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-1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 3 4 3 4 2 2 0 1 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 16 "Leobardo Rosales" }}{PARA 256 " " 0 "" {TEXT -1 5 "Final" }}{PARA 256 "" 0 "" {TEXT -1 9 "math 107B" } }{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 405 "Thi s project is concerned with several examples of sets and functions whi ch can be constructed iteratively. These sets and functions are import ant ones, for they have been the source of examples and counter-exampl es for analysts and topologists. As such, this project deals not only \+ with the technical challenges of reproducing the construction of these examples, but we will also discuss their relevance." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 " We start with an inter esting example, the Cantor Set and function." }}{SECT 0 {PARA 3 "" 0 " " {TEXT -1 10 "Cantor Set" }}{PARA 0 "" 0 "" {TEXT -1 589 "The Cantor \+ Set is a subset of the interval [0,1] which was discovered by Georg Ca ntor and presented in the appendix of a paper written in 1883. The set is constructed as follows, we start with the interval [0,1]. We remov e the open middle segment (1/3,2/3) from this interval, and we are lef t with two closed segements, [0,1/3] and [2/3,1]. We grab each of thes e segments and remove the middle thirds from them, so that we are left with the union of the closed intervals [0,1/9],[2/9,1/3],[2/3,7/9], a nd [8/9,1]. We then continue this process. What is left is what we cal l the Cantor Set. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 775 "This set has remarkable properties. Firstly, we should r emark that this set is non-empty. We always remove the middle thirds o f intervals in constructing the set, and as such, we will never remove the endpoints of any segment from the previous step of the constructi on. Thus, the points 0,1/9,2/9,1/3 and so on will be in the Cantor Set . However, much more is left than just these endpoints, in fact so muc h is left over as to make the Cantor Set uncountable. The proof is sim ple, once we identify that the Cantor Set is the set of all number in \+ [0,1] whose ternary decimal expansion contains only 0's and 1's. Such \+ a set is uncountable by a diagonalization argument. For supposing that we had such an enumeration of numbers, then we could list them as a s equence of points a" }{TEXT 292 3 "(n)" }{TEXT -1 43 " in their ternar y represenation as follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 293 1 "a" }{TEXT -1 21 "(0)= .020200020020..." }} {PARA 0 "" 0 "" {TEXT 294 1 "a" }{TEXT -1 21 "(1)= .220220020000..." } }{PARA 0 "" 0 "" {TEXT 295 1 "a" }{TEXT -1 21 "(2)= .000022222022..." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "and so \+ forth. However, then by letting " }{TEXT 296 1 "A" }{TEXT -1 21 " be t he number whose " }{TEXT 297 1 "n" }{TEXT -1 31 "th decimal place is a 0 if the " }{TEXT 298 1 "n" }{TEXT -1 20 "th decimal place of " } {TEXT 419 4 "a(n)" }{TEXT -1 35 " is 2, and vice versa, we get that " }{TEXT 300 1 "A" }{TEXT -1 48 " is in the Cantor Set but is different \+ from any " }{TEXT 420 4 "a(n)" }{TEXT -1 62 " in our enumeration. Thus , the Cantor Set must be uncountable." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 652 "The Cantor Set is also compact, perf ect, and its closure has an empty interior. That the Cantor Set is com pact is easy to see, for it is the countable intersection of the compa ct sets [0,1/3], [2/3,1], [0,1/9], [2/9,1/3], [2/3,7/9], [8/9,1], and \+ so forth. The Cantor Set is also totally disconnected, meaning that we cannot find two distinct points in the Cantor Set that can be connect ed by a continuous function on [0,1] whose image lies totally inside t he Cantor Set. Thus, the Cantor Set is in some respects very meaty. Ho wever, in other respects the Cantor Set is also very thin, in particul ar with respect to measure which we shall discuss later." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Now, how can we u se maple to better understand the Cantor Set? My first goal was to dev ise an algorithm which given and " }{TEXT 277 1 "i" }{TEXT -1 7 " and \+ a " }{TEXT 278 1 "k" }{TEXT -1 12 ", gives the " }{TEXT 279 1 "i" } {TEXT -1 58 "th point which is an endpoint of a segment removed at the " }{TEXT 280 4 " k" }{TEXT -1 75 "th step of the construction of th e Cantor Set. The algorithm is as follows:" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 196 "cantorpoint:= proc(i,k) \noption remember ; \nif i =1 then 0 else \nif i mod 2 = 0 \nthen cantorpoint(i-1,k)+(1/(3^k)) \+ ; \nelse cantorpoint(i-1,k)+((3^((ifactors(i-1)[2][1][2])-1))/(3^k)) ; \nfi fi end;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%,cantorpointGf*6$%\" iG%\"kG6\"6#%)rememberGF)@%/9$\"\"\"\"\"!@%/-%$modG6$F.\"\"#F0,&-F$6$, &F.F/F/!\"\"9%F/*&F/F/)\"\"$FF;F/F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 32 "The algori thm is recursive. The " }{TEXT 281 1 "i" }{TEXT -1 124 "th point is gi ven by adding an appropriate amount to the previous point, what you ad d depends highly on the divisibility of " }{TEXT 282 1 "i" }{TEXT -1 35 " by 2. In this algorithm, for each " }{TEXT 283 1 "k" }{TEXT -1 2 ", " }{TEXT 284 1 "i" }{TEXT -1 21 " ranges from 1 to 2^(" }{TEXT 291 1 "k" }{TEXT -1 4 "+1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "To verify the validity of this algorithm, let us mak e a list of the cantor points:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "cantorsequence:=(k)-> [seq(cantorpoint(i,k),i=1..2^(k+1))];" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%/cantorsequenceGf*6#%\"kG6\"6$%)oper atorG%&arrowGF(7#-%$seqG6$-%,cantorpointG6$%\"iG9$/F3;\"\"\")\"\"#,&F4 F7F7F7F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 95 "and let us thus list the points used in the construction of the cantor set for the steps 1,2,3 :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "cantorsequence(1);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#7&\"\"!#\"\"\"\"\"$#\"\"#F'F&" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "cantorsequence(2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*\"\"!#\"\"\"\"\"*#\"\"#F'#F&\"\"$#F)F+#\" \"(F'#\"\")F'F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "cantorse quence(3);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#72\"\"!#\"\"\"\"#F#\" \"#F'#F&\"\"*#F)F+#\"\"(F'#\"\")F'#F&\"\"$#F)F2#\"#>F'#\"#?F'#F.F+#F0F +#\"#DF'#\"#EF'F&" }}}{PARA 0 "" 0 "" {TEXT -1 175 "This all checks ou t. We see that on the second step of the construction, for example, af ter having removed the segment (1/3,2/3), we then remove (1/9,2/9) and (7/9,8/9). Thus " }{TEXT 285 14 "cantorsequence" }{TEXT -1 144 "(2) g ives all of the important points of our construction, including 0 and \+ 1. So, now let us make a procedure that gives us what is left at the \+ " }{TEXT 286 1 "k" }{TEXT -1 60 "th step of the Cantor Set constructio n. Specifically, given " }{TEXT 287 1 "k" }{TEXT -1 295 "=1 this proce dure will tell us that we have removed the segment (1/3,2/3) from [0,1 ], leaving us with the union of the closed segments [0,1/3] and [2/3,1 ]. We will do so by making a procedure that spits out a list of lists \+ which is to represent a union of closed intervals. Thus in the case fo r " }{TEXT 288 1 "k" }{TEXT -1 67 "=1, this procedure will give [[0,1/ 3], [2/3,1]] as the answer. So: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "cantorset:=(k)->[seq([cantorpoint(2*i-1,k), cantorpoint(2*i,k)],i=1..(2^k))];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %*cantorsetGf*6#%\"kG6\"6$%)operatorG%&arrowGF(7#-%$seqG6$7$-%,cantorp ointG6$,&%\"iG\"\"#\"\"\"!\"\"9$-F26$,$F5F6F9/F5;F7)F6F9F(F(F(" }}} {PARA 0 "" 0 "" {TEXT -1 57 "and now let us see what is left after ste ps 1,2,3,and 4: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "cantorset(1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$ \"\"!#\"\"\"\"\"$7$#\"\"#F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "cantorset(2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&7$\"\"!#\"\" \"\"\"*7$#\"\"#F(#F'\"\"$7$#F+F-#\"\"(F(7$#\"\")F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "cantorset(3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*7$\"\"!#\"\"\"\"#F7$#\"\"#F(#F'\"\"*7$#F+F-#\"\"(F(7$ #\"\")F(#F'\"\"$7$#F+F6#\"#>F(7$#\"#?F(#F1F-7$#F4F-#\"#DF(7$#\"#EF(F' " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "cantorset(4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#727$\"\"!#\"\"\"\"#\")7$#\"\"#F(#F'\"#F7$#F+ F-#\"\"(F(7$#\"\")F(#F'\"\"*7$#F+F6#\"#>F(7$#\"#?F(#F1F-7$#F4F-#\"#DF( 7$#\"#EF(#F'\"\"$7$#F+FG#\"#bF(7$#\"#cF(#F:F-7$#F=F-#\"#hF(7$#\"#iF(#F 1F67$#F4F6#\"#tF(7$#\"#uF(#FBF-7$#FEF-#\"#zF(7$#\"#!)F(F'" }}}{PARA 0 "" 0 "" {TEXT -1 465 "Now we may illustrate another remarkable fact ab out the Cantor Set, which is that it has Lebesgue measure zero. We can see this most readibly by investigating the measure of the complement of the Cantor Set on [0,1]. In the first step of the construction, we remove the open interval (1/3,2/3), which has measure 1/3. At the sec ond step, we remove two disjoint intervals of length 1/9, each of whic h is disjoint from the first interval removed. We proceed, at each " } {TEXT 289 1 "k" }{TEXT -1 20 "th step removing 2^(" }{TEXT 302 1 "k" } {TEXT -1 95 "-1) disjoint intervals that are also disjoint from anythi ng else removed, that are of length 3^" }{TEXT 290 1 "k" }{TEXT -1 274 ". Since all of the segments are mutually disjoint, we may get the measure of the complement of the Cantor Set by adding all of the meas ures of each of the removed segments, which in light of my previous st atements, can be evaluated as the infinite sum starting from 1 of 2^( " }{TEXT 303 1 "k" }{TEXT -1 6 "-1)/3^" }{TEXT 304 1 "k" }{TEXT -1 119 ". This is a geometric series which evaluates to 1, the measure of [0,1], so that the Cantor Set must have zero measure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 231 "Let us test this th rough data. First, we write a program which given a disjoint union of \+ sets represented by a list of two element lists, will give the measur e of that set by merely subtracting endpoints and totalling the result s:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "mea sure:= proc(x) \nL:=0 ; \nfor i from 1 to nops(x) do L:=L+(x)[i][2]-(x )[i][1] od ;\nL ; \nend;" }}{PARA 7 "" 1 "" {TEXT -1 65 "Warning, `L` \+ is implicitly declared local to procedure `measure`\n" }}{PARA 7 "" 1 "" {TEXT -1 65 "Warning, `i` is implicitly declared local to procedure `measure`\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(measureGf*6#%\"xG6$ %\"LG%\"iG6\"F+C%>8$\"\"!?