{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 "" 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 "" 0 1 0 0 0 0 0 1 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 } } {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 33 "The residue and local par ameters." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 221 "Let c be a complex number and let f be a function. Now f is analy tic at x=c if it has a convergent power series expansion at x=c. It is meromorphic at x=c when there is an integer n such that f*(x-c)^n is \+ analytic at x=c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 92 "When f is meromorphic at x=c we can calculate a series ex pansion at x=c. Here is an example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f:=x/(exp(x)-1-x-1/2*x^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG*&%\"xG\"\"\",*-%$expG6#F&\"\"\"!\"\"F,F&F-*$)F& \"\"#F'#F-F0!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "c:=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "series(f,x=c);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#++%\"xG\"\"'!\"##!\"$\"\"#!\"\"#\"\"$\"#S\"\"!-%\"OG6#\"\"\"\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "We see that f is meromorphic a t x=c (at x=0). It is also meromorphic at x=c for any complex number c ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "c:=1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "series(f,x=c,4); # compute 4 terms at x=c of f:" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#++,&%\"xG\"\"\"!\"\"F&*&\"\"\"F),&-%$expG6#F&F&! \"$F&!\"\"\"\"!,$*&,&F+#F&\"\"#F'F&F)*$)F*\"\"#F)F/F'\"\"#*&,&*&F+F)F* F/#F'\"\"'*&,&F+F&!\"#F&F)F*F/F4F)F*F/\"\"$-%\"OGF-\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "f is analytic at x=c (at x=1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "Note that at x= c, the series is presented as a power series in x-c." }}{PARA 0 "" 0 " " {TEXT 260 11 "Definition:" }{TEXT -1 47 " if c is a complex number t hen x-c is called a " }{TEXT 257 15 "local parameter" }{TEXT -1 8 " at x=c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 " The series expansion is given as a series in t, if t is the local para meter at x=c." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 11 "Definition:" }{TEXT -1 38 " If f is meromorphic at x=c, \+ then the " }{TEXT 259 20 "residue of f at x=c " }{TEXT -1 101 "is defi ned as the coefficient of t^(-1) in the series expansion of f, where t is the local parameter." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 48 "In the above example, f has residue -3/2 at x=0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "readlib(residue):" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "The readlib command loads the cod e into memory. Note: In Maple 6 this is not necessary, code will be lo aded automatically." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "resi due(f,x=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##!\"$\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "residue(f,x=1);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "Clearl y the residue is 0 at x=1, there were no negative powers of the local \+ parameter (there was no pole at x=1)." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 189 "The residue of f at x=c equals 1/(2* Pi*I) times the value of the integral of f*dx on a path that loops onc e around x=c, counterclockwise, and this path should not go around any other poles." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "F:=5/3*log (x^3-6*x-9);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG,$-%#lnG6#,(*$)% \"xG\"\"$\"\"\"\"\"\"F,!\"'!\"*F/#\"\"&F-" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "f:=normal(diff(F,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG,$*&,&*$)%\"xG\"\"#\"\"\"\"\"\"!\"#F-F,,(*$)F*\"\"$F,F-F*! \"'!\"*F-!\"\"\"\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "d:=d enom(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG,(*$)%\"xG\"\"$\"\" \"\"\"\"F(!\"'!\"*F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "fac tor(d);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&%\"xG\"\"\"!\"$F&F&,(*$ )F%\"\"#\"\"\"F&F%\"\"$F-F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "solve(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"\"$,&#!\"$\"\"#\" \"\"*&%\"IGF(-%%sqrtG6#F#\"\"\"#F(F',&F%F(F)#!\"\"F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "c:=-3/2+1/2*sqrt(-3);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"cG,&#!\"$\"\"#\"\"\"*&%\"IGF)-%%sqrtG6#\"\"$\"\" \"#F)F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "residue(f,x=c); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&*&%\"IG\"\"\"-%%sqrtG6#\"\"$\" \"\"!#:!\"&F'F,,&F%!\"*!\"$F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\" &\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "c:=3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "residue(f,x=c);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6## \"\"&\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 206 "As you can see, th e residue is precisely the number that goes in front of the logarithm \+ in the integral F of f. This is clear because if you have a series exp ansion at x=c, and t=x-c is the local parameter:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "add(a[i]*t^i,i=-3..5)+O(t^6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,6*&&%\"aG6#!\"$\"\"\"*$)%\"tG\"\"$F)!\"\"\"\"\"*& &F&6#!\"#F)*$)F,\"\"#F)F.F/*&&F&6#!\"\"F)F,F.F/&F&6#\"\"!F/*&&F&6#F/F/ F,F/F/*&&F&6#\"\"#F/)F,FDF)F/*&&F&6#\"\"$F/)F,FIF)F/*&&F&6#\"\"%F/)F,F NF)F/*&&F&6#\"\"&F/)F,FSF)F/-%\"OG6#*$)F,\"\"'F)F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "int(%,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #,6*&&%\"aG6#!\"$\"\"\"*$)%\"tG\"\"#F)!\"\"#!\"\"\"\"#*&&F&6#!\"#F)F,F .F0*&&F&6#F0\"\"\"-%#lnG6#F,F9F9*&&F&6#\"\"!F9F,F9F9*&&F&6#F9F9)F,F1F) #F9F1*&&F&6#F1F9)F,\"\"$F)#F9FJ*&&F&6#FJF9)F,\"\"%F)#F9FP*&&F&6#FPF9)F ,\"\"&F)#F9FV*&&F&6#FVF9)F,\"\"'F)#F9Ffn-%\"OG6#*$)F,\"\"(F)F9" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 124 "And indeed, you see the residue a [-1] appearing in front of the logarithm. This is also what the calcul us formula tells you:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "In t(a[-1]/(x-c),x) = int(a[-1]/(x-c),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$IntG6$*&&%\"aG6#!\"\"\"\"\",&%\"xG\"\"\"!\"$F/!\"\"F.*&F(F/- %#lnG6#F-F/" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "So the integrate \+ the rational functions with pole order at most 1, we need to introduce logarithms. These logarithms must be of the following form:" }}{PARA 0 "" 0 "" {TEXT -1 221 " a[-1] * log(d)\nwhere d is factor of the den ominator of f. This factor of d should be chosen in such a way that pr ecisely those poles of f where the residue is a[-1], those poles (and \+ no others) should be the roots of d." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "The question now is the following: Give n f, how to determine:" }}{PARA 0 "" 0 "" {TEXT -1 31 "*) the set of a ll residues of f" }}{PARA 0 "" 0 "" {TEXT -1 135 "*) for each residue, determine a polynomial d such that the poles that have that particula r residue are located at the roots of that d." }}}}{MARK "7 5 0" 46 } {VIEWOPTS 1 1 0 1 1 1803 }