On Jul 12, 3:48 am, MBALOVER <mbalov...@gmail.com> wrote: > Hi all, > > From WIKI, I learn that dirac delta function is not a true function. > However, I do not understand the explanation in WIKI and still get > confused, why it is not a TRUE function. > > Could you please help me to understand it? > > Thank you.
One way of looking at it is by analogy with vectors :
Let's take two 4 vectors :
[x1 , x2 , x3 ] = X [y1 , y2 , y3 ]= Y
The dot product for such vectors , X*Y is (x1 * y1) + (x2 * y2) + (x3 * x3) (geometrically it's the (length of of X * length of Y * (cosine of the angle between X and Y)) ) These vectors can be considered as having 3 components , indexed for example by : X(1) = x1 X(2) = x2 X(3) = x3
Similarly , a function f :R -> R can be considered a vector having R components , component p being f(p) . The dot product for functions f and g is (integral for x from -oo to +oo of f(x) * g(x) ) (the integral can be considered an infinitesimal summation of f(x) * g(x) for each component x)
This dot product for functions introduces some problems , since integrals , being infinite , need not converge ,and functions need not be integrable ) . Also , when treating functions as vectors , we can introduce some vectors that would not correspond to any normal function .
Let's get back to our 3-vector example . Assume we want a 3-vector X,such that for any Y Y*X = Y(2) . Our answer is [0,1,0] . Now , in the space of functions , assume we want a "function" G such that for any F , integral(F*G) = F(0) . Only one "function" satisfies the requirement . By looking at our vector example , [0,1,0] , for Y(2) , we notice that G must be 0 everywhere , except G(0) ,because integral (F*G) must always be F(0) . In 0 , G must have such a value that the overall integral of G is 1 . But G(x) is 0 for almost every x , G(0) cannot be an ordinary number . G(0) is a "number" so large that it cause the integral of G to become 1 , even though G(x) is 0 for all non-zero X . G is the dirac-delta "function" .