Finite Mathematical Models

Abstract

Three finite mathematical models are examined. The first deals with the discrete-time rabbit reproduction population dynamics model posed by Leonardo of Pisa which gives rise to the finite difference equation the solution of which produces the Fibonacci sequence. In the chapter this is solved both directly in scalar form and by placing it in a system formulation that is then solved by a Jordan canonical form method. Two other methods of solution are introduced in the problems for the system formulation: Namely a Cayley-Hamilton Theorem approach and a direct eigenvalue-eigenvector expansion method. The second model deals with the minimum fraction of the popular vote that can elect the President of the United States posed by George Pólya. The 1960 and 1996 Presidential elections are examined in the chapter while the 2008 election is considered in a problem. The third model deals with the financial mathematics problem of the repayment of a loan or mortgage. A loan shark example with 100% interest rate per pay period is examined in the chapter and a similar one with only a 50% interest rate per period is examined in the last problem.

Problems

21.1.

Consider the Fibonacci rabbit reproduction example of Section 21.1 in its $2\times 2$ matrix form

$$\begin{aligned} X_{k+1} = AX_k = \left( \begin{array}{cc} a &{} b \\ c &{} d \end{array}\right) \left[ \begin{array}{c} x_k \\ y_k \end{array}\right] \, \Rightarrow \, X_k = A^kX_0 \text{ with } X_0 = \left[ \begin{array}{c} x_0 \\ y_0 \end{array}\right] \text{ for } k \ge 0, \end{aligned}$$

where $a = x_0 = 0$ and $b = c = d = y_0 = 1$. The purpose of this problem is two-fold: Namely, to develop the alternate Cayley–Hamilton Theorem approach for deducing the representation of $A^k$ and to solve the original problem by a direct matrix method.

• (a) The Cayley–Hamilton Theorem states that a matrix satisfies its own characteristic polynomial. Since for a general $2 \times 2$ matrix that characteristic polynomial is given by

  $$\begin{aligned} \det (A - \lambda I) = \left| \begin{array}{cc} a - \lambda &{} b \\ c &{} d - \lambda \end{array}\right| = \lambda ^2 - (a+d)\lambda + ad-bc = g(\lambda ), \end{aligned}$$

  this implies

  $$\begin{aligned} g(A) = A^2 - (a+b)A + (ad-bc)I = \left( \begin{array}{cc} 0 &{} 0 \\ 0 &{} 0 \end{array}\right) \text{ where } A^2 = \left( \begin{array}{cc} a^2 + bc &{} ab + bd \\ ca + dc &{} cb + d^2 \end{array}\right) . \end{aligned}$$

  Show that result by direct computation.

• (b) The division algorithm for polynomials states that given another polynomial $p(\lambda ) = \lambda ^k$ with $\text{ deg }(\, p) = k > \text{ deg }(g) = 2$ there exists polynomials $q(\lambda )$ and $r(\lambda )$ such that

  $$\begin{aligned} p(\lambda ) = \lambda ^k = q(\lambda )g(\lambda ) + r(\lambda ) \text{ where } \text{ deg }(r) < \text{ deg }(g). \end{aligned}$$

  Hence, conclude that $r(\lambda )$ is of the form

  $$\begin{aligned} r(\lambda ) = \alpha + \beta \lambda . \end{aligned}$$

• (c) Given that $\lambda _{1,2} = (1\pm \sqrt{5})/2$, the eigenvalues of A, satisfy $g(\lambda _{1,2}) = 0$, deduce from part (b)

  $$\begin{aligned} \lambda _1^k = \alpha + \beta \lambda _1 \text{ and } \lambda _2^k = \alpha + \beta \lambda _2. \end{aligned}$$

• (d) Solving these equations of part (c) simultaneously determines that

  $$\begin{aligned} \alpha = \frac{\lambda _2\lambda _1^k - \lambda _1\lambda _2^k}{\lambda _2 - \lambda _1} \text{ and } \beta = \frac{\lambda _1^k - \lambda _2^k}{\lambda _1 - \lambda _2}. \end{aligned}$$

• (e) Finally, invoking the Cayley–Hamilton Theorem, obtain that

  $$\begin{aligned} p(A) = A^k = q(A)g(A) + r(A) = r(A) = \alpha I + \beta A = \left( \begin{array}{cc} \alpha &{} \beta \\ \beta &{} \alpha + \beta \end{array}\right) \text{ since } g(A) = \left( \begin{array}{cc} 0 &{} 0 \\ 0 &{} 0 \end{array}\right) , \end{aligned}$$

  as was to be demonstrated.

• (f) We now wish to use a direct matrix method analogous to the direct scalar method employed in Section 21.1 to solve the original Fibonacci matrix difference equation problem. That is, consider

  $$\begin{aligned} X_{k+1} = AX_k = \left( \begin{array}{cc} 0 &{} 1 \\ 1 &{} 1 \end{array}\right) \left[ \begin{array}{c} x_{k} \\ x_{k+1} \end{array}\right] \text{ for } k \ge 0 \text{ with } X_0 = \left[ \begin{array}{c} 0 \\ 1 \end{array}\right] ; \end{aligned}$$

  and, analogous to the method for solving a first-order matrix system of differential equations, show that seeking a vector solution of it of the form

  $$\begin{aligned} X_k = Z\lambda ^k \text{ where } Z = \left[ \begin{array}{c} z_1 \\ z_2 \end{array}\right] \end{aligned}$$

  yields the eigenvalue problem

  $$\begin{aligned} AZ = \lambda Z. \end{aligned}$$

• (g) Recalling from the eigenvalues and eigenvectors computed in Section 21.1 for A that

  $$\begin{aligned} \lambda = \lambda _{1,2} = \frac{1}{2}(1 \pm \sqrt{5}) \text{ and } Z = Z^{(1,2)} = \left[ \begin{array}{c} 1 \\ \lambda _{1,2} \end{array}\right] , \end{aligned}$$

  conclude from part (f) that this difference equation admits the solutions

  $$\begin{aligned} X_k^{(1)} = Z^{(1)}\lambda _1^k = \left[ \begin{array}{c} 1 \\ \lambda _1 \end{array}\right] \lambda _1^k \text{ and } X_k^{(2)} = Z^{(2)}\lambda _2^k =\left[ \begin{array}{c} 1 \\ \lambda _2 \end{array}\right] \lambda _2^k. \end{aligned}$$

• (h) Hence deduce from part (g) that this difference equation has a general solution composed of the linear combination

  $$\begin{aligned} X_k = c_1X_k^{(1)} + c_2 X_k^{(2)} = c_1\left[ \begin{array}{c} 1 \\ \lambda _1 \end{array}\right] \lambda _1^k + c_2 \left[ \begin{array}{c} 1 \\ \lambda _2 \end{array}\right] \lambda _2^k. \end{aligned}$$

• (i) Applying the initial condition to the general solution of part (h), obtain that

  $$\begin{aligned} X_0 = c_1\left[ \begin{array}{c} 1 \\ \lambda _1 \end{array}\right] + c_2\left[ \begin{array}{c} 1 \\ \lambda _2 \end{array}\right] = \left[ \begin{array}{c} 0 \\ 1 \end{array}\right] \, \Rightarrow \, c_1 + c_2 = 0 \\ \text{ and } \lambda _1c_1 + \lambda _2 c_2 = 1 \, \Rightarrow \, c_1 = -c_2 = \frac{1}{\lambda _1 - \lambda _2}. \end{aligned}$$

• ( j) Finally synthesizing the results of parts (h) and (j) again find that

  $$\begin{aligned} x_k = \frac{\lambda _1^k - \lambda _2^k}{\lambda _1 - \lambda _2} = \frac{(1 + \sqrt{5})^k - (1 - \sqrt{5})^k}{2^k\sqrt{5}} \end{aligned}$$

  and (equivalently)

  $$\begin{aligned} x_{k+1} = \frac{\lambda _1^{k+1} - \lambda _2^{k+1}}{\lambda _1 - \lambda _2} = \frac{(1 + \sqrt{5})^{k+1} - (1 - \sqrt{5})^{k+1}}{2^{k+1}\sqrt{5}}, \end{aligned}$$

  as was to be demonstrated.

21.2.

The purpose of this problem is to find the minimum fraction of popular votes necessary to elect a President of the US under the conditions prevailing in the 2008 election. The states and their corresponding electoral votes are listed in Table 21.5. Proceed in a manner similar to that developed in section 2 for the 1960 election by employing the same model and assumptions. First assume that D.C.’s electoral votes consist of 1 “virtual” representative r and 2 “virtual” senators just as all the other states that have an electoral vote $e = r + 2$. Then consider a second case in which D.C.’s electoral votes are assumed to consist of 3 “virtual” representatives but no “virtual” senators. Since there are 538 total electoral votes, a candidate must receive at least 270 of them to win in a two-candidate election. Note that for the second case the subcases in which D.C. is included as a winning “state” and the one in which it is not must be treated separately. Further take the total number of votes cast in this election to be 131.4 million.

