9.6: Matrix Exponential, Repeated Eigenvalues. Ex.: A = x 1 (t) = e t 2 F.M.: If we set

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1 9.6: Matrix Exponential, Repeated Eigenvalues x Ax, A : n n (1) Def.: If x 1 (t),...,x n (t) is a fundamental set of solutions (F.S.S.) of (1), then X(t) x 1 (t),...,x n (t) (n n) is called a fundamental matrix (F.M.) for (1). General solution: (c c 1,..., c n T ) x(t) c 1 x 1 (t) c n x n (t) X(t)c Thm.: If X(t) is a F.M. for (1) and C is a constant nonsingular matrix, then X(t)C is also a F.M. Proof: Each column of X(t)C is a linear combination of the columns of X(t) and so is a solution of (1), and X(0)C is nonsingular. Ex.: A Eigenvalues and eigenvectors: λ 1 1 λ 2 2 v 1 2,3 T v 2 1,1 T F.S.S.: x 1 (t) e t 2,x 3 2 (t) e 2t 1 1 F.M.: 2e t X(t) 3e t e 2t e 2t If we set y 1 (t) 2x 2 (t), y 2 (t) 3x 2 (t), y 1 (t),y 2 (t) are also F.S.S. with F.M. 3e 2t 4e Y (t) t 3e 2t 6e t 2e t e 2t 0 2 3e t e 2t 3 0 1

2 Matrix Exponential Consider IVP: x Ex.: A Ax, x(0) x 0 (2) Solution of IVP: If X(t) is a F.M., the general solution is x(t) X(t)c Match c to IC: x(0) X(0)c x 0 c (X(0)) 1 x 0 x(t) X(t)(X(0)) 1 x 0 Def.: Given a F.M. X(t), then e At def X(t)(X(0)) 1 is the matrix exponential of At. Thm.: The solution of (2) is x(t) e At x e t e F.M.: X(t) 2t 3e t e 2t 2 1 X(0), (X(0)) e At X(t)(X(0)) 1 2e t e 2t 1 1 3e t e 2t 3 2 3e 2t 2e t 2e t 2e 2t 3e 2t 3e t 3e t 2e 2t IVP: x x, x(0) 2 1 Solution: 3e x(t) 2t 2e t 2e t 2e 2t 2 3e 2t 3e t 3e t 2e 2t 1 4e 2t 2e t 4e 2t 3e t 2

3 Properties of the Matrix Exponential Exponential series (A 0 I): e At m0 (At) m /m! Convergence for any matrix A d dt eat Ae At e At A If D d ij is a diagonal matrix (d ij 0 for i j), then e Dt is a diagonal matrix with entries e diit. Ex.: exp ( a 0 t ) e at 0 0 b 0 e bt Special case (d ii r): If AB BA, then e (A+B)t e At e Bt e Bt e At Note: If AB BA, then in general e (A+B)t e At e Bt e Bt e At e At is nonsingular, and (e At ) 1 e At If V is nonsingular, then e (V AV 1 )t V e At V 1 If v is an eigenvector for an eigenvalue λ, then e (ri)t e rt I e At v e λt v 3

4 Matrices with only one eigenvalue Thm.: If A has only one eigenvalue λ, then there is an integer k, 0 < k n, such that (A λi) k 0 Use this to compute e At as follows. Write A λi +(A λi). Then e At e (λi)t+(a λi)t e (λi)t e (A λi)t e λt e (A λi)t e k 1 λt j0 (A λi) j (t j /j!) only k terms of exponential series required Ex.: A 1 4 : T 2, D p(λ) λ 2 Tλ + D λ 2 + 2λ + 1 (λ + 1) 2 only one eigenvalue λ A + I 1 2 (A + I) (k 2) 0 0 e (A+I)t I + (A + I)t 1 + 2t 4t t 1 2t e At e t e (A+I)t e t 1 + 2t 4t t 1 2t 4

