"Richard Hennessy" <rich.hennessy at verizon.net> writes:
> Yes, I know about distributions and I am thinking of diracdelta as a
> limit of very spiked distributions with an increasingly small
> variance. A function class I have used is
>
> assume(s>0);
> f[0](x) := exp( -x^2/(2*s) ) / sqrt( 2 * s * %pi);
> for i : 1 thru n do define( f[i](x), diff( f[i-1](x), x ));
>
> as s approaches zero this is a way to double check results, but
> f[0](x) is an even function and I was thinking that may not always
> have to be the case, in general. This function works at least for
> polynomials but it is pretty limited. You can't even do
> limit(integrate(f[0](x-a) * sin(x), x, minf, inf), which means I
> should find another function.
>
> limit(integrate(x^3*f[0](x-a),x,minf,inf),s,0)
>> a^3
> limit(integrate(x^3*f[1](x-a),x,minf,inf),s,0)
>> -3*a^2
I'm a little bit confused. Why are you using approximants to delta at
all? I haven't looked at your pw.mac code, but presumably you have some
rule that transforms
integrate(delta(x)*phi(x), x, a, b)
to something like
if is (a < 0 and b > 0) then <some expression with limits> else 0
Can't you just transform it to phi(0)? Or is there some other problem
you're solving?
As far as I understand it, the whole point of these "idealised
functions" (ie distributions) is to avoid the icky business of computing
with the approximants in the first place.
> "For delta, we can do slightly better and, if we define delta as a limit
> of functions in some space of distributions, I'm pretty certain that the
> value of (1) is independent of the approximating functions you choose as
> long as phi is continuous at zero. If phi is *not* continuous at zero,
> however, the answer depends on your approximating sequence."
>
> I have searched for other functions that may be used for this type of
> analysis without finding many good substitutes. Going the abstract way
> and defining a distribution class that is independent of the details
> of the chosen function is rather difficult. I am pretty certain you
> are right that it depends on the chosen distribution and if phi is
> discontinuous at zero the answer will change depending on f[n](x).
Yes, definitely. For example, take the approximants
dd (x, t, a) = { a*t when x ? [-1/t, 0]
{ (1-a)*t when x ? [0, 1/t]
{ 0 otherwise
The integral of dd is always 1 and the support is contained in [-1/t,
1/t], so for any a we get a sequence of functions approximating delta.
Now let f be a function that isn't necessarily continuous but has a
left and right limit at zero. Call these limits f(0-) and f(0+). I claim
that the limit of
integrate (dd(x,t,a)*f(x), x, -1, 1)
as t?? is a*f(0+) + (1-a)*f(0-).
To see this, you have to unpack the definition of right and left
limit and then pick t large enough that f(0-) and f(0+) give a close
approximation to f on [-1/t,0] and [0, 1/t] respectively.
Rupert
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