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Re: simplifying a resulted derivative expression

  • To: mathgroup at smc.vnet.net
  • Subject: [mg96074] Re: simplifying a resulted derivative expression
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Wed, 4 Feb 2009 05:20:12 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <gm9ckc$135$1@smc.vnet.net>

In article <gm9ckc$135$1 at smc.vnet.net>, kem <kemelmi at gmail.com> wrote:

> Hi, I have a function f[ux, uy, vx, vy, wx, wy, u, v, w, x, y]
> 
> I am doing different manipulations with this function and the results
> I get have expressions like that:
> 
> Derivative[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0][f][ux, uy, vx, vy, wx, wy,
> u, v, w, x, y]
> 
> which is difficult to understand. The question is how can I substitute
> automatically the above expression by: fu ?
> 
> (it means I need to see that the derivative is taken over the 7th
> argument which is 'u' and write that f is derived with respect to u
> which I want to be written as fu)
> 
> Can it be done?

AFAIR, Mathematica displays derivatives differently depending on how 
many variables the expression has and the order of differentiation (what 
are these thresholds, I cannot remember nor find them now). You may want 
to write your own "operator" like in In[2]/Ou[2].

In[1]:= fu = Derivative[0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0][f][ux, uy, vx,
vy, wx, wy, u, v, w, x, y]

pd[f @@ fu, fu[[Last[Flatten[Position[fu[[0]], 1]]]]]]

% /. pd -> D

D @@ %%

D[f[x], x]

Out[1]= 
 (0,0,0,0,0,0,1,0,0,0,0)
f                       [ux, uy, vx, vy, wx, wy, u, v, w, x, y]

Out[2]= pd[f[ux, uy, vx, vy, wx, wy, u, v, w, x, y], u]

Out[3]= 
 (0,0,0,0,0,0,1,0,0,0,0)
f                       [ux, uy, vx, vy, wx, wy, u, v, w, x, y]

Out[4]= 
 (0,0,0,0,0,0,1,0,0,0,0)
f                       [ux, uy, vx, vy, wx, wy, u, v, w, x, y]

Out[5]= f'[x]

HTH,
--Jean-Marc


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