(8%\"\"\"F2-%%nopsG6#9$%%trueG>F.,(F.F2&&F66 #F16#\"\"#F2&F;6#F2!\"\"F.F+F+F+" }}}{PARA 0 "" 0 "" {TEXT -1 132 "The n by calculating some examples, we can see the behavior of the measure of each step of the construction, which should go to zero:" } {MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "measure( cantorset(1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"#\"\"$" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "measure(cantorset(2));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"%\"\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "measure(cantorset(3));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\")\"#F" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(measure(cantorset(4)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$ \"+U'3`(>!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(measu re(cantorset(8)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+JU%=!R!#6" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "evalf(measure(cantorset(15 )));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+h#eOG#!#7" }}}{PARA 0 "" 0 "" {TEXT -1 141 "Now we would like a way to visualize the steps take n in constructing the Cantor Set. We would like to construct a functio n such that given a " }{TEXT 305 1 "k" }{TEXT -1 55 ", this function i s 1 on the segments removed up to the " }{TEXT 306 1 "k" }{TEXT -1 75 "th step of the construction of the Cantor Set, and is zero everywhere else." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 146 "Conceptually this is no problem at all, given that we have alread y devised a program which gives the cantor points. The idea is that at each step " }{TEXT 307 1 "k" }{TEXT -1 31 ", what is removed lies bet ween " }{TEXT 308 16 "cantorpoint(i,k)" }{TEXT -1 5 " and " }{TEXT 309 18 "cantorpoint(i+1,k)" }{TEXT -1 10 " for each " }{TEXT 311 1 "i " }{TEXT -1 18 " between 1 and 2^(" }{TEXT 310 1 "k" }{TEXT -1 85 "+1) -1 that is even. Thus, ideally we should define our function, which we shall call " }{TEXT 312 23 "ComplementFunction(x,k)" }{TEXT -1 4 " as \n" }}{PARA 0 "" 0 "" {TEXT 313 23 "ComplementFunction(x,k)" }{TEXT -1 31 "= 1 if for some even " }{TEXT 314 1 "i" }{TEXT -1 18 " between 1 and 2^(" }{TEXT 315 1 "k" }{TEXT -1 8 "+1)-1 , " }{TEXT 316 17 "cantorpoint(i,k) " }{TEXT -1 6 "< x < " }{TEXT 317 18 "cantorp oint(i+1,k)" }}{PARA 0 "" 0 "" {TEXT -1 73 " \+ 0 everywhere else." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "Now is where we run into some difficulties. Basically, we are attempting to define a sequence \+ of functions that are defined piecewise. However, the number of pieces varies with " }{TEXT 318 1 "k" }{TEXT -1 223 ". Even though maple doe s provide one with immediate tools in order to define piecewise functi ons, it does not allow for the pieces to vary. In other words, we can \+ only define functions where the pieces are set and definite." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 216 "This nothwiths tanding, there are two methods we can use to get around this obstructi on. One is involving sums, which is what we will use now, and the othe r involves do-loops. What we can do is define a function of x," } {TEXT 319 1 "i" }{TEXT -1 6 ", and " }{TEXT 320 1 "k" }{TEXT -1 15 " w hich is 1 if " }{TEXT 321 12 "cantorpoint(" }{TEXT -1 1 "2" }{TEXT 323 5 "*i,k)" }{TEXT -1 7 " < x < " }{TEXT 322 12 "cantorpoint(" } {TEXT -1 1 "2" }{TEXT 325 7 "*i+1,k)" }{TEXT -1 26 ". In essence, for \+ a fixed " }{TEXT 324 1 "k" }{TEXT -1 148 ", this function draws each o f the pieces where our ComplementFunction will be 1, individually. Eve rywhere else this complementfunction will be zero." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 112 "complementfunction:= proc(x,i,k) \nif canto rpoint(2*i,k) < x and x < cantorpoint(2*i+1,k)\nthen 1 else 0 \nfi end ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%3complementfunctionGf*6%%\"xG% \"iG%\"kG6\"F*F*@%32-%,cantorpointG6$,$9%\"\"#9&9$2F5-F/6$,&F2F3\"\"\" F:F4F:\"\"!F*F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 21 "So, for example, fo r " }{TEXT 326 1 "k" }{TEXT -1 79 "=1 we only removed one segment, and so there will be only one pice to graph at " }{TEXT 327 1 "i" }{TEXT -1 3 "=1:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "plot('complemen tfunction(x,1,1)',x=0..1,discont = true );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6&7S7$$\"\"!F)F(7$$\"+O U(\\E(!#7F(7$$\"+Z*='e8!#6F(7$$\"+7d]p?F1F(7$$\"+'p/^y#F1F(7$$\"+jDI( \\$F1F(7$$\"+`6gdTF1F(7$$\"+(4-8%[F1F(7$$\"+#f'Q[bF1F(7$$\"+eM?`iF1F(7 $$\"+PU>ypF1F(7$$\"+5lw;wF1F(7$$\"+WwkN$)F1F(7$$\"+t0[d!*F1F(7$$\"+_.5 `(*F1F(7$$\"+v*y%Q5!#5F(7$$\"+bHf86FYF(7$$\"+[VAx6FYF(7$$\"+'yU7D\"FYF (7$$\"+X;w;8FYF(7$$\"+=ik)Q\"FYF(7$$\"+Jt4d9FYF(7$$\"+'))=&G:FYF(7$$\" +oh5%f\"FYF(7$$\"+84&[m\"FYF(7$$\"+@[LQFYF(7$$\"+I!*e7?FYF(7$$\"+>'[,3#FYF(7$$\"+C?;b@FYF(7$$ \"+0\\cAAFYF(7$$\"+o=`%H#FYF(7$$\"+KYufBFYF(7$$\"+3%R5V#FYF(7$$\"+OA7) \\#FYF(7$$\"+@JCoDFYF(7$$\"+.\"*zOEFYF(7$$\"+i3d3FFYF(7$$\"+'f&pxFFYF( 7$$\"+@cQ[GFYF(7$$\"+(H!\\=HFYF(7$$\"+9#4H)HFYF(7$$\"+h0ucIFYF(7$$\"+* >wF7$FYF(7$$\"+iR=$>$FYF(7$$\"+vJdgKFYF(7$$\"+(3#*HL$FYF(7S7$$\"+S8cLL FY$\"\"\"F)7$$\"+p&3iS$FYF[u7$$\"+N&=%pMFYF[u7$$\"+_Z]SNFYF[u7$$\"+v@1 7OFYF[u7$$\"+)\\zKo$FYF[u7$$\"+sqI\\PFYF[u7$$\"+-[n&f@s%FYF[u7$$\" +j#31z%FYF[u7$$\"+Zt-i[FYF[u7$$\"+hBhF\\FYF[u7$$\"+fYN)*\\FYF[u7$$\"+C g$=2&FYF[u7$$\"+pD'FYF[u7$$\"+htO;jFYF[u7$$\"+bh>!R'FYF[u7$$\"+3&HiX'FYF[u7$$\"+ N[jElFYF[u7$$\"+<<-%f'FYF[u7$$\"+C\"Qkm'FYF[u7S7$$\"+0u+nmFYF(7$$\"+er lRnFYF(7$$\"+?$pG!oFYF(7$$\"+3!eR(oFYF(7$$\"+BWyFYF(7$$\"+#Q]#=zFYF(7$$\"+ ^#pP)zFYF(7$$\"+NQlb!)FYF(7$$\"+e\\5C\")FYF(7$$\"+Bl_&>)FYF(7$$\"+;Q6h #)FYF(7$$\"+s&e=L)FYF(7$$\"+\"\\U`S)FYF(7$$\"+e/Jp%)FYF(7$$\"+;wRQ&)FY F(7$$\"+49x4')FYF(7$$\"+Unfz')FYF(7$$\"+Tj:Z()FYF(7$$\"+e(p@#))FYF(7$$ \"+\\Ed*)))FYF(7$$\"+A'R:'*)FYF(7$$\"+'R_n-*FYF(7$$\"+$=Z!)4*FYF(7$$\" +A+8l\"*FYF(7$$\"+<4DN#*FYF(7$$\"+4p!QI*FYF(7$$\"+!oybP*FYF(7$$\"+DMqW %*FYF(7$$\"+gMR:&*FYF(7$$\"+Z\")\\&e*FYF(7$$\"+tq\"*\\'*FYF(7$$\"+K%[P s*FYF(7$$\"+!3%y*y*FYF(7$$\"+`=>g)*FYF(7$$\"+w5eF**FYF(7$F[uF(-%'COLOU RG6&%$RGBG$\"*++++\"!\")F(F(-%+AXESLABELSG6$Q\"x6\"Q!Fjgl-%%VIEWG6$;F( F[u%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 46.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 11 "Up to step " }{TEXT 328 1 "k" }{TEXT -1 82 "=2, we removed three segments. 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$\"+Y-nm**F`dnF(7$$\"+]\"\\\"p**F`dnF(7$$\"+>.ur**F`dnF(7$$\"+gOFu**F` dnF(7$$\"+Me#p(**F`dnF(7$$\"+#>![z**F`dnF(7$$\"+$R#4#)**F`dnF(7$$\"+jH o%)**F`dnF(7$$\"+GM1()**F`dnF(7$$\"+4B#***F`dnF(7$ $\"+'oL[***F`dnF(7$$\"+7RK(***F`dnF(7$F[uF(-%'COLOURG6&%$RGBG$\"*++++ \"!\")F(F(-%+AXESLABELSG6$Q\"x6\"Q!F__y-%%VIEWG6$;F(F[u%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 46.000000 45.000000 0 0 "Curve 1" }}}} {PARA 0 "" 0 "" {TEXT -1 96 "As we can see, the cantor set occupies le ss and less space on the interval [0,1], as advertised." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 447 "Our next topic of i nterest is the Cantor Function. This function is obtained from a limit of a sequence of functions directly obtained from the construction of the Cantor Set. The construction is as follows, we start with the fir st step of the construction of the Cantor Set where we have removed th e middle third of the interval [0,1]. We define our first function to \+ be 1/2 on that segment removed, (1/3,2/3). We define this function to \+ be 0 at " }{TEXT 340 1 "x" }{TEXT -1 13 "=0, and 1 at " }{TEXT 341 1 " x" }{TEXT -1 110 "=1. For the rest of the values, we simple make the s traight line connections, so that our function looks like:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7W7$$\"\"!F)F)7$$\"+;arz@!#6$\"+CJdpKF-7$$ \"+XTFwSF-$\"+=7T9hF-7$$\"+\"z_\"4iF-$\"+'=HPJ*F-7$$\"+S&phN)F-$\"+JaU `7!#57$$\"+*=)H\\5F?$\"+%GZRd\"F?7$$\"+[!3uC\"F?$\"+s?6r=F?7$$\"+J$RDX \"F?$\"+'**3)y@F?7$$\"+)R'ok;F?$\"+(fHq\\#F?7$$\"+1J:w=F?$\"+f'HU\"GF? 7$$\"+3En$4#F?$\"+7*309$F?7$$\"+/RE&G#F?$\"+ce*yU$F?7$$\"+D.&4]#F?$\"+ )[D9v$F?7$$\"+vB_Y2aF?Fip7$$\"+yXu9cF?Fip7$$\"+\\y))GeF ?Fip7$$\"+i_QQgF?Fip7$$\"+!y%3TiF?Fip7$$\"+O![hY'F?Fip7$$\"+4FEnlF?Fip 7$$\"+#Qx$omF?$\"+tgc-]F?7$$\"+y)Qjx'F?$\"+<$3X;&F?7$$\"+u.I%)oF?$\"+h 0XE`F?7$$\"+(pe*zqF?$\"+Y!Q*>cF?7$$\"+C\\'QH(F?$\"+'Q(zSfF?7$$\"+8S8& \\(F?$\"+?5qUiF?7$$\"+0#=bq(F?$\"+3tFelF?7$$\"+2s?6zF?$\"+53\"o'oF?7$$ \"+IXaE\")F?$\"+&z;)*=(F?7$$\"+l*RRL)F?$\"+[*44](F?7$$\"+`<.Y&)F?$\"+I w/>yF?7$$\"+8tOc()F?$\"+q4bM\")F?7$$\"+\\Qk\\*)F?$\"+udYC%)F?7$$\"+p0; r\"*F?$\"+a3uc()F?7$$\"+lxGp$*F?$\"+[;$R0*F?7$$\"+!oK0e*F?$\"+?!*zq$*F ?7$$\"+<5s#y*F?$\"+E:3u'*F?7$$\"\"\"F)Fdy-%'COLOURG6&%$RGBG$\"#5!\"\"F (F(-%+AXESLABELSG6$Q\"x6\"Q!Faz-%%VIEWG6$;F(Fdy%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}{TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 423 "This is the first function in our se quence. For the next function, we consider all of the segments removed up to the second step of the Cantor Set construction, which are (1/9, 2/9),(1/3,2/3), and (7/9,8/9). We define our function to be 1/4 on the first segment, 1/2 on the second, and 3/4 on the third. We let it be \+ 0 at 0 and 1 at 1, and make the straight line connections everywhere e lse, so that our function looks like:" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7 _p7$$\"\"!F)F)7$$\"+;arz@!#6$\"+'ofV!\\F-7$$\"+XTFwSF-$\"+Eohr\"*F-7$$ \"+\"z_\"4iF-$\"+y$fqR\"!#57$$\"+S&phN)F-$\"+Y\"Q,)=F:7$$\"+:ddC%*F-$ \"+O&H07#F:7$$\"+*=)H\\5F:$\"+D4#4O#F:7$$\"+A>1u5F:$\"+C$RmT#F:7$$\"+a c#))4\"F:$\"+AxNsCF:7$$\"+'Q*eB6F:$\"+++++DF:7$$\"+=JN[6F:FW7$$\"+$e!) y>\"F:FW7$$\"+[!3uC\"F:FW7$$\"+!pt*\\8F:FW7$$\"+J$RDX\"F:FW7$$\"+)R'ok ;F:FW7$$\"+1J:w=F:FW7$$\"+dG\"\\)>F:FW7$$\"+3En$4#F:FW7$$\"+K/dT@F:FW7 $$\"+c#o%*=#F:FW7$$\"+orT8AF:FW7$$\"+!3mtB#F:$\"+!otS`#F:7$$\"+#*\\JhA F:$\"+K(eze#F:7$$\"+/RE&G#F:$\"+%yV=k#F:7$$\"+9r5$R#F:$\"+15\\%)GF:7$$ \"+D.&4]#F:$\"+J#Qr7$F:7$$\"+vB_Y2aF:Fgs7$$\"+yXu9cF:F gs7$$\"+\\y))GeF:Fgs7$$\"+i_QQgF:Fgs7$$\"+!y%3TiF:Fgs7$$\"+O![hY'F:Fgs 7$$\"+4FEnlF:Fgs7$$\"+#Qx$omF:$\"+5\"\\Q+&F:7$$\"+y)Qjx'F:$\"+wCwY_F:7 $$\"+u.I%)oF:$\"+Uen*[&F:7$$\"+(pe*zqF:$\"+oq!*HfF:7$$\"+C\\'QH(F:$\"+ zg>6kF:7$$\"+8S8&\\(F:$\"+H:0koF:7$$\"+3hK+wF:$\"+VPt+rF:7$$\"+0#=bq(F :$\"+hfTPtF:7$$\"+I$H7t(F:$\"+#*fE&R(F:7$$\"+b/%pv(F:$\"+Cg6`uF:7$$\"+ =gzpxF:$\"+S5/#[(F:7$$\"+\"e^Ey(F:$\"+++++vF:7$$\"+Vr]&z(F:Ffz7$$\"+1F O3yF:Ffz7$$\"+d\\yfyF:Ffz7$$\"+2s?6zF:Ffz7$$\"+pe()=!)F:Ffz7$$\"+IXaE \")F:Ffz7$$\"+l*RRL)F:Ffz7$$\"+`<.Y&)F:Ffz7$$\"+K&*>^')F:Ffz7$$\"+8tOc ()F:Ffz7$$\"+Yko/))F:Ffz7$$\"+!e0I&))F:Ffz7$$\"+Z^;x))F:Ffz7$$\"+9ZK,* )F:$\"+11)z_(F:7$$\"+\"G%[D*)F:$\"+K'RBe(F:7$$\"+\\Qk\\*)F:$\"+g')pOwF :7$$\"+3ASg!*F:$\"+o\\!f)yF:7$$\"+p0;r\"*F:$\"+!G6^8)F:7$$\"+lxGp$*F:$ \"+ru*3e)F:7$$\"+!oK0e*F:$\"+I&)>c!*F:7$$\"+<5s#y*F:$\"+)GA6^*F:7$$\" \"\"F)Fh_l-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"x6\"Q!Fe` l-%%VIEWG6$;F(Fh_l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 44.000000 0 0 "Curve 1" }}{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 298 "We then proceed as such. What we will get is a sequence \+ of uniformly convergent continuous functions that must converge to a c ontinuous function. We call this limiting function the Cantor Function , which has very interesting properties. We have constructed our seque nce of functions so that at the " }{TEXT 342 1 "k" }{TEXT -1 142 "th s tep of the construction of the Cantor Set, all of our functions after \+ this step are all fixed constants on each segment removed up to the " }{TEXT 343 1 "k" }{TEXT -1 874 "th step of the construction. Thus, our limiting function will be constant on all of the segments removed in \+ the construction of the Cantor Set, which has measure 1. On each of th ese segments, the Cantor Function is differentiable and has derivative zero. Thus, the Cantor Function is differentiable almost everywhere o n [0,1] with derivative zero. However, the Cantor Function is a limit \+ of increasing functions, and so must be non-decreasing. In fact, each \+ function in our sequence leading up to the Cantor Function has value 0 at 0 and 1 at 1, thus the value of the Cantor Function is 0 at 0 and \+ 1 at 1. Thus, we have managed to produce a function that is differenti able almost everywhere on [0,1] with derivative 0, and yet it manages \+ to climb up from 0 all the way up to one! As Jason Lee remarked, if yo u blink on a set of measure zero, the Cantor Function is up there." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 170 "Let us n ow write a program which gives us our sequence of functions which lead up to the Cantor Function. The main obstacle is the same as the one i n the program for the " }{TEXT 344 18 "ComplementFunction" }{TEXT -1 204 ", which is that our sequence of functions are defined piecewise b ut how many pieces and where they start and end changes at each step. \+ However, we use the same technique of using sums here as was used in \+ " }{TEXT 345 19 "ComplementFunction." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "Thus, firstly we make a function which accounts for the places where our sequence of functions are constant: " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "cfsteps:=proc(x,i,k)\ni f cantorpoint(2*i,k) < x and x <= cantorpoint(2*i+1,k)\nthen i/(2^k) e lse 0 \nfi end ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(cfstepsGf*6%%\" xG%\"iG%\"kG6\"F*F*@%32-%,cantorpointG6$,$9%\"\"#9&9$1F5-F/6$,&F2F3\" \"\"F:F4*&F2F:)F3F4!\"\"\"\"!F*F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 9 "fo r each " }{TEXT 346 1 "k" }{TEXT -1 2 ", " }{TEXT 347 7 "cfsteps" } {TEXT -1 36 " draws each individual piece of our " }{TEXT 348 1 "k" } {TEXT -1 146 "th function where it is constant. To get a function that is constant on all of the appropriate segements, we simply do as befo re and take a sum: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "cfallsteps:= (x,k) -> sum('cfsteps(x,i,k)', 'i'=1..(( 2^k)-1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+cfallstepsGf*6$%\"xG% \"kG6\"6$%)operatorG%&arrowGF)-%$sumG6$.-%(cfstepsG6%9$%\"iG9%/.F5;\" \"\",&)\"\"#F6F:F:!\"\"F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 17 "Now, fo r a fixed " }{TEXT 349 1 "k" }{TEXT -1 231 ", we must make all of the \+ straight line connections. Again, we make each piece seperately. Since each piece is a line, we can readibly come up with a formula based on the cantor points. First we make some preliminary definitions: " } {MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "cflinepa rt1:= (x,i,k) -> ((3^k)/((2^k)(cantorpoint(2*i,k)-cantorpoint(2*i-1,k) )))*x\n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,cflinepart1Gf*6%%\"xG% \"iG%\"kG6\"6$%)operatorG%&arrowGF**()\"\"$9&\"\"\"-)\"\"#F16#,&-%,can torpointG6$,$9%F5F1F2-F96$,&F " 0 "" {MPLTEXT 1 0 130 "cflinepart2:= (i,k) ->\n(3^k)*(((i -1)*cantorpoint(2*i,k)-i*cantorpoint(2*i-1,k))/((2^k)(cantorpoint(2*i, k)-cantorpoint(2*i-1,k))));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,cfli nepart2Gf*6$%\"iG%\"kG6\"6$%)operatorG%&arrowGF)*()\"\"$9%\"\"\",&*&,& 9$F1F1!\"\"F1-%,cantorpointG6$,$F5\"\"#F0F1F1*&F5F1-F86$,&F5F;F1F6F0F1 F6F1-)F;F06#,&F7F1F=F6F6F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 29 " and n ow we define each line:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 135 "cfline:=proc(x,i,k) \nif cantorpoint(2*i-1,k) < x \+ and x <= cantorpoint(2*i,k)\nthen cflinepart1(x,i,k)+cflinepart2(i,k) \+ else 0 \nfi end ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'cflineGf*6%% \"xG%\"iG%\"kG6\"F*F*@%32-%,cantorpointG6$,&9%\"\"#\"\"\"!\"\"9&9$1F7- F/6$,$F2F3F6,&-%,cflinepart1G6%F7F2F6F4-%,cflinepart2G6$F2F6F4\"\"!F*F *F*" }}}{PARA 0 "" 0 "" {TEXT -1 51 "We are now ready to define our se quence. We define " }{TEXT 350 19 "CantorFunction(x,k)" }{TEXT -1 74 " to be our sequence of functions, and we use the same trick of using s ums:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "C antorFunction:=(x,k)->sum('cfline(x,i,k)','i'=1..(2^(k)))+'cfallsteps( x,k)' ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/CantorFunctionGf*6$%\"xG %\"kG6\"6$%)operatorG%&arrowGF),&-%$sumG6$.-%'cflineG6%9$%\"iG9%/.F6; \"\"\")\"\"#F7F;.-%+cfallstepsG6$F5F7F;F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 29 "Let us plot this for several " }{TEXT 351 1 "k" }{TEXT -1 2 ": " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot('CantorFunction(x,1)',x=0..1);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7W7$$\"\"!F)F)7$$\"+;arz@!#6$\" +CJdpKF-7$$\"+XTFwSF-$\"+=7T9hF-7$$\"+\"z_\"4iF-$\"+'=HPJ*F-7$$\"+S&ph N)F-$\"+JaU`7!#57$$\"+*=)H\\5F?$\"+%GZRd\"F?7$$\"+[!3uC\"F?$\"+s?6r=F? 7$$\"+J$RDX\"F?$\"+'**3)y@F?7$$\"+)R'ok;F?$\"+(fHq\\#F?7$$\"+1J:w=F?$ 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Graphically we can also be convin ced that these curves are converging uniformly. Indeed, each successiv e curve will be constant where the previous one was, so that the maxim um difference between one function in our sequence and another after i t must occur somewhere in between the places where the first curve is \+ constant. As such, then it is only a matter of getting our straight li ne connections to be sufficiently close to each other, which pictorial ly we can see is not such a far-fetched thing to do." }}}{PARA 0 "" 0 "" {TEXT -1 665 " One can now contemplate what would happen if inst ead of removing the middle thirds in the construction of the Cantor Se t, we remove some other length strictly between 0 and 1. If we do this , we get what is called the General Cantor Set. These sets have intere sing properties as well. They are all uncountable, perfect, compact, a nd are totally disconnected. However, it is possible to construct a Ge neral Cantor Set that has positive measure if we allow how much we rem ove at each step change. Moreover, given any number strictly between 0 and 1, we can produce a general cantor set that has that value as its measure. Let us investigate the General Cantor Set." }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 18 "General Cantor Set" }}{PARA 0 "" 0 "" {TEXT -1 142 "Our task will be to produce an algorithm which gives us the point s in the construction of the General Cantor Set. More specifically, gi ven an " }{TEXT 352 3 "i,k" }{TEXT -1 7 " and a " }{TEXT 256 1 "b" } {TEXT -1 34 ", this algorithm will give us the " }{TEXT 353 1 "i" } {TEXT -1 21 "th point used in the " }{TEXT 354 1 "k" }{TEXT -1 94 "th \+ step of the construction of the cantor set according to the rule that \+ we remove the middle " }{TEXT 257 1 "b" }{TEXT -1 41 " of each segment . For now, we regard the " }{TEXT 258 1 "b" }{TEXT -1 29 " as fixed at every step, and " }{TEXT 259 1 "b" }{TEXT -1 36 " is always strictly \+ between 0 and 1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 185 "This algorithm will be a doubly recursive one. We will f irst manually produce the four points that are used in the first step \+ of the general construction, which will directly depend on " }{TEXT 260 1 "b" }{TEXT -1 148 ". Then we will build on these points to get a ll the other points for every step beyond the first step. As such, the first set of points is given by:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "firstgencantset:= proc(b) [0,(1/2)*(1-b),(1/2)*(1+b), 1] end;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%0firstgencantsetGf*6#%\"b G6\"F(F(7&\"\"!,&9$#!