{VERSION 4 0 "HP RISC UNIX" "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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 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 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal " -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }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 1 }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 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE " " -1 -1 "Helvetica" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 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 1 0 0 0 0 0 0 0 0 1 }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 1 0 0 0 0 0 0 0 0 1 }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 1 0 0 0 0 0 0 0 0 1 }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 1 0 0 0 0 0 0 0 0 1 }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 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 265 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 266 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 267 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 268 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 269 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 270 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 271 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 272 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 273 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 274 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 275 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 276 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 277 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 278 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 279 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 280 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 41 "Themenvorschl\344ge f \374r den 30. Oktober 2002" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 29 "a) - e): Der Integrationsweg " }{XPPEDIT 18 0 "gamma;" "6#%&gammaG " }{TEXT -1 177 " ist offensichtlich eine Ellipse mit Halbachse 5 auf \+ der reellen Achse und Halbachse zwei auf der imagin\344ren. Alle Integ rale k\366nnen nach dem Residuensatz ausgerechnet werden:\na)" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(z^2+1);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6$^#\"\"\"^#!\"\"" }}}{EXCHG {PARA 258 "" 0 "" {TEXT -1 135 "Beides liegt in der Ellipse, mu\337 also ber\374cksichti gt werden. Die Residuen bestimmt man am einfachsten \374ber eine Parti albruchzerlegung: " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "z/(z^2+1)=convert(z/(z^2+1), parfrac, z, I);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/*&%\"zG\"\"\",&*$)F%\"\"#F&F&F&F&!\" \",&*&F&F&,&F%F&^#F&F&F+#F&F**&F0F&,&F%F&^#F+F&F+F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "residue(z/(z^2+1), z=I);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "residue(z/(z^2+1), z=-I);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 30 " Damit ist das Integral gleich " }{XPPEDIT 18 0 "2*Pi*I;" "6#*(\"\"#\" \"\"%#PiGF%%\"IGF%" }{TEXT -1 155 " .\nb) Hier sind die Polstellen bei 3i und -3i, was beides au\337erhalb der Ellipse liegt. Das Integral v erschindet also nach dem Cauchyschen Integralsatz.\nc) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "tan(z)=convert(tan(z), sincos);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$tanG6#%\"zG*&-%$sinGF&\"\"\"-%$cos GF&!\"\"" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 64 "Der Cosinus hat Nul lstellen bei den halbzahligen Vielfachen von " }{XPPEDIT 18 0 "Pi;" "6 #%#PiG" }{TEXT -1 61 ", tan(z/2) hat also im betrachteten Intervall Po lstellen bei " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 9 " und bei " }{XPPEDIT 18 0 "-Pi;" "6#,$%#PiG!\"\"" }{TEXT -1 91 " . Da beide Pole \+ einfach sind, k\366nnen die Residuen einfach als Grenzwerte berechnet \+ werden:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "(z-Pi)*sin(z/2)/ cos(z/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*&,&%\"zG\"\"\"%#PiG!\" \"F'-%$sinG6#,$F&#F'\"\"#F'F'-%$cosGF,F)" }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 8 "ist f\374r " }{XPPEDIT 18 0 "z = Pi;" "6#/%\"zG%#PiG" } {TEXT -1 130 " ein unbestimmter Ausdruck; nach de l'Hospital hat er de nselben Grenzwert wie der Quotient der Ableitungen von Z\344hler und N enner. " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "diff(numer(%), z)/diff(denom(%), z);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&-%$sinG6#,$%\"zG#\"\"\"\"\"#F,*(F+F,,&F*F,%#PiG!\"\"F,-%$c osGF(F,F,F,F&F1!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "subs (z=Pi, %);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!\"#" }}}{EXCHG {PARA 262 "" 0 "" {TEXT -1 40 "Genauso berechnet man das Residuum bei " } {XPPEDIT 18 0 "-Pi;" "6#,$%#PiG!\"\"" }{TEXT -1 1 ":" }{MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "limit(( z+Pi)*tan(z/2), z=-Pi);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!\"#" }}} {EXCHG {PARA 263 "" 0 "" {TEXT -1 55 "Die Residuensumme ist also -4 un d das Integral gleich " }{XPPEDIT 18 0 "-8*Pi*I;" "6#,$*(\"\")\"\"\"% #PiGF&%\"IGF&!\"\"" }{TEXT -1 58 ".\nd) Hier gibt es innerhalb der Ell ipse Polstellen bei 0, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 6 " \+ und " }{XPPEDIT 18 0 "-Pi;" "6#,$%#PiG!\"\"" }{TEXT -1 20 " . Da die \+ Sinusreihe" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "sin(z) = seri es(sin(z), z);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$sinG6#%\"zG++F' \"\"\"F)#!\"\"\"\"'\"\"$#F)\"$?\"\"\"&-%\"OG6#F)F," }}}{EXCHG {PARA 264 "" 0 "" {TEXT -1 66 "mit dem Term z beginnt, ist das Residuum bei \+ z=0 gleich eins. Bei " }{XPPEDIT 18 0 "z = Pi;" "6#/%\"zG%#PiG" } {TEXT -1 5 " und " }{XPPEDIT 18 0 "z = -Pi;" "6#/%\"zG,$%#PiG!\"\"" } {TEXT -1 51 " ist es damit gleich -1, denn bei Verschiebung um " } {XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 121 " ver\344ndern sowohl sin( z) als auch 1/sin(z) einfach ihr Vorzeichen. Also ist die Residuensumm e gleich -1 und das Integral " }{XPPEDIT 18 0 "-2*Pi*I;" "6#,$*(\"\"# \"\"\"%#PiGF&%\"IGF&!\"\"" }{TEXT -1 92 " .\ne) Die Nennernullstellen \+ sind hier nat\374rlich dieselben, allerdings ist der Integrand bei " } {XPPEDIT 18 0 "z = 0;" "6#/%\"zG\"\"!" }{TEXT -1 43 " holomorph; Pole \+ gibt es also nur noch bei " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 5 " und " }{XPPEDIT 18 0 "-Pi;" "6#,$%#PiG!\"\"" }{TEXT -1 117 " . Die Residuen von 1/sin(z) an diesen Stellen m\374ssen noch mit dem jeweil igen z-Wert multipliziert werden, sind also " }{XPPEDIT 18 0 "-Pi;" "6 #,$%#PiG!\"\"" }{TEXT -1 5 " und " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" } {TEXT -1 76 " . Residuensumme und Integral sind also null.\n\nf) Das i st einfach die durch " }{XPPEDIT 18 0 "z^5;" "6#*$%\"zG\"\"&" }{TEXT -1 94 " dividierte Taylor-Reihe des Sinus.\ng) Hier kann man z.B. mit \+ Partialbruchzerlegung arbeiten: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "1/(z^2+1)=convert(1/(z^2+1), parfrac, z, I);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/*&\"\"\"F%,&*$)%\"zG\"\"#F%F%F%F%!\" \",&*&^##F%F*F%,&F)F%^#F%F%F+F%*&^##F+F*F%,&F)F%^#F+F%F+F%" }}}{EXCHG {PARA 265 "" 0 "" {TEXT -1 22 "Der Hauptteil ist also" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "-I/2*1/(z-I);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #*&^##!\"\"\"\"#\"\"\",&%\"zGF(^#F&F(F&" }}}{EXCHG {PARA 266 "" 0 "" {TEXT -1 41 "der Rest ist die um i holomorphe Funktion" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "I/2*1/(z+I);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&^##\"\"\"\"\"#F&,&%\"zGF&^#F&F&!