Re: Mathematica question

*To*: mathgroup at smc.vnet.net*Subject*: [mg124799] Re: Mathematica question*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Tue, 7 Feb 2012 04:04:30 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com

On 2/6/12 at 2:39 AM, hcohl001 at gmail.com (Howie) wrote: >Let's say you've got an expression >expr = {Sin[x] + x^2 - x^2*Cos[x], x^3 (1 + x^2) - x^3, x^4 - x^3} >What does this mean? >Table[expr[[j]] // FullSimplify[#, $assumptions] & /@ # & // Expand >// @ # &, {j, 3}] Until $assumptions is defined, it doesn't mean much. But perhaps this is a typo and it should be the built-in symbol $Assumptions. If so and $Assumptions has not been modified from the default value the expression above has been made needlessly complex. That is Table[expr[[j]]//FullSimplify[#, $Assumptions]&, {j,3}] is exactly equivalent to Table[expr[[j]]//FullSimplify, {j,3] which is exactly equivalent to expr//FullSimplify That is In[5]:= (Table[ expr[[j]] // FullSimplify[#, $Assumptions] & /@ # & // Expand //@ # &, {j, 3}]) == (expr // FullSimplify // Expand) Out[5]= True >I am new to Mathematica as far as these more complicated expressions >are concerned. >Can you explain symbol by symbol? The # is a place holder in a pure function The & is used to define a pure function. So =46ullSimplify[#, $assumptions] & is a pure function that when used as =46ullSimplify[#, $assumptions] &[expr] or =46ullSimplify[#, $assumptions] &@ expr simply does a full simplification of expr The /@ is short hand for Map, i.e. =46ullSimplify[#, $assumptions] &/@expr is short hand for Map[FullSimplify[#, $assumptions] &, expr] But since FullSimplify does what it can on any expression, there is no need to Map FullSimplify to each element of a list. That is, Map[FullSimplify[#, $Assumptions] &, expr] does the same as =46ullSimpify[expr] or =46ullSimplify@expr (* pre-fix form *) or expr//FullSimplify (* post fix form *) The same observations apply to Expand