| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton-Raphson method |
| Type | Newton-Raphson error analysis |
| Difficulty | Challenging +1.8 This is a comprehensive Further Maths question requiring derivation of Newton-Raphson formula (routine), graphical analysis of convergence failure (requires insight into when |error| increases), investigation of iterative behavior at x₁=-1, and most challengingly, proving and verifying the quadratic convergence relationship d₄ ≈ d₃³/d₂². The error analysis and algebraic manipulation of successive differences elevate this significantly above standard Newton-Raphson applications, though it remains structured with clear parts guiding the solution. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{3x_n^3 + 5x_n^2 - x_n - 1}{9x_n^2 + 10x_n - 1}\) | B1 | Correct derivative seen |
| \(= \frac{x_n(9x_n^2 + 10x_n - 1) - (3x_n^3 + 5x_n^2 - x_n - 1)}{9x_n^2 + 10x_n - 1}\) | M1 | Combining terms as 1 fraction or 2 fractions with common denominator |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{6x_n^3 + 5x_n^2 + 1}{9x_n^2 + 10x_n - 1}\) | A1 | Line above seen ag. Must contain suffices. |
Total: 3
| Answer | Marks | Guidance |
|---|---|---|
| A suitable value is shown within range \([0.1, 0.25]\) | B1 | The point does not have to be labelled \(x_1\). Accept a tangent which shows this. |
Total: 1
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x_2 = 0 \Rightarrow x_3 = -1\), and statement that values alternate | B1 | Values seen either in words or on graph marked as these values |
| Clear diagram with tangents from \(-1\) to \(0\) and back to \(-1\) | B1 |
Total: 2
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \frac{d_4}{d_3} = \frac{kd_3^2}{kd_2^2} = \frac{d_3^2}{d_2^2} \Rightarrow d_4 = \frac{d_3^3}{d_2^2}\) | M1 | \(d_4\) and \(d_3\) and trying to combine them to eliminate \(k\) |
| A1 | Ag | |
| \(x_r\) | \(x_{r+1}\) | \(d_r\) |
| 1 | 0.666667 | \(-0.33333\) |
| 0.666667 | 0.517241 | \(-0.14943\) |
| 0.517241 | 0.481438 | \(-0.0358\) |
| 0.481438 | 0.479363 | \(-0.00207\) |
| 0.479363 | 0.479356 | \(-6.8\text{E-}06\) |
| 0.479356 | 0.479356 | |
| Sight of \(-0.0300\) | B1 | Condone 3 dp |
| Sight of \(-0.0358\) | B1 | 3sf or better |
Total: 4
| Answer | Marks | Guidance |
|---|---|---|
| Continuing the above | M1 | Or any other starting point that converges to the positive root |
| to give root \(0.47936\) | A1 | Cao |
Total: 2
## Question 6:
### Part (i):
$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{3x_n^3 + 5x_n^2 - x_n - 1}{9x_n^2 + 10x_n - 1}$ | **B1** | Correct derivative seen
$= \frac{x_n(9x_n^2 + 10x_n - 1) - (3x_n^3 + 5x_n^2 - x_n - 1)}{9x_n^2 + 10x_n - 1}$ | **M1** | Combining terms as 1 fraction or 2 fractions with common denominator
$= \frac{9x_n^3 + 10x_n^2 - x_n - 3x_n^3 - 5x_n^2 + x_n + 1}{9x_n^2 + 10x_n - 1}$
$= \frac{6x_n^3 + 5x_n^2 + 1}{9x_n^2 + 10x_n - 1}$ | **A1** | Line above seen ag. Must contain suffices.
Total: **3**
### Part (ii):
A suitable value is shown within range $[0.1, 0.25]$ | **B1** | The point does not have to be labelled $x_1$. Accept a tangent which shows this.
Total: **1**
### Part (iii):
$\Rightarrow x_2 = 0 \Rightarrow x_3 = -1$, and statement that values alternate | **B1** | Values seen either in words or on graph marked as these values
Clear diagram with tangents from $-1$ to $0$ and back to $-1$ | **B1** |
Total: **2**
### Part (iv):
$d_4 = kd_3^2$, $\quad d_3 = kd_2^2$
$\Rightarrow \frac{d_4}{d_3} = \frac{kd_3^2}{kd_2^2} = \frac{d_3^2}{d_2^2} \Rightarrow d_4 = \frac{d_3^3}{d_2^2}$ | **M1** | $d_4$ and $d_3$ and trying to combine them to eliminate $k$
| **A1** | Ag
| $x_r$ | $x_{r+1}$ | $d_r$ | $\frac{d_3^3}{d_2^2}$ |
|---|---|---|---|
| 1 | 0.666667 | $-0.33333$ | $d_2$ |
| 0.666667 | 0.517241 | $-0.14943$ | $d_3$ |
| 0.517241 | 0.481438 | $-0.0358$ | $d_4$ | $-0.030$ |
| 0.481438 | 0.479363 | $-0.00207$ | $d_5$ | |
| 0.479363 | 0.479356 | $-6.8\text{E-}06$ | | |
| 0.479356 | 0.479356 | | | |
Sight of $-0.0300$ | **B1** | Condone 3 dp
Sight of $-0.0358$ | **B1** | 3sf or better
Total: **4**
### Part (v):
Continuing the above | **M1** | Or any other starting point that converges to the positive root
to give root $0.47936$ | **A1** | Cao
Total: **2**
---6 It is given that the equation $3 x ^ { 3 } + 5 x ^ { 2 } - x - 1 = 0$ has three roots, one of which is positive.\\
(i) Show that the Newton-Raphson iterative formula for finding this root can be written
$$x _ { n + 1 } = \frac { 6 x _ { n } ^ { 3 } + 5 x _ { n } ^ { 2 } + 1 } { 9 x _ { n } ^ { 2 } + 10 x _ { n } - 1 } .$$
(ii) A sequence of iterates $x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots$ which will find the positive root is such that the magnitude of the error in $x _ { 2 }$ is greater than the magnitude of the error in $x _ { 1 }$. On the graph given in the Printed Answer Book, mark a possible position for $x _ { 1 }$.\\
(iii) Apply the iterative formula in part (i) when the initial value is $x _ { 1 } = - 1$. Describe the behaviour of the iterative sequence, illustrating your answer on the graph given in the Printed Answer Book.\\
(iv) A sequence of approximations to the positive root is given by $x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots$. Successive differences $x _ { r } - x _ { r - 1 } = d _ { r }$, where $r \geqslant 2$, are such that $d _ { r } \approx k \left( d _ { r - 1 } \right) ^ { 2 }$ where $k$ is a constant. Show that $d _ { 4 } \approx \frac { d _ { 3 } ^ { 3 } } { d _ { 2 } ^ { 2 } }$ and demonstrate this numerically when $x _ { 1 } = 1$.\\
(v) Find the value of the positive root correct to 5 decimal places.
\hfill \mbox{\textit{OCR FP2 2015 Q6 [12]}}