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MathGroup Archive 2006

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Re: Derivative[1] applied to numeric constants

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66780] Re: [mg66760] Derivative[1] applied to numeric constants
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 30 May 2006 05:48:23 -0400 (EDT)
  • References: <200605291006.GAA07577@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 29 May 2006, at 19:06, Andrew Moylan wrote:

> Hi,
>
> Do numeric constants have special behaviour under the Derivative[1]
> function?
>
> The number e.g. 2 is not defined as the function that always  
> returns 2:
>
> In[1]:=2[x]
> Out[1]=2[x]
>
> But Derivative[1][2] is defined:
>
> In[2]:=2'
> Out[2]=0&
>
> Could anyone explain why this is? Is this behaviour documented in the
> help system?
>
> Cheers,
>
> Andrew
>


I can't give a definitive answer but note that we also have this  
closely related behaviour,that is documented:

In[1]:=
SetAttributes[f,Constant]

In[2]:=
Derivative[1][f]

Out[2]=
0&

or

In[3]:=
D[f[#]&,#]

Out[3]=
0&

and

In[4]:=
Dt[f[#]&]

Out[4]=
0&


This is consistent with the documentation for Constant:

Functions f[ ? ] are taken to have zero total derivative if f has  
attribute Constant.

If a symbol f with attribute Constant is treated in this way, it  
seems reasonable that genuine constants like 2 also are, although I  
can't find any explicit mention of this (perhaps nobody has  
considered the possibility that anyone might ask ;-)). But note that  
(from the Help for Derivative):

Whenever Derivative[n][f] is generated, Mathematica rewrites it as D[f 
[#]&, {#, n}].

So Derivative[1][2] is D[2[#]&,{#,1}] and presumably, for the same  
reason as f[#]& when f has attribute Constant, this is  taken to be 0.

As for the "deeper reasons" why this is so: at the moment I can't  
think of one.  I might play some role in the mechanism of functional  
differentiation or it could simply be a side-effect of something that  
does.

(Of course, the function that always returns 2 is 2& and it has the  
same derivative as the abnormal "function" 2[#]& (or simple 2):

Derivative[1][2&]

0&


Derivative[1][2]


0&

Andrzej Kozlowski


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