From ecb409cc2f9f5accb2a8f310a38263b82690aabe Mon Sep 17 00:00:00 2001 From: tomit4 Date: Fri, 10 Jul 2026 05:50:47 -0700 Subject: [PATCH] :construction: Still mid 5.7 --- chapter_5/exercises.md | 361 ++++++++++++++++++++++++++++++++++++++++- 1 file changed, 360 insertions(+), 1 deletion(-) diff --git a/chapter_5/exercises.md b/chapter_5/exercises.md index 6ca5c92..79202ab 100644 --- a/chapter_5/exercises.md +++ b/chapter_5/exercises.md @@ -9061,19 +9061,160 @@ $$ = 3^n - 1 $$ integer $k \geq 1$. Use mathematical induction to prove that $a_n = a_0 + nd$, for every integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $d$ be any fixed constant, and let $a_0, a_1, a_2, \dots$ be the sequence +defined recursively by $a_k = a_{k - 1} + d$ for each integer $k \geq 1$. + +Let $P(n)$ be the equation: + +$$ a_n = a_0 + nd $$ + +We must show by mathematical induction that $P(n)$ is true for every integer +$n \geq 0$. + +_Basis Step:_ + +Prove that $P(0)$ is true. That is: + +$$ a_0 = a_0 + (0)d $$ + +$$ a_0 = a_0 $$ + +This equality holds, and therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer such that $k \geq 0$. + +Suppose $P(k)$. That is: + +$$ a_k = a_0 + kd $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$. That is: + +$$ a_{k + 1} = a_0 + (k + 1)d $$ + +By the definition of the given sequence: + +$$ a_{k + 1} = a_k + d $$ + +By substitution of the inductive hypothesis: + +$$ a_{k + 1} = (a_0 + kd) + d $$ + +By algebra: + +$$ a_{k + 1} = a_0 + (k + 1)d $$ + +Which is the equality that was to be shown. Therefore $P(k + 1)$ is true. + +Q.E.D. + 19. A worker is promised a bonus if he can increase his productivity by 2 units a day for a period of 30 days. If on day 0 he produces 170 units, how many units must he produce on day 30 to qualify for the bonus? +Let $U_n$ be the number of units produced on day $n$. Then: + +$$ U_k = U_{k - 1} + 2 $$ + +for every integer $k \geq 1$, and: + +$$ U_0 = 170 $$ + +Hence $U_0, U_1, U_2, \dots$ is an arithmetic sequence with a fixed constant +$2$. It then follows that when $n = 30$: + +$$ U_n = U_0 + n \cdot 2 $$ + +$$ U_{30} = 170 + (30) \cdot 2 = 170 + 60 = 230 \text{ units} $$ + +Thus, in order to qualify for the bonus, the worker must produce 230 units on +day 30. + 20. A runner targets herself to improve her time on a certain course by 3 seconds a day. If on day 0 she runs the course in 3 minutes, how fast must she run it on day 14 to stay on target? +First, let's convert 3 minutes to seconds for ease of evaluation: + +$$ 3 \text{ minutes } \cdot 60 \frac{\text{seconds}}{\text{minute}} = 180 \text{ seconds} $$ + +Let $R_n$ be the number of seconds the runner ran on day $n$. Then: + +$$ R_k = R_{k - 1} - 3 $$ + +for every integer $k \geq 1$, and: + +$$ R_0 = 180 $$ + +Hence $R_0, R_1, R_2, \dots$ is an arithmetic sequence with a fixed constant +$3$. It follows then that when $n = 14$: + +$$ U_n = U_0 - n \cdot 3 $$ + +$$ U_{14} = (180) - 14 \cdot 3 = 180 - 42 = 138 \text{ seconds} $$ + +Therefore, the runner must run the certain course in 138 seconds (approximately +2.3 minutes) on day 14 in order to stay on target. + 21. Suppose $r$ is a fixed constant and $a_0, a_1, a_2, \dots$ is a sequence - that