{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 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 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 "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 -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 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 260 42 "Integration of rational f unctions, part 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "Suppose f is a rational function that we want to integrat e. Lets just take an example. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "F := nor mal((x+1)/((x^2+1)^2*(x^3+x+1)*x^3),expanded)-2*ln(x^2+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG,&*&,&%\"xG\"\"\"F)F)\"\"\",0*$)F(\"#5 F*F)*$)F(\"\")F*\"\"$*$)F(\"\"(F*F)*$)F(\"\"'F*F2*$)F(\"\"&F*\"\"#*$)F (\"\"%F*F)*$)F(F2F*F)!\"\"F)-%#lnG6#,&*$)F(F " 0 "" {MPLTEXT 1 0 30 "f:=normal(diff(F,x),expanded);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&,@*$)%\"xG\"#:\"\"\"!\"%*$)F) \"#8F+!#;*$)F)\"#7F+!\")*$)F)\"#6F+!#C*$)F)\"#5F+F8*$)F)\"\"*F+!#?*$)F )\"\")F+F8*$)F)\"\"(F+!#7*$)F)\"\"'F+!#<*$)F)\"\"&F+!#9*$)F)\"\"%F+FF* $)F)\"\"$F+F?*$)F)\"\"#F+!#5F)!\"'!\"$\"\"\"F+,:FOFenFKFWFGFQFCFBF@FBF 5FBF " 0 "" {MPLTEXT 1 0 14 "den:=denom(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$denG*&)%\"xG\"\"%\"\"\",:\"\"\"F+F'\"\"#*$)F'F,F)F(*$)F'\"\"$ F)\"\")*$F&F)F2*$)F'\"\"(F)F2*$)F'\"\"&F)\"#7*$)F'\"#5F)F9*$)F'F2F)F=* $)F'\"\"'F)\"#6*$)F'\"\"*F)F,*$)F'F:F)F+F+" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 16 "v:=sqrfree(den);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%\"vG7$\"\"\"7%7$,(*$)%\"xG\"\"$\"\"\"F&F,F&F&F&\"\"#7$,&*$)F,F/F.F& F&F&F-7$F,\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "So in this exa mple we can write:" }}{PARA 0 "" 0 "" {TEXT -1 26 " den = p2^2 * p3^3 * p4^4" }}{PARA 0 "" 0 "" {TEXT -1 186 "where p2, p3, p4 are square-f ree polynomials (so they do not have multiple roots) and the polynomia ls p2,p3,p4 are relatively prime (means they have gcd=1, so they have \+ no common roots)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 232 "We then know that the pole orders of f are: 2,3,4. The p ole orders of the integral F that we are aiming to find are 1 lower, s o they are 1,2,3. For now, let's focus on the highest pole order of f, which is 4. We would like to write:" }}{PARA 0 "" 0 "" {TEXT -1 17 " \+ f = a1 + f1" }}{PARA 0 "" 0 "" {TEXT -1 19 "in such a way that:" }}{PARA 0 "" 0 "" {TEXT -1 55 " 1) a1 is a function that we know \+ how to integrate" }}{PARA 0 "" 0 "" {TEXT -1 41 " 2) f1 is a func tion with poles < 4." }}{PARA 0 "" 0 "" {TEXT -1 259 "How can we be su re that a1 is a function that we are able to integrate? Well, function s that we can certainly integrate are the functions that are derivativ es of known functions. So in other words: a1=A1' for some function A1 \+ that we construct. So write f as:" }}{PARA 0 "" 0 "" {TEXT -1 23 " \+ f = A1' + f1" }}{PARA 0 "" 0 "" {TEXT -1 17 "That way, we have " }}{PARA 0 "" 0 "" {TEXT -1 32 " int(f,x) = A1 + int(f1,x)" }} {PARA 0 "" 0 "" {TEXT -1 578 "and then we have reduced the problem of \+ integrating a function f (max pole order = 4) to another integration p roblem, namely the problem of integrating function f1 (max pole order \+ is at most 3). This way, we can keep on decreasing the highest pole or der that occurs in our integration problem. The method given below wor ks until the highest pole order equals 1. At that point we'll need to \+ do something different. If you remember the calculus formulas, then th at will be no surprise because int(x^n,x) has a special case for n=-1, namely at that point we need to use logarithms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "f;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,@*$)% \"xG\"#:\"\"\"!