From cd7142bb314fb606a49adec7eacac32838117cdc Mon Sep 17 00:00:00 2001 From: tomit4 Date: Tue, 14 Jul 2026 05:44:22 -0700 Subject: [PATCH] :construction: Fin 5.8 --- appendix_b.txt | 2 +- chapter_5/exercises.md | 1213 ++++++++++++++++++++++++++++++++++++ chapter_5/test_yourself.md | 10 + leftoff.txt | 2 +- 4 files changed, 1225 insertions(+), 2 deletions(-) diff --git a/appendix_b.txt b/appendix_b.txt index be1c55a..8a8022a 100644 --- a/appendix_b.txt +++ b/appendix_b.txt @@ -1 +1 @@ -952 +954 diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index 3c110d5..c6da977 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -10902,31 +10902,60 @@ Page 385 a. $a_k = 2a_{k - 1} - 5a_{k - 2}$ +Yes, $A = 2$ and $B = -5$ + b. $b_k = kb_{k - 1} + b_{k - 2}$ +No, $A = k$, which is not a constant coefficient. + c. $c_k = 3c_{k - 1} \cdot c_{k - 2}^2$ +No, this does not take the form of a linear homogeneous recurrence relation with +constant coefficients. The first term $3c_{k - 1}$ is fine, but then there is +multiplication instead of addition/subtraction, and $c_{k - 2}^2$ violates the +homogeneous rule (each term must have the same total degree). + d. $d_k = 3d_{k - 1} + d_{k - 2}$ +Yes, $A = 3$, $B = 1$. + e. $r_k = r_{k - 1} - r_{k - 2} - 2$ +No, this violates the "second-order" definition, as $r_k$ must only contain the +two previous terms $r_{k - 1}$ and $r_{k - 2}$, the $-2$ at the end of the +equation violates this definition. + f. $s_k = 10s_{k - 2}$ +Yes. + 2. Which of the following are second-order linear homogeneous recurrence relations with constant coefficients? a. $a_k = (k - 1)a_{k - 1} + 2ka_{k - 2}$ +No. + b. $b_k = -b_{k - 1} + 7b_{k - 2}$ +Yes. + c. $c_k = 3c_{k - 1} + 1$ +No. + d. $d_k = 3d_{k - 1}^2 + d_{k - 2}$ +No. + e. $r_k = r_{k - 1} + 6r_{k - 3}$ +Yes. + f. $s_k = s_{k - 1} + 10s_{k - 2}$ +Yes. + 3. Let $a_0, a_1, a_2, \dots$ be the sequence defined by the explicit formula $$ a_n = C \cdot 2^n + D \quad \text{ for every integer } n \geq 0 $$ @@ -10935,8 +10964,92 @@ where $C$ and $D$ are real numbers. a. Find $C$ and $D$ so that $a_0 = 1$ and $a_1 = 3$. What is $a_2$ in this case? +$$ a_0 = 1 = C \cdot 2^0 + D = C(1) + D = C + D $$ + +$$ a_1 = 3 = C \cdot 2^1 + D = 2C + D $$ + +So we have: + +$$ C + D = 1 \quad \text{ and } 2C + D = 3 $$ + +Let's solve: + +$$ C = 1 - D $$ + +$$ 2(1 - D) + D = 3 $$ + +$$ 2 - 2D + D = 3 $$ + +$$ 2 - D = 3 $$ + +$$ 2 = 3 + D $$ + +$$ -1 = D $$ + +$$ C = 1 - (-1) $$ + +$$ C = 2 $$ + +So: + +$$ a_n = C \cdot 2^n + D $$ + +$$ a_n = 2 \cdot 2^n + (-1) $$ + +$$ a_n = 2^{n + 1} - 1 $$ + +And: + +$$ a_2 = 2^{2 + 1} - 1 $$ + +$$ a_2 = 2^{3} - 1 $$ + +$$ a_2 = 8 - 1 $$ + +$$ a_2 = 7 $$ + b. Find $C$ and $D$ so that $a_0 = 0$ and $a_1 = 2$. What is $a_2$ in this case? +$$ a_0 = 0 = C \cdot 2^0 + D = C + D $$ + +$$ a_1 = 2 = C \cdot 2^1 + D = 2C + D $$ + +So we have: + +$$ C + D = 0 \quad \text{ and } \quad 2C + D = 2 $$ + +Let's solve: + +$$ C = -D $$ + +$$ 2(-D) + D = 2 $$ + +$$ -2D + D = 2 $$ + +$$ -D = 2 $$ + +$$ D = -2 $$ + +$$ C = -(-2) $$ + +$$ C = 2 $$ + +Then: + +$$ a_n = 2 \cdot 2^n + (-2) $$ + +$$ a_n = 2^{n + 1} - 2 $$ + +Then: + +$$ a_2 = 2^{2 + 1} - 2 $$ + +$$ a_2 = 2^3 - 2 $$ + +$$ a_2 = 8 - 2 $$ + +$$ a_2 = 6 $$ + 4. Let $b_0, b_1, b_2, \dots$ be the sequence defined by the explicit formula $$ b_n = C \cdot 3^n + D(-2)^n \quad \text{ for each integer } n \geq 0 $$ @@ -10945,8 +11058,90 @@ where $C$ and $D$ are real numbers. a. Find $C$ and $D$ so that $b_0 = 0$ and $b_1 = 5$. What is $b_2$ in this case? +$$ b_0 = 0 = C \cdot 3^0 + D(-2)^0 = C + D $$ + +$$ b_1 = 5 = C \cdot 3^1 + D(-2)^1 = 3C - 2D $$ + +So we have: + +$$ C + D = 0 \quad \text{ and } \quad 3C - 2D = 5 $$ + +Let's solve: + +$$ C = -D $$ + +$$ 3(-D) - 2D = 5 $$ + +$$ -3D - 2D = 5 $$ + +$$ -5D = 5 $$ + +$$ D = -1 $$ + +$$ C = -(-1) $$ + +$$ C = 1 $$ + +So we have: + +$$ b_n = 1 \cdot 3^n + (-1)(-2)^n $$ + +$$ b_n = 3^n - (-2)^n $$ + +Then: + +$$ b_2 = 3^2 - (-2)^2 $$ + +$$ b_2 = 9 - (4) $$ + +$$ b_2 = 5 $$ + b. Find $C$ and $D$ so that $b_0 = 3$ and $b_1 = 4$. What is $b_2$ in this case? +$$ b_0 = 3 = C \cdot 3^0 + D(-2)^0 = C + D $$ + +$$ b_1 = 4 = C \cdot 3^1 + D(-2)^1 = 3C - 2D $$ + +So we have: + +$$ C + D = 3 \quad \text{ and } \quad 3C - 2D = 4 $$ + +Let's solve: + +$$ C = 3 - D $$ + +$$ 3(3 - D) - 2D = 4 $$ + +$$ 9 - 3D - 2D = 4 $$ + +$$ 9 - 5D = 4 $$ + +$$ 9 = 4 + 5D $$ + +$$ 5 = 5D $$ + +$$ 1 = D $$ + +$$ C = 3 - 1 $$ + +$$ C = 2 $$ + +Then we have: + +$$ b_n = 2 \cdot 3^n + (1)(-2)^n $$ + +$$ b_n = 2 \cdot 3^n + (-2)^n $$ + +Then we have: + +$$ b_2 = 2 \cdot 3^2 + (-2)^2 $$ + +$$ b_2 = 2 \cdot 9 + (4) $$ + +$$ b_2 = 18 + 4 $$ + +$$ b_2 = 22 $$ + 5. Let $a_0, a_1, a_2, \dots$ be the sequence defined by the explicit formula $$ a_n = C \cdot 2^n + D \quad \text{ for each integer } n \geq 0 $$ @@ -10955,6 +11150,52 @@ where $C$ and $D$ are real numbers. Show that for any choice of $C$ and $D$, $$ a_k = 3a_{k - 1} - 2a_{k - 2} \quad \text{ for every integer } k \geq 2 $$ +**Proof:** + +Let $a_0, a_1, a_2, \dots$ be the sequence defined by the explicit formula +$a_n = C \cdot 2^n + D$ for each integer $n \geq 0$, where $C$ and $D$ are real +numbers. + +Let $k$ be any integer such that $k \geq 2$. It follows that: + +$$ a_k = C \cdot 2^k + D $$ + +$$ a_{k - 1} = C \cdot 2^{k - 1} + D $$ + +$$ a_{k - 2} = C \cdot 2^{k - 2} + D $$ + +We must show that for any choice of $C$ and $D$, that: + +$$ a_k = 3a_{k - 1} - 2a_{k - 2} $$ + +By substitution: + +$$ = 3(C \cdot 2^{k - 1} + D) - 2(C \cdot 2^{k - 2} + D) $$ + +Then, by algebra: + +$$ = 3C \cdot 2^{k - 1} + 3D - 2C \cdot 2^{k - 2} - 2D $$ + +$$ = 3C \cdot 2^{k - 1} - 2C \cdot 2^{k - 2} + D $$ + +$$ = 3C \cdot 2^{k - 1} - C \cdot 2 \cdot 2^{k - 2} + D $$ + +$$ = 3C \cdot 2^{k - 1} - C \cdot 2^{k - 1} + D $$ + +$$ = 2C \cdot 2^{k - 1} + D $$ + +$$ = C \cdot 2 \cdot 2^{k - 1} + D $$ + +$$ = C \cdot 2^k + D $$ + +By the definition of the given equation: + +$$ = a_k $$ + +This is what was to be shown. + +Q.E.D. + 6. Let $b_0, b_1, b_2, \dots$ be the sequence defined by the explicit formula $$ b_n = C \cdot 3^n + D(-2)^n \quad \text{ for every integer } n \geq 0 $$ @@ -10963,10 +11204,136 @@ where $C$ and $D$ are real numbers. Show that for any choice of $C$ and $D$, $$ b_k = b_{k - 1} + 6b_{k - 2} \quad \text{ for each integer } k \geq 2 $$ +**Proof:** + +Let $b_0, b_1, b_2, \dots$ be the sequence defined by the explicit formula +$b_n = C \cdot 3^n + D(-2)^n$ for every integer $n \geq 0$, where $C$ and $D$ +are real numbers. + +Let $k$ be any integer where $k \geq 2$. It then follows that: + +$$ b_k = C \cdot 3^k + D(-2)^k $$ + +$$ b_{k - 1} = C \cdot 3^{k - 1} + D(-2)^{k - 1} $$ + +$$ b_{k - 2} = C \cdot 3^{k - 2} + D(-2)^{k - 2} $$ + +We must show that for any choice $C$ and $D$: + +$$ b_k = b_{k - 1} + 6b_{k - 2} $$ + +By substitution: + +$$ b_{k - 1} + 6b_{k - 2} = (C \cdot 3^{k - 1} + D(-2)^{k - 1}) + 6(C \cdot 3^{k - 2} + D(-2)^{k - 2}) $$ + +By algebra: + +$$ = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 6C \cdot 3^{k - 2} + 6D(-2)^{k - 2} $$ + +$$ = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 2C \cdot 3 \cdot 3^{k - 2} + (-3)D \cdot (-2) \cdot (-2)^{k - 2} $$ + +$$ = C \cdot 3^{k - 1} + D(-2)^{k - 1} + 2C \cdot 3^{k - 1} + (-3)D \cdot (-2)^{k - 1} $$ + +$$ = 3C \cdot 3^{k - 1} + (-2)D(-2)^{k - 1} $$ + +$$ = C \cdot 3 \cdot 3^{k - 1} + D \cdot (-2) \cdot (-2)^{k - 1} $$ + +$$ = C \cdot 3^k + D(-2)^k $$ + +$$ = b_k $$ + +This is what was to be shown. + +Q.E.D. + 7. Solve the system of equations in Example 5.8.4 to obtain $$ C = \frac{1 + \sqrt{5}}{2\sqrt{5}} \quad \text{ and } \quad D = \frac{-(1 - \sqrt{5})}{2\sqrt{5}} $$ +**Proof:** + +The initial conditions are: + +$$ F_0 = F_1 = 1 $$ + +5.8.4 has established that the Fibonacci relation is a second-order linear +homogeneous recurrence relation with constant coefficients. It also established +that the explicit formula for the Fibonacci sequence is: + +$$ F_n = C\left(\frac{1 + \sqrt{5}}{2}\right)^n + D\left(\frac{1 - \sqrt{5}}{2}\right)^n $$ + +It follows then that: + +$$ F_0 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right)^0 + D\left(\frac{1 - \sqrt{5}}{2}\right)^0 $$ + +$$ F_0 = 1 = C + D $$ + +$$ F_1 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right)^1 + D\left(\frac{1 - \sqrt{5}}{2}\right)^1 $$ + +$$ F_1 = 1 = C\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) $$ + +So we have: + +$$ C + D = 1 $$ + +$$ C\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) = 1 $$ + +Evaluating for both $C$ and $D$: + +$$ C = 1 - D $$ + +$$ (1 - D)\left(\frac{1 + \sqrt{5}}{2}\right) + D\left(\frac{1 - \sqrt{5}}{2}\right) = 1 $$ + +$$ (1 - D)\left(\frac{1 + \sqrt{5}}{2}\right) = 1 - D\left(\frac{1 - \sqrt{5}}{2}\right) $$ + +$$ \frac{(1 - D)(1 + \sqrt{5})}{2} = \frac{2}{2} - \left(\frac{D(1 - \sqrt{5})}{2}\right) $$ + +$$ \frac{(1 - D)(1 + \sqrt{5})}{2} = \left(\frac{2 - D(1 - \sqrt{5})}{2}\right) $$ + +$$ (1 - D)(1 + \sqrt{5}) = 2 - D(1 - \sqrt{5}) $$ + +$$ (1 \cdot 1) + (-D)(1) + (1)(\sqrt{5}) + (-D)(\sqrt{5}) = 2 - D(1 - \sqrt{5}) $$ + +$$ 1 - D + \sqrt{5} - D\sqrt{5} = 2 - D(1 - \sqrt{5}) $$ + +$$ (1 + \sqrt{5}) - D(1 + \sqrt{5}) = 2 - D(1 - \sqrt{5}) $$ + +$$ -D(1 + \sqrt{5}) + D(1 - \sqrt{5}) = 2 - (1 + \sqrt{5}) $$ + +$$ D(-1(1 + \sqrt{5}) + (1 - \sqrt{5})) = 2 - 1 - \sqrt{5} $$ + +$$ D((1 - \sqrt{5}) - (1 + \sqrt{5})) = 1 - \sqrt{5} $$ + +$$ D(1 - \sqrt{5} - 1 - \sqrt{5}) = 1 - \sqrt{5} $$ + +$$ D(-\sqrt{5} - \sqrt{5}) = 1 - \sqrt{5} $$ + +$$ D(-2\sqrt{5}) = 1 - \sqrt{5} $$ + +$$ D = \frac{1 - \sqrt{5}}{-2\sqrt{5}} $$ + +$$ D = \frac{\sqrt{5} - 1}{2\sqrt{5}} $$ + +$$ D = \frac{5 - \sqrt{5}}{2 \cdot 5} $$ + +$$ D = \frac{5 - \sqrt{5}}{10} $$ + +Then: + +$$ C = 1 - \left(\frac{5 - \sqrt{5}}{10}\right) $$ + +$$ C = \frac{10}{10} - \left(\frac{5 - \sqrt{5}}{10}\right) $$ + +$$ C = \frac{10 - (5 - \sqrt{5})}{10} $$ + +$$ C = \frac{10 - 5 + \sqrt{5}}{10} $$ + +$$ C = \frac{5 + \sqrt{5}}{10} $$ + +The closed form solution then is: + +$$ F_n = \left(\frac{5 + \sqrt{5}}{10}\right)\left(\frac{1 + \sqrt{5}}{2}\right)^n + \left(\frac{5 - \sqrt{5}}{10}\right)\left(\frac{1 - \sqrt{5}}{2}\right)^n $$ + In each of 8-10: (a) suppose a sequence of the form $1, t, t^2, t^3, \dots, t^n, \dots$ where $t \neq 0$, satisfies the given recurrence relation (but not necessarily the initial conditions), and find all @@ -10977,33 +11344,643 @@ the sequence. 8. $a_k = 2a_{k - 1} + 3a_{k - 2}$, for every integer $k \geq 2$ $a_0 = 1, a_1 = 2$ +a. + +Since the given second-order homogeneous recurrence relation with constant +coefficient, $a_k$, is satisfied by the sequence +$1, t, t^2, t^3, \dots t^n, \dots$. $t^k$ can be expressed as: + +$$ t^k = 2t^{k - 1} + 3t^{k - 2} $$ + +It follows that: + +$$ t^2 = 2t^{2 - 1} + 3t^{2 - 2} $$ + +$$ t^2 = 2t^1 + 3t^0 $$ + +$$ t^2 = 2t + 3 $$ + +By algebra then: + +$$ t^2 - 2t - 3 = 0 $$ + +$$ (t - 3)(t + 1) = 0 $$ + +Therefore the possible values of $t$ are: + +$$ t = -1 \quad \text{ and } \quad t = 3 $$ + +b. + +It follows from (a) and the distinct roots theorem that the explicit formula for +the given sequence follows the form of: + +$$ a_n = Cr^n + Ds^n $$ + +where $r$ and $s$ are the roots found in part (a): + +$$ a_n = C \cdot 3^n + D \cdot (-1)^n $$ + +for every integer $n \geq 0$. + +By the Distinct roots theorem we can find $C$ and $D$ by looking at the values +of $a_0$ and $a_1$ and evaluating for them: + +$$ a_0 = 1 = C \cdot 3^0 + D \cdot (-1)^0 = C + D $$ + +$$ a_1 = 2 = C \cdot 3^1 + D \cdot (-1)^1 = 3C - D $$ + +So we have: + +$$ C + D = 1 \quad \text{ and } \quad 3C - D = 2 $$ + +Evaluating: + +$$ C = 1 - D $$ + +$$ 3(1 - D) - D = 2 $$ + +$$ 3 - 3D - D = 2 $$ + +$$ 3 - 4D = 2 $$ + +$$ 3 = 2 + 4D $$ + +$$ 1 = 4D $$ + +$$ \frac{1}{4} = D $$ + +$$ C = 1 - \frac{1}{4} $$ + +$$ C = \frac{4}{4} - \frac{1}{4} $$ + +$$ C = \frac{3}{4} $$ + +So, our explicit formula for the given sequence is: + +$$ a_n = \frac{3}{4} \cdot 3^n + \frac{1}{4} \cdot (-1)^n $$ + 9. $b_k = 7b_{k - 1} - 10b_{k - 2}$, for every integer $k \geq 2$ $b_0 = 2, b_1, = 2$ +a. + +Since the given second-order homogeneous recurrence relation with constant +coefficient, $b_k$, is satisfied by the sequence +$1, t, t^2, t^3, \dots t^n, \dots$. $t^k$ can be expressed as: + +$$ t^k = 7t^{k - 1} - 10t^{k - 2} $$ + +It follows that: + +$$ t^2 = 7t^{2 - 1} - 10t^{2 - 2} $$ + +$$ t^2 = 7t^1 - 10t^0 $$ + +$$ t^2 = 7t - 10 $$ + +By algebra: + +$$ t^2 - 7t + 10 = 0 $$ + +$$ (t - 5)(t - 2) = 0 $$ + +The possible values of $t$ then are: + +$$ t = 2 \quad \text{ and } \quad t = 5 $$ + +b. + +It follows from (a) and the distinct roots theorem that the explicit formula for +the given sequence follows the form of: + +$$ b_n = Cr^n + Ds^n $$ + +where $r$ and $s$ are the roots found in part (a): + +$$ b_n = C \cdot 2^n + D \cdot 5^n $$ + +for every integer $n \geq 0$. + +By the Distinct roots theorem we can find $C$ and $D$ by looking at the values +of $b_0$ and $b_1$ and evaluating for them: + +$$ b_0 = 2 = C \cdot 2^0 + D \cdot 5^0 = C + D $$ + +$$ b_1 = 2 = C \cdot 2^1 + D \cdot 5^1 = 2C + 5D $$ + +So we have: + +$$ C + D = 2 \quad \text{ and } \quad 2C + 5D = 2 $$ + +Evaluating: + +$$ C = 2 - D $$ + +$$ 2(2 - D) + 5D = 2 $$ + +$$ 4 - 2D + 5D = 2 $$ + +$$ 4 + 3D = 2 $$ + +$$ 3D = 2 - 4 $$ + +$$ 3D = -2 $$ + +$$ D = -\frac{2}{3} $$ + +$$ C = 2 - \left(-\frac{2}{3}\right) $$ + +$$ C = \frac{6}{3} + \frac{2}{3} $$ + +$$ C = \frac{8}{3} $$ + +So the explicit formula for the given sequence is: + +$$ b_n = \frac{8}{3} \cdot 2^n + \left(-\frac{2}{3}\right) \cdot 5^n $$ + 10. $c_k = c_{k - 1} + 6c_{k - 2}$, for every integer $k \geq 2$ $c_0 = 0, c_1 = 3$ +a. + +Since the given second-order homogeneous recurrence relation with constant +coefficient, $c_k$, is satisfied by the sequence +$1, t, t^2, t^3, \dots t^n, \dots$. $t^k$ can be expressed as: + +$$ t^k = t^{k - 1} + 6t^{k - 2} $$ + +It follows that: + +$$ t^2 = t^{2 - 1} + 6t^{2 - 2} $$ + +$$ t^2 = t^1 + 6t^0 $$ + +$$ t^2 = t + 6 $$ + +By algebra: + +$$ t^2 - t - 6 = 0 $$ + +$$ (t - 3)(t + 2) = 0 $$ + +So the possible values of $t$ are: + +$$ t = -2 \quad \text{ and } \quad t = 3 $$ + +b. + +It follows from (a) and the distinct roots theorem that the explicit formula for +the given sequence follows the form of: + +$$ c_n = Cr^n + Ds^n $$ + +where $r$ and $s$ are the roots found in part (a): + +$$ c_n = C \cdot 3^n + D \cdot (-2)^n $$ + +for every integer $n \geq 0$. + +By the Distinct roots theorem we can find $C$ and $D$ by looking at the values +of $c_0$ and $c_1$ and evaluating for them: + +$$ c_0 = 0 = C \cdot 3^0 + D \cdot (-2)^0 = C + D $$ + +$$ c_1 = 3 = C \cdot 3^1 + D \cdot (-2)^1 = 3C - 2D $$ + +So we have: + +$$ C + D = 0 \quad \text{ and } \quad 3C - 2D = 3 $$ + +Evaluating: + +$$ C = -D $$ + +$$ 3(-D) - 2D = 3 $$ + +$$ -3D - 2D = 3 $$ + +$$ -5D = 3 $$ + +$$ D = -\frac{3}{5} $$ + +$$ C = -\left(-\frac{3}{5}\right) $$ + +$$ C = \frac{3}{5} $$ + +So our explicit formula for the given sequence is: + +$$ c_n = \frac{3}{5} \cdot 3^n + \left(-\frac{3}{5}\right) \cdot (-2)^n $$ + In each of 11-16 suppose a sequence satisfies the given recurrence relation and initial conditions. Find an explicit formula for the sequence. 11. $d_k = 4d_{k - 2}$ , for each integer $k \geq 2$ $d_0 = 1, d_1 = -1$ +A lot of this is similar to 8-10. Let's streamline it: + +$$ t^k = 4t^{k - 2} $$ + +$$ t^2 = 4t^{2 - 2} $$ + +$$ t^2 = 4t^0 $$ + +$$ t^2 = 4 $$ + +$$ t = \pm 2 $$ + +$$ d_n = C \cdot r^n + D \cdot s^n $$ + +$$ d_n = C \cdot 2^n + D \cdot (-2)^n $$ + +$$ d_0 = 1 = C \cdot 2^0 + D \cdot (-2)^0 = C + D $$ + +$$ d_1 = -1 = C \cdot 2^1 + D \cdot (-2)^1 = 2C - 2D $$ + +$$ C + D = 1 \quad \text{ and } \quad 2C - 2D = -1 $$ + +$$ C = 1 - D $$ + +$$ 2(1 - D) - 2D = -1 $$ + +$$ 2 - 2D - 2D = -1 $$ + +$$ 2 - 4D = -1 $$ + +$$ 2 = -1 + 4D $$ + +$$ 3 = 4D $$ + +$$ \frac{3}{4} = D $$ + +$$ C = 1 - \frac{3}{4} $$ + +$$ C = \frac{4}{4} - \frac{3}{4} $$ + +$$ C = \frac{1}{4} $$ + +$$ \boxed{d_n = \frac{1}{4} \cdot 2^n + \frac{3}{4} \cdot (-2)^n} $$ + 12. $e_k = 9e_{k - 1}$, for each integer $k \geq 2$ $e_0 = 0, e_1 = 2$ +$$ t^k = 9t^{k - 1} $$ + +$$ t^2 = 9t^{2 - 1} $$ + +$$ t^2 = 9t^1 $$ + +$$ t^2 = 9t $$ + +$$ t^2 - 9t = 0 $$ + +$$ t(t - 9) = 0 $$ + +This is a single root: $t = 9$. And so takes the form: + +$$ e_n = Cr^n + Dnr^n $$ + +$$ e_n = C \cdot 9^n + Dn \cdot 9^n $$ + +$$ e_0 = 0 = C \cdot 9^0 + D(0) \cdot 9^0 = C $$ + +$$ e_1 = 2 = C \cdot 9^1 + D(1) \cdot 9^1 = 9C + 9D $$ + +$$ C = 0 \quad \text{ and } 9C + 9D = 2 $$ + +$$ 9(0) + 9D = 2 $$ + +$$ 0 + 9D = 2 $$ + +$$ 9D = 2 $$ + +$$ D = \frac{2}{9} $$ + +$$ e_n = 0 \cdot 9^n + \left(\frac{2}{9}\right)n \cdot 9^n $$ + +$$ \boxed{e_n = \frac{2}{9}n \cdot 9^n} $$ + 13. $r_k = 2r^{k - 1} - r^{k - 2}$, for each integer $k \geq 2$ $r_0 = 1, r_1 = 4$ +$$ t^k = 2t^{k - 1} - t^{k - 2} $$ + +$$ t^2 = 2t^{2 - 1} - t^{2 - 2} $$ + +$$ t^2 = 2t^1 - t^0 $$ + +$$ t^2 = 2t - 1 $$ + +$$ t^2 - 2t + 1 = 0 $$ + +$$ (t - 1)(t - 1) = 0 $$ + +$$ t = 1 $$ + +$$ r_n = Cs^n + Dns^n $$ + +$$ r_n = C \cdot 1^n + D(n) \cdot 1^n $$ + +$$ r_0 = 1 = C \cdot 1^0 + D(0) \cdot 1^0 = C $$ + +$$ r_1 = 4 = C \cdot 1^1 + D(1) \cdot 1^1 = C + D $$ + +$$ C = 1 \quad \text{ and } \quad C + D = 4 $$ + +$$ 1 + D = 4 $$ + +$$ D = 3 $$ + +$$ r_n = 1 \cdot 1^n + 3(n) \cdot 1^n $$ + +$$ r_n = 1^n + 3n \cdot 1^n $$ + +$$ \boxed{r_n = 1 + 3n} $$ + 14. $s_k = -4s_{k - 1} - 4s_{k - 2}$, for every integer $k \geq 2$ $s_0 = 0, s_1 = -1$ +$$ t^k = -4t^{k - 1} - 4t^{k - 2} $$ + +$$ t^2 = -4t^{2 - 1} - 4t^{2 - 2} $$ + +$$ t^2 = -4t^1 - 4t^0 $$ + +$$ t^2 = -4t - 4 $$ + +$$ t^2 + 4t + 4 = 0 $$ + +$$ (t + 2)(t + 2) = 0 $$ + +$$ t = -2 $$ + +$$ s_n = Cr^n + Dnr^n $$ + +$$ s_n = C \cdot (-2)^n + Dn \cdot (-2)^n $$ + +$$ s_0 = 0 = C \cdot (-2)^0 + D(0) \cdot (-2)^0 = C $$ + +$$ s_1 = -1 = C \cdot (-2)^1 + D(1) \cdot (-2)^1 = -2C - 2D $$ + +$$ C = 0 \quad \text{ and } \quad -2C - 2D = -1 $$ + +$$ -2(0) - 2D = -1 $$ + +$$ -2D = -1 $$ + +$$ D = \frac{1}{2} $$ + +$$ s_n = 0 \cdot (-2)^n + \frac{1}{2}n \cdot (-2)^n $$ + +$$ \boxed{s_n = \frac{1}{2}n \cdot (-2)^n} $$ + 15. $t_k = 6t_{k - 1} - 9t_{k - 2}$, for each integer $k \geq 2$ $t_0 = 1, t_1= 3$ +$$ x^k = 6x^{k - 1} - 9x^{k - 2} $$ + +$$ x^2 = 6x^{2 - 1} - 9x^{2 - 2} $$ + +$$ x^2 = 6x^1 - 9x^0 $$ + +$$ x^2 = 6x - 9 $$ + +$$ x^2 - 6x + 9 = 0 $$ + +$$ (x - 3)(x - 3) = 0 $$ + +$$ x = 3 $$ + +$$ t_n = Cr^n + Dnr^n $$ + +$$ t_n = C \cdot 3^n + Dn \cdot 3^n $$ + +$$ t_0 = 1 = C \cdot 3^0 + D(0) \cdot 3^0 = C $$ + +$$ t_1 = 3 = C \cdot 3^1 + D(1) \cdot 3^1 = 3C + 3D $$ + +$$ C = 1 \quad \text{ and } \quad 3C + 3D = 3 $$ + +$$ 3(1) + 3D = 3 $$ + +$$ 3 + 3D = 3 $$ + +$$ 3D = 0 $$ + +$$ D = 0 $$ + +$$ t_n = 1 \cdot 3^n + (0)n \cdot 3^n $$ + +$$ \boxed{t_n = 3^n} $$ + 16. $s_k = 2s_{k - 1} + 2s_{k - 2}$, for every integer $k \geq 2$ $s_0 = 1, s_1 = 3$ +$$ t^k = 2t^{k - 1} + 2t^{k - 2} $$ + +$$ t^2 = 2t^{2 - 1} + 2t^{2 - 2} $$ + +$$ t^2 = 2t^1 + 2t^0 $$ + +$$ t^2 = 2t + 2 $$ + +$$ t^2 - 2t - 2 = 0 $$ + +$$ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} $$ + +$$ t = \frac{2 \pm \sqrt{12}}{2} $$ + +$$ t = \frac{2 \pm 2\sqrt{3}}{2} $$ + +$$ t = 1 \pm \sqrt{3} $$ + +$$ s_n = Cx^n + Dy^n $$ + +$$ s_n = C \cdot \left(1 + \sqrt{3}\right)^n + D \cdot \left(1 - \sqrt{3}\right)^n $$ + +$$ s_0 = 1 = C \cdot \left(1 + \sqrt{3}\right)^0 + D \cdot \left(1 - \sqrt{3}\right)^0 = C + D $$ + +$$ s_1 = 3 = C \cdot \left(1 + \sqrt{3}\right)^1 + D \cdot \left(1 - \sqrt{3}\right)^1 = C \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) $$ + +$$ C + D = 1 \quad \text{ and } \quad C \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) = 3 $$ + +$$ C = 1 - D $$ + +$$ (1 - D) \cdot \left(1 + \sqrt{3}\right) + D \cdot \left(1 - \sqrt{3}\right) = 3 $$ + +$$ (1 - D) \cdot \left(1 + \sqrt{3}\right) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ (1 - D) \cdot \left(1 + \sqrt{3}\right) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ (1)(1) + (-D)(1) + (1)(\sqrt{3}) + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ 1 + (-D)(1) + \sqrt{3} + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ (1 + \sqrt{3}) + (-D)(1) + (-D)(\sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ (1 + \sqrt{3}) + (-D)(1 + \sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ (1 + \sqrt{3}) + D(-1 - \sqrt{3}) = 3 - \left[D \cdot \left(1 - \sqrt{3}\right)\right] $$ + +$$ D(-1 - \sqrt{3}) + D \cdot \left(1 - \sqrt{3}\right) = 3 - (1 - \sqrt{3}) $$ + +$$ D\left[(-1 - \sqrt{3}) + \left(1 - \sqrt{3}\right)\right] = 3 - 1 + \sqrt{3} $$ + +$$ D\left[-1 - \sqrt{3} + 1 - \sqrt{3}\right] = 2 + \sqrt{3} $$ + +$$ D\left[-\sqrt{3} - \sqrt{3}\right] = 2 + \sqrt{3} $$ + +$$ D(-2\sqrt{3}) = 2 + \sqrt{3} $$ + +$$ D = \frac{2 + \sqrt{3}}{-2\sqrt{3}} $$ + +$$ D = \frac{2\sqrt{3} + 3}{-2 \cdot 3} $$ + +$$ D = \frac{2\sqrt{3} + 3}{-6} $$ + +$$ C = 1 - \left(\frac{2\sqrt{3} + 3}{-6}\right) $$ + +$$ C = \frac{-6}{-6} - \left(\frac{2\sqrt{3} + 3}{-6}\right) $$ + +$$ C = \frac{-6 - (2\sqrt{3} + 3)}{-6} $$ + +$$ C = \frac{-6 - 2\sqrt{3} - 3}{-6} $$ + +$$ C = \frac{-9 - 2\sqrt{3}}{-6} $$ + +$$ C = \frac{9 + 2\sqrt{3}}{6} $$ + +$$ s_n = \frac{9 + 2\sqrt{3}}{6} \cdot \left(1 + \sqrt{3}\right)^n + \frac{2\sqrt{3} + 3}{-6} \cdot \left(1 - \sqrt{3}\right)^n $$ + 17. Find an explicit formula for the sequence of exercise 39 in Section 5.6. +The recurrence relation found in exercise 39 is: + +$$ c_k = c_{k - 1} + c_{k - 2} $$ + +$$ t^k = t^{k - 1} + t^{k - 2} $$ + +$$ t^2 = t^{2 - 1} + t^{2 - 2} $$ + +$$ t^2 = t^1 + t^0 $$ + +$$ t^2 = t + 1 $$ + +$$ t^2 - t - 1 = 0 $$ + +$$ t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} $$ + +$$ t = \frac{1 \pm \sqrt{5}}{2} $$ + +$$ c_n = Cr^n + Ds^n $$ + +$$ c_n = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^n + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^n $$ + +$$ c_1 = 1 = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^1 + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^1 $$ + +$$ c_2 = 2 = C \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^2 + D \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^2 $$ + +Let $\phi = \dfrac{1 + \sqrt{5}}{2}$ and $\psi = \dfrac{1 - \sqrt{5}}{2}$. Then: + +$$ c_1 = 1 = C \cdot \phi + D \cdot \psi $$ + +$$ c_2 = 2 = C \cdot \phi^2 + D \cdot \psi^2 $$ + +$$ C \cdot \phi + D \cdot \psi = 1 $$ + +$$ C\phi = 1 - D\psi $$ + +$$ C = \frac{1 - D\psi}{\phi} $$ + +$$ C \cdot \phi^2 + D \cdot \psi^2 = 2 $$ + +$$ \frac{1 - D\psi}{\phi} \cdot \phi^2 + D \cdot \psi^2 = 2 $$ + +$$ (1 - D\psi) \cdot \phi + D \cdot \psi^2 = 2 $$ + +$$ \phi - D\psi\phi + D\psi^2 = 2 $$ + +$$ -D\psi\phi + D\psi^2 = 2 - \phi $$ + +$$ D(-\psi\phi + \psi^2) = 2 - \phi $$ + +$$ D = \frac{2 - \phi}{-\psi\phi + \psi^2} $$ + +$$ D = \frac{1}{\psi}\frac{2 - \phi}{-\phi + \psi} $$ + +Now back-substitution yields: + +$$ D = \frac{1}{\dfrac{1 - \sqrt{5}}{2}} \cdot \frac{2 - \dfrac{1 + \sqrt{5}}{2}}{-\dfrac{1 + \sqrt{5}}{2} + \left(\dfrac{1 - \sqrt{5}}{2}\right)} $$ + +$$ D = \frac{2}{1 - \sqrt{5}} \cdot \frac{\dfrac{4 - (1 + \sqrt{5})}{2}}{\dfrac{-(1 + \sqrt{5}) + (1 - \sqrt{5})}{2}} $$ + +$$ D = \frac{2}{1 - \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{-1 - \sqrt{5} + 1 - \sqrt{5}} $$ + +$$ D = \frac{2}{1 - \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{-2\sqrt{5}} $$ + +$$ D = \frac{2(3 - \sqrt{5})}{(1 - \sqrt{5})(-2\sqrt{5})} $$ + +$$ D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + (-2\sqrt{5})(-\sqrt{5})} $$ + +$$ D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + (2 \cdot 5)} $$ + +$$ D = \frac{6 - 2\sqrt{5}}{-2\sqrt{5} + 10} $$ + +$$ D = \frac{3 - \sqrt{5}}{-\sqrt{5} + 10} $$ + +$$ D = \frac{3 - \sqrt{5}}{10 - \sqrt{5}} $$ + +$$ D = \frac{3 - \sqrt{5}}{10 - \sqrt{5}} \cdot \frac{10 + \sqrt{5}}{10 + \sqrt{5}} $$ + +$$ D = \frac{(3 - \sqrt{5})(10 + \sqrt{5})}{(10 - \sqrt{5})(10 + \sqrt{5})} $$ + +$$ D = \frac{(3)(10) + (-\sqrt{5})(10) + (3)(\sqrt{5}) + (-\sqrt{5})(\sqrt{5})}{(10)(10) + (-\sqrt{5})(10) + (10)(\sqrt{5}) + (-\sqrt{5})(\sqrt{5})} $$ + +$$ D = \frac{30 - 10\sqrt{5} + 3\sqrt{5} - 5}{100 - 10\sqrt{5} + 10\sqrt{5} - 5} $$ + +$$ D = \frac{25 - 7\sqrt{5}}{95} $$ + +$$ C = \frac{1 - \left(\dfrac{25 - 7\sqrt{5}}{95}\right)\left(\dfrac{1 - \sqrt{5}}{2}\right)}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{1 - \left(\dfrac{(25 - 7\sqrt{5})(1 - \sqrt{5})}{190}\right)}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{\dfrac{190 - (25 - 7\sqrt{5})(1 - \sqrt{5})}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{\dfrac{190 - ((25)(1) + (-7\sqrt{5})(1) + (25)(-\sqrt{5}) + (-7\sqrt{5})(-\sqrt{5}))}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{\dfrac{190 - (25 - 7\sqrt{5} - 25\sqrt{5} + 35)}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{\dfrac{190 - (60 - 32\sqrt{5})}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{\dfrac{190 - 60 + 32\sqrt{5}}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{\dfrac{130 + 32\sqrt{5}}{190}}{\dfrac{1 + \sqrt{5}}{2}} $$ + +$$ C = \frac{130 + 32\sqrt{5}}{190} \cdot \frac{2}{1 + \sqrt{5}} $$ + +$$ C = \frac{(130 + 32\sqrt{5})(2)}{(190)(1 + \sqrt{5})} $$ + +$$ C = \frac{(130 + 32\sqrt{5})(2)}{2(95)(1 + \sqrt{5})} $$ + +$$ C = \frac{130 + 32\sqrt{5}}{95(1 + \sqrt{5})} $$ + +$$ C = \frac{130 + 32\sqrt{5}}{95 + 95\sqrt{5}} $$ + +$$ C = \frac{130 + 32\sqrt{5}}{95 + 95\sqrt{5}} \cdot \frac{1 - \sqrt{5}}{1 - \sqrt{5}} $$ + +$$ C = \frac{(130 + 32\sqrt{5})(1 - \sqrt{5})}{(95 + 95\sqrt{5})(1 - \sqrt{5})} $$ + +$$ C = \frac{(130)(1) + (32\sqrt{5})(1) + (130)(-\sqrt{5}) + (32\sqrt{5})(-\sqrt{5})}{(95)(1) + (95\sqrt{5})(1) + (95)(-\sqrt{5}) + (95\sqrt{5})(-\sqrt{5})} $$ + +$$ C = \frac{130 + 32\sqrt{5} - 130\sqrt{5} - 160}{95 + 95\sqrt{5} - 95\sqrt{5} - 475} $$ + +$$ C = \frac{98\sqrt{5} - 30}{-380} $$ + +$$ C = \frac{30 - 98\sqrt{5}}{380} $$ + +$$ C = \frac{15 - 49\sqrt{5}}{190} $$ + +$$ c_n = \frac{15 - 49\sqrt{5}}{190} \cdot \left(\frac{1 + \sqrt{5}}{2}\right)^n + \frac{25 - 7\sqrt{5}}{95} \cdot \left(\frac{1 - \sqrt{5}}{2}\right)^n $$ + 18. Suppose that the sequences $s_0, s_1, s_2, \dots$ and $t_0, t_1, t_2, \dots$ both satisfy the same second-order linear homogeneous recurrence relation with constant coefficients: @@ -11019,6 +11996,24 @@ $$ 2s_k + 3t_k = 5(2s_{k - 1} + 3t_{k - 1}) - 4(2s_{k - 2} + 3t_{k - 2}) $$ for each integer $k \geq 2$. Do _not_ use Lemma 5.8.2. +**Proof:** + +$$ 2s_k + 3t_k $$ + +By substitution of the given recurrence relations: + +$$ 2s_k + 3t_k = 2(5s_{k - 1} - 4s_{k - 2}) + 3(5t_{k - 1} - 4t_{k - 2}) $$ + +$$ = 10s_{k - 1} - 8s_{k - 2} + 15t_{k - 1} - 12t_{k - 2} $$ + +$$ = 10s_{k - 1} + 15t_{k - 1} - 8s_{k - 2} - 12t_{k - 2} $$ + +$$ = 5(2s_{k - 1} + 3t_{k - 1}) - 4(2s_{k - 2} + 3t_{k - 2}) $$ + +This is what was to be shown. + +Q.E.D. + 19. Show that if $r, s, a_0$, and $a_1$ are numbers with $r \neq s$, then there exist unique numbers $C$ and $D$ so that @@ -11026,6 +12021,47 @@ $$ C + D = a_0 $$ $$ Cr + Ds = a_1 $$ +**Proof:** + +Suppose $r$, $s$, $a_0$, and $a_1$ are numbers with $r \neq s$. + +Consider the system of equations: + +$$ C + D = a_0 $$ + +$$ Cr + Ds = a_1 $$ + +By solving for $D$ and substituting: + +$$ D = a_0 - C $$ + +$$ Cr + (a_0 - C)s = a_1 $$ + +$$ Cr + a_0s - Cs = a_1 $$ + +$$ Cr - Cs = a_1 - a_0s $$ + +$$ C(r - s) = a_1 - a_0s $$ + +Since $r \neq s$, both sides can be divided by $(r - s)$: + +$$ C = \frac{a_1 - a_0s}{r - s} $$ + +Then: + +$$ D = a_0 - \frac{a_1 - a_0s}{r - s} $$ + +$$ D = \frac{a_0(r - s) - (a_1 - a_0s)}{r - s} $$ + +$$ D = \frac{a_0r - a_0s - a_1 + a_0s}{r - s} $$ + +$$ D = \frac{a_0r - a_1}{r - s} $$ + +Since $r \neq s$, division by $r - s$ is valid, yielding a unique value for $C$, +which in turn yields a unique value for $D$. + +Q.E.D. + 20. Show that if $r$ is a nonzero real number, $k$ and $m$ are distinct integers, and $a_k$ and $a_m$ are any real numbers, then there exist unique real numbers $C$ and $D$ so that @@ -11034,9 +12070,176 @@ $$ Cr^k + kDr^k = a_k $$ $$ Cr^m + mDr^m = a_m $$ +**Proof:** + +Suppose $r$ is a nonzero real number, $k$ and $m$ are distinct integers, and +$a_k$ and $a_m$ are any real numbers. + +Consider the system of equations: + +$$ Cr^k + kDr^k = a_k $$ + +$$ Cr^m + mDr^m = a_m $$ + +By solving for $D$ and substituting: + +$$ kDr^k = a_k - Cr^k $$ + +Since $r$ is a nonzero real number, we can divide by $r^k$: + +$$ D = \frac{a_k - Cr^k}{kr^k} $$ + +Then: + +$$ Cr^m + m \cdot \frac{a_k - Cr^k}{kr^k} \cdot r^m = a_m $$ + +$$ Cr^m + \frac{mr^m(a_k - Cr^k)}{kr^k} = a_m $$ + +$$ \frac{Cr^m(kr^k) + mr^m(a_k - Cr^k)}{kr^k} = a_m $$ + +$$ \frac{Ckr^{k + m} + mr^ma_k - mr^mCr^k}{kr^k} = a_m $$ + +$$ \frac{Ckr^{k + m} - Cmr^mr^k + mr^ma_k}{kr^k} = a_m $$ + +$$ \frac{Ckr^{k + m} - Cmr^{k + m} + mr^ma_k}{kr^k} = a_m $$ + +$$ \frac{Cr^{k + m}(k - m) + mr^ma_k}{kr^k} = a_m $$ + +$$ Cr^{k + m}(k - m) + mr^ma_k = a_m(kr^k) $$ + +$$ Cr^{k + m}(k - m) = a_m(kr^k) - mr^ma_k $$ + +$$ C = \frac{a_m(kr^k) - mr^ma_k}{r^{k + m}(k - m)} $$ + +Now we solve for $D$: + +$$ D = \frac{a_k - \left(\frac{a_m(kr^k) - mr^ma_k}{r^{k + m}(k - m)}\right)r^k}{kr^k} $$ + +$$ D = \frac{a_k - \left(\frac{(a_m(kr^k) - mr^ma_k)r^k}{r^{k + m}(k - m)}\right)}{kr^k} $$ + +$$ D = \frac{a_k - \left(\frac{a_m(kr^k)r^k - mr^ma_kr^k}{r^{k + m}(k - m)}\right)}{kr^k} $$ + +$$ D = \frac{a_k - \left(\frac{a_mkr^{2k} - mr^{k + m}a_k}{r^{k + m}(k - m)}\right)}{kr^k} $$ + +$$ D = \frac{\frac{a_k(r^{k + m}(k - m)) - (a_mkr^{2k} - mr^{k + m}a_k)}{r^{k + m}(k - m)}}{kr^k} $$ + +$$ D = \frac{r^k(a_kr^k(k - m) - a_mkr^k + mr^ma_k)}{r^{k + m}(k - m)kr^k} $$ + +$$ D = \frac{a_kr^k(k - m) - a_mkr^k + mr^ma_k}{r^{k + m}(k - m)k} $$ + +$$ D = \frac{r^k(a_k(k - m) - a_mk) + mr^ma_k}{r^{k + m}(k - m)k} $$ + +$$ D = \frac{r^k(a_k(k - m) - a_mk)}{r^{k + m}(k - m)k} + \frac{mr^ma_k}{r^{k + m}(k - m)k} $$ + +$$ D = \frac{r^k(a_k(k - m) - a_mk)}{r^k(r^m(k - m)k)} + \frac{mr^ma_k}{r^m(r^k(k - m)k)} $$ + +$$ D = \frac{a_k(k - m) - a_mk}{r^m(k - m)k} + \frac{ma_k}{r^k(k - m)k} $$ + +$$ D = \frac{(a_k(k - m) - a_mk)(r^k) + (ma_k)(r^m)}{r^{m + k}(k - m)k} $$ + +$$ D = \frac{a_k(k - m)r^k - a_mk(r^k) + ma_kr^m}{r^{m + k}(k - m)k} $$ + +$$ D = \frac{a_k[(k - m)r^k + mr^m] - a_mk(r^k)}{r^{m + k}(k - m)k} $$ + +Since $r \neq 0$ and $k$ and $m$ are distinct integers, division by +$r^{m + k}(k - m)$ is valid, yielding a unique value for $C$, which in turn +yields a unique value for $D$. + +Q.E.D. + 21. Prove Theorem 5.8.5 for the case where the values of $C$ and $D$ are determined by $a_0$ and $a_1$. +**Theorem 5.8.5 Single-Root Theorem** + +Suppose a sequence $a_0, a_1, a_2, \dots$ satisfies a recurrence relation + +$$ a_k = Aa_{k - 1} + Ba_{k - 2} $$ + +for some real numbers $A$ and $B$ with $B \neq 0$ and for every integer +$k \geq 2$. If the characteristic equation $t^2 - At - B = 0$ has a single +(real) root $r$, then $a_0, a_1, a_2, \dots$ is given by the explicit formula + +$$ a_n = Cr^n + Dnr^n $$ + +where $C$ and $D$ are the real numbers whose values are determined by the values +of $a_0$ and any other known value of the sequence. + +**Proof (by