1(i)
(ii)
h
est f '(2)
0.4
3.195
differences
0.2
2.974
0.1
2.871
M1A1A1A1 [4]
-0.221
-0.103
M1A1
differences approximately halving so extrapolate to 2.871 - 0.103 = 2.768.
Last figure unreliable so 2.77.
Accept argument to 2.8.
M1A1
[4]
2(i)
E.g. 2/3 rounded to 0.666 666 7, chopped to 0.666 666 6
E1
[1]
(ii)
2/3 stored as 0.666 666 7
mpe is 0.000 000 05
mpre is greatest when x is least
mpre is 0.000 000 05 / 0.1 =
A1A1
A1
M1
M1A1
[6]
3
4
5
x
1.4
1.5
Absolute error 0.000 000 033...
5 * 10^
-7
5E-07
f(x)
-0.82176
1.09375
root in the interval (1.4, 1.5)
r
0
Xr
1.4
1
2
1.5
1.4429
3
1.446535
4
1.446859
Root at 1.447 seems secure.
x
f(x)
2
1.553774
3
1.652892
4
1.732051
B1
f(Xr)
0.82176
1.09375
0.07436
0.00609
3.88E05
M1
M1A1
A1
M1A1
B1
M=
T=
S=
(2*M + T) / 3
=
[8]
3.305783 M1A1
3.285825 A1
3.299130 M1A1
S(h=2)
3.299130 diffs
S(h=1)
3.299231 0.0001
S(h=0.5)
3.299238 7 E -06
Differences reducing very rapidly. 3.29924 seems secure.
Computations of this type contain rounding errors
The rounding errors will be different when the two sums are computed
Adding from large to small loses precision (the small number is lost)
Adding from small to large allows each number to contribute to the sum
Hence the second sum is likely to be more accurate
M1A1A1
E1
E1
E1
E1
E1
[8]
6(i) x
1
Δf(x)
f(x)
4
Δ²f(x)
Δ³f(x)
–3
2
1
6
3
3
4
a – 13
a–7
a–4
4
a
87 – 3a
80 – 2a
A4
(-1 each
error)
76 – a
5
76
[4]
87 – 3a = a – 13 gives a = 25
M1
f(x) = 4 - 3(x-1) + 6(x-1)(x-2)/2 + 12(x-1)(x-2)(x-3)/6
= 4 - 3x + 3 + 3x2 - 9x + 6 + 2x3 - 12x2 + 22x -12
= 2x3 - 9x2 + 10x + 1
M1A1A1A1A1
A1
A1
[8]
(ii)
Algebra may appear in
(iii)
rather than (ii) for full
credit
f '(x) = 6x2 - 18x + 10 = 0
(iii)
x=
2.26
f(2.26...) = 0.718
(2.26376)
A1
A1
(iv) f(x) = 4 (x - 2)(x - 3)(x - 5)/(1 - 2)(1 - 3)(1 - 5) + three similar terms
7(i) 0
0.785398
1.570796
2.356194
3.141593
3.926991
4.712389
5.497787
6.283185
4
3
2.828427
2.45E-16
-2.82843
-4
-2.82843
-7.4E-16
2.828427
4
2.214602
1.429204
0.643806
-0.14159
-0.92699
-1.71239
-2.49779
-3.28319
M1
M1A1A1
[3]
[3]
Total
18
G1
G1
shows
two
roots
E1
[3]
(ii)
E.g.:
r
Xr
0
1
1
1.04720
2
3
4
5
1.06077 1.06465 1.06576 1.06608
alpha = 1.066 correct to 3 decimal places
1
(iii) 0
1
1.04720
diffs
0.04720
ratio of diffs
2
1.06077
0.01357
0.28756
3
1.06465
0.00388
0.28615
4
1.06576
0.00111
0.28575
5
1.06608
0.00032
0.28564
ratios (approx) constant so first order convergence.
(iv) Obtain N-R iteration (beware printed answer)
E.g.:
r
Xr
0
5
1
4.35177
0
5
diffs
ratio of diffs
1
4.35177
-0.64823
2
4.36435
0.01258
2
3
4
4.36435 4.36432 4.36432
-0.01940
3
4.36432
0.00002
0.00184
M1A1
E1
[3]
M1A1
beta = 4.3643 correct to 4 decimal places
(v)
6
1.06617
0.00009
0.28561
6
7
1.06617
1.06620
M1A1A1
A1
[4]
M1A1
A1
[5]
4
4.36432
0.00000
0.00000
ratios getting (much) smaller so faster than first order
M1A1
E1
[3]
Total
18
4776
Mark Scheme
January 2006
MEI Numerical Methods 4776
1
2(i)
(ii)
3
Use binomial expansion of (1 + r)-1 or sum of GP, or 1 - r2
with r2 taken to be zero to obtain given result.
Relative error in reciprocal is of same magnitude but opposite sign
E.g.
10 is approx 2% greater than 9.8
1/10 = 0.1 is approx 2% less than 1/9.8 = 0.10204
xr+1 = 1/sin(xr)
r
0
1
2
3
10
11
12
xr
1
1.188395 1.077852 1.135147 1.113855 1.114323 1.114067
root is 1.11 to 3 sf.
x
1/x sin(x)
2.7725
M
2.60242
2.56982
x
f(x)
1
3
h
2
1
(i)
f '(3)
0.925
0.85
[M1]
[M1A1]
[A1]
[subtotal
4]
[M1A1A1]
[subtotal
3]
[TOTAL
7]
T
2.44866
T2 = (M1 + T1)/2 =
T4 = (M2 + T2)/2 =
S1 = (2M1 + T1)/3
=
S2 = (2M2 + T2)/3
=
2.52554
2.54768
[M1A1]
[A1]
2.55117
[M1A1]
2.55506
[A1]
I = 2.56 (or 2.555) is justified
4
[M1A1]
[TOTAL
6]
2.7735
-8.4E-05 0.000719 change of sign, so 2.773 correct to 3 dp
h
2
1
[M1A1]
[E1E1]
2
4.5
3
5.4
4
6.2
5
6.7
(ii)
min
0.9
0.8
[A1]
[TOTAL
7]
max
0.95
0.9
Larger h gives smaller interval
(or 0.9 is the only common value)
f '(3) = 0.9 is the value that seems justified.
(Or 0.8 seems to be the limit the process is tending to. [E1E1])
(i)
[M1A1A1]
(ii)
[M1A1A1]
[E1]
[E1]
[TOTAL
8]
4776
5
x
g(x)
Mark Scheme
January 2006
1
4
3
1
4
11
L(2) = 4(2-3)(2-4)/(1-3)(1-4) + 1(2-1)(2-4)/(3-1)(3-4) + 11(2-1)(2-3)/(4-1)(4-3)
= -11/3
x
1
2
3
4
g(x)
Δg(x)
Δ2g(x)
4 -5.33333 7.666666
1.33333 2.333333 7.666667
1
10
11
2nd differences constant
so correct for a quadratic
[M1A1A1A1]
[A1]
[M1A1E1]
[TOTAL 8]
6 (i)
y ' = 10x9 - 10 = 0 only when x = 1, hence at most one turning point
tenth degree polynomial is positive as x tends to plus or minus infinity
(hence exactly one turning point)
(or other methods)
x
0
1
2
f(x)
1
-8
1005
changes of sign so roots in [0,1] and [1,2]
since only one turning point cannot be any more roots
(ii)
NR:
r
xr
(iii)
xr+1 = xr - (xr10 - 10xr + 1)/(10xr9 - 10)
0
1
2
3
4
5
1.2 1.315589 1.284353 1.280004 1.279928 1.279928
1.2799
xr+1 = (xr10 +
1)/10
If x0 = 0.1 then x1 = (0.110 + 1)/10 = 0.111 + 0.1
This is so close to 0.1 that further iterations are unnecessary
[M1A1]
[E1]
[M1A1]
[E1]
[E1]
[subtotal 7]
[M1A1A1]
[M1A1A1]
[subtotal 6]
[M1A1]
[M1A1]
[E1]
[subtotal 5]
[TOTAL 18]
4776
7 (i)
(ii)
Mid-point rule with h=1 and 4 strips to obtain given result.
1.60943
loge(5) =
8
Mid-pt 1.57460 error is (=
3 )
0.034835
N
loge(N)
(iii)(iv)
Mark Scheme
January 2006
N
10
20
40
80
[M1A1]
[M1A1A1]
[subtotal 5]
10
20
40
80
2.30258
5 2.995732 3.688879 4.382027
Mid-pt
2.26651
1
2.95934
6
3.65241
6
4.34554
3
ln(N)
est k
0.036074
[M1A1A1]
[subtotal 3]
diffs
ratio of
diffs
estimates
2.302585
[M1A1A1]
0.036386
2.995732
0.000312
0.036463
3.688879
[subtotal 3]
0.247231
7.72E-05
0.036484
4.382027
diffs
[M1A1]
ratio
[M1A1]
0.261472
2.02E-05
(approx 0.25)
extrapolating:
(or equivalent)
0.036489
0.036490
0.036490
5.05E-06
1.26E-06
3.15E-07
0.03649(0) seems secure
[M1A1]
[A1]
[subtotal 7]
[TOTAL
18]
4776
1
Mark Scheme
June 2006
Sketch, with explanation.
