- Start with the given value at index \(0\) and apply the rule once. - Each new row depends on the value in the immediately preceding row. - Interpret the subscript as the number of recurrence steps from the starting index.

1. Apply the recurrence to \(a_0\): \(a_1=3\cdot2-1=5\). 2. Use the newly found value: \(a_2=3\cdot5-1=14\). 3. Continue once more: \(a_3=3\cdot14-1=41\). 4. The value \(a_2=14\) is the result after applying the recurrence rule twice starting from \(a_0\). It is also the input used to produce \(a_3\).

\(a_1=5\), \(a_2=14\), and \(a_3=41\). The term \(a_2\) is the value after two recurrence steps from the starting term.

- A fixed point gives the same value before and after the recurrence update. - Represent that unchanged value with one variable on both sides of the update rule. - Check the result by putting it back into the recurrence.

1. At a fixed point, the current and next values are equal, so \(L=\frac{1}{4}L+9\). 2. Thus \(\frac{3}{4}L=9\), giving \(L=12\). 3. Substitution verifies the result: \(\frac{1}{4}(12)+9=12\).

The fixed point is \(12\).

- Match index \(0\) with the situation before any update has happened. - Identify the change that happens during one day. - Check that the recurrence moves from the value after one day to the value after the next day.

1. Before any days of sales have passed, \(a_0=120\). 2. Each day reduces the number remaining by \(15\), so \(a_{n+1}=a_n-15\) for \(0\le n<5\). 3. Applying the recurrence twice gives \(a_1=105\) and \(a_2=90\). 4. The initial condition \(a_0=120\) means there are \(120\) tickets remaining before the first day of sales.

a) \(a_0=120\) and \(a_{n+1}=a_n-15\) for \(0\le n<5\). b) \(a_2=90\). c) \(a_0=120\) is the number of tickets remaining before any sales day has occurred.

- Apply the rule to the latest value, not the starting value each time. - Compare the differences and ratios after calculating the terms.

1. The first update gives \(a_1=2^2-1=3\). 2. The next two terms are \(a_2=3^2-1=8\) and \(a_3=8^2-1=63\). 3. The changes and ratios do not stay constant. Each update squares the previous term and subtracts \(1\).

\(a_1=3\), \(a_2=8\), \(a_3=63\). The rule uses neither a constant addition nor a constant multiplier.

- What happens if you try to find \(b_2\) using only one of the starting values? - Work forward one index at a time, keeping track of the two most recent terms.

1. Use the two preceding terms: \(b_2=2+1=3\). 2. Continue: \(b_3=3+2=5\) and \(b_4=5+3=8\). 3. Every update depends on two earlier values, so the first new term cannot be found from one starting value alone.

\(b_2=3\), \(b_3=5\), \(b_4=8\). Two starting values are needed because each new term uses the preceding two.

- Read the coordinates of the earliest plotted point before comparing terms. - How do the values at consecutive indices compare? - Apply that operation one more time to find the next term.

1. The plotted terms at \(n=0,1,2,3\) are \(2,4,8,16\). 2. Each term is twice the previous one, so \(a_0=2\) and \(a_{n+1}=2a_n\) for \(n\ge0\). 3. The next term is \(a_4=2\cdot16=32\).

\(a_0=2\), common ratio \(2\), \(a_{n+1}=2a_n\) for \(n\ge0\), and \(a_4=32\).

- Determine what fraction of the concentration remains after one \(6\)-hour interval. - Find a value that would remain unchanged by one complete recurrence step. - Compare each term's distance from that stable value with the previous term's distance.

1. After a \(6\)-hour interval, \(60\%\) of the previous concentration remains. The next dose then adds \(80\,\text{mg/L}\), so \(c_{n+1}=0.6c_n+80\), with \(c_0=0\). 2. A stable value \(L\) satisfies \(L=0.6L+80\), giving \(L=200\). 3. Subtract \(200\) from both sides of the recurrence: \(c_{n+1}-200=0.6(c_n-200)\). Thus the distance from \(200\) is multiplied by \(0.6\) after every dose. Since \(0<0.6<1\), this geometric difference approaches \(0\). 4. Therefore \(c_n\) approaches \(200\,\text{mg/L}\).

a) \(c_0=0\) and \(c_{n+1}=0.6c_n+80\). b) \(200\,\text{mg/L}\); the difference from \(200\) is multiplied by \(0.6\) at each step, so it approaches \(0\).

