In article <email@example.com>, firstname.lastname@example.org (Michael Stemper) wrote:
> In article > <email@example.com>, MBALOVER > <firstname.lastname@example.org> writes: > >On Jul 11, 7:36=A0pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email> wrote: > >> In article > >> <4c1e3a16-63c5-470a-b617-8528f4005...@u6g2000prc.googlegroups.com>, > >> =A0MBALOVER <mbalov...@gmail.com> wrote: > > >> > From WIKI, I learn that dirac delta function is not a true function. > > >> > Could you please help me to understand it? > > >> A function from a set A to a set B is an assignment of > >> an element of B to each element of A. > >> > >> If A and B are not specified, it is often assumed that both > >> are taken to be the set R of real numbers. > >> > >> 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".
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). The point of delta is what it *does* (as just given), not what its values may or may not be, and in that way it differs from a "true function".
-- Gerry Myerson (email@example.com) (i -> u for email)