{VERSION 4 0 "IBM INTEL LINUX22" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 12 0 0 0 0 0 2 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 12 0 0 0 0 0 2 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 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 "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }1 0 0 0 8 4 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 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 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 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 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 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 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 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 77 "The two \+ highest coefficients of a symmetric power of a second order operator. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "by:" } }{PARA 0 "" 0 "" {TEXT -1 24 " Mark van Hoeij" }}{PARA 0 "" 0 "" {TEXT -1 34 " Florida State University" }}{PARA 0 "" 0 " " {TEXT -1 19 " May 2001." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "with(DEtools):\n_Envdiffopd omain:=[Dx,x]: # means d/dx is denoted by Dx" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 39 "(n-1)'th symmetric power of Dx^2 + a*Dx" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "L = Dx^2 + a*Dx" }} {PARA 0 "" 0 "" {TEXT -1 32 "Basis of solutions of L: 1, y." }} {PARA 0 "" 0 "" {TEXT -1 30 "Then y' is a solution of Dx+a." }}{PARA 0 "" 0 "" {TEXT -1 46 "Now Y := 1/y' is a solution of Dx-a so Y'=a*Y. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "Let \+ Ln = symmetric_power(L,n-1) = Dx^n + An * Dx^(n-1) + Bn * Dx^(n-2) + . .. where the dots stand for lower order terms." }}{PARA 0 "" 0 "" {TEXT -1 37 "Solutions of Ln: 1,y,y^2,..,y^(n-1)." }}{PARA 0 "" 0 "" {TEXT -1 77 "Define L_n as (d/dy)^n. Solutions of L_n are the same as \+ the solutions of Ln." }}{PARA 0 "" 0 "" {TEXT -1 27 "d/dy = 1/y' * d/d x = Y*d/dx" }}{PARA 0 "" 0 "" {TEXT -1 35 "L_n = (Y*d/dx)^n = Y^n * Dx ^n + ..." }}{PARA 0 "" 0 "" {TEXT -1 70 "Since L_n has the same soluti ons as Ln we have L_n = Y^n * Ln and so:" }}{PARA 0 "" 0 "" {TEXT -1 79 "L_n = (Y*d/dx)^n = Y^n * Dx^n + Y^n * An * Dx^(n-1) + Y^n * Bn * D x^(n-1) + ..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 228 "Now write L_n = sum(c[i,n]*Dx^i, i=0..n). Then sum(c[i, n+1]*Dx^i, i=0..n+1) = L_(n+1) = Y*Dx*L_n = Y*sum( (Dx*c[i,n])*Dx^i,i= 0..n) = Y*sum( (c[i,n]*Dx + c[i,n]')*Dx^i,i=0..n) = Y*sum( (c[i-1,n]+c [i,n]')*Dx^i,i=0..n+1) and thus:" }}{PARA 0 "" 0 "" {TEXT -1 39 " c [i,n+1] = Y * (c[i-1,n] + c[i,n]')" }}{PARA 0 "" 0 "" {TEXT -1 76 "Now c[0,n]=0 and c[n,n]=Y^n for all n>0. Let d[n]=c[n-1,n]. Then d[1]=0 \+ and" }}{PARA 0 "" 0 "" {TEXT -1 78 " d[n+1] = Y * (d[n] + c[n,n]') \+ = Y * (d[n] + (Y^n)') = Y * (d[n] + a*n*Y^n)" }}{PARA 0 "" 0 "" {TEXT -1 67 "It is easy to solve this recurrence:\n d[n] = n*(n-1)/2 * a \+ * Y^n" }}{PARA 0 "" 0 "" {TEXT -1 6 "Hence:" }}{PARA 0 "" 0 "" {TEXT -1 23 " An = n*(n-1)/2 * a." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 144 "Now define e[n] = c[n-2,n]. Then e[1]=e[ 2]=0 and\n e[n+1] = Y * (c[n-2,n] + c[n-1,n]') = Y * (e[n] + d[n]') \+ = Y * (e[n] + n*(n-1)/2 * (a*Y^n)')" }}{PARA 0 "" 0 "" {TEXT -1 105 "A nd (a*Y^n)' = a'*Y^n + a*(Y^n)' = Y^n * (a' + n*a^2) so\n e[n+1] = \+ Y*(e[n] + n*(n-1)/2*Y^n*(a'+n*a^2))" }}{PARA 0 "" 0 "" {TEXT -1 86 "Th e solution of the recurrence is:\n e[n]=Y^n * sum(i*(i-1)/2*(a' + i* a^2), i=0..n-1)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "e[n]/Y^n \+ = factor(sum(i*(i-1)/2 * (ap + i*a^2),i=0..n-1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&&%\"eG6#%\"nG\"\"\")%\"YGF(!\"\",$**F(F),&F(F)F)F,F) ,&F(F)\"\"#F,F),(*&F(F))%\"aGF1F)\"\"$*$F4F)F,*&\"\"%F)%#apGF)F)F)#F) \"#C" }}}{PARA 0 "" 0 "" {TEXT -1 11 "Conclusion:" }}{PARA 0 "" 0 "" {TEXT -1 52 " Bn = 1/24 * n*(n-1)*(n-2) * (4*a' + (3*n-1)*a^2)" }}} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 8 "(Dx+b)^n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Now define L_n as (Dx+b)^n. Wri te L_n = sum(c[i,n]*Dx^i,i=0..n). Then:" }}{PARA 0 "" 0 "" {TEXT -1 196 "sum(c[i,n+1]*Dx^i, i=0..n+1) = L_(n+1) = (Dx+b)*L_n = sum( (Dx+b) *c[i,n]*Dx^i, i=0..n) = sum( (b*c[i,n] + c[i,n]' + c[i,n]*Dx)*Dx^i, i= 0..n) = sum( (b*c[i,n]+c[i,n]'+c[i-1,n])*Dx^i,i=0..n) and so:" }} {PARA 0 "" 0 "" {TEXT -1 43 " c[i,n+1] = b*c[i,n] + c[i,n]' + c[i-1, n]" }}{PARA 0 "" 0 "" {TEXT -1 72 "Now: c[0,1]=b and c[n,n]=1 for all \+ n. Define d[n]=c[n-1,n] so d[1]=b and" }}{PARA 0 "" 0 "" {TEXT -1 38 " d[n+1] = b*1 + 1' + d[n] = b + d[n]" }}{PARA 0 "" 0 "" {TEXT -1 6 " Hence:" }}{PARA 0 "" 0 "" {TEXT -1 13 " d[n] = n*b" }}{PARA 0 "" 0 " " {TEXT -1 130 "Now define e[n]=c[n-2,n] so e[1]=0 and\n e[n+1] = b* c[n-1,n] + c[n-1,n]' + c[n-2,n] = b*d[n] + d[n]' + e[n] = n*b^2 + n*b' + e[n]" }}{PARA 0 "" 0 "" {TEXT -1 7 "and so:" }}{PARA 0 "" 0 "" {TEXT -1 59 " e[n] = sum(i,i=0..n-1)*(b^2+b') = n*(n-1)/2 * (b^2 + b ')" }}{PARA 0 "" 0 "" {TEXT -1 3 "So:" }}{PARA 0 "" 0 "" {TEXT -1 75 " (Dx+b)^n = Dx^n + n*b * Dx^(n-1) + n*(n-1)/2 * (b^2+b') * Dx^(n-2) \+ + ..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 60 "(n-1)'th symmetric power of Dx^2 + (a+2*b)*Dx + a*b+b'+b^ 2\n\n" }{TEXT 258 45 "If L=sum(c[i]*Dx^i, i=0..n) with c[n]=1 then:" } {TEXT -1 5 "\n " }{TEXT 257 55 "symmetric_product(L, Dx+b) = sum(c[ i]*(Dx+b)^i, i=0..n)" }}{PARA 0 "" 0 "" {TEXT -1 104 "and we can evalu ate the highest order terms of each (Dx+b)^i with the formula from the previous section." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "L := s ymmetric_product(Dx^2+a(x)*Dx, Dx+b(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,,*$)%#DxG\"\"#\"\"\"F**&,&-%\"bG6#%\"xGF)-%\"aGF /F*F*F(F*F**&F1F*F-F*F*-%%diffG6$F-F0F**$)F-F)F*F*" }}}{PARA 0 "" 0 " " {TEXT -1 106 "Let L = symmetric_product(Dx^2 + a*Dx, Dx+b)\nL = (Dx +b)^2 + a*(Dx+b) = Dx^2 + (a+2*b)*Dx + a*b + b' + b^2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "symmetric_power(L, n-1 )\n = symmetric_product(\n symmetric_power(Dx^2 + a*Dx, n-1 )," }}{PARA 0 "" 0 "" {TEXT -1 165 " symmetric_power(Dx+b, n- 1) )\n = symmetric_product(\n Dx^n + An*Dx^(n-1) + Bn*D x^(n-2) + ... ,\n Dx + b1 ) where b1 = (n-1)*b" }} {PARA 0 "" 0 "" {TEXT -1 216 " = (Dx + b1)^n + An * (Dx + b1)^(n-1) \+ + Bn * (Dx + b1)^(n-2) + ...