5 Generalized Eigenvectors and Associated Solutions If A has repeated eigenvalues, n linearly independent eigenvectors may not exist need generalized eigenvectors Def.: Let λ be eigenvalue of A. (a) The algebraic multiplicity, m, of λ is the multiplicity of λ as root of the characteristic polynomial (CN Sec. 9.5). (b) The geometric multiplicity, m g, of λ is dimnull(a λi). Need: m linearly independent solutions of x Ax associated with λ. If m g m m linearly independent eigenvector solutions. What if m g < m? Thm.: If λ is an eigenvalue with algebraic multiplicity m, then there is an integer k, 0 < k m, such that dimnull((a λi) k ) m dimnull((a λi) k 1 ) < m Def.: Any nonzero vector v in null((a λi) k ) is a generalized eigenvector for λ. Solution associated with v: e At v e (A λi) k v 0 k 1 λt j0 (t j /j!)(a λi) j v Thm.: Let v 1,...,v m be a basis of null(a λi) k. Then the x i (t) e k 1 λt j0 (t j /j!)(a λi) j v i, 1 i m, are m linearly independent solutions of x Ax. 5

6 2d Systems: (Sec. 9.2) { a b T a + d A c d D ad bc Assume T 2 4D 0 p(λ) (λ λ 1 ) 2, λ 1 T/2 (a) If A λ 1 I m g 2 x(t) e λ 1t x(0) (any vector is eigenvector) (b) If A λ 1 I m g 1: Compute eigenvector v Pick vector w that is not a multiple of v (A λ 1 I)w av for some a 0 (any w R 2 is generalized eigenvector) F.S.S.: x 1 (t) e λ 1t v x 2 (t) e λ 1t (w + avt) } Ex.: A 1 4 : T 2, D T 2 4D 0 eigenvalue λ A + I 1 2 eigenvector v 2,1 T Choose w 1,0 T (simple form) (A+I)w v F.S.S.: x 1 (t) e t v e t 2,1 T x 2 (t) e t (w vt) e t (1,0 T t 2,1 T ) e t 1 + 2t, t T Other Method: Compute (c.f. p.4) e At e t (I+(A+I)t) e t 1 + 2t 4t t 1 2t Columns of e At are also F.S.S. 6

7 Ex.: A p(λ) λ λ λ 1 λ λ ( 1 λ) ( 1 λ)(1 + λ)(3 + λ) + 1 (λ + 1)(λ 2 + 4λ + 4) (λ + 1)(λ + 2) 2 eigenvalues λ 1 1, m 1 1 λ 2 2, m 2 2 Compute A λ 1 I A + I: A+I / Set x 3 2 eigenvector v 1 1,1, 2 T Since m 1 1 one (eigenvector) solution: x 1 (t) e t 1,1, 2 T m 2 2 check A λ 2 I, (A λ 2 I) 2 : A+2I (A+2I) A+2I eigenvector v 2 1,0, 1 T eigenvector solution: x 2 (t) e 2t 1,0, 1 T For x 3 (t) use generalized eigenvector v 3 that is not an eigenvector. { Basis of null((a+2i) 2 u1 1,0,0 ): T u 2 0,0,1 T Note: v 2 u 1 u 2 (A + 2I)u 1 (A + 2I)u 2 v 3 can be any vector c 1 u 1 + c 2 u 2 s.t. (A+2I)(c 1 u 1 +c 2 u 2 ) (c 1 +c 2 )v 2 0 Choose v 3 u 2 0,0,1 T (text: u 1 ) x 3 (t) e 2t (Iv 3 + t(a + 2I)v 3 ) e 2t (v 3 + tv 2 ) e 2t t,0,1 t T 7

8 Ex.: A Matlab p(λ) ((λ + 1) 2 + 1) 2 single complex pair of eigenvalues λ i, λ 2 λ 1 (m 2). 1. Check B A λ 1 I A ( 1+i)I: 7 i i 2 0 B i i Matlab basis for null(b): v 1 2,0,4, 1 + i T Complex eigenvector solution: 2. Check B 2 (A λ 1 I) 2 : 2 14i 3 12i 2 + 6i 4i B 2 8i 2 + 6i 4i i 6 14i 6 + 6i 8i 2 2i i 2 + 2i Matlab basis for null(b 2 ): u 1 2,0,4, 1 + i T v 1 u 2 3 i,4,0, 2 + 2i T u 2 is generalized eigenvector that is not an eigenvector. Pick v 3 u 2 3 i,4,0, 2 + 2i T Need: Bv 3 2,0, 4,1 i T v 2 Complex solution associated with v 3 : z 2 (t) e ( 1+i)t (Iv 3 + tbv 3 ) e ( 1+i)t (v 3 tv 2 ) e ( 1+i)t 3 i 2t,4, 4t, 2 + 2i + (1 i)t T z 1 (t) e ( 1+i)t 2,0,4, 1 + i T 3. Take real and imaginary parts of z 1 (t) and z 2 (t) to obtain F.S.S: x 1 (t) Rez 1 (t) e t 2cos t,0,4cos t, cos t sin t T x 2 (t) Imz 1 (t) e t 2sin t,0,4sin t,cos t sin t T x 3 (t) Rez 2 (t) e t sin t (3 + 2t)cos t,4cos t, 4tcos t,(t 2)(cos t + sin t) T x 4 (t) Imz 2 (t) e t cos t (3 + 2t)sin t, 4sin t, 4tsin t,(t 2)(sin t cos t) T 8

9 Ex.: A Matlab p(λ) (λ + 1)(λ 1) 3 eigenvalues λ 1 1, m 1 1 λ 2 1, m 2 3 Find eigenvector for λ 1 : A + I Matlab eigenvector (basis vector for null(a + I)): v 1 1,0,2, 1 T Associated eigenvector solution: x 1 (t) e t 1,0,2, 1 T For λ 2 1 check powers of A I: B A I Matlab basis of null(b): 1,0,2,0 T v 2 v 3 1, 2,0,2 T Associated eigenvector solutions: x 2 (t) e t 1,0,2,0 T x 3 (t) e t 1, 2,0,2 T To find 4th solution check B 2 : B RREF(B 2 ) has only one nonzero row 1, 1, 1/2, 1/2. Construct basis of null(b 2 ) by setting x 2, x 3 0, x 4 2 u 1 1,0,0,2 T x 2, x 4 0, x 3 2 u 2 1,0,2,0 T x 3, x 4 0, x 2 1 u 3 1, 1,0,0 T Check which are not eigenvectors: Bu 1 2v 2, Bu 2 0, Bu 3 v 2 Can choose v 4 u 1 (simple). Associated solution: x 4 (t) e t (Iv 4 + tbv 4 ) e t (u 1 2tv 2 ) e t 1 2t,0, 4t,2 T 9

10 Advanced Theory: Chains of Generalized Eigenvectors Thm.: Let λ be an eigenvalue of a n n-matrix A with algebraic multiplicity m geometric multiplicity m g Let B A λi and k be s.t. dimnull(b k ) m dimnull(b k 1 ) < m There are m g chains of vectors v (i) 1,...,v(i) r i, 1 i m g s.t. r 1 + r r mg m, Bv (i) j+1 v(i) j, 1 j < r i Bv (i) 1 0 and all the vectors v (i) j basis of null(b k ). are a Computation of chains: Assume i 1 chains have been computed. Let q be the largest integer for which there is a vector v in null(b q ) s.t. B q 1 v 0, and v and all previously computed chain-vectors are linearly independent. Set r i q. Then the ith chain is computed as v v (i) r i v (i) j Bv (i) j+1 for r i > j 1 Note: Bv i 1 0 vi 1 Solutions of x Ax: x (i) j (t) e λt v (i) j + x (i) 1 (t) eλt v (i) 1 is eigenvector. j 1 (t l /l!)v (i) l if j > 1 l1 Single chain: If k m only one chain v 1,...,v m, v j Bv j+1 (j < m) 10

11 Ex.: A Matlab p(λ) (λ + 1) 3 (λ 2) eigenvalues λ 1 1, m 1 3 λ 4 2, m 4 1 Eigenvector for λ 4 : v 4 1,1,1,2 T x 4 (t) e 2t 1,1,1,2 T λ 1 1: Set B A + I. Matlab dimnull(b 2 ) 2 k 3 1 chain Matlab s null basis for null(b 3 ): 1,0,0,0 T u 1 u 2 u 3 0,0,1,0 T 0,1,0,1 T Find B 2 u 1 0 chain can be generated by u 1 (simple form). Set v 3 u 1 1,0,0,0 T Bv 3 4, 3,2, 3 T v 2 v 1 Bv 2 2,1, 2,1 T Solutions associated with 3-chain: x 1 (t) e t v 1 e t 2,1, 2,1 T x 2 (t) e t (v 2 + tv 1 ) e t 4 + t, 3,2, 3 T x 3 (t) e t (v 3 + tv 2 + t 2 v 1 /2) e t 1 + 4t t 2, 3t + t 2 /2, 2t t 2, 3t + t 2 /2 T Ex.: In example on p.8: m g 2, k 2, m 3 2 chains: u 1, 2v 2 is 2-chain v 3 is 1-chain Note: Approach via chains is useful if m >> m g, especially if m g 1 and k m >> 1. Summary: 1. For any matrix A, a F.S.S x 1 (t),...,x n (t) for x Ax can be computed using eigenvalues and (generalized) eigenvectors. 2. X(t) x 1 (t),...,x n (t) is a F.M: x(t) X(t)c is gen. sol. 3. M.E.: e At X(t)(X(0)) 1 x(t) e At x(0) 11

