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Re: Defining derivatives
*To*: mathgroup at smc.vnet.net
*Subject*: [mg84499] Re: Defining derivatives
*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
*Date*: Sat, 29 Dec 2007 20:02:23 -0500 (EST)
*References*: <fl4uhv$9jb$1@smc.vnet.net>
Hi,
if x is symbolic, Abs[x]>1.0 can't not evaluated and
you should define:
QuarticSpline[x_?NumericQ] := Module[{r}, r = Abs[x];
If[r > 1.0,
0.0,
1 + r^2 (-6 + r (8 - 3 r))
]
]
QuarticSplineD1[x_?NumericQ] := Module[{r}, r = Abs[x];
If[r > 1.0,
0.0,
Sign[x]*r (-12 + (24 r - 12))
]
]
Derivative[1][QuarticSpline][x_] := QuarticSplineD1[x]
Regards
Jens
KF 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
>
>
>
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