\"%*$)F'\"#8F)!#;*$)F'\"#7F)!\")*$)F'\"#6F)!#C*$)F'\"#5 F)F6*$)F'\"\"*F)!#?*$)F'\"\")F)F6*$)F'\"\"(F)!#7*$)F'\"\"'F)!#<*$)F'\" \"&F)!#9*$)F'\"\"%F)FD*$)F'\"\"$F)F=*$)F'\"\"#F)!#5F'!\"'!\"$\"\"\"F), :FMFYFIFUFEFOFAF@F>F@F3F@F:F1*$)F'\"#9F)FKF/F9F7F5F+FU*$)F'\"#;F)FY!\" \"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "A1 := c0/x^3;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A1G*&%#c0G\"\"\"*$)%\"xG\"\"$F'!\" \"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "a1 := diff(A1,x);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#a1G,$*&%#c0G\"\"\"*$)%\"xG\"\"%F(! \"\"!\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 190 "Now A1 has a pole of order 3, a1 has a pole of order 4, and we hope to get rid of the pole of order 4 in f by substracting a1 from f, resulting in a new functio n f1=f-a1 with pole orders <4." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "f1:=f - a1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1G,&*&,@*$) %\"xG\"#:\"\"\"!\"%*$)F*\"#8F,!#;*$)F*\"#7F,!\")*$)F*\"#6F,!#C*$)F*\"# 5F,F9*$)F*\"\"*F,!#?*$)F*\"\")F,F9*$)F*\"\"(F,!#7*$)F*\"\"'F,!#<*$)F* \"\"&F,!#9*$)F*\"\"%F,FG*$)F*\"\"$F,F@*$)F*\"\"#F,!#5F*!\"'!\"$\"\"\"F ,,:FPFfnFLFXFHFRFDFCFAFCF6FCF=F4*$)F*\"#9F,FNF2F " 0 "" {MPLTEXT 1 0 15 "f1:=normal(f1);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>% #f1G,$*&,X\"\"$\"\"\"%\"xG\"\"'*$)F*\"\"#\"\"\"\"#5*$)F*F(F/\"#?*$)F* \"\"%F/\"#7*$)F*\"\"(F/F7*$)F*\"\"&F/\"#9*$)F*F0F/\"#C*$)F*\"\")F/FA*$ )F*F+F/\"#<*$)F*\"\"*F/F3*$)F*F7F/FD*$)F*\"#6F/FA*$)F*\"#:F/F6*$)F*\"# 8F/\"#;%#c0G!\"$*&FWF)F*F)!\"'*&FWF/F-F/!#7*&FWF/F2F/!#C*&FWF/F5F/Fhn* &FWF/F9F/Fhn*&FWF/F " 0 "" {MPLTEXT 1 0 17 "normal(f1 * x^4);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,$*&,X\"\"$\"\"\"%\"xG\"\"'*$)F(\"\"#\"\"\"\"#5*$)F( F&F-\"#?*$)F(\"\"%F-\"#7*$)F(\"\"(F-F5*$)F(\"\"&F-\"#9*$)F(F.F-\"#C*$) F(\"\")F-F?*$)F(F)F-\"#<*$)F(\"\"*F-F1*$)F(F5F-FB*$)F(\"#6F-F?*$)F(\"# :F-F4*$)F(\"#8F-\"#;%#c0G!\"$*&FUF'F(F'!\"'*&FUF-F+F-!#7*&FUF-F0F-!#C* &FUF-F3F-Ffn*&FUF-F7F-Ffn*&FUF-F:F-!#O*&FUF-F>F-!#:*&FUF-FAF-!#I*&FUF- FDF-!#L*&FUF-FGF-FX*&FUF-FJF-FVF-,:F'F'F(F,F*F4F/FBF2FBF6FBF9F5F=F;F@F .FCFMFFF,FIF'!\"\"!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 " should_be_zero := subs(x=0,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/s hould_be_zeroG,&!\"$\"\"\"%#c0G\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "solve(should_be_zero, \{c0\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/%#c0G\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "A1 := subs(%,A1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A1G*&\" \"\"F&*$)%\"xG\"\"$F&!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Now the integral of f equals A1 plus the integral of:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "f1 := normal(f - diff(A1,x),expanded);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1G*&,<\"\"#\"\"\"%\"xG\"\"%*$)F)F' \"\"\"\"#7*$)F)\"\"&F-F.*$)F)\"\"$F-\"#A*$)F)\"\")F-!\"**$)F)\"\"'F-F< *$)F)F*F-\"#;*$)F)\"\"(F-!#9*$)F)\"#5F-!\"&*$)F)\"\"*F-!#C*$)F)\"#8F-! \"%*$)F)\"#6F-!#;F-,:F+F(F2F'F=F*F/F8F:F8FHF8F@F.*$)F)F.F-F1FDFFF6FRFP F'*$)F)\"#9F-F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Now we'll do the same game again:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "sqrfree(denom(f1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"7%7$,( *$)%\"xG\"\"$\"\"\"F$F*F$F$F$\"\"#7$F*F-7$,&*$)F*F-F,F$F$F$F+" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "So the highest pole order is 3. T he poles of order 3 are precisely the roots of the square-free polynom ial:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "d := x^2+1;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,&*$)%\"xG\"\"#\"\"\"\"\"\"F+F+ " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Now again, we want to write: " }}{PARA 0 "" 0 "" {TEXT -1 21 " f1 = A2' + f2" }}{PARA 0 "" 0 "" {TEXT -1 206 "for some known function A2, and some function f2 th at has pole orders <3. Then the integral of f1 equals A2 plus the inte gral of f2. We take A2 as follows, where c0,c1 are constants that we d o not yet know:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "A2 := (c 0 + c1*x)/d^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G*&,&%#c0G\"\" \"*&%#c1GF(%\"xGF(F(\"\"\"*$),&*$)F+\"\"#F,F(F(F(\"\"#F,!\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "Why would it be enough to conside r only this kind of function A2? Keep in mind the following: The only \+ purpose of function A2 is that when we compute:" }}{PARA 0 "" 0 "" {TEXT -1 22 " f2 = f1 -A2'" }}{PARA 0 "" 0 "" {TEXT -1 35 "th en we end up with pole orders <3." }}{PARA 0 "" 0 "" {TEXT -1 174 "For this objective, all poles of A2' of order <3 are irrelevant, and sinc e the pole orders of A2' are 1 higher than those of A2, the poles of A 2 with order <2 are irrelevant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 40 "So suppose we used function of the form: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "A2 := ( r + q*d ) /d^2 \+ + R;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G,&*&,&%\"rG\"\"\"*&%\"qG F),&*$)%\"xG\"\"#\"\"\"F)F)F)F)F)F1*$)F,\"\"#F1!\"\"F)%\"RGF)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "where q and r are polynomials and R is a function with pole orders <2. And suppose that A2 \"does the \+ job\", in other words:" }}{PARA 0 "" 0 "" {TEXT -1 20 " f2 = f 1-A2'" }}{PARA 0 "" 0 "" {TEXT -1 33 "has pole orders <3. Now A2 equal s" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "A2 := r/d^2 + q/d^1 + \+ R;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G,(*&%\"rG\"\"\"*$),&*$)%\" xG\"\"#F(\"\"\"F0F0\"\"#F(!\"\"F0*&%\"qGF(F+F2F0%\"RGF0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "and the derivative of the terms:" }} {PARA 0 "" 0 "" {TEXT -1 17 " q/d^1 + R" }}{PARA 0 "" 0 "" {TEXT -1 117 "have pole orders <3. So if we would just throw away thes e terms, it would not change the fact that the pole orders of" }} {PARA 0 "" 0 "" {TEXT -1 21 " f2 = f1 - A2'" }}{PARA 0 "" 0 "" {TEXT -1 109 "are <3. Therefore, whenever we have an A2 of the above f orm, only the part r/d^2 is relevant. Suppose we had:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "A2 := randpoly(x,degree=5)/d^2;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G*&,.*$)%\"xG\"\"&\"\"\"!#&)*$)F) \"\"%F+!#b*$)F)\"\"$F+!#P*$)F)\"\"#F+!#NF)\"#(*\"#]\"\"\"F+*$),&F5F;F; F;\"\"#F+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "P:=numer( A2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,.*$)%\"xG\"\"&\"\"\"!#& )*$)F(\"\"%F*!#b*$)F(\"\"$F*!#P*$)F(\"\"#F*!#NF(\"#(*\"#]\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "q:=quo(P,d,x,'r'); # this as signs q=quotient and r=remainder" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% \"qG,**$)%\"xG\"\"$\"\"\"!