4777
Mark Scheme
June 2006
MEI Numerical Computation (4777) June 2006
1
(i)
(ii)
(x2 - α) / (x1 - α) = (x1 - α) / (x0 - α)
convincing algebra to required result.
x
exp(x) - tan(x)
1
1.160874
Examples of divergence:
r
0
1
(iii)
to eliminate k
[M1A1A1]
[A1A1]
[subtotal 5]
1.5
-9.61973
change of sign (and no asymptote)
2
3
4
5
6
#NUM!
#NUM!
#NUM!
#NUM!
-0.21274
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
xr
1
0.443023
xr
1.25
1.101797
-0.74554
0.67982
1
xr
1.5
2.646275
#NUM!
xr
1.25
1.101797
xr
1.330227
1.405193
xr
1.312029
1.329149
xr
1.306628
1.307521
xr
1.306328
1.30633
xr
1.306327
x
exp(-x) - tan(x)
Mark scheme
0.67982
1
1.78895
9
[M1A1]
[M1A1A1]
α:
1.330227
[M1A1]
α:
1.312029
[M1A1]
1.40054
1.31106
9
α:
1.306628
α:
1.306328
α:
1.306327
1.306327
1.30634
1.30632
7
α:
1.306328
1.30633 to 5 dp
3.142
0.042789
3.2
-0.01771
Examples of divergence:
r
0
1
2
xr
3.142
7.805847
xr
3.2
xr
[M1]
[A1]
[subtotal 11]
change of sign
[M1A1]
4
5
6
-3.03297
3
2.21594
3
#NUM!
#NUM!
#NUM!
2.839176
#NUM!
#NUM!
#NUM!
#NUM!
#NUM!
3.18
3.259015
2.13737
α:
xr
3.1852
3.131898
#NUM!
α:
#NUM!
xr
#NUM!
#NUM!
#NUM!
α:
#NUM!
xr
#NUM!
#NUM!
#NUM!
α:
#NUM!
xr
3.184
3.159834
α:
3.183327
xr
3.183327
3.17584
α:
3.183054
xr
3.183054
3.182409
α:
3.183029
xr
3.183029
3.183024
α:
3.183029
3.18303 to 5 dp
[M1A1]
Eg:
3.1852
[M1A1]
but:
4.00395
6
3.37375
3
3.19812
2
3.18313
7
[M1A1]
[subtotal 8]
4777
Mark Scheme
June 2006
[TOTAL 24]
2
(i)
(ii)
Divided differences do not require data to be equally spaced (as ordinary differences
do). Divided differences allow additional data to be added (unlike Lagrange).
x
1
2
2.5
3.5
4
4.5
f
-3
-6.5
-8.03
-6.66
-2.25
5.65
1
-3
-3.5
2
-6.5
-3.06
2.5
3.5
4
-8.03
-6.66
-2.25
1.37
8.82
1.5
-3
-4.75
linear
0.29333
3
2.95333
3
4.96666
7
-4.82333
quadratic
1.064
1.00666
7
-4.55733
cubic
-0.01911
3.5
-6.66
1.37
2.5
4
2
4.5
3
-8.03
-2.25
-6.5
5.65
-6.66
3.853333
2.125
4.86
-7.345
linear
4.96666
7
3.45666
7
5.47
1.00666
7
1.00666
7
-8.58667
quadratic
-8.335
cubic
-4.54778
quartic
5.65
15.8
4
-2.25
8.82
3.5
2.5
2
5
-6.66
-8.03
-6.5
5.65
1.37
-3.06
13.55
linear
6.98
4.96666
7
2.95333
3
17.04
quadratic
1.00666
7
1.00666
7
17.795
cubic
estimates
[M1A1A1A1A1
]
[E1E1]
[subtotal 10]
-4.4E-16
-8.335
quartic
2nd dp unreliable (from data), 1st dp seems reliable: -8.3
4.5
table
[M1A1A1]
2nd dp unreliable (from data), 1st dp uncertain: could be -4.5 or -4.6
(iii)
[E1E1]
[subtotal 2]
rearrange
data and
re-run
[M1A1]
[E1]
-6.2E-16
17.795
quartic
rearrange
data and
re-run
[M1A1]
4777
Mark Scheme
June 2006
2nd dp unreliable (from data), 1st dp seems reliable: 17.8
(iv)
4
-2.25
15.8
6.98
4.5
5.65
12.31
3.5
2.5
2
4.16
-6.66
-8.03
-6.5
-2.25
1.37
-3.06
5.47
2.95333
3
0.278
4.17
-2.25
0.436
4.165
5.65
1.52615
4.175
-2.25
0.515
linear
Hence root is 4.17 to 2 dp
-0.10171
0.04442
2
0.30757
1
0.11801
2
quadratic
1.00666
7
1.00666
7
-0.13786
0.00658
4
-0.06582
0.07936
6
cubic
[E1]
[subtotal 6]
-4.4E-16
-0.13786
rearrange
data and
re-run
[M1A1]
0.006584
[M1A1]
-0.06582
0.079366
quartic
[A1]
[A1]
[subtotal 6]
[TOTAL 24]
4777
3
(i)
Mark Scheme
Substitute central difference formulae for y' and y" to obtain given result (*)
[M1A1]
Central difference formula for y' at x=0 to show y1 = y-1
[M1A1]
2
Use of (*) to show y1 = (2h - (1 + 2h)y-1)/(1 - 2h)
[M1A1]
2
Hence y1 = h as given
h
0.1
h
0.1
0.05
[M1]
x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
y
0
0.01
0.047618
0.124458
0.25785
0.473034
0.805379
1.301401
2.015508
2.996344
4.253311
β
4.253311
4.190790
diffs
k
1
[M1A1]
as required
ratio of
diffs
[A1]
extrapolated
value
-0.06252
re-runs
0.24040
0.025 4.175759
-0.01503
6
0.012
0.24760 4.17079
5 4.172037
-0.00372
4 7
ratio of differences approximately 0.25 so second order
4.17 to 2 dp is secure
(ii)
June 2006
k
-5
-4
-3
β
18.4
13.1
9.7
[A1A1A1]
[M1A1E1]
[A1]
[subtotal 17]
mods
[M1A1]
131
4777
Mark Scheme
-2
-1
0
1
2
3
4
5
June 2006
7.4
6
4.9
4.2
3.6
3.2
2.9
2.6
values
[A1A1A1]
graph
[G2]
[subtotal 7]
[TOTAL 24]
4
(i)
Diagonal dominance: modulus of diagonal element is greater than or equal
to sum of moduli of other elements on the same row.
