Aimathic – Model with recursive sequences

Model with recursive sequences

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Difficulty: 1 – 5

55062811

A plant is $6\,\text{cm}$ tall at the end of week $1$. In a simple model, its height increases by $4\,\text{cm}$ each week. The model is $h_1=6$ and $h_n=h_{n-1}+4$ for $n\ge2$. Find $h_2,h_3,h_4,h_5$, and explain what the $+4$ means in the recurrence.

Hints

- Start with the given initial value before using the recursive rule. - Each new value depends on the immediately preceding value. - Interpret the added constant in the units and time interval of the model.

Solution

Apply the recurrence one week at a time: $h_2=10$, $h_3=14$, $h_4=18$, and $h_5=22$. The $+4$ means the model adds $4\,\text{cm}$ to the previous week's height to obtain the next week's height.

Answer

$h_2=10\,\text{cm}$, $h_3=14\,\text{cm}$, $h_4=18\,\text{cm}$, $h_5=22\,\text{cm}$. The $+4$ represents weekly growth of $4\,\text{cm}$.

55062911

An account starts with $\$1000$ and earns $5\%$ interest at the end of each year, with no deposits or withdrawals. Let $B_n$ be the balance after $n$ years. a) Write a recursive model with an initial value. b) Write an equivalent explicit formula. c) Find $B_3$ to the nearest cent.

Hints

- Convert the percent increase into the multiplier applied to the previous balance. - In the recursive form, distinguish the starting balance from the update rule. - In the explicit form, the exponent counts how many annual multiplications have occurred.

Solution

a) The initial balance is $B_0=1000$. Each year the balance is multiplied by $1.05$, so $B_n=1.05B_{n-1}$ for $n\ge1$. b) Repeated multiplication by $1.05$ gives $B_n=1000(1.05)^n$. c) $B_3=1000(1.05)^3=1157.625$, so the balance is $\$1157.63$.

Answer

a) $B_0=1000$, $B_n=1.05B_{n-1}$ for $n\ge1$ b) $B_n=1000(1.05)^n$ c) $\$1157.63$

52844811

A patient takes a $20\,\text{mg}$ tablet once each day at the same time. During each day, the body eliminates $30\%$ of the medication currently present. a) Make a table of the amount in the body immediately after each dose for days $1$ through $5$. Before the first dose, the amount is $0\,\text{mg}$. b) On what day does the amount first exceed $50\,\text{mg}$? c) Find the theoretical long-term amount in the body immediately after a dose.

Hints

- At each step, keep the stated fraction of the previous amount before adding the new dose. - Compare the generated values with the threshold only after the recurrence is applied. - At equilibrium, one full recurrence step leaves the amount unchanged.

Solution

a) If $x_n$ is the amount immediately after the dose on day $n$, then $x_n=0.7x_{n-1}+20$, with $x_0=0$. Thus, $x_1=20$, $x_2=34$, $x_3=43.8$, $x_4=50.66$, and $x_5=55.462$, all in milligrams. b) The first value greater than $50\,\text{mg}$ is $x_4=50.66\,\text{mg}$, so the amount first exceeds the threshold on day $4$. c) At equilibrium, $L=0.7L+20$. Thus $0.3L=20$, so $L=\frac{200}{3}\approx66.67\,\text{mg}$.

Answer

a) <table> <tr> <th>Day</th> <th>Amount after dose (mg)</th> </tr> <tr><td>$1$</td><td>$20$</td></tr> <tr><td>$2$</td><td>$34$</td></tr> <tr><td>$3$</td><td>$43.8$</td></tr> <tr><td>$4$</td><td>$50.66$</td></tr> <tr><td>$5$</td><td>$55.462$</td></tr> </table> b) Day $4$ c) $\frac{200}{3}\,\text{mg}\approx66.67\,\text{mg}$

52845011

Immediately after the final dose of an earlier treatment, a patient has $40\,\text{mg}$ of a medication in the body at day $0$. The patient now takes $5\,\text{mg}$ once daily. Before each new dose, the body eliminates $20\%$ of the medication present. a) Find the amount immediately after the dose on days $1$ and $2$. b) Show that the amount decreases over time despite the daily dose, and find the value to which it stabilizes. c) How would the limiting amount change if the body eliminated $25\%$ each day instead of $20\%$?

Hints

- Compare the amount removed during a day with the fixed amount added afterward. - At equilibrium, one recurrence step produces no net change. - In part c, change only the retention factor in the equilibrium equation.

Solution

a) The recurrence is $A_{n+1}=0.8A_n+5$, with $A_0=40$. Thus, $A_1=0.8\cdot40+5=37\,\text{mg}$ and $A_2=0.8\cdot37+5=34.6\,\text{mg}$. b) The daily change is $A_{n+1}-A_n=5-0.2A_n$, which is negative whenever $A_n>25$. Also, $A_{n+1}-25=0.8(A_n-25)$, so the terms remain above $25$ and approach it. The limiting amount is $25\,\text{mg}$. c) With a $25\%$ daily elimination, the limit satisfies $L=0.75L+5$, giving $L=20\,\text{mg}$. The limiting amount decreases by $5\,\text{mg}$.

Answer

a) Day $1$: $37\,\text{mg}$; day $2$: $34.6\,\text{mg}$ b) The amount decreases toward $25\,\text{mg}$. c) The limiting amount would decrease to $20\,\text{mg}$.

52845711

A newly planted forest can support at most $4000$ trees. Initially, $800$ young trees are planted. Each year, the number of trees increases by $25\%$ of the remaining capacity, the difference between $4000$ and the current number of trees. a) Write a recurrence for $B_{n+1}$ in terms of $B_n$. b) Find the model values for the first five years after planting. Round to the nearest whole tree. c) Explain why the model represents bounded growth and describe how the annual increases behave over time.

Hints

- Express the remaining capacity as the difference between the maximum and the current amount. - Add the stated fraction of that remaining capacity to the current value. - For the long-run behavior, focus on what happens to the gap from capacity.

Solution

a) The annual increase is $0.25(4000-B_n)$, so $B_{n+1}=B_n+0.25(4000-B_n)=0.75B_n+1000$, with $B_0=800$. b) Without rounding intermediate values, $B_1=1600$, $B_2=2200$, $B_3=2650$, $B_4=2987.5\approx2988$, and $B_5=3240.625\approx3241$. c) The population is bounded above by $4000$. As the gap to capacity becomes smaller, the annual increase $0.25(4000-B_n)$ also becomes smaller and approaches $0$.

Answer

a) $B_{n+1}=B_n+0.25(4000-B_n)$, or $B_{n+1}=0.75B_n+1000$ b) $B_1=1600$, $B_2=2200$, $B_3=2650$, $B_4\approx2988$, $B_5\approx3241$ c) The capacity $4000$ is an upper bound, and the annual increases approach $0$.

55063011

Two recursive models both start at $10$: Model I: $x_0=10$, $x_n=x_{n-1}+4$ Model II: $y_0=10$, $y_n=1.5y_{n-1}$ Panels a) and b) show the first six values of the two models, but the panels are not named. a) Match each panel to Model I or Model II. b) Explain how the spacing between consecutive plotted values supports your match.

Hints

- Compare the vertical gaps between neighboring points in each panel. - An additive recursive rule produces a constant difference from one step to the next. - A multiplicative recursive rule produces a constant ratio, so its absolute increases need not stay constant.

Solution

a) Panel a) is Model I, and panel b) is Model II. b) In panel a), the vertical change from one point to the next stays constant at $4$, matching the additive recurrence. In panel b), the vertical changes get larger because each value is $1.5$ times the previous value, matching the multiplicative recurrence.

Answer

a) Panel a): Model I; panel b): Model II. b) Panel a) has constant additive changes, while panel b) has multiplicative growth with increasing vertical changes.

52844911

An artificial pond begins empty. Each day, $15\%$ of the water evaporates, and then $60\,\text{L}$ of fresh water is added. a) Find the amount of water immediately after the addition on days $2$, $3$, and $4$. b) Find the theoretical upper limit of the water amount. c) On what day does the amount first exceed $250\,\text{L}$?

Hints

- Write one recurrence step using the retained water and the fixed addition. - At the limiting amount, one full day leaves the amount unchanged. - For the threshold, compare consecutive day values near the target.

Solution

a) Let $W_n$ be the amount immediately after the addition on day $n$. Then $W_{n+1}=0.85W_n+60$, with $W_0=0$. This gives $W_1=60$, $W_2=111$, $W_3=154.35$, and $W_4=191.1975\approx191.20$, in liters. b) At equilibrium, $L=0.85L+60$. Therefore, $0.15L=60$, so $L=400\,\text{L}$. c) The explicit form is $W_n=400(1-0.85^n)$. It gives $W_6\approx249.14\,\text{L}$ and $W_7\approx271.77\,\text{L}$, so the amount first exceeds $250\,\text{L}$ on day $7$.

Answer

a) Day $2$: $111\,\text{L}$; day $3$: $154.35\,\text{L}$; day $4$: about $191.20\,\text{L}$ b) $400\,\text{L}$ c) Day $7$

53351411

A town has $20{,}000$ residents. A new car-sharing service begins with $2000$ registered users. Each month, the company expects to register $20\%$ of the residents who are not yet users. a) Find the number of users after each of the first three months and complete the table. <table> <tr><th>Month $x$</th><td>$0$</td><td>$1$</td><td>$2$</td><td>$3$</td></tr> <tr><th>Users $N(x)$</th><td>$2000$</td><td>...</td><td>...</td><td>...</td></tr> </table> b) Write an explicit function for the number of users after $x$ months. c) About how many users does the model predict after one year? In the graph, the green points show the monthly user model and the dashed line $s$ shows the saturation level. Use the graph to explain why the model never exceeds $20{,}000$ users.

Hints

- Each month, first determine how many residents are still nonusers. - Track the remaining gap to the saturation level; that gap changes by a constant factor each month. - In the graph, compare the discrete monthly points with the horizontal saturation line.

Solution

a) Each month, add $20\%$ of the remaining nonusers. Month $1$: $2000+0.20\cdot(20{,}000-2000)=5600$. Month $2$: $5600+0.20\cdot(20{,}000-5600)=8480$. Month $3$: $8480+0.20\cdot(20{,}000-8480)=10{,}784$. b) The number of nonusers begins at $18{,}000$ and is multiplied by $0.8$ each month. Therefore, $N(x)=20{,}000-18{,}000(0.8)^x$ for whole-number months $x\ge0$. c) $N(12)=20{,}000-18{,}000(0.8)^{12}\approx18{,}763$. The plotted monthly points stay below the dashed line $y=20{,}000$ and move closer to it. Algebraically, $18{,}000(0.8)^x>0$ for every finite whole-number $x$, so $N(x)<20{,}000$.

Answer

a) Month $1$: $5600$; month $2$: $8480$; month $3$: $10{,}784$ b) $N(x)=20{,}000-18{,}000(0.8)^x$ for whole-number $x\ge0$ c) About $18{,}763$ users; the monthly points approach $20{,}000$ from below and never reach or exceed it.

55063111

A recursive model is $P_{n+1}=0.8P_n+c$, with $P_0=100$. The constant $c$ is to be chosen so that the model has an equilibrium value of $500$. a) Find $c$. b) Write an explicit formula for $P_n$. c) What is the smallest nonnegative integer $n$ for which $P_n>450$?

Hints

- At an equilibrium, replacing the current value by the equilibrium must produce the same value again. - Re-center the recurrence around the equilibrium so that the constant addition disappears. - Once the explicit form is available, translate the target level into an inequality involving the exponential factor. - Because the index is an integer, verify the first candidate above the logarithmic threshold and the preceding index.

Solution

a) At equilibrium, the value stays at $500$, so $500=0.8(500)+c$. Thus $c=100$. b) Subtract the equilibrium value from both sides of the recurrence: $P_{n+1}-500=0.8(P_n-500)$. Since $P_0-500=-400$, $P_n-500=-400(0.8)^n$, so $P_n=500-400(0.8)^n$. c) We need $500-400(0.8)^n>450$, which is equivalent to $(0.8)^n<0.125$. This gives $n>\frac{\ln(0.125)}{\ln(0.8)}\approx9.32$. Checking consecutive integer indices, $P_9\approx446.31$ and $P_{10}\approx457.05$. Therefore, the smallest value is $n=10$.

Answer

a) $c=100$ b) $P_n=500-400(0.8)^n$ c) $n=10$

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