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Re: dirac delta function is not a TRUE function??
Posted:
Jul 18, 2011 1:53 PM
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In article <gerry-E6603D.09354915072011@news.eternal-september.org>, Gerry Myerson <gerry@maths.mq.edi.ai.i2u4email> writes:
>In article <ivn53v$7e1$1@dont-email.me>, mstemper@walkabout.empros.com (Michael Stemper) wrote:
>> In article <9f2192a5-9776-4c55-9a2d-eeb93f0559ab@g3g2000prf.googlegroups.com>, MBALOVER <mbalover9@gmail.com> writes:
>> >On Jul 11, 7:36=A0pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote:
>> >> The Dirac delta function does not assign a real number to zero,
>> >> and is thus not a true function.
>>
>> >Thanks Gerry. But I am still confused. I think at x =3D 0, delta(x) =
>> >infinity.
>>
>> That's one way of looking at it -- in fact it's the one that I was
>> taught while studying electical engineering. However, look more
>> carefully at what was said: "... does not assign a real number to
>> zero ...".
>>
>> There is no real number called "infinity".
>
>Quite.
>
>You could try to think of delta as a function
>from the reals to the extended reals, the extended
>reals being "the reals union infinity", but that
>won't help you to understand why
>integral from minus infinity to infinity of f(x) delta(x) dx
>is f(0).
It's been a long time since I studied this, but isn't that true for
the integral from minus epsilon to epsilon (for any positive epsilon),
as well?
--
Michael F. Stemper
#include <Standard_Disclaimer>
Life's too important to take seriously.
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