\n = Dx^n + n*b1 * Dx^(n-1) + n*(n-1)/2 \+ * (b1^2+b1') * Dx^(n-2) + ...\n + An * (Dx^(n-1) + (n-1)*b1 * Dx^(n- 2) + ... )\n + Bn * (Dx^(n-2) + ... )" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 220 "b1 := (n-1)*b(x):\nAn := n*(n-1)/2 * a(x):\nBn := 1/ 24*n*(n-1)*(n-2)*(4*diff(a(x),x)+(3*n-1)*a(x)^2):\nsp :=\nDx^n + n*b1* Dx^(n-1)+n*(n-1)/2*(b1^2+diff(b1,x))*Dx^(n-2)\n+ An * ( Dx^(n-1) + (n- 1)*b1*Dx^(n-2) )\n+ Bn * Dx^(n-2);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6# >%#spG,,)%#DxG%\"nG\"\"\"**F(F),&F(F)F)!\"\"F)-%\"bG6#%\"xGF))F'F+F)F) *,#F)\"\"#F)F(F)F+F),&*&)F+F4F))F-F4F)F)*&F+F)-%%diffG6$F-F0F)F)F))F', &F(F)F4F,F)F)*,F3F)F(F)F+F)-%\"aGF/F),&F1F)*(F7F)F-F)F=F)F)F)F)*.#F)\" #CF)F(F)F+F)F>F),&-F;6$F@F0\"\"%*&,&F(\"\"$F)F,F))F@F4F)F)F)F=F)F)" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "CF_n_1 := factor(coeff(sp,D x^(n-1)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'CF_n_1G,$*(%\"nG\"\" \",&-%\"bG6#%\"xG\"\"#-%\"aGF,F(F(,&F'F(F(!\"\"F(#F(F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "CF_n_2 := collect(coeff(sp,Dx^(n-2) ), \{a(x),b(x)\},factor);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'CF_n_2 G,**,%\"nG\"\"\",&F'F(F(!\"\"F(,&F'F(\"\"#F*F(,&F'\"\"$F(F*F()-%\"aG6# %\"xGF,F(#F(\"#C*,#F(F,F(F'F()F)F.F(F0F(-%\"bGF2F(F(**F7F(F'F(F8F()F9F ,F(F(**#F(\"\"'F(F'F(F)F(,**&F'F(-%%diffG6$F9F3F(F.*&F'F(-FC6$F0F3F(F( *&F.F(FBF(F**&F,F(FFF(F*F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "W e can now write" }}{PARA 0 "" 0 "" {TEXT -1 27 " L = Dx^2 + A*D x + B" }}{PARA 0 "" 0 "" {TEXT -1 5 "where" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 19 "A(x)=coeff(L,Dx,1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"AG6#%\"xG,&-%\"bGF&\"\"#-%\"aGF&\"\"\"" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 19 "B(x)=coeff(L,Dx,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"BG6#%\"xG,(*&-%\"aGF&\"\"\"-%\"bGF&F,F,-%%diffG6$F -F'F,*$)F-\"\"#F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "S:=d solve(\{%,%%\}, \{a(x),b(x)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"SG7$<#/-%%diffG6$-%\"bG6#%\"xGF.,(-%\"BGF-\"\"\"*&F+F2-%\"AGF-F2!\" \"*$)F+\"\"#F2F2<#/-%\"aGF-,&F4F2*&F9F2F+F2F6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "map(collect,subs(op(S),expand(subs(op(S),[CF_n_1 ,CF_n_2]))), \{A(x),B(x)\}, factor);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#7$,$*(%\"nG\"\"\"-%\"AG6#%\"xGF',&F&F'F'!\"\"F'#F'\"\"#,(*,F&F'F,F', &F&F'F/F-F',&F&\"\"$F'F-F')F(F/F'#F'\"#C*,#F'\"\"'F'F&F'F,F',&F&F'F'F' F'-%\"BGF*F'F'*,F9F'F&F'-%%diffG6$F(F+F'F,F'F2F'F'" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 26 "We conclude the following:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 261 "" 0 "" {TEXT 261 14 "Proposition: " }{TEXT -1 2 "If" }}{PARA 260 "" 0 "" {TEXT -1 22 " L = Dx^2 + A*Dx + B" }}{PARA 258 "" 0 "" {TEXT -1 4 "Then" }}{PARA 259 "" 0 "" {TEXT -1 80 " symmetric_power(L, n-1) = Dx^n + An * D x^(n-1) + Bn * Dx^(n-2) + ...