\"\"\"\"##\"\"\"F/F1,&F0F1*&F0F1F,F1F1F1F(F(F(" } }}{PARA 0 "" 0 "" {TEXT -1 41 "one can check that the segment ((1/2)*( 1-" }{TEXT 264 1 "b" }{TEXT -1 11 "),(1/2)*(1+" }{TEXT 265 1 "b" } {TEXT -1 14 ")) has length " }{TEXT 263 1 "b" }{TEXT -1 144 ". Now we \+ define a preliminary function, which we also used in the original Cant or Set, but that we describe explicitly here for simplicity. " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "factorsoftwo:= (i)-> ifactor s(i)[2][1][2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-factorsoftwoGf*6# %\"iG6\"6$%)operatorG%&arrowGF(&&&-%)ifactorsG6#9$6#\"\"#6#\"\"\"F3F(F (F(" }}}{PARA 0 "" 0 "" {TEXT -1 26 "Now we define our points: " } {MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 465 "gencant point:=proc(i,k,b) \nif k=0 then 1 ; else \nif k=1 then (firstgencants et(b))[i] ; else\nif i=1 then 0 else\nif i mod 2 = 0 then\nif factorso ftwo(i)=1 then gencantpoint(i-1,k,b)+(((1/2)*(1-b))^k) ;\nelse gencant point(i*(2^(1-factorsoftwo(i))),k+1-factorsoftwo(i),b) ; fi ;\nelse if factorsoftwo(i-1)=1 then gencantpoint(i-1,k,b)+b*(((1/2)*(1-b))^(k-1) ) ;\nelse gencantpoint((i-1)*(2^(1-factorsoftwo(i-1)))+1,k+1-factorsof two(i-1),b) ;\nfi ; fi ; fi ; fi ; fi ; end ;\n\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%-gencantpointGf*6%%\"iG%\"kG%\"bG6\"F*F*@%/9%\"\"!\" \"\"@%/F-F/&-%0firstgencantsetG6#9&6#9$@%/F8F/F.@%/-%$modG6$F8\"\"#F.@ %/-%-factorsoftwoGF7F/,&-F$6%,&F8F/F/!\"\"F-F6F/),&#F/F@F/*&#F/F@F/F6F /FIF-F/-F$6%*&F8F/)F@,&F/F/FCFIF/,(F-F/F/F/FCFIF6@%/-FD6#FHF/,&FFF/*&F 6F/)FK,&F-F/F/FIF/F/-F$6%,&*&FHF/)F@,&F/F/FWFIF/F/F/F/,(F-F/F/F/FWFIF6 F*F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 85 "This is a formidable algorithm , one that uses a heavy recursion. In order to get the " }{TEXT 355 1 "i" }{TEXT -1 21 "th point used in the " }{TEXT 356 1 "k" }{TEXT -1 68 "th step, this algorithm not only investigates the the nature of th e " }{TEXT 357 1 "i" }{TEXT -1 45 ", but it also looks at the divisibi lity of " }{TEXT 358 1 "i" }{TEXT -1 84 "-1 by 2, and then draws on \+ the points used in the previous step of the construction." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 116 "Now we define as \+ before, a procedure which lists out the sets that are left over at eac h step in our construction: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 92 "gencantset:=proc(k,b) [seq([gencantpoint(2*i-1 ,k,b),gencantpoint(2*i,k,b)],i=1..(2^k))] end;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+gencantsetGf*6$%\"kG%\"bG6\"F)F)7#-%$seqG6$7$-%-genc antpointG6%,&%\"iG\"\"#\"\"\"!\"\"9$9%-F06%,$F3F4F7F8/F3;F5)F4F7F)F)F) " }}}{PARA 0 "" 0 "" {TEXT -1 71 "Now we can check that algorithm is c orrect, by putting in the value of " }{TEXT 261 1 "b" }{TEXT -1 62 "=1 /3. This should give us the regular Cantor Set, and we list " }{TEXT 360 11 "gencantset " }{TEXT -1 4 "and " }{TEXT 361 10 "cantorset " } {TEXT -1 22 "for several values of " }{TEXT 359 1 "k" }{TEXT -1 17 " t o verify this: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "gencantset(1,1/3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #7$7$\"\"!#\"\"\"\"\"$7$#\"\"#F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "cantorset(1,1/3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# 7$7$\"\"!#\"\"\"\"\"$7$#\"\"#F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "gencantset(2,1/3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #7&7$\"\"!#\"\"\"\"\"*7$#\"\"#F(#F'\"\"$7$#F+F-#\"\"(F(7$#\"\")F(F'" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "cantorset(2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&7$\"\"!#\"\"\"\"\"*7$#\"\"#F(#F'\"\"$7$#F+F-# \"\"(F(7$#\"\")F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "genc antset(3,1/3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*7$\"\"!#\"\"\"\"#F 7$#\"\"#F(#F'\"\"*7$#F+F-#\"\"(F(7$#\"\")F(#F'\"\"$7$#F+F6#\"#>F(7$#\" #?F(#F1F-7$#F4F-#\"#DF(7$#\"#EF(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "cantorset(3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*7$ \"\"!#\"\"\"\"#F7$#\"\"#F(#F'\"\"*7$#F+F-#\"\"(F(7$#\"\")F(#F'\"\"$7$# F+F6#\"#>F(7$#\"#?F(#F1F-7$#F4F-#\"#DF(7$#\"#EF(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "gencantset(4,1/3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#727$\"\"!#\"\"\"\"#\")7$#\"\"#F(#F'\"#F7$#F+F-#\"\"(F(7 $#\"\")F(#F'\"\"*7$#F+F6#\"#>F(7$#\"#?F(#F1F-7$#F4F-#\"#DF(7$#\"#EF(#F '\"\"$7$#F+FG#\"#bF(7$#\"#cF(#F:F-7$#F=F-#\"#hF(7$#\"#iF(#F1F67$#F4F6# \"#tF(7$#\"#uF(#FBF-7$#FEF-#\"#zF(7$#\"#!)F(F'" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 13 "cantorset(4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#727$\"\"!#\"\"\"\"#\")7$#\"\"#F(#F'\"#F7$#F+F-#\"\"(F(7$#\"\")F(#F' \"\"*7$#F+F6#\"#>F(7$#\"#?F(#F1F-7$#F4F-#\"#DF(7$#\"#EF(#F'\"\"$7$#F+F G#\"#bF(7$#\"#cF(#F:F-7$#F=F-#\"#hF(7$#\"#iF(#F1F67$#F4F6#\"#tF(7$#\"# uF(#FBF-7$#FEF-#\"#zF(7$#\"#!)F(F'" }}}{PARA 0 "" 0 "" {TEXT -1 59 "It seems to check out. Now out of curiosity, let us put in " }{TEXT 262 1 "b" }{TEXT -1 32 "=1/2 and list some of the steps:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "gencantset(1,1/2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#7$7$\"\"!#\"\"\"\"\"%7$#\"\"$F(F'" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "gencantset(2,1/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7&7$\"\"!#\"\"\"\"#;7$#\"\"$F(#F'\"\"%7$#F+F -#\"#8F(7$#\"#:F(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "genc antset(3,1/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*7$\"\"!#\"\"\"\"#k 7$#\"\"$F(#F'\"#;7$#F+F-#\"#8F(7$#\"#:F(#F'\"\"%7$#F+F6#\"#\\F(7$#\"#^ F(#F1F-7$#F4F-#\"#hF(7$#\"#jF(F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "gencantset(4,1/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #727$\"\"!#\"\"\"\"$c#7$#\"\"$F(#F'\"#k7$#F+F-#\"#8F(7$#\"#:F(#F'\"#;7 $#F+F6#\"#\\F(7$#\"#^F(#F1F-7$#F4F-#\"#hF(7$#\"#jF(#F'\"\"%7$#F+FG#\"$ $>F(7$#\"$&>F(#F:F-7$#F=F-#\"$0#F(7$#\"$2#F(#F1F67$#F4F6#\"$T#F(7$#\"$ V#F(#FBF-7$#FEF-#\"$`#F(7$#\"$b#F(F'" }}}{PARA 0 "" 0 "" {TEXT -1 91 " Now, let us measure take the measure the intervals left in each step o f this construction: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "measure(gencantset(1,1/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "measure(gencantset(2,1/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\" \"\"\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "measure(gencan tset(3,1/2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"\")" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "measure(gencantset(4,1/2)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"\"\"\"#;" }}}{PARA 0 "" 0 "" {TEXT -1 116 "We see that the measure of this cantor set will be zero. Indeed, any general cantor set will have measure zero when " }{TEXT 266 1 "b" }{TEXT -1 173 " is fixed. This is again more apparent when w e investigate the measure of the compleement of such a general cantor \+ set on [0,1]. The length of the first interval removed is " }{TEXT 267 1 "b" }{TEXT -1 52 ". We are left with two intervals of length (1/ 2)*(1-" }{TEXT 268 1 "b" }{TEXT -1 62 "). At the second step, we remov e two intervals each of length " }{TEXT 270 2 "b*" }{TEXT -1 9 "(1/2)* (1-" }{TEXT 269 1 "b" }{TEXT -1 83 "). All of the segments removed are disjoint. As such, at the kth step, we remove 2^" }{TEXT 362 1 "k" } {TEXT -1 20 " segments of length " }{TEXT 271 1 "b" }{TEXT -1 11 "*((1 /2)*(1-" }{TEXT 272 1 "b" }{TEXT -1 3 "))^" }{TEXT 363 1 "k" }{TEXT -1 216 ". All of these segements will be disjoint, so to get the measu re of the complement of this Cantor Set, we sum the lengths of each of the pieces. Including the first step, we get the sum from 0 to infini ty of (2^k)*" }{TEXT 273 1 "b" }{TEXT -1 11 "*((1/2)*(1-" }{TEXT 274 1 "b" }{TEXT -1 3 "))^" }{TEXT 364 1 "k" }{TEXT -1 28 ". The summa nd simplifies to " }{TEXT 275 1 "b" }{TEXT -1 4 "*(1-" }{TEXT 276 1 "b " }{TEXT -1 2 ")^" }{TEXT 365 1 "k" }{TEXT -1 58 ". This is again a ge ometric series, which will sum to 1. " }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 124 "At this point one can continue as in the Cantor Set and define a general Devil's Staircase based on the Ge neral Cantor Set. " }}}{PARA 0 "" 0 "" {TEXT -1 448 "Our next example \+ deals with a classic question which comes to the mind of most undergra duate mathematics students. Is every continuous function differentiabl e? The answer is no, since the absolute value function is continuous a t zero but not differentiable there. However, we can ask, is there a f unction which is continuous everywhere but differentiable nowhere? The answer is in the affirmative, and we now explore an example of such a function. " }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 47 "Van der Waerden No where Differentiable Function" }}{PARA 0 "" 0 "" {TEXT -1 276 "Our exa mple of a continuous nowhere differentiable function is due to Bartel \+ Leendert van der Waerden, and it will be obtained from taking the sum \+ of a sequence of functions. This sequence of functions is best describ ed through graphs. The first function is a simple sawtooth:" }}{PARA 0 "" 0 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7U7$$\"\"!F )F(7$$\"3emmm;arz@!