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1 }{PSTYLE "" -1 271 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 0 0 1 }1 1 0 0 0 0 2 0 2 0 2 2 -1 1 }{CSTYLE "" -1 205 "Times" 1 10 0 0 0 1 2 2 2 2 2 2 0 0 0 1 }} {SECT 0 {EXCHG {PARA 206 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 207 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 208 "" 0 "" {TEXT -1 12 "Eigensystems" }}{PARA 209 "" 0 "" {TEXT -1 18 "1. What Are They??" }}{PARA 210 "" 0 "" {TEXT -1 327 "Good question. Let's get right to the point. Let \+ A be a square n x n matrix (these are the only kind of matrices fo r which eigenvalues and eigenvectors will be defined). An eigenvector of A is a NONZERO vector x in Rn (or Cn, if we are working o ver complex nos), such that for some scalar number l, we have\n" } {TEXT -1 1 "\n" }{TEXT -1 17 " \+ A x = l x\n" }{TEXT -1 1 "\n" }{TEXT -1 376 "The number l is called an eigenvalue of the matrix A, and the vector x is called an eig envector belonging to (or corresponding to) the eigenvalue l. Obser ve that the eigenvalue l is allowed to be the 0 scalar, but eige nvectors x are, by definition, never the 0 vector. (Question: \+ what does it tell you about a matrix that 0 is an eigenvalue of it?) \n" }{TEXT -1 1 "\n" }{TEXT -1 80 "Maple knows all about eigenvalue s and eigenvectors. Click the following cells." }}{PARA 211 "> " 0 "" {MPLTEXT 1 0 25 "with(linalg):with(plots):" }}{PARA 212 "" 0 "" {TEXT -1 307 "The reason for using a decimal point in the following matrix \+ in one entry is to tell Maple that we want numerical (but approximate ) results; otherwise, Maple will try to give symbolic (but exact) ans wers that can get quite complicated. With 2x2's it isn't so bad, but \+ anything bigger can get complicated." }}{PARA 213 "> " 0 "" {MPLTEXT 1 0 43 "A := matrix([[4.0,0,1],[-2,1,0],[-2,0,1]]);" }}{PARA 214 "> " 0 "" {MPLTEXT 1 0 28 "B := matrix([[3,2],[-1,0]]);" }}{PARA 215 "> " 0 "" {MPLTEXT 1 0 31 "C := matrix([[-2.0,-1],[5,2]]);" }}{PARA 216 "> " 0 "" {MPLTEXT 1 0 13 "eigenvals(A);" }}{PARA 217 "> " 0 "" {MPLTEXT 1 0 13 "eigenvals(B);" }}{PARA 218 "> " 0 "" {MPLTEXT 1 0 13 "eigenval s(C);" }}{PARA 219 "" 0 "" {TEXT -1 58 "So how many eigenvalues do you expect from an nxn matrix?\n" }{TEXT -1 1 "\n" }{TEXT -1 132 "You'll \+ find out more technical details in class lecture, but here are some o f the basic points about eigenvalues and eigenvectors:\n" }{TEXT -1 1 "\n" }{TEXT -1 23 "A has eigenvalue l \n" }{TEXT -1 65 " \+ iff A x = l x, for some nonzero vector x,\n" }{TEXT -1 8 " \n" }{TEXT -1 41 " i ff A x - l x = 0, x nonzero\n" }{TEXT -1 9 " \+ \n" }{TEXT -1 40 " iff (A - l I ) x = 0, x nonzero\n" }{TEXT -1 9 " \+ \n" }{TEXT -1 70 " iff x is a nonzero element of the Null Space of (A - l I)\n" }{TEXT -1 10 " \+ \n" }{TEXT -1 44 " iff rank (A - l I) \+ < n = size of A\n" }{TEXT -1 10 " \+ \n" }{TEXT -1 30 " iff det (A - l I) = 0.\n" } {TEXT -1 1 "\n" }{TEXT -1 239 "But observe that the det above is just \+ a polynomial of degree n in the variable l. The last equation abo ve is called the characteristic equation of A. It will have exactly \+ n roots, if we allow for repetitions and complex numbers.