tomit4/discrete_mathematics_with_applications
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

🚧 Still mid 5.7

Browse source
This commit is contained in:
tomit4 2026-07-10 05:50:47 -07:00
parent fc358579a9
commit ecb409cc2f
1 changed files with 360 additions and 1 deletions
Download patch file Download diff file Expand all files Collapse all files

361
chapter_5/exercises.md
View file

@ -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 = $\dfrac{1}{m}$<sup>th</sup> 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.