Re: concept of test function in calculus
- From: Virgil <virgil@xxxxxxxxxxx>
- Date: Thu, 18 Oct 2007 16:08:13 -0600
In article <1192743852.218647.30050@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
luca.pamparana@xxxxxxxxx wrote:
On Oct 18, 10:21 pm, luca.pampar...@xxxxxxxxx wrote:
On Oct 18, 10:03 pm, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1192740679.519486.213...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
luca.pampar...@xxxxxxxxx wrote:
Hi everyone,
I was reading about the dirac delta function which is defined as an im
pulse which has infinite magnitude at x = 0 and is 0 everywhere else.
However, when I look at the mathematical equation describing it on
wikipedia and other books, it is as follows:
Integral (-inf, +inf) f(x) del(x) dx = f(0)
I thought Integral (-inf, +inf) del(x) dx = 1 should be sufficient.
What is this f(x) function and why is it used? Wikipedia says it is
some sort of a "test function" but I am not sure as to what it means!
I would be grateful if someone could explain this to me.
Thanks,
Luca
I think that what they are saying is that the del function has the
property that that for every function f(x) defined on R,
Integral (-inf, +inf) f(x) del(x) dx = f(0)
Or at least that it holds for every function on R which is bounded on
some neighborhood of 0.
Hello Virgil,
Thanks for the reply. I guess that makes sense.
I have one more question and that might sound stupid and it is
probably due to my dodgy calculus knowledge.
I can see why the intergral of f(x) at x = 0 is 1 because the area of
the dirac function is 1. However, how come the area of the curve
everywhere else has become zero as well? So, how come the curve only
has an area under the origin?
I actually wonder if this question makes sense at all...
Thanks again,
Luca
I think it is because the area of del(x) is 0 everywhere except the
origin that makes the integral disappear everywhere else.
I am just very weak at integrals..
So, we have
Integral (-inf, +inf) f(x)del(x)dx
= f(x)del(x) - integral(del(x)dx
Is that correct or have I messed it up completely?
Integral (-inf, +inf) del(x) dx = 1 if you believe that there is such a
function at all.
See my other post to see that it can be sort of justified as a limit of
a sequence of integrals.
.
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