\n" }{TEXT -1 1 "\n" }{TEXT -1 451 "Once we have found all eigenvalues of A, we can always find the appropriate Null Space and hence all eigenvectors belonging to each eigenvalue. By the way, this Null Space has a nam e: it is called the eigenspace corresponding to the eigenvalue l. A s an example, click the following cells. The first is just an ident ity matrix builder function. The next one actually calculates the cha racteristic polynomial of A as a function of lambda. " }}{PARA 220 "> " 0 "" {MPLTEXT 1 0 31 "Id := n -> diag(seq(1,i=1..n));" }}{PARA 221 "> " 0 "" {MPLTEXT 1 0 27 "p := det(A - lambda*Id(3));" }}{PARA 222 "> " 0 "" {MPLTEXT 1 0 17 "fsolve(p,lambda);" }}{PARA 223 "" 0 "" {TEXT -1 161 "Maple has a \"nullspace\" command. Unless you are worki ng with exact data, like B or C, don't take the output as always true. It is only approximate and may err." }}{PARA 224 "> " 0 "" {MPLTEXT 1 0 23 "nullspace(A-2.0*Id(3));" }}{PARA 225 "" 0 "" {TEXT -1 26 "Fina lly, try this command:" }}{PARA 226 "> " 0 "" {MPLTEXT 1 0 14 "eigenve cts(A);" }}{PARA 227 "" 0 "" {TEXT -1 463 "Notice that this gives us a little more than we bargained for: actually, this is an eigensystem, i.e., a complete description of all eigenvalues and eigenvectors. Fo r example, the first row says that 3.0 is an eigenvalue of A with alg ebraic multiplicity (number of times it occurs as a root to the charac teristic polynomial) 1 and basis for the eigenspace corresponding to l ambda = 3.0 given by the vectors in the list, of which there is only o ne in this case." }}{PARA 228 "" 0 "" {TEXT -1 17 "2. Why Are They??" }}{PARA 229 "" 0 "" {TEXT -1 175 "This is a good question and the ans wer could go on and on. We will try to indicate their importance for \+ one special class of problems, namely, discrete dynamical systems.\n " }{TEXT -1 2 " \n" }{TEXT -1 151 "Here is how such a system is described: one starts with an initial state vector x0. Now to pass \+ from one state vector to the next, one uses the rule\n" }{TEXT -1 2 " \+ \n" }{TEXT -1 22 " x(n+1) = A x(n),\n" }{TEXT -1 6 " \n" }{TEXT -1 375 "where x(k) means the kth state vector. The matrix A is fixe d for the dynamical system. It is called the transition matrix of the system. You have already seen one such example, namely, Markov ch ains. There are others, from fields like biology and economics. Let 's try to understand how state vectors change in the case of the matri x A from the previous section." }}{PARA 230 "> " 0 "" {MPLTEXT 1 0 22 "x0 := vector([1,1,1]);" }}{PARA 231 "> " 0 "" {MPLTEXT 1 0 15 "evalm( A &* x0);" }}{PARA 232 "> " 0 "" {MPLTEXT 1 0 17 "evalm(A^2 &* x0);" } }{PARA 233 "> " 0 "" {MPLTEXT 1 0 17 "evalm(A^3 &* x0);" }}{PARA 234 " > " 0 "" {MPLTEXT 1 0 17 "evalm(A^4 &* x0);" }}{PARA 235 "" 0 "" {TEXT -1 59 "What do you predict will be the next state, approximately?\n" }{TEXT -1 1 "\n" }{TEXT -1 74 "What do you get if you divide the above vector by its largest component? \n" }{TEXT -1 1 "\n" }{TEXT -1 60 "N ow compare these to the eigenvectors and eigenvalues of A.\n" }{TEXT -1 1 "\n" }{TEXT -1 63 "Now do the whole sequence above with another r andom vector x0.