12 Eigenvalues & Eigenvectors eig(a): vector of eigenvalues of A V,Deig(A) outputs V: matrix of eigenvectors (columns) D: diagonal matrix of eigenvalues Symbolic Computation: >> A1 1;-1 1;V,Deig(sym(A)) V D 1, 1 1+i, 0 i, -i 0, 1-i >> Asym(-2 1-1;1-3 0;3-5 0); >> V,Deig(A) V D 1, -2-2, 0, 0 1, -1 0, -2, 0 1, 1 0, 0, -1 Numerical Computation: >> A-2 1-1;1-3 0;3-5 0; >> V,Deig(A) V D Note: no generalized eigenvectors Matlab Tools Matrix Exponential expm(a): matrix exponential of A Symbolic Computation: >> A1 1;-1 1;syms t;expm(sym(a)*t) ans exp(t)*cos(t), exp(t)*sin(t) -exp(t)*sin(t), exp(t)*cos(t) Note: t must be declared symbolically >> Asym(-2 1-1;1-3 0;3-5 0);syms t; >> expm(a*t);ans(:,1) ans 3*exp(-2*t)-2*exp(-t)+2*t*exp(-2*t) 2*t*exp(-2*t)-exp(-t)+exp(-2*t) 2*t*exp(-2*t)+exp(-t)-exp(-2*t) In this example only first column of e At is displayed Numerical Computation: >> A-2 1-1;1-3 0;3-5 0;t2;expm(A*t) ans Use loop to compute solution array: >> tlinspace(0,1,20);x01;0;0;x ; >> for n1:20;xx expm(a*t(n))*x0;end First entry of solution can be plotted via >> plot(t,x(1,:)) 12

13 Worked Out Examples from Exercises Ex : Find general solution of y Ay for A T 4, D 4 T 2 4D single eigenvalue λ A 2I eigenvector v Pick w (A 2I)w F.S.S.: y 1 (t) e 2t v e 2t 1,1 T General solution: v y 2 (t) e 2t (w + tv) e 2t (1,0 T + t1,1 T ) e 2t 1 + t, t T y(t) c 1 y 1 (t) + c 2 y 2 (t) Y (t)c with F.M. Y (t) e 2t t 1 t Ex : Find solution of system of Ex. 31 with IC y(0) 2, 1 T 1 1 c1 2 1st method: Match c to IC: Y (0)c 1 0 c c c y(t) e 2t t 1 e 1 t 3 2t 2 + 3t 3t 1 2nd method: y(t) e At y(0) e 2t (I + (A 2I)t)y(0) 1 + t t e 2t t e t 1 t 1 2t 3t 1 13

14 2 4 Ex : Compute e A for A using the exponential series 1 2 A A m 0 if m > 1 e A 1 4 I + A cos t sin t Ex : Let A. Show that e 1 0 At sin t cos t { } cos t 1 t using 2 /2! + t 4 /4! t 6 /6! + sin t t t 3 /3! + t 5 /5! t 7. Compute: /7! + A I, A A( I) A, A 4 A 2 ( I) I A 4m I, A 4m+1 A, A 4m+2 I, A 4m+3 A for m 0,1,2,... e At (At) n /n! n0 m0 (4m)! + A t4m+1 (4m + 1)! I t 4m+2 (4m + 2)! A t4m+3 (4m + 3)! ) (I t4m I(1 t 2 /2! + t 4 /4! t 6 /6! + ) + A(t t 3 /3! + t 5 /5! t 7 /7! + ) cos t sin t I cos t + Asin t sin t cos t 14