\"\" " }}}{EXCHG {PARA 267 "" 0 "" {TEXT -1 36 "Das kann auch geschrieben w erden als" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "(1/4)*(1/(1 + ((z-I)/(2*I))));" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#,$*&\"\"\"F%,&F%F%*&^##!\"\"\"\"#F%,&%\"zGF%^#F*F%F%F%F*#F%\"\"%" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Probe:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*& \"\"\"F%,&!\"\"F%*&^#F%F%%\"zGF%F%F'#F'\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "simplify(% - I/2*1/(z+I));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 268 "" 0 "" {TEXT -1 70 "Damit g ibt die Formel f\374r die geometrische Reihe die restlichen Terme." } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "series( 1/(z^2+1), z=I, 10);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#+9,&%\"zG\"\" \"^#!\"\"F&^##F(\"\"#F(#F&\"\"%\"\"!^##F&\"\")F&#F(\"#;F+^##F(\"#K\"\" $#F&\"#kF-^##F&\"$G\"\"\"&#F(\"$c#\"\"'^##F(\"$7&\"\"(#F&\"%C5F1-%\"OG 6#F&\"\"*" }}}{EXCHG {PARA 269 "" 0 "" {TEXT -1 177 "h) Hier gibt es e inen Pol zweiter Ordnung bei z=-1, wir k\366nnen das Residuum also nic ht einfach als Grenzwert berechnen, sondern m\374ssen beispielsweise z un\344chst den Koeffizienten " }{XPPEDIT 18 0 "a[-2];" "6#&%\"aG6#,$\" \"#!\"\"" }{TEXT -1 71 " der Laurent-Reihe bestimmen. Dieser ist der G renzwert f\374r z -> -1 von " }{XPPEDIT 18 0 "(z+1)^2*f(z);" "6#*&,&% \"zG\"\"\"F&F&\"\"#-%\"fG6#F%F&" }{TEXT -1 118 " , also -1 + 2 = 1. Na ch Subtraktion des entsprechenden Terms erhalten wir eine Funktion mit einem Pol erster Ordnung." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "(z+2)/(z+1)^2 - 1/(z+1)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*& ,&%\"zG\"\"\"\"\"#F'F'*$),&F&F'F'F'F(F'!\"\"F'*&F'F'*$F*F'F,F," }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&\"\"\"F$,&%\"zGF$F$F$!\"\"" }}}{EXCHG {PARA 270 "" 0 "" {TEXT -1 63 "Also ist das Residuum gleich eins, wie man anhand der Zerlegung" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "(z+2)/(z+1)^2 = 1/(z+1)^2 + (z+1)/(z+1)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&,&%\"zG\"\"\"\"\"#F'F'*$),&F&F'F'F'F(F'! \"\",&*&F'F'*$F*F'F,F'*&F'F'F+F,F'" }}}{EXCHG {PARA 271 "" 0 "" {TEXT -1 27 "auch direkt sehen kann.\ni) " }{XPPEDIT 18 0 "2*Pi*I;" "6#*(\" \"#\"\"\"%#PiGF%%\"IGF%" }{TEXT -1 95 " nach dem Residuensatz.\nj) Auc h hier haben wir wieder Pole zweiter Ordnung. Aus der Darstellung" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "1/(z^2 + a^2)^2 = 1/((z+I*a) ^2*(z-I*a)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&\"\"\"F%*$),&*$)% \"zG\"\"#F%F%*$)%\"aGF,F%F%F,F%!\"\"*&F%F%*&),&F+F%*&^#F%F%F/F%F%F,F%) ,&F+F%*&^#F0F%F/F%F%F,F%F0" }}}{EXCHG {PARA 272 "" 0 "" {TEXT -1 31 "f olgt, da\337 der Koeffizient f\374r " }{XPPEDIT 18 0 "z = i*a;" "6#/% \"zG*&%\"iG\"\"\"%\"aGF'" }{TEXT -1 8 " gleich" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 16 "1/(I*a + I*a)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"\"F%*$)%\"aG\"\"#F%!\"\"#F*\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 16 "ist; genauso f\374r" }{TEXT -1 1 " " }{XPPEDIT 18 0 "z = -I*a;" "6#/%\"zG,$*&%\"IG\"\"\"%\"aGF(!\"\"" }{TEXT -1 2 " :" } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "1/(-I*a - I*a)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"\"F%*$)%\"aG\"\" #F%!\"\"#F*\"\"%" }}}{EXCHG {PARA 273 "" 0 "" {TEXT -1 31 "Subtrahiere den f\374hrenden Term:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 61 "factor(1/(z^2 + a^2)^2, I) + factor(1/(4*a^2)*1/(z \+ - I*a)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&\"\"\"F%*&),&%\"zGF% *&^#F%F%%\"aGF%F%\"\"#F%),&F)F%*&^#!