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
>
>
>

```

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