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Re: Define a function and its derivatives

  • To: mathgroup at smc.vnet.net
  • Subject: [mg43412] Re: Define a function and its derivatives
  • From: "Martin Manscher" <reversed-email-rehcsnam at kd.utd.tac>
  • Date: Fri, 29 Aug 2003 07:16:20 -0400 (EDT)
  • References: <bi7mvh$p1h$1@smc.vnet.net> <bihovv$b3o$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Paul Abbott" <paul at physics.uwa.edu.au> wrote

> But these derivatives are _not_ correct over the same domain as the
> function (you have z Real which includes z == 0)! The DiracDelta's arise
> (correctly here) because you are taking derivatives of a step function.

I understand where the DiracDeltas come from (and that they are in principle
correct), and that the derivative is undefined in z==0. But since my purpose
of defining the function is using it with NDSolve, this is of minor
importance. But of course you are right; I should somehow take care of z==0.

What I really want is the continuous function whose derivative is E^(-z^n)
for z real. This function exists (or at least, so it seems to me :-), and is
something like

Sign[z] (Gamma[1/n] - Gamma[1/n, Abs[z]^n])/n

The problem seems to arise from Mathematica's integration. I started out
from

In[1]:= P[z_] = E^(-z^n);

-- and then integrated:

In[2]:= PInt[z_] = Integrate[P[z],z]

Out[2]= -z (z^n)^(-1/n) Gamma[1/n, z^n])/n.

This function is discontinuous in z==0. Mathematica seems to know the value
of this function for z==0:

In[3]:= PInt[0]

Out[3]= 0

-- but not for z==0. :

In[4]:=PInt[0.] /. n -> 4
Power::infy : Infinite expression 1/0.^(1/4) encountered. More...

Infinity::indet : Indeterminate expression 0. ComplexInfinity encountered.
More...

Out[4]= Indeterminate


Martin



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