Table 21.5 The states and their electoral votes $e = r + 2$ for the 2008 Presidential election.

Hints: Assume the winning candidate carries s states. Consider two cases: (1) in which D.C.’s electoral votes consist of 1 “virtual” representative and 2 “virtual” senators just the same as all the other states with $e = r + 2$; (2) in which D.C.’s electoral votes $e_k$ are assumed to consist of 3 “virtual” representatives $r_k$ but no “virtual” senators. The latter case has subcases (2a) when D.C. is not included as a winning “state” and (2b) when it is. The number of voters in each state is proportional to the $r_i$ value for that state with proportionality constant $N_j$, a large positive even integer, where $j=1$ and 2, for the two cases respectively. Let W denote the number of popular votes obtained by the winning candidate and T, the total number of votes cast in the election.

Then

$$\begin{aligned} W&\ge {\left\{ \begin{array}{ll} \displaystyle \sum _{i=1}^{s}\left( \frac{r_iN_1}{2} + 1\right) = \frac{N_1}{2}\sum _{i=1}^{s}r_i + s &{} \text{ for } \text{ Case } \text{(1) } \\ \displaystyle \sum _{i=1}^{s}\left( \frac{r_iN_2}{2} + 1 \right) = \frac{N_2}{2}\sum _{i=1}^{s}r_i + s &{} \text{ for } \text{ Case } \text{(2) } \end{array}\right. },\\&T = {\left\{ \begin{array}{ll} 436N_1 &{} \text{ for } \text{ Case } \text{(1) } \\ 438N_2 &{} \text{ for } \text{ Case } \text{(2) } \end{array}\right. } \approx 131.4 \text{ million. } \end{aligned}$$

Cases (1) & (2a):

$$\begin{aligned} \sum _{i=1}^{s}e_i = \sum _{i=1}^{s}(r_i + 2) = \sum _{i=1}^{s}r_i + 2s \ge 270 \, \Rightarrow \, \sum _{i=1}^{s}r_i \ge 270 - 2s; \end{aligned}$$

$$\begin{aligned} \Rightarrow \, \frac{W}{T} \ge {\left\{ \begin{array}{ll} \displaystyle \frac{1}{872}\sum _{i=1}^{s}r_i + \frac{s}{436 N_1} &{} \text{ for } \text{ Case } \text{(1) } \\ \displaystyle \frac{1}{876}\sum _{i=1}^{s}r_i + \frac{s}{438 N_2} &{} \text{ for } \text{ Case } \text{(2a) } \end{array}\right. } \ge {\left\{ \begin{array}{ll} \displaystyle \frac{270-2s}{872} + \frac{s}{436N_1} &{} \text{ for } \text{ Case } \text{(1) } \\ \displaystyle \frac{270-2s}{876} + \frac{s}{438N_2} &{} \text{ for } \text{ Case } \text{(2a) } \end{array}\right. }. \end{aligned}$$

Case (2b):

$$\begin{aligned} \sum _{i=1}^{s}e_i = \sum _{i=1}^{k-1}(r_i + 2) + r_k + \sum _{i=k+1}^{s}(r_i + 2) = \sum _{i=1}^{s}r_1 + 2(s-1) \ge 270 \, \Rightarrow \, \sum _{i=1}^{s}r_i \ge 272 - 2s; \end{aligned}$$

$$\begin{aligned} \Rightarrow \, \frac{W}{T} \ge \frac{1}{876} \sum _{i=1}^{s}r_i + \frac{s}{438 N_2} \ge \frac{272 - 2s}{876} + \frac{s}{438 N_2}. \end{aligned}$$

21.3.

The purpose of this problem is to consider a loan shark example in which $1300 is borrowed and repaid over a 4 month period with a payment each month consistent with an interest rate of 200% per the whole period.

• (a) Conclude that then

  $$\begin{aligned} A = 1300,\, n = 4, \text{ and } r = \frac{2}{4} = \frac{1}{2}. \end{aligned}$$

• (b) Show that under the conditions of part (a) the monthly mortgage payment $P = 810$.

• (c) Now construct a table analogous to Table 21.4 for this situation.

• (d) Finally demonstrate that $V_4 = C_4 = 6,581.25$.