| Exam Board | OCR MEI |
|---|---|
| Module | Further Numerical Methods (Further Numerical Methods) |
| Session | Specimen |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Fixed Point Iteration |
| Type | Apply iteration to find root (pure fixed point) |
| Difficulty | Challenging +1.2 This is a Further Maths numerical methods question requiring understanding of fixed point iteration and relaxation techniques. Part (i) involves computing a few iterations to demonstrate divergence (straightforward calculation), while part (ii) applies a given relaxation formula iteratively until convergence. The concepts are moderately advanced but the execution is computational rather than requiring novel insight—students follow prescribed procedures with a calculator. |
| Spec | 1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (i) | 1(cid:16)sinh1 |
| Answer | Marks |
|---|---|
| − 0.175.., − 6.7128…, − 61.442… so fails | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 2.2a | At least 3 iterates BC |
| N | Alternative Method α ≈ 1 and |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (ii) | 0.706199, 0.612353, 0.601606, 0.601403, 0.601402,.. |
| 0.601402 cao | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | E |
| Answer | Marks |
|---|---|
| BC | Answer only does not score |
Question 3:
3 | (i) | 1(cid:16)sinh1
soi
1
− 0.175.., − 6.7128…, − 61.442… so fails | M1
A1
[2] | 1.1
2.2a | At least 3 iterates BC
N | Alternative Method α ≈ 1 and
substitution in
(cid:16)xcoshx(cid:16)1(cid:14)sinhx
gʹ(x) =
x2
g'(x) (cid:124)1.368(cid:33)1 so iteration
fails
3 | (ii) | 0.706199, 0.612353, 0.601606, 0.601403, 0.601402,..
0.601402 cao | M1
A1
[2] | 1.1
1.1 | E
Must see at least 3 iterates
BC | Answer only does not score3 The equation $\sinh x + x ^ { 2 } - 1 = 0$ has a root, $\alpha$, such that $0 < \alpha < 1$.\\
(i) Verify that the iteration $x _ { r + 1 } = \frac { 1 - \sinh x _ { r } } { x _ { r } }$ with $x _ { 0 } = 1$ fails to converge to this root.\\
(ii) Use the relaxed iteration $x _ { r + 1 } = ( 1 - \lambda ) x _ { r } + \lambda \left( \frac { 1 - \sinh x _ { r } } { x _ { r } } \right)$ with $\lambda = \frac { 1 } { 4 }$ and $x _ { 0 } = 1$ to find $\alpha$ correct to 6 decimal places.
\hfill \mbox{\textit{OCR MEI Further Numerical Methods Q3 [4]}}