If diagonal dominance exists (with at least one inequality strict) convergence
of Gauss-Seidel is assured.
G-S using the given non-dominant diagonal:
x
0
0.2
0.192381
0.186262
…
0.180325
0.180327
0.180328
0.180328
(ii)
a=3
x
0
0.333333
0.447619
0.54449
…
1
1
1
y
0
-0.06667
-0.12
-0.16267
…
-0.33333
-0.33333
-0.33333
z
0
-0.04762
-0.09116
-0.12987
…
-0.33333
-0.33333
-0.33333
a=4
x
0
0.5
0.9375
1.583333
2.543403
3.976273
6.119551
9.329443
14.1401
21.35259
…
2.6E+18
3.91E+1
8
[E1]
[E1]
y
0
0.02857
1
0.04199
5
0.04733
5
…
0.04918
0.04918
0.04918
0.04918
z
0
y
0
-0.25
-0.64583
-1.25694
-2.18808
-3.59684
-5.72002
-8.91317
-13.7099
-20.9107
…
-2.6E+18
z
0
-0.04167
-0.07639
-0.10532
-0.12944
-0.14953
-0.16628
-0.18023
-0.19186
-0.20155
…
0
-3.9E+18
0
-0.01587
[M1]
[M1A1]
-0.0191
-0.01866
…
-0.01639
-0.01639
-0.01639
-0.01639
[M1A1]
[subtotal 7]
mods
[M1A1]
a=3
[M1A1]
a=4
[M1A1]
4777
Mark Scheme
June 2006
5.86E+1
8
G-S scheme converges for a=3.3
diverges for a=3.4
(diverges for
a=3.35)
So a=3.3 (to 1dp) is required value
(iii)
Gauss-Jacobi
a=0
x
0
0.166667
0.054167
0.19375
0.067708
0.200521
…
0.202778
0.072222
0.202778
0.072222
0.202778
0.072222
-5.9E+18
0
[M1A1]
[M1A1]
[A1]
[subtotal 11]
y
0
0.125
-0.00833
0.12083
3
-0.01042
0.11979
2
…
0.11944
4
-0.01111
0.11944
4
-0.01111
0.11944
4
-0.01111
Diverges: diagonal dominance not strict.
z
0
0.1
-0.04583
0.07708
3
-0.05729
0.07135
4
…
0.06944
4
-0.06111
0.06944
4
-0.06111
0.06944
4
-0.06111
[M1A1]
[M1A1]
[A1]
[E1]
[subtotal 6]
[TOTAL 24]
4777
1(i)
Mark Scheme
June 2007
Convincing algebra to k = (x2 - x1)/(x1 - x0)
[M1A1]
Convincing algebra to α = (x2 - k x1)/(1 - k) or equivalent
(ii)
(iii)
x
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
y=x
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
5.5
6
y=f(x)
1.5
1.527842
1.612144
1.755252
1.961151
2.235574
2.586161
3.022674
3.557265
4.204819
4.983366
5.914581
7.024391
converges
slowly
to
root
near
2
2
1.961151
1.942783
1.934241
1.9303
1.928489
1.927657
1.927276
diverges
from
root
near
5
x0
2
1.92631
1.926953
x1
1.961151
1.926659
1.926953
x0
5
5.023872
5.023461
x0
4.6
5.216066
5.047555
5.02388
5.023461
[M1A1A1]
[subtotal 5]
[G2]
4.5
4.204819
3.807921
3.339412
2.872419
2.488967
2.228729
2.07777
5
4.983366
4.95514
4.90763
4.828739
4.70068
4.500432
4.205432
x2
1.942783
1.926818
1.926953
k
0.472807
0.458143
0.45827
new x0
1.92631
1.926953
1.926953
x1
4.983366
5.024167
5.023461
x2
4.95514
5.024673
5.023461
k
1.696813
1.71656
1.716217
new x0
5.023872
5.023461
5.023461
x1
4.349412
5.365628
5.064991
5.024181
5.023461
x2
3.996895
5.647933
5.095267
5.024697
5.023461
k
1.406756
1.887551
1.73646
1.716567
1.716217
new x0
5.216066
5.047555
5.02388
5.023461
5.023461
5.5
5.914581
6.820878
9.3175
21.8726
1466.344
1.9E+212
#NUM!