#&)*$)F(\"\"#F*!#bF(\"#[\"#?\"\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "'P' = 'q'*'d' + 'r';" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"PG,&*&%\"qG\"\"\"%\"dGF(F(%\"rGF( " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "%;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,.*$)%\"xG\"\"&\"\"\"!#&)*$)F'\"\"%F)!#b*$)F'\"\"$F)!# P*$)F'\"\"#F)!#NF'\"#(*\"#]\"\"\",(*&,*F/F*F3F.F'\"#[\"#?F9F9,&F3F9F9F 9F9F9\"#IF9F'\"#\\" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "If d is a p olynomial, then every polynomial P can be written uniquely as:" }} {PARA 0 "" 0 "" {TEXT -1 10 " " }{TEXT 258 11 "P = r + q*d" } }{PARA 0 "" 0 "" {TEXT -1 38 "where q,r are polynomials such that: " }{TEXT 259 25 "degree(r,x) < degree(d,x)" }}{PARA 0 "" 0 "" {TEXT -1 21 "Here q is called the " }{TEXT 256 27 "quotient of P divided by d. " }}{PARA 0 "" 0 "" {TEXT -1 20 "And r is called the " }{TEXT 257 28 " remainder of P divided by d." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Therefore, in the above scenario, if we have:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "P:='P';" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"PGF$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "A2 := P/d^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G*&%\"PG\"\" \"*$),&*$)%\"xG\"\"#F'\"\"\"F/F/\"\"#F'!\"\"" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 47 "then we can always write:\n P = r + q*d" }} {PARA 0 "" 0 "" {TEXT -1 270 "where: degree(r,x) " 0 "" {MPLTEXT 1 0 12 "deg ree(d,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Now the most general polynomial of degree < degre e(d,x) is:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "P:=c0+c1*x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,&%#c0G\"\"\"*&%#c1GF'%\"xGF'F '" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "A2:=P/d^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A2G*&,&%#c0G\"\"\"*&%#c1GF(%\"xGF(F(\"\" \"*$),&*$)F+\"\"#F,F(F(F(\"\"#F,!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "So we may assume that A2 has this form. We do not yet kno w the right values for c0,c1." }}{PARA 0 "" 0 "" {TEXT -1 80 "Back to \+ integration. We want that f2=f1-A2' has pole order <3 at the roots of \+ d." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "f2 := normal(f1 - dif f(A2,x)):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "So we want that f2* d^2 has no poles at the roots of d, in other words we want that f2*d^3 has roots at the roots of d." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "normal(f2*d^3);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#*&,X*&%#c0G\" \"\")%\"xG\"\"$\"\"\"\"\"%*&F&F+)F)F,F+\"\")*&%#c1GF'F.F+\"\"#*&F1F+)F )\"#5F+F*F2F'*&F1F+)F)F2F+!\"\"F)F,*$F7F+\"#7*$F(F+\"#A*$F.F+\"#;*$)F) \"\"(F+!#9*$)F)\"\"&F+F:*$F4F+!\"&*$)F)F/F+!\"**$)F)\"\"'F+FM*$)F)\"\" *F+!#C*$)F)\"#6F+!#;*$)F)\"#8F+!\"%*&F&F+FOF+F,*&F&F+F@F+F/*&F&F+FLF+F /*&F&F+FDF+F,*&F1F+F(F+!\"#*&F1F+FLF+F'*&F1F+F@F+FM*&F1F+FIF+FE*&F1F+F DF+F,F+*&,.FKF'F=F2F;F2F9F'F)F2F'F'\"\"\")F)\"\"#F+!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "S:=numer(%);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"SG,X*&%#c0G\"\"\")%\"xG\"\"$\"\"\"\"\"%*&F'F,)F*F-F ,\"\")*&%#c1GF(F/F,\"\"#*&F2F,)F*\"#5F,F+F3F(*&F2F,)F*F3F,!\"\"F*F-*$F 8F,\"#7*$F)F,\"#A*$F/F,\"#;*$)F*\"\"(F,!#9*$)F*\"\"&F,F;*$F5F,!\"&*$)F *F0F,!\"**$)F*\"\"'F,FN*$)F*\"\"*F,!#C*$)F*\"#6F,!