Aimathic – Recurrence relations – Free Math Worksheets

Recurrence relations

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Difficulty: 1 – 5

52885812

A patient receives a dose of medication every $6$ hours. Immediately after each dose, the concentration in the blood increases by $80\,\text{mg/L}$. During each $6$-hour interval, the body removes $40\%$ of the current concentration. At the start of treatment, $c_0=0$. a) Write a recursive formula for the concentration $c_n$ immediately after the $n$th dose. b) Find the limiting concentration immediately after a dose.

Hints

- Determine what percentage remains after the body removes $40\%$. - Each new value combines the remaining concentration with a fixed added amount. - At the limit, two consecutive concentration values are equal.

Solution

1. After a $6$-hour interval, $60\%$ of the previous concentration remains, so the retained amount is $0.6c_n$. 2. The next dose adds $80\,\text{mg/L}$. Therefore, $c_{n+1}=0.6c_n+80$, with $c_0=0$. 3. If the sequence approaches a limit $L$, then $L=0.6L+80$. 4. Solving gives $0.4L=80$, so $L=200$. 5. The limiting concentration immediately after a dose is $200\,\text{mg/L}$.

Answer

a) $c_0=0$; $c_{n+1}=0.6c_n+80$ b) $200\,\text{mg/L}$

52888012

Use fixed-point iteration to approximate a solution of $2x-\cos(x)-1=0$. 1. Rewrite the equation as $x=g(x)$, where $g(x)=\frac{1}{2}(\cos(x)+1)$. 2. Starting with $x_0=0.8$, calculate $x_1$, $x_2$, and $x_3$. Use radians and round each value to four decimal places.

Hints

- Move the non-$x$ terms to the other side, then divide by $2$. - Set the calculator to radians. - Repeatedly apply $x_{n+1}=g(x_n)$.

Solution

1. Rearrange the equation: $2x=\cos(x)+1$, so $x=\frac{1}{2}(\cos(x)+1)$. 2. In radians, $x_1=\frac{1}{2}(\cos(0.8)+1)\approx0.8484$. 3. Then $x_2=\frac{1}{2}(\cos(x_1)+1)\approx0.8306$. 4. Finally, $x_3=\frac{1}{2}(\cos(x_2)+1)\approx0.8372$.

Answer

1. $x=\frac{1}{2}(\cos(x)+1)$ 2. $x_1\approx0.8484$; $x_2\approx0.8306$; $x_3\approx0.8372$

52846312

A student is learning vocabulary for a language course. The first module contains $1200$ words. At the start, the student already knows $300$ words. Each week, the student learns $10\%$ of the words not yet known. a) How many words does the student know after $8$ weeks? b) After how many weeks will the student first know more than $1000$ words?

Hints

- The weekly increase is based on the remaining gap to $1200$. - Write a recursion using the current number of known words. - The initial gap to the goal is $900$ words. - Use logarithms to find the boundary time, then check whole weeks.

Solution

1. The recursive model is $V_{n+1}=V_n+0.10(1200-V_n)$, with $V_0=300$. 2. The equivalent explicit model is $V_n=1200-900(0.9)^n$. 3. a) $V_8=1200-900(0.9)^8\approx812.58$, so the student knows about $813$ words. 4. b) Find the boundary by solving $1200-900(0.9)^n=1000$. Then $0.9^n=\frac{2}{9}$, so $n\approx14.28$. 5. After $14$ weeks, the student knows fewer than $1000$ words; after $15$ weeks, the student knows more than $1000$.

Answer

a) About $813$ words b) After $15$ weeks

52847312

A town has $8000$ households. At the start of a fiber-optic campaign, $400$ households already have service. Each month, $12\%$ of the households that do not yet have service sign up. After how many months will at least $75\%$ of all households have fiber-optic service?

Hints

- First find the number of households represented by $75\%$. - Each month, the increase is based on the households that still lack service. - You may use either a recursion or an explicit formula for the remaining gap. - Use logarithms to find the boundary time, then check whole months.

Solution

1. The target is $0.75\cdot8000=6000$ households. 2. The recursive model is $H_{n+1}=H_n+0.12(8000-H_n)$, with $H_0=400$. 3. The equivalent explicit model is $H_n=8000-7600(0.88)^n$. 4. Solve the boundary equation $8000-7600(0.88)^n=6000$. Then $0.88^n=\frac{5}{19}$, so $n\approx10.44$. 5. After $10$ months, about $5883$ households have service; after $11$ months, about $6137$ do. Therefore, the target is first reached after $11$ months.

Answer

After $11$ months

52886712

A cup of coffee begins at $80^\circ\text{C}$ in a room kept at $21^\circ\text{C}$. Each minute, the difference between the coffee temperature and the room temperature decreases by $12\%$. a) Write a recursive rule for $T_n$, the coffee temperature after $n$ minutes. b) Write an explicit formula for $T_n$. c) After how many minutes is the coffee first cooler than $45^\circ\text{C}$?

Hints

- Focus on how the difference between the coffee and room temperatures changes. - Identify the temperature the sequence approaches. - Determine the multiplier for the portion of the difference that remains. - Evaluate terms until the temperature first falls below the target.

Solution

1. Since $88\%$ of the temperature difference remains each minute, $T_{n+1}-21=0.88(T_n-21)$. Therefore, $T_{n+1}=0.88T_n+2.52$, with $T_0=80$. 2. The initial difference from room temperature is $80-21=59$. Thus, $T_n=21+59\cdot0.88^n$. 3. Evaluate consecutive terms near the threshold: $T_7\approx45.112^\circ\text{C}$ and $T_8\approx42.218^\circ\text{C}$. 4. The coffee is first cooler than $45^\circ\text{C}$ after $8$ minutes.

Answer

a) $T_0=80$; $T_{n+1}=0.88T_n+2.52$ b) $T_n=21+59\cdot0.88^n$ c) $8$ minutes

52886812

A patient receives a $100\,\text{mg}$ medication dose every $8$ hours. Before each new dose, $75\%$ of the previous amount has been eliminated, so $25\%$ remains. Let $a_n$ be the amount in the body immediately after the $(n+1)$th dose, with $a_0=100$. a) Write a recursive rule for $a_n$. b) Find an explicit formula for $a_n$. c) After which dose does the amount first exceed $133\,\text{mg}$?

Hints

- Determine how much of the previous amount remains before the next dose. - Find the fixed point that the sequence approaches. - Remember that $a_0$ represents the amount after the first dose. - Evaluate terms until the amount first exceeds $133\,\text{mg}$.

Solution

1. One-fourth of the previous amount remains, and the next dose adds $100\,\text{mg}$. Thus, $a_{n+1}=0.25a_n+100$, with $a_0=100$. 2. The fixed point satisfies $S=0.25S+100$, so $S=\frac{400}{3}$. 3. The explicit form is $a_n=S-(S-a_0)0.25^n=\frac{400}{3}-\frac{100}{3}\cdot0.25^n$. 4. The relevant terms are $a_3=132.8125\,\text{mg}$ and $a_4=133.203125\,\text{mg}$. 5. Because $a_0$ is the amount after the first dose, $a_4$ corresponds to the fifth dose. The amount first exceeds $133\,\text{mg}$ after dose $5$.

Answer

a) $a_0=100$; $a_{n+1}=0.25a_n+100$ b) $a_n=\frac{400}{3}-\frac{100}{3}\cdot0.25^n$ c) After the fifth dose

52886912

A startup needs $\$150{,}000$ for an expansion and is comparing two loans. Offer A: The loan principal is $\$150{,}000$, the annual interest rate is $5.5\%$, and the company makes a $\$20{,}000$ payment at the end of each year. Offer B: The loan principal is $\$165{,}000$, but $\$15{,}000$ is withheld as an upfront financing charge, so the company receives $\$150{,}000$. The annual interest rate is $3.0\%$, and the annual payment is also $\$20{,}000$. Iterate each loan balance for five years. Which offer has the smaller remaining balance, and which should the company choose based on this comparison?

Hints

- Build each year's balance from the previous balance, interest, and end-of-year payment. - Use the actual loan principal, not only the amount received, as the initial balance. - Calculate one year at a time. - Compare the two balances after year $5$.

Solution

1. For Offer A, use $K_{n+1}=1.055K_n-20{,}000$, with $K_0=150{,}000$. 2. The balances after years $1$ through $5$ are $\$138{,}250.00$, $\$125{,}853.75$, $\$112{,}775.71$, $\$98{,}978.37$, and $\$84{,}422.18$. 3. For Offer B, use $K_{n+1}=1.03K_n-20{,}000$, with $K_0=165{,}000$. 4. The balances after years $1$ through $5$ are $\$149{,}950.00$, $\$134{,}448.50$, $\$118{,}481.96$, $\$102{,}036.41$, and $\$85{,}097.51$. 5. After five years, Offer A has the smaller balance by about $\$675.33$. Based only on the remaining balance after five years, the company should choose Offer A.

Answer

After five years, Offer A has a balance of approximately $\$84{,}422.18$, and Offer B has a balance of approximately $\$85{,}097.51$. Offer A has the smaller remaining balance.

52887012

An investor deposits $\$50{,}000$ and withdraws $\$4000$ at the end of each year after interest is credited. The investor compares two plans. Plan 1 earns $3.5\%$ each year. Plan 2 earns $2.0\%$ in the first year, and the annual rate then increases by $0.5$ percentage point each year: $2.5\%$ in year $2$, $3.0\%$ in year $3$, and so on. Iterate the account balance for four years. Which plan leaves more money after that time?

Hints

- Make a year-by-year table for each plan. - In Plan 2, the growth factor changes each year. - Apply the withdrawal after adding that year's interest.

Solution

1. For Plan 1, use $K_{n+1}=1.035K_n-4000$, with $K_0=50{,}000$. 2. The balances after years $1$ through $4$ are $\$47{,}750.00$, $\$45{,}421.25$, $\$43{,}010.99$, and $\$40{,}516.38$. 3. For Plan 2, apply the yearly factors $1.02$, $1.025$, $1.03$, and $1.035$, subtracting $\$4000$ after each year. 4. The balances after years $1$ through $4$ are $\$47{,}000.00$, $\$44{,}175.00$, $\$41{,}500.25$, and $\$38{,}952.76$. 5. Plan 1 leaves more money after four years.

Answer

After four years, Plan 1 has approximately $\$40{,}516.38$, and Plan 2 has approximately $\$38{,}952.76$. Plan 1 leaves more money.

52887112

At the beginning of each month, $500$ young fish are added to a pond. During the following month, $80\%$ of the total fish population survives. Let $a_n$ be the number of fish immediately after fish are added in month $n$, with $a_1=500$. a) Write a recurrence of the form $a_{n+1}=qa_n+c$. b) Use $a_n=q^{n-1}a_1+c\cdot\frac{q^{n-1}-1}{q-1}$ to find the fish population in month $5$. c) What value does the population approach in the long run? Justify your answer using a stable value of the recurrence.

Hints

- Identify the proportion that survives and the fixed number added. - Substitute $q$, $c$, and $a_1$ carefully into the given formula. - At a stable population, $a_{n+1}=a_n$.

Solution

1. Since $80\%$ of the fish survive and $500$ new fish are added, $a_{n+1}=0.8a_n+500$, with $a_1=500$. 2. Substitute $q=0.8$, $c=500$, and $n=5$: $a_5=0.8^4\cdot500+500\cdot\frac{0.8^4-1}{0.8-1}$. 3. Thus, $a_5=0.4096\cdot500+500\cdot2.952=204.8+1476=1680.8$. The model gives approximately $1681$ fish. 4. A stable value $L$ satisfies $L=0.8L+500$. Therefore, $0.2L=500$, so $L=2500$.

Answer

a) $a_{n+1}=0.8a_n+500$, with $a_1=500$ b) $a_5=1680.8$, or approximately $1681$ fish c) The population approaches $2500$ fish.

52887312

A patient takes a daily tablet containing $12\,\text{mg}$ of a medication. Overnight, the body eliminates $20\%$ of the amount currently present. At the beginning of treatment, no medication is in the body. a) Find the amount present immediately after each dose for the first five days. b) Find the long-term limiting amount if treatment continues. c) For medical reasons, the amount must not exceed $50\,\text{mg}$. Determine whether this limit will be maintained. d) How would the limiting amount change if the daily dose were doubled to $24\,\text{mg}$?

Hints

- Determine what fraction remains after one night. - Express each day's amount using the previous day's amount. - At a stable value, the next amount equals the current amount. - Decide whether the sequence increases and what value it approaches.

Solution

1. Since $80\%$ remains overnight, the recurrence is $a_{n+1}=0.8a_n+12$, with $a_0=0$. 2. The first five amounts are $a_1=12$, $a_2=21.6$, $a_3=29.28$, $a_4=35.424$, and $a_5=40.3392$, all in milligrams. 3. A limiting value $L$ satisfies $L=0.8L+12$. Therefore, $0.2L=12$, so $L=60\,\text{mg}$. 4. The sequence increases toward $60\,\text{mg}$, so it eventually exceeds $50\,\text{mg}$. 5. With a $24\,\text{mg}$ daily dose, $L=0.8L+24$, giving $L=120\,\text{mg}$.

Answer

a) Day 1: $12\,\text{mg}$; Day 2: $21.6\,\text{mg}$; Day 3: $29.28\,\text{mg}$; Day 4: $35.424\,\text{mg}$; Day 5: $40.3392\,\text{mg}$ b) $60\,\text{mg}$ c) No. The amount eventually exceeds $50\,\text{mg}$. d) The limiting amount doubles to $120\,\text{mg}$.

52887412

A remote mountain lake initially contains $1500\,\text{kg}$ of a certain salt. At the end of each year, $10\%$ of the salt flows out, and then natural inflow adds $200\,\text{kg}$ of salt. a) Write a recurrence for the salt amount and calculate the amount after two years and after four years. b) Find the long-term limiting amount. c) Find the new limiting amount if the annual inflow is reduced to $150\,\text{kg}$. d) Describe what happens if the inflow stops completely.

Hints

- Relate each year's amount to the previous year's amount. - Compare the initial amount with the limiting value. - Recalculate the fixed-point equation after changing the inflow. - A quantity multiplied by the same factor less than $1$ each year follows exponential decay.

Solution

1. Since $90\%$ remains and $200\,\text{kg}$ is added, $s_{n+1}=0.9s_n+200$, with $s_0=1500$. 2. The successive values are $s_1=1550$, $s_2=1595$, $s_3=1635.5$, and $s_4=1671.95$, in kilograms. 3. A limiting value satisfies $L=0.9L+200$. Thus, $0.1L=200$, so $L=2000\,\text{kg}$. 4. With an inflow of $150\,\text{kg}$, $L=0.9L+150$, so $L=1500\,\text{kg}$. Because the initial amount is already $1500\,\text{kg}$, it remains constant. 5. With no inflow, $s_{n+1}=0.9s_n$. This geometric sequence decreases exponentially toward $0$.

Answer

a) $s_{n+1}=0.9s_n+200$, with $s_0=1500$; $s_2=1595\,\text{kg}$; $s_4=1671.95\,\text{kg}$ b) $2000\,\text{kg}$ c) $1500\,\text{kg}$ d) The amount decreases exponentially toward $0\,\text{kg}$.

52887912

Consider the equation $x^3+3x-5=0$. 1. Use suitable function values to show that a solution lies in $[1, 2]$. 2. Rewrite the equation in the fixed-point form $x=\frac{5}{x^2+3}$. 3. Starting with $x_0=1$, use $x_{n+1}=\frac{5}{x_n^2+3}$ to calculate $x_1$, $x_2$, and $x_3$. Round each value to four decimal places.

Hints

- A sign change for a continuous function guarantees a zero between the inputs. - Factor out $x$ before isolating it. - Substitute each new value into the same iteration rule. - Keep enough precision before rounding each reported approximation.

Solution

1. Let $f(x)=x^3+3x-5$. Since $f(1)=-1$, $f(2)=9$, and $f$ is continuous, the Intermediate Value Theorem guarantees a solution in $(1, 2)$. 2. Rearrange: $x^3+3x=5$, so $x(x^2+3)=5$, and therefore $x=\frac{5}{x^2+3}$. 3. The iterations are $x_1=\frac{5}{1^2+3}=1.2500$, $x_2=\frac{5}{1.25^2+3}\approx1.0959$, and, using the unrounded value of $x_2$, $x_3\approx1.1902$.

Answer

1. $f(1)=-1<0$ and $f(2)=9>0$, so a solution lies in $[1, 2]$. 2. $x=\frac{5}{x^2+3}$ 3. $x_1=1.2500$; $x_2\approx1.0959$; $x_3\approx1.1902$

52888112

A stormwater retention basin has a capacity of $2000\,\text{m}^3$. At the beginning of a storm, it contains $400\,\text{m}^3$ of water. Each hour, $350\,\text{m}^3$ flows in. If the volume at the beginning of an hour is at least $1200\,\text{m}^3$, a pump also removes $120\,\text{m}^3$ during that hour. a) Write a piecewise recursive rule $V_{n+1}=f(V_n)$. b) Calculate $V_1$ through $V_4$. c) After how many complete hours is the basin first more than $75\%$ full?

Hints

- Determine the hourly change with the pump off and with the pump on. - Choose the correct case based on the volume at the beginning of each hour. - Convert $75\%$ of capacity to a volume. - Calculate the values in order.

Solution

1. If $V_n<1200$, the pump is off and $V_{n+1}=V_n+350$. If $V_n\geq1200$, the pump is on and $V_{n+1}=V_n+350-120=V_n+230$. 2. Starting with $V_0=400$, the values are $V_1=750$, $V_2=1100$, $V_3=1450$, and $V_4=1680$, all in cubic meters. 3. Seventy-five percent of capacity is $0.75\cdot2000=1500\,\text{m}^3$. Since $V_3=1450$ and $V_4=1680$, the basin is first more than $75\%$ full after $4$ hours.

Answer

a) $V_{n+1}=\begin{cases}V_n+350,&V_n<1200\\V_n+230,&V_n\geq1200\end{cases}$ b) $V_1=750\,\text{m}^3$; $V_2=1100\,\text{m}^3$; $V_3=1450\,\text{m}^3$; $V_4=1680\,\text{m}^3$ c) After $4$ hours

52888312

A patient takes $40\,\text{mg}$ of a medication each day, and the body eliminates $25\%$ of the amount present each day. Let $W_0=0$ represent the amount before treatment. Two models are considered: Model A: $W_n=0.75W_{n-1}+40$ Model B: $W_n=0.75(W_{n-1}+40)$ a) Explain how the timing of the dose and elimination differs between the models. b) Find the long-term limiting amount for each model. c) In Model A, on which day is at least $90\%$ of the limiting amount first reached?

Hints

- Use the order of operations to interpret each model. - At the limit, consecutive amounts are equal. - You may iterate the recurrence or derive an explicit formula. - Start from $W_0=0$ before the first dose.

Solution

1. In Model A, the existing amount is reduced first, and then the new dose is added. In Model B, the dose is added first, and then $25\%$ of the total is eliminated. 2. For Model A, $L=0.75L+40$, so $L_A=160\,\text{mg}$. 3. For Model B, $L=0.75(L+40)$, so $L_B=120\,\text{mg}$. 4. Ninety percent of $160\,\text{mg}$ is $144\,\text{mg}$. In Model A, $W_8\approx143.982\,\text{mg}$ and $W_9\approx147.986\,\text{mg}$. Therefore, at least $90\%$ of the limiting amount is first reached on day $9$.

Answer

a) Model A applies elimination before the dose; Model B applies the dose before elimination. b) Model A: $160\,\text{mg}$; Model B: $120\,\text{mg}$ c) Day $9$

52888412

A pond initially contains $10\,\text{kg}$ of algae. Natural growth adds $3\,\text{kg}$ each day, and a filter then removes $15\%$ of the total daily amount. a) Write a recursive rule for the algae mass $A_n$ after $n$ days. b) Find the long-term algae mass. c) After how many days is the algae mass within $1\%$ of the long-term value for the first time?

Hints

- Follow the order of events during one day. - At a stable amount, the next day's value equals the current value. - Convert $1\%$ of the limit to an allowable absolute error. - Continue until the error is first below that amount.

Solution

1. Because the filter acts after the daily growth, $A_n=0.85(A_{n-1}+3)=0.85A_{n-1}+2.55$, with $A_0=10$. 2. A limiting value satisfies $L=0.85L+2.55$. Thus, $0.15L=2.55$, so $L=17\,\text{kg}$. 3. One percent of the limit is $0.17\,\text{kg}$. The sequence gives $A_{22}\approx16.8040$, whose error is about $0.1960\,\text{kg}$, and $A_{23}\approx16.8334$, whose error is about $0.1666\,\text{kg}$. 4. The algae mass is first within $1\%$ of the limiting value after $23$ days.

Answer

a) $A_n=0.85(A_{n-1}+3)$, with $A_0=10$ b) $17\,\text{kg}$ c) After $23$ days

52846112

A new fish species is introduced into a lake with a carrying capacity of $3000$ fish. Initially, $600$ fish are released. Each year, the population increases by $18\%$ of the difference between the carrying capacity and the current population. a) Create a table showing the fish population for years $0$ through $5$. Round to the nearest fish. b) After how many years will the population first exceed $2500$ fish?

Hints

- Each year, find $18\%$ of the difference between $3000$ and the current population. - Add that increase to the current population to get the next term. - An explicit form can be written using the initial distance from the carrying capacity. - The answer to part b must be a whole year that satisfies the condition.

Solution

1. The recursive model is $B_{n+1}=B_n+0.18(3000-B_n)=0.82B_n+540$, with $B_0=600$. 2. Repeatedly applying the recursion gives $B_1=1032$, $B_2\approx1386$, $B_3\approx1677$, $B_4\approx1915$, and $B_5\approx2110$. 3. The equivalent explicit model is $B_n=3000-2400(0.82)^n$. 4. Find the boundary by solving $3000-2400(0.82)^n=2500$. Then $0.82^n=\frac{500}{2400}$, so $n=\frac{\ln\left(\frac{500}{2400}\right)}{\ln(0.82)}\approx7.90$. 5. Therefore, the population first exceeds $2500$ fish in year $8$.

Answer

a) <table> <tr><td>Year</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr> <tr><td>Population</td><td>600</td><td>1032</td><td>1386</td><td>1677</td><td>1915</td><td>2110</td></tr> </table> b) Year $8$

52846812

A battery is connected to a charger at a charge level of $12\%$. Every $5$ minutes, the charge level increases by $15\%$ of the amount still needed to reach $100\%$. a) Write a function $C(n)$ for the charge level after $n$ complete $5$-minute intervals. b) Find the charge level after $30$ minutes. c) After how many complete minutes will the battery first reach at least $90\%$?

Hints

- Focus on the percentage that is still missing from full charge. - The missing amount is multiplied by the same factor every $5$ minutes. - Convert $30$ minutes to a number of charging intervals. - After finding a noninteger number of intervals, check the next complete interval.

Solution

1. The initial gap to full charge is $100-12=88$ percentage points. Each interval leaves $85\%$ of the previous gap. 2. Therefore, $C(n)=100-88(0.85)^n$. 3. b) Thirty minutes is $6$ intervals, so $C(6)=100-88(0.85)^6\approx66.81\%$. 4. c) The boundary equation is $100-88(0.85)^n=90$, so $0.85^n=\frac{10}{88}$. 5. This gives $n\approx13.38$. Since the model changes only at complete $5$-minute intervals, check $n=13$ and $n=14$. The levels are about $89.36\%$ and $90.96\%$, respectively. 6. Therefore, the battery first reaches at least $90\%$ after $14$ intervals, or $70$ minutes.

Answer

a) $C(n)=100-88(0.85)^n$ b) About $66.81\%$ c) $70$ minutes

52887212

A sequence is defined by $a_{n+1}=1.5a_n-100$, with initial value $a_1$. a) Substitute $q=1.5$ and $c=-100$ into $a_n=q^{n-1}a_1+c\frac{q^{n-1}-1}{q-1}$ and show that $a_n=200+(a_1-200)1.5^{n-1}$. b) Find the initial value that makes the sequence constant. c) Describe the long-term behavior when $a_1>200$ and when $a_1<200$.

Hints

- Simplify division by $0.5$. - A constant sequence cannot depend on $n$. - Consider the behavior of a power with base greater than $1$. - Track the sign of the coefficient multiplying the power.

Solution

1. Substitute into the explicit formula: $a_n=1.5^{n-1}a_1-100\frac{1.5^{n-1}-1}{0.5}$. 2. Since dividing by $0.5$ multiplies by $2$, $a_n=a_1\cdot1.5^{n-1}-200(1.5^{n-1}-1)$. 3. Therefore, $a_n=200+(a_1-200)1.5^{n-1}$. 4. The sequence is constant when $a_1-200=0$, so $a_1=200$. 5. Because $1.5^{n-1}$ grows without bound, $a_n\to\infty$ when $a_1>200$, and $a_n\to-\infty$ when $a_1<200$.

Answer

a) $a_n=200+(a_1-200)1.5^{n-1}$ b) $a_1=200$ c) If $a_1>200$, then $a_n\to\infty$. If $a_1<200$, then $a_n\to-\infty$.

52887612

A fish population in a managed lake has a carrying capacity of $G=2000$. Its weekly population follows the discrete logistic model $B_{n+1}=B_n+kB_n(G-B_n)$. Initially, $B_0=400$, and after one week, $B_1=528$. 1. Find the growth constant $k$. 2. Use the model to predict the populations in weeks $2$ and $3$, rounded to whole fish. 3. Without further calculation, explain what happens to the weekly increase $B_{n+1}-B_n$ as $B_n$ approaches the carrying capacity.

Hints

- Substitute $B_0$, $B_1$, and $G$ into the model to solve for $k$. - Use the unrounded population from one week in the next calculation. - Examine the factor $G-B_n$ when the population is close to capacity.

Solution

1. Substitute the given values: $528=400+k\cdot400(2000-400)$. Therefore, $128=640{,}000k$, so $k=0.0002$. 2. For week $2$, $B_2=528+0.0002\cdot528(2000-528)=683.4432\approx683$. 3. Using the unrounded value of $B_2$, $B_3=683.4432+0.0002\cdot683.4432(2000-683.4432)\approx863.4016\approx863$. 4. As $B_n$ approaches $G$, the factor $G-B_n$ approaches $0$. Therefore, the weekly increase approaches $0$.

Answer

1. $k=0.0002$ 2. $B_2\approx683$; $B_3\approx863$ 3. The weekly increase approaches $0$.

52887712

A lake initially contains $B_0=800$ fish and has carrying capacity $K=5000$. After one year without harvesting, the population is $B_1=912$. The population follows $B_{n+1}=B_n+rB_n\frac{K-B_n}{K}$. 1. Find the yearly growth-rate parameter $r$. 2. With an annual harvest fraction $q$, the model becomes $B_{n+1}=B_n+rB_n\frac{K-B_n}{K}-qB_n$. Find the minimum harvest fraction for which every initial population $0<B_0\leq K$ decreases in the long run toward $0$.

Hints

- Use the first-year increase to solve for the unknown growth-rate parameter. - Compare the harvest term with the largest possible natural growth rate. - Examine the model for very small positive populations.

Solution

1. Substitute $B_0=800$, $B_1=912$, and $K=5000$: $112=r\cdot800\cdot\frac{4200}{5000}=672r$. Hence, $r=\frac{1}{6}\approx0.1667$. 2. The population change is $B_{n+1}-B_n=B_n\left(r\frac{K-B_n}{K}-q\right)$. 3. For $0<B_n\leq K$, the largest possible value of $\frac{K-B_n}{K}$ is approached as $B_n$ approaches $0$, and it is less than $1$. If $q\geq r$, the expression in parentheses is negative for every positive population, so the population decreases toward $0$. 4. If $q<r$, sufficiently small positive populations have positive growth. Therefore, the critical harvest fraction is $q=r=\frac{1}{6}\approx16.67\%$.

Answer

1. $r=\frac{1}{6}\approx0.1667$ 2. The minimum harvest fraction is $q=\frac{1}{6}$, or approximately $16.67\%$.

52888512

The equation $x^3-5x+1=0$ can be rewritten in each of the following fixed-point forms: (1) $x=\frac{x^3+1}{5}$ (2) $x=\sqrt[3]{5x-1}$ (3) $x=\frac{1}{5-x^2}$ a) Verify algebraically that all three forms are equivalent to the original equation. b) For each function $g$, begin with $x_0=0$ and calculate the first four terms of $x_{n+1}=g(x_n)$. Determine whether the sequence approaches the solution in $[0, 1]$. c) Beginning with $x_0=2$, calculate several terms for each form and determine which one approaches the solution in $[2, 3]$.

Hints

- Rearrange each proposed equation back to the original polynomial equation. - Use the previous result as the input for the next iteration. - Observe whether the values stabilize, move away, or approach a different solution. - Local convergence depends on the behavior of $g$ near the target fixed point.

Solution

1. Each equation rearranges to $x^3-5x+1=0$: multiplying form (1) by $5$, cubing form (2), and multiplying form (3) by $5-x^2$ all recover the original equation. 2. With $x_0=0$, form (1) gives $0.2$, $0.2016$, $0.2016387$, and $0.2016397$, so it approaches the solution $x\approx0.2016$. 3. With $x_0=0$, form (2) gives $-1$, $-1.8171206$, $-2.1605648$, and $-2.2768197$. It does not approach the solution in $[0, 1]$; it moves toward the negative solution. 4. With $x_0=0$, form (3) gives $0.2$, $0.2016129$, $0.2016392$, and $0.2016397$, so it approaches the solution $x\approx0.2016$. 5. With $x_0=2$, forms (1) and (3) move toward the small positive solution. Form (2) gives $2.0800838$, $2.1104857$, $2.1218007$, and $2.1259814$, approaching the solution $x\approx2.1284$. Therefore, only form (2) is suitable for the solution in $[2, 3]$.

Answer

a) All three fixed-point forms are algebraically equivalent to $x^3-5x+1=0$. b) Forms (1) and (3) approach $x\approx0.2016$. Form (2) moves toward the negative solution instead. c) Only form (2), $x=\sqrt[3]{5x-1}$, approaches the solution in $[2, 3]$ from $x_0=2$.

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