[M1A1E1]
f '(x) = 1/(2√x)
[M1A1]
0.01581
1
hence mpe is approx 0.05 / (2√2.5) =
0.01597
(or 0.05 / 2√2.45 =
2
or 0.05 / 2√2.55 =
(0.016)
0.01565
6
[M1A1]
)
[TOTAL 7]
2
x
f(x)
0
1
1
-3
a
0
b
1
f(a)
1
f(b)
-3
0
0.25
1
-0.24902
x
0.1995
0.00281
6
0.2005
f(x)
-0.00218
change of sign so root
x
f(x)
0.25 -0.24902
0.20015
6 -0.00046
0.200 to 3 dp
sign change so root is correct to 3 dp
[M1A1]
[M1A1]
[A1]
[A1]
[M1A1]
[TOTAL 8]
3
h
M
3.46410
2
3.51041
1
2
1
T
3.65028
2
3.55719
2
S
3.52616
2
3.52600
4
values
3.526(0) appears to be justified
[A1A1A1A1A1]
evidence of efficient
formulae for T and S
[M1M1]
[A1]
[TOTAL 8]
4
h
f(2 + h)
est f
'(2)
0
1.4427
0.1
1.3478
-0.949
0.01
1.4324
-1.03
0.001
1.4416
-1.1
[M1A1A1A1]
Clear loss of significant figures as h is reduced
Impossible to know which estimate is most accurate
[E1]
[E1]
[TOTAL 6]
5
x
1
2
3
4
5
6
g(x)
3.2
12.8
28.4
50.2
77.9
111.6
Δg
9.6
15.6
21.8
27.7
33.7
Δ2g
6
6.2
5.9
6
table
second differences nearly
constant
so approximately quadratic
123
[M1A1]
[E1]
4776
Mark Scheme
g(1.5) =
3.2 + 0.5*9.6 + 0.5*(-0.5)*6/2 =
June 2006
7.25
[M1A1A1A1]
[TOTAL 7]
6 (i)
x
-58.6228
a
4.6
4.65
b
4.7
4.7
sign f(a)
1
1
sign f(b)
-1
-1
4.65
4.675
1
-1
x2-tan(x)
(ii)
NB: 3 pi /2 =4.71 (not
reqd)
4.6
12.2998
3
x
x2-tan(x)
7.7
52.8471
3
4.7
7.9
84.1251
1
change of sign, so root
x
4.65
4.675
[M1A1]
sign f(x)
1
-1
4.6625
root is 4.6625 with mpe
0.0125
mpe
0.05
0.025
0.012
5
no change of sign, so no evidence of root
Sketch showing asymptote for tan(x) at 5pi/2 = 7.854
So x2 curve is above tan(x) at both end points
(iii)
best possible estimate is 7.8
x
x2-tan(x)
7.75
50.4801
7.85
-189.529
[M1A1]
[M1A1]
[M1A1]
[A1]
[subtotal 9]
[M1A1]
[G2]
[E1]
[subtotal 5]
[A1]
change of sign so 7.8 is correct to 1 dp
[M1]
[A1E1]
[subtotal 4]
[TOTAL 18]
7
(i)
(ii)
D = (36 - 8) / (4 - 2) = 14
I = 0.5 (-3 + 8) + (8 + 36) = 46.5
[M1A1]
[M1A1]
[subtotal 4]
q(x) = -3 (x-2)(x-4)/(1-2)(1-4) + 8 (x-1)(x-4)/(2-1)(2-4) + 36 (x-1)(x-2)/(4-1)(4-2)
= - (x2-6x+8) - 4 (x2-5x+4) + 6 (x2-3x+2)
= x2 + 8x - 12
q'(x) = 2x + 8 so D = 12
[M1A1]
∫ q(x) dx = x3/3 + 4x2 - 12x so I = 45
(iii)
[M1A1A1A1]
[A1]
[A1]
[M1A1A1]
[subtotal 11]
Large relative difference between estimates of D
Small relative difference in estimates of I
To be expected as integration is a more stable process than differentiation
[E1]
[E1]
[E1]
[subtotal 3]
[TOTAL 18]
124
4776
Mark Scheme
Jan 2007
MEI Numerical Methods (4776) January 2007
mpe:
mpre:
1
0.000 000 05 x 1098
Mark scheme
= 5 x 1090
0.000 000 05 / 1.7112245 =
[M1A1]
[M1A1]
2.92 x 10-8
Extra digits are used internally so that rounding errors will not (usually)
show in the displayed answer
[E1]
[TOTAL 5]
2
(i)
(ii)
3
tan 0.2=
error:
0.20271
0
-4.3E-05
approx =
rel error:
0.20266
7
-0.00021
k 0.2^5=
4.34E-05
hence k=
0.13552
8
accept 0.13 or 0.14
2
0.35646
2
0.00169
5
0.35557
0
3
0.35706
7
0.00060
5
0.35693
2
r
0
xr
0.35
1
0.354767
Differences
0.004767
Ratio of differences
root =
=
4
[M1A1A1]
[subtotal 3]
[TOTAL 7]
[M1A1]
[M1A1]
0.35706
7
+0.000605 (0.356932 + 0.3569322 + …)
0.35740
3
0.3574 seems justified
[M1A1]
[A1]
[A1]
[TOTAL 8]
Graph of y = cos x and y = x2 showing one intersection for x > 0. (Or equivalent.)
x
cos x -x2
r
xr
f(x)
5
[A1A1]
[A1A1]
[subtotal 4]
x
f(x)
0.7
0.27484
2
0.9
-0.18839
0
1
0.7
0.27484
2
0.9
-0.18839
2
0.81866
3
0.01298
9
0
1.1105
0.25
1.2446
0.5
1.4065
[G2]
change of sign so root
[M1A1]
3
0.82390
9
0.00053
1
4
0.82413
3
root
0.824
[M1A1A1]
-1.6E-06
to 3dp
[A1]
[TOTAL 8]
h
0.5
0.25
f '(0)
0.5920
0.5364
poor accuracy: estimates very different, at most 1 dp reliable
h
0.25
f '(0.25)
0.5920
accuracy likely to be better because of the use of central difference formula;
90
[M1A1A1]
[E1]
[subtotal 4]
[M1A1]
[E1]
4776
Mark Scheme
Jan 2007
nothing more than 1 dp because there is nothing to compare the answer with.
6
(i)
(ii)
x
f(x)
0.9
-0.43
1.1
-0.09
1.2
0.15
1.4
0.78
[E1]
[subtotal 4]
[TOTAL 8]
1.5
1.15
y = -0.09 (x - 1.2) / (1.1 - 1.2) + 0.15 (x - 1.1) / (1.2 - 1.1) = 2.4 x - 2.73
Estimate of α:
1.1375
[M1A1A1A1]
[A1]
Using values -0.085 and 0.155 gives α as 1.1354
Using values -0.095 and 0.145 gives α as 1.1396
Hence quote 1.14
[M1A1]
[M1A1]
[A1]
[subtotal 10]
y = -0.09 (x - 1.2) (x - 1.4) / (1.1 - 1.2) (1.1 - 1.4) + two similar terms
y = -3 (x2 - 2.6x + 1.68) - 7.5 (x2 - 2.5x +1.54) + 13 (x2 - 2.3x + 1.32)
= 2.5 x2 - 3.35x + 0.57
y = 0 gives α = 1.14 (reject other root)
[M1A1A1A1]
[A1]
[A1]
[M1A1]
[subtotal 8]
[TOTAL 18]
7
(i)
x
1
2
1.5
1.25
1.75
a=
(ii)
b=
c=
x^-x
1
0.25
0.54433
1
0.75659
3
0.37556
4
0.57122
1
0.57227
4
0.57234
4
M
T
S
0.544331
0.625
0.58466
6
0.57122
1
0.566078
0.57537
2
0.57227
4
b-a=
0.00105
3
(h=0.5)
[M1A1A1]
(h=0.25)
ratio =
0.06647
c-b=
7E-05
7
Theoretically 1/16 (= 0.0625): good agreement with theory.
[M1A1A1A1]
[subtotal 7]
[M1A1A1]
[E1E1]
[subtotal 5]
0.572344 + 0.0000699 (1/16 + 1/162 + …)
0.572349
[M1A1]
[A1]
0.57235 appears completely secure from the rate of convergence
but there may be rounding errors in the 6th dp
[A1E1]
[E1]
[subtotal 6]
(iii)
=
[TOTAL 18]
91
4777
1
Mark Scheme
June 2007
x
1
1.4
f(x)
2.414214
3.509193
<3
>3
change of sign hence root in (1, 1.4)
1.2
1.3
1.25
2.92324
3.206575
3.0625
<3
>3
>3
root in (1.2, 1.4)
root in (1.2, 1.3)
root in (1.2, 1.25)
est 1.3
est 1.25
est 1.225
[M1A1]
mpe 0.1
mpe 0.05
mpe 0.025
mpe
mpe reduces by a factor of 2, 4, 8, …
Better than a factor of 5 after 3 more iterations
2
x
0
0.25
0.5
1/(1+x^4)
1
0.996109
0.941176
[M1A1]
[TOTAL 8]
values:
M=
0.498054
T=
0.485294
S = (2M + T) / 3 =
0.493801
h
S
ΔS
0.5
0.493801
0.25
0.493952 0.000151
/ one term enough
0.493962
Extrapolating:
0.493952 + 0.000151 (1/16 + 1/162 +...) =
0.49396 appears reliable. (Accept 0.493962)
3
Cosine rule:
Approx formula:
Absolute error:
Relative error:
4(i)
r represents the relative error in X
(ii)
Xn = xn(1 + r)n ≈ xn(1 + nr) for small r
hence relative error is nr
(iii)
pi =
22/7 =
5.204972
5.205228
0.000255
0.000049
3.141593
3.142857
[M1A1A1]
[A1]
[A1]
[A1]
[A1]
[M1]
[M1]
[M1A1]
[A1]
[TOTAL 8]
[M1A1]
[A1]
[B1]
[B1]
[TOTAL 5]
[E1]
(abs error:
rel error:
[A1E1]
0.001264
0.000402
approx relative error in π2 (multiply by 2):
approx relative error in sqrt(π) (multiply by 0.5):
)
0.000805
0.000201
[M1]
[A1]
(0.0008)
(0.0002)
[M1M1A1]
[TOTAL 8]
5
x
-1
0
4
f(x)
3
2
9
f(x) =
3 x (x-4) / (-1)(-5) +
2 (x+1)(x-4) / (1)(-4) +
9 (x+1) x / (5)(4)
f(x) =
0.55 x2 - 0.45 x + 2
f '(x) = 1.1 x - 0.45
Hence minimum at x = 0.45 / 1.1 = 0.41
[M1A1]
[A1]
[A1]
[A1]
[B1]
[A1]
[TOTAL 7]
4777
6(i)
Mark Scheme
June 2007
Sketch showing curve, root, initial estimate, tangent, intersection of tangent
with x-axis as improved estimate
[E1E1E1]
[subtotal 3]
(ii)
0
0.35
0.7
1.05
1.4
0
-0.33497
-0.55771
-0.35668
2.997884
Sketch showing root, α
E.g. starting values just to the
left of the root can produce an
x1 that is the wrong side of the
asymptote
E.g. starting values further left
can converge to zero.
[G2]
[M1]
[E1]
[M1]
[E1]
[subtotal 6]
(iii)
Convincing algebra to obtain the N-R formula
r
xr
0
1
2
1.2 1.169346 1.165609
root is 1.1656 to 4 dp
[M1A1]
3
1.165561
4
1.165561
[M1A1A1]
[A1]
differences from root
-0.03065
-0.00374 -4.8E-05 Accept diffs of successive terms
ratio of differences
0.1219
0.012877
[M1A1]
ratio of differences is decreasing (by a large factor), so faster than first order
[E1]
[subtotal 9]
[TOTAL 18]
7
(i)
x
1
2
3
4
5
g(x)
2.87
4.73
6.23
7.36
8.05
Δg
Δ2g
1.86
1.50
1.13
0.69
-0.36
-0.37
-0.44
[M1A1A1]
Not quadratic
Because second differences not constant
(ii)
x
1
3
5
g(x)
2.87
6.23
8.05
Δg
Δ2g
3.36
1.82
-1.54
[E1]
[E1]
[subtotal 5]
[B1]
Q(x) = 2.87 + 3.36 (x - 1)/2 - 1.54 (x - 1)(x - 3)/8
=
0.6125 + 2.45 x - 0.1925 x2
(iii)
x
2
4
Q(x)
4.7425
7.3325
g(x)
4.73
7.36
error
0.0125
-0.0275
rel error
0.002643
-0.00374
[M1A1A1A1]
[A1A1A1]
[subtotal 8]
Q:
errors:
rel errors:
[A1A1]
[A1]
[M1A1]
[subtotal 5]
[TOTAL 18]
4776
Mark Scheme
4776
1
x
f(x)
Eg:
2
0.24
January 2008
Numerical Methods
root = (2 x 0.03 - 3 x 0.24) / (0.03 - 0.24)
= 3.142857
3
0.03
graph showing turning point at x = 3 with root some way
to the left or the right.
[M1A1]
[A1]
[G2]
[TOTAL 5]
2
x
0
1
0.5
f(x)
1
0.333333
0.477592
T1 =
M=
hence
and
0.666667
0.477592
T2 = (T1 + M)/2 =
S = (T1 + 2*M)/3 =
0.572129
0.540617
[M1A1]
[M1A1]
[M1A1]
[M1A1]
[TOTAL 8]
3
x
f(x)
f(2) =
=
0
2
1
2.57
3
3.85
3 terms:
form:
use x=2:
2(2-1)(2-3)/(0-1)(0-3) + 2.57(2-0)(2-3)/(1-0)(1-3) + 3.85(2-0)(2-1)/(3-0)(3-1)
3.186667
(3.19)
[M1]
[M1]
[M1]
[A1A1A1]
[A1]
[TOTAL 7]
4
x
x3(2-x)-1
1.5
0.6875
2
-1
a
1.5
1.75
1.75
b
2
2
1.875
x
1.75
1.875
1.8125
change of sign, so root (may be implied)
x3(2-x)-1
0.339844
-0.17603
mpe
0.25
0.125
0.0625
4 further iterations reqd: mpe 0.0325, 0.015625, 0.0078125, 0.00390625
[M1A1]
[M1A1]
[A1]
[A1]
[M1A1]
[TOTAL 8]
5
Sketch showing curve, tangent, chord, h. Makes clear that tangent and chord
have substantially different gradients.
h
g(2 + h)
est g '(2)
0
3.61
0.1
3.849
2.39
0.01
3.633
2.3
Clear loss of significant figures as h is reduced
0.001
3.612
2
[G3]
[M1A1A1A1]
[E1]
[TOTAL 8]
64
4776
6
(i)
Mark Scheme
x
3
4
5
6
quadratic
f(x)
1
3
-1
-10
Δf
Δ2f
Δ3f
2
-4
-9
-6
-5
1
= 1 + 2(x-3) - 6(x-3)(x-4)/2
= 1 + 2x-6 - 3x2+21x-36
= -3x2 +23x -41
[M1A1A1]
[M1A1]
[A1]
[A1]
q'(x) = -6x + 23 = 0 at x = 23/6 (= 3.833...)
[M1A1]
q(x) = 0 at x = 4.847(127); also at 2.81954 - not reqd.
[M1A1]
q(6) = -11 (or point out that the second differences not constant)
(ii)
January 2008
cubic est
= 1 + 2(4.5-3) - 6(4.5-3)(4.5-4)/2 + 1(4.5-3)(4.5-4)(4.5-5)/6
= 1.6875
S = 1.5/3 (1 + 4x1.6875 -10) =
-1.125
[A1]
[subtotal
12]
[M1A1A1]
[A1]
[M1A1]
[subtotal 6]
[TOTAL 18]
7
(i)
mpe 0.000 000 5
mpre 0.000 000 5 / 2.506 628
=
[B1]
1.99 x 10-7
[M1A1]
[subtotal 3]
(ii)
mpe 1000 x 0.000 000 5 = 0.000 5
In practice the positive and negative errors will tend to cancel out
[M1A1]
[E1]
[subtotal 3]
(iii)
mpe 1000 x 0.000 001 = 0.001
In practice 1000 x 0.000 000 5 = 0.000 5
because average error in chopping will be 0.000 000 5
[M1A1]
[M1A1]
[E1]
[subtotal 5]
(iv)
L to R:
1 (or 1.000 000)
R to L:
1.000 001
L to R requires 8 sf, (R to L doesn't)
[B1]
[B1]
[E1]
[subtotal 3]
(v)
Reverse order more accurate
as that way allows the very small terms at the end of the series
to contribute to the sum.
The spreadsheet is likely to work to greater accuracy
The spreadsheet works to more sf than are displayed
[E1]
[E1]
[E1]
[E1]
[subtotal 4]
[TOTAL 18]
65
4776 Numerical Methods
x
f(x)
1
3
0.5
3.5
-0.8
root = (3 x (-0.8) - 3.5 x 0.5) / (-0.8 - 0.5)
= .192308 (3.192, 3.19)
[M1A1A1]
[A1]
(-) mpe is 3.5 - 3.192308 = 0.307602 (0.308, 0.31)
[M1A1]
[TOTAL 6]
1
3
5
7
9
2
2
1
5
k
2
-1
4
k-5
2-k
16-3k = k-14
3
4
5
k-9
7-2k
k-14
16-3k
[M1A1A1A1]
hence k = 7.5
[M1A1]
[TOTAL 6]
h
f(2+h)
f(2-h)
f '(2)
derivatives
0.2
0.1
0.05
.494507
.323418
.241636
.867869
.010586
.085281
.566594
.564163
.563555
[M1A1A1A1]
-0.00243
-0.00061
differences reducing by a factor 4 so next estimate about 1.56340.
1.563 secure to 3 dp.
[M1]
[B1]
[TOTAL 8]
f(x) = x3-25
f '(x) = 3x2
xr+1 = xr - (xr3-25)/3xr2
(a.g.)
[M1A1A1]
r
xr
diffs
ratios
0
4
1
3.1875
2
.945197
3
2.92417
-0.8125
-0.2423
.298219
-0.02103
.086783
differences reducing at an increasing rate (hence faster than first order)
5
(i)
(ii)
differences
[M1A1]
0.001 369 352
(accept 0.001 369 4)
sin 86o = 0.997
564
sin 86o - sin 86o = 0.001 369
sin 85o = 0.996
195
[M1A1]
[B1]
[B1]
[E1]
[TOTAL 8]
[B1]
[B1B1]
[A1]
(iii)
2 x 0.0784591 x 0.008 726 54
= 0.00136935
[M1]
[A1]
(iv)
Rounding has different effects in the two expressions (may be implied)
First method involves subtraction of nearly equal numbers and so loses accuracy
[E1]
[E1]
[TOTAL 8]
6
h
M
T
2 2.763547 2.425240
1 2.677635 2.594393
0.5 2.656743 2.636014
(i)
M:
(ii)
2.763547
2.677635
2.656743
diffs
-0.08591
-0.02089
mid-point:
trapezium:
reducing by a factor 4 (may be implied)
[M1A1E1]
Differences in T reduce by a factor 4, too
(iii)
M
T
S
2.763547
2.677635
2.656743
2.425240
2.594393
2.636014
2.650778
2.649888
2.649833
[M1A1A1]
[M1A1A1A
1]
[subtotal 7]
[B1]
[subtotal 4]
-0.00089033
-0.000054333
S values:
diffs
Differences in S reducing fast e.g by a factor of (about) 16
How this leads to an answer, e.g:
Next difference about -0.0000034 and/or next answer about 2.649830
Accept 2.6498 or 2.64983
[M1]
[A1A1]
[A1]
[E1]
[E1]
[A1]
[subtotal 7]
[TOTAL 18]
7
(i)
Eg: graph of x2 and 4 + 1/x for x > 0 showing single intersection
Change of sign to find interval (2,3) - i.e. a = 2
r
xr
0
1
2
3
2.5 2.097618 2.115829 2.114859
[G2]
[B1]
4
5
2.11491 2.114907
2.1149 secure to 4 dp
(ii)
[A1]
[subtotal 7]
The iteration gives positive values only.
