{VERSION 3 0 "SGI MIPS UNIX" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "N ormal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } 0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 " " 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 25 "Equation f or the evolute." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 586 "Given an algebraic curve C, and a point P on C, one take s the circle that touches the curve C at the point P with the highest \+ possible multiplicity, so nearby P the circle and the curve will be as close as possible. Let R be the center of that curve. If we compute t he point R=[R[1],R[2]] for all points P on the curve C, we get another curve called the evolute. The question is now, compute an algebraic r elation F in Q[X,Y] between R[1] and R[2], i.e. compute a polynomial F such that subs(X=R[1],Y=R[2], F)=0. Then the evolute, the set of poin ts R, is just the set of solutions of F." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 403 "The method of constructing R is t he following. We take a point P=[x,y], and a point [x+delta_x, y+delta _y] that is infinitely nearby (the delta_x and delta_y depend on a sma ll epsilon, and we will take the limit epsilon -> 0, simply by substit ing epsilon=0 at the end). Then we take the normal lines L_1 and L_2 t hrough those two points, and compute the intersection using Maple's so lve. Then we find R." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "f:= x^2+2*y^2-1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,(*$)%\"xG\"\"# \"\"\"\"\"\"*$)%\"yGF)F*F)!\"\"F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "The curve C is given by the solutions of f. We see that C is an el lipse." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 311 "L_1:=[x,y] + t_1 * [diff(f,x), diff(f,y)];\ndelta_x:=epsilon*diff(f,y) + error1 * epsi lon^2;\ndelta_y:=-epsilon*diff(f,x) + error2 * epsilon^2;\nL_2:=subs(x =x+delta_x,y=y+delta_y, t_1=t_2, L_1);\nzero:=expand(L_1-L_2);\nR:=map (normal,subs(solve(\{op(zero)\},\{t_1,t_2\}),expand(L_1)));\nR:=map(no rmal,subs(epsilon=0,R));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$L_1G, &7$%\"xG%\"yG\"\"\"*&%$t_1GF)7$,$F'\"\"#,$F(\"\"%F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(delta_xG,&*&%(epsilonG\"\"\"%\"yGF(\"\"%*&%'err or1GF()F'\"\"#\"\"\"F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(delta_yG, &*&%(epsilonG\"\"\"%\"xGF(!\"#*&%'error2GF()F'\"\"#\"\"\"F(" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$L_2G,&7$,(%\"xG\"\"\"*&%(epsilonG\"\"\",( %\"yGF)*&F+F)F(F)!\"#*&%'error2GF))F+\"\"#F,F)F)\"\"%*&%'error1GF)F3F, F)F-F)*&%$t_2GF)7$,(F(F4F*\"\")F6F4,(F.F5F/!\")F1F5F)F)" }}{PARA 12 " " 1 "" {XPPMATH 20 "6#>%%zeroG7$,6*&%\"xG\"\"\"%$t_2GF)!\"#*(%(epsilon GF)%\"yGF)F*\"\"\"!\")*()F-\"\"#F/F(F/F*F/\"#;*(%'error2GF))F-\"\"$F/F *F/F0*(%'error1GF)F2F/F*F/F+*&F(F/%$t_1GF)F3*&F-F/F.F/!\"%*&F2F/F(F/\" \")*&F6F/F7F/F>*&F:F/F2F/!\"\",.*&F.F/F*F/F>*(F-F/F(F/F*F/F@*(F6F/F2F/ F*F/F>*&F.F/F%\"RG7$,$*&*&%\"xG\"\"\",4*(%(epsilonG\"\"\"F)F.%\"yGF .!\")*(%'error2GF.)F-\"\"#F*F/F*\"\"%*&F3F*)F)F4F*\"#;*&)F2F4F*)F-F5F* F5*$F7F*!\"#*(F)F*F2F*F-F*F.*()F-\"\"$F*F)F*F2F*!#;*(F3F*F)F*%'error1G F.F=*(F2F*F@F*FDF*F.F.F*,.F!\"\"*$)F/F4F*F5F,F0F1F5*(FDF*F-F*F/F* F.!\"\"#FGF4,$*&*&F/F*,4F,FBF1\"\")FJF.F6F8F9F5F?FBFCF=FEF.FHF5F.F*FFF KFG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG7$,$*&*$)%\"xG\"\"$\"\"\" F,,&*$)F*\"\"#F,\"\"\"*$)%\"yGF0F,F0!\"\"#F1F0,$*&*$)F4F+F,F,F-F5!\"# " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Now we have the following equ ations:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "f = 0" }}{PARA 0 "" 0 "" {TEXT -1 10 "X-R[1] = 0" }}{PARA 0 "" 0 "" {TEXT -1 10 "Y-R[2] = 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 200 "These are equations consisting of the variables x,y ,X,Y. The goal is to have just one equation F=0 involving X and Y, so \+ we want to eliminate x and y from these equations. Do this using the f ollowing:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 257 18 "General Principle:" }}{PARA 257 "" 0 "" {TEXT 258 41 "If: you want to get rid of some variable." }}{PARA 258 "" 0 " " {TEXT 259 24 "Then: use the resultant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 155 "So remember: when ever there's one variable too many, the question is not: \"should I ap ply the resultant?\". The question is: \"what should I take as input? \"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "As signment: Find the polynomial F in Q[X,Y] and e-mail it to hoeij@math. fsu.edu" }}{PARA 0 "" 0 "" {TEXT -1 63 "Make your answer squarefree, b ecause when F^2=0 then also: F=0." }}}}{MARK "4 6 0" 10 }{VIEWOPTS 1 1 0 1 1 1803 }