15 Ex : Compute e At by diagonalizing A for A A upper triangular eigenvalues are diagonal entries: λ 1 2, λ A ( 2)I v ,0 T. A ( 1)I v ,1 T Set V v 1,v 1 V Verify V AV is diagonal: V AV D A V DV 1 e At V e Dt V e 2t e t 0 1 e 2t 6e t 1 6 e 2t 6e 0 e t t 6e 2t e t 2 1 Ex : Compute e At for A 1 0 C.P.: p(λ) 2 λ 1 ( 2 λ)( λ) + 1 λ 1 λ 2 + 2λ + 1 (λ + 1) eigenvalue λ 1. Set B A + I. Compute: 1 1 B e At e t (I+Bt) e t 1 t t t 1 + t 15

16 Ex : Compute e At for A C.P.: p(λ) 1 λ λ λ (λ + 1) 1 λ λ (λ + 1)(λ 1)(λ + 3) + 4 (λ + 1)(λ 2 + 2λ + 1) (λ + 1) 3 single eigenvalue λ 1. Set B A + I B B B 2 0 (k 2) e At e t (I + Bt) e t t 1 + 2t t 2t 4t 1 2t Note: That B 2 0 follows also directly from dimnull(b) 2 16

17 Ex : Compute e At for A C.P.: p(λ) 1 λ λ λ λ (1 λ) 4 single eigenvalue λ 1. Set B A I B e At e t (I + Bt) e t (1 λ) t 2t t 2t 1 1 λ λ λ

18 Ex : Do the 6 tasks below for A by hand Find eigenvalues: 2 λ 1 1 p(λ) 1 3 λ λ ( 1) λ + ( 1)2+2 ( 3 λ) 2 λ 1 3 λ ( λ 5) (λ + 3)(λ + 2)λ + 3 λ + 5 (λ + 3)(λ 2 + 2λ + 3) λ + 5 (λ 3 + 4λ 2 + 9λ + 9) (λ 3 + 5λ 2 + 8λ + 4) (λ + 1)(λ + 2) 2 eigenvalues λ 1 and λ 2 2. Find algebraic (m) and geometric (m g ) multiplicities for each eigenvalue: λ 1 m 1; λ 2 m 2 (from p(λ)) Find geometric multiplicities: λ 1: A + I m g 1 λ 2: A + 2I B m g 1 18

19 Ex continued 1 3. For each eigenvalue find smallest k s.t. dim null((a λi) k ) m For λ 1: k 1 (since m m g 1) For λ 2: B dimnull(b 2 ) 2 m. Since dimnull(b) 1 k 2 4. For each eigenvalue, find m linearly independent generalized eigenvectors. For λ 1: From RREF(A + I) eigenvector v 1 2, 1,1 T For λ 2: From RREF(B) eigenvector v 2 1,1,1 T ; (B A + 2I) m 2 need solution of B 2 v 0 that is not a multiple of v 2 Use RREF(B 2 ): set y 2 0, y 3 1 y 1 1 v 3 1,0,1 T is in null(b 2 ) Since v 2,v 3 are linearly independent, they are linearly independent generalized eigenvectors for λ Verify linear independence of all generalized eigenvectors from Set V v 1,v 2,v Compute determinant of V : det(v ) ( 1) 2+1 ( 1) ( 1) v 1,v 2,v 3 are linearly independent. 19

20 Ex continued 2 6. Find fundamental set of solutions for y Ay λ 1, v 1 eigenvector solution: y 1 (t) e t 2, 1,1 T λ 2, v 2 eigenvector solution: y 2 (t) e 2t 1,1,1 T λ 2, v 3 generalized eigenvector solution y 3 (t) e 2t (v 3 + tbv 3 ) Compute Bv v y 3 (t) e 2t (v 3 tv 2 ) e 2t ( 1,0,1 T t1,1,1 T ) e 2t 1 t, t,1 t T and y 1 (t),y 2 (t),y 3 (t) are F.S.S. for y Ay. 20

