{VERSION 5 0 "IBM INTEL LINUX" "5.0" } {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 Output" 2 20 "" 0 1 0 0 255 1 0 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 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 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 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 31 "with(plots): read \" values_5.m\";" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 661 "#Below I'm graphing the nor mal derivative of the heat kernel over time for the 5-series. You can check the difference between using 256 points or 512 points by changi ng the range in \"summd\". This graph takes a while to plot, so be pa tient. The reason the 4/3 appears in the equation is due to our need \+ for an orthonormal basis. On the five series, the eigenfunctions are n ot orthonormal, and so a new basis is created. This results in only a constant multiplication of the normal derivative. The graph is logar ithmic scaled, so as times goes on, you move down and to the left. Yo u can change the time range by manipulating the x range in the plot co mmand." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "summd:=log(sum('(exp(-(El5[j])*x))*(4*((d n5[j])^2)/3)','j'=1..256)):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "plot([log(1/x),summd,x=10^(-9)..1],title=\"Plot of function with \+ norm at same point on j=256 for 5-series\");" }}{PARA 13 "" 1 "" {GLPLOT2D 382 382 382 {PLOTDATA 2 "6&-%'CURVESG6$7gn7$$\"IToyFf@4B>;c6 TYp$eEB2#!#Q$\"Ixw:]\"*oJf%G;18nLFR9D8#F*7$$\"Ih?vax]^hN02.^7C'H5Ob %)o`.$=RsWmpJ313E(F07$$\"INqvKAy=-Q5Kf@KH')*)4$>'F0$\"ID^Ey87>Y+&>@LmA 2(z.HnF07$$\"I6-rCuH=lS(Q\\!=:=OqT0fF0$\"IC0[3s'f>bl!*=0/fF07$$\"I/=R- ]XC #*\\!o![F0$\"Il#Rrs_K\"zt(z\"o\"3jK'HQ&Q&F07$$\"IZ.u@qP>*e69ni&oK]H7>X F0$\"I#*f9$)4dVDLg@m*p@LRly8&F07$$\"IRY:KK^\"f^,n))>s.G(yl8TF0$\"IUk%* 3HVpN1@],i%*3I;F,ZF07$$\"Ig+$)Hs7Wv'fxjkXWc\"e(f#QF0$\"I3rXs?HsYQbBou[ kb/->VF07$$\"IZ)HE437.IxutPRN2+yZY$F0$\"It\">8*=i()y\"*fm&G;Z;Hyhs$F07 $$\"I._5#Q%=.(*))\\$R5?63z')**>$F0$\"IM&=\">d(y9#*y>,)e%3f!=[uJF07$$\" IJPYJZB.]!o]X#yqq5d9zFF0$\"IlF'oxx-sM^!>:*3e?\")pC(>F07$$\"I;gh)>8T``' [*\\'*o?\\(3v76m8\"z.!*R8-kM NYaAF0$!I#zL8U&p2=nEG^M!y(48&\\A%F^p7$$\"IeB3:I_T9#*\\k\\q`8\\s^\"3#F0 $!IN[EHo$RIY'pZTo*z[4a'H:F07$$\"I4&e0ThJfF`!3>z*=%*zr#H>F0$!IA1T$)Gy=2 /_aCB!*G4y/wEF07$$\"I4iz'y**QMplv!Rgp/FG;Ot;F0$!Iq_'=KJOh-zWRb-Is\\[M/&F07$$\"I4 X]34xbPFjN:XP\\O`mj:F0$!I]qcpIUD$RgRnV]FDBu!fiF07$$\"I!))pP[K)yF`#fQ)4 hxxN5w9F0$!I;G\"y*)**z2N.Z:Lo_:/)zHtF07$$\"Il9[m+D))p(\\&))y9DN+V\"fQ \"F0$!IyLovBj;%e)[e!)y5C]%z^`)F07$$\"I.*o\"[3ray!Q \"F*7$$\"I'3_(eMJ&)G&**)e6Vl,R(yi4\"F0$!I$GPP%o&pMkh?SA@xqAaIK\"F*7$$ \"IZPTvtV\\UTccFj@)G+62/\"F0$!Iq0qO2-vm#zj:WS5>%zuH9F*7$$\"IAmQ&yxEq@Z ![LU`+t8L(z*F^p$!IkX=(4oJBsDE#>e'ek4eQb\"F*7$$\"I'pWejVwLwC_**[!z^.+&p G*F^p$!Ij,!yS'*)f8H&4>\"[-*Gh;Pm\"F*7$$\"Id/ab(H9oD$Go%4%er]4Tb()F^p$! 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Since the graph was parametric , if you would like to find a value at a certain spot on the x-axis, s ay x=a, you'll need to check the value at exp(-a)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "intvalue5:= x -> log(sum('(exp(-(El5[j])*x))*(4*((dn5[j])^2)/3)',' j'=1..256));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*intvalue5Gf*6#%\"xG 6\"6$%)operatorG%&arrowGF(-%$logG6#-%$sumG6$.,$*&#\"\"%\"\"$\"\"\"*&-% $expG6#,$*&&%$El5G6#%\"jGF89$F8!\"\"F8)&%$dn5GFA\"\"#F8F8F8/.FB;F8\"$c #F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "evalf(intvalue5( 1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!I(4&y-LKV6n(fYhZ3H!RUW]!#Q" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-14)) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"I%\\hEf2sV;q_^tPK4?F!)p\"!#Q " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-15 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"I3*4-xSb&e^3Z&GGR8yy#H=!#Q " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-16 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"IQL?e(H3d)R_mB6&=51vY'>!#Q " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-17 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"Ina=$\\'4!4P[A,:<'[h3Jk?!#Q " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-18 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"IR\"Q\"e3 " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-1 9)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"Izx6[0PspnF`f=.8WVTC@!#Q" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-20)) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"I\\hkc[W\"fW_#)RMvd5HS18#!#Q " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(intvalue5(exp(-21 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"Im\\[.yg%y3rhmf]\\r)=%H8#!# Q" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "evalf(intvalue5(10^(-9 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"Ixw:]\"*oJf%G;18nLFR9D8#!#Q " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 44 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