\n" }{TEXT -1 1 "\n" }{TEXT -1 78 "Next try the same t hing with the matrix C. Repeat the above type of sequence." }}{PARA 236 "> " 0 "" {MPLTEXT 1 0 20 "x0 := vector([1,1]);" }}{PARA 237 "" 0 "" {TEXT -1 58 "What's going on here? Here is a start on an explanati on:\n" }{TEXT -1 1 "\n" }{TEXT -1 35 "Notice that x2 = A x1 = A A x 0,\n" }{TEXT -1 1 "\n" }{TEXT -1 38 " x3 = A x2 = A A \+ A x0, etc.\n" }{TEXT -1 1 "\n" }{TEXT -1 18 "See the pattern? \n" } {TEXT -1 18 " xn = An x0.\n" }{TEXT -1 5 " \n" }{TEXT -1 173 "So, to under stand how a dynamical system works, what we really need to know is ho w the powers of the transition matrix A behave. But in general, t his is very hard!!!\n" }{TEXT -1 1 "\n" }{TEXT -1 55 "Here is a case w e can handle: what if A is diagonal?" }}{PARA 238 "> " 0 "" {MPLTEXT 1 0 39 "D := matrix([[1,0,0],[0,2,0],[0,0,3]]);" }}{PARA 239 "> " 0 "" {MPLTEXT 1 0 11 "evalm(D^2);" }}{PARA 240 "> " 0 "" {MPLTEXT 1 0 11 "evalm(D^3);" }}{PARA 241 "" 0 "" {TEXT -1 46 "See what's goin g on? It's easy to show this\n" }{TEXT -1 1 "\n" }{TEXT -1 162 "FACT : For a (square) diagonal matrix D and positive integer n the mat rix Dn is computed from D by raising each diagonal entry of D to \+ the nth power. \n" }{TEXT -1 1 "\n" }{TEXT -1 183 "By the way, one \+ can use this fact to answer a question that we will be interested in i n the next section: When do the positive powers of D tend to a cons tant matrix? Answer....?\n" }{TEXT -1 1 "\n" }{TEXT -1 151 "Now for a more general A. Let's take a 3 \357\277\275 3 matrix A. What if we could find three linearly independent eigenvectors v1, v2, and v3? W e would have\n" }{TEXT -1 1 "\n" }{TEXT -1 18 " \+ A v1 = l1 v1\n" }{TEXT -1 18 " \+ A v2 = l2 v2\n" }{TEXT -1 19 " A v3 \+ = l3 v3,\n" }{TEXT -1 1 "\n" }{TEXT -1 23 " \+ l1 0 0 \n" }{TEXT -1 53 "Or: \+ A [v1, v2, v3] = [v1, v2, v3] . 0 l2 0 \n" }{TEXT -1 22 " 0 0 l3 \n" } {TEXT -1 1 "\n" }{TEXT -1 188 "Now set P = [v1, v2, v3] and set \+ D = DiagonalMatrix[l1,l2, l3]. Then P is invertible . (Why?) More over, the above equation, if multiplied on the left by P-1, gives the \+ equation\n" }{TEXT -1 1 "\n" }{TEXT -1 21 " \+ P-1 A P = D.\n" }{TEXT -1 1 "\n" }{TEXT -1 62 "Consider D D \+ = (P-1 A P ) (P-1 A P ) =P-1 A A P = P-1 A2 P.\n" }{TEXT -1 126 "In g eneral, the same holds for the nth powers of A and D. Finally, multi ply on the left and right by P and P-1 to obtain:\n" }{TEXT -1 1 "\n " }{TEXT -1 98 "KEY FACT: If there is a matrix P such that P-1 A \+ P = D, then for all positive integers n,\n" }{TEXT -1 1 "\n" }{TEXT -1 23 " An = P Dn P-1.\n" }{TEXT -1 1 "\n" }{TEXT -1 327 "Call a matrix A diagonalizable if there e xists an invertible matrix P such that P-1 A P is a diagonal matrix . What is the story here? Notice that all the steps we performed ar e reversible. So what we have really shown (for the case n = 3, b ut the same argument works in general) is the following important fact :\n" }{TEXT -1 1 "\n" }{TEXT -1 266 "KEY FACT (Diagonalization Theorem ): Let A be an n\357\277\275n matrix. Then A is diagonalizable if and only if there exists a basis of eigenvectors v1, v2, ... , vn of A (i.e. v1, v2, ... , vn are linearly independent) for Rn. M oreover, in this case the matrix \n" }{TEXT -1 127 "P = [v1, v2, ... , vn] is nonsingular and P-1 A P is a diagonal matrix with the eige nvalues of A along its main diagonal.\n" }{TEXT -1 1 "\n" }{TEXT -1 59 "Just as an aside, here is a very handy fact worth knowing:\n" } {TEXT -1 1 "\n" }{TEXT -1 139 "FACT: A set of eigenvectors of the m atrix A which correspond to distinct eigenvalues of A must necessa rily be linearly independent.\n" }{TEXT -1 1 "\n" }{TEXT -1 18 "Nice a pplication:\n" }{TEXT -1 1 "\n" }{TEXT -1 96 "APPLICATION: If the n\3 57\277\275n matrix A has n distinct eigenvalues, then A is dia gonalizable.\n" }{TEXT -1 5 " \n" } {TEXT -1 75 "Let's carry out this whole program on the matrix a of the previous section." }}{PARA 242 "> " 0 "" {MPLTEXT 1 0 31 "esys := arr ay([eigenvects(A)]);" }}{PARA 243 "> " 0 "" {MPLTEXT 1 0 41 "D := diag (esys[1,1],esys[2,1],esys[3,1]);" }}{PARA 244 "> " 0 "" {MPLTEXT 1 0 19 "v1 := esys[1,3][1];" }}{PARA 245 "> " 0 "" {MPLTEXT 1 0 42 "v2 := \+ evalm(esys[2,3][1]/esys[2,3][1][1]);" }}{PARA 246 "> " 0 "" {MPLTEXT 1 0 42 "v3 := evalm(esys[3,3][1]/esys[3,3][1][1]);" }}{PARA 247 "> " 0 "" {MPLTEXT 1 0 53 "P := map(round,evalm(transpose(matrix([v1,v2,v3]) )));" }}{PARA 248 "> " 0 "" {MPLTEXT 1 0 27 "evalm((1/P) &* A &* P - D );" }}{PARA 249 "> " 0 "" {MPLTEXT 1 0 31 "evalm(A^4 - P &* D^4 &* (1/ P));" }}{PARA 250 "" 0 "" {TEXT -1 405 "The behavior of powers of the \+ matrix C was rather strange. Can diagonalization explain it? Remembe r that the eigenvalues of C are complex. Nonetheless, everything we have said above works perfectly well for complex numbers. So C is di agonalizable to a matrix with diagonal entries I and -I. This explai ns why the powers of C bounce around but do not grow. Here is how t o get a formula for C^k:" }}{PARA 251 "> " 0 "" {MPLTEXT 1 0 38 "esys := evalf(array([eigenvects(C)]));" }}{PARA 252 "> " 0 "" {MPLTEXT 1 0 31 "D := diag(esys[1,1],esys[2,1]);" }}{PARA 253 "> " 0 "" {MPLTEXT 1 0 42 "v1 := evalm(esys[1,3][1]/esys[1,3][1][1]);" }}{PARA 254 "> " 0 "" {MPLTEXT 1 0 42 "v2 := evalm(esys[2,3][1]/esys[2,3][1][1]);" }} {PARA 255 "> " 0 "" {MPLTEXT 1 0 50 "P := map(round,evalm(transpose(ma trix([v1,v2]))));" }}{PARA 256 "> " 0 "" {MPLTEXT 1 0 27 "evalm((1/P) \+ &* C &* P - D);" }}{PARA 257 "> " 0 "" {MPLTEXT 1 0 31 "C4 := evalm(P \+ &* D^4 &* (1/P));" }}{PARA 258 "> " 0 "" {MPLTEXT 1 0 11 "evalm(C^4);" }}{PARA 259 "" 0 "" {TEXT -1 21 "3. Foxes and Chickens" }}{PARA 260 " " 0 "" {TEXT -1 492 "Yes, foxes and chickens!! Consider the followin g problem: a population of foxes and chickens enjoys a predator/prey \+ relationship which is modelled as a linear system. The coefficients t hat model the relationship are determined experimentally, and one para meter, the kill rate k (of chickens by foxes ) varies with different \+ populations and locales. If F_n and C_n are the fox and chicken popu lations at the beginning of year n, then the populations in the follow ing year are given by \n" }{TEXT -1 1 "\n" }{TEXT -1 31 " \+ F_\{n+1\} = 0.6*F_n + 0.5*C_n\n" }{TEXT -1 32 " C_ \{n+1\} = -k*F_n + 1.2*C_n\n" }{TEXT -1 3 " \n" } {TEXT -1 151 "Can you attach a meaning to each of the coefficients? \+ In matrix form we can write the column vector [F_i, C_i] as x(i) an d express the formula as\n" }{TEXT -1 1 "\n" }{TEXT -1 22 " \+ x(i+1) = A . x(i)\n" }{TEXT -1 1 "\n" }{TEXT -1 77 "where the m atrix A = [[0.6, 0.5], [-k, 1.2]]. Click the following cell:\n" } }{PARA 261 "> " 0 "" {MPLTEXT 1 0 42 "A := k -> matrix([[0.6, 0.5], [- k, 1.2]]);" }}{PARA 262 "" 0 "" {TEXT -1 216 "Now here is the question : for what values of the kill rate k will the fox and chicken popul ations die out? For what values will it grow without bound? Is there a value for which the populations will stabilize?\n" }{TEXT -1 1 "\n " }{TEXT -1 689 "Here is where Maple really shines as an experimental \+ tool! We recognize this as an example of a discrete linear dynamical \+ system; that is, the system is described at each discrete time step by a state vector, and one gets the subsequent state vector for the curr ent one by multiplying the current state by a fixed transition matrix. Later on, we will analyze this phenonemon thoroughly in terms of eig envectors and eigenvalues. For the time being, we are going to attack the problem directly with experimental calculations. Suppose we have the following initial population (state vector). Let's see what happ ens over a period of time with certain values of k. (Try k = 1.0, 0.5, 0.1)." }}{PARA 263 "> " 0 "" {MPLTEXT 1 0 14 "evalm(A(1.0));" }} {PARA 264 "> " 0 "" {MPLTEXT 1 0 25 "x := vector([100,10000]);" }} {PARA 265 "> " 0 "" {MPLTEXT 1 0 24 "x := evalm(A(1.0) &* x);" }} {PARA 266 "" 0 "" {TEXT -1 42 "Now let's introduce a function as follo ws:" }}{PARA 267 "> " 0 "" {MPLTEXT 1 0 43 "s := x -> norm([eigenvals( A(x))],infinity);" }}{PARA 268 "> " 0 "" {MPLTEXT 1 0 41 "plot([seq([i *0.01,s(i*0.01)],i=0..100)]);" }}{PARA 269 "" 0 "" {TEXT -1 13 "Conclu sions??" }}{PARA 270 "" 0 "" {TEXT -1 1 "\n" }{TEXT -1 9 "PROJECT:\n" }}{PARA 271 "" 0 "" {TEXT -1 216 "Read at least the Foxes and Chickens part of this notebook. Now here is a population problem that is simi lar to the Fox/Chicken model, except that the population growth matrix directly involves only one population:\n" }{TEXT -1 1 "\n" }{TEXT -1 830 "A species of bird has been transported into a new ecosystem (by a ccident or design). The birds can be divided into three age groups, s ay birds of age less than 2 years for group 1, birds of age between 2 \+ and 4 years for group 2, and birds of age between 4 and 6 years for t he third group. Although a few of the birds survive beyond the sixth year, they are relatively few in number, so we will assume that for a ll practical purposes, these birds have at most a 6 year life span. F rom empirical data, it is determined that in this new ecosystem the su rvival rates for group 1 and 2 birds are 50% and 25%, respectively. On the other hand, it is found that group 1 birds produce no offspring, \+ while group 2 birds average 4 offspring in any biennium (period of 2 y ears) , and birds in group 3 average 3 offspring in a biennium. \n" } {TEXT -1 1 "\n" }{TEXT -1 516 "Model this bird population as a discret e dynamical system, where a state vector (p1, p2, p3) for a given bi ennium means that there are p1 birds in group 1, p2 birds in group 2 a nd p3 birds in group 3. For example, notice that the number of group \+ 1 birds in a new biennium will be 4 times the number of group 2 birds \+ in the last biennium plus 3 times the number of group 3 birds in the l ast biennium. Groups 2 and 3 get their new populations from the survi ving birds of an earlier group in the previous biennium.