#>F+7$$\"3[LL$e9ui2%F-F/7$$\"3nmmm\"z_\"4iF-F27$$\" 3[mmmT&phN)F-F57$$\"3CLLe*=)H\\5!#=F87$$\"3gmm\"z/3uC\"F:F<7$$\"3%)*** \\7LRDX\"F:F?7$$\"3]mm\"zR'ok;F:FB7$$\"3w***\\i5`h(=F:FE7$$\"3WLLL3En$ 4#F:FH7$$\"3qmm;/RE&G#F:FK7$$\"3\")*****\\K]4]#F:FN7$$\"3$******\\PAvr #F:FQ7$$\"3)******\\nHi#HF:FT7$$\"3jmm\"z*ev:JF:FW7$$\"3?LLL347TLF:FZ7 $$\"3,LLLLY.KNF:Fgn7$$\"3w***\\7o7Tv$F:Fjn7$$\"3'GLLLQ*o]RF:F]o7$$\"3A ++D\"=lj;%F:F`o7$$\"31++vV&RY2aF:$\"3]++D\"3QDf%F:7$$\"39mm;zXu9cF:$\"3'QLL3Ub_Q%F:7$$\"3l****** \\y))GeF:$\"3O+++]@6rTF:7$$\"3'*)***\\i_QQgF:$\"3/,+]PZhhRF:7$$\"3@*** \\7y%3TiF:$\"3y++v=_\"*ePF:7$$\"35****\\P![hY'F:$\"3*3++D'>&Q`$F:7$$\" 3kKLL$Qx$omF:$\"3Pnmm;EiJLF:7$$\"3!)*****\\P+V)oF:$\"3?+++D'*p:JF:7$$ \"3?mm\"zpe*zqF:$\"3zLL3-8/?HF:7$$\"3%)*****\\#\\'QH(F:$\"3<+++v]81FF: 7$$\"3GKLe9S8&\\(F:$\"3snmT&)f'[]#F:7$$\"3R***\\i?=bq(F:$\"3g++v$z\"[% H#F:7$$\"3\"HLL$3s?6zF:$\"33nmm\"z#z)3#F:7$$\"3a***\\7`Wl7)F:$\"3Y++vo aXt=F:7$$\"3#pmmm'*RRL)F:$\"33LLLL+1m;F:7$$\"3Qmm;a<.Y&)F:$\"3iLL$eCoR X\"F:7$$\"3=LLe9tOc()F:$\"3$om;aoKOC\"F:7$$\"3u******\\Qk\\*)F:$\"3E++ +]hN]5F:7$$\"3CLL$3dg6<*F:$\"3anmm\"H%R)G)F-7$$\"3ImmmmxGp$*F:$\"30PLL LB72jF-7$$\"3A++D\"oK0e*F:$\"3i(***\\(=tY>%F-7$$\"3A++v=5s#y*F:$\"3#y* **\\7)*ys@F-7$$\"\"\"F)F(-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELS G6$Q\"x6\"Q!Fgx-%%VIEWG6$;F(Fjw%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 58 "the second function is another saw tooth, with fou r teeth:" }}{PARA 0 "" 0 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CU RVESG6$7]q7$$\"\"!F)F(7$$\"+3x&)*3\"!#6$\"+;arz@F-7$F.$\"+K3VfVF-7$$\" +!y%*z7$F-$\"+g&*)fD'F-7$$\"+XTFwSF-$\"+!H[D:)F-7$$\"+oMrU^F-$\"+%pU&G 5!#57$$\"+\"z_\"4iF-$\"+e0$=C\"FB7$$\"+m6m#G(F-$\"+LA`c9FB7$$\"+S&phN) F-$\"+3RBr;FB7$$\"+:ddC%*F-$\"+V^\"\\)=FB7$$\"+*=)H\\5FB$\"+yjf)4#FB7$ $\"+=JN[6FB$\"+Oiq'H#FB7$$\"+[!3uC\"FB$\"+'4;[\\#FB7$$\"+!pt*\\8FB$\"+ ?E0+BFB7$$\"+J$RDX\"FB$\"+Q8#\\4#FB7$$\"+kGhe:FB$\"+sUx#)=FB7$$\"+)R'o k;FB$\"+/siq;FB7$$\"+_(>/x\"FB$\"+'\\g\"f9FB7$$\"+1J:w=FB$\"+)y$pZ7FB7 $$\"+dG\"\\)>FB$\"+'Gu,.\"FB7$$\"+3En$4#FB$\"+SyaE\")F-7$$\"+c#o%*=#FB $\"+!)[j5iF-7$$\"+/RE&G#FB$\"+?>s%H%F-7$$\"+9r5$R#FB$\"+?x&y8#F-7$$\"+ D.&4]#FB$\"+++l+>!#87$$\"+]jB4EFB$\"++qs%=#F-7$$\"+vB_/\"FB7$$\"+(*ev:JFB$\"+%z6:B\"FB7$$\"+-%Q%GKFB$\"+/o(oX\"FB7$$\"+34 7TLFB$\"+;=C#o\"FB7$$\"+qxdOMFB$\"+Sb:t=FB7$$\"+LY.KNFB$\"+m#pS1#FB7$$ \"+cO2VOFB$\"+7t9'G#FB7$$\"+\"o7Tv$FB$\"+QYx\"\\#FB7$$\"+K5S_QFB$\"+Oz >&H#FB7$$\"+$Q*o]RFB$\"+M7i)4#FB7$$\"+#GF&eSFB$\"+Oa%H)=FB7$$\"+\"=lj; %FB$\"+Q'psm\"FB7$$\"+iB0pUFB$\"+w_*=Y\"FB7$$\"+V&RY2aFB$\"+g$Q#\\\")F-7$$\"+[K56bFB$\"+'\\1A- \"FB7$$\"+yXu9cFB$\"+c\"*[H7FB7$$\"+8i\"=s&FB$\"+ECjV9FB7$$\"+\\y))GeF B$\"+)pvxl\"FB7$$\"+bljLfFB$\"+5JFn=FB7$$\"+i_QQgFB$\"+C0xw?FB7$$\"+@] tRhFB$\"+U+ZzAFB7$$\"+!y%3TiFB$\"+g&p@[#FB7$$\"+3kh`jFB$\"+%=nFH#FB7$$ \"+O![hY'FB$\"+GRqn?FB7$$\"+4FEnlFB$\"+#eua'=FB7$$\"+#Qx$omFB$\"+O_Cj; FB7$$\"+y)Qjx'FB$\"+WAKZ9FB7$$\"+u.I%)oFB$\"+_#*RJ7FB7$$\"+N&H@)pFB$\" +I4uN5FB7$$\"+(pe*zqFB$\"+gg#3S)F-7$$\"+5=\"p=(FB$\"++QwhiF-7$$\"+C\\' QH(FB$\"+?:qATF-7$$\"+o%*\\%R(FB$\"+S1,5@F-7$$\"+8S8&\\(FB$\"++S(>t*F \\s7$$\"+3hK+wFB$\"+g@_1?F-7$$\"+0#=bq(FB$\"++TO5TF-7$$\"+1FO3yFB$\"+? TDnhF-7$$\"+2s?6zFB$\"+ST9C#)F-7$$\"+pe()=!)FB$\"+Q^')FB$\"+ k!*R-BFB7$$\"+8tOc()FB$\"+u`E([#FB7$$\"+!e0I&))FB$\"+S)))RH#FB7$$\"+\\ Qk\\*)FB$\"+-Br+@FB7$$\"+3ASg!*FB$\"+%e&>z=FB7$$\"+p0;r\"*FB$\"+i)ywl \"FB7$$\"+mTAq#*FB$\"+o;bf9FB7$$\"+lxGp$*FB$\"+qWUh7FB7$$\"+A-\"\\Z*FB $\"+c&z,0\"FB7$$\"+!oK0e*FB$\"++kM*Q)F-7$$\"+[oi\"o*FB$\"+SIYnjF-7$$\" +<5s#y*FB$\"+g'zbM%F-7$$\"+30O\"*)*FB$\"+S)*ys@F-7$$\"\"\"F)F(-%'COLOU RG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"x6\"Q!F^jl-%%VIEWG6$;F(Fail %(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "C urve 1" }}{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 94 "However, the he ight of each sawtooth is reduced by a half to 1/4. We proceed as such, at each " }{TEXT 367 1 "k" }{TEXT -1 38 "th step getting a function t hat has 2^" }{TEXT 366 1 "k" }{TEXT -1 30 " sawteeth, each of height 1 /2^" }{TEXT 368 1 "k" }{TEXT -1 91 ". The sum of these functions is un iformly convergent by the Weierstrauss M-test, since the " }{TEXT 369 1 "k" }{TEXT -1 30 "th function is bounded by 1/2^" }{TEXT 370 1 "k" } {TEXT -1 261 ". Thus, we call the resulting sum the Van der Waerden Fu nction, and it is continuous since it is the uniformly convergent sum \+ of a sequence of continuous functions. However, it fails to be differe ntiable anywhere on [0,1]. I shall give some reasoning why later." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 "Now let u s write a program that gives us the " }{TEXT 371 1 "n" }{TEXT -1 236 " th partial sum of our sequence of sawtooth functions. First, we must g et this sequence of functions. This task is easy enough given our stra tegies. We first define each of the pieces of each sawtooth, and then \+ sum them all in the end: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "nowherediffeven:= proc(x,i,k)\nif 2*i/(2^(k+1)) <= x and x < (2*i+1)/(2^(k+1)) \nthen 2*x-(2*i/(2^k)) else 0 \nfi end ; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%0nowherediffevenGf*6%%\"xG%\" iG%\"kG6\"F*F*@%31,$*&9%\"\"\")\"\"#,&9&F1F1F1!\"\"F39$2F7*&,&F0F3F1F1 F1F2F6,&F7F3*(F3F1F0F1)F3F5F6F6\"\"!F*F*F*" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 124 "nowherediffodd:= proc(x,i,k)\nif (2*i+1)/(2^(k+1)) <= x and x < (2*i+2)/(2^(k+1)) \nthen -2*x+((2*i+2)/(2^k)) else 0 \nf i end ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/nowherediffoddGf*6%%\"xG %\"iG%\"kG6\"F*F*@%31*&,&9%\"\"#\"\"\"F2F2)F1,&9&F2F2F2!\"\"9$2F7*&,&F 0F1F1F2F2F3F6,&F7!\"#*&F:F2)F1F5F6F2\"\"!F*F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 114 "These describe the left and right pieces of each sawtoot h. 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The main idea \+ is that at each step of the sum, we add more peaks. The function will \+ certainly not be differentiable at the top of each of these peaks. How ever, for any other point, what occurs is that we add so many peaks wh ich occur on smaller and smaller intervals that we can get this point \+ to be between two peaks, so just like the absolute value function at 0 , the difference quotient at this point will have different limits dep ending on whether we take it from the left or the right." }}}{PARA 0 " " 0 "" {TEXT -1 362 "We now change gears and move on to the topic of s pace filling curves. We know the interval of points [0,1] is an uncoun table set. As such, it is impossible to find a function defined on the natural numbers that is onto the unit interval. However, it is a simp le task to come up with a function that maps the unit interval onto th e whole real line. We could take:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 7 "f(x):= " }{XPPEDIT 18 0 "-cot(Pi*x);" "6#, $-%$cotG6#*&%#PiG\"\"\"%\"xGF)!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 17 "with f(0)=f(1)=1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 607 "This shows that the card inality of the unit segment is the same as the real number line. Now t he question is, can we find a map from the unit segment [0,1] that is \+ onto the unit square [0,1]x[0,1]? The answer is in the affirmative, as was shown by Giuseppe Peano and David Hilbert. The remarkable conclus ion is that [0,1] and [0,1]x[0,1] are of the same size. Furthermore, t hrough our discussion of the Cantor Function, it is trivial to see tha t Cantor Set can be mapped onto [0,1]. Thus, our misleadingly thin set the Cantor Set in reality has so many points in it that we can map th e set onto [0,1]x[0,1]!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 356 "Such a curve that maps the unit interval onto the u nit square is called a space-filling curve, and for good reason. Howev er, Eugen Netto proved that any continuous map from [0,1] to [0,1]x[0, 1] (or the unit cube) that is onto must fail to be injective. All of o ur examples will be onto and continuous, and we will readily see how t hey fail to be injective." }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 27 "Pean o's Space Filling Curve" }}{PARA 0 "" 0 "" {TEXT -1 203 "Peano gave a \+ wondeful example of a space-filling curve. Like the Cantor Function, t he Peano Curve can be constructed as the limit of a sequence of curves . The first curve in the sequence maps [0,1] onto:" }}{PARA 0 "" 0 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7U7$$\"\"!F)F(7$$\"+; arz@!#6F+7$$\"+XTFwSF-F/7$$\"+!z_\"4iF-F27$$\"+S&phN)F-F57$$\"+*=)H\\5 !#5F87$$\"+[!3uC\"F:F<7$$\"+J$RDX\"F:F?7$$\"+)R'ok;F:FB7$$\"+1J:w=F:FE 7$$\"+3En$4#F:FH7$$\"+/RE&G#F:FK7$$\"+D.&4]#F:FN7$$\"+vB_Y2aF:$\"+!3QDf%F:7$$\" +!eWZh&F:$\"+?aD&Q%F:7$$\"+]y))GeF:$\"+]@6rTF:7$$\"+g_QQgF:$\"+SZhhRF: 7$$\"+!y%3TiF:$\"+?_\"*ePF:7$$\"+N![hY'F:$\"+l>&Q`$F:7$$\"+!Qx$omF:$\" +?EiJLF:7$$\"+v.I%)oF:$\"+D'*p:JF:7$$\"+&pe*zqF:$\"+08/?HF:7$$\"+D\\'Q H(F:$\"+v]81FF:7$$\"+:S8&\\(F:$\"+&)f'[]#F:7$$\"+0#=bq(F:$\"+&z\"[%H#F :7$$\"+0s?6zF:$\"+&z#z)3#F:7$$\"+IXaE\")F:$\"+qaXt=F:7$$\"+l*RRL)F:$\" +N+1m;F:7$$\"+b<.Y&)F:$\"+X#oRX\"F:7$$\"+:tOc()F:$\"+&oKOC\"F:7$$\"+]Q k\\*)F:$\"+]hN]5F:7$$\"+q0;r\"*F:$\"*I%R)G)F:7$$\"+lxGp$*F:$\"*NArI'F: 7$$\"+!oK0e*F:$\"*?tY>%F:7$$\"+:5s#y*F:$\"*&)*ys@F:7$$\"\"\"F)F(-%'COL OURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q!6\"Ffx-%%VIEWG6$%(DEFAULTG F[y" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 44.000000 0 0 "Curve 1 " }}{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 168 "again, we reiterate \+ that this is a vector-valued function, so that the above red line is t he image of our function. The image of the second function in our sequ ence is:" }}{PARA 0 "" 0 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CU RVESG6$7[o7$$\"\"!F)F(7$$\"+K3VfV!