{VERSION 5 0 "IBM INTEL NT" "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 "" -1 264 "Helvetica" 1 14 128 0 0 1 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 24 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "Helvetica" 0 1 128 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 267 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 1 12 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 273 "" 1 12 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 256 1 {CSTYLE "" -1 -1 "Time s" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }1 1 0 0 12 12 1 0 1 0 2 2 19 1 } {PSTYLE "Author" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 8 8 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 258 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 259 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 } {PSTYLE "Author" -1 260 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 1 2 2 2 1 1 1 1 }3 1 0 0 8 8 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 261 1 {CSTYLE "" -1 -1 "Courier" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 262 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Normal" -1 263 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 264 1 {CSTYLE "" -1 -1 "Courier" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 265 1 {CSTYLE "" -1 -1 "Ti mes" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 2" -1 266 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 265 30 "MLC Lab Visit - Lab 04 \+ - Maple" }}{PARA 0 "" 0 "" {TEXT 264 46 "Mth 355 (a.k.a. Mth 399) Jan \+ 29, 2003 Maple 7" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 266 16 "Bent E. Petersen" }{TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT 268 17 "bent@al um.mit.edu" }}{PARA 264 "" 0 "" {TEXT 269 22 "petersen@math.orst.edu" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT 267 180 "The re are 6 problems below. Problem solutions are due Feb 5, 2003. Email \+ your solutions to me as Maple worksheet attachments. Your worksheet mu st execute correctly for full credit." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 162 "This worksheet contains a few commen ts on some of Maple's recurrence relations support. If you find a seri ous error be sure to mention it in your solution report." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 38 "Linear Re currence Equations - Examples" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT 273 27 "Example. Fibon acci sequence" }{TEXT -1 0 "" }}{EXCHG {PARA 263 "" 0 "" {TEXT -1 0 " " }{TEXT 270 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 190 "The M aple function rsolve() is used to solve linear recurrence relations. H ere is an example of an order 2 recurrence relation with initial condi tions. The solution is the Fibonacci sequence" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "eqn1:=a(n)= a(n-1)+a(n-2); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "init1:=a (0)=0,a(1)=1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 99 "To solve the recurrence equation we use rsolve(). Not e how we specify for which variable to solve." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "soln1:=rsol ve(\{eqn1,init1\},a); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 146 "We can also use rsolve() to find the g enerating function for the solution. Note the single quotes surroundin g genfunc in the following command." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "gen1:=rsolve(\{eqn1,init1 \},a,'genfunc'(z));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "A very nice feature of Maple is that we can eve n define a procedure which will compute the a(n):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "fun1:=rsolv e(\{eqn1,init1\},a,'makeproc'):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Let's check it:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "fun1(3); fun1(4); fun1(5); fun1(6); fun1( 7); fun1(8);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 60 "The fun1() procedure can be use to plot the sequence \+ a(n)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "seq1:=seq([n,fun1(n)],n=0..15);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "plot([seq1],style=point,symbol=diamond,symbol size=20,title=\"fun1 - Fibonacci sequence\",color=blue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 139 "If we om it the \"style=point\" directive Maple will produce the plot of the pi ecewise linear interpolation rather than the individual points:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "plot([seq1],title=\"fun1 - Fibonacci sequence\" ,color=blue, thi ckness=2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 265 "" 0 "" {TEXT 274 21 "Example. Derangements" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 271 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "In class we obtained a recurrence equat ion for the number of derangements." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "eqn2:=d(n)=n*d(n-1)+(-1)^n ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "init2:=d(2)=1;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 98 "Th e initial condition just means the number of derangements of the set \+ \{1,2\} is 1, which is clear." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "soln2:=rsolve(\{eqn2,init2\} ,d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 105 "Hmm. Maple's response is in terms of the Incomplete Gamma func tion - not a familiar object to most of us." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 184 "It is usually more convenient \+ to have the initial condition at the origin, especially for computing \+ the generating function in Maple, so let's see what the initial condit ion should be." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 40 "fun2:=rsolve(\{eqn2,init2\},d,'makeproc'):" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "simplify(fun2(0));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 195 "T his condition is a bit difficult to interpret. We can view it as sayin g there is one permutation of the empty set, the identity permutation, and it is a derangement since it has no fixed points." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "Let's look \+ at the exponential generating function. Maple can not compute it direc tly, so we compute the ordinary generating function for a(n)=d(n)/n! \+ nstead:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "eqn2a:=a(n)=a(n-1)+(-1)^n/n!;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "init2a:=a(0)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "gen2:=rsolve(\{eqn2a,init2a\},a,'genfunc'(z)); " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "value(gen2): simplify(%);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Well, we had to coax Maple a bit , but we got the right answer." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 266 "" 0 "" {TEXT 275 49 "Example. A recurr ence relation with complex roots" }{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 62 "Here's a recurrence \+ equation demonstrating damped oscillation:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "eqn3:=a(n)=a(n-1) -a(n-2)/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "init3:=a(0)=1 ,a(1)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "soln3:=rsolve( \{eqn3,init3\},a); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "gen3 :=rsolve(\{eqn3,init3\},a,'genfunc'(z));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "fun3:=rsolve(\{eqn3,init3\},a,'makeproc'):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "seq3:=seq([n,fun3(n)],n=0..2 0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "plot([seq3],style=po int,symbol=box,symbolsize=20,color=blue,title=\"Damped vibration\");" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 55 "Linear Homogeneous Recurrence Equations of Finite Or der" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "For the l inear homogeneous recurrence equation of order 3" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "eqn:=a(n)=A [1]*a(n-1)+A[2]*a(n-2)+A[3]*a(n-3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 " the characteristic polynomial is given by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "charpoly:=s ubs(a(n-3)=1,a(n-2)=x,a(n-1)=x^2,a(n)=x^3,eqn);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 82 "We can write a simpl e (not very robust) procedure to do this substitutuion for us:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "cp:=proc(recureqn,a,n,order,var)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 " local k,sq;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "sq:=seq(a( n-k)=var^(order-k),k=0..order);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 " subs(sq,recureqn);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 16 "cp(eqn,a,n,3,z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 1" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 439 "L et a(n) be the number of strings of 0's and 1's of length n wh ich do not contain the bit pattern 11. If a such a string starts wit h 0 the remainder can be any string of length n-1 which does not c ontain 11. If on the other hand it starts with 1 then the second c haracter must be 0 and after that we can have any string of length \+ n-2 which does not contain 11. Thus (note we just have a shifted Fi bonacci sequence) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eqn4:=b(m)=b(m-1)+b(m-2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "init4:=b(0)=1,b(1)=2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "soln4:=rsolve(\{eqn4,init4\},b(m));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 118 "L et's compute the characteristic polynomial and the characteristic root s to see how they match up to Maple's solution:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "cpoly:=cp(e qn4,b,m,2,z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "solve(cpol y,z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "fun4:=rsolve(\{eqn 4,init4\},b,'makeproc'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "for k from 1 to 8 do fun4(k); od;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 2" } }{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 53 "H ere's another homogeneous linear recurrence relation" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eqn5:=a(n )=-a(n-2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "init5:=a(0)=2 ,a(1)=-1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "soln5:=rsolve( \{eqn5,init5\},a); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "fun5 :=rsolve(\{eqn5,init5\},a,'makeproc'):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "for k from 1 to 8 do fun5(k); od;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "gen5:=rsolve(\{eqn5,init5\},a,'genfunc'(z)) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "seq5:=seq([n,fun5(n)], n=0..20);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "plot([seq5],st yle=point,symbol=circle,symbolsize=20,color=blue,title=\"Periodic exam ple\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 80 "This plot is hard to interpret. Let's look at the piecewi se linear plot instead:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "plot([seq5],color=blue,title=\"Peri odic example\");" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 38 "The characteristic polynomial here is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "cpo ly5:=cp(eqn5,a,n,2,z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 57 "Linear Inhomogeneous Recurre nce Equations of Finite Order" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "E xample 1" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "eqn6:=c(n)=c(n-1 )+c(n-2)-n;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "init6:=c(0)= 0,c(1)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "soln6:=rsolve( \{eqn6,init6\},c); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "gen6 :=rsolve(\{eqn6,init6\},c,'genfunc'(z)); simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "fun6:=rsolve(\{eqn6,init6\},c,'make proc');" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "seq6tmp:=seq([n, fun6(n)],n=0..2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 75 "Unless we really need it, this expression is too c omplicated. Try 21 terms!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "seq6:=seq([n,round(0,evalf(fun6(n)) )],n=0..20);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot([seq6] ,style=point,symbol=diamond,symbolsize=20,color=blue);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 9 "Example 2" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "eqn7:=a(n )=3*a(n-1)+4*a(n-2)-12*a(n-3)-n^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "init7:=a(0)=2,a(1)=5,a(2)=13;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 31 "soln7:=rsolve(\{eqn7,init7\},a); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "gen7:=rsolve(\{eqn7,init7\},a,'genf unc'(z)); simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "f un7:=rsolve(\{eqn7,init7\},a,'makeproc'):" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 34 "for k from 1 to 10 do fun7(k); od;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "unassign('k');" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 27 "Ex ample 3 Nonparallel lines" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 313 "Consider n nonparallel lines in the pl ane and suppose no three of them meet in a point. The lines divide the plane into a(n) regions. If we already have n-1 lines and add one more line, it cuts each of the existing n-1 lines and so creates n new regions. (This statement may require some argument.) Thus" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eqn8:=a(n)=a(n-1)+n;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "init8:=a(0)=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "soln8 :=rsolve(\{eqn8,init8\},a(n));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 8 "Problems" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Solve the following recursive equations. \+ In each case find the generating function as well." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 276 9 "Problem 1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "a(n) = 2*a(n-1)-7; a(0)=8;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 277 9 "Pr oblem 2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 42 "a(n ) = 7*a(n-1)-11*a(n-2); a(0)=3,a(1)=-2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 278 9 "Problem 3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "a(n) = a(n-1) + n^3; a(1)=1;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "What does this problem have to do \+ with" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Sum(k^3,k=1..n);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "What does it have to do with" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Sum(k,k=1..n)^2;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 279 9 "Pr oblem 4" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "a(n) = a(n-1) + \+ n^5; a(1)=1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "What does this pr oblem have to do with" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Su m(k^5,k=1..n);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 280 9 "Problem 5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "a(n) = 2*a(n-1)+11*a(n-2)-12*a(n-3)+5*2^n; a(0)=-1,a(1)=3,a(2)=5; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}}}{MARK "0 9 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }