Never mind, it?s the chain rule. I just thought of it.
I am not always on the ball.
Rich
From: Richard Hennessy
Sent: Wednesday, July 17, 2013 5:03 PM
To: Barton Willis ; Maxima List
Subject: Re: [Maxima] pw.mac problem
From: Barton Willis
Sent: Sunday, July 07, 2013 9:47 AM
To: Richard Hennessy ; Maxima List
Subject: RE: [Maxima] pw.mac problem
Let F be integrable and continuous at 0. Then F(x) delta(x) = F(0). Specializing this to F(x) = x yields x delta(x) = 0.
Thus x delta'(x) = -delta(x) and x delta''(x) = -2 delta'(x).
How did you get that result I changed to blue, I have been racking my brain over it?
Let F(x) = F(0) + F'(0) x + G(x) x^2, where G is continuous at zero. Then G(0) = (1/2) F''(0); further
F(x) delta''(x) = (F(0) + F'(0) x + G(x) x^2) delta''(x) = ... = F(0) delta''(x) - 2 F'(0) delta'(x) + F''(0) delta(x).
This calculation extends to any positive integer n. The generalization of this to any positive integer n is listed at
http://functions.wolfram.com/GeneralizedFunctions/DiracDelta/20/01/ . Integrating this identity gives the relation you want.
Incidentally, Maxima's general simplifier does x * delta(x) / x --> delta(x). In the context of distributions,
cancelling the common factor isn't desired.
--Barton
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Hi List,
Someone pointed out that there is an error in pw.mac, for integration of delta functions and their derivatives.
Currently,
integrate(f(x)*diff_pwdelta(m, x-a), x);
gives the wrong answer.
This person says it should be.
integrate(f(x) * pwdelta(x-a, n), x);
sum((-1)^k*binomial(n,k)*at(diff(f(x),x,k),x=a) * pwdelta(x-a,n-k-1),k,0,n)
where the two argument form of pwdelta(x,n) = diff_pwdelta(n,x) and pwdelta(x,0) = pwdelta(x) and pwdelta(x,-1)=unit_step(x).
Does anyone know exceptions to this answer? It is beyond me to verify the math. (Maybe I am not qualified to write pw.mac).
Rich
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