#6F+7$$\"+!H[D:)F-F/7$$\"+e0$=C\"!#5 F27$$\"+3RBr;F4F67$$\"+yjf)4#F4F97$$\"+Oiq'H#F4F<7$$\"+'4;[\\#F4F?7$$ \"+?E0+BF4$\"+!QZ**p#F47$$\"+Q8#\\4#F4$\"+i'y]!HF47$$\"+0siq;F4$\"+&zs $HLF47$$\"+)y$pZ7F4$\"+7iI_PF47$$\"*&yaE\")F4$\"+:_M(=%F47$$\"*#>s%H%F 4$\"+3y_qXF47$$\"*sdy8#F4$\"+GU@'y%F47$$\"+++l+>!#8$\"+]1!>+&F47$$\"++ qs%=#F-$\"++FZ=_F47$$\"++vW]VF-$\"+]Z/NaF47$$\"++NfC&)F-$\"+]$fC&eF47$ $\"+&z6:B\"F4$\"+&z6:B'F47$$\"+:=C#o\"F4$\"+:=C#o'F47$$\"+l#pS1#F4$\"+ l#pS1(F47$$\"+7t9'G#F4$\"+7t9'G(F47$$\"+i`A3DF4$\"+QYx\"\\(F47$$\"+l?! 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We then took the image of our original function, halfed it , then we rotated it 90 degrees clockwise and drew it on the first squ are. This gives us the first one fourth of our above image. To get the second fourth, we again half the size of our first image, and this ti me translate the image up by a half. We get the last two fourths again by translations and rotations of the first image." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "To get the image of the \+ next function, we apply the same transformations to the second image. \+ We will get:" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7]s7$$\"\"!F)F(7$$\"+K3VfV!#6F+7$$\"+k;')=( )F-F/7$$\"+R?ah5!#5F27$$\"+7z>^7F4$\"+)3-)[7F47$$\"+&y`3W\"F4$\"+:i9f5 F47$$\"+e'40j\"F4$\"+DM!\\p)F-7$$\"+)Q&3d?F4$\"+Dh9HWF-7$$\"+;6m$[#F4$ \")%))Qj\"F47$$\"+mW18HF4$\"+iYkITF-7$$\"+;yYULF4$\"+i\"yYU)F-7$$\"+^! \\hb$F4$\"+^!\\h0\"F47$$\"+9(p,t$F4$\"+'GI)p7F47$$\"+z%)[;NF4$\"+@:^$[ \"F47$$\"+Ws!GI$F4$\"+cF>(p\"F47$$\"+Gve1HF4$\"+sCT$4#F47$$\"+3yO5DF4$ \"+#>K'*[#F47$$\"+gZ*)**GF4Fap7$$\"+Ct:5LF4Fdp7$$\"+#R/B_$F4Fgp7$$\"+c 9XMPF4Fjp7$$\"+w9S`NF4$\"+C&)fYRF47$$\"+3WDTLF4$\"+#fX(eTF47$$\"+#*4K= HF4$\"+3!z;e%F47$$\"+wvQ&\\#F4$\"+wvQ&*\\F47$$\"+s&[.1#F4$\"+s&[.c%F47 $$\"+o&4`i\"F4$\"+o&4`7%F47$$\"+s#=PV\"F4$\"+s#=P$RF47$$\"+wp7U7F4$\"+ CI(yv$F47$$\"+!oN00\"F4$\"+?VY\\RF47$$\"+QQW*e)F-$\"+;c0TTF47$$\"+Qarv UF-$\"+c%GCd%F47$$\"+++I,Q!#8$\"++8!Q+&F47$$\"++SXpVF-$\"++a%pV&F47$$ \"++]*3q)F-$\"++&*3qeF47$$\"+/ozy5F4$\"+/ozygF47$$\"+'*e\\77F4$\"+/T]( G'F47$$\"+'f)y.5F4$\"+/9@'\\'F47$$\"++I\"3&zF-$\"++(=\\q'F47$$\"+i&)Gg TF-$\"+W6(R3(F47$$\")7k(p$F4$\"+)eBIY(F47$$\"+Dh`PTF-$\"+7Ov8zF47$$\"+ vj$[k)F-$\"+QO[k$)F47$$\"+]tRb5F4$\"+]tRb&)F47$$\"+v5JY7F4$\"+v5JY()F4 7$$\"++[AP9F4$\"++_xi&)F47$$\"+Q&Q\"G;F4$\"+i9'=P)F47$$\"+DYHs?F4$\"+v `qFzF47$$\"+D2X;DF4$\"+D2X;vF47$$\"+DTg4HF4$\"+DTg4zF47$$\"+Qvv-LF4$\" +Qvv-$)F47$$\"+QLV=NF4$\"+QLV=&)F47$$\"+D\"4Tt$F4$\"+D\"4Tt)F47$$\"+D \\y\\RF4$\"+v]@]&)F47$$\"+D2YlTF4$\"+v#RXL)F47$$\"+]%4id%F4$\"+]0zBzF4 7$$\"+v\"ep)\\F4$\"+D=/8vF47$$\"+D'oWe%F4$\"+D'oW3(F47$$\"+Da*e:%F4$\" +Da*el'F47$$\"+DM6fRF4$\"+DM6fkF47$$\"+Q9LiPF4$\"+Q9LiiF47$$\"+i0XMRF4 $\"+Q%\\b1'F47$$\"+]DBJTF4$\"+]uwoeF47$$\"+)[Wdb%F4$\"+7bDWaF47$$\"+7k D!)\\F4$\"+)eV(>]F47$$\"++]?@aF4Fj^l7$$\"+)e`@'eF4F]_l7$$\"++r2agF4F`_ l7$$\"+D1+YiF4Fc_l7$$\"+]e2igF4$\"+]T#zV'F47$$\"+DB:qeF4$\"+vw%)HmF47$ $\"+7qebaF4$\"+))HTWqF47$$\"+)o@5/&F4$\"+7$y*euF47$$\"+][E(Q&F4$\"+][E ()yF47$$\"++9b:eF4$\"++9b:$)F47$$\"+7)[]-'F4$\"+7)[]_)F47$$\"+DiaMiF4$ \"+DiaM()F47$$\"+QO/WkF4$\"+ij&fb)F47$$\"+]5a`mF4$\"+]*ekM)F47$$\"+)3S *eqF4$\"+7*f5%zF47$$\"+D\"RVY(F4$\"+v3mNvF47$$\"+QcY9zF4Fbcl7$$\"+]@fk $)F4Fecl7$$\"+)[@oc)F4Fhcl7$$\"+Q30p()F4$\"+i\"\\4t)F47$$\"+v,Gr*)F4$ \"+D)>(G&)F47$$\"+D&4N<*F4$\"+v/\\E$)F47$$\"+7bN0'*F4$\"+)[WY*yF47$$\" ++&)zi**F4$\"++&)ziuF47$$\"+i=[r&*F4$\"+i=[rqF47$$\"+7_;!=*F4$\"+7_;!o 'F47$$\"+))*ei'*)F4$\"+))*eiY'F47$$\"+iFN_()F4$\"+iFN_iF47$$\"+iMbh*)F 4$\"+QlWQgF47$$\"++(fa<*F4$\"++.aCeF47$$\"+vy*zd*F4$\"+D@+AaF47$$\"+]g `!)**F4$\"+]RY>]F47$$\"+ibp)f*F4$\"+ibp)f%F47$$\"+vr#z<*F4$\"+vr#z<%F4 7$$\"+v\"QA(*)F4$\"+v\"QA(RF47$$\"+v\"\\lw)F4$\"+v\"\\lw$F47$$\"+v,'3c )F4$\"+D)R\"RRF47$$\"+v6%F-7$$\"+i5:xu F4$\")Q*[G#F47$$\"+))3k**yF4$\"+v)3k*RF-7$$\"+D28A$)F4$\"+]sI@#)F-7$$ \"+]!>V_)F4$\"+]!>V-\"F47$$\"+)Q2ls)F4$\"+)Q2lA\"F47$$\"+DdpG*)F4$\"+v UIr5F47$$\"+iS)38*F4$\"+v$f6p)F-7$$\"+Q?Wl&*F4$\"+D'zbM%F-7$$\"\"\"F)F (-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q!6\"Fcbm-%%VIEWG6$%( DEFAULTGFhbm" 1 2 0 1 10 0 2 9 1 4 1 1.000000 45.000000 45.000000 0 0 "Curve 1" }}{MPLTEXT 1 0 0 "" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 105 "One will note that the first fourth of t his image is the previous one rotated clockwise by a right angle." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 233 "If we pr oceed as such, we will get a sequence of continuous functions which wi ll converge uniformly to another continuous function. This limiting fu nction, which we will call the Peano Curve, will turn out to be onto t he unit square." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 442 "Our task now is to come up with an algorithm which will \+ allow us to graph the sequence of functions which lead up to the Peano Curve. Our strategy is simple, we will first derive a procedure which lists for us the \"peano points.\" We note from above that our sequen ce of functions consist of straight lines pieced together at a number \+ of vertices. We call this sequence of vertices the peano points. If we can come up with a procedure that at " }{TEXT 373 1 "k" }{TEXT -1 35 " will list out the peano points at " }{TEXT 374 2 "k," }{TEXT -1 20 " then producing the " }{TEXT 375 1 "k" }{TEXT -1 165 "th function in o ur sequence is merely connecting each of these vertices in the right o rder. We start by first manually producing the peano points for the fi rst step:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "peanopoints1:=[ [0,0],[1/2,1/2],[1,0]];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-peanopoi nts1G7%7$\"\"!F'7$#\"\"\"\"\"#F)7$F*F'" }}}{PARA 0 "" 0 "" {TEXT -1 113 "let us plot these points and see that indeed, if we join them in \+ order we get the first function in our sequence:" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "pointplot(peanopoints1,color=red);" }}{PARA 13 "" 1 " " {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%'POINTSG6%7$$\"\"!F(F'7$$\"++ +++]!#5F*7$$\"\"\"F(F'-%'COLOURG6&%$RGBG$\"*++++\"!\")F'F'" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 150 "Now, to get the next set of peano points, we mer ely apply the four transformations to the previous points. We define t hese transformations now using: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "rotate:=(x,y)->[(x[1])*(cos(y))-(x[2])*(sin (y)),(x[1])*(sin(y))+(x[2])*(cos(y))];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'rotateGf*6$%\"xG%\"yG6\"6$%)operatorG%&arrowGF)7$,&*&&9$6#\" \"\"F3-%$cosG6#9%F3F3*&&F16#\"\"#F3-%$sinGF6F3!\"\",&*&F0F3F " 0 "" {MPLTEXT 1 0 52 "peanotransform1:=(x )->(1/2)*rotate(x,-Pi/2)+[0,1/2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#> %0peanotransform1Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&-%'rotateG6$9$, $%#PiG#!\"\"\"\"##\"\"\"F57$\"\"!F6F7F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "peanotransform2:=(x)->(1/2)*x+[0,1/2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%0peanotransform2Gf*6#%\"xG6\"6$%)operatorG %&arrowGF(,&9$#\"\"\"\"\"#7$\"\"!F.F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "peanotransform3:=(x)->(1/2)*x+[1/2,1/2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%0peanotransform3Gf*6#%\"xG6\"6$%)operatorG %&arrowGF(,&9$#\"\"\"\"\"#7$F.F.F/F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "peanotransform4:=(x)->(1/2)*rotate(x,Pi/2)+[1,0];" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%0peanotransform4Gf*6#%\"xG6\"6$%)ope ratorG%&arrowGF(,&-%'rotateG6$9$,$%#PiG#\"\"\"\"\"#F37$F4\"\"!F4F(F(F( " }}}{PARA 0 "" 0 "" {TEXT -1 159 "As a check, we apply the first tran sformation to the first peano points, which should give us the first t hree vertices of the second function in our sequence: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "pointplot(map(peanotr ansform1,peanopoints1),color=red);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%'POINTSG6%7$$\"\"!F($\"+++++]!#57$$\"+++++DF+ F-7$F'F'-%'COLOURG6&%$RGBG$\"*++++\"!\")F'F'" 1 2 0 1 10 0 2 6 1 4 1 1.000000 52.000000 89.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 34 "Now, let me display my algorithm: " }{MPLTEXT 1 0 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 544 "peanopoint:=proc(i,k)\nopti on remember;\nif k=1 then (peanopoints1)[i+1];\nelse if i=0 then [0,0] ; \nelse if i <= (2^(2*k-3)) then\nmap(peanotransform1,[seq(peanopoint (j,k-1),j=0..(2^(2*k-3)))])[(2^(2*k-3))-i+1];\nelse if i <= (2^(2*k-2) ) then \nmap(peanotransform2,[seq(peanopoint(j,k-1),j=0..(2^(2*k-3)))] )[i-(2^(2*k-3))+1];\nelse if i <= 3*(2^(2*k-3)) then\nmap(peanotransfo rm3,[seq(peanopoint(j,k-1),j=0..(2^(2*k-3)))])[i-(2^(2*k-2))+1];\nelse \nmap(peanotransform4,[seq(peanopoint(j,k-1),j=0..(2^(2*k-3)))])[(2^(2 *k-1))-i+1];\nfi; fi; fi; fi; fi; end;\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%+peanopointGf*6$%\"iG%\"kG6\"6#%)rememberGF)@%/9%\"\"\"&%-pean opoints1G6#,&9$F/F/F/@%/F4\"\"!7$F7F7@%1F4)\"\"#,&F.F<\"\"$!\"\"&-%$ma pG6$%0peanotransform1G7#-%$seqG6$-F$6$%\"jG,&F.F/F/F?/FK;F7F;6#,(F;F/F 4F?F/F/@%1F4)F<,&F.F&-FB6$%0peanotransform3GFE6#,(F4F/FSF?F/F/&-FB6$%0peanotransfo rm4GFE6#,()F<,&F.F " 0 "" {MPLTEXT 1 0 60 "peanopointset:=(k)->[seq(peanopoint(i,k),i=0..(2^(2*k -1)))];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.peanopointsetGf*6#%\"kG6 \"6$%)operatorG%&arrowGF(7#-%$seqG6$-%+peanopointG6$%\"iG9$/F3;\"\"!) \"\"#,&F4F9\"\"\"!