set up
iteration
[M1A1]
near 2
[A1]
near 5
[A1A1]
(theoretical arguments
involving f 'acceptable)
[subtotal 7]
=alpha
= beta
range
4.6 to 5.7
k
est of root
use as x0
iterate
[M1A1]
[M1A1]
[M1]
[M1A1]
alpha
[A1]
beta
[A1]
[M1A1A1]
[subtotal 12]
[TOTAL 24]
119
4777
2 (i)
Mark Scheme
June 2007
Substitute f(x) = 1, x2, x4, x6 into the integration fomula
Obtain
a+b=h
aα2 + bβ2 = h3/3
aα4 + bβ4 = h5/5
(aα6 + bβ6 = h7/7)
[M1M1M1M1]
[A1]
[A1]
[A1]
[subtotal 7]
(ii)
(ii)
E.g.
x
0
0.5
1
1.5
2
2.5
3
3.5
x
sin(x) / x
0.1
0.998334
0.01
0.999983
0.001
1
[B1]
sin(x) / x
1
0.958851
0.841471
0.664997
0.454649
0.239389
0.04704
-0.10022
[G2]
Single
application
of Gaussian 4-pt
rule
Subdividing the
interval
m=
α, β
-0.86114
-0.33998
0.339981
0.861136
m=
m=
(iii)
1.570796
x
0.218127
1.036755
2.104837
2.923466
h=
f(x)
0.992089
0.830241
0.408942
0.074022
0.785398
h=
2.356194
h=
By trial and error
m= 0.53242
α, β
x
-0.86114 0.073934
-0.33998 0.351407
0.339981 0.713433
0.861136 0.990906
h=
f(x)
0.999089
0.979546
0.917302
0.8442
Hence t = 2m =
1.570796
a, b
0.347855
0.652145
0.652145
0.347855
sum:
integral:
0.785398
gives
0.785398
gives
sum
0.345103
0.541438
0.26669
0.025749
1.17898
1.851937
set up
[M4]
1.370762
[M1A1]
0.481175
1.851937
[M1A1]
( = 6dp)
[A1]
[subtotal 13]
[A1]
0.53242
a, b
0.347855
0.652145
0.652145
0.347855
1.065
0.347538
0.638806
0.598214
0.293659
1
(1.06484)
trial
and
error
[M1A1]
[M1A1]
[subtotal 4]
[TOTAL 24]
120
4777
Mark Scheme
3 (i) Euler
(ii)
Modified
Euler
h
0.2
0.2
0.2
0.2
0.2
0.2
x
0
0.2
0.4
0.6
0.8
1
y
0
0.02
0.080404
0.182079
0.326073
0.513783
y'
0.1
0.30202
0.508372
0.719971
0.938552
h
0.2
0.1
0.05
0.025
y(1)
0.513783
0.569802
0.598337
0.612748
diffs
ratio of
diffs
0.056019
0.028535
0.014411
h
0.2
0.2
0.2
0.2
0.2
0.2
x
0
0.2
0.4
0.6
0.8
1
y
0
0.040202
0.121675
0.245483
0.413051
0.626446
k1
0.02
0.06082
0.102588
0.145565
0.190228
h
0.2
0.1
0.05
0.025
y(1)
0.626446
0.627065
0.627213
0.627249
diffs
ratio of
diffs
0.000619
0.000147
3.58E-05
0.509387
0.505038
0.238113
0.242993
June 2007
new y
0.02
0.080404
0.182079
0.326073
0.513783
setup
[M2]
estimates
[A1A1]
differences
[M1A1]
approx 0.5, so first order
k2
0.060404
0.102126
0.145028
0.189571
0.236562
[E1]
[subtotal 7]
new y
0.040202
0.121675
0.245483
0.413051
0.626446
setup
[M2]
estimates
[A1A1]
differences
[A1]
approx 0.25, so second order
[E1]
[subtotal 6]
(iii)
(iv)
predictor-corrector
h
x
0.2
0
0.2
0.2
0.2
0.4
0.2
0.6
0.2
0.8
0.2
1
y
0
0.040412
0.122124
0.246214
0.414137
0.628006
h
0.2
0.1
0.05
0.025
y(1)
0.628006
0.627447
0.627307
0.627272
-0.00056
-0.00014
-3.5E-05
0.250462
0.250113
mod Euler
0.626446
0.627065
0.627213
0.627249
pre-corr
0.628006
0.627447
0.627307
0.627272
average
0.627226
0.627256
0.62726
0.627261
y'
0.1
0.304124
0.512989
0.727917
0.951306
pred
0.02
0.101237
0.224722
0.391798
0.604398
corr1
0.040202
0.12189
0.245942
0.413802
0.627569
corr2
0.04041
0.122121
0.246211
0.414132
0.627998
corr3
0.040412
0.122124
0.246214
0.414137
0.628006
setup
[M3]
estimates
[A1A1]
differences
[A1]
Still second order. Differences very
similar in magnitude to modified Euler.
diffs
3.04E-05
3.78E-06
4.21E-07
ratio of
diffs
0.124555
0.111332
approx 0.125
so third order
[E1E1]
[subtotal 8]
values
[A1]
differences
[A1]
[E1]
[subtotal 3]
[TOTAL 24]
121
4777
4 (i)
Mark Scheme
0
3
2
1
1
0
3
2
1
3
2
2
1
0
3
2
-0.66667
2.666667
2.222222
3.111111
3
2
1
0
3
-0.33333
-0.66667
3.111111
-0.44444
3.428571
1
2
3
4
1
1.666667
3.333333
0.444444
2.222222
-1.14286
June 2007
x1 =
0.666667
elimin'n
[M1M1M1]
[A1A1]
x2 =
0.666667
x3 =
x4 =
0.666667
-0.33333
back sub
[M1]
solutions
[A1A1A1A1]
pivot (shaded) is element
of largest magnitude in column
Demonstrate check by substituting values back into equations.
(ii)
Apply to
v=
To get
M-1 =
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
-0.20833
0.04167
0.04167
0.29167
0.29167
-0.20833
0.04167
0.04167
0.04167
0.29167
-0.20833
0.04167
0.04167
0.04167
0.29167
-0.20833
[M1]
[E1]
[B1]
[subtotal 13]
NB: clear
evidence
required
that own
routine
is used
at least one v
[M1]
other three
[M1]
columns
[A1A1A1A1]
[subtotal 6]
(iii)
The product of the pivots is
96
In each of the first three cases, the pivot is in the second row
of the reduced matrix. This is equivalent to three row
interchanges. Hence multiply by (-1)3.
i.e. determinant is -96
[M1A1]
[M1E1]
[A1]
[subtotal 5]
[TOTAL 24]
122
4777
Mark Scheme
June 2008
4777 Numerical Computation
1
(i)
(ii)
Eg: er+1 is approximately ker
[E2]
Uses y0 = α + e0, y1 = α + ke0, y2 = α + k2e0
or equivalent
Convincing algebra to eliminate k hence given result
[M1A1]
[A1A1]
[subtotal 6]
Convincing re-arrangment
[A1]
extrap
(new yo)
new y1
0.917409 0.916644
0.916648
y0
y1
1
0.908662
y2
x
y0
y1
y2
extrap
(new yo)
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
1
0.916647
0.856948
0.811835
0.776292
0.747335
0.723087
0.70231
0.684155
0.668023
0.908662
0.845937
0.799744
0.763904
0.734953
0.7108
0.690112
0.671996
0.655831
0.641175
0.917409
0.858962
0.815042
0.780556
0.752555
0.729213
0.70934
0.692131
0.677026
0.663627
0.916644
0.856936
0.811814
0.776263
0.747298
0.723043
0.702258
0.684095
0.667954
0.653402
(iii)
x
0.653483
0.668023
0.684155
0.70231
0.723087
0.747335
0.776292
0.811835
0.856948
0.916647
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
y
2
1.9
1.8
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0.916647
0.856948
0.811835
0.776292
0.747335
0.723087
0.70231
0.684155
0.668023
0.653483
new y2
extrap
0.916647 0.916647
4 or 5 sf looks secure
new y1
new y2
0.916648 0.916647
0.85695 0.856947
0.81184 0.811833
0.776302 0.776288
0.747351 0.747329
0.72311 0.723076
0.70234 0.702292
0.684194 0.684128
0.668075 0.667985
0.65355 0.653427
3 or 4 sf looks secure
once
[M1A1]
twice
[M1A1]
[A1]
[subtotal 6]
extrap
0.916647
0.856948
0.811835
0.776292
0.747335
0.723087
0.70231
0.684155
0.668023
0.653483
set up
SS
[M2A2]
values
[A3]
[A1]
2
1.8
1.6
1.4
organise
1.2
1
data
0.8
[M1A1]
0.6
graph
0.4
G2
0.2
0
0
0.5
1
1.5
2
Sub Total 12
TOTAL 24
101
4777
2
(i)
Mark Scheme
June 2008
Tn - I = A2h2 + A4h4 + A6h6 + …
T2n - I = A2(h/2)2 + A4(h/2)4 + A6(h/2)6 + …
[M1A1]
4(T2n - I) - (Tn - I) = b4h4 + b6h6 + …
[M1]
4T2n - Tn - 3 I = b4h4 + b6h6 + …
[A1]
4
6
(4T2n - Tn)/3 - I = B4h + B6h + …
(Tn* = (4T2n - Tn)/3
[A1]
has error of order h4 as given)
Tn** = (16T2n* - Tn*)/15 has error of order h6
[B1]
[subtotal 6]
(ii)
x
0
2
1
0.5
1.5
0.25
0.75
1.25
1.75
0.125
0.375
0.625
0.875
1.125
1.375
1.625
1.875
f(x)
0
3.523188
0.731059
0.155615
1.839543
0.035136
0.382038
1.214531
2.609105
0.0083
0.083344
0.254435
0.540367
0.955439
1.509072
2.206199
3.048173
T
T*
3.523188
2.492653
2.149141
2.243905
2.160989
2.182155
2.166744
2.161572
2.161606
T**
(T***)
f:
[A1]
T:
[M1A2]
T*:
[M1A1]
T**:
[M1A1]
answer:
[A1]
2.161779
2.161611
2.161609
2.161608
2.161609
[subtotal 9]
(iii)
k
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
I
0
0.002847
0.024686
0.089495
0.225935
0.466242
0.845007
1.398068
2.161609
2.5
modify SS
[M2]
2
1.5
values of I
[A2]
1
graph
[G2]
0.5
0
[subtotal 6]
0
(iv)
k
1.57
1.58
1.579
I
0.980739
1.001291
0.999223
0.5
accept 1.57
or 1.58
(or in between)
1
1.5
2
evidence of t&e:
result:
[M2]
[A1]
[subtotal 3]
[TOTAL 24]
102
4777
3
(i)
Mark Scheme
June 2008
h
x
y
k1
k2
k3
k4
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
2.8
3
0
0.125024
0.189763
0.221666
0.229182
0.217146
0.188783
0.146433
0.091887
0.026567
-0.04836
-0.13194
-0.22334
-0.32186
-0.42689
-0.53789
0.2
0.085978
0.046408
0.018708
-0.0029
-0.02065
-0.03569
-0.04871
-0.06015
-0.0703
-0.0794