#;*$)F*\"#8F,!\"%*&F 'F,FPF,F-*&F'F,FAF,F0*&F'F,FMF,F0*&F'F,FEF,F-*&F2F,F)F,!\"#*&F2F,FMF,F (*&F2F,FAF,FN*&F2F,FJF,FF*&F2F,FEF,F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 112 "So S should have roots at the roots of d, and because d \+ has no multiple roots it means that S can be written as:" }}{PARA 0 " " 0 "" {TEXT -1 23 " S = q*d (*)" }}{PARA 0 "" 0 "" {TEXT -1 71 "for some polynomial q. Now every polynomial S can always be wri tten as:" }}{PARA 0 "" 0 "" {TEXT -1 60 " S = r + q*d with d egree(r,x) \+ " 0 "" {MPLTEXT 1 0 44 "r := rem(S,d,x); # rem = remainder in Maple" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG,(*&,&!\"%\"\"\"%#c0GF(F)%\"xG F)F)F(F)%#c1G\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "So conditio n (*) only holds when r=0. Now r=0 only when its coefficients of r are zero:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "should_be_zero:= \{ coeffs(r,x) \};" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/should_be_zer oG<$,&!\"%\"\"\"%#c1G\"\"%,&F'F(%#c0GF'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "solve(should_be_zero, \{c0,c1\} );" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#<$/%#c1G\"\"\"/%#c0G!\"\"" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 15 "A2:=subs(%,A2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#A2G*&,&!\"\"\"\"\"%\"xGF(\"\"\"*$),&*$)F)\"\"#F*F(F(F(\"\"#F*!\"\" " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "f2:=normal(f1 - diff(A2 ,x),expanded);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f2G*&,6*$)%\"xG\" #6\"\"\"!\"%*$)F)\"\"*F+!#7*$)F)\"\")F+!\"#*$)F)\"\"(F+!#;*$)F)\"\"'F+ F4*$)F)\"\"%F+\"\"\"*$)F)\"\"$F+\"#7*$)F)\"\"#F+F/F)F>FFF?F+,6*$)F)FCF +F?*$)F)\"#5F+F>F-FFF1F;F5F;F9\"\"&*$)F)FMF+F;F " 0 "" {MPLTEXT 1 0 19 "sqrfree(denom(f2));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"\"7$7$,,*$)%\"xG\"\"&\"\"\"F$*$) F*\"\"$F,\"\"#*$)F*F0F,F$F*F$F$F$F07$F*F0" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 33 "d:=expand(x*(x^5+2*x^3+x^2+x+1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,,*$)%\"xG\"\"'\"\"\"\"\"\"*$)F(\"\"%F*\"\"#*$ )F(\"\"$F*F+*$)F(F/F*F+F(F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "A3:=(c0+c1*x+c2*x^2+c3*x^3+c4*x^4+c5*x^5)/d;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A3G*&,.%#c0G\"\"\"*&%#c1GF(%\"xGF(F(*&%#c2GF()F+\"\" #\"\"\"F(*&%#c3GF()F+\"\"$F0F(*&%#c4GF()F+\"\"%F0F(*&%#c5GF()F+\"\"&F0 F(F0,,*$)F+\"\"'F0F(*$F7F0F/*$F3F0F(*$F.F0F(F+F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "Now d has degree 6, so for the numerator it is sufficient to consider only a polynomial of degree <6, so the n umerator can be written in that form A3." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f3:=normal(f2-diff(A3,x));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#f3G*&,do*&%#c0G\"\"\")%\"xG\"\"$\"\"\"\"\")*&F(F-)F+ \"\"#F-F,*&%#c1GF))F+\"\"%F-\"\"'F1F)*&F3F-F0F-F)F+F5*&F(F-F+F)F1*$F0F -\"\"**$F*F-\"#7*$F4F-F)*$)F+\"\"(F-!#;*$)F+F.F-!\"#*$)F+F6F-FD*$)F+F: F-!#7*$)F+\"#6F-!\"%*&F(F-)F+\"\"&F-F6F(F)*&F3F-F*F-F1*&F3F-FFF-FP*&%# c2GF)F0F-!\"\"*&%#c3GF)F*F-FD*&%#c4GF)F4F-!\"$*&%#c5GF)FOF-FM*&FTF-F?F -F5*&FTF-FOF-F5*&FTF-F4F-F)*&FWF-FCF-F,*&FWF-FFF-F1*&FWF-F4F-FU*&FYF-F HF-F1*&FYF-FFF-FU*&FYF-FOF-FD*&FfnF-)F+\"#5F-F)*&FfnF-FCF-FD*&FfnF-F?F -FD*&FfnF-FFF-FZF-*&),,*$FOF-F)F;F1F9F)F+F)F)F)\"\"#F-)F+\"\"#F-!\"\" " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "r:=rem(numer(normal(f3* d^2)),d,x);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"rG,0*&,.