From 1af31a3b2a9a05fd9098b7dc79f5f8b72cdcb415 Mon Sep 17 00:00:00 2001 From: tomit4 Date: Tue, 16 Jun 2026 17:56:24 -0700 Subject: [PATCH] :construction: Fin 5.1 --- chapter_5/exercises.md | 574 ++++++++++++++++++++++++++++++++++++- chapter_5/notes.md | 2 +- chapter_5/test_yourself.md | 14 + 3 files changed, 587 insertions(+), 3 deletions(-) diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index e9a0f29..68b256f 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -6,220 +6,725 @@ Write the first four terms of the sequences defined by the formulas 1-6. 1. $a_k = \dfrac{k}{10 + k}$, for every integer $k \geq 1$. +$$ a_1 = \frac{1}{10 + 1} = \frac{1}{11} $$ + +$$ a_2 = \frac{2}{10 + 2} = \frac{2}{12} $$ + +$$ a_3 = \frac{3}{10 + 3} = \frac{3}{13} $$ + +$$ a_4 = \frac{4}{10 + 4} = \frac{4}{14} $$ + 2. $b_j = \dfrac{5 - j}{5 + j}$, for every integer $j \geq 1$. +$$ b_1 = \dfrac{5 - 1}{5 + 1} = \frac{4}{6} $$ + +$$ b_2 = \dfrac{5 - 2}{5 + 2} = \frac{3}{7} $$ + +$$ b_3 = \dfrac{5 - 3}{5 + 3} = \frac{2}{8} $$ + +$$ b_4 = \dfrac{5 - 4}{5 + 4} = \frac{1}{9} $$ + 3. $c_i = \dfrac{(-1)^i}{3^i}$, for every integer $i \geq 0$. +$$ c_0 = \dfrac{(-1)^0}{3^0} = \frac{1}{1} $$ + +$$ c_1 = \dfrac{(-1)^1}{3^1} = \frac{-1}{3} $$ + +$$ c_2 = \dfrac{(-1)^2}{3^2} = \frac{1}{9} $$ + +$$ c_3 = \dfrac{(-1)^3}{3^3} = \frac{-1}{27} $$ + 4. $d_m = 1 + \left(\dfrac{1}{2}\right)^m$ for every integer $m \geq 0$. -5. $e_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 2$, for every integer +$$ d_0 = 1 + \left(\dfrac{1}{2}\right)^0 = 1 $$ + +$$ d_1 = 1 + \left(\dfrac{1}{2}\right)^1 = \frac{1}{2} $$ + +$$ d_2 = 1 + \left(\dfrac{1}{2}\right)^2 = \frac{1}{4} $$ + +$$ d_3 = 1 + \left(\dfrac{1}{2}\right)^3 = \frac{1}{8} $$ + +5. $e_n = \left\lfloor \dfrac{n}{2} \right\rfloor \cdot 2$, for every integer $n \geq 0$. +$$ e_0 = \left\lfloor \dfrac{0}{2} \right\rfloor \cdot 2 = 0 $$ + +$$ e_1 = \left\lfloor \dfrac{1}{2} \right\rfloor \cdot 2 = 0 $$ + +$$ e_2 = \left\lfloor \dfrac{2}{2} \right\rfloor \cdot 2 = 2 $$ + +$$ e_3 = \left\lfloor \dfrac{3}{2} \right\rfloor \cdot 2 = 2 $$ + 6. $f_n = \left\lfloor \dfrac{n}{4} \right\rfloor \cdot 4$, for every integer $n \geq 1$. +$$ f_1 = \left\lfloor \dfrac{1}{4} \right\rfloor \cdot 4 = 0 $$ + +$$ f_2 = \left\lfloor \dfrac{2}{4} \right\rfloor \cdot 4 = 0 $$ + +$$ f_3 = \left\lfloor \dfrac{3}{4} \right\rfloor \cdot 4 = 0 $$ + +$$ f_4 = \left\lfloor \dfrac{4}{4} \right\rfloor \cdot 4 = 4 $$ + 7. Let $a_k = 2k + 1$ and $b_k = (k - 1)^3 + k + 2$ for every integer $k \geq 0$. Show that the first three terms of these sequences are identical but that their fourth terms differ. +$$ a_0 = 2(0) + 1 = 1 $$ + +$$ a_1 = 2(1) + 1 = 3 $$ + +$$ a_2 = 2(2) + 1 = 5 $$ + +$$ a_3 = 2(3) + 1 = 7 $$ + +$$ b_0 = (0 - 1)^3 + 0 + 2 = 1 $$ + +$$ b_1 = (1 - 1)^3 + 1 + 2 = 3 $$ + +$$ b_2 = (2 - 1)^3 + 2 + 2 = 5 $$ + +$$ b_3 = (3 - 1)^3 + 3 + 2 = 13 $$ + Compute the first fifteen terms of each of the sequences in 8 and 9, and describe the general behavior of these sequences in words. (A definition of logarithm is given in Section 7.1.) 8. $g_n = \lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$. +$$ g_1 = \lfloor \log_{2}(1) \rfloor = 0 $$ + +$$ g_2 = \lfloor \log_{2}(2) \rfloor = 1 $$ + +$$ g_3 = \lfloor \log_{2}(3) \rfloor = 1 $$ + +$$ g_4 = \lfloor \log_{2}(4) \rfloor = 2 $$ + +$$ g_5 = \lfloor \log_{2}(5) \rfloor = 2 $$ + +$$ g_6 = \lfloor \log_{2}(6) \rfloor = 2 $$ + +$$ g_7 = \lfloor \log_{2}(7) \rfloor = 2 $$ + +$$ g_8 = \lfloor \log_{2}(8) \rfloor = 3 $$ + +$$ g_9 = \lfloor \log_{2}(9) \rfloor = 3 $$ + +$$ g_{10} = \lfloor \log_{2}(10) \rfloor = 3 $$ + +$$ g_{11} = \lfloor \log_{2}(11) \rfloor = 3 $$ + +$$ g_{12} = \lfloor \log_{2}(12) \rfloor = 3 $$ + +$$ g_{13} = \lfloor \log_{2}(13) \rfloor = 3 $$ + +$$ g_{14} = \lfloor \log_{2}(14) \rfloor = 3 $$ + +$$ g_{15} = \lfloor \log_{2}(15) \rfloor = 3 $$ + +The general behavior of this sequence is that it increments in binary +increments, as in it increments every 1, then 2, then 4, then 8 iterations of +the index $n$. + 9 $h_n = n\lfloor \log_{2}n \rfloor$ for every integer $n \geq 1$. +$$ h_1 = (1)\lfloor \log_{2}(1) \rfloor = 0 $$ + +$$ h_2 = (2)\lfloor \log_{2}(2) \rfloor = 2 $$ + +$$ h_3 = (3)\lfloor \log_{2}(3) \rfloor = 3 $$ + +$$ h_4 = (4)\lfloor \log_{2}(4) \rfloor = 8 $$ + +$$ h_5 = (5)\lfloor \log_{2}(5) \rfloor = 10 $$ + +$$ h_6 = (6)\lfloor \log_{2}(6) \rfloor = 12 $$ + +$$ h_7 = (7)\lfloor \log_{2}(7) \rfloor = 14 $$ + +$$ h_8 = (8)\lfloor \log_{2}(8) \rfloor = 24 $$ + +$$ h_9 = (9)\lfloor \log_{2}(9) \rfloor = 27 $$ + +$$ h_{10} = (10)\lfloor \log_{2}(10) \rfloor = 30 $$ + +$$ h_{11} = (11)\lfloor \log_{2}(11) \rfloor = 33 $$ + +$$ h_{12} = (12)\lfloor \log_{2}(12) \rfloor = 36 $$ + +$$ h_{13} = (13)\lfloor \log_{2}(13) \rfloor = 39 $$ + +$$ h_{14} = (14)\lfloor \log_{2}(14) \rfloor = 42 $$ + +$$ h_{15} = (15)\lfloor \log_{2}(15) \rfloor = 45 $$ + +The sequence finds the minimal (floor) power of $log_{2}n$ and then multiplies +it by $n$, which is why there are sudden "jumps" when the floor calculates a +jump to the next power of $2$. For example, at $n = 7$ to $n = 8$, there is a +noticeable jump because $\lfloor \log_{2}7 \rfloor$ is $2$, and then +$\lfloor \log_{2}8 \rfloor$ is $3$. + Find explicit formulas for sequences of the form $a_1, a_2, a_3, \dots$ with the initial terms given in 10-16. 10. $-1, 1, -1, 1, -1, 1$ +$a_n = (-1)^n$ where $n$ is an integer such that $n \geq 1$. + 11. $0, 1, -2, 3, -4, 5$ +$a_n = (n - 1)(-1)^{n}$ where $n$ is an integer such that $n \geq 1$. + 12. $\dfrac{1}{4}, \dfrac{2}{9}, \dfrac{3}{16}, \dfrac{4}{25}, \dfrac{5}{36}, \dfrac{6}{49}$ +$a_n = \dfrac{n}{(n + 1)^2}$ where $n$ is an integer such that $n \geq 1$. + 13. $1 - \dfrac{1}{2}, \dfrac{1}{2} - \dfrac{1}{3}, \dfrac{1}{3} - \dfrac{1}{4}, \dfrac{1}{4} - \dfrac{1}{5}, \dfrac{1}{5} - \dfrac{1}{6}, \dfrac{1}{6} - \dfrac{1}{7}$ +$a_n = \dfrac{1}{n} - \dfrac{1}{n + 1}$ where $n$ is an integer such that +$n \geq 1$. + 14. $\dfrac{1}{3}, \dfrac{4}{9}, \dfrac{9}{27}, \dfrac{16}{81}, \dfrac{25}{243}, \dfrac{36}{729}$ +$a_n = \dfrac{n^2}{3^n}$ where $n$ is an integer such that $n \geq 1$. + 15. $0, -\dfrac{1}{2}, \dfrac{2}{3}, -\dfrac{3}{4}, \dfrac{4}{5}, -\dfrac{5}{6}, \dfrac{6}{7}$ +$a_n = \dfrac{(n - 1)(-1)^{n + 1}}{n}$ where $n$ is an integer such that +$n \geq 1$. + 16. $3, 6, 12, 24, 48, 96$ +$a_n = 3 \cdot 2^{n - 1}$ where $n$ is an integer such that $n \geq 1$. + 17. Consider the sequence defined by $a_n = \dfrac{2n + (-1)^n - 1}{4}$ for every integer $n \geq 0$. Find an alternative explicit formula for $a_n$ that uses the floor notation. +Omitted. + 18. Let $a_0 = 2, a_1 = 3, a_2 = -2, a_3 = 1, a_4 = 0, a_5 = -1, \text{ and } a_6 = -2$. Compute each of the summations and products below. a. $\sum_{i = 0}^{6}{a_i}$ +$$ \sum_{i = 0}^{6}{a_i} = 2 + 3 + (-2) + 1 + 0 + (-1) + (-2) = 1 $$ + b. $\sum_{i = 0}^{0}{a_i}$ +$$ \sum_{i = 0}^{0}{a_i} = 2 $$ + c. $\sum_{j = 1}^{3}{a_{2j}}$ +$$ \sum_{j = 1}^{3}{a_{2j}} = (-2) + 0 + (-2) = -4 $$ + d. $\prod_{k = 0}^{6}{a_k}$ +$$ \prod_{k = 0}^{6}{a_k} = 2 \cdot 3 \cdot (-2) \cdot 1 \cdot 0 \cdot (-1) \cdot (-2) = 0 $$ + e. $\prod_{k = 2}^{2}{a_k}$ +$$ \prod_{k = 2}^{2}{a_k} = -2 $$ + Compute the summations and products in 19-28. 19. $\sum_{k = 1}^{5}{(k + 1)}$ +$$ \sum_{k = 1}^{5}{(k + 1)} = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1) + (5 + 1) = 20 $$ + 20. $\prod_{k = 2}^{4}{k^2}$ +$$ \prod_{k = 2}^{4}{k^2} = (2)^2 \cdot (3)^2 \cdot (4)^2 = 576 $$ + 21. $\sum_{k = 1}^{3}{(k^2 + 1)}$ +$$ \sum_{k = 1}^{3}{(k^2 + 1)} = ((1)^2 + 1) + ((2)^2 + 1) + ((3)^2 + 1) = 17 $$ + 22. $\prod_{j = 0}^{4}{(-1)^j}$ +$$ \prod_{j = 0}^{4}{(-1)^j} = (-1)^{(0)} \cdot (-1)^{(1)} \cdot (-1)^{(2)} \cdot (-1)^{(3)} \cdot (-1)^{(4)} = 1 $$ + 23. $\sum_{i = 1}^{1}{i(i + 1)}$ +$$ \sum_{i = 1}^{1}{i(i + 1)} = (1)((1) + 1) = 2 $$ + 24. $\sum_{j = 0}^{0}{(j + 2) \cdot 2^j}$ +$$ \sum_{j = 0}^{0}{(j + 2) \cdot 2^j} = (0 + 2) \cdot 2^0 = 2 $$ + 25. $\prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)}$ +$$ \prod_{k = 2}^{2}{\left(1 - \dfrac{1}{k}\right)} = \left(1 - \dfrac{1}{2}\right) = \frac{1}{2} $$ + 26. $\sum_{k = -1}^{1}{(k^2 + 3)}$ +$$ \sum_{k = -1}^{1}{(k^2 + 3)} = ((-1)^2 + 3) + ((0)^2 + 3) + ((1)^2 + 3) = 11 $$ + 27. $\sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)}$ +$$ \sum_{n = 1}^{6}{\left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right)} = \left(\dfrac{1}{(1)} - \dfrac{1}{(1) + 1}\right) + \left(\dfrac{1}{(2)} - \dfrac{1}{(2) + 1}\right) + \left(\dfrac{1}{(3)} - \dfrac{1}{(3) + 1}\right) + \left(\dfrac{1}{(4)} - \dfrac{1}{(4) + 1}\right) + \left(\dfrac{1}{(5)} - \dfrac{1}{(5) + 1}\right) + \left(\dfrac{1}{(6)} - \dfrac{1}{(6) + 1}\right) = \frac{6}{7} $$ + 28. $\prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}}$ +$$ \prod_{i = 2}^{5}{\dfrac{i(i + 2)}{(i - 1) \cdot (i + 1)}} = \dfrac{(2)((2) + 2)}{((2) - 1) \cdot ((2) + 1)} + \dfrac{(3)((3) + 2)}{((3) - 1) \cdot ((3) + 1)} + \dfrac{(4)((4) + 2)}{((4) - 1) \cdot ((4) + 1)} + \dfrac{(5)((5) + 2)}{((5) - 1) \cdot ((5) + 1)} = \frac{35}{3} $$ + Write the summations in 29-32 in expanded form. 29. $\sum_{i = 1}^{n}{(-2)^i}$ +$$ \sum_{i = 1}^{n}{(-2)^i} = (-2)^1 + (-2)^2 + (-2)^3 + \dots + (-2)^{n} $$ + 30. $\sum_{j = 1}^{n}{j(j + 1)}$ +$$ \sum_{j = 1}^{n}{j(j + 1)} = ((1)((1) + 1)) + ((2)((2) + 1)) + ((3)((3) + 1)) + \dots + ((n)((n) + 1)) = 2 + 6 + 12 + \dots n(n + 1) $$ + 31. $\sum_{k = 0}^{n + 1}{\dfrac{1}{k!}}$ +$$ \sum_{k = 0}^{n + 1}{\dfrac{1}{k!}} = \dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dots + \dfrac{1}{(n + 1)!} $$ + 32. $\sum_{i = 1}^{k + 1}{i(i!)