-0.08761
-0.09507
-0.10187
-0.1081
-0.11382
0.110557
0.063177
0.031125
0.007021
-0.01239
-0.02863
-0.04256
-0.05472
-0.06547
-0.07506
-0.08369
-0.0915
-0.0986
-0.1051
-0.11107
-0.11656
0.121189
0.064854
0.032033
0.007628
-0.01194
-0.02828
-0.04228
-0.05449
-0.06527
-0.07488
-0.08353
-0.09136
-0.09849
-0.105
-0.11097
-0.11647
0.086653
0.046393
0.018694
-0.00291
-0.02066
-0.0357
-0.04872
-0.06015
-0.07031
-0.07941
-0.08762
-0.09507
-0.10187
-0.1081
-0.11382
-0.1191
setup
[M3]
values
[A3]
0.3
0.2
0.1
0
-0.1 0
0.5
1
1.5
2
2.5
3
[G2]
3.5
-0.2
-0.3
-0.4
-0.5
-0.6
Maximum about (0.8, 0.23)
root about 1.8
[A1A1A1]
[subtotal11]
(ii)
Eg:
h = 0.01 gives (p, q) as (0.77, 0.22743)
hence (0.77, 0.23)
h = 0.01 gives root as between 1.87 and 1.88
accept either
(iii)
Eg:
s
1
1
1
1
1
h
0.01
0.01
0.01
0.01
0.01
x
0
0.01
0.02
0.03
0.04
y
0
0.009112
0.017437
0.025286
0.032757
k1
0.01
0.008618
0.008065
0.007649
0.007303
s = 0.715, h = 0.01 gives root closest to x = 1
k2
0.009
0.008314
0.007844
0.007468
0.007147
[M2]
[A1A1]
[A1]
[subtotal5]
k3
0.009025
0.008319
0.007847
0.00747
0.007148
accept 0.71 to 0.72
k4
0.008621
0.008065
0.007649
0.007303
0.007002
Mods
[M3]
t&e
[M3]
[A2]
[subtotal8]
[TOTAL 24]
103
4777
4
(i)
Mark Scheme
Q = Σ (y - a - bx - cx2)2
Σy=
dQ/da = 0
gives
other
equations:
(ii)
June 2008
X
0
0.5
1
1.5
2
2.5
3
[M1]
na + b Σ x + c Σ x2
as given
Σ xy =
a Σ x + b Σ x2 + c Σ x3
Σ x2y =
a Σ x2 + b Σ x3 + c Σ x4
Y
1.02
2.08
2.73
3.14
2.87
2.22
1.43
[M1A1]
[B1]
[B1]
[subtotal 5]
3.50
3.00
2.50
2.00
[G2]
1.50
1.00
0.50
0.00
0
1
2
3
roughly
parabolic
(quadratic) in shape
(iii)
[E1]
x
y
xy
x2y
x2
x3
x4
0
0.5
1
1.5
2
2.5
3
10.5
1.02
2.08
2.73
3.14
2.87
2.22
1.43
15.49
0
1.04
2.73
4.71
5.74
5.55
4.29
24.06
0
0.52
2.73
7.065
11.48
13.875
12.87
48.54
0
0.25
1
2.25
4
6.25
9
22.75
0
0.125
1
3.375
8
15.625
27
55.125
0
0.0625
1
5.0625
16
39.0625
81
142.1875
7
10.5
10.5
22.75
22.75
55.125
15.49
24.06
22.75
55.125
-6.46154
142.1875
-21
48.54
0.554615
-2.69231
-10.5
1.656923
-1.75
1.425833
[subtotal 3]
[M2]
[A2]
normal
equations:
x
y
y fitted
residual
0
0.5
1
1.5
2
2.5
3
1.02
2.08
2.73
3.14
2.87
2.22
1.43
1.017619
2.095
2.765
3.027619
2.882857
2.330714
1.37119
0.002381
-0.015
-0.035
0.112381
-0.01286
-0.11071
0.05881
-3.6E-15
res
form equations
[M1A1]
a= 1.017619
b= 2.562143
c=
solution
[M2A2]
-0.81476
2
5.67E-06
0.000225
0.001225
0.012629
0.000165
0.012258
0.003459
0.029967
residual sum is zero (except for rounding errors) as it should be
residual sum of squares is 0.029967
104
y fitted
[M1A1]
residuals
[M1A1]
[E1]
[A1]
[subtotal 16]
[TOTAL 24]
4777
Mark Scheme
June 2009
4777 MEI Numerical Computation
1(i)
-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
In practice use an initial estimate x0 in place of α
[M1A1]
[M1A1]
[A1]
[A1]
[subtotal 7]
x
0
0.5
1
1.5
2
2.5
3
(iii)
x 3sinx - 0.5
0
-0.5
0.5
0.938277
1
2.024413
1.5
2.492485
2
2.227892
2.5
1.295416
3
-0.07664
[G3
]
[B1B1]
Roots approximately 0.25, 2.1
Eg:
r
0
1
2
3
4
5
6
7
8
9
10
xr
0
-0.5
-1.93828
-3.29971
-0.02763
-0.58289
-2.15131
-3.00855
-0.89795
-2.84615
-1.37349
xr
0.2
0.096008
-0.21242
-1.13247
-3.21639
-0.2758
-1.31696
-3.40387
0.277847
0.322857
0.451832
xr
0.4
0.668255
1.358852
2.432871
1.452591
2.479066
1.345334
2.424072
1.472555
2.485535
1.329994
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
1.925489
2.31326
1.710416
2.470807
1.364805
2.436576
1.444139
2.475969
1.352649
2.4289
xr
2.4
1.52639
2.497043
1.302517
2.392685
1.542517
2.498801
1.298298
2.389305
1.549934
2.499347
No convergence in each case
[M1A1A1]
Let g(x) = 3 sinx − 0.5
Then g'(x) = 3 cosx
So λ = 1 / (1 − 3 cosα)
Smaller root:
λ=
[M1A1]
-0.52446
(approx -0.5)
r
0
1
2
Larger root:
xr
0.25
0.253894
0.254078
NB: must
be using
relaxatio
95
λ=
0.397687
(approx 0.4)
r
0
1
2
xr
2.1
2.095851
2.095866
[M1A1A1]
4777