\nwhere" }}{PARA 0 "" 0 "" {TEXT 260 110 " An = 1/2*n*(n-1) * A\nand\n Bn = 1/24*n*(n-1)*(n-2)*(3*n-1)*A^2 \+ + 1/6*(n-1)*n*(n+1)*B+1/6*n*(n-1)*(n-2)*A'\n" }}{PARA 0 "" 0 "" {TEXT -1 487 "We proved this for the case where A=a+2*b, B=a*b+b'+b^2 for \+ some functions a,b. Since all A,B can be written in this form, it foll ows that the proposition is true in general.\nNote that it is trivial \+ to verify for any particular value of n that the formula is correct, f or example, if you want to verify the case n=4 then just type:\n n: =4; with(DEtools): symmetric_power(Dx^2 + a(x)*Dx + b(x), n-1, [Dx,x] );\nin Maple. However, this approach does not provide a formula for ge neral n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 69 "Application: Detection of symmetr ic powers of second order operators." }{TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 263 1 " " }{TEXT -1 1 " " }{TEXT 265 6 "Input:" }{TEXT -1 52 " L = Dx^n + An * Dx^(n-1) + Bn * Dx^(n-2) + ...\n " }{TEXT 266 18 " Output is either:\n" }{TEXT -1 60 " \"L is not a symmetric power o f a 2nd order operator\"\n " }{TEXT 267 4 "or:\n" }{TEXT -1 68 " \+ A,B such that L is (n-1)-th symmetric power of Dx^2 + A*Dx + B" }} {PARA 0 "" 0 "" {TEXT 264 10 "Algorithm:" }}{PARA 0 "" 0 "" {TEXT -1 31 "Step 1: A := 2/(n*An/(n*(n-1))" }}{PARA 0 "" 0 "" {TEXT -1 97 "St ep 2: B := 6/((n-1)*n*(n+1)) * Bn - (n-2)/(n+1) * Ap - (3*n-1)*(n-2)/ (n^2*(n-1)^2*(n+1)) *An^2" }}{PARA 0 "" 0 "" {TEXT -1 102 "Step 3: Co mpute (n-1)'th symmetric power of Dx^2+A*Dx+B with the algorithm of Br onstein/Mulders/Weil." }}{PARA 0 "" 0 "" {TEXT -1 133 "Step 4: Test i f L equals this symmetric power. If so, then output A,B, otherwise L i s not a symmetric power of a 2nd order operator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 268 14 "Implementation" }{TEXT -1 15 " (in MapleV 5)." }}{PARA 0 "" 0 "" {TEXT -1 228 "Note 1: m=n- 1.\nNote 2: The formulas on which the following algorithm is based w ere computed by the author of this document in 1997 but the computatio n was not written down until now. The implementation was done by G. La bahn." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 947 "proc(A, x)\nlocal i, a, DF, newA, b, ans, m, compare, y;\noption `Cop yright (c) 1997 George Labahn. All rights reserved.`;\n m := nops(A ) - 2;\n newA := [seq(A[i]/A[m + 2], i = 1 .. m + 2)]; # Make L m onic.\n a := 2*newA[m + 1]/(m*(m + 1));\n b := normal(1/4*(-3*m^ 4*a^2 - 2*m^3*a^2 - 4*m^3*diff(a, x) + 3*m^2*a^2\n + 2*m*a^2 + 4*m*diff(a, x) + 24*newA[m])/(m*(m^2 + 3*m + 2)));\n compare := DE tools[symmetric_power](DF^2 + a*DF + b, m, [DF, x]);\n for i from 0 to m do\n if simplify(coeff(compare, DF, i) - newA[i + 1]) <> \+ 0 then\n RETURN([])\n fi\n od;\n ans := readli b('`dsolve/diffeq/linearODE`')(b*y + a*_.y[1] + _.y[2], x, y);\n if ans <> [] and nops(ans) = 2 and not has(ans, 'DESol') then\n # ans[1] and ans[2] are the solutions of the 2nd order operator, so\n \+ # their products are the solutions of L:\n RETURN([seq(an s[1]^i*ans[2]^(m - i), i = 0 .. m)])\n fi;\n []\nend\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 11 "Reference:\n" }{TEXT -1 87 "Bro nstein, Mulders, Weil. \"On Symmetric Powers of Differential Operator s\", ISSAC'1997." }}}}{MARK "5 19 0" 322 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }