{VERSION 3 0 "IBM INTEL NT" "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 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 14 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 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 1 14 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 }{CSTYLE "" -1 261 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "versuch" 20 262 "Copperpla te Gothic Bold" 0 0 255 0 255 1 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 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart;\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1 "\n" }{TEXT -1 26 "Differentia lgleichung " }{XPPEDIT 18 0 "diff(y,t) = -y;" "6#/-%%diffG6$%\"yG% \"tG,$F'!\"\"" }{MPLTEXT 1 0 2 " " }{TEXT -1 37 "und ihre exakte L \366sung an der Stelle " }{XPPEDIT 18 0 "t = 1;" "6#/%\"tG\"\"\"" } {TEXT -1 2 ": " }{MPLTEXT 1 0 75 "\ndsolve(\{diff(y(t),t)=-y(t),y(0)=1 \}, y(t));\nExakte_lsg:=subs(t=1,rhs(%));\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "\n Die numerischen L\366sungen nach der Euler-Met hode " }{TEXT 256 4 "E[j]" }{TEXT -1 11 " mit den Sc" }{TEXT 18 0 "" } {TEXT 262 0 "" }{TEXT -1 12 "hrittweiten " }{XPPEDIT 18 0 "h = 1;" "6# /%\"hG\"\"\"" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "1/2;" "6#*&\"\"\"\" \"\"\"\"#!\"\"" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "1/4;" "6#*&\"\"\" \"\"\"\"\"%!\"\"" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "1/8;" "6#*&\"\" \"\"\"\"\"\")!\"\"" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "1/16;" "6#*&\" \"\"\"\"\"\"#;!\"\"" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "1/32;" "6#*& \"\"\"\"\"\"\"#K!\"\"" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "1/64;" "6#* &\"\"\"\"\"\"\"#k!\"\"" }{TEXT -1 32 " und die dazugeh\366rigen Fehl er " }{TEXT 257 11 " " }{TEXT 258 16 " EFehler[j] " } {TEXT -1 1 ":" }{MPLTEXT 1 0 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 231 "for j from 0 to 6 do\n g:=dsolve(\{diff(y(t),t)=-y(t),y(0)=1\}, \+ y(t), type=numeric, method=classical, stepsize=1/(2^j),output=lis tprocedure);\n gy := subs(g,y(t)):\n E[j]:=gy(1);\n EFehler[j]:=\na bs(evalf(Exakte_lsg-E[j]));\nod:\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 " Die numerischen L\366sungen mit der Richardson-Extrapolat ion " }{TEXT 259 4 "R[j]" }{TEXT -1 31 " und die dazugeh\366rigen Fe hler " }{TEXT 260 0 "" }{TEXT 261 11 "RFehler[j] " }{TEXT -1 1 ":" } {MPLTEXT 1 0 1 "\n" }{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "for j from 0 to 5 do\n R[j]:=2*E[j+1]-E[j];\n RFehler[j]:=abs(ev alf(Exakte_lsg-R[j]));\nod:\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 " Ausgabe:" }{MPLTEXT 1 0 1 "\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 1016 "printf(`Tabelle 1: Richardson-Extrapolation`);\nprintf(`--- ---------------------------------------------------------------------- -------------------------------------\\n`);\nprintf(` h Eu ler Richardson Euler-Fehler Richardson- Euler-Fehler /h Richardson-\\n`);\nprintf(` \+ Fehler Fehler/h^ 2\\n`);\nprintf(`----------------------------------------------------- ---------------------------------------------------------\\n`);\nfor j from 0 to 5 do\n printf(` 1/%2d %10.8f %10.8f %10.8f \+ %10.8f %10.8f %10.8f \\n`, 2^j,E[j], R[j], EFehler [j], RFehler[j], EFehler[j]*(2^j), RFehler[j]/((1/(2^j))^2) );\nod;\n printf(` 1/64 %10.8f - %10.8f - \+ %10.8f - \\n`, E[6], EFehler[6], EFehler[6]*(2^6) );\nprintf(`--------------------------------------------------------- -----------------------------------------------------\\n`);\n\n\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 9 } {VIEWOPTS 1 1 0 1 1 1803 }