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manual:chapter3:symbolic [2018/12/14 11:32] claudio |
manual:chapter3:symbolic [2018/12/14 11:42] claudio [Algebraic Rules] |
| ''%%'%%.nN*.vX:->.nN*(.vX-.nN)%%'%%'' | ''%%'%%√3*X^2+2.5*(X-2.5)+5*(X-5)*(Y-5)+(A+B)*Y^2%%'%%'' | It finds ''2.5*X'' and replaces, then finds ''5*X'', replaces with ''5*(X-5)'' then on the same term finds ''5*Y'' and replaces it with ''5*(Y-5)'', reporting a total of 3 replacements | | | ''%%'%%.nN*.vX:->.nN*(.vX-.nN)%%'%%'' | ''%%'%%√3*X^2+2.5*(X-2.5)+5*(X-5)*(Y-5)+(A+B)*Y^2%%'%%'' | It finds ''2.5*X'' and replaces, then finds ''5*X'', replaces with ''5*(X-5)'' then on the same term finds ''5*Y'' and replaces it with ''5*(Y-5)'', reporting a total of 3 replacements | |
| ''%%'%%.nN*.vX^2:->.nN*(.vX-1)^2%%'%%'' | ''%%'%%√3*X^2+2.5*X+5*X*Y+(A+B)*Y^2%%'%%'' | Does not make any replacements, since ''√3'' is not a single number but a numeric expression | | | ''%%'%%.nN*.vX^2:->.nN*(.vX-1)^2%%'%%'' | ''%%'%%√3*X^2+2.5*X+5*X*Y+(A+B)*Y^2%%'%%'' | Does not make any replacements, since ''√3'' is not a single number but a numeric expression | |
| ''%%'%%.NN*.vX^2:->.nN*(.vX-1)^2%%'%%'' | ''%%'%%√3*(X-1)^2+2.5*X+5*X*Y+(A+B)*Y^2%%'%%'' | Using ''.NN'' instead, matches ''√3'' | | | ''%%'%%.NN*.vX^2:->.NN*(.vX-1)^2%%'%%'' | ''%%'%%√3*(X-1)^2+2.5*X+5*X*Y+(A+B)*Y^2%%'%%'' | Using ''.NN'' instead, matches ''√3'' and does the replacement | |
| | ''%%'%%.xN*.vX^2:->.xN*(.vX-1)^2%%'%%'' | ''%%'%%√3*(X-1)^2+2.5*X+5*X*Y+(A+B)*(Y-1)^2%%'%%'' | Using ''.xN'' instead, matches ''√3'' and ''(A+B)'' as well | |
| | ''%%'%%.NN*.vX^2+.XR:->.XR%%'%%'' | ''%%'%%2.5*X+5*X*Y+(A+B)*Y^2%%'%%'' | Finds ''√3*X^2'+...'' where ''.XR'' represents the rest of the expression, and removes the term from the expression | |
| | ''%%'%%.xN*.vX^2+.XR:->.XR%%'%%'' | ''%%'%%2.5*X+5*X*Y%%'%%'' | Finds ''√3*X^2'+...'' and removes the term from the expression, then also finds ''(A+B)*Y^2'' and removes it as well | |
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