Re: Factorising operators??
- To: mathgroup at smc.vnet.net
- Subject: [mg30990] Re: Factorising operators??
- From: "Andrzej Kozlowski" <andrzej at bekkoame.ne.jp>
- Date: Sat, 29 Sep 2001 20:43:34 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I think the easiest way is as follows:
In[1]:=
Unprotect[Plus];
In[2]:=
(m_:1) a[x] +(n_:1)b[x_]+c_:0 :=(m a + n b+c)[x]
In[3]:=
Protect[Plus];
In[4]:=
3a[x]+5b[x]+2
Out[4]=
(2+3 a+5 b)[x]
In[5]:=
a[x]+5b[x]
Out[5]=
(a+5 b)[x]
One should always think twice before modifying such basic built in
functions as Plus. You might therefore prefer to use a different
approach. You might for example define a function f, that will transform
your expression only when you apply f to it:
In[2]:=
f[(m_:1) a[x] +(n_:1)b[x_]+c_:0] :=(m a + n b+c)[x]
In[3]:=
f[3a[x]+5b[x]+2]
Out[3]=
(2+3 a+5 b)[x]
Or you might even find it convenient to use f as a transformation
function in Simplify:
In[6]:=
Simplify[3a[x]+5b[x]+2,TransformationFunctions->{Automatic,f}]
Out[6]=
(2+3 a+5 b)[x]
Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/
On Friday, September 28, 2001, at 04:55 PM, Mat Bowen wrote:
> Hello,
>
> I'm new to mathematica and looking for help with the following:
>
> I have two operators, a and b, that function as approximations to
first
> and
> second order derivatives. I can write (a+b)[x] and then use
> Through[%,Plus]
> to generate a[x]+b[x] but I want to do the reverse, ie. Get
mathematica
> to
> output (a+b)[x] if I give it a[x]+b[x]. Is there any way (preferably
> simple)
> to do this which will also work with more general cases ie.
> a[x]+5b[x]+const
> c[x] should produce (a+5b+const)[x]
>
> Thanks,
> Mat Bowen
>
>
>