Re: Some Easy (hopefully) Questions

*To*: mathgroup at christensen.cybernetics.net*Subject*: [mg2038] Re: Some Easy (hopefully) Questions*From*: David Harvatin <dtharvat at sprint.uccs.edu>*Date*: Sat, 16 Sep 1995 01:41:41 -0400*Organization*: University of Colorado at Boulder

Rob Carscadden <carscadd at pps.pubpol.duke.edu> wrote: >I'm trying to run a model using Mathematica. I've scoured all our >libraries and all the books are gone. Working with all too basic >Mathematica By Example, I haven't been able to find how to do some rather >easy things. > >The first thing I want to do is find a max (I need to plug the max and >where it occurs into anther equation). I have a function, and I have >taken both the first and second derivatives. I solve for f'(x) = 0. But >here's where the trouble comes in. I want to plug these points back into >my f(x). How can I do this. Look at the following example: > >possmax = Solve[(x - 1)^2 == 0,x] // N > >I get {{x -> 1.},{x ->1.}} >and when I try > >possmax[[1]] > >I get {x ->1.} > >but I want the number 1 instead. > >I know this if probably extremely basic, but due to lack of available >documentation here, I can not figure it out. > > >P.S. another useful trick to know would be how many objects are in a >list. My function a bit more complicated (compositions, logs etc.), and I >don't exactly how many solutions I will have from the Solve[ ] statement. > >-- > >Rob Carscadden >carscadd at pps.duke.edu >http://www.duke.edu/~cars >49% of all Statistics are made up on the spot! > > > > > Rob, To answer your first question, suppose you want to put both solutions into an expression called exp1 that is also a function of x : exp1 = 5 x -3. The ReplaceAll operator (forward slash then period) will replace all occurrences of x in exp1 with the values you obtain for possmax if you do the following : exp1 /. possmax Your answer will be in the form of a list, with each element of the list corresponding to one of your x values in possmax. If you want only one solution, then do the following : exp1 /. possmax[[1]] or exp1 /. possmax[[2]] To answer your second question, use the Length command : Length[possmax] gives you an answer of 2. For a list such as : list = { {1,2},{3,4} }, Length[list] also gives an answer of 2. Good luck. I had to learn Mathcad without the benefit of the book, so I know how you feel. BTW, unlike Mathcad, Stephen Wolfram's MMA book is readily available in most large bookstores for around $50. Dave Harvatin dtharvat at sprint.uccs.edu