}$ +$$ \sum_{i = 1}^{k + 1}{i(i!)} = 1(1!) + 2(2!) + 3(3!) + \dots + (k + 1)((k + 1)!) $$ + Evaluate the summations and products in 33-36 for the indicated values of the variable. 33. $\dfrac{1}{1^2} + \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dots + \dfrac{1}{n^2}; n = 1$ +$$ \frac{1}{1^2} = 1 $$ + 34. $1(1!) + 2(2!) + 3(3!) + \dots + m(m!); m = 2$ +$$ 1(1!) + 2(2!) = 5 $$ + 35. $\left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) \dots \left(\dfrac{k}{k + 1}\right); k = 3$ +$$ \left(\dfrac{1}{1 + 1}\right)\left(\dfrac{2}{2 + 1}\right)\left(\dfrac{3}{3 + 1}\right) = \frac{1}{4} $$ + 36. $\left(\dfrac{1 \cdot 2}{3 \cdot 4}\right)\left(\dfrac{4 \cdot 5}{6 \cdot 7}\right)\left(\dfrac{6 \cdot 7}{8 \cdot 9}\right) \dots \left(\dfrac{m \cdot (m + 1)}{(m + 2) \cdot (m + 3)}\right); m = 1$ +$$ \left(\frac{1 \cdot 2}{3 \cdot 4}\right) = \frac{2}{12} = \frac{1}{6} $$ + Write each of 37-39 as a single summation. 37. $\sum_{i = 1}^{k}{i^3 + (k + 1)^3}$ +$$ \sum_{i = 1}^{k}{i^3 + (k + 1)^3} = \sum_{i = 1}^{k + 1}{i^3} $$ + 38. $\sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}}$ +$$ \sum_{k = 1}^{m}{\dfrac{k}{k + 1} + \dfrac{m + 1}{m + 2}} = \sum_{k = 1}^{m + 1}{\dfrac{k}{k + 1}} $$ + 39. $\sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}}$ +$$ \sum_{m = 0}^{n}{(m + 1)2^m + (n + 2)2^{n + 1}} = \sum_{m = 0}^{n + 1}{(m + 1)2^m} $$ + Rewrite 40-42 by separating off the final term. 40. $\sum_{i = 1}^{k + 1}{i(i!)}$ +$$ \sum_{i = 1}^{k + 1}{i(i!)} = \sum_{i = 1}^{k}{i(i!) + (k + 1)(k + 1)!} $$ + 41. $\sum_{k = 1}^{m + 1}{k^2}$ +$$ \sum_{k = 1}^{m + 1}{k^2} = \sum_{k = 1}^{m}{k^2 + (m + 1)^2} $$ + 42. $\sum_{m = 1}^{n + 1}{m(m + 1)}$ +$$ \sum_{m = 1}^{n + 1}{m(m + 1)} = \sum_{m = 1}^{n}{m(m + 1) + (n + 1)(n + 2)} $$ + Write each of 43-52 using summation or product notation. 43. $1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2$ +$$ 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 $$ + +$$ \sum_{k = 1}^{7}{(-1)^{k + 1}k^2} $$ + 44. $(1^3 - 1) - (2^3 - 1) + (3^3 - 1) - (4^3 - 1) + (5^3 - 1)$ +$$ \sum_{k = 1}^{5}{(-1)^{k + 1}(k^3 - 1)} $$ + 45. $(2^2 - 1) \cdot (3^2 - 1) \cdot (4^2 - 1)$ +$$ \prod_{k = 2}^{4}{(k^2 - 1)} $$ + 46. $\dfrac{2}{3 \cdot 4} - \dfrac{3}{4 \cdot 5} + \dfrac{4}{5 \cdot 6} - \dfrac{5}{6 \cdot 7} + \dfrac{6}{7 \cdot 8}$ +$$ \sum_{k=2}^{6}{\dfrac{(-1)^k \cdot k}{(k + 1) \cdot (k + 2)}} $$ + 47. $1 - r + r^2 - r^3 + r^4 - r^5$ +$$ \sum_{k = 0}^{5}{(-1)^kr^{k + 1}} $$ + 48. $(1 - t) \cdot (1 - t^2) \cdot (1 - t^3) \cdot (1 - t^4)$ +$$ \prod_{k = 1}^{4}{(1 - t^k)} $$ + 49. $1^3 + 2^3 + 3^3 + \dots + n^3$ +$$ \sum_{k}^{n}{k^3} $$ + 50. $\dfrac{1}{2!} + \dfrac{2}{3!} + \dfrac{3}{4!} + \dots + \dfrac{n}{(n + 1)!}$ +$$ \sum_{k = 1}^{n}{\frac{k}{(k + 1)!}} $$ + 51. $n + (n - 1) + (n - 2) + \dots + 1$ +$$ \sum_{k = 0}^{n - 1}{(n - k)} $$ + 52. $n + \dfrac{n - 1}{2!} + \dfrac{n - 2}{3!} + \dfrac{n - 3}{4!} + \dots + \dfrac{1}{n!}$ +$$ \sum_{k = 0}^{n - 1}{\frac{n - k}{(k + 1)!}} $$ + Transform each of 53 and 54 by making the change of variable $i = k + 1$. +$$ i = k + 1 $$ + +$$ i - 1 = k $$ + 53. $\sum_{k = 0}^{5}{k(k - 1)}$ +$$ \sum_{k = 0}^{5}{k(k - 1)} = \sum_{i = 1}^{6}{(i - 1)(i - 2)} $$ + 54. $\prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}}$ +$$ \prod_{k = 1}^{n}{\dfrac{k}{k^2 + 4}} = \prod_{i = 2}^{n + 1}{\frac{i - 1}{(i - 1)^2 + 4}} $$ + Transform each of 55-58 by making the change of variable $j = i - 1$. +$$ j = i - 1 $$ + +$$ i = j + 1 $$ + 55. $\sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}}$ +$$ \sum_{i = 1}^{n + 1}{\dfrac{(i - 1)^2}{i \cdot n}} = \sum_{j = 0}^{n}{\frac{j^2}{jn + n}} $$ + 56. $\sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}}$ +$$ \sum_{i = 3}^{n}{\dfrac{i}{i + n - 1}} = \sum_{j = 2}^{n - 1}{\frac{j + 1}{j + n}} $$ + 57. $\sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}}$ +$$ \sum_{i = 1}^{n - 1}{\dfrac{i}{(n - i)^2}} = \sum_{j = 0}^{n - 2}{\frac{j + 1}{(n - j - 1)^2}} $$ + 58. $\prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}}$ +$$ \prod_{i = n}^{2n}{\dfrac{n - i + 1}{n + i}} = \prod_{j = n - 1}^{2n - 1}{\frac{n - j}{n + j + 1}} $$ + Write each of 59-61 as a single summation or product. 59. $3 \cdot \sum_{k = 1}^{n}{(2k - 3)} + \sum_{k = 1}^{n}{(4 - 5k)}$ +$$ \sum_{k = 1}^{n}{3(2k - 3) + (4 - 5k)} $$ + +$$ \sum_{k = 1}^{n}{(6k - 9 + 4 - 5k)} $$ + +$$ \sum_{k = 1}^{n}{(k - 5)} $$ + 60. $2 \cdot \sum_{k = 1}^{n}{(3k^2 + 4)} + 5 \cdot \sum_{k = 1}^{n}{(2k^2 - 1)}$ +$$ \sum_{k = 1}^{n}{2(3k^2 + 4) + 5(2k^2 - 1)} $$ + +$$ \sum_{k = 1}^{n}{(6k^2 + 8 + 10k^2 - 5)} $$ + +$$ \sum_{k = 1}^{n}{(16k^2 + 3)} $$ + 61. $\left(\prod_{k = 1}^{n}{\dfrac{k}{k + 1}}\right) \cdot \left(\prod_{k = 1}^{n}{\dfrac{k + 1}{k + 2}}\right)$ +$$ \prod_{k = 1}^{n}{\left(\frac{k}{k + 1}\right)\left(\frac{k + 1}{k + 2}\right)} $$ + +$$ \prod_{k = 1}^{n}{\left(\frac{k(k + 1)}{(k + 1)(k + 2)}\right)} $$ + +$$ \prod_{k = 1}^{n}{\left(\frac{k\cancel{(k + 1)}}{\cancel{(k + 1)}(k + 2)}\right)} $$ + +$$ \prod_{k = 1}^{n}{\left(\frac{k}{k + 2}\right)} $$ + Compute each of 62-76. Assume the values of the variables are restricted so that the expressions are defined. 62. $\dfrac{4!}{3!}$ +$$ \frac{4!}{3!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1} = 4 $$ + 63. $\dfrac{6!}{8!}$ +$$ \frac{6!}{8!} = \frac{6!}{8 \cdot 7 \cdot 6!} = \frac{1}{56} $$ + 64. $\dfrac{4!}{0!}$ +$$ \frac{4!}{0!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{1} = 24 $$ + 65. $\dfrac{n!}{(n - 1)!}$ +$$ \frac{n!}{(n - 1)!} = \frac{n(n - 1)!}{(n - 1)!} = n $$ + 66. $\dfrac{(n - 1)!}{(n + 1)!}$ +$$ \frac{(n - 1)!}{(n + 1)!} = \frac{(n - 1)!}{(n + 1)(n)(n - 1)!} = \frac{1}{n(n + 1)} $$ + 67. $\dfrac{n!}{(n - 2)!}$ +$$ \dfrac{n!}{(n - 2)!} = \frac{n(n - 1)(n - 2)!}{(n - 2)!} = n(n - 1) $$ + 68. $\dfrac{((n + 1)!)^2}{(n!)^2}$ +$$ \frac{((n + 1)!)^2}{(n!)^2} = \frac{((n + 1)(n!))^2}{(n!)^2} = (n + 1)^2 $$ + 69. $\dfrac{n!}{(n - k)!}$ +$$ (n - k)! = (n - k)(n - k - 1)(n - k - 2) \dots (2)(1) $$ + +$$ n! = n(n - 1)(n - 2) \dots (n - k + 1)(n - k)(n - k - 1)(n - k - 2) \dots (2)(1) $$ + +$$ \frac{n!}{(n - k)!} = n(n - 1)(n - 2) \dots (n - k + 1) $$ + 70. $\dfrac{n!}{(n - k + 1)!}$ +$$ (n - k + 1)! = (n - k + 1)(n - k)(n - k - 1) \dots (2)(1) $$ + +$$ n! = n(n - 1)(n - 2) \dots (n - k + 2)(n - k + 1)(n - k)(n - k - 1) \dots (2)(1)$$ + +$$ \frac{n!}{(n - k + 1)!} = n(n - 1)(n - 2) \dots (n - k + 2) $$ + 71. $\dbinom{5}{3}$ +$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$ + +$$ \binom{5}{3} = \frac{5!}{3!(5 - 3)!} $$ + +$$ = \frac{5 \cdot 4 \cdot 3!}{3!(2)!} $$ + +$$ = \frac{20}{2 \cdot 1} $$ + +$$ = 10 $$ + 72. $\dbinom{7}{4}$ +$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$ + +$$ \binom{7}{4} = \frac{7!}{4!(7 - 4)!} $$ + +$$ = \frac{7 \cdot 6 \cdot 5 \cdot 4!}{4!(3)!} $$ + +$$ = \frac{210}{3!} $$ + +$$ = \frac{210}{6} $$ + +$$ = 35 $$ + 73. $\dbinom{3}{0}$ +$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$ + +$$ \binom{3}{0} = \frac{3!}{0!(3 - 0)!} $$ + +$$ = \frac{3!}{1(3)!} $$ + +$$ = 1 $$ + 74. $\dbinom{5}{5}$ +$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$ + +$$ \binom{5}{5} = \frac{5!}{5!(5 - 5)!} $$ + +$$ = \frac{1}{1(0)!} $$ + +$$ = 1 $$ + 75. $\dbinom{n}{n - 1}$ +$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$ + +$$ \binom{n}{n - 1} = \frac{n!}{(n - 1)!(n - (n - 1))!} $$ + +$$ = \frac{n!}{(n - 1)!(n - n + 1)!} $$ + +$$ = \frac{n!}{(n - 1)!(1)!} $$ + +$$ = \frac{n(n - 1)!}{(n - 1)!(1)!} $$ + +$$ = \frac{n}{1} $$ + +$$ = n $$ + 76. $\dbinom{n + 1}{n - 1}$ +$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$ + +$$ \binom{n + 1}{n - 1} = \frac{(n + 1)!}{(n - 1)!((n + 1) - (n - 1))!} $$ + +$$ = \frac{(n + 1)!}{(n - 1)!(n + 1 - n + 1)!} $$ + +$$ = \frac{(n + 1)!}{(n - 1)!(2)!} $$ + +$$ = \frac{(n + 1)(n)(n - 1)!}{(n - 1)!(2)!} $$ + +$$ = \frac{n(n + 1)}{2} $$ + 77. a. Prove that $n! + 2$ is divisible by $2$, for every integer $n \geq 2$. +**Proof:** + +Suppose that $n$ is any integer such that $n \geq 2$. + +By the definition of a factorial: + +$$ n! = n \cdot (n - 1) \dots 3 \cdot 2 \cdot 1 $$ + +Since $n \geq 2$, this can be represented as: + +$$ +n! = +\begin{cases} +2 & \text{if } n = 2 \\ +3 \cdot 2 \cdot 1& \text{if } n = 3 \\ +n \cdot (n - 1) \dots \cdot 2 \cdot 1 & \text{if } n > 3 \\ +\end{cases} +$$ + +In each case, $n!$ has a factor of $2$. Then: + +$$ n! + 2 = 2k + 2 $$ + +$$ n! + 2 = 2(k + 1) $$ + +for some integer $k$. + +Now, $k + 1$ is an integer by the sum of integers. + +Therefore $n! + 2$ is divisible by $2$. + +Q.E.D. + b. Prove that $n! + k$ is divisible by $k$, for every integer $n \geq 2$ and $k = 2, 3, \dots, n$. +**Proof:** + +Suppose $n$ is any integer such that $n \geq 2$, and $k$ is any integer such +that $2 \leq k \leq n$. + +Since $2 \leq k \leq n$, it follows that $k$ is one of the factors of $n!$. +Then: + +$$ n! = km $$ + +for some integer $m$. + +By substitution: + +$$ n! + k = km + k $$ + +$$ = k(m + 1) $$ + +Now, $m + 1$ is an integer by the sum of integers. + +Therefore $n! + k$ is divisible by $k$. + +Q.E.D. + c. Given any integer $m \geq 2$, is it possible to find a sequence of $m - 1$ consecutive positive integers none of which is prime? Explain your answer. +**Proof:** + +Suppose $m$ is any integer such that $m \geq 2$. + +Consider the sequence + +$$ m! + 2, m! + 3, \dots, m! + m $$ + +This is a sequence of $m - 1$ consecutive positive integers. + +Let $k$ be any integer such that $2 \leq k \leq m$. The $k - 1$th +term of the sequence is $m! + k$. + +Since $k \leq m$, it follows that $k \mid m!$ (by part b). Then: + +$$ m! = kt $$ + +for some integer $t$. + +Then: + +$$ m! + k = kt + k = k(t + 1) $$ + +Now, $t + 1$ is an integer by the sum of integers. Thus $k$ divides $m! + k$ and +since $k \geq 2$ and $(t + 1) > 1$ are both factors greater than or equal to +$1$, it follows that $m! + k$ is composite. + +Therefore every term in the sequence is not prime, so there exists a sequence of +$m - 1$ consecutive positive integers none of which is prime. + +Q.E.D. + 78. Prove that for all nonnegative integers $n$ and $r$ with $$ r + 1 \leq n, \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} $$ +**Proof:** + +Suppose $n$ and $r$ are any nonnegative integers such that $r + 1 \leq n$. + +The given equation shown is: + +$$ \frac{n - r}{r + 1}\binom{n}{r} = \frac{n - r}{r + 1}\left(\frac{n!}{r!(n - r)!}\right) $$ + +$$ = \frac{n!(n - r)}{r!(r + 1)(n - r)!}$$ + +$$ = \frac{n!(n - r)}{r!(r + 1)(n - r)(n - r - 1)!}$$ + +$$ = \frac{n!}{r!(r + 1)(n - r - 1)!}$$ + +$$ = \frac{n!}{r!(r + 1)(n - (r + 1))!}$$ + +Notice that this in the form of a "$n$ choose $r + 1$": + +$$ \binom{n}{r + 1} $$ + +Therefore, it has been shown that: + +$$ \binom{n}{r + 1} = \frac{n - r}{r + 1}\binom{n}{r} $$ + +Q.E.D. + 79. Prove that if $p$ is a prime number and $r$ is an integer with $0 < r < p$, then $\dbinom{p}{r}$ is divisible by $p$. +**Proof:** + +Suppose that $p$ is any prime number and $r$ is any integer such that +$0 < r < p$. + +_[We need to show that $p \mid \dbinom{p}{r}$.]_ + +Consider: + +$$ \binom{p}{r} = \frac{p!}{r!(p - r)!} $$ + +Since $0 < r < p$, both $r!$ and $(p - r)!$ are less than $p$. Thus, the +denominator $r!(p - r)!$ can never have a factor of $p$. + +The numerator can be expressed as $p! = p(p - 1)!$: + +$$ \binom{p}{r} = \frac{p(p - 1)!}{r!(p - r)!} $$ + +Factoring $p$ out of the numerator gives: + +$$ \binom{p}{r} = p \cdot \frac{(p - 1)!}{r!(p - r)!} $$ + +Therefore it has been shown that: + +$$ p \mid \binom{p}{r} $$ + +Q.E.D. + 80. Suppose $a[1], a[2], a[3], \dots, a[m]$ is a one-dimensional array and consider the following algorithm segment: @@ -232,38 +737,103 @@ a. $\text{sum } := 0\\ \text{\textbf{for }} i := 0 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} i$ +$m - 1$; $\text{sum } + a[i + 1]$ + b. $\text{sum } := 0\\ \text{\textbf{for }} j := 2 \text{\textbf{ to \_\_\_\_}}\\ \ \ \text{sum } := \text{\_\_\_\_}\\ \text{\textbf{next }} j$ +$m + 1$; $\text{sum } + a[j - 1]$ + Use repeated division by $2$ to convert (by hand) the integers in 81-83 from base 10 to base 2. 81. $90$ +$$ 90_{10} = 1011010_2 $$ + 82. $98$ +$$ 98_{10} = 1100010_2 $$ + 83. $205$ +$$ 205_{10} = 11001101_2 $$ + Make a trace table to trace the action of Algorithm 5.1.1 on the input in 84-86. 84. $23$ +| | 0 | 1 | 2 | 3 | 4 | 5 | +| ------ | -- | -- | - | - | - | - | +| $a$ | 23 | | | | | | +| $r[i]$ | | 1 | 1 | 1 | 0 | 1 | +| $q$ | 23 | 11 | 5 | 2 | 1 | 0 | +| $i$ | 0 | 1 | 2 | 3 | 4 | 5 | + +Outputs: 10111, which is $23_{10} = 10111_2$. + 85. $28$ +| | 0 | 1 | 2 | 3 | 4 | 5 | +| ------ | -- | -- | - | - | - | - | +| $a$ | 28 | | | | | | +| $r[i]$ | | 0 | 0 | 1 | 1 | 1 | +| $q$ | 28 | 14 | 7 | 3 | 1 | 0 | +| $i$ | 0 | 1 | 2 | 3 | 4 | 5 | + +Outputs: 11100, which is $28_{10} = 11100_2$. + 86. $44$ +| | 0 | 1 | 2 | 3 | 4 | 5 | 6 | +| ------ | -- | -- | -- | - | - | - | - | +| $a$ | 44 | | | | | | | +| $r[i]$ | | 0 | 0 | 1 | 1 | 0 | 1 | +| $q$ | 44 | 22 | 11 | 5 | 2 | 1 | 0 | +| $i$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | + +Outputs: 101100, which is $44_{10} = 101100_2$ + 87. Write an informal description of an algorithm (using repeated division by 16) to convert a nonnegative integer from decimal notation to hexadecimal notation (base 16). +**Input:** $a$ _[a nonnegative integer]_ + +**Algorithm Body:** + +$q := a, i := 0$ + +_[Repeatedly perform the integer division of $q$ by $16$ until $q$ becomes $0$. +Store successive remainders in a one-dimensional array +$r[0], r[1], r[2], \dots r[k]$. Even if the initial-value of $q$ equals $0$, the +loop should execute one time (so that $r[0]$ is computed). Thus the guard +condition for the **while** loop is $i = 0$ or $q \neq 0$.]_ + +$\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 16\\ \ \ q := q \text{ div } 16\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$ + +_[After execution of this step, the values of $r[0], r[1], \dots, r[i - 1]$ are +all $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F$, and +$a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_{16}$.]_ + +**Output:** $r[0], r[1], r[2], \dots, r[i - 1]$ _[a sequence of integers]_ + Use the algorithm you developed for exercise 87 to convert the integers in 88-90 to hexadecimal notation. 88. $287$ +$$ 287_{10} = 11F_{16} $$ + 89. $693$ -90. $2,301$ +$$ 693_{10} = 1BF_{16} $$ + +91. $2,301$ + +$$ 2301_{10} = 8FD_{16} $$ 91. Write a formal version of the algorithm you developed for exercise 87. + +Already done. diff --git a/chapter_5/notes.md b/chapter_5/notes.md index a9a1aa1..3882628 100644 --- a/chapter_5/notes.md +++ b/chapter_5/notes.md @@ -111,7 +111,7 @@ $r[0], r[1], r[2], \dots r[k]$. Even if the initial-value of $q$ equals $0$, the loop should execute one time (so that $r[0]$ is computed). Thus the guard condition for the **while** loop is $i = 0$ or $q \neq 0$.]_ -$\text{\textbf{while }}(i = 0 \text{ or } q \new 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$ +$\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$ _[After execution of this step, the values of $r[0], r[1], \dots, r[i - 1]$ are all $0$'s and $1$'s, and diff --git a/chapter_5/test_yourself.md b/chapter_5/test_yourself.md index 68d23a0..dc554ab 100644 --- a/chapter_5/test_yourself.md +++ b/chapter_5/test_yourself.md @@ -4,15 +4,29 @@ Page 296 1. The notation $\sum_{k = m}^{n}{a_k}$ is read "_____." +The summation from $k$ equals $m$ to $n$ of $a$ sub $k$. + 2. The expanded form of $\sum_{k = m}^{n}{a_k}$ is _____. +$$ a_m + a_{m + 1} + a_{m + 2} + \dots + a_n $$ + 3. The value of $a_1 + a_2 + a_3 + \dots + a_n$ when $n = 2$ is "_____." +$$ a_1 + a_2 $$ + 4. The notation $\prod_{k = m}^{n}{a_k}$ is read "_____." +The product from $k$ equals $m$ to $n$ of $a$ sub $k$. + 5. If $n$ is a positive integer, then $n! =$ _____. +$$ n \cdot (n - 1) \dots \cdot 3 \cdot 2 \cdot 1 $$ + 6. $\sum_{k = m}^{n}{a_k} + c\sum_{k = m}^{n}{b_k} =$ _____. +$$ \sum_{k = m}^{n}{a_k + cb_k} $$ + 7. $\left(\prod_{k = m}^{n}{a_k}\right)\left(\prod_{k = m}^{n}{b_k}\right) =$ _____. + +$$ \prod_{k = m}^{n}{a_kb_k} $$