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