Re: questions about delayed expression.
- To: mathgroup at smc.vnet.net
- Subject: [mg19891] Re: [mg19862] questions about delayed expression.
- From: "David Park" <djmp at earthlink.net>
- Date: Sun, 19 Sep 1999 18:47:38 -0400
- Sender: owner-wri-mathgroup at wolfram.com
Wen-Feng Hsiao wrote: >Hi, > > The following is the process that I run in my notebook. > >In[1]:= >a[x_] := x + 4 >b[x_] := -3 x + 8 > >In[2]:= >sola = x /. Solve[{a[x] == c}, x]; >solb = x /. Solve[{b[x] == c}, x]; > >In[3]:= >Inter[c_] := Interval[{sola[[1]], solb[[1]]}] > >In[4]:= >Inter[.3] >Inter[.4] > >Out[4]= >\!\(Interval[{\(-4\) + c, \(8 - c\)\/3}]\) > >Out[5]= >\!\(Interval[{\(-4\) + c, \(8 - c\)\/3}]\) > >My questions are: >1. Why Inter[.3] and Inter[.4] cannot be evaluated? Their results should >not be the same. This is not my intention. > >2. I don't know if there is any better way to extract the 'root(s)' from >the output of 'Solve' command. The output form is {{x->root1}, {x- >>root2}, ...{}}. If I use 'ReplaceAll'(/.) command, it will remain a list >of solutions of x. It seems I can only use element operation to extract >the root(s) from the solution list? In my case, I use sola[[1]] and >solb[[1]]. > >Please give me suggestions. Thanks for your help. > > There are two things that can help. First, you can pick off the two solutions by: sola = x /. Solve[{a[x] == c}, x][[1,1]] -4 + c solb = x /. Solve[{b[x] == c}, x][[1,1]] (8 - c)/3 Then, define Inter with immediate evaluation: Inter[c_] = Interval[{sola, solb}] Interval[{-4 + c, (8 - c)/3}] To get Inter[c_] defined as a function of c, it is necessary to get c explicitly on the right hand side. Then wherever c occurs, it is replaced by the argument as the first step in evaluation. But with the delayed definition, there is no c there, so the argument never gets substituted. Inter[.3] Interval[{-3.7, 2.56667}] Inter[.4] Interval[{-3.6, 2.53333}] David Park djmp at earthlink.net http://home.earthlink.net/~djmp/