MATH 56A: STOCHASTIC PROCESSES CHAPTER 0 50.3. systems of first order equations. I explained that differentialequations involving second and higher order derivatives can be reducedto a system of first order equations by introducing more variables. ThenI did the following example.y!= zz!= 6y − zIn matrix form this is:ddt!yz"=!0 16 −1"!yz"Which we can write as Y = AY withY =!yz", A =!0 16 −1"0.3.1. exponential of a matrix. The solution of this equations isY = etAY0where Y0=!y0z0"andetA:= I2+ tA +t2A22+t3A33!+ ···This works because the derivative of each term is A times the previousterm:ddttkAkk!=ktk−1Akk!=tk−1Ak(k −1)!= Atk−1Ak−1(k −1)!So,ddtetA= AetA0.3.2. diagonalization (corrected). Then I explained how to computertA. You have to diagonalize A. This meansA = QDQ−1where D is a diagonal matrix D =!d100 d2".I should have explained this formula so that I get it right: If X1, X2are eigenvalues of A with eigenvalues d1, d2then AX1= X1d1, AX2=X2d2andA(X1X2) = (X1d1X2d2) = (X1X2)!d100 d2"6 MATH 56A: STOCHASTIC PROCESSES CHAPTER 0Solve for A givesA = (X1X2)!d100 d2"(X1X2)−1= QDQ−1where Q = (X1, X2).This is a good idea because A2= QDQ−1QDQ−1= QD2Q−1andmore generally, tkAk= tkQDkQ−1. Divide by k! and sum over k toget:etA= QetDQ−1= Q!etd100 etd2"Q−10.3.3. eigenvectors and eigenvalues. The diagonal entries d1, d2are theeigenvalues of the matrix A and Q = (X1X2) where Xiis the eigenvec-tor corresponding to di. This works if the eigenvalues of A are distinct.The eigenvalues are defined to be the solutions of the equationdet(A − λI) = 0but there is a trick to use for 2 × 2 matrices. The determinant of amatrix is always the product of its eigenvalues:det A = d1d2= −6The trace (sum of diagonal entries) is equal to the sum of the eigenval-ues:trA = d1+ d2= −1So, d1= 2, d2= −3. The eigenvalues are X1=!12"and X2=!1−3".SoQ = (X1X2) =!1 12 −3"Q−1=1det Q!−3 −1−2 1"=!3/5 1/52/5 −1/5"The solution to the original equation is!yz"=!1 12 −3"!e2t00 e−3t"!3/5 1/52/5 −1/5"!y0z0"MATH 56A: STOCHASTIC PROCESSES CHAPTER 0 70.4. Linear difference equations. We are looking for a sequenceof numbers f(n) where n ranges over all the integers from K to N(K ≤ n ≤ N) so that(0.3) f(n) = af(n − 1) + bf(n + 1)I pointed out that the solution set is a vector space of dimension 2.So we just have to find two linearly independent solutions. Then Ifollowed the book.The solution has the form f(n) = cnwhere you have to solve for c:cn= acn−1+ bcn+1(0.4) bc2− c + a = 0c =1 ±√1 − 4ab2bThere were two cases.Case 1: (4ab %= 1) When the quadratic equation (0.4) has two rootsC1, c2then the linear combinations of cn1and cn2give all the solutionsof the homogeneous linear recursion (0.3).Case 2: (4ab = 1) In this case there is only one root c =12band thetwo independent solutions are f(n) = cnand ncn. The reason we geta factor of n is because when a linear equation has a double root thenthis root will also be a root of the derivative. This gives f (n) = ncn−1as a solution. But then you can multiply by the constant c since theequation is homogeneous.Example 0.4. (Fibonacci numbers) These are given by f(0) = 1, f(1) =1 and f(n + 1) = f(n) + f(n − 1) or:f(n) = f(n + 1) − f(n − 1)This is a = −1, b = 1. The roots of the quadratic equation are c =1±√52.So,f(n) =1√5#1 +√52$n−1√5#1 −√52$nThis is a rational number since it is Galois invariant (does not changeif you switch the sign of√5). However, it is not clear from the formulawhy it is an
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