Re: Coefficient Question
- To: mathgroup at smc.vnet.net
- Subject: [mg104056] Re: Coefficient Question
- From: Bill Rowe <readnews at sbcglobal.net>
- Date: Sat, 17 Oct 2009 07:03:08 -0400 (EDT)
On 10/16/09 at 7:19 AM, sguthery at gmail.com (Scott) wrote:
>I've got the expression
>8(3+5k+4x1+6x2+45x3)
>and I want to extract the constant 24.
>Yes, I can define a function and then set k and x1 and x2 and x3 to
>0 but there has to be easier way, no?
>Any suggestions?
Is this acceptable to you?
In[5]:= expr = 8 (3 + 5 k + 4 x1 + 6 x2 + 45 x3);
expr /. {8 (3 + b_) -> 8 b}
Out[6]= 8 (5 k+4 x1+6 x2+45 x3)
Note, using replacement rules in this fashion isn't truly a
subtraction operation. There is no mathematics being done by the
replacement rule. It is up to the user to determine the
mathematical validity of doing replacements like this.