Mark Scheme
June 2009
n
3
4
5
0.254087
0.254088
0.254088
3
4
5
2.095866
2.095866
2.095866
[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
4
f(x) = x
2h5/5 = 2aα4
Convincing algebra to verify given results
L
0
(ii)
R
0.785398
function values
weights
integral
L
0
function values
weights
integral
0.392699
function values
R
0.392699
0.785398
weights
integral
[M1A1]
[M1A1]
[A1]
[A1]
[A1A1]
[subtotal 8]
m
0.392699
1.189207
0.349066
0.415112
h
0.392699
m
0.19635
1.094949
0.174533
0.191105
0.589049
1.29158
h
0.19635
0.19635
0.174533
0.225423
×1
0.088516
1.043431
0.218166
0.227641
×2
0.696882
1.35535
0.218166
0.295691
×1
0.044258
1.021903
0.109083
0.111472
0.436957
1.211226
×2
0.348441
1.167589
0.109083
0.127364
0.74114
1.383901
0.109083
0.132124
0.109083
0.15096
setup:
[M3A3]
0.938444 [A1]
0.429941
repeat:
[M2
]
0.508508
0.938449 [A1]
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)
[M1A1]
or [E1A1]
[subtotal 12]
Use routine known to deliver 6dp and vary k:
L
0
R
0.392699
function values
weights
integral
0.392699
function values
weights
integral
0.785398
k
integral
m
0.19635
1.136464
0.174533
0.19835
0.589049
1.406898
0.174533
0.24555
h
0.19635
1.465
1.466
0.999908 1.000036
hence k = 1.466
1.467
1.000163
0.19635
×1
0.044258
1.031946
0.109083
0.112568
0.436957
1.297918
0.109083
0.141581
k = 1.46572
×2
0.348441
1.237918
0.109083
0.135036
0.445954
0.74114
modify
1.530164
[M1A1]
0.109083
0.166915 0.554046
1.000000
find k
[M1A1]
[subtotal 4]
[TOTAL 24]
96
4777
Mark Scheme
Use central difference formulae for 2nd and 1st derivatives to obtain first given result
Hence obtain y1 = h2 − y−1
Use central difference to obtain y1 − y−1 = 2h
Hence given result for y1
3(i)
(ii)
h
x
y
0.1
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
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
97
June 2009
[M1A1A1]
[M1A1]
[M1A1]
[M1]
[subtotal 8]
4777
Mark Scheme
4.7
4.8
4.9
5
June 2009
0.179352
0.185502
0.191424
0.196725
setup
[M3]
(ii)
Obtain formula y1 = ah + 0.5h2
Modify routine
Trial on a to obtain a = -1.4 or -1.5
h
x
y
0.1
a
-1.4
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
numbers
[A3]
graph
[A3]
[subtotal 9]
[M1A1]
[M1A1]
[M1A1G1]
0
-0.135
-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
98
4777
Mark Scheme
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5
June 2009
0.296623
0.27349
0.252909
0.235026
0.219875
0.207386
0.197404
0.189706
[subtotal 7]
[TOTAL24]
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]
(iii)
No convergence when a = 2, b = 0
Indicates that non-strict diagonal dominance is not sufficient
99
[M1A1]
[E1E1]
4777
Mark Scheme
June 2009
[subtotal 4]
(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 0,0,0,1
-0.15625
-0.03125
0.34375
-0.03125
-0.03125
-0.15625
-0.03125
0.34375
100
[M1]
[A1]
[A1]
[A1]
[A1]
[subtotal 5]
[TOTAL 24]
GCE
Mathematics (MEI)
Advanced GCE 4777
Numerical Computation
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4777
1
(i)
(ii)
Mark Scheme
June 2010
The data are not evenly spaced so (ordinary) differences will not work
Lagrange's method is not well suited to increasing the degree of the
approximating polynomial because it requires complete recalculation
x
0.09
0.93
1.91
4.10
4.91
6.04
[E1]
[E1]
[E1]
[subtotal 3]
f
1.076
0.897
0.498
-0.544
-0.740
-0.900
[G2]
extrap
(iii)
x
1.91
4.10
4.91
0.93
0.09
6.04
f
0.498
-0.544
-0.740
0.897
1.076
-0.900
1DD
2DD
-0.4758
-0.24198
-0.41131
-0.2131
-0.3321
0.077941
0.053417
-0.04112
-0.02329
f(3)
=
+
+
+
+
0.498
-0.51862
-0.09345
0.057309
0.003774
-0.021
-0.114
-0.057
-0.053
3DD
[subtotal 2]
4DD
5DD
re-order:
table:
0.025025
0.023576
0.015782
0.000796
-0.00402
-0.00117
linear
quadratic
cubic
quartic
[M1A1]
[M1A1]
[M1A1]
[M1A1]
f(3) approximately zero, but difficult to say whether -0.05 or -0.06, -0.1 or 0.0.