\n" }{TEXT -1 1 "\n" }{TEXT -1 505 " Now apply the theory of eigenvectors/values dev eloped above to the transition matrix for this model to show that, as things stand, the population of these birds will grow without bound. \+ In the long run, how much do you expect the population to grow per ye ar? Finally, if you could modify the ecosystem in such a way as to ch ange the survival rate for group 1 birds, is there a survival rate tha t would stabilize the whole population? Answer the same question for \+ group 2 birds and for group 3 birds.\n" }{TEXT -1 1 "\n" }{TEXT -1 258 "Organize your solutions as though you were project manager for th e team of researchers who gathered and analyzed this data. (Pick a d epartment name, if you like. Maybe you work for the EPA.) This repor t goes to your boss, a reasonably informed audience.\n" }{TEXT -1 1 " \n" }{TEXT -1 201 "About project writing: You should use the Edit pro gram in Nextstep or a copy of this notebook to create your document. \+ Edit produces a more presentable document, but Maple is ok, if you are careful.\n" }{TEXT -1 1 "\n" }{TEXT -1 370 "For one thing, get rid of my comments and instructions. Put your own title in the notebook and be sure to write in the style of a paper: your project should have a \+ beginning, middle and end. Use simple sentences and avoid flowery phr ases or other such nonsense. Do NOT write the project for me, your in structor. Your target audience should be your fellow students. " }} {PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\"\"$F%E\\[l*6$F&F&$\"#S!\" \"6$F'F&!\"#6$\"\"#F'\"\"!6$F'F0F16$F0F0F&6$F'F'F&6$F0F&F.6$F&F'F&6$F& F0F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\"\"#F%E\\[l%6$F&F& \"\"$6$F'F'\"\"!6$F'F&!\"\"6$F&F'F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "= 6\"6$;\"\"\"\"\"#F%E\\[l%6$F&F&$!#?!\"\"6$F'F'F'6$F'F&\"\"&6$F&F'F," } }{PARA 11 "" 1 "" {XPPMATH 20 "6%$\"\"\"\"\"!$\"\"$F%$\"\"#F%" }} {PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"#\"\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$^$$\"\"!F%$\"\"\"F%^$F$$!\"\"F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "f*6#I\"nG6\"F%6$I)operatorGF%I&arrowGF%F%-_I'linalgG6$%*p rotectedGI(_syslibGF%I%diagGF%6#-I$seqGF-6$\"\"\"/I\"iGF%;F49$F%F%F%" }}{PARA 11 "" 1 "" {XPPMATH 20 ",*$\"#g!\"\"\"\"\"I'lambdaG6\"$!$5\"F% *$F'\"\"#F#*$F'\"\"$F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%$\"\"\"\"\"! $\"\"#F%$\"\"$F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "<#=6\"6#;\"\"\"\"\"$ E\\[l$\"\"\"$!+LLLLL!#5\"\"#$\"+nmmmm!#5\"\"$$\"+nmmmm!#5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%7%$\"+++++?!\"*\"\"\"<#=6\"6#;F'\"\"$E\\[l$F'$! +jN@99F&\"\"#$\"+CrUGGF&F-$\"+DrUGGF&7%$F'\"\"!F'<#=F*F+E\\[l$F'F8F1F' F-F87%$\"+++++IF&F'<#=F*F+E\\[l$F'$\"+7y1rq!#5F1$!+5y1rqFDF-$!+7y1rqFD " }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E\\[l$F&F&\"\"#F&F 'F&" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E\\[l$F&$\"#]! \"\"\"\"#F+F'F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E \\[l$F&$\"%+>!\"#\"\"#$!$5\"!\"\"F'F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E\\[l$F&$\"&+]'!