- After the first update, compare the result with the starting value. What might that tell you about the next update? - Do not use rounded displayed values when calculating later terms. - Once you have an equation for the unchanged value, how can you rewrite it with \(0\) on one side?

1. The first update gives \(x_1=\frac{1}{2}(\cos(0.8)+1)\approx0.8484\). 2. Use each unrounded result as the next input to get \(x_2\approx0.8306\) and \(x_3\approx0.8372\). 3. If \(L\) stays unchanged, then \(L=\frac{1}{2}(\cos(L)+1)\). Multiplying by \(2\) and rearranging gives \(2L-\cos(L)-1=0\).

1. \(x_1\approx0.8484\), \(x_2\approx0.8306\), \(x_3\approx0.8372\). 2. At a fixed point, \(L=\frac{1}{2}(\cos(L)+1)\), so \(2L-\cos(L)-1=0\).

- Translate the percentage drained into the percentage that remains. - Preserve the stated order: drain first, then add the fixed amount. - Use each day's result as the input for the next recurrence step.

1. Draining \(15\%\) leaves \(85\%\) of the current amount, so after the drain the tank contains \(0.85V_n\). 2. The \(40\)-gallon addition occurs afterward. Therefore \(V_{n+1}=0.85V_n+40\), with \(V_0=300\). 3. Iterating gives \(V_1=0.85\cdot300+40=295\), \(V_2=0.85\cdot295+40=290.75\), and \(V_3=0.85\cdot290.75+40=287.1375\).

a) \(V_0=300\) and \(V_{n+1}=0.85V_n+40\). b) \(V_1=295\) gallons, \(V_2=290.75\) gallons, and \(V_3=287.1375\) gallons.

- Iterate the recurrence several times before trying to describe the long-term behavior. - Separate even and odd indices when looking for an explicit rule. - Boundedness only limits where the terms can lie; convergence requires the terms to approach one value.

1. Repeated substitution gives the terms \(2,4,2,4,2,4\). 2. The parity pattern is described by \(a_n=3-(-1)^n\). 3. The sequence is bounded because every term is either \(2\) or \(4\). 4. It does not converge because the even-indexed terms are always \(2\) while the odd-indexed terms are always \(4\), so the sequence never approaches one common value.

a) \(2,4,2,4,2,4\). b) \(a_n=3-(-1)^n\). c) The sequence is bounded but does not converge; it alternates forever between \(2\) and \(4\).

- Notice how many earlier terms appear on the right side of the recurrence. - Compute new terms in order so that every required earlier value is already known. - Ask what information would be missing when trying to compute the first new term if only one starting value were given.

1. Use the two preceding terms each time: \(a_2=5+2(2)=9\). 2. Then \(a_3=9+2(5)=19\), \(a_4=19+2(9)=37\), and \(a_5=37+2(19)=75\). 3. The recurrence for \(a_n\) depends on both \(a_{n-1}\) and \(a_{n-2}\). To compute the first new term \(a_2\), two earlier values are therefore required; one initial value alone would not determine it.

a) \(a_2=9\), \(a_3=19\), \(a_4=37\), \(a_5=75\). b) Each update uses the two preceding terms, so two initial values are needed to start the recurrence uniquely.

- Each pair of neighboring table entries gives an equation in the two unknown constants. - Compare two equations to eliminate one unknown. - Use another row to check the proposed rule before predicting the next term.

1. The first two updates give \(8=3m+c\) and \(18=8m+c\). 2. Subtract the equations: \(10=5m\), so \(m=2\) and \(c=8-3\cdot2=2\). 3. The third update checks: \(2\cdot18+2=38\). Thus \(a_4=2\cdot38+2=78\).

\(m=2\), \(c=2\), and \(a_4=78\).

- Bring the terms involving the fixed value together before solving its equation. - Check any values you find by putting them back into the original update. - Start the updates at the given initial term and check whether a later term changes.

1. A fixed point \(L\) satisfies \(L=L(5-L)\), so \(L(4-L)=0\). 2. The fixed points are \(L=0\) and \(L=4\). Substituting either value leaves it unchanged. 3. Starting from \(a_0=1\), the first update gives \(a_1=1(5-1)=4\), and the next gives \(a_2=4(5-4)=4\).

a) \(0\) and \(4\). b) \(a_1=4\), \(a_2=4\); the sequence reaches the fixed point \(4\).

- Apply each recurrence separately after the shared first update. - Compare the operations in the two rules, not just the first matching values. - Think about whether squaring and doubling always have the same effect.

1. Sequence A has terms \(a_0=2\), \(a_1=4\), \(a_2=16\), and \(a_3=256\). 2. Sequence B has terms \(b_0=2\), \(b_1=4\), \(b_2=8\), and \(b_3=16\). 3. Squaring a number is different from doubling it. The rules agree at the first update only because the current term is \(2\).

A: \(2,4,16,256\). B: \(2,4,8,16\). They agree at indices \(0\) and \(1\), but A squares each term while B doubles it.

- The weekly increase depends on the current number of words still unknown. - After writing one update rule, use each new value as the input for the next week. - For the threshold, compare consecutive recurrence values around \(1000\) rather than solving an explicit exponential equation.

1. The recurrence is \(V_0=300\) and \(V_{n+1}=V_n+0.10(1200-V_n)=0.9V_n+120\). 2. Repeated updates give \(V_1=390\), \(V_2=471\), \(V_3=543.9\), \(V_4=609.51\), \(V_5=668.559\), \(V_6=721.7031\), \(V_7=769.53279\), and \(V_8=812.579511\). Thus after \(8\) weeks she knows about \(813\) words. 3. Continuing the same recurrence gives \(V_{14}\approx994.1089\) and \(V_{15}\approx1014.6980\). Therefore week \(15\) is the first week in which the count exceeds \(1000\).

a) \(V_0=300\), \(V_{n+1}=0.9V_n+120\). b) About \(813\) words. c) Week \(15\).

- In the original recurrence, the monthly increase is based on the households still lacking service. - Subtract the next service count from \(8000\) and simplify before trying to write an explicit formula for the remaining households. - Translate the \(75\%\) service target into a maximum allowed value of \(U_n\).

1. The recurrence is \(H_0=400\) and \(H_{n+1}=H_n+0.12(8000-H_n)=0.88H_n+960\). 2. Since \(U_n=8000-H_n\), \(U_{n+1}=8000-H_{n+1}=8000-(0.88H_n+960)=0.88(8000-H_n)=0.88U_n\). Also, \(U_0=7600\), so \(U_n=7600(0.88)^n\). 3. At least \(75\%\) with service means at most \(2000\) households remain without service. Solve \(7600(0.88)^n\le2000\). The boundary is \(n\approx10.44\). At month \(10\), about \(2116.61\) households remain without service, while at month \(11\), about \(1862.61\) remain. Therefore month \(11\) is the first qualifying month.

a) \(H_0=400\), \(H_{n+1}=0.88H_n+960\). b) \(U_0=7600\), \(U_{n+1}=0.88U_n\), so \(U_n=7600(0.88)^n\). c) Month \(11\).

- 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.

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.

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

- 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 \(133\,\text{mg}\).

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 \(5\).

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

- Build each year's balance from the previous balance, interest, and end-of-year payment. - For each offer, which amount does the lender charge interest on? - Calculate one year at a time. - Compare the two balances after year \(5\).

1. For Offer A, use \(K_{n+1}=1.055K_n-20{,}000\), with \(K_0=150{,}000\). 2. The year-end balances for Offer A 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 year-end balances for Offer B are \(\$149{,}950.00\), \(\$134{,}448.50\), \(\$118{,}481.96\), \(\$102{,}036.41\), and \(\$85{,}097.51\). 5. Offer A has the smaller remaining balance after five years, by about \(\$675.33\). Based on this comparison, choose Offer A.

Offer A's year-end balances are approximately \(\$138{,}250.00\), \(\$125{,}853.75\), \(\$112{,}775.71\), \(\$98{,}978.37\), and \(\$84{,}422.18\). Offer B's are approximately \(\$149{,}950.00\), \(\$134{,}448.50\), \(\$118{,}481.96\), \(\$102{,}036.41\), and \(\$85{,}097.51\). Offer A has the smaller balance after five years; choose A based on this comparison.

- 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.

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.

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

- Identify the surviving fraction and the fixed number added each month. - Substitute the recurrence parameters into the supplied explicit formula without rounding early. - After finding a stable value, compare the next term's distance from it with the current term's distance.

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 \(n=5\): \(a_5=0.8^4\cdot500+500\cdot\frac{0.8^4-1}{0.8-1}=1680.8\). The model gives approximately \(1681\) fish. 3. A stable value \(L\) satisfies \(L=0.8L+500\), so \(L=2500\). 4. Subtracting \(2500\) from the recurrence gives \(a_{n+1}-2500=0.8(a_n-2500)\). Therefore the distance from \(2500\) is multiplied by \(0.8\) each month. Since \(0<0.8<1\), that distance approaches \(0\), so \(a_n\) approaches \(2500\).

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 because its distance from \(2500\) is multiplied by \(0.8\) each month.

- Separate the overnight retention from the new dose when writing one recurrence update. - A fixed point is an amount that would be unchanged by one full recurrence update. - To justify convergence from the recurrence itself, compare successive distances from the fixed point. - For the doubled dose, identify which constant in the recurrence changes and which multiplier stays the same.

1. Since \(80\%\) remains overnight, the recurrence is \(a_{n+1}=0.8a_n+12\), with \(a_0=0\). 2. Repeated updates give \(a_1=12\), \(a_2=21.6\), \(a_3=29.28\), \(a_4=35.424\), and \(a_5=40.3392\), all in milligrams. 3. A fixed point satisfies \(L=0.8L+12\), so \(L=60\,\text{mg}\). 4. Subtracting \(60\) from the recurrence gives \(a_{n+1}-60=0.8(a_n-60)\). Thus the distance from the fixed point is multiplied by \(0.8\) each day and approaches \(0\), so \(a_n\to60\,\text{mg}\). 5. Because the sequence approaches \(60\,\text{mg}\), which is above \(50\,\text{mg}\), the \(50\,\text{mg}\) limit will eventually be exceeded. 6. With a \(24\,\text{mg}\) dose, the recurrence is \(a_{n+1}=0.8a_n+24\). Its fixed point satisfies \(L=0.8L+24\), giving \(L=120\,\text{mg}\).

a) \(a_0=0\), \(a_{n+1}=0.8a_n+12\); the first five post-dose amounts are \(12\,\text{mg}\), \(21.6\,\text{mg}\), \(29.28\,\text{mg}\), \(35.424\,\text{mg}\), and \(40.3392\,\text{mg}\). b) \(L=60\,\text{mg}\), with \(a_{n+1}-60=0.8(a_n-60)\), so \(a_n\to60\,\text{mg}\). c) No. The amount eventually exceeds \(50\,\text{mg}\). d) \(a_{n+1}=0.8a_n+24\), with limiting amount \(120\,\text{mg}\).

- Relate each year's amount to the previous year's amount using the retained fraction and the inflow. - After finding a stable value, rewrite the recurrence in terms of the difference from that value. - For the reduced inflow, compare the new stable value with the stated initial amount. - With no inflow, identify the familiar type of sequence that remains.

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 stable value satisfies \(L=0.9L+200\), so \(L=2000\,\text{kg}\). Also, \(s_{n+1}-2000=0.9(s_n-2000)\). Because \(0<0.9<1\), the difference from \(2000\) approaches \(0\), so the salt amount approaches \(2000\,\text{kg}\). 4. With an inflow of \(150\,\text{kg}\), the stable value is \(1500\,\text{kg}\). Since the initial amount is already \(1500\,\text{kg}\), the recurrence keeps it constant. 5. With no inflow, \(s_{n+1}=0.9s_n\), so \(s_n=1500(0.9)^n\), which approaches \(0\).

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 geometrically toward \(0\,\text{kg}\).

- Use each unrounded term as the input for the next update. - To check an interval, look for the smallest and largest possible output when the input is inside it. - Compare what the update does to a value smaller than \(L\) with what it does to \(L\) itself.

1. Iterating from \(x_0=1\) gives \(x_1=1.2500\), \(x_2\approx1.0959\), and \(x_3\approx1.1902\), using unrounded values at each step. 2. If \(1\le x_n\le\frac{5}{4}\), then \(4\le x_n^2+3\le\frac{73}{16}\). Hence \(\frac{80}{73}\le x_{n+1}\le\frac{5}{4}\). Since \(\frac{80}{73}>1\), the next term stays in the stated interval. The starting term lies in it, so induction shows that every term does. 3. On this interval, \(x\mapsto\frac{5}{x^2+3}\) decreases as \(x\) increases. Since \(\frac{5}{L^2+3}=L\), any \(x_n<L\) gives \(x_{n+1}>L\), and any \(x_n>L\) gives \(x_{n+1}<L\). The terms therefore alternate from one side of the fixed point to the other.

1. \(x_1=1.2500\), \(x_2\approx1.0959\), \(x_3\approx1.1902\). 2. The update sends \([1,\frac{5}{4}]\) into \([\frac{80}{73},\frac{5}{4}]\), which is inside the original interval. All terms stay there. 3. The update is decreasing on the interval, so values below a fixed point move above it and values above move below it. The terms alternate around \(L\).

- Determine the hourly net change separately when the pump is off and when it is on. - Find the beginning-of-hour volume above which adding the pumped-case net amount would exceed capacity. - Apply the appropriate case in order, beginning with \(V_0=400\). - Convert \(75\%\) of the capacity to a volume before comparing the iterates.

1. If \(0\le V_n<1200\), the pump is off and \(V_{n+1}=V_n+350\). This case cannot exceed capacity because \(V_n+350<1550\). 2. If \(1200\le V_n\le1770\), the pump is on and \(V_{n+1}=V_n+350-120=V_n+230\). 3. If \(1770<V_n\le2000\), the pump is on but \(V_n+230>2000\), so overflow leaves \(V_{n+1}=2000\). Therefore, \(V_{n+1}=\begin{cases}V_n+350,&0\le V_n<1200\\V_n+230,&1200\le V_n\le1770\\2000,&1770<V_n\le2000\end{cases}\), with \(V_0=400\). 4. The first values are \(V_1=750\), \(V_2=1100\), \(V_3=1450\), and \(V_4=1680\), all in cubic meters. 5. 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 \(4\) complete hours.

a) \(V_0=400\) and \(V_{n+1}=\begin{cases}V_n+350,&0\le V_n<1200\\V_n+230,&1200\le V_n\le1770\\2000,&1770<V_n\le2000\end{cases}\). 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 \(4\) complete hours.

- Use the order of operations in each recurrence to interpret the daily timing. - Find a stable value for each model, then compare how the distance from that value changes after one recurrence step. - For part c, convert \(90\%\) of Model A's justified limit to a target amount. - Check the two consecutive days around the first crossing of that target.

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, a stable value satisfies \(L=0.75L+40\), so \(L_A=160\,\text{mg}\). Subtracting \(160\) from the recurrence gives \(W_n-160=0.75(W_{n-1}-160)\). Since \(0<0.75<1\), the difference from \(160\) approaches \(0\). 3. Model B can be written \(W_n=0.75W_{n-1}+30\). Its stable value is \(L_B=120\,\text{mg}\), and \(W_n-120=0.75(W_{n-1}-120)\). Therefore Model B also converges to its stable value. 4. Ninety percent of \(160\,\text{mg}\) is \(144\,\text{mg}\). For 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 \(9\).

a) Model A applies elimination before the dose; Model B applies the dose before elimination. b) Model A approaches \(160\,\text{mg}\); Model B approaches \(120\,\text{mg}\). In each model, the difference from its fixed point is multiplied by \(0.75\) each day. c) Day \(9\).

- Follow the stated order of growth and filtering when forming one recurrence step. - After finding a stable value, rewrite the recurrence in terms of the difference from that value. - Express the difference after \(n\) days as a geometric factor times the initial difference. - Translate \(1\%\) of the limiting value into an allowable absolute error.

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 stable value satisfies \(L=0.85L+2.55\), so \(L=17\,\text{kg}\). 3. Subtracting \(17\) from the recurrence gives \(A_n-17=0.85(A_{n-1}-17)\). Hence \(A_n-17=(10-17)(0.85)^n=-7(0.85)^n\), which approaches \(0\). Therefore \(A_n\) approaches \(17\,\text{kg}\). 4. One percent of the limit is \(0.17\,\text{kg}\). The error is \(|A_n-17|=7(0.85)^n\). At \(n=22\), the error is approximately \(0.1960\,\text{kg}\); at \(n=23\), it is approximately \(0.1666\,\text{kg}\). 5. Therefore the algae mass is first within \(1\%\) of the long-term value after \(23\) days.

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

- Classify every complete tiling by what appears at its right-hand end. - Removing a final unit square shortens the strip by one, while removing a final domino shortens it by two. - Account separately for the two possible colors of a final unit square.

1. A tiling can end with a red unit square or a blue unit square. Removing that last square leaves any tiling of length \(n-1\), giving \(2t_{n-1}\) possibilities. 2. A tiling can instead end with a domino. Removing it leaves any tiling of length \(n-2\), giving \(t_{n-2}\) possibilities. 3. Therefore \(t_n=2t_{n-1}+t_{n-2}\) for \(n\ge2\), with \(t_0=1\) and \(t_1=2\). 4. The recurrence gives \(t_2=5\), \(t_3=12\), \(t_4=29\), and \(t_5=70\).

\(t_0=1\), \(t_1=2\), and \(t_n=2t_{n-1}+t_{n-2}\) for \(n\ge2\). Also, \(t_5=70\).

- At what kinds of points are the horizontal and vertical coordinates equal? - Follow each cobweb step vertically to the function and horizontally to the line to see how the next input is produced. - For the exact value, set the next value equal to the current value and check any solutions against the original square-root equation.

1. The dashed straight line labeled \(q\) represents \(y=x\). It meets the recurrence curve near \(x=2\). 2. Iteration gives \(x_1\approx1.581\), \(x_2\approx1.892\), \(x_3\approx1.973\), and \(x_4\approx1.993\). 3. The fixed-point equation \(x=\sqrt{x+2}\) gives \(x^2-x-2=0\), or \((x-2)(x+1)=0\). Only \(x=2\) satisfies the original square-root equation. 4. The displayed cobweb steps and iterates appear to approach \(2\).

a) The dashed line \(q\) is \(y=x\). The fixed point is approximately \(2\). b) \(x_1\approx1.581\), \(x_2\approx1.892\), \(x_3\approx1.973\), \(x_4\approx1.993\). c) The exact fixed point is \(2\), and the displayed iterates appear to converge to it.

- Keep only the remainder after each update, not the full result before division. - Watch for the first value that returns; the same sequence of updates follows it. - How can you match a distant index with a position in the repeating cycle?

1. Applying the rule gives \(1,4,5,0,1,4,5,0\) for indices \(0\) through \(7\). 2. The same value \(1\) returns after four steps. Since the rule always gives the same next value from the same current value, the cycle \(1,4,5,0\) repeats with length \(4\). 3. We have \(2026=4\cdot506+2\). The term at index \(2026\) therefore occupies the same cycle position as \(a_2=5\).

a) \(1,4,5,0,1,4,5,0\). b) The cycle has length \(4\). c) \(a_{2026}=5\).

- One recurrence update should start with the current population and add a fraction of the current gap to \(3000\). - For the table, feed each unrounded result back into the recurrence before rounding the displayed value. - To derive an explicit form, track the gap between the population and the carrying capacity from one year to the next. - The threshold time must be converted to the first whole-number year that actually satisfies the condition.

1. The recurrence is \(B_0=600\) and \(B_{n+1}=B_n+0.18(3000-B_n)=0.82B_n+540\). 2. Repeated updates give \(B_1=1032\), \(B_2=1386.24\), \(B_3=1676.7168\), \(B_4=1914.907776\), and \(B_5=2110.22437632\). Rounded to the nearest fish, the table entries are \(600,1032,1386,1677,1915,2110\). 3. Let \(G_n=3000-B_n\). Then \(G_{n+1}=3000-(0.82B_n+540)=0.82(3000-B_n)=0.82G_n\). Since \(G_0=2400\), \(G_n=2400(0.82)^n\), so \(B_n=3000-2400(0.82)^n\). 4. The boundary equation is \(3000-2400(0.82)^n=2500\), so \(n=\frac{\ln(500/2400)}{\ln(0.82)}\approx7.90\). Therefore the first whole-number year exceeding \(2500\) is year \(8\).

a) \(B_0=600\), \(B_{n+1}=0.82B_n+540\). b) <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> c) \(B_n=3000-2400(0.82)^n\). d) Year \(8\).

- Build one update from the current charge plus a fraction of the amount still missing. - Convert elapsed minutes to complete charging intervals before iterating the recurrence. - The distance from \(100\%\) has a simpler update rule than the charge level itself. - After solving for a noninteger number of intervals, test the neighboring complete intervals and convert back to minutes.

1. The recurrence is \(C_0=12\) and \(C_{n+1}=C_n+0.15(100-C_n)=0.85C_n+15\). 2. Thirty minutes is \(6\) complete intervals. Repeated updates give \(C_6\approx66.8108\), so the charge level is about \(66.81\%\). 3. For \(D_n=100-C_n\), \(D_{n+1}=100-(0.85C_n+15)=0.85(100-C_n)=0.85D_n\). Since \(D_0=88\), \(D_n=88(0.85)^n\), so \(C_n=100-88(0.85)^n\). 4. Solve the boundary equation \(100-88(0.85)^n=90\). This gives \(n=\frac{\ln(10/88)}{\ln(0.85)}\approx13.38\). At \(n=13\), \(C_{13}\approx89.36\%\), while at \(n=14\), \(C_{14}\approx90.96\%\). Therefore the first qualifying time is \(14\) intervals, or \(70\) minutes.

a) \(C_0=12\), \(C_{n+1}=0.85C_n+15\). b) About \(66.81\%\). c) \(C_n=100-88(0.85)^n\). d) \(70\) minutes.

- 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.

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 \(a_1<200\).

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 \(a_n\to-\infty\).

- 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.

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 \(0\).

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

- Use the first-year increase to determine the unknown growth parameter. - When the lake has very few fish, compare its natural growth rate with the fraction being harvested. - At the critical harvest fraction, simplify the recurrence and examine the multiplier applied to a positive population. - Consider what the limiting value must satisfy if a positive, decreasing, bounded sequence converges.

1. The first-year increase gives \(112=r\cdot800\cdot\frac{4200}{5000}=672r\), so \(r=\frac{1}{6}\). 2. The change per year is \(B_{n+1}-B_n=B_n\left(r\frac{K-B_n}{K}-q\right)\). If \(q<r\), this change is positive whenever \(0<B_n<K(1-q/r)\). A positive population approaching zero would eventually enter this interval, where it must increase instead. Therefore such a population cannot tend to zero, and the minimum possible harvest fraction is at least \(r\). 3. At \(q=r=\frac{1}{6}\), \(B_{n+1}=B_n\left(1-\frac{B_n}{6K}\right)\). Whenever \(0<B_n\le K\), the factor is between \(\frac{5}{6}\) and \(1\), so the terms stay positive and decrease. 4. This positive decreasing sequence has a limit \(L\ge0\). Taking the limit in the recurrence gives \(L=L\left(1-\frac{L}{6K}\right)\), which implies \(L=0\). 5. Therefore the minimum harvest fraction is \(q=\frac{1}{6}\approx16.67\%\).

1. \(r=\frac{1}{6}\approx0.1667\). 2. The minimum harvest fraction is \(q=\frac{1}{6}\approx16.67\%\). Smaller fractions allow sufficiently small populations to grow. At \(q=\frac{1}{6}\), the recurrence becomes \(B_{n+1}=B_n\left(1-\frac{B_n}{6K}\right)\); positive populations at or below \(K\) decrease and can only approach \(0\).

- For each rule, compare the input and output when they have the same value. Keep track of any denominator restriction. - Use each rule's newly calculated term for its next update, rather than starting over from \(x_0\). - For each starting value, track which interval the updates stay in and how much the terms change. - What kind of evidence would you need to make a claim about every later term?

1. For A, the fixed-point equation is \(5x=x^3+1\). For B, cubing gives \(x^3=5x-1\). For C, multiplying by the nonzero denominator gives \(5x-x^3=1\). Each is equivalent to \(x^3-5x+1=0\). Values with \(x^2=5\) do not satisfy that polynomial, so C loses no fixed points through its denominator restriction. The rules still act differently on values that are not fixed points. 2. Starting at \(0\), A gives \(x_1=0.2000\), \(x_2=0.2016\), and \(x_3\approx0.2016\). B gives \(x_1=-1.0000\), \(x_2\approx-1.8171\), and \(x_3\approx-2.1606\). 3. Starting at \(0\), C gives \(x_1=0.2000\), \(x_2\approx0.2016\), and \(x_3\approx0.2016\). A and C stay in \([0,1]\) during these steps; B leaves it. However, A's exact second term is \(0.2016\), while C's is \(\frac{25}{124}\approx0.2016129\). The displayed rounded values hide a real difference. 4. Starting at \(2\), A gives \(x_1=1.8000\), \(x_2=1.3664\), and \(x_3\approx0.7102\). B gives \(x_1\approx2.0801\), \(x_2\approx2.1105\), and \(x_3\approx2.1218\). 5. Starting at \(2\), C gives \(x_1=1.0000\), \(x_2=0.2500\), and \(x_3\approx0.2025\). Only B remains in \([2,3]\); its successive changes decrease from about \(0.0801\) to \(0.0304\) to \(0.0113\). Three updates show different behavior but do not prove convergence.

a) At a fixed point, A gives \(5x=x^3+1\), B gives \(x^3=5x-1\), and C gives \(5x-x^3=1\). All reduce to \(x^3-5x+1=0\), but the rules can give different terms away from fixed points. b) From \(x_0=0\): A: \(0.2000,0.2016,0.2016\); B: \(-1.0000,-1.8171,-2.1606\); C: \(0.2000,0.2016,0.2016\). A and C stay in \([0,1]\). Their second terms are not equal: A gives \(0.2016\), while C gives \(\frac{25}{124}\). c) From \(x_0=2\): A: \(1.8000,1.3664,0.7102\); B: \(2.0801,2.1105,2.1218\); C: \(1.0000,0.2500,0.2025\). Only B stays in \([2,3]\), and its step sizes decrease. This does not prove long-term convergence.

- At an equilibrium, replacing the current value by the equilibrium must produce the same value again. - How does the distance from a term to the equilibrium change in one step? - A pattern in those distances may be easier to express than a pattern in the original terms. - For the threshold, check which whole-number index first passes the stated value.

1. At equilibrium, \(500=0.8(500)+c\), so \(c=100\). 2. Subtract the equilibrium: \(P_{n+1}-500=0.8(P_n-500)\). Since \(P_0-500=-400\), it follows that \(P_n=500-400(0.8)^n\). 3. Solve \(500-400(0.8)^n>450\), equivalent to \((0.8)^n<0.125\). Thus \(n>\frac{\ln(0.125)}{\ln(0.8)}\approx9.32\). 4. Checking neighboring indices gives \(P_9\approx446.31\) and \(P_{10}\approx457.05\), so the first possible index is \(n=10\).

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

- First identify a value that would remain unchanged after one recurrence update. - Compare each term with that fixed value rather than studying the original recurrence directly. - Once the difference from the fixed point has a multiplicative form, separate cases according to the behavior of powers of the multiplier. - Treat boundary multipliers separately from values whose absolute value is strictly less than or greater than \(1\).

1. For \(m\ne1\), a fixed point satisfies \(L=mL+c\), so \(L=\frac{c}{1-m}\). 2. Because \(c=(1-m)L\), subtracting \(L\) from the recurrence gives \(a_{n+1}-L=m(a_n-L)\). Repeating this relation yields \(a_n-L=m^n(a_0-L)\). 3. If \(|m|<1\), then \(m^n\to0\), so every sequence converges to \(L\). 4. If \(m=-1\), then \(a_n-L=(-1)^n(a_0-L)\). When \(a_0=L\), the sequence is constant. Otherwise it alternates between two values, so it is bounded but does not converge. 5. If \(|m|>1\), then \(|m|^n|a_0-L|\) grows without bound unless \(a_0=L\). Thus the fixed initial value gives a constant sequence, while every other initial value produces an unbounded sequence. 6. If \(m=1\), then \(a_{n+1}=a_n+c\), so \(a_n=a_0+nc\). If \(c=0\), the sequence is constant; if \(c\ne0\), it is unbounded and does not converge to a finite limit.

a) \(L=\frac{c}{1-m}\) for \(m\ne1\), and \(a_n-L=m^n(a_0-L)\). b) If \(|m|<1\), then \(a_n\to L\). If \(m=-1\), the sequence is constant when \(a_0=L\) and otherwise is bounded but nonconvergent. If \(|m|>1\), it is constant when \(a_0=L\) and otherwise unbounded. c) If \(m=1\), then \(a_n=a_0+nc\); it is constant when \(c=0\) and otherwise unbounded.