satisfies the recurrence 4elation $a_k = ra_{k - 1}$, for each integer + that satisfies the recurrence relation $a_k = ra_{k - 1}$, for each integer $k \geq 1$ and $a_0 = a$. Use mathematical induction to prove that $a_n = ar^n$, for every integer $n \geq 0$. +**Proof (by mathematical induction):** + +Let $P(n)$ be the equation: + +$$ a_n = ar^n $$ + +_Basis Step:_ + +Prove $P(0)$, that is: + +$$ a_0 = ar^0 $$ + +$$ a_0 = a(1) $$ + +$$ a_0 = a $$ + +The given problem statement tells us that $a_0 = a$. Since this matches the +equality found for $P(0)$, we can conclude therefore that $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer such that $k \geq 1$. + +Suppose $P(k)$, that is: + +$$ a_k = ar^k $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$, that is: + +$$ a_{k + 1} = ar^{k + 1} $$ + +By the given recurrence relation: + +$$ a_{k + 1} = r \cdot a_k $$ + +By substitution with the inductive hypothesis: + +$$ a_{k + 1} = r \cdot (a \cdot r^k) $$ + +By algebra: + +$$ a_{k + 1} = ar^{k + 1} $$ + +This equality is what was to be shown, therefore $P(k + 1)$ is true. + +Q.E.D. + 22. As shown in Example 5.6.8, if a bank pays interest at a rate of $i$ compounded $m$ times a year, then the amount of money $P_k$ at the end of $k$ time periods (where one time period = $\dfrac{1}{m}$th of a @@ -9082,10 +9223,41 @@ $$ = 3^n - 1 $$ condition $P_0 = \text{ the initial amount deposited}$. Find an explicit formula for $P_n$. +$$ P_0 = P_0 $$ + +$$ P_1 = \left[1 + \frac{i}{m}\right] \cdot P_0 = \left[1 + \frac{i}{m}\right]^1 \cdot P_0 $$ + +$$ P_2 = \left[1 + \frac{i}{m}\right] \cdot P_1 = \left[1 + \frac{i}{m}\right] \cdot \left[1 + \frac{i}{m}\right] \cdot P_0 = \left[1 + \frac{i}{m}\right]^2 + P_0 $$ + +Guess: + +$$ P_n = \left[1 + \frac{i}{m}\right]^n \cdot P_0 $$ + 23. Suppose the population of a country increases at a steady rate of 3% per year. If the population is 50 million at a certain time, what will it be 25 years later? +Let $P_n$ be the population of the country at year $n$. Then: + +$$ P_{k + 1} = 1.03 \cdot P_k $$ + +for every integer $k \geq 1$, and: + +$$ P_0 = 50000000 $$ + +The explicit formula then is: + +$$ P_n = (1.03)^n \cdot P_0 $$ + +Then: + +$$ P_{25} = (1.03)^{25} \cdot 50000000 $$ + +$$ \approx 104688896 $$ + +Therefore, the population of the country 25 years later will be approximately +104,688896. + 24. A chain letter works as follows: One person sends a copy of the letter to five friends, each of whom sends a copy to five friends, each of whom sends a copy to five friends, each of whom sends a copy to five friends, and so @@ -9093,12 +9265,32 @@ $$ = 3^n - 1 $$ twentieth reception of this process, assuming no person receives more than one copy? +$$ \sum_{k = 0}^{20}{5^k} = \frac{5^{21} - 1}{4} $$ + +$$ \approx 1.192092896 \cdot 10^{14} \text{ people} $$ + 25. A certain computer algorithm executes twice as many operations when it is run with an input size $k$ as when it is run with an input size $k - 1$ (where $k$ is an integer that is greater than $1$). When the algorithm is run with an input size $1$, it executes seven operations. How many operations does it execute when it is run with an input size of $25$? +Let $P_k$ be the