Re: A question of simplification

• To: mathgroup at smc.vnet.net
• Subject: [mg21001] Re: [mg20953] A question of simplification
• From: eexiong <eexiong at ee.ust.hk>
• Date: Thu, 2 Dec 1999 21:41:14 -0500 (EST)
• References: <B46B3B34.35C1%andrzej@tuins.ac.jp>
• Sender: owner-wri-mathgroup at wolfram.com

```Thank you!
Indeed, I have made a mistake. The right answer for me is D[c*f[x],x]=c f[x].
For multiplier c other than a vector, I should use " * ".

Green

Andrzej Kozlowski wrote:

> I assume you do mean to the dot product when you use "." don't you? The
> reason why you get left with 0.f[x] is that Mathematica considers all your
> symbols as scalars rather than vectors and does not know what the dot
> product of the scalar 0 with something else is. So you have two possible
> approaches. Either work explicitly with vectors, e.g.:
> In[21]:=
> D[{c1, c2, c3}.{f1[x], f2[x], f3[x]}, x]
> Out[21]=
> c1 f1'[x] + c2 f2'[x] + c3 f3'[x]
>
> or teach Mathematica the following new rule:
>
> In[23]:=
> In[92]:=
> Unprotect[Dot];
> Dot /: Dot[___, 0, ___] := 0
> Protect[Dot];
>
> so that now:
>
> In[26]:=
> D[c.f[x], x]
> Out[26]=
> c . f'[x]
>
> > From: Xiong Zhenhua <eexiong at ee.ust.hk>
To: mathgroup at smc.vnet.net
> > Date: Wed, 1 Dec 1999 01:50:20 -0500 (EST)
> > To: mathgroup at smc.vnet.net
> > Subject: [mg21001] [mg20953] A question of simplification
> >
> > In Mathematica, when using partial differentiation, we get
> > D[c.f[x]]=0.f[x]+c.f'[x], not c.f'[x]. So, when the equation is long,
> > the result will be very tedious.
> > How can I simply the result to c.f'[x]? I have tested Simplify and
> > FullSimplify. It does not work.