[E1]
r
0
1
2
3
4
5
xr
-2
-1.87083
-1.86158
-1.86087
-1.86081
-1.86081
-1.8608 secure to 4 dp
(iii)
Eg
r
0
1
2
3
4
xr
-0.5
-1.41421
-1.81463
-1.85713
-1.86052
xr+1 = 1 / (xr2 - 4)
r
xr
0
-0.5
1
-0.26667
[M1A1A1]
[A1]
[subtotal 5]
not converging to required root (converging to previous root)
Eg
[M1A1A1]
[M1A1]
[M1]
2
-0.25452
-0.2541 secure to 4 dp
3
-0.25412
4
-0.2541
5
-0.2541
[M1A1]
[A1]
[subtotal 6]
[TOTAL 18]
4776
Mark Scheme
January 2009
4776 Numerical Methods
1(i)
(ii)
2(i)
(ii)(A)
(B)
x
-3
-1
1
3
y
-16
-2
4
2
1st diff
2nd diff
14
6
-2
-8
-8
2nd difference constant so quadratic fits
f(x) = -16 + 14(x + 3)/2 - 8(x + 3)(x + 1)/8
= -16 + 7x + 21 - x2 - 4x - 3
= 2 + 3x - x2
[M1A1]
[E1]
[M1A1A1A1]
[A1]
[TOTAL 8]
Convincing algebra to demonstrate result
Direct subtraction:
0.0022
Using (*):
1/(223.6090+223.6068) = 0.002236057
Second value has many more significant figures ("more accurate") -- may be implied
Subtraction of nearly equal quantities loses precision
[M1A1]
[B1]
[M1A1]
[E1]
[E1]
[TOTAL 7]
3(i)
x
0
0.8
0.4
f(x)
1
0.819951
0.994867
T1 =
M1 =
hence S1 =
0.72798
0.795893
0.773256
all values
(ii)
4(i)
T2 = 0.761937
M2 = 0.784069
so S2 = 0.776692
S2 will be much more accurate than S1 so 0.78 or 0.777 would be justified
x
0.3
cosx
0.955336
1 - 0.5x2
0.955
error
-0.000336
rel error
-0.000352
want k 0.34 = 0.000336
gives k = 0.041542 (0.0415, 0.042, 1/24)
(ii)
condone signs here
but require correct
sign for k
[M1]
[M1]
[M1]
[A1]
[B1]
[M1A1]
[A1]
[TOTAL 8]
[M1A1A1A1]
[M1]
[A1]
[TOTAL 6]
5
r
xr
0
3
1
3
2
3
xr
2.99
2.9701
2.911194
xr
3.01
3.0301
3.091206
Derivative is 2x - 3. Evaluates to 3 at x = 3
3 is clearly a root, but the iteration does not converge
Need -1 < g'(x) < 1 at root for convergence
66
[M1A1A1]
[M1A1]
[E1]
[E1]
[TOTAL 7]
4776
6(i)
(ii)
Mark Scheme
January 2009
Demonstrate change of sign (f(a), f(b) below) and hence existence of root
a
b
f(a)
f(b)
x
mpe
f(x)
0.2
0.3 -0.06429 0.021031
0.25
0.05
-0.01827
0.25
0.3 -0.01827 0.021031
0.275
0.025 0.002134
-0.00787
0.25
0.275
0.2625 0.0125
r
0
1
2
3
xr
0.2
0.3
0.275352
0.272161
fr
-0.06429
0.021031
0.00241
-0.0001
accept 0.27 or 0.272 as secure
secant method much faster
(iii)
r
0
1
2
3
xr
1.4
1.314351
1.298887
1.298504
er
0.101496
0.015847
0.000383
= root
[B1]
[M1]
[M1]
[A1A1A1]
[subtotal 6]
[M1A1]
[M1A1]
[A1]
[E1]
[subtotal 6]
er+1/er2
e col:
1.538329
e/e2 col:
1.525122
equal values show 2nd order convergence
second order convergence: each error is
proportional to the square of the previous error
[M1A1]
[M1A1]
[E1]
[E1]
[subtotal 6]
[TOTAL 18]
7(i)
(ii)
(iii)
(iv)
fwd diff:
cent diff:
h
f '(0)
diffs
0.4
0.444758
h
f '(0)
diffs
0.4
0.491631
0.2
0.473525
0.028768
0.1
0.48711
0.013585
0.2
0.498315
0.006684
0.1
0.50008
0.001765
approx halved
[M1A1A1]
[B1]
[subtotal 4]
reduction greater than
for forward difference
[M1A1A1]
[B1]
[subtotal 4]
(D2 - d) = 0.5 (D1 - d)
convincing algebra to d = 2D2 - D1
(D2 - d) = 0.25 (D1 - d)
convincing algebra to d = (4D2 - D1)/3
fwd diff:
cent diff:
2(0.48711) - 0.473525 =
(4(0.50008) - 0.498315) / 3 =
[M1A1]
[M1A1A1]
[subtotal 5]
0.500695
[M1A1]
0.500668
[M1A1]
0.5007 seems secure
67
[E1]
[subtotal 5]
[TOTAL 18]
4776
Mark Scheme
June 2009
4776 Numerical Methods
1(i)
(ii)
f(x) = 1.6(x - 0.4)(x - 1)/(-0.4)(-1) + 2.4x(x - 1)/0.4(0.4 - 1) + 1.8x(x - 0.4)/1(1 - 0.4)
= 4(x2 - 1.4x + 0.4) - 10(x2 - x) + 3(x2 - 0.4x)
2
= - 3x + 3.2x + 1.6
Newton's formula requires equally spaced data
[M1A1,1,1]
[A1]
[A1]
[E1]
[TOTAL 7]
x
x2 + 1/x - 3
2
f(x) = x2 + 1/x - 3
r
xr
1
-1
so
2
1.5
f '(x) = 2x - 1/x2
0
1
1.5
1.532609
(change of sign so root)
[M1A1]
hence NR formula
2
3
[M1A1]
1.532089
1.532089
1.53209
[M1A1A1]
[TOTAL 7]
3(i)
(ii)
term
mpe
term
mpre
X
0.0005
X+Y
0.001
X
0.00018
4
X-Y
10X + 20Y
0.001
0.015
Y
XY
X/Y
0.000159
0.000343
0.000343
[B1B1B1B
1]
[B1B1B1B
1]
[TOTAL 8]
4(i)
(ii)
(iii)
to 6 dp:
sin A
0.84683
2
sin B
0.841471
LHS
0.5361
RHS
0.536088
[B1B1]
It is an approximate equality. LHS involves subtraction of nearly equal numbers.
LHS involves 2 trig functions, RHS just 1.
Subtraction of nearly equal quantities is a bigger problem as the difference decreases.
RHS involves no such problem.
[E1E1]
[E1E1]
[TOTAL 6]
0
0
1
0.25
0.5
0.75
1
0.996094
0.9375
0.683594
0
r
0
1
2
3
5
0.25
0.5
0.75
1
xr
0.6
0.8704
0.426048
0.967052
cobweb diagram showing
spiralling out from root
[G2]
[M1A1A1]
[M1A1A1]
[TOTAL 8]
99
4776
6(i)
Mark Scheme
x
0
0.8
0.4
f(x)
1.732051
1.777639
1.8
M1 =
1.44
0.2
0.6
1.777639
1.8
M2 =
1.431056
June 2009
T1 =
T2 =
1.403876
1.421938
M
T
v
a
l
u
e
s
[M1]
[M1]
[A1,1,1,1]
[subtotal
6]
(ii)
S1 =
S2 =
1.427959
1.428016
(a.g.)
(iii)
S4 =
(2 M4 + T4) / 3 =
(iv)
M
diffs
ratio
1.44
S
diffs
ratio
1.427959
[M1]
[M1A1]
[subtotal
3]
1.428020
1.431056
-0.00894
1.428016
5.77E-05
[M1A1]
[subtotal
2]
1.428782
-0.00227
0.254186
approx 0.25
1.428020
3.99E-06
0.069037
(approx 0.0625)
Reasoning to: integral is secure as 1.42802(0)
7(i)
x
1
1.2
1.4
f(x)
0.6
-0.1
0.4
1st diff
2nd diff
-0.7
0.5
1.2
f(x) = 0.6 + (-0.7)(x - 1) / 0.2 + 1.2(x - 1)(x - 1.2) /(2 (0.2)2)
2
= 0.6 - 3.5x + 3.5 +15x - 33x + 18
2
= 15x - 36.5x + 22.1
(ii)
(iii)
[M1A1]
[M1A1A1]
[M1B1]
[subtotal
7]
[TOTAL
18]
[M1A1]
[M1A1A1A
1
[M1A1]
[subtotal
8]
f '(x) = 30x - 36.5
f '(1.2) = 36 - 36.5 = -0.5
Central difference:
(0.4 - 0.6)/(1.4 - 1) = -0.2/0.4 = -0.5
Suggests central difference is accurate for quadratics.
f '(1) = 30 - 36.5 = -6.5
100
[M1A1]
[M1A1]
[E1]
[subtotal
5]
[B1]
4776
Mark Scheme
Forward difference:
(-0.1 - 0.6)/(1.2 - 1) = -0.7/0.2 = -3.5
Shows that forward difference is not exact for quadratics.
Quadratic estimate (-6.5) is likely to be more accurate. (Allow comments
saying that we cannot be sure.)
101
June 2009
[M1A1]
[E1]
[E1]
[subtotal
5]
[TOTAL
18]
4777
1(i)
Mark Scheme
June 2009
-1 < g'(α) < 1
[B1]
E.g. Multiply both sides of x = g(x) by λ and add (1 - λ)x to both sides.