\"\"F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 39 "and now \+ we list the points for several " }{TEXT 386 1 "k" }{TEXT -1 1 ":" } {MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "peanopoi ntset(1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%7$\"\"!F%7$#\"\"\"\"\"# F'7$F(F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "peanopointset(2 );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7+7$\"\"!F%7$#\"\"\"\"\"%F'7$F%# F(\"\"#7$F'#\"\"$F)7$F+F+7$F.F.7$F(F+7$F.F'7$F(F%" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 17 "peanopointset(3);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7C7$\"\"!F%7$#\"\"\"\"\")F'7$#F(\"\"%F%7$#\"\"$F)F'7$F+ F+7$F.F.7$F+#F(\"\"#7$F'F.7$F%F37$F'#\"\"&F)7$F%#F/F,7$F'#\"\"(F)7$F+F ;7$F.F=7$F3F;7$F.F87$F3F37$F8F8FA7$F8F=7$F;F;7$F=F=7$F(F;7$F=F87$F(F37 $F=F.7$F;F37$F8F.7$F;F+7$F8F'7$F;F%7$F=F'7$F(F%" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 17 "peanopointset(4);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7]s7$\"\"!F%7$#\"\"\"\"#;F'7$F%#F(\"\")7$F'#\"\"$F)7$F+ F+7$F.F.7$#F(\"\"%F+7$F.F'7$F3F%7$#\"\"&F)F'7$#F/F,F%7$#\"\"(F)F'7$F;F +7$F=F.7$F;F37$F8F.7$F3F37$F8F8FA7$F=F87$F;F;7$F=F=7$F;#F(\"\"#7$F8F=7 $F3FI7$F.F=7$F3F;7$F.F87$F+F;7$F'F87$F%F;7$F'F=7$F%FI7$F'#\"\"*F)7$F+F I7$F.FV7$F+#F9F,7$F.#\"#6F)7$F+#F/F47$F'Fgn7$F%Fjn7$F'#\"#8F)7$F%#F>F, 7$F'#\"#:F)7$F+Fao7$F.Fco7$F3Fao7$F.F^o7$F3Fjn7$F8F^oFgo7$F8Fco7$F;Fao 7$F=Fco7$FIFao7$F=F^o7$FIFjn7$F=Fgn7$F;Fjn7$F8Fgn7$F;Fen7$F8FVFH7$F=FV 7$FIFI7$FVFV7$FenFI7$FgnFV7$FenFen7$FgnFgn7$FenFjn7$FVFgnF`p7$FVF^oF^p 7$FVFco7$FenFao7$FgnFco7$FjnFao7$FgnF^o7$FjnFjn7$F^oF^oFcq7$F^oFco7$Fa oFao7$FcoFco7$F(Fao7$FcoF^o7$F(Fjn7$FcoFgn7$FaoFjn7$F^oFgn7$FaoFen7$F^ oFV7$FaoFI7$FcoFV7$F(FI7$FcoF=7$F(F;7$FcoF87$FaoF;7$F^oF87$FjnF;7$F^oF =7$FjnFI7$FgnF=Fip7$FVF=7$FenF;7$FVF87$FenF37$FgnF87$FjnF37$FgnF.Fas7$ FVF.7$FenF+7$FVF'7$FenF%7$FgnF'7$FjnF%7$F^oF'7$FjnF+7$F^oF.7$FaoF+7$Fc oF.7$F(F+7$FcoF'7$F(F%" }}}{PARA 0 "" 0 "" {TEXT -1 22 "and we also pl ot them:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "pointplot(peanopointset(1),color=red);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%'POINTSG6%7$$\"\"!F(F'7$$\"++++ +]!#5F*7$$\"\"\"F(F'-%'COLOURG6&%$RGBG$\"*++++\"!\")F'F'" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "pointplot(peanopointset(2),color=re d);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%'POINTS G6+7$$\"\"!F(F'7$$\"+++++D!#5F*7$F'$\"+++++]F,7$F*$\"+++++vF,7$F.F.7$F 1F17$$\"\"\"F(F.7$F1F*7$F6F'-%'COLOURG6&%$RGBG$\"*++++\"!\")F'F'" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "pointplot(peanopointset(3),c olor=red);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-% 'POINTSG6C7$$\"\"!F(F'7$$\"++++]7!#5F*7$$\"+++++DF,F'7$$\"++++]PF,F*7$ F.F.7$F1F17$F.$\"+++++]F,7$F*F17$F'F67$F*$\"++++]iF,7$F'$\"+++++vF,7$F *$\"++++]()F,7$F.F>7$F1FA7$F6F>7$F1F;7$F6F67$F;F;FE7$F;FA7$F>F>7$FAFA7 $$\"\"\"F(F>7$FAF;7$FMF67$FAF17$F>F67$F;F17$F>F.7$F;F*7$F>F'7$FAF*7$FM F'-%'COLOURG6&%$RGBG$\"*++++\"!\")F'F'" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 38 "pointplot(peanopointset(4),color=red);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%'POINTSG6]s7$$\"\"!F(F' 7$$\"++++]i!#6F*7$F'$\"++++]7!#57$F*$\"++++v=F07$F.F.7$F2F27$$\"+++++D F0F.7$F2F*7$F7F'7$$\"++++DJF0F*7$$\"++++]PF0F'7$$\"++++vVF0F*7$F?F.7$F BF27$F?F77$F " 0 "" {MPLTEXT 1 0 38 "pointplot(peanopointset(5),c olor=red);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-% 'POINTSG6][m7$$\"\"!F(F'7$$\"++++DJ!#6F*7$$\"++++]iF,F'7$$\"++++v$*F,F *7$F.F.7$F1F17$F.$\"++++]7!#57$F*F17$F'F67$F*$\"+++]i:F87$F'$\"++++v=F 87$F*$\"+++](=#F87$F.F?7$F1FB7$F6F?7$F1F<7$F6F67$F " 0 "" {MPLTEXT 1 0 79 "peanocurvepart:=(t,i,k)->(peanopoint(i+1,k)-peanopoin t(i,k))*t+peanopoint(i,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%/peano curvepartGf*6%%\"tG%\"iG%\"kG6\"6$%)operatorG%&arrowGF*,&*&,&-%+peanop ointG6$,&9%\"\"\"F6F69&F6-F26$F5F7!\"\"F69$F6F6F8F6F*F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 90 "With each piece defined as above, we will not be \+ able to simply sum each piece to get our " }{TEXT 390 1 "k" }{TEXT -1 173 "th function. Thus, we will now illustrate the second solution in \+ defining a sequence of piecewise defined functions, which involves usi ng a do loop. Here is our algorithm: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 203 "peanocurve:=proc(t,k)\nif t=1 \nth en [1,0] else\nfor i from 0 to (2^(2*k-1)) do \nif (i/(2^(2*k-1))) <= \+ t and t < ((i+1)/(2^(2*k-1))) then\npeanocurvepart((t*(2^(2*k-1)))-i,i ,k); break; else fi; od; \nfi; end;" }}{PARA 7 "" 1 "" {TEXT -1 68 "Wa rning, `i` is implicitly declared local to procedure `peanocurve`\n" } }{PARA 12 "" 1 "" {XPPMATH 20 "6#>%+peanocurveGf*6$%\"tG%\"kG6#%\"iG6 \"F+@%/9$\"\"\"7$F/\"\"!?(8$F1F/)\"\"#,&9%F5F/!\"\"%%trueG@%31*&F3F/F4 F8F.2F.*&,&F3F/F/F/F/F4F8C$-%/peanocurvepartG6%,&*&F.F/F4F/F/F3F8F3F7[ F+F+F+F+" }}}{PARA 0 "" 0 "" {TEXT -1 48 "This algorithm works as foll ows, we are given a " }{TEXT 391 1 "t" }{TEXT -1 7 " and a " }{TEXT 392 1 "k" }{TEXT -1 30 ", and we want to evaulate the " }{TEXT 393 1 " k" }{TEXT -1 31 "th function in our sequence at " }{TEXT 394 1 "t" } {TEXT -1 11 ". Now, the " }{TEXT 395 1 "k" }{TEXT -1 47 "th function i n our sequence is composed of 2^(2" }{TEXT 396 1 "k" }{TEXT -1 139 "-1 ) straight lines which are consecutive peano points joined together. W hat this algorithm does is divide the unit interval [0,1] into 2^(2" } {TEXT 397 1 "k" }{TEXT -1 82 "-1) pieces. Then in a do loop, this algo rithm determines in which of these pieces " }{TEXT 398 1 "t" }{TEXT -1 134 " lies in. Once it has determined that t lies in the ith interv al, it assigns to it the value corresponding to the line connecting th e " }{TEXT 399 1 "i" }{TEXT -1 7 "th and " }{TEXT 400 1 "i" }{TEXT -1 95 "+1th peano point. Let us see the algorithm at work and graph some \+ curves. First we must define:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "peanocurve1:=(t,k)->peanocurve(t,k)[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,peanocurve1Gf*6$%\"tG%\"kG6\"6$%)operatorG%&arrowGF) &-%+peanocurveG6$9$9%6#\"\"\"F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "peanocurve2:=(t,k)->peanocurve(t,k)[2];" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%,peanocurve2Gf*6$%\"tG%\"kG6\"6$%)operatorG%&a rrowGF)&-%+peanocurveG6$9$9%6#\"\"#F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 74 "So that we can parametrically graph our curves. Let us see the \+ curves for " }{TEXT 401 1 "k" }{TEXT -1 12 " up to six: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot(['peanocurve1( t,1)','peanocurve2(t,1)',t=0..1]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7U7$$\"\"!F)F(7$$\"+;arz@!#6F+7$$\" +XTFwSF-F/7$$\"+!z_\"4iF-F27$$\"+S&phN)F-F57$$\"+*=)H\\5!#5F87$$\"+[!3 uC\"F:F<7$$\"+J$RDX\"F:F?7$$\"+)R'ok;F:FB7$$\"+1J:w=F:FE7$$\"+3En$4#F: FH7$$\"+/RE&G#F:FK7$$\"+D.&4]#F:FN7$$\"+vB_Y2aF:$\"+!3QDf%F:7$$\"+!eWZh&F:$\"+? aD&Q%F:7$$\"+]y))GeF:$\"+]@6rTF:7$$\"+g_QQgF:$\"+SZhhRF:7$$\"+!y%3TiF: $\"+?_\"*ePF:7$$\"+N![hY'F:$\"+l>&Q`$F:7$$\"+!Qx$omF:$\"+?EiJLF:7$$\"+ v.I%)oF:$\"+D'*p:JF:7$$\"+&pe*zqF:$\"+08/?HF:7$$\"+D\\'QH(F:$\"+v]81FF :7$$\"+:S8&\\(F:$\"+&)f'[]#F:7$$\"+0#=bq(F:$\"+&z\"[%H#F:7$$\"+0s?6zF: $\"+&z#z)3#F:7$$\"+IXaE\")F:$\"+qaXt=F:7$$\"+l*RRL)F:$\"+N+1m;F:7$$\"+ b<.Y&)F:$\"+X#oRX\"F:7$$\"+:tOc()F:$\"+&oKOC\"F:7$$\"+]Qk\\*)F:$\"+]hN 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Thus, \+ our limiting function will not be injective. We also note the increasi ng amount of iterations needed to produce a respectable graph." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 223 "Hopefull y, one is convinced from the graphs that the Peano Curve will be onto \+ and continuous. However, is it differentiable? The answer is in the ex treme negative. The Peano Curve as it turns out, is nowhere differenti able." }}}{PARA 0 "" 0 "" {TEXT -1 310 "Our next example of a space-fi lling curve is due to Hilbert. It can also be constructed as the limit of a sequence of function. Like the Peano Curve, the Hilbert Curve wi ll be continuous and nowhere differentiable, and onto but not injectiv e. Let us explore the algorithm derived to produce the Hilbert Curve. \+ " }{MPLTEXT 1 0 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 29 "Hilbert's \+ Space Filling Curve" }}{PARA 0 "" 0 "" {TEXT -1 65 "As in the Peano Cu rve, we first start with a small set of points:" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 82 "hilbertpoints1:=[[0,0],[1/2,1/2],[0,1/2],[1/2, 1],[1/2,0],[1,1/2],[1/2,1/2],[1,1]];" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%/hilbertpoints1G7*7$\"\"!F'7$#\"\"\"\"\"#F)7$F'F)7$F)F*7$F)F'7$F*F )F(7$F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 157 "Now, if we connect these p oints in order we will get the first function in our sequence of funct ions leading up to the Hilbert Curve. Let us plot the points:" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "pointplot(hilbertpoints1,color=red) ;" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6$-%'POINTSG6 *7$$\"\"!F(F'7$$\"+++++]!#5F*7$F'F*7$F*$\"\"\"F(7$F*F'7$F/F*F)7$F/F/-% 'COLOURG6&%$RGBG$\"*++++\"!\")F'F'" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 56 "Now , let us define the following three transformations: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "hilberttransform1:=(x )->(1/2)*x+[0,1/2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%2hilberttrans form1Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&9$#\"\"\"\"\"#7$\"\"!F.F/F( F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "hilberttransform2:= (x)->(1/2)*x+[1/2,0];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%2hilberttra nsform2Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&9$#\"\"\"\"\"#7$F.