\"#5\"\"\"% #c4G\"\"%%#c5G\"\"&%#c2G!\"%%#c3G!\"$%#c0G\"\"'F))%\"xGF-\"\"\"F)*&,.F ,!\"\"\"\"$F)F0F+F*F-%#c1GF/F.F1F))F5F+F6F)*&,0F,F:\"#7F)F0F9F*\"\"(F; F1F.F/F2\"\")F))F5F:F6F)*&,0F,F9\"#6F)F0F+F*F@F;F/F.!\"&F2F:F))F5\"\"# F6F)*&,.FHF)F2FHF,F/F0F+F*F:F;FFF)F5F)F)FHF)F2F)" }}}{EXCHG {PARA 12 " " 1 "" {TEXT -1 262 "As before, we take the numerator, then the remain der of that after division by d. And, as before, r should be equal to \+ 0, because then numer(normal(f3*d^2)) is divisible by d, hence it vani shes at the roots of d, hence the pole order of f3 at these points is \+ <2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "coeffs(r,x);" }} {PARA 12 "" 1 "" {XPPMATH 20 "6(,&\"\"#\"\"\"%#c0GF%,.F$F%F&F$%#c5G!\" %%#c3G\"\"%%#c4G\"\"$%#c1G!\"&,0F(!\"\"\"#6F%F*F+F,\"\"(F.F)%#c2GF/F&F -,0F(F-\"#7F%F*F1F,F3F.!\"$F4F)F&\"\"),.F(F1F-F%F*F+F,\"\"&F.F)F4F7,. \"#5F%F,F+F(F:F4F)F*F7F&\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(\{%\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<(/%#c1G!\"#/%# c2G\"\"\"/%#c5G\"\"!/%#c3GF&/%#c4GF,/%#c0GF&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "A3:=subs(%,A3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#A3G*&,*!\"#\"\"\"%\"xGF'*$)F)\"\"#\"\"\"F(*$)F)\"\"$F-F'F-,,*$)F )\"\"'F-F(*$)F)\"\"%F-F,F.F(F*F(F)F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "f3:=normal(f2-diff(A3,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f3G,$*&%\"xG\"\"\",&*$)F'\"\"#F(\"\"\"F-F-!\"\"!\"% " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Now: int(f,x) = A1+int(f1,x)= A1+A2+int(f2,x)=A1+A2+A3+int(f3,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "Int('f',x) = 'A1'+'A2'+'A3'+Int('f3',x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$%\"fG%\"xG,*%#A1G\"\"\"%#A2GF+%#A3GF+ -F%6$%#f3GF(F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "%;" }} {PARA 12 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&,@*$)%\"xG\"#:\"\"\"!\"%*$ )F+\"#8F-!#;*$)F+\"#7F-!\")*$)F+\"#6F-!#C*$)F+\"#5F-F:*$)F+\"\"*F-!#?* $)F+\"\")F-F:*$)F+\"\"(F-!#7*$)F+\"\"'F-!#<*$)F+\"\"&F-!#9*$)F+\"\"%F- FH*$)F+\"\"$F-FA*$)F+\"\"#F-!#5F+!\"'!\"$\"\"\"F-,:FQFgnFMFYFIFSFEFDFB FDF7FDF>F5*$)F+\"#9F-FOF3F=F;F9F/FY*$)F+\"#;F-Fgn!\"\"F+,**&F-F-*$)F+ \"\"$F-F_oFgn*&,&!\"\"FgnF+FgnF-*$),&FWFgnFgnFgn\"\"#F-F_oFgn*&,*!\"#F gnF+F^pFWFgnFTF^pF-,,FIFgnFQFYFTFgnFWFgnF+FgnF_oFgn-F%6$,$*&F+F-FjoF_o F.F+Fgn" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "Lets check if this is really so. We calculate the left-hand side minus the right-hand side, then compute the derivative, and that should be 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "normal(diff(lhs(%) - rhs(%),x));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "As you can see, integration of a rational function f has been reduced to integration of f3, where f3 is a rational function wi th pole order at most 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 " f3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"xG\"\"\",&*$)F%\"\"#F&\" \"\"F+F+!\"\"!\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "So now we st ill need to know how to integrate rational functions that have only po les of order 1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "int(f3,x );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#,&*$)%\"xG\"\"#\"\"\" \"\"\"F-F-!\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "How to do this \+ will be the topic of the next Maple worksheet." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "19 0 0" 31 }{VIEWOPTS 1 1 0 1 1 1803 }