Mathematica 9 is now available
Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Defining derivatives

  • To: mathgroup at smc.vnet.net
  • Subject: [mg84491] Re: [mg84484] Defining derivatives
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sat, 29 Dec 2007 19:58:10 -0500 (EST)
  • Reply-to: hanlonr at cox.net

Clear[QuarticSpline];
QuarticSpline[x_] := Module[{r},
   r = Sqrt[x^2]; Piecewise[{{1 + r^2 (-6 + r (8 - 3 r)), r <= 1}}]];

QuarticSpline'[x] // FullSimplify

Piecewise[
   {{-12*(x^3 - 2*Abs[x]*x + x), 
       Sqrt[x^2] <= 1}}]

Plot[{QuarticSpline[x], QuarticSpline'[x]}, {x, -1.1, 1.1}]


Bob Hanlon

---- KF <fresl at grad.hr> wrote: 
> (I am sorry if this or similar question was already answered. I did 
> archive search, but didn't find anything.)
> 
> I have a function:
> 
>    QuarticSpline[x_] :=
>      Module[
>        { r },
>        r = Abs[x];
>        If [r > 1.0,
>            0.0,
>            1 + r r (-6 + r (8 - 3 r))
>        ]
>      ]
> 
> It's first derivative should be (at least for my purposes :o) :
> 
>    QuarticSplineD1 [x_] :=
>      Module[
>        { r },
>        r = Abs [x];
>        If [r > 1.0,
>            0.0,
>            Sign[x] * r (-12 + (24 r - 12))
>        ]
>      ]
> 
> but D[QuarticSpline[x], x] returns
> 
>    If[Abs[x] > 1., 0, 0]
> 
> Now, Mathematica Tutorial, in chapter Defining Derivatives, says
> 
> "You can define the derivative in Mathematica of a function f of one 
> argument simply by an assignment like f'[x_]=fp[x]."
> 
> But if I try to write
> 
>     QuarticSpline'[x_] = QuarticSplineD1[x]
> 
> I got
> 
>     Set::write: "Tag Function in (If[Abs[#1]>1.,0,0]&)[x_] is Protected."
> 
> Is it possible to define QuarticSpline' to "return" QuarticSplineD1 ?
> 
> Regards,
> 
> K. Fresl
> 
> 
> 



  • Prev by Date: Re: Solving stiff differential equations
  • Next by Date: Re: Defining derivatives
  • Previous by thread: Defining derivatives
  • Next by thread: Re: Defining derivatives