Re: help with dirac delta function ?



On 2007-12-22 08:01:08 -0500, pagcal <pagcal@xxxxxxxxxx> said:

Can someone offer help understanding properties of the dirac delta function?

given

integ(delta(alpha*x)) = integ(delta(u)du/|alpha|) = 1/|alpha|

In the second term, how did magnitude alpha appear and in the
denominator?

integ is integral +infinity to -infin

You can also use the Fourier transform's property (scaling) that says
u(ax) =F.T.=> 1/|a| U(w/a)
where F.T.[u] = U.

Since F.T.[delta] = 1, therefore
delta(ax) = 1/|a| delta(x).

--

-kira

.



Relevant Pages

  • help with dirac delta function ?
    ... In the second term, how did magnitude alpha appear and in the ... integ is integral +infinity to -infin ...
    (sci.math)
  • Re: help with dirac delta function ?
    ... how did magnitude alpha appear and in the ... integ is integral +infinity to -infin ...
    (sci.math)
  • Re: limit of an equation
    ... >>> infinity, so I thought I'd be able to solve using L'Hopitals rule. ... >>> original function! ... >> Why not divide numerator and denominator by something (the same for ... >> both numerator and denominator, that is) to make the whole thing ...
    (sci.math)
  • Re: limit of an equation
    ... > Mike wrote: ... >> Evaluating the limit of fresults in an infinity over an infinity, ... > Why not divide numerator and denominator by something (the same for ... > both numerator and denominator, that is) to make the whole thing ...
    (sci.math)
  • Re: limit of an equation
    ... >> infinity, so I thought I'd be able to solve using L'Hopitals ... >> of the original function! ... > Why not divide numerator and denominator by something (the same for ... > both numerator and denominator, that is) to make the whole thing ...
    (sci.math)