\"$\"\"#$!%+\\!\"#F'F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E\\[l$F&$\"(++6#!\"%\"\"#$!'+!z \"!\"$F'F-" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"#E\\[l#F& F&F'F&" }}{PARA 8 "" 1 "" {TEXT 206 53 "Error, attempting to assign to `D` which is protected" }}{PARA 11 "" 1 "" {XPPMATH 20 "*$I\"DG6$%*pr otectedGI(_syslibG6\"\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "*$I\"DG6$% *protectedGI(_syslibG6\"\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$; \"\"\"\"\"$F%E\\[l*6$F&F&$\"+++++?!\"*6$F'F&$\"+++++IF,6$\"\"#F'<#=F#6 #F%E\\[l$F&\"\"!F1F&F'F66$F'F1F&6$F1F1F&6$F'F'<#=F#F4E\\[l$F&$\"+7y1rq !#5F1$!+5y1rqF?F'$!+7y1rqF?6$F1F&$F&F66$F&F'<#=F#F4E\\[l$F&$!+jN@99F,F 1$\"+CrUGGF,F'$\"+DrUGGF,6$F&F1F&" }}{PARA 8 "" 1 "" {TEXT 206 53 "Err or, attempting to assign to `D` which is protected" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E\\[l$F&$!+jN@99!\"*\"\"#$\"+CrUGGF+F' $\"+DrUGGF+" }}{PARA 8 "" 1 "" {TEXT 206 42 "Error, numeric exception: division by zero" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"$E \\[l$F&$\"+)*********!#5\"\"#$!+&*********F+F'$!+)*********F+" }} {PARA 8 "" 1 "" {TEXT 206 55 "Error, (in linalg:-transpose) unassigned matrix element" }}{PARA 8 "" 1 "" {TEXT 206 78 "Error, (in linalg:-mu ltiply) non matching dimensions for vector/matrix product" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\"\"$F%E\\[l*6$F&F&,&$\"(++Y\"!\"%F& -I#&*GF#6%=F#6$;F&\"\"#F3E\\[l%F)F&6$F4F4^$!\"#!\"\"6$F4F&^$F8F&6$F&F4 F&*$I\"DG6$%*protectedGI(_syslibGF#\"\"%=F#F2E\\[l%F)^$#F&F4F9F6^#FFF: ^$FFF&F<^##F9F4F96$F'F&$!'++8!\"$6$F4F'$!%+]F86$F'F4$\"\"!FTF6,&$F&FTF &F.F96$F'F',&$!%+\\F8F&F.F9F:FL6$F&F'$\"&+]'FNF6 $F'F&^$F-$\"+++++5F16$F&F)<#=F#F5E\\[l#F&^$F8$\"+++++SF:F'^$F-$!\"\"F. 6$F&F'F>" }}{PARA 8 "" 1 "" {TEXT 206 53 "Error, attempting to assign \+ to `D` which is protected" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\" \"\"\"#E\\[l#F&^$$\"+++++5!\"*$\"\"!F.F'^$$!+++++?F,F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"#E\\[l#F&^$$\"+++++5!\"*$\"\"!F.F'^ $$!+++++?F,$!+++++5F," }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\" \"#F%E\\[l%6$F&F&F&6$F'F'^$!\"#!\"\"6$F'F&^$F,F&6$F&F'F&" }}{PARA 11 " " 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\"\"#F%E\\[l%6$F&F&,&^$$\"\"!F-$!++++ +5!\"*F&I\"DG6$%*protectedGI(_syslibGF#!\"\"6$F'F',&^$F,$\"+++++5F0F&F 1F56$F'F&^$F,F,6$F&F'F<" }}{PARA 11 "" 1 "" {XPPMATH 20 "-I#&*G6\"6%=F $6$;\"\"\"\"\"#F(E\\[l%6$F)F)F)6$F*F*^$!\"#!\"\"6$F*F)^$F/F)6$F)F*F)*$ I\"DG6$%*protectedGI(_syslibGF$\"\"%=F$F'E\\[l%F,^$#F)F*F0F-^#F=F1^$F= F)F3^##F0F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\"\"#F%E\\[l %6$F&F&$\"&++\"!\"%6$F'F'$F&\"\"!6$F'F&$!\"!F/6$F&F'F1" }}{PARA 11 "" 1 "" {XPPMATH 20 "f*6#I\"kG6\"F%6$I)operatorGF%I&arrowGF%F%-_I'linalgG 6$%*protectedGI(_syslibGF%I'matrixGF,6#7$7$$\"\"'!\"\"$\"\"&F57$,$9$F5 $\"#7F5F%F%F%" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6$;\"\"\"\"\"#F%E \\[l%6$F&F&$\"\"'!\"\"6$F'F'$\"#7F,6$F'F&$!#5F,6$F&F'$\"\"&F," }} {PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"#E\\[l#F&\"$+\"F'\"&++ \"" }}{PARA 11 "" 1 "" {XPPMATH 20 "=6\"6#;\"\"\"\"\"#E\\[l#F&$\"&+1&! 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