\"\"F%F,F%F%F-F%F2F%*&#F%\"\"%F%*& )F,F-F%F.F%F2F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "normal(% );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,(*$)%\"aG\"\"#\"\"\"\"\"$*$ )%\"zGF)F*F**(^#F)F*F.F*F(F*F*F**(),&F.F**&^#F*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 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,(*$)%\"aG \"\"#\"\"\"\"\"$*$)%\"zGF)F*F**(^#F)F*F.F*F(F*F*F**(),&F.F**&^#F*F*F(F *F*F)F*),&F.F**&^#!\"\"F*F(F*F*F)F*F'F*F:#F*\"\"%" }}}{EXCHG {PARA 274 "" 0 "" {TEXT -1 68 "Zur Berechnung des Residuums mu\337 dies mit \+ z-ia multipliziert werden:" }{MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "factor((z - I*a)*%);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&%\"zG\"\"\"*&^#\"\"$F'%\"aGF'F'F '*&),&F&F'*&^#F'F'F+F'F'\"\"#F')F+F1F'!\"\"#F'\"\"%" }}}{EXCHG {PARA 275 "" 0 "" {TEXT -1 25 "Einsetzen von z=ia ergibt" }{MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs(z=I*a, %);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*&^##!\"\"\"\"%\"\"\"*$)%\"aG\"\"$F(F& " }}}{EXCHG {PARA 276 "" 0 "" {TEXT -1 51 "Genauso berechnet man auch \+ das Residuum bei -ia als" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "residue(1/(z^2 + a^2)^2, z=-I*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&^##\"\"\"\"\"%F&*$)%\"aG\"\"$F&!\"\"" }}}{EXCHG {PARA 277 "" 0 "" {TEXT -1 31 "Der Hauptteil bei z=ia ist also" } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "-1/(4*a ^2) * 1/(z - I*a)^2 - I/(4*a^3) * 1/(z - I*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&\"\"\"F%*&)%\"aG\"\"#F%),&%\"zGF%*&^#!\"\"F%F(F%F%F )F%F/#F/\"\"%*&^#F0F%*&)F(\"\"$F%F+F%F/F%" }}}{EXCHG {PARA 278 "" 0 " " {TEXT -1 13 "der bei z=-ia" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 53 "-1/(4*a^2) * 1/(z + I*a)^2 + I/(4*a^3) * 1/(z \+ + I*a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&\"\"\"F%*&)%\"aG\"\"#F% ),&%\"zGF%*&^#F%F%F(F%F%F)F%!\"\"#F/\"\"%*&^##F%F1F%*&)F(\"\"$F%F+F%F/ F%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Probe:" }{MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "series(1/(z^2 + a^2)^2, z= I*a, 2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#++,&%\"zG\"\"\"*&^#!\"\"F& %\"aGF&F&,$*&F&F&*$)F*\"\"#F&F)#F)\"\"%!\"#*&^#F0F&*$)F*\"\"$F&F)F),$* &F&F&*$)F*F1F&F)#F7\"#;\"\"!-%\"OG6#F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "series(1/(z^2 + a^2)^2, z=-I*a,2);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#++,&%\"zG\"\"\"*&^#F&F&%\"aGF&F&,$*&F&F&*$)F)\"\"#F&! \"\"#F/\"\"%!\"#*&^##F&F1F&*$)F)\"\"$F&F/F/,$*&F&F&*$)F)F1F&F/#F8\"#; \"\"!-%\"OG6#F&F&" }}}{EXCHG {PARA 279 "" 0 "" {TEXT -1 245 "k) Beim R adius a/2 ist die Funktion im Innern holomorph, das Integral verschwin det also.\nBeim Radius 2a liegen beide Pole im Innern; da sich ihre Re siduen aufheben, verschwindet das Integral auch hier.\nl) Man betracht e denselben Integrationsweg " }{XPPEDIT 18 0 "gamma[R];" "6#&%&gammaG6 #%\"RG" }{TEXT -1 69 " wie in der Vorlesung. Das Integral dar\374ber i st nach dem Residuensatz" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "2*Pi*I * (-1/4*I/(a^3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%#PiG \"\"\"*$)%\"aG\"\"$F&!\"\"#F&\"\"#" }}}{EXCHG {PARA 280 "" 0 "" {TEXT -1 50 "Das Integral \374ber den Kreisbogen verschwindet f\374r " } {XPPEDIT 18 0 "proc (R) options operator, arrow; infinity end;" "6#R6# %\"RG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT -1 96 " , da der Integrand dann auch noch nach Einsetzen des Integrationswegs gege n null geht. Also ist" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "as sume(a>0):\nInt(1/(x^2+a^2)^2, x=-infinity..infinity)\n= int(1/(x^2+a^ 2)^2, x=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$I ntG6$*&\"\"\"F(*$),&*$)%\"xG\"\"#F(F(*$)%#a|irGF/F(F(F/F(!\"\"/F.;,$%) infinityGF3F7,$*&%#PiGF(*$)F2\"\"$F(F3#F(F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "1 0 0" 8 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }