Re: Re: Factorising operators??
- To: mathgroup at smc.vnet.net
- Subject: [mg30992] Re: [mg30978] Re: [mg30959] Factorising operators??
- From: Andrzej Kozlowski <andrzej at bekkoame.ne.jp>
- Date: Sat, 29 Sep 2001 20:43:36 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Looking again at my message below I can see now a error or rather a
misprint. In fact there are two possibilities. If you want the variable
always to be x you should define f as:
f[(m_:1) a[x] +(n_:1)b[x]+c_:0] :=(m a + n b+c)[x]
If on the other hand you want to allow arbitrary variable names you
could use:
f[(m_:1) a[x_Symbol] +(n_:1)b[x_Symbol]+c_:0] :=(m a + n b+c)[x]
Andrzej
On Saturday, September 29, 2001, at 05:18 PM, Andrzej Kozlowski wrote:
> 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
>>
>>
>>
>
>
>