(iv)
x
1.91
4.10
4.91
0.93
0.09
6.04
[E1E1]
[subtotal 14]
f
0.498
-0.544
-0.740
0.897
1.076
-0.900
1DD
2DD
3DD
4DD
5DD
-0.4758
-0.24198
-0.41131
-0.2131
-0.3321
0.077941
0.053417
-0.04112
-0.02329
0.025025
0.023576
0.015782
0.000796
-0.00402
-0.00117
user-specified x:
2.89
0.498
-0.46628
-0.09242
0.056679
0.003738
0.032
-0.061
-0.004
0.000
1
[M1A1]
[M1A1]
adjust SS to allow
user-specified x:
trial and error:
answer:
[M1A1]
[M1A1]
[A1]
[subtotal 5]
[TOTAL 24]
4777
2 (i)
Mark Scheme
June 2010
Tn - I = A2h2 + A4h4 + A6h6 + …
T2n - I = A2(h/2)2 + A4(h/2)4 + A6(h/2)6 + …
4
[M1A1]
6
4(T2n - I) - (Tn - I) = b4h + b6h + …
4
[M1]
6
4T2n - Tn - 3 I = b4h + b6h + …
4
[A1]
6
(4T2n - Tn)/3 - I = B4h + B6h + …
(Tn* = (4T2n - Tn)/3
[A1]
4
has error of order h as given)
Tn** = (16T2n* - Tn*)/15 has error of order h6
(ii)
(iii)
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0
0.391654
0.610565
0.691894
0.646735
0.469048
0.13201
-0.43199
-1.41388
-3.79558
x
0
3.141593
1.570796
0.785398
2.356194
0.392699
1.178097
1.963495
2.748894
0.19635
0.589049
0.981748
1.374447
1.767146
2.159845
2.552544
2.945243
f(x)
0
2.22E-16
0.693147
0.5348
0.5348
0.324026
0.654344
0.654344
0.324026
0.178222
0.441842
0.605119
0.683493
0.683493
0.605119
0.441842
0.178222
[B1]
[subtotal 6]
[G2]
[subtotal 2]
T
T*
3.49E-16
1.088793
1.451724
1.384458
1.483014
1.460639
1.479855
1.486033
1.48626
T**
T***
(T****)
f:
[A1]
T:
[M1A2]
T*:
T**:
T***
[M1A1]
[M1A1]
[M1A1]
answer:
[A1]
1.485099
1.486234
1.486275
1.486252
1.486276
1.486276
[subtotal 11]
(iv)
Spreadsheet as above, but seen to work for user-specified c in place of 3.141593
Sequence of values representing trial and error towards solution:
c
4
4.5
4.4
4.45
4.44
I
0.977343 -0.20713 0.133659 -0.02687 0.006681
Answer 4.442 to 3 decimal places
4.442
0.00003
[M2]
[M1A1]
[A1]
[subtotal 5]
[TOTAL 24]
2
4777
3 (i)
Mark Scheme
Modified Euler method
h
x
y
0.1
1
1
1.1 1.145803
1.2 1.300904
1.3 1.466056
1.4 1.641997
1.5 1.829446
1.6 2.029112
1.7
2.24169
1.8 2.467869
1.9 2.708328
2 2.963739
h
0.1
0.05
0.025
0.0125
0.00625
α
2.963739
2.964219
2.964341
2.964372
2.964380
diffs
0.000480
0.000122
0.000031
0.000008
k1
0.141421
0.150346
0.160034
0.170466
0.181626
0.193499
0.206072
0.219337
0.233284
0.247908
k2
0.150185
0.159856
0.170271
0.181415
0.193273
0.205833
0.219085
0.23302
0.247633
0.262916
June 2010
new y
1.145803
1.300904
1.466056
1.641997
1.829446
2.029112
2.24169
2.467869
2.708328
2.963739
setup:
[M2]
first run:
[A2]
ratio
of diffs
further runs:
0.254789
0.252418
0.251215
differences:
ratios:
Correct to 4 dp, α = 2.9644
Ratio of differences indicates 2nd order convergence
(ii)
Predictor corrector method
h
x
y
0.1
1
1
1.1 1.145885
1.2
1.30108
1.3 1.466338
1.4 1.642397
1.5 1.829978
1.6 2.029786
1.7
2.24252
1.8 2.468866
1.9 2.709504
2 2.965107
h
0.1
0.05
0.025
0.0125
0.00625
(iii)
α
2.965107
2.964564
2.964428
2.964394
2.964385
diffs
-0.000543
-0.000136
-0.000034
-0.000008
y pred
1.141421
1.296234
1.46112
1.636815
1.824039
2.023497
2.235885
2.461889
2.702189
2.957457
y corr1
1.145803
1.300989
1.466239
1.64229
1.829862
2.029664
2.242392
2.468732
2.709364
2.964961
[A1A1A1]
[M1]
[M1A1]
[A1]
[E1]
[subtotal 12]
y corr2
1.145884
1.301078
1.466336
1.642395
1.829975
2.029784
2.242518
2.468864
2.709501
2.965104
y corr3
1.145885
1.30108
1.466338
1.642397
1.829978
2.029786
2.24252
2.468866
2.709504
2.965107
setup:
[M2]
first run:
[A2]
ratio
of diffs
further runs:
0.250154
0.250039
0.25001
these -->
may appear in (iii)
The rate of convergence (see ratio of differences) is the same for both methods.
