Further Maths B (MEI) · OCR H645 · Numerical Methods (minor option - choose one with a major, or three alone) · June 2025
The table shows some values of x and the associated values of f (x).
| x | 1.9 | 2.0 | 2.1 |
| f(x) | 0.216 92 | 0.2 | 0.184 84 |
Determine an estimate of f ′ (2.0) using the forward difference method.
[2]
Determine an estimate of f ′ (2.0) using the central difference method.
[2]
The numbers p and q are approximated by
P = 323 and Q = 162.
P has been found by rounding p to the nearest whole number.
Q has been found by chopping q to the nearest whole number.
Find the maximum possible relative error in using P to approximate p.
[1]
Find the maximum possible relative error in using Q to approximate q.
[1]
Determine the range of possible values of R = 
[3]
Explain why your answer to part (b) is so large.
[1]
with n = 1 and n = 2 are shown in the table. The table is incomplete.
Complete the copy of the table in the Printed Answer Booklet. Give your answers to
6 decimal places.
[4]
[1]
The Newton-Raphson method is to be used to find the positive root of the equation
tanh x - x2 + 4 = 0.
The diagram shows part of the graph of y = tanh x - x2 + 4.
On the copy of the diagram in the Printed Answer Booklet, show how the Newton-Raphson
method works to find x1 using the starting value x0 = 1.
[1]
Use the Newton-Raphson method using the starting value x0 = 1 to determine the values of
x1 and x2 correct to 8 decimal places.
[4]
Continue the iteration to determine the value of the positive root of the equation
tanh x - x2 + 4 = 0 correct to 7 decimal places.
[2]
The diagram shows part of the graph of y = sinh x - 3 x + 1.
The largest positive root, α, of the equation sinh x - 3 x + 1 = 0 is to be found using the secant method with x0 = 2 and x1 = 3.
Table 5.1 shows the associated spreadsheet output.
Table 5.1
 | A | B | C | D | E |
| 2 | r | xr | f(xr) | xr+1 | f(xr+1) |
| 3 | 0 | 2 | –1.37314 | 3 | 2.01787 |
| 4 | 1 | 3 | 2.017875 | 2.404935 | –0.7211 |
| 5 | 2 | 2.404935 | –0.72109 | 2.561598 | –0.2451 |
| 6 | 3 | 2.561598 | –0.24513 | 2.642285 | 0.06018 |
| 7 | 4 | 2.642285 | 0.060175 | 2.626382 | –0.0035 |
| 8 | 5 | 2.626382 | –0.00348 | 2.627252 | –5E-05 |
| 9 | 6 | 2.627252 | –4.5E-05 | 2.627264 | 3.5E-08 |
Write down a suitable formula for cell D4.
[2]
State the most accurate approximation to α in the spreadsheet output.
[1]
Determine whether this approximation is accurate to 6 decimal places.
[2]
Explain what would happen if the initial entry in cell D3 were to be changed from 3 to 1.
[1]
Table 5.2 shows some further analysis of the sequence of approximations to α.
Table 5.2
| A | B | C | D | E |
| r | xr | xr+1 - xr | xr+2 - xr+1 xr+1 - xr | xr+2 - xr+1 (xr+1 - xr)2 |
| 1 | 3 | –0.5951 | –0.2633 | 0.44242 |
| 2 | 2.404935 | 0.15666 | 0.51504 | 3.28756 |
| 3 | 2.561598 | 0.08069 | –0.1971 | –2.4427 |
| 4 | 2.642285 | –0.0159 | –0.0547 | 3.44193 |
| 5 | 2.626382 | 0.00087 | 0.01315 | 15.1113 |
| 6 | 2.627252 | 1.1E-05 | ||
| 7 | 2.627264 |
Explain what may be inferred from the entries in column D about the order of convergence of
this sequence of approximations to α.
[2]
Explain what may be inferred from the entries in column E about the order of convergence of
this sequence of approximations to α.
[2]
The associated spreadsheet output, together with some further analysis, is shown in the table.
| n | Mn | difference | ratio |
| 1 | 0.6924056 | ||
| 2 | 0.6615057 | –0.0308999 | |
| 4 | 0.6498274 | –0.0116782 | 0.37794 |
| 8 | 0.6454773 | –0.0043501 | 0.3725 |
| 16 | 0.6438811 | –0.0015962 | 0.36692 |
| 32 | 0.643302 | –0.0005791 | 0.36281 |
| 64 | 0.6430936 | –0.0002085 | 0.35996 |
| 128 | 0.6430189 | –7.463E-05 | 0.35801 |
Explain what the entries in the ratio column tell you about the order of the midpoint rule in
this case.
[2]
Use the information in the table to determine the most accurate estimate of
[5]
[2]
[1]
Explain why the values displayed in cells F5 and F6 are the same as each other, but the
values in cells G5 and G6 are different to each other.
[2]
The negative root of the equation 8 ln (x + 2) - 3 x + 1 = 0, δ, is close to –1.
It is found that g′(δ) ≈ 13.9.
Explain why the iterative formula
xn+1 = g(xn) with x0 = –1
may not be used to find δ.
[1]
[2]
State what would happen if the iteration in part (e)(i) were to be continued.
[1]
Use the relaxed iteration
xn+1 = (1 - λ)xn + λg(xn) with λ = –0.078 and x0 = –1
to determine the value of δ correct to 7 decimal places.
[2]
Following an injury an athlete returns to training in preparation for a regional competition. At the end of each week the athlete’s trainer records the time, t seconds, the athlete takes to run 100 metres.
Table 8.1 shows some of the times recorded at the end of week W.
Table 8.1
| W | 0 | 2 | 3 |
| t | 18.00 | 17.68 | 16.83 |
The trainer uses the information in the table to construct a polynomial of degree 2 to model these data.
Use Lagrange’s interpolating polynomial of degree 2 to obtain the trainer’s model.
[3]
Subsequently it is found that when W = 4, t = 15.44.
Determine whether the model is a good fit for these values.
[1]
The trainer decides to use the three most recent data points, when W = 2, 3 and 4, to construct a new model using Newton’s forward difference interpolation formula.
Table 8.2 shows the difference table for these data.
Table 8.2
The trainer obtains the interpolation formula
t = –0.27W2 + BW + C, where B and C are constants.
• Determine the value of B.
• Determine the value of C.
[3]
Explain why the new model is a refinement of the model found in part (a).
[1]
At the regional competition, the athlete needs to run 100 metres in less than 11 seconds in order to qualify for a national competition. The regional competition takes place when W = 6.
Determine whether, according to the new model, the athlete will qualify for the national
competition.
[1]
Explain whether it would be appropriate to use the new model to predict the athlete’s times
for W ⩾ 7.
[1]
The paper as printed
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