{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 1 12 255 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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "2D Output" 2 20 "" 1 12 0 0 255 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple O utput" 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 "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "R3 Font 0" -1 256 1 {CSTYLE " " -1 -1 "Helvetica" 1 16 0 0 255 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Couri er" 1 10 0 0 0 1 2 1 2 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 "Newtonverfahren" {TEXT -1 64 "Dr. M. Komm a Isolde-Kurz-Gymnasium Reutlingen 1.2.96" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 18 "" 0 "" {TEXT -1 15 "Newtonverfahren" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "res tart: with(plots):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "Formeln wi e die \"Mitternachtsformel\" stellen Ausnahmef\344lle dar. In der Rege l lassen sich die Nullstellen von Funktionen nur numerisch bestimmen. " }}{PARA 0 "" 0 "" {TEXT -1 109 "Die Idee des Newtonverfahrens ist es , die Funktion durch eine Gerade zu ersetzen und linear zu extrapolier en." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "plot(\{x^2-1,4*(x-2)+3\},x=- 1..3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 256 "Statt der Nullstelle d er gegebenen Funktion kann man auch den Schnittpunkt einer Tangente an die Kurve mit der x-Achse nehmen, wenn der Tangentenpunkt nahe genug \+ an der Nullstelle liegt. Aber wie kommt man nahe genug an die Nullstel le, die man nicht kennt?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Man kann ja f\374r alle F\344lle e inmal die Tangentengleichung aufstellen:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "tgl:=(yt-f(x))/(xt-x)=fs(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$tglG/*&,&%#ytG\"\"\"-%\"fG6#%\"xG!\"\"F),&%#xtGF)F-F .F.-%#fsGF," }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 2 "yt" }{TEXT -1 5 " \+ und " }{TEXT 257 2 "xt" }{TEXT -1 42 " sind die Variablen der Geradeng leichung, " }{TEXT 258 1 "y" }{TEXT -1 5 " und " }{TEXT 259 1 "x" } {TEXT -1 30 " geben den Tangentenpunkt an, " }{TEXT 260 1 "f" }{TEXT -1 22 " ist die Funktion und " }{TEXT 261 2 "fs" }{TEXT -1 15 " die Ab leitung." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 2 "" 1 "" {TEXT -1 1 "\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Wo liegt der Sc hnittpunkt der Tangente mit der x-Achse?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "yt:=0:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "r s:=expand(solve(tgl,xt));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#rsG,&* &-%#fsG6#%\"xG!\"\"-%\"fGF)\"\"\"F+F*F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "Dieser " }{TEXT 262 1 "x" }{TEXT -1 135 "-Wert liegt jeden falls n\344her an der Nullstelle, als der Startwert. Also lohnt es sic h wohl, das Ganze zu wiederholen. Mit der Funktion " }{XPPEDIT 18 0 "p hi(x)" "-%$phiG6#%\"xG" }{TEXT -1 37 " ordnen wir also dem urspr\374n glichen " }{TEXT 263 1 "x" }{TEXT -1 26 "-Wert einen Nachfolger zu:" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "phi:=unapply(rs,x);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%$phiG:6#%\"xG6\"6$%)operatorG%&arrowGF(,&*&-%# fsG6#9$!\"\"-%\"fGF0\"\"\"F2F1F5F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Wir sollten noch die Ableitung definieren" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "fs:=D(f);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fsG-% \"DG6#%\"fG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "und eine Funktion \+ angeben:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f:=x->x^3-2*x-5;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG:6#%\"xG6\"6$%)operatorG%&arrowG F(,(*$9$\"\"$\"\"\"F.!\"#!\"&F0F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "So sieht die Funktion aus:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "fpl:=plot(f,-2..3):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "fpl;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "So sieht " }{XPPEDIT 18 0 "phi" "I$ phiG6\"" }{TEXT -1 5 " aus:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "x:='x ':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "phi(x);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,&*&,&*$%\"xG\"\"#\"\"$!\"#\"\"\"!\"\",(*$F'F)F+ F'F*!\"&F+F+F,F'F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Und wenn wi r phi dreimal anwenden, bekommen wir diesen Ausdruck" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "(phi@@3)(x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# ,**&,&*$,(*&,&*$,&*&,&*$%\"xG\"\"#\"\"$!\"#\"\"\"!\"\",(*$F/F1F3F/F2! \"&F3F3F4F/F3F0F1F2F3F4,**$F+F1F3F,F0F/F2F7F3F3F4F,F4F/F3F0F1F2F3F4,,* $F'F1F3F(F0F,F0F/F2F7F3F3F4F(F4F,F4F/F3" }}}{EXCHG {PARA 0 "" 0 "Itera tion" {TEXT -1 13 "Und wenn wir " }{XPPEDIT 18 0 "phi" "I$phiG6\"" } {TEXT -1 152 " zehnmal anwenden brauchen wir schon ein CAS zur Darstel lung des Terms. Wir k\366nnen es aber auch numerisch versuchen. Mit n \+ Schritten und dem Startwert " }{XPPEDIT 18 0 "x[1]" "&%\"xG6#\"\"\"" }{TEXT -1 14 " erhalten wir:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "n:=5 :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "x[1]:=3:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "for i to n do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "x[i+1]:=evalf(phi(x[i]));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 " od;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"# $\"++++gB!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"$$\"+!y' >F@!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"%$\"+Pg8&4#!\" *" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"&$\"+u;b%4#!\"*" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"xG6#\"\"'$\"+#[^X4#!\"*" }}} {EXCHG {PARA 0 "" 0 "Fixpunkt" {TEXT -1 39 "Schon beim f\374nften Schr itt \344ndert sich " }{TEXT 264 1 "x" }{TEXT -1 56 " praktisch nicht m ehr, wir sind sehr nahe am Fixpunkt " }{XPPEDIT 18 0 "x=phi(x)" "/% \"xG-%$phiG6#F#" }{TEXT -1 2 " ." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}{PARA 2 "" 1 "" {TEXT -1 1 "\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "Zur graphischen Veranschaulichung bilden wir die Tangenten des i-ten Schrittes mit der Punkt-Steigungsform:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "i:='i':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 " psf:=(i,x)->fs(x[i])*(x-x[i])+f(x[i]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$psfG:6$%\"iG%\"xG6\"6$%)operatorG%&arrowGF),&*&-%#fsG6#&9%6#9 $\"\"\",&F3F6F2!\"\"F6F6-%\"fGF1F6F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "i:='i':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 " tangenten:=seq(psf(i,x),i=1..n);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>% *tangentenG6',&%\"xG\"#D!#f\"\"\",&F'$\"+++)3Z\"!\")$!++7&)GJF.F*,&F'$ \"+U)*[d6F.$!+\"o)4DCF.F*,&F'$\"+/&yo6\"F.$!+S(f$RBF.F*,&F'$\"+:S9;6F. $!++6#yL#F.F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Testplot der ers ten Tangente:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plot(\{f(x),tangen ten[1]\},x=x[n]-1..x[n]+1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Au schnitt der Funktion in dem interessierenden Bereich:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "rfpl:=replot(fpl,view=[min(x[n],x[1])-0.2..max(x [n],x[1])+0.2,min(seq(f(x[i]),i=1..n))-1.." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "max(seq(f(x[i]),i=1..n))+1]):rfpl;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Darstellung der Iteration - statisch:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 96 "plseq:=seq(display(\{rfpl,plot(tangenten[ i],x=min(x[n],x[1])-0.2..max(x[n],x[1])+0.2,color=blue)," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "plot([[x[i],0],[x[i],f(x[i])]],color=red)\}), i=1..n):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(\{plseq\});" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "bewegt:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "display([plseq],insequence=true);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "So geht es auch:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "zickzack:=seq(display(\{rfpl,plot([[x[i],f(x[i])],[x[i+1],0]],co lor=blue)," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot([[x[i],0],[x[i], f(x[i])]],color=red)\}),i=1..n-1):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "display(\{zickzack\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "display([zickzack],insequence=true);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Aufgaben: " }}{PARA 0 "" 0 "" {TEXT -1 29 "1. Neue Wahl d es Startpunkts " }{XPPEDIT 18 0 "x[1]" "&%\"xG6#\"\"\"" }{TEXT -1 103 ". Gibt es Startpunkte, mit denen man nie nahe genug an die Nullstelle kommt - keine Konvergenz erzielt?" }}{PARA 0 "" 0 "" {TEXT -1 26 "2. \+ Neue Wahl der Funktion " }{TEXT 265 4 "f(x)" }{TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "U nd welche L\366sung schl\344gt Maple vor?" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Digits: =100:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "fsolve(f(x));" }} {PARA 12 "" 1 "" {XPPMATH 20 "6#$\"_qex:x " 0 "" {MPLTEXT 1 0 11 "Digits:=10:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Darstellung des Fixpunkts" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot(\{x,phi(x),f(x)\},x=-5..5,-10..10);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 109 "Kann man dem letzten Plot entnehmen, f\374r welch e Startpunkte es Schwierigkeiten mit der Konvergenz geben kann?" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "to be continu ed..." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 2 "" 1 "" {TEXT -1 1 "\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "komma@oe.uni-tuebinge n.de" }}}}{MARK "0 0 0" 21 }{VIEWOPTS 1 1 0 1 1 1803 }