{VERSION 2 3 "IBM INTEL NT" "2.3" } {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 Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 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 1 1 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 1 14 0 0 0 0 1 0 1 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 "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 6 6 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 "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }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 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 39 "Autoren: Jan K\344stle und Andreas Pf\344ffle " }}}{EXCHG {PARA 256 "" 0 "" {TEXT 256 22 "Vollst\344ndige Induktion " }}{PARA 0 "" 0 "" {TEXT -1 82 "Die vollst\344ndige Induktion dient d azu, eine mathematische Behauptung zu beweisen. " }}}{SECT 1 {PARA 3 " " 0 "" {TEXT -1 9 "Erkl\344rung" }}{PARA 0 "" 0 "" {TEXT -1 249 "Die V orraussetzung f\374r eine vollst\344ndige Induktion ist, da\337 das er ste Glied der Behauptung zutrifft. Somit wird angenommen, da\337 die B ehauptung f\374r alle weiteren Folgenglieder richtig ist. Dies gillt e s durch eine vollst\344ndige Induktion zu beweisen. " }}{PARA 0 "" 0 " " {TEXT -1 63 "Bsp.: Als erstes stellt man eine Behauptung f\374r eine Folge auf." }}{PARA 0 "" 0 "" {TEXT -1 76 " Dann pr\374ft man , ob diese Behauptung f\374r das erste Element zutrifft." }}{PARA 0 " " 0 "" {TEXT -1 64 " Nun wird angenommen, da\337 die Behauptun g A(k) wahr sei. " }}{PARA 0 "" 0 "" {TEXT -1 57 " Es gillt zu beweisen, da\337 A(k+1) auch wahr ist. " }}{PARA 0 "" 0 "" {TEXT -1 48 "Damit ist bewiesen, da\337 die Annahme richtig ist." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Beispiele" }}{PARA 0 "" 0 "" {TEXT -1 8 "A nnahme:" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 " 1+2+3" ",(\"\"\"\"\"\"\"\"# F$\"\"$F$" }{TEXT -1 4 "+..." }{XPPEDIT 18 0 "+n=(1/2)*n*(n+1)" "/%\"n G**\"\"\"\"\"\"\"\"#!\"\"F#F&,&F#F&\"\"\"F&F&" }}{PARA 0 "" 0 "" {TEXT -1 5 "also " }{XPPEDIT 18 0 " Sum(i,i=0..n" "-%$SumG6$%\"iG/F%; \"\"!%\"nG" }{TEXT -1 2 " =" }{XPPEDIT 18 0 " 1/2" "*&\"\"\"\"\"\"\"\" #!\"\"" }{TEXT -1 8 " n (n+1)" }}{PARA 0 "" 0 "" {TEXT -1 53 "Pr\374fe n, ob diese Behauptung f\374r ein Element zutrifft:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart: with (plots):" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 36 "Summe(n):=factor(sum('i','i'=0..n));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>-%&SummeG6#%\"nG,$*&F'\"\"\",&F'F*F*F *F*#F*\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "evalb(Summe( n)=subs(n=1,(1/2)*n*(n+1)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%true G" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Da die Annahme f\374r n=1 st immt, gehen wir einen Schritt weiter:" }}{PARA 0 "" 0 "" {TEXT -1 52 " Wir pr\374fen, ob die Behauptung auch f\374r n+1 zutrifft." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Summe(n+1):=subs(n=n+1,Summe(n));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%&SummeG6#,&%\"nG\"\"\"F)F),$*&,&F (F)\"\"#F)F)F'F)#F)F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 230 "Die Sum me von n+1 kann man mit der Summe von n gleichsetzen, wenn man zu dies er Summe noch das n\344chste Element dazuaddiert, also n+1. Wenn die b eiden Summen gleich sind, ist der Beweis erbracht, da\337 die Behauptu ng f\374r alle n gilt." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "A ddition:=subs(n=n+1,n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)Addition G,&%\"nG\"\"\"F'F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "evalb (expand(Summe(n)+Addition)=expand(Summe(n+1)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Ein an derer Weg ist auch, die beiden Summen gleichzusetzen und nach n aufzul \366sen. Bei einer wahren Behauptung gilt die Gleichung f\374r alle n. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "solve(\{Summe(n)+Additi on=Summe(n+1)\},n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/%\"nGF%" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart: with (plots):" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 257 11 "2. Beispiel" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 11 "Annahme:\n " }{XPPEDIT 18 0 " 1^3+2^3+3^3" ",(*$ \"\"\"\"\"$\"\"\"*$\"\"#\"\"$F&*$\"\"$\"\"$F&" }{TEXT -1 6 "+....+" } {XPPEDIT 18 0 " n^3 " "*$%\"nG\"\"$" }{TEXT -1 3 " = " }{XPPEDIT 18 0 " sum(i^3,i=0..n)" "-%$sumG6$*$%\"iG\"\"$/F&;\"\"!%\"nG" }{TEXT -1 2 " =" }{XPPEDIT 18 0 " (1/4) " "*&\"\"\"\"\"\"\"\"%!\"\"" }{TEXT -1 1 " \+ " }{XPPEDIT 18 0 " n^2*(n+1)^2" "*&%\"nG\"\"#,&F#\"\"\"\"\"\"F&\"\"#" }{TEXT -1 1 " " }{MPLTEXT 1 0 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Summe(n):=factor(sum('i^3','i'=0..n));" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>-%&SummeG6#%\"nG,$*&F'\"\"#,&F'\"\"\"F,F,F*#F, \"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "f\374r n=1 \374berpr\374 fen" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "evalb(1^3=subs(n=1,S umme(n)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Summe von n+1" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 33 "Summe(n+1):=subs(n=n+1,Summe(n));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%&SummeG6#,&%\"nG\"\"\"F)F),$*&F'\"\"#,&F(F)F,F)F, #F)\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "Um von der Summe von n auf die Summe von n+1 zu kommen, mu\337 man zu der Summe von n das \+ n+1 te Folgenglied addieren:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Addition:=subs(n=n+1,n^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% )AdditionG*$,&%\"nG\"\"\"F(F(\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 50 "Nachweis, da\337 die Behauptung allgemein g\374ltig ist:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "evalb(expand(Summe(n+1))=exp and(Summe(n)+Addition));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Zusatz: allgemein g\374ltig f \374r alle n:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "solve(\{ex pand(Summe(n+1))=expand(Summe(n)+Addition)\});" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<#/%\"nGF%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "Allgemein" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "restart: with (plots):\nSumme(n):=factor( sum(\n" }{TEXT -1 50 "Bitte geben sie die Summe einer Folge von 1..n a n:" }{MPLTEXT 1 0 56 "\n'1/(2^i)','i'=1..n));\nSumme(n+1):=subs(n=n+1, Summe(n));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%&SummeG6#%\"nG,&)#\" \"\"\"\"#,&F'F+F+F+!\"#F+F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%&Sum meG6#,&%\"nG\"\"\"F)F),&)#F)\"\"#,&F(F)F-F)!\"#F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Addition:=subs(n=n+1,\n" }{TEXT -1 119 "B itte geben Sie das-zu-addierende Folgenglied an (Bsp.: 1+2+3+4+....n, \+ dann ist n das n\344chste Glied das addiert wird): " }{MPLTEXT 1 0 8 " 1/2^n);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)AdditionG*$)\"\"#,&%\" nG\"\"\"F*F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 216 "ant:= evalb(simplify(expand(Summe(n+1)))=simplify(expand(Summe(n)+Addition)) ): if ant=true then `Ihre Behauptung ist richtig und f\374r alle n g \374ltig` else `Ihre Behautung ist nicht richtig und gilt nicht f\374r alle n` fi;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%RIhre~Behauptung~ist~ richtig~und~f|gzr~alle~n~g|gzltigG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }