Iterative formulae, roots and growth models

(a) $3.700$

1. $$\sqrt{13}$$
2. $$\sqrt{10+\sqrt{13}}$$
3. Substitute each new value into the same rule, retaining calculator precision.
4. x₁ = $\sqrt{13}$ = 3.605551… and x₂ = $\sqrt{}$(10 + 3.605551…) = 3.688570….
5. x₃ = $\sqrt{}$(10 + 3.688570…) = 3.699806…, which rounds to 3.700.

• P1 Establishing $\sqrt{13}$ or an equivalent valid method.
• P1 Establishing $\sqrt{10+\sqrt{13}}$ or an equivalent valid method.
• A1 Correct answer: $3.700$

(a) $f(2) = -1 < 0$ and $f(3) = 13 > 0$: a sign change.

1. $f(2) = 8 - 10 + 1 = -1$ and $f(3) = 27 - 15 + 1 = 13$.
2. The value changes from negative to positive, and $f$ is continuous, so $f(x) = 0$ somewhere between 2 and 3.

• M1 Working out $f(2) = -1$ and $f(3) = 13$.
• A1 Stating that there is a sign change, so there is a solution between 2 and 3.

(b) $2.1218$

1. $$x_1 = \sqrt[3]{9} = 2.080083\ldots$$
2. $$x_2 = \sqrt[3]{5 \times 2.080083\ldots - 1} = 2.110485\ldots$$
3. $$x_3 = \sqrt[3]{5 \times 2.110485\ldots - 1} = 2.121800\ldots \approx 2.1218$$

• M1 Finding $x_1 = 2.0800\ldots$.
• M1 Finding $x_2 = 2.1104\ldots$ using the unrounded $x_1$.
• A1 The correct answer, $x_3 = 2.1218$.

(a) $2.707$

1. $$\sqrt{4+9/3}$$
2. $$\sqrt{4+9/\sqrt{7}}$$
3. For positive x, divide $x^{3}$ = 4x + 9 by x and take the positive square root to obtain the given iteration.
4. Retaining calculator precision gives x₁ = 2.645751…, x₂ = 2.720602… and x₃ = 2.703347….
5. The fourth iteration gives x₄ = $\sqrt{}$(4 + 9/2.703347…) = 2.707250….
6. To 3 decimal places, x₄ = 2.707.

• P1 Establishing $\sqrt{4+9/3}$ or an equivalent valid method.
• P1 Establishing $\sqrt{4+9/\sqrt{7}}$ or an equivalent valid method.
• A1 Correct answer: $2.707$

(a) £$1105.43$

1. $$1200\times 1.05-90$$
2. $$1170\times 1.05-90$$
3. $$1138.5\times 1.05-90$$
4. Each year changes balance B to 1.05B $-$ 90.
5. Year 1: £1170. Year 2: £1138.50.
6. Year 3: 1.05 $\times$ 1138.50 $-$ 90 = £1105.425, which rounds to £1105.43.

• P1 Establishing $1200\times 1.05-90$ or an equivalent valid method.
• P1 Establishing $1170\times 1.05-90$ or an equivalent valid method.
• P1 Establishing $1138.5\times 1.05-90$ or an equivalent valid method.
• A1 Correct answer: £$1105.43$

(a) $3.701$

1. Use the previous output as the next input; first compute x1.
   $$\sqrt{10+3}$$
2. The next values are 3.688570357 and 3.699806800; use the third for the fourth update.
   $$\sqrt{10+3.69980679989958}$$
3. Therefore $3.701$.

• M1 Use the previous output as the next input; first compute x1.
• M1 The next values are 3.688570357 and 3.699806800; use the third for the fourth update.
• A1 Correct answer: $3.701$

(b) $x^{2}-x-10=0$

1. At a fixed point the input and next output are equal.
   $$x=\sqrt{10+x}$$
2. Therefore $x^{2}-x-10=0$.

• M1 At a fixed point the input and next output are equal.
• A1 Correct answer: $x^{2}-x-10=0$

(a) £$1243.81$

1. Apply interest before each withdrawal; calculate the first year.
   $$1300\times 1.04-70$$
2. Repeat the entire update twice more.
   $$((1300\times 1.04-70)\times 1.04-70)\times 1.04-70$$
3. Therefore £$1243.81$.

• P1 Apply interest before each withdrawal; calculate the first year.
• P1 Repeat the entire update twice more.
• A1 Correct answer: £$1243.81$

(b) $y=1.04x-70$

1. The next balance is 1.04 times the previous balance, less 70.
   $$y=1.04x-70$$
2. Therefore $y=1.04x-70$.

• P1 The next balance is 1.04 times the previous balance, less 70.
• A1 Correct answer: $y=1.04x-70$

(a) $x^3 = 5x - 1 \Rightarrow x = \sqrt[3]{5x - 1}$

1. $$x^3 - 5x + 1 = 0 \Rightarrow x^3 = 5x - 1 \Rightarrow x = \sqrt[3]{5x - 1}$$

• M1 Writing $x^3 = 5x - 1$.
• A1 Taking the cube root of both sides to reach the result.

(b) $2.13$

1. Iterate.
   $$2.0801,\ 2.1105,\ 2.1218,\ 2.1260,\ 2.1275,\ 2.1281,\ \ldots$$
2. The values settle at $2.128\ldots$, which suggests $2.13$.
3. Check with a sign change at the limits of 2.13.
   $$f(2.125) = -0.029 < 0, \qquad f(2.135) = 0.057 > 0$$
4. So the solution is 2.13 to 2 decimal places.

• P1 Carrying out at least four iterations correctly.
• P1 Checking the sign of $f$ at $2.125$ and $2.135$.
• A1 $2.13$ with the sign-change check.

(a) $4.12$

1. $$x_2 = 3 + 4 = 7, \quad x_3 = 3 + \tfrac{4}{7} = 3.5714\ldots, \quad x_4 = 3 + \tfrac{4}{3.5714\ldots} = 4.12$$

• M1 Finding $x_2 = 7$ and $x_3 = 3.571\ldots$.
• A1 The correct answer, $4.12$.

(b) $L = 3 + \frac{4}{L} \Rightarrow L^2 - 3L - 4 = 0 \Rightarrow L = 4$ (positive)

1. In the limit, $x_{n+1}$ and $x_n$ are both $L$.
   $$L = 3 + \frac{4}{L}$$
2. Multiply by $L$ and rearrange.
   $$L^2 - 3L - 4 = 0 \Rightarrow (L - 4)(L + 1) = 0$$
3. The iterates are all positive, so $L = 4$ (not $-1$).

• P1 Writing $L = 3 + \frac{4}{L}$.
• P1 Forming $L^2 - 3L - 4 = 0$.
• C1 Solving to $L = 4$ or $-1$ and rejecting $-1$ because every iterate is positive.