Re: defining a function with D
- To: mathgroup at smc.vnet.net
- Subject: [mg37175] Re: defining a function with D
- From: "Allan Hayes" <hay at haystack.demon.co.uk>
- Date: Tue, 15 Oct 2002 04:17:48 -0400 (EDT)
- References: <aodnti$mlg$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Jason,
We have
dfx[x_,t_]= D[f[x,t],x]
t Cos[t x]
One advantage of using = rather than := is that it differentiates once, when
the definition is stored,
Definition[dfx]
t Cos[t x]
With := we get
Clear[dfx]
dfx[x_,t_]:= D[f[x,t],x]
Definition[dfx]
dfx[x_, t_] := D[f[x, t], x]
So the differentiation is done each time that the function dfx is evaluated.
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Jason Miller" <millerj at truman.edu> wrote in message
news:aodnti$mlg$1 at smc.vnet.net...
> Dear Listers,
>
> I find myself defining functions in terms of differentiation. For
example,
>
> f[x_,t_]:=Sin[x*t]
> dfx[x_,t]:=D[Sin[y,t],y]/.y->x
>
> This works well, but it seems to me that there should be a better way
> to do this. That is, there should be a better way to define a
> 'derivative' of a previous function without going through the
> replacement contortions. I can't find the answer in the archive.
> Can someone tell me the most straightforward way to do this? Will it
> work to define a gradient vector or Jacobian matrix? A Hessian
> matrix?
>
> Thanks in advance.
> --
> Jason Miller, Ph.D.
> Division of Mathematics and Computer Science
> Truman State University
> 100 East Normal St.
> Kirksville, MO 63501
> http://vh216801.truman.edu
> 660.785.7430
>