0. Homework 0 answers 0.1. 0.2. 0.3. 0.4.MATH 56A: STOCHASTIC PROCESSESANSWERS TO HOMEWORK0. Homework 0 answers0.1. In the Kermack-McKendrick model, prove that the number of infected reaches itshighest point when the number of susceptibles is equal to the threshold (or at t = 0 at thebeginning of the recorded/modeled time period).This is easy. You just need to look at the notes to see what the equations are.The number of infected isy = N − x0e−z/ρ− zwhere N is the total population, x0is the initial number of infected, z is the number ofremoved and ρ is the threshold, a positive constant.Since d2y/dz2< 0, the infected curve is concave down and reaches its highest point eitherat one endpoint or at the point where dy/dz = 0. The epidemic cannot stop when y > 0because the infected are continually being removed. The numbers cannot stabilize untily = 0. So, the maximum is either at the beginning (at t = 0, z = 0) or when dy/dz = 0. Butdydz=x0ρe−z/ρ− 1 =xρ− 1since the number of susceptible isx = x0e−z/ρ.So, dy/dz = 0 if and only if x = ρ.If the maximum occurs at the beginning then the epidemic is in its final stages and themaximum actually occurred earlier before we were recording the data.0.2. Find all functions x(t), y(t) so thatx0(t) = 5x − y, y0(t) = 3x + yFind the particular solution with initial position (x0, y0) = (1, 3).The matrix isA =5 −13 1This has eigenvalues 4, 2 with corresponding eigenvectors X1=11, X2=13. So A =QDQ−1whereQ =1 11 3, D =4 00 2, Q−1=123 −1−1 1AndetA= QetDQ−1=121 11 3e4t00 e2t3 −1−1 1=123e4t− e2t−e4t+ e2t3e4t− 3e2t−e4t+ 3e2t1The general solution is X = etAX0orx =x023e4t− e2t+y02−e4t+ e2ty =x023e4t− 3e2t+y02−e4t+ 3e2tWhen x0= 1, y0= 3 then the e4tterms all cancel and we getx = e2t, y = 3e2t.0.3. Find all functions f from integers to complex numbers so thatf(n + 1) = 4f(n) − 5f(n − 1).Now find the solution when f(0) = f(1) = 2 and explain why it is real.To solve the homogeneous equation, try f = cn. If c is a double root then the secondsolution is f (n) = ncn. The equation isc2− 4c + 5 = 0This has two complex roots:c = 2 ± iSo, the two linearly independent complex solutions aref+(n) = (2 + i)nf−(n) = (2 − i)nThe general complex solution is therefore given byf(n) = a(2 + i)n+ b(2 − i)nwhere a, b ∈ C.Given the initial conditions f(0) = f (1) = 2 we get:f(0) = a + b = 2or: b = 2 − a.f(1) = a(2 + i) + b(2 − i) = a(2 + i) + (2 − a)(2 − i) = 2The solution is:a = 1 + i, b = 1 − iSo,f(n) = (1 + i)(2 + i)n+ (1 − i)(2 − i)n.This is a sequence of real number, in fact integers, since they are Gaussian integers which areequal to their own conjugates. (Gaussian integers are complex numbers of the form a + ibwhere a, b ∈ Z.)0.4. Find the function f(n), n = 0, 1, 2, 3, ··· so that f(0) = 0f(n) =13[f(n − 1) + f(n + 1) + f(n + 2)] , n ≥ 1limn→∞f(n) = 1.The general solutions are f(n) = cnwhere c is a solution ofc3+ c2− 3c + 1 = 0After factoring:c3+ c2− 3c + 1 = (c − 1)(c2+ 2c − 1).So, c = 1, −1 ±√2. The general solution is:f(n) = A + B(−1 +√2)n+ C(−1 −√2)nThe condition f (0) = 0 implies thatA + B + C = 0.Since limn→∞f(n) = 1 it also follows thatlimn→∞f(2n) = 1but,limn→∞(−1 −√2)2n= ∞since | − 1 −√2| > 1 andlimn→∞(−1 +√2)2n= 0since | − 1 +√2| < 1. Therefore, in order to have limn→∞f(n) = 1, we must have C = 0and A = 1. So, B = −1 andf(n) = 1 − (−1
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