XPost: comp.soft-sys.math.maple   
      
   Axel Vogt schrieb:   
   >   
   > On 14.08.2015 18:45, Peter Luschny wrote:   
   > >   
   > > OK. So what about these?   
   > >   
   > > [1] -1/7+x-(2/7)*cos((2/7)*Pi)+(2/7)*cos((3/7)*Pi)+(2/7)*cos((1/7)*Pi)   
   > >   
   > > [2] (4/7)*x*cos((1/7)*Pi)-(2/7)*cos((1/7)*Pi)-(4/7)*x*cos((2   
   7)*Pi)+(2/7)*cos((2/7)*Pi)+(4/7)*x*cos((3/7)*Pi)-(2/7)*cos((3/7)   
   Pi)+1/7-(2/7)*x+x^2   
   > >   
   > > [3] (2/7)*cos((1/7)*Pi)+(6/7)*x^2*cos((1/7)*Pi)-(6/7)*x*cos(   
   1/7)*Pi)-(2/7)*cos((2/7)*Pi)-(6/7)*cos((2/7)*Pi)*x^2+(6/7)*x*cos   
   (2/7)*Pi)+(2/7)*cos((3/7)*Pi)+(6/7)*x^2*cos((3/7)*Pi)-(6/7)*x*co   
   ((3/7)*Pi)-1/7+(3/7)*x-(3/7)*x^2+x^3   
   > >   
   > > [4] -(2/7)*cos((1/7)*Pi)-(12/7)*x^2*cos((1/7)*Pi)+(8/7)*x*co   
   ((1/7)*Pi)+(8/7)*x^3*cos((1/7)*Pi)-(8/7)*cos((2/7)*Pi)*x^3+(2/7)   
   cos((2/7)*Pi)+(12/7)*cos((2/7)*Pi)*x^2-(8/7)*x*cos((2/7)*Pi)-(2/   
   )*cos((3/7)*Pi)-(12/7)*x^2*cos((3/7)*Pi)+(8/7)*x*cos((3/7)*   
   Pi)+(8/7)*x^3*cos((3/7)*Pi)+1/7-(4/7)*x+(6/7)*x^2-(4/7)*x^3+x^4   
   > >   
   >   
   > evalf[20](L): fnormal(%): identify(%); # to have a guess   
   >   
   >                                  2   3   4   
   >                             [x, x , x , x ]   
   >   
   > convert(L, RootOf): # nun aber in echt ...   
   > simplify(%);   
   >                                  2   3   4   
   >                             [x, x , x , x ]   
   >   
   > I think it is also "what is intended by simplify (and should trig   
   > survive)?" Thus I included sci.math.symbolic for further answers.   
   >   
   > PS: would you mind to post as list   
   >   
   > PPS: well, it may break down at some degree   
      
      
   Derive 6.10 doesn't need any teaching: your quadruple expression   
      
   [-1/7+x-2/7*COS(2/7*pi)+2/7*COS(3/7*pi)+2/7*COS(1/7*pi),4/7*x*CO~   
   S(1/7*pi)-2/7*COS(1/7*pi)-4/7*x*COS(2/7*pi)+2/7*COS(2/7*pi)+4/7*~   
   x*COS(3/7*pi)-2/7*COS(3/7*pi)+1/7-2/7*x+x^2,2/7*COS(1/7*pi)+6/7*~   
   x^2*COS(1/7*pi)-6/7*x*COS(1/7*pi)-2/7*COS(2/7*pi)-6/7*COS(2/7*pi~   
   )*x^2+6/7*x*COS(2/7*pi)+2/7*COS(3/7*pi)+6/7*x^2*COS(3/7*pi)-6/7*~   
   x*COS(3/7*pi)-1/7+3/7*x-3/7*x^2+x^3,-2/7*COS(1/7*pi)-12/7*x^2*CO~   
   S(1/7*pi)+8/7*x*COS(1/7*pi)+8/7*x^3*COS(1/7*pi)-8/7*COS(2/7*pi)*~   
   x^3+2/7*COS(2/7*pi)+12/7*COS(2/7*pi)*x^2-8/7*x*COS(2/7*pi)-2/7*C~   
   OS(3/7*pi)-12/7*x^2*COS(3/7*pi)+8/7*x*COS(3/7*pi)+8/7*x^3*COS(3/~   
   7*pi)+1/7-4/7*x+6/7*x^2-4/7*x^3+x^4]   
      
   is automatically simplified to   
      
   [x,x^2,x^3,x^4]   
      
   within a fraction of a second. These are the reduction steps for the   
   first expression:   
      
   -1/7+x-2/7*COS(2/7*pi)+2/7*COS(3/7*pi)+2/7*COS(1/7*pi)   
      
   " COS(n*pi) -> SIN((1/2-n)*pi) "   
      
   -1/7+x-2*SIN(3*pi/14)/7+2*COS(3*pi/7)/7+2*COS(pi/7)/7   
      
   " COS(n*pi) -> SIN((1/2-n)*pi) "   
      
   -1/7+x-2*SIN(3*pi/14)/7+2*SIN(pi/14)/7+2*COS(pi/7)/7   
      
   " SIN(3*pi/14)-SIN(pi/14) -> COS(pi/7)-1/2 "   
      
   -1/7+x-2*(COS(pi/7)-1/2)/7+2*COS(pi/7)/7   
      
   " one final step "   
      
   x   
      
   I expect the remainder to be handled in the same manner. But I don't see   
   why Derive should not fail to simplify similar expressions whose   
   trigonometric arguments involve larger denominators, as the rule to   
   handle SIN(3*pi/14) - SIN(pi/14) is not generic.   
      
   Martin.   
      
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