Derivative of rhs set to zero at root: λg'(α) + 1 - λ =
0
algebra to obtain given result
[M1A1]
[M1A1]
[A1]
In practice use an initial estimate x0 in place of α
(iii)
x
0
0.5
1
1.5
2
2.5
3
x
0
0.5
1
1.5
2
2.5
3
[A1]
[subtotal
7]
3sinx - 0.5
-0.5
0.938277
2.024413
2.492485
2.227892
1.295416
-0.07664
[G3]
Roots approximately 0.25, 2.1
[B1B1]
Eg:
r
0
xr
0
xr
0.2
1
-0.5
0.096008
2
-1.93828
-0.21242
3
-3.29971
-1.13247
4
-0.02763
-3.21639
5
-0.58289
-0.2758
6
-2.15131
-1.31696
7
-3.00855
-3.40387
8
-0.89795
0.277847
9
-2.84615
0.322857
10
-1.37349
0.451832
xr
0.4
0.668
255
1.358
852
2.432
871
1.452
591
2.479
066
1.345
334
2.424
072
1.472
555
2.485
535
1.329
994
No convergence in each case
xr
2
2.227892
1.875308
2.36198
1.609012
2.49781
1.300676
2.391217
1.545741
2.499058
1.297679
xr
2.2
1.92
5489
2.31
326
1.71
0416
2.47
0807
1.36
4805
2.43
6576
1.44
4139
2.47
5969
1.35
2649
2.42
89
xr
2.4
1.52639
2.497043
1.302517
2.392685
1.542517
2.498801
1.298298
2.389305
1.549934
2.499347
[M1A1A1]
Let g(x) = 3 sinx - 0.5
Then g'(x) = 3
cosx
So λ = 1 / (1 - 3 cosα)
[M1A1]
102
4777
Mark Scheme
Smaller root:
λ=
Larger root:
=
-0.52446
(approx -0.5)
r
June 2009
xr
NB:
must
be
using
relaxat
ion
λ
0.397687
(approx 0.4)
r
xr
0
2.1
2.09585
1
2.09586
6
2.09586
6
2.09586
6
2.09586
6
0
0.25
1
0.253894
2
0.254078
3
0.254087
3
4
0.254088
4
5
0.254088
5
1
2
[M1A1A1]
[M1A1]
[M1A1]
[subtotal
17]
[TOTAL
24]
2(i)
f(x) = 1
2h = 2a + b
f(x) = x, x3 give
0=0
f(x) =
x2
2h3/3 = 2aα2
f(x) =
2h5/5 = 2aα4
x4
Convincing algebra to verify given
results
(ii)
L
R
0
0.785398
function values
weig
hts
inte
gral
L
R
0
0.392699
function values
weig
hts
inte
gral
0.392699
function values
weig
0.785398
[M1A1]
[M1A1]
[A1]
[A1]
[A1A1]
[subtotal
8]
m
0.392
699
1.189
207
0.349
066
0.415
112
h
0.3926
99
m
0.196
35
1.094
949
0.174
533
0.191
105
0.589
049
1.291
58
0.174
h
0.1963
5
0.1963
5
103
x1
0.08851
6
1.04343
1
0.21816
6
0.22764
1
x1
0.04425
8
1.02190
3
0.10908
3
0.11147
2
0.43695
7
1.21122
6
0.10908
x2
0.69688
2
1.35535
0.21816
6
0.29569
1
0.9384
44
x2
0.34844
1
1.16758
9
0.10908
3
0.12736
4
0.4299
41
0.74114
1.38390
1
0.10908
setup:
[M3A3]
[A1]
repeat:
[M2]
4777
Mark Scheme
hts
inte
gral
533
0.225
423
June 2009
3
0.13212
4
3
0.15096
0.5085
08
0.9384
49
Either repeat with h halved to verify that 0.938449 is correct to 6 dp
Or observe that the method is converging so rapidly that 0.938449 will be correct
to 6dp
(iii)
[A1]
[M1A1]
or [E1A1]
[subtotal
12]
Use routine known to deliver 6dp and vary k:
L
R
m
0.196
35
1.136
464
0.174
533
0.198
35
0.589
049
1.406
898
0.174
533
0.245
55
h
0.1963
5
0
0.392699
1.466
1.000
0.999908
036
hence k = 1.466
1.467
1.0001
63
function values
weig
hts
inte
gral
0.392699
0.785398
function values
weig
hts
inte
gral
k
integral
1.465
0.1963
5
x1
0.04425
8
1.03194
6
0.10908
3
0.11256
8
0.43695
7
1.29791
8
0.10908
3
0.14158
1
k=
x2
0.34844
1
1.23791
8
0.10908
3
0.13503
6
0.74114
1.53016
4
0.10908
3
0.16691
5
1.4657
2
0.4459
54
modify
[M1A1]
0.5540
46
1.0000
00
find k
[M1A1]
[subtotal
4]
[TOTAL
24]
104
4777
3(i)
Mark Scheme
Use central difference formulae for 2nd and 1st derivatives to obtain first given result
0.1
[M1A1A1
]
Hence obtain y1 = h2 - y-1
[M1A1]
Use central difference to obtain y1 - y-1 = 2h
[M1A1]
Hence given result for y1
(ii)
h
June 2009
x
y
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
0
0.105
0.216472
0.332426
0.450961
0.570174
0.68815
0.802981
0.912793
1.015786
1.11027
1.194705
1.26774
1.328248
1.375354
1.40846
1.42726
1.431751
1.42223
1.399287
1.363785
1.316838
1.259773
1.194096
1.121445
1.04354
0.962141
0.878993
0.79578
0.714082
0.635337
0.560807
0.491549
0.428404
0.371982
0.322662
0.280597
0.245729
0.217808
0.196416
0.180999
0.170894
0.165365
0.163635
0.164915
0.168435
0.173469
0.179352
0.185502
[M1]
[subtotal
8]
105
4777
Mark Scheme
4.9
5
June 2009
0.191424
0.196725
setup
[M3]
(ii)
h
0.1
a
1.4
Obtain formula y1 = ah + 0.5h2
Modify routine
0
-0.135
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4
4.1
4.2
4.3
4.4
-0.25582
-0.36107
-0.44993
-0.5219
-0.57677
-0.6146
-0.63565
-0.64047
-0.6298
-0.60462
-0.56614
-0.51572
-0.45494
-0.3855
-0.3092
-0.22792
-0.14356
-0.05802
0.026884
0.109408
0.187962
0.26113
0.327696
0.386672
0.437316
0.479135
0.51189
0.535589
0.550471
0.556986
0.555768
0.547604
0.533401
0.514147
0.490876
0.464631
0.43643
0.40724
0.377942
0.349319
0.322033
0.296623
0.27349
graph
[A3]
[subtotal
9]
[M1A1]
[M1A1]
[M1A1G1
]
Trial on a to obtain a = -1.4 or -1.5
x
y
0
0.1
numb
ers
[A3]
106
4777
Mark Scheme
4.5
4.6
4.7
4.8
4.9
5
June 2009
0.252909
0.235026
0.219875
0.207386
0.197404
0.189706
[subtotal
7]
[TOTAL2
4]
4(i)
Diagonal dominance: the magnitude of the diagonal element in any row is greater
than or equal to the sum of the magnitudes of the other element.
| a | > | b | + 2 will ensure convergence. ( > required as dominance has to be strict)
(ii)
4
1
2
1
1
4
1
2
2
1
4
1
1
2
1
4
0
0.25
0.321289
0.340733
0.344469
0.344515
0.344124
0.343886
0.343789
0.343758
0.34375
0.343749
0.34375
0.34375
0
-0.0625
-0.05103
-0.03941
-0.03388
-0.0319
-0.03134
-0.03123
-0.03123
-0.03124
-0.03125
-0.03125
-0.03125
-0.03125
0
-0.10938
-0.14691
-0.15599
-0.15715
-0.15681
-0.15648
-0.15633
-0.15627
-0.15625
-0.15625
-0.15625
-0.15625
-0.15625
0
-0.00391
-0.01808
-0.02648
-0.02989
-0.03098
-0.03124
-0.03127
-0.03127
-0.03126
-0.03125
-0.03125
-0.03125
-0.03125
2
1
4
1
1
2
1
4
4
1
2
1
1
4
1
2
0
0.5
2.03125
6.033203
12.69934
3.347054
-156.613
-1153.81
-5937.67
0
-0.25
-1.95313
-10.5127
-46.9236
-183.278
-632.515
-1881.51
-4365.6
0
-0.875
-3.42969
-9.11279
-13.2195
37.89147
456.5137
2690.835
12560.88
0
0.6875
4.605469
22.56519
94.10735
345.9377
1115.079
2994.509
5419.593
1
0
0
0
a
4
[E1]
[E1E1]
[subtotal
3]
b
2
setup
[M3A3]
values
[A3]
1
0
0
0
a
2
b
4
values
[A3]
[subtotal
12]
107
4777
Mark Scheme
(iii)
No convergence when a = 2, b = 0
Indicates that non-strict diagonal dominance is not sufficient
(iv)
Use RHSs 1,0,0,0 0,1,0,0
to obtain inverse as
0.34375
-0.03125
-0.15625
-0.03125
-0.03125
0.34375
-0.03125
-0.15625
0,0,1,0
-0.15625
-0.03125
0.34375
-0.03125
0,0,0,1
-0.03125
-0.15625
-0.03125
0.34375
108
June 2009
[M1A1]
[E1E1]
[subtotal
4]
[M1]
[A1]
[A1]
[A1]
[A1]
[subtotal
5]
[TOTAL
24]
4776
Mark Scheme
January 2010
4776 Numerical Methods
1
x
1.3
1.5
1.4
1.35
1.375
1.3875
mpe:
2
h
1
0.5
LHS
2.868415 < 3
3.181981 > 3
[M1A1]
mpe (may be implied)
0.1
0.05
0.025
0.0125
3.017945
2.941413
2.979232
0.00625
0.003125
0.001563
0.000781 < 0.001
finishing at this point:
[M1]
[A1]
[A1]
[A1]
so 4 more iterations
[M1A1]
[TOTAL 8]
T
S
[M1A1]
[M1A1A1]
S
M
T
2.579768 2.447490
2.535675
2.547350 2.513629
2.536110
2.536 secure by comparison of S values.
[E1A1]
[TOTAL 7]
3(i)
f '(x) = 3 x2 – 2 x
so f '(0.5) = –0.25
f(0.5) = 0.875 hence given result
[B1B1]
[B1]
(ii)
Require –0.0005 < 0.25 h < 0.0005
Hence –0.002 < h < 0.002
And so 0.498 < x < 0.502
4(i)
Convincing algebra to given result
(ii)
Eg
[M1A1]
[A1]
[B1]
[TOTAL 7]
[M1A1]
k = 1000 correct evaluation to 2
k = 1000000 incorrect evaluation to zero (NB some will need larger k)
Mathematically equivalent expressions do not always evaluate equally
(because calculators do not store (large) numbers exactly)
Subtraction of nearly equal quantities often causes problems
64
[B1]
[B1]
[E1]
[E1]
[TOTAL 6]
4776
x
0
1
2
3
4
5(i)
(ii)
Mark Scheme
f(x)
1.883
2.342
2.874
3.491
4.206
Δf(x)
0.459
0.532
0.617
0.715
Δ2f(x)
January 2010
Δ3f(x)
1st diff:
2nd, 3rd
0.073
0.085
0.012
0.098
0.013
3rd diffs almost constant
[M1A1]
[F1]
[E1]
f(1.5) = 1.883 + 0.459 × 1.5 + 0.073 × 1.5 × 0.5 / 2! + 0.012 × 1.5 × 0.5 × (–0.5) / 3!
f(1.5) = 2.598125 or 2.598 to 3 dp
6 (i) Sketch of smooth curve and its tangent.
Forward and central difference chords.
Clear statement or implication that the central difference chord has gradient
closer to that of the tangent
[M1A1A1]
[A1]
[TOTAL 8]
[G1]
[G1G1]
[E1]
[subtotal 4]
(ii)
(iii)
h
2
1
0.5
tan 60º
1.732051
1.732051
1.732051
h
2
1
0.5
(iv) forward difference:
tan (60 + h)º
1.880726
1.804048
1.767494
tan (60 + h)º tan (60 – h)º
1.880726
1.600335
1.804048
1.664279
1.767494
1.697663
derivative
0.074338
0.071997
0.070886
central difference:
derivative
0.074338
0.071997
0.070886
derivative
0.070098
0.069884
0.069831
diffs
[M1A1]
[A1]
[A1]
[subtotal 4]
derivative
0.070098
0.069884
0.069831
[M1A1]
[A1]
[A1]
[subtotal 4]
ratio
of diffs
–0.00234
–0.00111 0.474407
diffs
(about 0.5, may be implied)
[M1A1A1]
ratio
of diffs
–0.00021
–5.3E–05 0.24896
(about 0.25, less than
forward difference, hence faster)
65
[M1A1E1]
[subtotal 6]
[TOTAL 18]
4776
7 (i)
(ii)
Mark Scheme
January 2010
Sketch showing y = 3 sin x and y = x with intersection in (½π, π)
State or show that there is only one other non-zero root
r
xr
0
1
2
3
2 2.727892 1.206001
2.80259
clearly not converging
Cobweb diagram to illustrate process
4
0.997639
[G1G1]
[E1]
[subtotal 3]
5
2.52058
(iii) Convincing algebra to given result.
1
2
r 0
xr 2 2.242631 2.277768
Root appears to be 2.27886 to 5 dp
sin x + ⅔ x
x
2.278855 < 2.2788625
2.278865 > 2.2788627
[M1A1A1]
[B1]
[G3]
[subtotal 7]
[M1]
3
2.278844
4
2.278862
5
2.278863
hence result is correct to 5 dp
[M1A1A1]
[A1]
[M1A1E1]
[subtotal 8]
[TOTAL 18]
66
GCE
Mathematics (MEI)
Advanced GCE 4776
Numerical Methods
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4776
1(i)
Mark Scheme
LHS
1
0.5
x
1
2
0
(ii)
1
(Change of sign implies root.)
(or equivalent)
2
3
5
6
0.347352
E.g. xr+1 = √(3 - 1/x)
r
xr
0
1
2
3
1.5
1.527525
1.531452
4
1.532077
1.532
5
1.532087
2(i)
Forward difference:
(0.9996 - 0.9854)/0.2 = 0.071
Central difference:
(0.9996 - 0.9508)/0.4 = 0.122
Central difference expected to be more accurate.
(ii)
Forward difference maximum:
Central difference maximum:
3(i)
r is the relative error (in X as an approximation to x)
E.g. xr+1 = 3/x - 1/x2
0
1
1.5
1.555556
4
1.523326
2
1.515306
5
1.538438
[M1A1]
[E1]
[B1]
3
1.544287
[M1A1]
[TOTAL 8]
[M1A1]
[M1A1]
[E1]
(0.99965 - 0.98535)/0.2 = 0.0715
(0.99965 - 0.95075)/0.4 = 0.12225
[B1]
[B1]
[TOTAL 7]
(1 + r)n = 1 + nr (provided r is small)
[E1]
[M1M1A1]
G2 (= 0.332 929, not required) is about 0.08% smaller than g2
√G (= 0.795 605, not required) is about 0.02% smaller than √g
[M1A1A1]
Xn = xn (1 + r)n
(ii)
4
[M1A1]
1.5 1.333333 0.818182 0.429078 0.355127
0.347961
State or clearly imply convergence outside the interval (1, 2)
r
xr
RHS
2
-1
<
>
June 2010
[TOTAL 7]
4(i)
x
0.2
0.1
sin + tan
0.401379
0.200168
2x
0.4
0.2
(ii)
2 × 0.23 / k = 0.00138 gives k = 11.59
2 × 0.13 / k = 0.00017 gives k = 11.76
5
Data not equally spaced in x
error
-0.00138
-0.00017
rel error
-0.00344
-0.00084
accept:
+ve, +ve
-ve, +ve
-ve, -ve
Either of these (or other methods)
to suggest k = 12
[M1A1A1A1]
[M1A1]
[B1]
[TOTAL 7]
[E1]
f(x) = - 10(x - 3)(x - 6) / (1 - 3)(1 - 6) - 12(x - 1)(x - 6) / (3 - 1)(3 - 6) + 30(x - 1)(x - 3) / (6 - 1)(6 - 3)
f(x) = - (x2 - 9x +18) + 2(x2 - 7x + 6) + 2(x2 - 4x + 3)
= 3x2 - 13x
1
[M1A1A1A1]
[A1]
[A1]
[TOTAL 7]
4776
6(i)
(ii)
(iii)
Mark Scheme
h
0.8
0.4
0.2
T
1.611209
1.579581
1.571610
S
1.569038
1.568953
1.568949
M:
T:
S:
1.56895 appears justified
Comparison of last two S values, e.g.:
last change in S is -0.000004; next change negligible
h
0.8
0.4
0.2
(A)
(B)
7(i)
M
1.547953
1.563639
1.567619
June 2010
M error
-0.02100
-0.00531
-0.00133
T error
0.04226
0.01063
0.00266
accept consistent
use of other sign
convention
[M1A1A1]
M errors are about half the T errors so M is twice as accurate as T
Errors for both T and M reduce by a factor of 4 as h is halved so
the rates of convergence are the same, both second order
[E1A1]
[E1]
[A1A1]
[subtotal 8]
[TOTAL 17]
f(0) = 5, f(1) = -2. (Change of sign implies root.)
[M1A1]
f '(x) = 5x4 - 8 hence N-R formula
[M1A1]
r
0
1
2
3
4
xr
0.5 0.634146 0.638232 0.638238 0.638238
differences
0.134146 0.004086 5.98E-06 1.29E-11
ratios
0.030457 0.001462 2.17E-06
The ratios of differences are decreasing (fast) so process is faster than first order
(ii)
[M1A1A1]
[M1A1]
[M1A1]
[subtotal 7]
[B1]
[E1]
[subtotal 2]
r
0
1
2
3
4
xr
1.4
-0.82176
1.5
0.59375
1.458054
-0.0747
1.462741
-0.00559
1.46312
5.99E-05
f(xr)
root is 1.46 correct to 3 sf
differences
0.1 -0.04195 0.004687 0.000379
ratios
-0.41946 -0.11175 0.080876
The ratios of differences are decreasing (fast) so process is faster than first order
accept 'second order'
2
[M1A1A1]
[A1]
[M1A1]
[E1]
[subtotal
11]
[M1A1A1]
[A1]
[A1]
[M1A1]
[E1]
[subtotal 8]
[TOTAL 19]
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4776: Numerical Methods
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4776
1
Mark Scheme
Question
(i)
x
1
1.2
LHS
2
2.2
>
<
Answer
RHS
1.557408
2.572152
January 2011
Marks
M1 A1
Guidance
no explicit explanation required
[2]
(ii)
(iii)
2
r
xr
0
1.1
1
0.96476
2
0.442927
e.g. re-arrange to x = arctan(1 + x)
0
1
2
r
xr 1.1 1.126377 1.131203
h
2
1
0.5
M
T
1.987467 1.354440
1.830595 1.670954
1.750774
3
–0.52564
4
–1.58007
(i)
3
1.132076
4
1.132233
5
1.132261
1.132
Simpson's rule (2M + T) / 3
1.776458
1.777381
M1 A1
A1
[4]
T: M1A1A1
S: M1A1A1
E1 A1
[8]
h=1
g′(0) = (2.0100 – 1.4509)/1 = 0.5591
h = 0.5 g′(0) = (1.6799 – 1.4509)/0.5 = 0.458
Estimate with smaller h (0.458) likely to be more accurate:
smaller h is more accurate (provided there is no great loss of significant
figures)
B1
B1
B1
E1
[4]
(ii)
r = 3 required
B1
Reference to justification/accuracy : 1.777 or 1.78
3
M1 A1
[2]
h = 0.5 g′(0.5) = (2.0100 – 1.4509)/1 = 0.5591
This estimate, central diff, likely to be more accurate than either of the
forward diffs
M1
E1
[2]
1
Lose 1 for any additional 'answer'(s)
but do not penalise extrapolation
4776
4
Mark Scheme
Question
(i)
(ii)
Answer
Max poss loss: 365 (or 366) times 0.01 pence: = 3.65 (or 3.66) pence
Arises if each daily amount would round up but gets chopped down
Average loss 1.825 (or 1.83) pence, because average is half of max.
Marks
B1
E1
B1 E1
[4]
£150 000 divided by 1.825 pence: about 8.2 million (8 million) accounts
M1 A1
[2]
5
6
x
–1
1
3
5
7
9
11
(i)
P(x)
–11
–10
3
44
129
274
495
f
ΔP(x)
Δ2P(x)
(i) bold:
1
13
41
85
145
221
g
12
28
44
60
76
16
16
16
16
Diff table
3rd diffs constant
so cubic
(ii) italic:
working forwards
working backwards
A1
M1 A1
E1
B1
M1 A1
M1 A1
[4] + [4]
h
A1
A1
A1
A1
Errors in g and h are of opposite sign; g is about 4 times as accurate as h.
x
f
(4g + h)/5
abs err
rel err
0.2
0.013351
0.013351
–2.5E–08
–1.9E–06
0.1
0.003334
0.003334
–4E–10
–1.2E–07
A1
(iii)
Guidance
Δ3P(x)
abs err g
rel err g
abs err h
rel err h
–
0.2 0.013351 0.013333 0.013423 0.0000179 –0.0013424 0.0000716 0.0053600
–
0.1 0.003334 0.003333 0.003339 0.0000011 –0.0003339 0.0000045 0.0013350
x
A1
(ii)
January 2011
A1
A1
A1
abs M1
rel M1
[9]
E1 E1
M1
[6]
x / sin x ≈ 1.000 000 002 ≈ 1
g(10-4) = 3.33 × 10-9
Subtraction of nearly equal quantities
B1
B1
E1
[3]
2
f, g, h values may be implied
4776
7
Mark Scheme
Question
(i)
Answer
January 2011
Marks
B1
M1 A1
B1
E1
f(0) = –1 f(1) = 1 (hence root)
f '(x) = 7x6 + 5x4 which is zero only at x = 0.
Convincing argument that this is not a turning point
No turning points implies no other roots.
G2
6
5
4
3
2
1
0
0
0.5
1
1.5
-1
-2
[7]
(ii)
NR iteration: xr+1 = xr – (xr7 + xr5 – 1) / (7xr6 + 5xr4)
r
0
1
2
xr
0.6
1.51756
1.289164
On graph: tangent at 0.6, intersection at 1.5, ordinate & tangent, intersection
at 1.3
B1
A1 A1
G4
[7]
(iii)
r
0
1
2
xr
0.3
22.1703
19.00128
Comment: e.g. converging but initially very slow (or difficult to tell with only
2 iter'ns)
r
0
1
2
xr
0.9
0.890174
0.889891
Comment: e.g. almost converged, root very close to 0.89
3
A1
E1
A1
E1
[4]
Guidance
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4776: Numerical Methods
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
4776
Mark Scheme
1(i) Best estimate: 1.1
June 2011
Maximum possible error: 0.7
[B1B1]
Bisecting mpe: 0.7 → 0.35 → 0.175 → 0.0875 → 0.04375 so 4 iterations
(ii) False position: estimate is (0.4 × 0.5 – 1.8 × (–0.2) / (0.5 – (–0.2))
= 0.8
[M1A1]
BoD for 0.8 alone
[M1A1]
[A1]
[TOTAL 7]
2(i) (3.02 – 2.66) / (2.20 – 1.80) =
0.9
[M1A1]
(ii) max:
min:
(3.025 – 2.655) / (2.20 – 1.80) =
(3.015 – 2.665) / (2.20 – 1.80) =
0.925
0.875
[M1A1]
[A1]
(iii) max:
min:
(3.025 – 2.655) / (2.195 – 1.805) =
(3.015 – 2.665) / (2.205 – 1.795) =
0.94872
0.85366
3(i) Linear interpolation: Q(0) = –8
(ii) Lagrange:
x
0.7
0.8
<
>
4
3
(ii) Derivative of 1 – x is –4x
x abs(–4x3)
0.7
1.372 > 1
0.8
2.048 > 1
5(i)
(ii)
x
3.162278
x4
[B1]
Lagrange form
three terms
terms
[M1]
[DM1]
[A1,1,1]
cao
1 – x4
0.7599
0.5904
X abs err
3.162 –0.00028
[M1A1]
[A1]
[TOTAL 8]
write down or any method
Q(x) = (–4)(x – 1)(x – 5) / (–1 – 1)(–1 – 5) +
(–12)(x + 1)(x – 5) / (1 + 1)(1 – 5) +
20(x + 1)(x – 1) / (5 + 1)(5 – 1)
(iii) Hence by substitution Q(0) = –10
4(i)
[M1] either denominator
correct. Max [2] if 3 dp
(hence root)
[M1A1]
[TOTAL 8]
[M1A1]
[M1]
(so for all [0.7, 0.8] abs gradient or abs RHS > 1)
rel err
–0.0000878
X4
(abs err
rel err
100 99.96488
–0.03512)
–0.0003512
[M1] for grad anywhere
in interval
[M1A1E1]
[TOTAL 6]
do not insist on sign
but do require consistency
between parts (i) and (ii)
[M1A1A1]
allow multiplying answer in (i) by 4
[M1A1A1]
(iii) Relative error will be approximately k × (–)0.0000878
[B1]
[TOTAL 7]
5
4776
6(i)
(ii)
Mark Scheme
x
f(x)
2
3
2.4 3.850195
2.8 4.790825
M1 =
T1 =
T2 =
S1 =
x
f(x)
2.2 3.412917
2.6 4.309988
3.0801558
3.1163298
3.0982428 (=(M1+T1)/2)
3.0922138 (=(2M1+T1)/3)
June 2011
Lose A1, 2, 3, 4
if 6, 5, 4, 3 dp
M2 = 3.0891620
S2 = 3.0921890 (=(2M2+T2)/3)
or ... 889
M4 = 3.0914298
T4 = 3.0937024
S4 = 3.0921873
(iii)
ratio of
diffs
approx 1/16 so fourth order
0.0648518
(FT their precision)
Consider rate of convergence of S: conclude I = 3.09219
(or 3.092187 if using extrapolation)
[B2] for 3.09219 alone
x
1
3
5
f(x)
4
–2
10
Δf(x)
(iii)
f(2) =
f(6) =
x
1
3
5
7
–6
12
18
Δf(x)
–6
12
1
[E1A1]
[subtotal 7]
[TOTAL 18]
[M1A1]
[M1A1A1A1]
[subtotal 6]
–1.25 likely to be more accurate (interpolation)
22.75 than this (extrapolation)
f(x)
4
–2
10
11
[M1A1E1]
Δ2f(x)
f(x) = 4 + (–6)(x – 1) / 2 + 18(x – 1)(x – 3) / (22 × 2!)
(ii)
[M1A1]
[B1]
[subtotal 3]
[B1]
[B1]
S1 = 3.0922138
diffs
S2 = 3.0921890 –2.5E–05
S4 = 3.0921873 –1.6E–06
7(i)
[M1A1]
[M1A1]
[M1A1]
[M1A1]
[subtotal 8]
Δ2f(x)
Δ3f(x)
18
–11
–29
[A1A1E1]
[subtotal 3]
← extend table
new f(x) = old f(x) + (–29)(x – 1)(x – 3)(x – 5) / (23 × 3!)
FT incorrect old f(x) here
f(2) = –1.25 + (–1.8125) =
f(6) = 22.75 + (–9.0625) =
[M1] for substituting
cao for each [A1]
–3.0625
13.6875
(iv) Absolute change greater in f(6), relative change greater in f(2)
Must have all 4 values correct
[E1], [E1] for intelligent comments on absolute, relative changes
6
[M1A1]
[M1A1]
[M1A1]
[A1]
[subtotal 7]
[E1E1]
[subtotal 2]
[TOTAL 18]
4776
Mark Scheme
Question
1
Answer
y = 3 x (x  2) / (1) (3) + 6 (x + 1) (x  2) / (1) (2)  4 (x + 1) x / (3) (2)
y = (x2  2x)  3 (x2  x  2)  ⅔ (x2 + x)
y = ⅓ (8x2 + x + 18)
y ' = ⅓ (16x + 1) = 0 when x = 1/16
2
(i)
rearrange convincingly to r = (X  x) / x
2
(ii)
(1 + r)n = 1 + nr + n(n1)r2 / 2 + …
result given
2
(iii)
(A) n=3
(B) n=1
3
3
(i)
(ii)
6%
2%
LHS
x
1
1
2
32
Or equivalent
f(x) = x5  x4 – 2
r
xr
0
1.5
<
>
RHS
3
18
f '(x) = 5x4  4x3
1
1.455026
hence N-R formula
2
1.451113
3
1.4510851
4
1.4510851
hence root is 1.451085 to 6 decimal places
4
(i)
&
(ii)
h
g'(5)
0.4
0.2
0.1
0.37203
0.39065
0.40040
June 2012
Marks
B1B1B1
A1
A1
M1A1
[7]
M1A1
[2]
M1
E1
[2]
B1
B2
[3]
Guidance
M1 for diffn, A1 cao
No further explanation required
Need to see correct r2 term
For saying r2 negligible
B1 for 2%
M1A1
No explanation required
[2]
M1A1
May be implied by subsequent work
M1A1A1
M1 1st application, A1 2nd, A1 for 2
successive values agreeing to 6 dp
A1
[6]
0.01863
0.00975 0.52349 (or 'the differences approx. halve')
approx 0.5 (as h is halved) so first order
5
M1A1A1
[3]
M1A1
E1
Estimates
M1 diffs, A1 ratio (may be implied)
4776
Mark Scheme
Question
4
5
(iii)
(i)
5
(ii)
6
(i)
Marks
[3]
Answer
Some way off convergence, so accept 0.4
(or an argument by extrapolation to 0.41)
It means 5.5 × 10-17
It is not zero because of rounding errors
in the representation of numbers like 0.6, 0.4, 0.2 (or in the calculations)
Cell B4 displays 1 either because the spreadsheet does not show enough dp
or because adding 1 pushes the error beyond the sf the spreadsheet stores.
The zero in C4 shows that the error must have been lost in B4
T1
T2
T4
T8
2.357070
2.394855
2.404253
2.406599
diffs
0.037785
0.009397
0.002346
ratios
0.248710
0.249667
both approximately 0.25
indicates 2nd order method
6
(ii)
(iii)
E1A1
[2]
E1
E2,1,0
[3]
E1
E1
E1
[3]
M1A1
A1
A1
B1
E1
E1
[7]
M1
A1
A1
B1
Using ⅓ (4*T2n  Tn) to obtain Simpson's rule estimates
S1
2.407450
diffs
ratios
S2
2.407385
6.5E05
S4
2.407381
4.2E06
0.064103
approximately 1/16
indicates 4th order method
6
June 2012
E1
E1
[6]
Convergence of the Simpson's rule estimates suggests 2.40738
(or even 2.407381 by extrapolation)
If data are rounded to 5 dp, there is an error in the range ± 0.000 005 in
each value Since the integral is over a range of 2 units the correct value
will lie within ± 0.000 01 of the value given previously.
6
A1E1
B1
A1E1
[5]
Guidance
A1for 0.4 or for 0.41 with
extrapolation, E1 for a reason
(Or 0.6 was not exact)
Any T value
2 further T values
All T values
Ratios
explanation
(Or via M values)
Any S value
All S values
Ratio
Explanation
Either answer
Sight of f(x) + or  0.000 005
4776
Mark Scheme
Question
7
(i)
Answer
x
1
0
1
2
7
(ii)
a
1
1
1
LHS
0
1
0
3
b
0
0.54304
0.62662
>
<
<
>
RHS
-0.84147
0
0.841471
0.909297
f(a)
0.841471
0.841471
0.841471
June 2012
Marks
or equivalent
no explicit explanation reqd.
f(b)
1
0.18836
0.02093
x
0.54304
0.62662
0.63568
f(x)
0.18836
0.02093
The sequence of values of x has not converged: 0.6 is the only safe
estimate.
Guidance
M1A1A1
[3]
M1A1
M1A1
A1
M1 is for correct end-points
E1A1
[7]
7
(iii)
x
f(x)
1
0.84147
2
2.090703
1.286979
0.30368
1.377411
0.0841
1.412046
0.006448
The sequence of values of x has not converged, but 1.4 appears safe
7
(iv)
The rules converge at about the same rate
The rules converge slowly
Secant rule involves less calculation at each step (but may be harder to
implement on a calculator)
M1A1A1A1
E1A1
[6]
Reward other appropriate comments
E2,1
[2]
7
4776
Mark Scheme
Question
1 (i)
Answer
x
0
1
2
1
(ii)
r
xr
1.4934 to 4dp
2
3
0
1.5
[2]
M1
A1
( x  4 x4  1)
(4 xr3  4)
4
r
1
1.493421
2
1.493359
(ii)
20
21
exact
1
8
approx
0.723607
8.024922
6765.0000296
10945.9999817
error
−0.27639
0.024922
hence
hence
Guidance
Values
Reference to 2 changes of
sign
B1 for derivative only
M1
A1
3
1.493359
A1
[5]
B1
M1
A1
A1
(i)
r
1
6
2
Marks
B1
E1
LHS
1
−2
9
N-R formula xr 1  xr 
January 2013
rel error
−0.27639
0.003115
[4]
M1
A1
A1
dep
[3]
M1
A1
A1
6765
10946
(i)
h
f '(x)
0.2
(1.569−1.464)/0.2 0.525
(1.516−1.464)/0.1 0.52
0.1
Number of sf reduces as h decreases (subtracting nearly equal
quantities)
E1
[4]
5
Approximations
Errors
Errors: condone positive
Relative errors: must be
opposite sign
Approximate values
integers
4776
Question
3 (ii)
4 (i)
(ii)
4 (iii)
Mark Scheme
Answer
Change in f '(x) is −0.005. If this is half the error in the first estimate
then a better estimate is 0.52 − 0.005 = 0.515
h
1
0.5
T
1.332375
1.354935
M
1.377495
1.366179
S
1.362455
1.362431
I = 1.3624 with comment that second S more accurate than first
January 2013
Marks
M1
A1
E1
[3]
M1
A1
[2]
M1
A1
A1
[3]
B1
E1
OR
I = 1.36243 with comment on the next difference in S
[2]
5 (i)
g(x)
x
1
−15
2
−14
3
k
4
54
5
145
27 – 3k = 3k − 3
k=5
∆
1
k + 14
54 − k
91
∆2
∆3
M1
A1
k + 13
40 − 2k
37 + k
27 – 3k
3k − 3
M1
A1
[4]
5 (ii)
x
0
g(x)
−10
∆
∆2
∆3
−5
M1
6
12
g(0) = −10
A1
[2]
6
Guidance
Or sum a gp
For T
For S
4776
Question
5 (iii)
Mark Scheme
Answer
Marks
2  5  3   14 
= 2.736842
f( x ) 
Guidance
M1
5   14 
6 (i)
January 2013
7.5( x  2)( x  4)
( 3)( 5)

9.0( x  1)( x  4)
(3)( 2)

2.2( x  1)( x  2)
(5)(2)
 0.78 x 2  1.28 x  9.56
f(0)  9.56
Solve f( x)  0
Positive root is 4.4(16295)
Interpolation (f(0)) generally more reliable than extrapolation (f(x) = 0)
6 (ii)
TL  TR  0.5  3(7.5  9.0)  0.5  2(9.0  2.2)
= 35.95
6 (iii)
Simpson's rule requires equal spacing in x
f(1.5)  9.725
S  13  2.5(7.5  4  9.725  2.2)
= 40.5
A1
[2]
M1
A1
A1
A1
M1
A1
A1
A1
M1
A1
E1
[11]
M1
A1
[2]
E1
M1
A1
M1
Convincing algebra
First coefficient
Second coefficient
CAO
1.5 used
A1
[5]
7 (i)
G2
G1
E1
3
2.5
2
1.5
1
0.5
0
0
1
2
3
4
5
6
7
[4]
7
One graph
Second graph
Comment on roots
4776
Question
7 (ii)
Mark Scheme
Answer
January 2013
Marks
Guidance
Eg
x0
x1
x2
x3
x4
x5
0.5 0.675938 0.615146 0.634084 0.627967 0.629921
0.629921 0.629295 0.629495 0.629431 0.629451 0.629445
  0.629 to 3dp
M1
A1
A1
[3]
7 (iii)
x
3.9
4.1
LHS
0.25641
0.24390
<
>
RHS
0.312234
0.181723
B1
Eg
x0
x1
x2
4 4.111884 5.715937
May offer more iterations, or just x1 with explanation.
a
3.9
3.9
3.95
b
4.1
4
4
m
4
3.95
3.975
f(m)
0.006802
−0.02365
M1
A1
B1
M1
A1
A1
mpe
0.1
0.05
0.025
7 (iv)
5.2
x
f(x) 0.075762
  5.33 to 3sf
5.4
−0.04205
5.328615
0.003732
5.334433
0.00015
5.334677
−5.9E−07
8
[7]
M1
A1
A1
A1
[4]
Mpe s
One iteration
m
4776
Mark Scheme
Question
1 (i)
1
(ii)
Answer
Convincing sketches of x and cos x.
Single intersection.
Estimate of root in [0.5, 1]
Iteration xr +1 = (cos xr )0.5
0
1
0.8
0.83469
xr
0.82 correct to 2 dp
2
Marks
G2
G1 for each graph
2
r
(i)
n
5
10
2
0.819395
3
0.826235
4
0.823195
5
0.82455
exact
approx
error rel error
252 258.3688 6.36877 0.025273
184756
187079 2322.973 0.012573
errors increase but relative errors decrease with n
2
(ii)
=
10k
1
= 79.5548
0.01257
k = 8 to nearest integer
OR
5k
=
(i)
x
0.1
0.2
0.3
0.4
f(x)
1.641
1.990
1.840
1.192
∆
Accept π/4. Accept an interval in [0.5, 1]
M1
A1
Max 1 for a diverging iteration
A1 requires agreement to 2 dp
Dependent on previous A1
A1
[5]
M1
For any valid rearrangement
For writing it as an iteration (soi)
Requires method for abs and rel error
B1
B1
B1
B1
[5]
M1
Approximations
Errors
Relative errors
A1
Must be an integer
0.349
-0.150
-0.648
A1
[2]
M1
A1
E1
∆2
-0.499 these almost equal
-0.498 (so approx qdratic)
[3]
7
Guidance
B1
[3]
M1
A1
M1
1
= 39.5726
0.0257
k = 8 to nearest integer
3
June 2013
Must be an integer
4776
Question
3 (ii)
Mark Scheme
Answer
0.349(0.15 − 0.1) 0.499(0.15 − 0.1)(0.15 − 0.2)
f (1.5) =
1.641 +
−
0.1
2(0.1) 2
Marks
M1
A1
A1
A1
[4]
G2/E2
= 1.878 to 3dp
4
(i)
Sketch or convincing argument to an increasing function
hence a single root
x
function
(Hence root)
4
(ii)
5
(i)
5
(ii)
a
0.7
0.759329
0.76 to 2dp
h
0.2
0.1
0.05
M1
A1
0.7 3.782174 < 4
0.8 4.149326 > 4
(i)
Guidance
For recognizable attempt at correct formula
Either second or third term correct
All three terms correct. Accept cubic.
Accept any awrt 1.878
2 out 3 for
formula with
x and no 0.15
If comparing with zero: -0.2178, 0.1493
Max 1 if function sign but not values
[4]
f(a)
-0.21783
-0.00431
g(-h)
1.1292
1.1766
1.1974
b
0.8
0.8
f(b)
0.149326
0.149326
x
0.759329
0.760469
f(x)
-0.00431
M1
A1
M1
A1
[4]
M1
A1
A1
A1
g(h)
g '(0)
1.2745 0.36325
1.2489 0.3615
1.2335
0.361
0.36 because last figure still changing and so unreliable
Or 0.361 if some argument about convergence or extrapolation is used
6
June 2013
f(x)
x
T
M
S
0
1
0.5 1.243504
0.25 1.12042 0.560876 0.560210 0.560432
[4]
A1
E1
[2]
Full marks for h = 0.15, 0.1, 0.05.
Max 3 if other values of h used
Any sensible comment or attempt to analyse
errors
M1
M Award these marks for a correct answer
M1
S
or a correct method with wrong answer
M1
T
Do not penalise no. of sf
[3]
8
Allow a maximum of 3 out 4 for a solution
which goes wrong but self corrects
For correct interval and calculating x
Must follow from false position
h = 0.15 gives
0.361667
4776
Question
6 (ii)
6
7
7
(iii)
(i)
(ii)
Mark Scheme
Answer
June 2013
Marks
Guidance
A6
Values
Lose 1 for each error
Lose 1 overall if no. of sf is not 6
FT sensible but incorrect M and/or T to S
f(x)
x
T
M
S
0.125 1.060969
0.375 1.18052 0.560543 0.560372 0.560429
0.0625 1.030816
0.1875 1.090747
0.3125 1.150255
0.4375 1.211499 0.560458 0.560415 0.560429
0.560429 is justified
(for information only: 0.5604289 is justified if more sf used)
A1
diffs
ratio
diffs
ratio
T
M
0.560876
0.560210
0.560543 -0.000333
0.560372 0.000162
0.560458 -0.000085 0.256788 0.560415 0.000043 0.262091
Ratios about 0.25 in each case;
indicates both have 2nd order convergence
But M is more accurate than T;
smaller differences so nearer the correct answer
In the first 100 terms the positive rounding errors exceed the negative
rounding errors
The opposite occurs in the first 200 terms.
Chopping will reduce the sum
by an average of 0.00005 per term ie by 0.005 and 0.01 in S 100 and S 200
Hence estimate as 18.5846 (18.585) and 26.8493 (26.85)
9
[7]
M1
A1
M1
A1
E1
E1
E1
E1
[8]
E1
E1
[2]
E1
M1A1
A1
[4]
T Allow small errors that still give ratios …
M … of approximately 0.25
Allow correct explanations using 0.25 if the
ratios come out wrong
Allow correct statements about M and T
even if not supported by the numbers
Allow E1 for an incomplete explanation that
shows some understanding
M1 for 0.00005, A1 rest
4776
Question
7 (iii)
Mark Scheme
Answer
∫
k + 0.5
k − 0.5
Marks
k + 0.5
1
dx =  2 x 
k − 0.5
x
(iv)
S 100
S 200
7
(v)
Guidance
M1
= RHS
Midpoint rule
Gives LHS
7
June 2013
approx
exact
error
18.63572 18.5896 0.046124
26.90539 26.8593 0.046091
A1
M1
A1
[4]
B1
B1
Answer given
Must be convincing
Answer given
M1
A1
Errors
Approximations
Errors almost exactly equal
E1
[5]
assumed
approx
error estimate
S 1000 61.84715
0.046 61.80115
(61.801 or 61.80)
(For information, correct sum is 61.80101 to 5dp)
B1
Approx
M1
A1
Correction using 0.046 (or similar)
Penalize more dp
[3]
10