strong mathematical induction):** + +Suppose a sequence $a_0, a_1, a_2, \dots$ satisfies a recurrence relation + +$$ a_k = Aa_{k - 1} + Ba_{k - 2} $$ + +for some real numbers $A$ and $B$ with $B \neq 0$ and for every integer +$k \geq 2$. Assume the characteristic equation $t^2 - At - B = 0$ has a single +(real) root $r$. + +It follows then that the sequence $a_0, a_1, a_2, \dots$ is given by the +explicit formula + +$$ a_n = Cr^n + Dnr^n $$ + +where $C$ and $D$ are the real numbers whose values are determined by the values +of $a_0$ and any other known value of the sequence. + +Let $P(n)$ be the recurrence relation: + +$$ a_n = Cr^n + Dnr^n $$ + +_Basis Step:_ + +Prove $P(0)$ and $P(1)$, that is: + +$$ a_0 = Cr^0 + D(0)r^0 = C $$ + +$$ a_1 = Cr^1 + D(1)r^1 = Cr + Dr $$ + +Therefore $P(0)$ and $P(1)$ are true by the assumption that $C$ and $D$ are +determined by the values of $a_0$ and $a_1$. + +_Inductive Step:_ + +Let $k$ be any integer where $k \geq 2$. Let $i$ be some integer such that +$0 \leq i \leq k$. + +Suppose $P(i)$, that is: + +$$ a_i = Cr^i + Dir^i $$ + +Prove $P(k + 1)$, that is: + +$$ a_{k + 1} = Cr^{k + 1} + D(k + 1)r^{k + 1} $$ + +By the definition of a recurrence relation: + +$$ a_{k + 1} = Aa_k + Ba_{k - 1} $$ + +By the inductive hypothesis: + +$$ = A(Cr^k + Dkr^k) + B(Cr^{k - 1} + D(k - 1)r^{k - 1}) $$ + +By algebra: + +$$ = ACr^k + ADkr^k + BCr^{k - 1} + BD(k - 1)r^{k - 1} $$ + +$$ = ACr^k + BCr^{k - 1} + ADkr^k + BD(k - 1)r^{k - 1} $$ + +$$ = C(Ar^k + Br^{k - 1}) + D(Akr^k + B(k - 1)r^{k - 1}) $$ + +By Lemma 5.8.4, we know that because $r$ satisfies the characteristic equation, +it follows that $1, r, r^2, \dots, r^n, \dots$ and +$0, r, 2r^2, \dots, nr^n, \dots$ also satisfy the same recurrence relation as +$a_k$. + +Therefore: + +$$ = Cr^{k + 1} + D(k + 1)r^{k + 1} $$ + +This is what was to be shown. + +Q.E.D. + Exercises 22 and 23 are intended for students who are familiar with complex numbers. @@ -11047,6 +12250,8 @@ $$ a_k = 2a_{k - 1} - 2a_{k - 1} \quad \text{ for every integer } k \geq 2 $$ with initial conditions $a_0 = 1$ and $a_1 = 2$. +Omitted. + 23. Find an explicit formula for a sequence $b_0, b_1, b_2, \dots$ that satisfies @@ -11054,6 +12259,8 @@ $$ b_k = 2b_{k - 1} - 5b_{k - 2} \quad \text{ for each integer } k \geq 2 $$ with initial conditions $b_0 = 1$ and $b_1 = 1$. +Omitted. + 24. The numbers $\dfrac{1 + \sqrt{5}}{2}$ and $\dfrac{1 - \sqrt{5}}{2}$ that appear in the explicit formula for the Fibonacci sequence are related to a quantity called the _golden ratio_ in Greek mathematics. Consider a @@ -11074,8 +12281,14 @@ $$ \frac{\phi}{1} = \frac{1}{\phi - 1} $$ a. Show that if $\phi$ satisfies the equation above, then it also satisfies the quadratic equation: $t^2 - t - 1 = 0$. +Omitted. + b. Find the two solutions of $t^2 - t - 1 = 0$ and call them $\phi_1$ and $\phi_2$. +Omitted. + c. Express the explicit formula for the Fibonacci sequence in terms of $\phi_1$ and $\phi_2$. + +Omitted. diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md index 1243935..b281789 100644 --- a/chapter_5/test_yourself.md +++ b/chapter_5/test_yourself.md @@ -229,17 +229,27 @@ Page 385 coefficients is a recurrence relation of the form _____ for every integer $k \geq$ _____, where _____. +$a_k = Aa_{k - 1} + Ba_{k - }$; 2; $A$ and $B$ are fixed real numbers with +$B \neq 0$. + 2. Given a recurrence relation of the form $a_k = Aa_{k - 1} + Ba_{k - 2}$ for every integer $k \geq 2$, the characteristic equation of the relation is _____. +$t^2 - At - B = 0$ + 3. If a sequence $a_1, a_2, a_3, \dots$ is defined by a second-order linear homogeneous recurrence relation with constant coefficients and the characteristic equation for the relation has two distinct roots $r$ and $s$ (which could be complex numbers), then the sequence is given by an explicit formula of the form _____. +$a_n = Cr_n + Ds_n$ for every integer $n \geq 0$ where $C$ and $D$ are real or +complex numbers. + 4. If a sequence $a_1, a_2, a_3, \dots$ is defined by a second-order linear homogeneous recurrence relation with constant coefficients and the characteristic equation for the relation has only a single root $r$, then the sequence is given by an explicit formula of the form _____. + +$a_n = Cr^n + Dnr^n$ where $C$ and $D$ are real numbers. diff --git a/leftoff.txt b/leftoff.txt index d9061d9..32890db 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -375 +387