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MathGroup Archive 2000

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Re: ? D[f,{x,n}]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg25554] Re: ? D[f,{x,n}]
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Sat, 7 Oct 2000 03:35:59 -0400 (EDT)
  • References: <8rjioe$oea@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Jack,

It seems that D[f,x] and D[f, {x,1}] are independently definable

Clear[D]

Unprotect[D];

D[f_, x_] := A
D[f_, {x_, 1}] := B

DownValues[D]

{HoldPattern[D[f_, {x_, 1}]] :> B, HoldPattern[D[f_, x_]] :> A}

Never mind the math its the matching that counts.

D[f, x]

        A

D[f, {x, 1}]

        B

--
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

"Jack Goldberg" <jackgold at math.lsa.umich.edu> wrote in message
news:8rjioe$oea at smc.vnet.net...
> Hi folks,
>
> I am touching up a package I have worked on for a number of years.
> Rather than send it off as is, I keep tweeking it to improve its
> usefulness.  I should know better...  My latest tweek leads me to
> a problem with  D[f,{x,n}].  Here's the story:
>
> In my package I define a constant "flag" which I set true: flag = True.
>
> Then I Unprotect D:  Unprotect[D]:  I want to modify f before the built-in
> D fires so I write a little program:
>
> D[f_,x_]/;flag := Block[ {flag=False},
>   D[ mySimplification[f],x]
> ]
>
> So, what I think happens is this. Since flag = True and since Mathematica
uses
> my definition of D before its own, this little snippet fires.  Block sets
> flag = False, mySimplification[f] fires, and my definition of D does not
> fire again (or I would be caught in a loop) so Mathematica's definition of
D
> now works and I have accomplished my aim which is to simplify f before
> taking the derivative.  This seems to work.  However, this idea fails to
> work for  D[f,{x,n}]!  Here's my code:
>
> D[f_,{x_,n_}]/;flag := Block[ {flag=False},
>   D[ mySimplification[f],{x,n} ]
> ]
>
> (Of course, the nature of mySimplification should be irrelevant.)
>
> While trying to work out this problem, I discovered that I don't
> understand the difference between  D[f,x] and D[f,{x,1}].  If I
> alter the definition of D[f,x] have I also altered the definition
> of  D[f,{x,1}]?  How about the other way around?  If I alter the
> definiton of  D[f,{x,1}] does this alter the definition of D[f,x]?
> It seems as though they should be linked internally but if so,
> I can find no reference to this linkage. This not a mute point since
> I found that in trying to use mySimplification in both  D[f,x] and
> D[f,{x,n}] it is necessary to know what happens if a user chooses to
> write D[f,x] in the form  D[f,{x,n}].
>
> Help is needed.  Thanks.
>
>
>






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