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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