21 Ex : 6 tasks as in Ex. 26 for A Find eigenvalues: >> A ; ; ; ; ; ; λ 1,2 >> Asym(A);factor(poly(A)) ans both with (x-1)^3*(x-2)^3 m 3 2. Find algebraic (m) and geometric (m g ) multiplicities: From p(λ): m 3 for λ 1 and λ 2. Find m g : >> B1A-eye(6);null(B1) ans -1, 0, -2, -1, 1, 1 >> B2A-2*eye(6);null(B2) ans 1, 7, -6, 0, 0, 0 0, 3, -3, 0, 1, 0 0, -6, 5, 1, 0, 0 m g 1 for λ 1 m g 3 m for λ 2 3. For each eigenvalue find smallest k s.t. dim null((a λi) k ) m For λ 2 we are done: from 2. k 1. Check λ 1: >> null(b1^2) ans 1, -2, 0, -1, -3, 1 0, 1, 1, 1, 1, -1 >> null(b1^3) ans -1, 2, 1, 0, 0, 0-1, 3/2, 0, 0, 1, 0-2, 5/2, 0, -1, 0, 1 k 3 21

22 Ex continued 1 4. For each eigenvalue, find m linearly independent generalized eigenvectors. (a) For λ 2 we are done: m g m 3 every generalized eigenvector is in null(a 2I) and so is an eigenvector. Assign variables to the eigenvectors in Matlab and denote them by v 1,v 2,v 3 : >> null(b2);v1ans(:,1);v2ans(:,2);v3ans(:,3); >> v1 v2 v3 ans 0, 3, -3, 0, 1, 0 1, 7, -6, 0, 0, 0 0, -6, 5, 1, 0, 0 v 1 v 2 v 3 0,3, 3,0,1,0 T 1,7, 6,0,0,0 T 0, 6,5,1,0,0 T (b) For λ 1 we need a basis of null((a I) 3 ). In view of task 6 it makes sense to determine a chain of 3 generalized eigenvectors. Let s check the chains for each of the basis vectors: >> null(b1^3);u1ans(:,1);u2ans(:,2);u3ans(:,3); >> u1 B1*u1 B1^2*u1,u2 B1*u2 B1^2*u2,u3 B1*u3 B1^2*u3 ans -1,2, 1, 0,0,0-1,1,-1, 0,2,0-1,0,-2,-1,1,1 ans -1,3/2, 0, 0, 1, 0-1,1/2,-3/2,-1/2,3/2,1/2-1/2, 0, -1,-1/2,1/2,1/2 ans -2,5/2, 0, -1, 0, 1-1,3/2,-1/2, 1/2,5/2,-1/2-3/2, 0, -3,-3/2,3/2, 3/2 All three basis vectors generate full chains (this needs not to be the case in general). Let s choose the simplest chain which is the first. Assign names to chain vectors in Matlab; denote them v 6,v 5,v 4 : 22

23 Ex continued 2 >> v6u1;v5b1*v6;v4b1*v5; >> v6 v5 v4 ans -1, 2, 1, 0, 0, 0-1, 1, -1, 0, 2, 0-1, 0, -2, -1, 1, 1 v 6 v 5 v 4 1,2,1,0,0,0 T (A I)v 6 1,1, 1,0,2,0 T (A I)v 5 1,0, 2, 1,1,1 T 0 (A I)v 4 (v 4 is eigenvector) 5. Verify linear independence of all generalized eigenvectors from 4. >> det(v1 v2 v3 v4 v5 v6) ans -1 Since det(v 1,v 2,v 3,v 4,v 5,v 6 ) 1 0, the six generalized eigenvectors are linearly independent. 6. Grand Finale: Find fundamental set of solutions for y Ay Vectors v 1,v 2,v 3 are linearly independent eigenvectors for λ 2 three linearly independent eigenvector solutions: y 1 (t) e 2t v 1 e 2t 0,3, 3,0,1,0 T y 2 (t) e 2t v 2 e 2t 1,7, 6,0,0,0 T y 3 (t) e 2t v 3 e 2t 0, 6,5,1,0,0 T Solutions associated with chain v 4,v 5,v 6 : y 4 (t) e t v 4 e t 1,0, 2, 1,1,1 T y 5 (t) e t (v 5 + tv 4 ) e t 1 t,1, 1 2t, t,2 + t, t T y 6 (t) e t (v 6 + tv 5 + t 2 v 4 /2) e t 1 t t 2 /2,2 + t,1 t t 2, t 2 /2,2t + t 2 /2, t 2 /2 T Note: Ex require same tasks as Ex , except task 5. 23

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