CAS command question
01-04-2017, 03:23 AM (This post was last modified: 01-04-2017 01:04 PM by compsystems.)
Post: #8
 compsystems Senior Member Posts: 1,380 Joined: Dec 2013
RE: CAS command question
What is the reason for "reverse" expression (transposing terms and reverse the operator)? (some help for the parser or interpreter of expressions), in my case I do not like as the hp-prime shows, books do not show this notation, is somewhat confusing for to read.
(x+y)< 4 →
4>(x+y)

Examples of how to use the command LEFT & RIGHT

PHP Code:
// simplify none#cas    testleftcas():=  begin   local expr1;    //local expr1:='(x+y)<4';  =(    expr1:='(x+y)<4'      return([ left(expr1), left(eval(expr1)) ]);  end;  // testleftcas() -> [ x+y, 4 ] // ok  testleftcas2(expr2):=  begin    return [ left(expr2), left(eval(expr2)) ];  end;  // testleftcas2( '(x+y)<4' ) -> [ x+y, 4 ] // ok  // testleftcas2( (x+y)<4 ) -> [ 4, 4 ]    // ok  testcas( id0 ):=  begin    local id1:= id0;    print; //print("");    print( [ left(id1), right(id1) ] ); wait;    return ( [ left(id1), right(id1) ] );  end;  // testcas( (x+y)^(1+1)-3 ) -> [(x+y)^2,-3] // ok  // testcas( '(x+y)^(1+1)-3' ) -> [(x+y)^(1+1),-3] // ok#end
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