number of operations the algorithm when the input size is $k$, +and $P_0 = 7$. The recurrence relation is: + +$$ P_k = 2P_{k - 1} $$ + +So: + +$$ P_n = 2^n \cdot P_0 $$ + +$$ P_{25} = 2^{25} \cdot 7 $$ + +$$ = 234881024 $$ + +So the algorithm executes 234881024 operations when it is run with an input size +of 25. + 26. A person saving for retirement makes an initial deposit of $1,000 to a bank account earning interest at a rate of 3% per year compounded monthly, and each month she adds an addition $200 to the account. @@ -9107,30 +9299,197 @@ a. For each nonnegative integer $n$, let $A_n$ be the amount in the account at the end of $n$ months. Find the recurrence relation relating $A_k$ to $A_{k - 1}$. +$$ A_k = \left[1 + \left(\frac{1}{12}\right)\left(\frac{3}{100}\right)\right] \cdot A_{k - 1} + 200 $$ + +$$ = \left[1 + \left(\frac{3}{1200}\right)\right] \cdot A_{k - 1} + 200 $$ + +$$ = \left[1 + \left(\frac{1}{400}\right)\right] \cdot A_{k - 1} + 200 $$ + +$$ = \frac{401}{400} \cdot A_{k - 1} + 200 $$ + +$$ = 1.0025 \cdot A_{k - 1} + 200 $$ + b. Use iteration to find an explicit formula for $A_n$. +$$ A_0 = 1000 $$ + +$$ A_1 = 1.0025 \cdot A_0 + 200 = 1.0025 \cdot (1000) + 200 $$ + +$$ A_2 = 1.0025 \cdot A_1 + 200 = 1.0025 \cdot (1.0025 \cdot (1000) + 200) + 200 $$ + +$$ A_3 = 1.0025 \cdot A_3 + 200 = 1.0025 \cdot (1.0025 \cdot (1.0025 \cdot (1000) + 200) + 200) + 200 $$ + +Guess: + +$$ A_n = 200 + 200(1.0025) + 200(1.0025)^2 + \cdots + 200(1.0025)^{n - 1} + 1000(1.0025)^n $$ + +Explicit formula: + +$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$ + c. Use mathematical induction to prove the correctness of the formula you obtained in part (b). +**Proof (by mathematical induction):** + +Let $P(n)$ be the equation: + +$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$ + +_Basis Step:_ + +Prove $P(0)$, that is: + +$$ A_0 = 200 \cdot \left(\frac{1.0025^0 - 1}{1.0025 - 1}\right) + 1000(1.0025)^0 $$ + +$$ A_0 = 200 \cdot \left(\frac{1 - 1}{0.0025}\right) + 1000(1) $$ + +$$ A_0 = 200 \cdot 0 + 1000 $$ + +$$ A_0 = 0 + 1000 $$ + +$$ A_0 = 1000 $$ + +This equality holds as $A_0$ was established as being equal to $1000$ in the +given problem statement. Therefore $P(0)$ is true. + +_Inductive Step:_ + +Let $k$ be any integer such that $k \geq 0$. + +Suppose $P(k)$, that is: + +$$ A_k = 200 \cdot \left(\frac{1.0025^k - 1}{1.0025 - 1}\right) + 1000(1.0025)^k $$ + +This is the inductive hypothesis. + +Prove $P(k + 1)$, that is: + +$$ A_{k + 1} = 200 \cdot \left(\frac{1.0025^{k + 1} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{k + 1} $$ + +By the recurrence relation in part (a), we have: + +$$ A_{k + 1} = 1.0025 \cdot A_k + 200 $$ + +By substitution with the inductive hypothesis: + +$$ = 1.0025 \cdot (200 \cdot \left(\frac{1.0025^k - 1}{1.0025 - 1}\right) + 1000(1.0025)^k) + 200 $$ + +$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025}{1.0025 - 1}\right) + 1000(1.0025)^{k + 1} + 200 $$ + +$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025}{0.0025}\right) + 1000(1.0025)^{k + 1} + \frac{0.5}{0.0025} $$ + +$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1.0025 + 0.0025}{0.0025}\right) + 1000(1.0025)^{k + 1} $$ + +$$ = 200 \cdot \left(\frac{1.0025^{k + 1} - 1}{0.0025}\right) + 1000(1.0025)^{k + 1} $$ + +Q.E.D. + d. How much will the account be worth at the end of 20 years? At the end of 40 years? +We can just use the explicit formula: + +$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$ + +$$ A_{20} = 200 \cdot \left(\frac{1.0025^{20} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{20} $$ + +$$ \approx \$5147.65 $$ + +$$ A_{40} = 200 \cdot \left(\frac{1.0025^{40} - 1}{1.0025 - 1}\right) + 1000(1.0025)^{40} $$ + +$$ \approx \$9507.67 $$ + e. In how many years will the account be worth $10,000? +$$ A_n = 200 \cdot \left(\frac{1.0025^n - 1}{1.0025 - 1}\right) + 1000(1.0025)^n $$ + +$$ 10000 = 200 \cdot \left(\frac{1.0025^n - 1}{0.0025}\right) + 1000(1.0025)^n $$ + +$$ 10000 = \left(\frac{200 \cdot (1.0025^n - 1)}{0.0025}\right) + 1000(1.0025)^n $$ + +$$ 10000 = 80000(1.0025^n - 1) + 1000(1.0025)^n $$ + +$$ 10000 = 80000(1.0025^n) - 80000 + 1000(1.0025)^n $$ + +$$ 10000 = 81000(1.0025^n) - 80000 $$ + +$$ 90000 = 81000(1.0025^n) $$ + +$$ \frac{10}{9} = 1.0025^n $$ + +$$ \ln\left(\frac{10}{9}\right) = \ln(1.0025^n) $$ + +$$ \ln\left(\frac{10}{9}\right) = n\ln(1.0025) $$ + +$$ \frac{\ln\left(\frac{10}{9}\right)}{\ln(1.0025)} = n $$ + +$$ n \approx 42 \text{ months} $$ + +$$ \frac{42}{12} = 3.5 \text{ years} $$ + 27. A person borrows $3,000 on a bank credit card at a nominal rate of 18% per year, which is actually charged at a rate of 1.5% per month. a. What is the annual percentage yield (APY) for the card? (See Example 5.6.8 for a definition of APY.) +$$ \text{APY} = \left(1 + \frac{r}{n}\right)^n - 1 $$ + +$$ = \left(1 + \frac{0.18}{12}\right)^{12} - 1 $$ + +$$ \approx 0.1956181715 $$ + +$$ \approx 19.6\% $$ + b. Assume that the person does not place any additional charges on the card and pays the bank $150 each month to pay off the loan. Let $B_n$ be the balance owed on the card after $n$ months. Find an explicit formula for $B_n$. +$$ B_0 = 3000 $$ + +$$ B_n = 1.015 \cdot B_{n - 1} - 150 \quad n \geq 1 $$ + +$$ B_1 = 1.015 \cdot B_0 - 150 = 1.015 \cdot 3000 - 150 $$ + +$$ B_2 = 1.015 \cdot B_1 - 150 = 1.015 \cdot (1.015 \cdot 3000 - 150) - 150 $$ + +$$ B_3 = 1.015 \cdot B_2 - 150 = 1.015 \cdot (1.015 \cdot (1.015 \cdot 3000 - 150) - 150) - 150 $$ + +Guess: + +$$ B_n = 1.015^n \cdot B_0 - 150 \cdot \left(\frac{1.015^n - 1}{1.015 - 1}\right) $$ + +$$ = 1.015^n \cdot 3000 - 150 \cdot \left(\frac{1.015^n - 1}{0.015}\right) $$ + +$$ = 1.015^n \cdot 3000 - 10000(1.015^n - 1) $$ + +$$ = 1.015^n \cdot 3000 - 10000(1.015^n) + 10000 $$ + +$$ = 10000 - 7000(1.015^n) $$ + c. How long will be required to pay off the debt? +$$ 0 = 10000 - 7000(1.015^n) $$ + +$$ 7000(1.015^n) = 10000 $$ + +$$ 1.015^n = \frac{10000}{7000} $$ + +$$ 1.015^n = \frac{10}{7} $$ + +$$ \ln(1.015^n) = \ln\left(\frac{10}{7}\right) $$ + +$$ n\ln(1.015) = \ln\left(\frac{10}{7}\right) $$ + +$$ n = \frac{\ln\left(\frac{10}{7}\right)}{\ln(1.015)} $$ + +$$ n \approx 24 \text{ months } = 2 \text{ years} $$ + d. What is the total amount of money the person will have paid for the loan? +$$ 24 \cdot 150 = \$3600 $$ + In 28-42 use mathematical induction to verify the correctness of the formula you obtained in the referenced exercise.