\"\"!F/ F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "hilberttransform3 :=(x)->(1/2)*x+[1/2,1/2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%2hilber ttransform3Gf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&9$#\"\"\"\"\"#7$F.F.F /F(F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 60 "Using these transformations, \+ define the following algorithm:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 497 "hilbertpoint:=proc(i,k)\noption remember;\n if k=1 then (hilbertpoints1)[i]\nelse\nif i <= 2^(2*k-1)\nthen ((1/2)* [seq(hilbertpoint(j,k-1),j=1..(2^(2*k-1)))])[i] ;\nelse\nif i <= 2^(2* k)\nthen map(hilberttransform1,[seq(hilbertpoint(j,k-1),j=1..(2^(2*k-1 )))])[i-(2^(2*k-1))] ;\nelse\nif i <= 3*(2^(2*k-1))\nthen map(hilbertt ransform2,[seq(hilbertpoint(j,k-1),j=1..(2^(2*k-1)))])[i-(2^(2*k))] ; \nelse\nmap(hilberttransform3,[seq(hilbertpoint(j,k-1),j=1..(2^(2*k-1) ))])[i-(3*(2^(2*k-1)))] ;\nfi; fi; fi; fi; end;\n\n" }}{PARA 12 "" 1 " " {XPPMATH 20 "6#>%-hilbertpointGf*6$%\"iG%\"kG6\"6#%)rememberGF)@%/9% \"\"\"&%/hilbertpoints1G6#9$@%1F3)\"\"#,&F.F7F/!\"\"&7#,$-%$seqG6$-F$6 $%\"jG,&F.F/F/F9/FB;F/F6#F/F7F2@%1F3)F7,$F.F7&-%$mapG6$%2hilberttransf orm1G7#F=6#,&F3F/F6F9@%1F3,$F6\"\"$&-FM6$%2hilberttransform2GFP6#,&F3F /FIF9&-FM6$%2hilberttransform3GFP6#,&F3F/*&FVF/F6F/F9F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 374 "One can compare this algorithm with the algori thm for the Peano points, and note the great similarities. We start wi th a small set of points, then we basically divide the unit square int o quadrants, and apply transformations to our first set of points to g et some sort of image of them into each of the subdivisions of the squ are. We then proceed inductively. Thus, at each " }{TEXT 403 1 "k" } {TEXT -1 102 ", hilbertpoint defines a sequence of points from applyin g transformations to the set of points at the " }{TEXT 404 1 "k" } {TEXT -1 61 "-1th step. We first start with the set I manually defined as " }{TEXT 405 14 "hilbertpoints1" }{TEXT -1 53 ". For the curious, \+ let us list these points for some " }{TEXT 406 1 "k" }{TEXT -1 1 ":" } {MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "hilbertp ointset:=(k)->[seq(hilbertpoint(i,k),i=1..(2^(2*k+1)))];" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%0hilbertpointsetGf*6#%\"kG6\"6$%)operatorG%&ar rowGF(7#-%$seqG6$-%-hilbertpointG6$%\"iG9$/F3;\"\"\")\"\"#,&F4F9F7F7F( F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "hilbertpointset(1); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*7$\"\"!F%7$#\"\"\"\"\"#F'7$F%F'7 $F'F(7$F'F%7$F(F'F&7$F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "hilbertpointset(2);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#7B7$\"\"!F% 7$#\"\"\"\"\"%F'7$F%F'7$F'#F(\"\"#7$F'F%7$F,F'F&7$F,F,7$F%F,7$F'#\"\"$ F)7$F%F37$F'F(F+7$F,F3F27$F,F(7$F,F%7$F3F'F/7$F3F,7$F3F%7$F(F'F:7$F(F, F07$F3F3F77$F3F(F;7$F(F3F?7$F(F(" }}}{PARA 0 "" 0 "" {TEXT -1 102 "The first point is always (0,0) while the last is always (1,1). Also, let us plot some of the points: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 30 "pointplot(hilbertpointset(1));" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6#-%'POINTSG6*7$$\"\"!F(F'7$$ \"+++++]!#5F*7$F'F*7$F*$\"\"\"F(7$F*F'7$F/F*F)7$F/F/" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pointplot(hilbertpointset(2));" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6#-%'POINTSG6B7$$\"\"!F(F'7 $$\"+++++D!#5F*7$F'F*7$F*$\"+++++]F,7$F*F'7$F/F*F)7$F/F/7$F'F/7$F*$\"+ ++++vF,7$F'F67$F*$\"\"\"F(F.7$F/F6F57$F/F:7$F/F'7$F6F*F27$F6F/7$F6F'7$ F:F*F?7$F:F/F37$F6F6F<7$F6F:F@7$F:F6FD7$F:F:" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "pointplot(hilbertpointset(3));" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6#-%'POINTSG6\\s7$$\"\"!F(F'7$$\"++ ++]7!#5F*7$F'F*7$F*$\"+++++DF,7$F*F'7$F/F*F)7$F/F/7$F'F/7$F*$\"++++]PF ,7$F'F67$F*$\"+++++]F,F.7$F/F6F57$F/F:7$F/F'7$F6F*F27$F6F/7$F6F'7$F:F* F?7$F:F/F37$F6F6F<7$F6F:F@7$F:F6FD7$F:F:7$F'F:7$F*$\"++++]iF,7$F'FJ7$F *$\"+++++vF,F97$F/FJFI7$F/FN7$F'FN7$F*$\"++++]()F,7$F'FT7$F*$\"\"\"F(F M7$F/FTFS7$F/FXF=7$F6FJFP7$F6FNFE7$F:FJFfn7$F:FNFQ7$F6FTFZ7$F6FXFgn7$F :FTFjn7$F:FX7$F:F'7$FJF*FB7$FJF/7$FJF'7$FNF*F_o7$FNF/FC7$FJF6FF7$FJF:F `o7$FNF6Fdo7$FNF:7$FNF'7$FTF*Fbo7$FTF/7$FTF'7$FXF*Fio7$FXF/Fco7$FTF6Ff o7$FTF:Fjo7$FXF6F^p7$FXF:FG7$FJFJFhn7$FJFNFeo7$FNFJFbp7$FNFNFin7$FJFTF \\o7$FJFXFcp7$FNFTFfp7$FNFXFgo7$FTFJFdp7$FTFNF_p7$FXFJFjp7$FXFNFep7$FT FTFhp7$FTFXF[q7$FXFTF^q7$FXFX" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "pointplot(hilbertpointset(4));" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6#-%'POINTSG6\\[m7$$\"\"!F(F'7$$\"+ +++]i!#6F*7$F'F*7$F*$\"++++]7!#57$F*F'7$F/F*F)7$F/F/7$F'F/7$F*$\"++++v =F17$F'F77$F*$\"+++++DF1F.7$F/F7F67$F/F;7$F/F'7$F7F*F37$F7F/7$F7F'7$F; F*F@7$F;F/F47$F7F7F=7$F7F;FA7$F;F7FE7$F;F;7$F'F;7$F*$\"++++DJF17$F'FK7 $F*$\"++++]PF1F:7$F/FKFJ7$F/FO7$F'FO7$F*$\"++++vVF17$F'FU7$F*$\"+++++] F1FN7$F/FUFT7$F/FYF>7$F7FKFQ7$F7FOFF7$F;FKFgn7$F;FOFR7$F7FUFen7$F7FYFh n7$F;FUF[o7$F;FY7$F;F'7$FKF*FC7$FKF/7$FKF'7$FOF*F`o7$FOF/FD7$FKF7FG7$F KF;Fao7$FOF7Feo7$FOF;7$FOF'7$FUF*Fco7$FUF/7$FUF'7$FYF*Fjo7$FYF/Fdo7$FU F7Fgo7$FUF;F[p7$FYF7F_p7$FYF;FH7$FKFKFin7$FKFOFfo7$FOFKFcp7$FOFOFjn7$F KFUF]o7$FKFYFdp7$FOFUFgp7$FOFYFho7$FUFKFep7$FUFOF`p7$FYFKF[q7$FYFOFfp7 $FUFUFip7$FUFYF\\q7$FYFUF_q7$FYFY7$F'FY7$F*$\"++++DcF17$F'Feq7$F*$F+F1 FX7$F/FeqFdq7$F/Fiq7$F'Fiq7$F*$\"++++voF17$F'F^r7$F*$\"+++++vF1Fhq7$F/ F^rF]r7$F/FbrFfn7$F7FeqFjq7$F7FiqF\\o7$F;FeqFfr7$F;FiqF[r7$F7F^rFdr7$F 7FbrFgr7$F;F^rFjr7$F;Fbr7$F'Fbr7$F*$\"++++D\")F17$F'F`s7$F*$\"++++]()F 1Far7$F/F`sF_s7$F/Fds7$F'Fds7$F*$\"++++v$*F17$F'Fjs7$F*$\"\"\"F(Fcs7$F /FjsFis7$F/F^tFer7$F7F`sFfs7$F7FdsF[s7$F;F`sFbt7$F;FdsFgs7$F7FjsF`t7$F 7F^tFct7$F;FjsFft7$F;F^tF^o7$FKFeqFhr7$FKFiqFhp7$FOFeqFjt7$FOFiqFir7$F KF^rF\\s7$FKFbrF[u7$FOF^rF^u7$FOFbrFjp7$FUFeqF\\u7$FUFiqF`q7$FYFeqFbu7 $FYFiqF]u7$FUF^rF`u7$FUFbrFcu7$FYF^rFfu7$FYFbrF]s7$FKF`sFdt7$FKFdsF_u7 $FOF`sFju7$FOFdsFet7$FKFjsFht7$FKF^tF[v7$FOFjsF^v7$FOF^tFau7$FUF`sF\\v 7$FUFdsFgu7$FYF`sFbv7$FYFdsF]v7$FUFjsF`v7$FUF^tFcv7$FYFjsFfv7$FYF^t7$F YF'7$FeqF*F]p7$FeqF/7$FeqF'7$FiqF*F[w7$FiqF/F^p7$FeqF7Fap7$FeqF;F\\w7$ FiqF7F`w7$FiqF;7$FiqF'7$F^rF*F^w7$F^rF/7$F^rF'7$FbrF*Few7$FbrF/F_w7$F^ rF7Fbw7$F^rF;Ffw7$FbrF7Fjw7$FbrF;Fbp7$FeqFKF]q7$FeqFOFaw7$FiqFKF^x7$Fi qFOF^q7$FeqFUFaq7$FeqFYF_x7$FiqFUFbx7$FiqFYFcw7$F^rFKF`x7$F^rFOF[x7$Fb rFKFfx7$FbrFOFax7$F^rFUFdx7$F^rFYFgx7$FbrFUFjx7$FbrFY7$FbrF'7$F`sF*Fhw 7$F`sF/7$F`sF'7$FdsF*F_y7$FdsF/Fiw7$F`sF7F\\x7$F`sF;F`y7$FdsF7Fdy7$Fds F;7$FdsF'7$FjsF*Fby7$FjsF/7$FjsF'7$F^tF*Fiy7$F^tF/Fcy7$FjsF7Ffy7$FjsF; Fjy7$F^tF7F^z7$F^tF;F]x7$F`sFKFhx7$F`sFOFey7$FdsFKFbz7$FdsFOFix7$F`sFU F\\y7$F`sFYFcz7$FdsFUFfz7$FdsFYFgy7$FjsFKFdz7$FjsFOF_z7$F^tFKFjz7$F^tF OFez7$FjsFUFhz7$FjsFYF[[l7$F^tFUF^[l7$F^tFYFbq7$FeqFeqFdu7$FeqFiqFcx7$ FiqFeqFb[l7$FiqFiqFeu7$FeqF^rFhu7$FeqFbrFc[l7$FiqF^rFf[l7$FiqFbrFex7$F ^rFeqFd[l7$F^rFiqF[y7$FbrFeqFj[l7$FbrFiqFe[l7$F^rF^rFh[l7$F^rFbrF[\\l7 $FbrF^rF^\\l7$FbrFbrFiu7$FeqF`sFdv7$FeqFdsFg[l7$FiqF`sFb\\l7$FiqFdsFev 7$FeqFjsFhv7$FeqF^tFc\\l7$FiqFjsFf\\l7$FiqF^tFi[l7$F^rF`sFd\\l7$F^rFds F_\\l7$FbrF`sFj\\l7$FbrFdsFe\\l7$F^rFjsFh\\l7$F^rF^tF[]l7$FbrFjsF^]l7$ FbrF^tF]y7$F`sFeqF\\\\l7$F`sFiqFgz7$FdsFeqFb]l7$FdsFiqF]\\l7$F`sF^rF` \\l7$F`sFbrFc]l7$FdsF^rFf]l7$FdsFbrFiz7$FjsFeqFd]l7$FjsFiqF_[l7$F^tFeq Fj]l7$F^tFiqFe]l7$FjsF^rFh]l7$FjsFbrF[^l7$F^tF^rF^^l7$F^tFbrFa\\l7$F`s F`sF\\]l7$F`sFdsFg]l7$FdsF`sFb^l7$FdsFdsF]]l7$F`sFjsF`]l7$F`sF^tFc^l7$ FdsFjsFf^l7$FdsF^tFi]l7$FjsF`sFd^l7$FjsFdsF_^l7$F^tF`sFj^l7$F^tFdsFe^l 7$FjsFjsFh^l7$FjsF^tF[_l7$F^tFjsF^_l7$F^tF^t" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{PARA 0 "" 0 "" {TEXT -1 223 "Hopefully, one is convinced that as before, the union of all o f these points is a dense set. Thus, our Hilbert Curve which will be c ontinuous and which will pass through all of these points, will hence \+ be onto [0,1]x[0,1]. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 128 "Let us now define our sequence of curves leading up t o the Hilbert Curve. As before, we define our curves in a piecewise ma nner:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "hilbertcurvepart:=( t,i,k)->(hilbertpoint(i+1,k)-hilbertpoint(i,k))*t+hilbertpoint(i,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%1hilbertcurvepartGf*6%%\"tG%\"iG% \"kG6\"6$%)operatorG%&arrowGF*,&*&,&-%-hilbertpointG6$,&9%\"\"\"F6F69& F6-F26$F5F7!\"\"F69$F6F6F8F6F*F*F*" }}}{PARA 0 "" 0 "" {TEXT -1 106 "S o that then we can employ the do-loop method to define a sequence of f unctions that are piecewise linear:" }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 225 "hilbertcurve:=proc(t,k)\nif t=1 \n then [1,1] else\nfor i from 1 to ((2^(2*k+1))-1) do \nif ((i-1)/((2^(2 *k+1))-1)) <= t and t < (i/((2^(2*k+1))-1)) then\nhilbertcurvepart((t* ((2^(2*k+1))-1))+1-i,i,k); break; else fi; od; \nfi; end;" }}{PARA 7 " " 1 "" {TEXT -1 70 "Warning, `i` is implicitly declared local to proce dure `hilbertcurve`\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%-hilbertcur veGf*6$%\"tG%\"kG6#%\"iG6\"F+@%/9$\"\"\"7$F/F/?(8$F/F/,&)\"\"#,&9%F5F/ F/F/F/!\"\"%%trueG@%31*&,&F2F/F/F8F/F3F8F.2F.*&F2F/F3F8C$-%1hilbertcur vepartG6%,(*&F.F/F3F/F/F/F/F2F8F2F7[F+F+F+F+" }}}{PARA 0 "" 0 "" {TEXT -1 13 "So, using: " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "hilbertcurve1:=(t,k)->hilbertcurve(t,k)[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.hilbertcurve1Gf*6$%\"tG%\"kG6\"6$%)operatorG%&arrowG F)&-%-hilbertcurveG6$9$9%6#\"\"\"F)F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "hilbertcurve2:=(t,k)->hilbertcurve(t,k)[2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.hilbertcurve2Gf*6$%\"tG%\"kG6\"6$%)operat orG%&arrowGF)&-%-hilbertcurveG6$9$9%6#\"\"#F)F)F)" }}}{PARA 0 "" 0 "" {TEXT -1 62 "We can now finally present what our Hilbert Curves look l ike: " }{MPLTEXT 1 0 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 " plot(['hilbertcurve1(t,1)','hilbertcurve2(t,1)',t=0..1]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7gq7$$\"\"!F) 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