous personal remarks, merely because you disagree with an opinion expressed.

Shame on you - you should know better.

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Isn't 1/n _pretty close_ to zero for n -> oo ?

Depends. I routinely replace delta functions by bell-shapes (or even triangular shapes _/\_) in my numerical work. And got no problems at all.

Han de Bruijn

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Not a personal remark at all. And the idea that these things are matters of opinion is hilarious. You're entitled to whatever "opinion" you wish. But this is math - the fact that you're entitled to your opinion doesn't change the fact that your "opinions" are ludicrously wrong.

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David C. Ullrich

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No. It may be pretty close for a particular value of n, and it gets arbitrarily close as n goes towards infinity, but you can't, say, claim that the limit for

1/n

------ n^{-2}

has to be 0, since 1/n is pretty close to zero, and the limit of

0/n^{-2} is 0 as n->oo.

Just tell me what kind of bridges are designed with that kind of mathematics so that I can remember to avoid them.

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Indeed, it was a personal remarks, and followed the several that you have indulged yourself with; the several that showed that you were incapable of partaking in an adult discussion or even an acadaemic discussion.

All mathematics is a matter of interpretation by practitioners. What is bizarre about your quote below is that your opinions are ludicrously wrong and yet you say that my opinions are ludicrously wrong.

You say that the Diracian is not continuous, but the Gaussian form is continuous, differentiable and integrable using classical mathematics. You see - your opinions are ludicrously wrong.

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That's because those exponential things do indeed converge to a delta function, _in_ the sense of distributions. Which fact has no relevance whatever to the question of whether the antiderivative of a delta function is continuous.

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David C. Ullrich

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The antiderivative of a delta function is a heaviside function h(x) which is not continuous for x = 0, in the classical sense:

h(x) = 0 for x < 0 h(x) = 1 for x > 0

When it comes to constructivism, there are problems with heaviside's definition at x = 0. Constructivists would say that a heaviside is not defined at x = 0. Hence it is _not a function_, in the proper sense of their words. As physicists, we wouldn't go that far. While mainstream mathematics defines something like h(0) = 1/2, we would rather say that h(x) is _multi-valued_ for x = 0. Huh?

A heaviside function may be not continuous, but h(x) is quite close to an error function Erf(x/eps) which still _is_ continuous for x = 0, as eps approaches zero. Therefore "continuous or not" for h(0) seems like nitpicking for the *real* Applied among us. Like radio amateurs ...

Han de Bruijn

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Poor you! Almost every bridge in modern times is designed with just that kind of mathematics, as embedded in some Finite Element Method.

Han de Bruijn

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This NG is an international public forum wherein one expects to see technical excellence.

It should be a place for enlightenment and reasoned technical discussion. It certainly used to be like that.

For some reason, best known only to yourself, you have chosen to turn it into an outpost of the infants' school playground using behaviour and emotional stances left behind years ago by mature contributors.

Such an exhibition by you does nothing for your reputation, nor for the well-being of this NG.

Shame on you.

You should know better.

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Ok. Here's a personal remark, just so you can see the difference: If you're not just trolling then you're either an idiot, a lunatic or both.

Here's a proof that 0 > 0, using the same reasoning as you're using to show that a delta function is continuous: 1/n > 0, and 1/n is a form of 0.

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David C. Ullrich

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Uh, everyone except Airy understands that.

That's nonsense. What the value of this function is at 0 has nothing whatever to do with constructivism.

In fact if you define h(x) as above for x 0 and define h(0) to be anything you like, then the derivative of h, _in the sense of distributions_, is delta.

Or, if you want a stranger example: define h(x) = 42 if x is rational, 1 if x is irrational and x > 0, and 0 if x is irrational and x < 0. Then h is continuous nowhere. Its distribution derivative is still delta.

(The actual explanation for this mystery is that we're talking about distributions, which are in fact not functions at all - a distribution does not have a value at _any_ point, it's a certain linear functional. All those various versions of h define the same distribution.)

So 0 is quite close to 1/2? Fine.

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David C. Ullrich

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It's _no nonsense_. It's called "Brouwer's Continuity Theorem". Don't know if it is accepted by all constructivist mathematicians, admittedly. That theorem says that h(x) is not even defined for x = 0, if h(x) is not continuous there. Look it up!

Ah, that's the good news!

[ .. even better news following, but skipped here .. ]

No, but 0 is quite close to our multivalued 0 < h(0) < 1 . (I know that a picture is no proof, but it may be convincing once in a while ...)

Han de Bruijn

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Oops. I meant that Erf(0) = 1/2 is close to our multivalued 0 < h(0) < 1

Han de Bruijn

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Then go away.

Then go away.

It will be a step in that direction if you go away.

You are still confusing first and second person pronouns.

If you care for the well-being of this NG, go away.

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But by people that have a clue about just what they are doing. Numerics is not "we'll try something and hope it will be close enough". Numerics is about _knowing_ what kind and size of errors you introduce.

Anyway, most bridges are calculated completely with a beam model under load where finite elements do not really come into play.

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Stupid boy.

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Rubbish. I use the same server and had no problems.

out cold after Saturday?

and by the time I got in, most of

You mean you filtered out all the lost arguments.

Is that because you don't have enough of a brain to make such decisions?

When is your keeper next available to read them to you?

Why? It's all you're fit for. ...(_!_)...

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Stop posting on usenet.

Bob

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Yes, you good folks on .dsp. ;-)

You should have heard his nonsense on alt.folklore.computers (don't recall if the nut posted to comp.arch too)! He has a long history of infesting newsgroups. Trust me, it's best to deal with vermin as they show up!

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