Magnitude of errors about the same for a given h
More programming required for predictor-corrector
Modified Euler (at least in this case) is preferable
3
differences
and ratios:
[A1A1A1]
[M1]
[subtotal 8]
[E1]
[E1]
[E1]
[E1]
[subtotal 4]
[TOTAL 24]
4777
4
(i)
Mark Scheme
7.1
6
5
4
6
5.1
4
3
0.029577
-0.22535
-0.38028
product of pivots:
(ii)
α=
7.01
6
5
4
-0.18390
0.01
6
5.01
4
3
-0.12552
-0.2796
-0.42368
product of pivots:
α=
0.01
x1
x2
x3
x4
5
4
3.1
2
-0.22535
-0.42113
-0.8169
-0.28889
0.062963
(A) β = 0
0.302
0.100
-0.101
-0.303
4
3
2
1.1
-0.38028
-0.8169
-1.15352
-0.47
-0.13333
-0.23577
-0.00198
x1 =
0.320827
x2 =
x3 =
0.103317
-0.11419
x4 =
-0.3317
magnitude of determinant:
β=
5
4
3.01
2
-0.2796
-0.55633
-0.85307
-0.02687
0.006633
1
1
1
1
0.15493
0.295775
0.43662
0.188889
0.037037
0.078205
June 2010
4
3
2
1.01
-0.42368
-0.85307
-1.27245
-0.0467
-0.01333
-0.02486
0.18390
Gauss elim:
[M2A2]
pivoting:
[M1A2]
back subn:
[M1A2]
solutions:
[A2]
[M1A1]
[subtotal 14]
0.01
1.01
1
1
1
0.135521
0.279601
0.42368
0.01
0
0.002469
magnitude of determinant:
x1 =
0.599796
x2 =
x3 =
-0.2999
-0.1996
x4 =
-0.09929
0.001984
(B)β =
0.1
0.600
-0.300
-0.200
-0.099
[M1A1]
solutions:
[M1A1]
[M1A1]
Very large changes in the solution for small change in one coefficient.
The determinant is very small in relation to the magnitude of the coefficients.
4
[E1E1]
[E1E1]
[subtotal 10]
[TOTAL 24]
4777
Mark Scheme
June 2011
1(i)
-2
-1
0
1
2
-2
-1
0
1
2
0.018316
0.367879
1
0.367879
0.018316
r
xr
0
0.7
1
0.612626
2
0.687075
3
0.623708
4
0.677726
5
0.631718
6
0.670946
7
0.637521
8
0.666022
- slow convergence
[G2]
[M1A1]
Derivative of exp(-x2) is -2x exp(-x2). Value at 0.6 (0.7) about -0.83 (-0.86).
Less than 1 in magnitude so converges, but not close to zero so slow.
(ii)
Multiply both sides by λ, then add (1 - λ) x to both sides.
x0
x1
λ
0.4
0.7
0.6651
0.5
0.7
0.6563
0.6
0.7
0.6476
λ = 0.5 (to 1 dp) seems fastest
(iii)
x2
0.6561
0.6532
0.6535
x3
0.6537
0.6529
0.6529
[M1A1]
x4
0.6531
0.6529
0.6529
x5
0.6530
0.6529
0.6529
Differentiate RHS, set to zero at x = α and solve for λ
Best λ evaluates to about
(iv)
[M1A1A1]
[E1E1]
[subtotal 9]
[M1A1A1]
[A1]
[subtotal 6]
[M1A1]
0.53978
[B1]
[subtotal 3]
r
0
1
2
3
4
xr
0.7
0.652966
0.652919
0.652919
0.652919
λr
0.538307
0.539778
0.53978
0.53978
0.53978
-0.04703
-4.8E-05
-1.2E-10
0
Δx r
x6
0.6529
0.6529
0.6529
Δx r+1 /Δx r
0.001011 2.47E-06
0
Ratio of differences tending (rapidly) to zero so (much) faster than first order
1
[M1A1A1]
[M1A1]
[E1]
[subtotal 6]
[TOTAL 24]
4777
2(i)
(ii)
Mark Scheme
June 2011
set up RHS as
f(x) = 1:
f(x) = x:
f(x) = x2
a(f(-) + f())
2h = 2a
0=0
2h3/3 = 2a2
f(x) = x3
f(x) = x4
0=0
2h5/5 = 2a4 = 2h5/9
error is 8h5/45
so no error
so local error of order h5, global error h4
m-h/sqrt3
0.211325
0.105662
0.605662
0.052831
0.302831
0.552831
0.802831
0.026416
0.151416
0.276416
0.401416
0.526416
0.651416
0.776416
0.901416
f(m-h/sqrt3)
1.486525
1.451022
1.602906
1.432762
1.515989
1.589056
1.649038
1.423521
1.466573
1.507617
1.546196
1.581909
1.614408
1.643389
1.668594
h
0.5
0.25
0.25
0.125
0.125
0.125
0.125
0.0625
0.0625
0.0625
0.0625
0.0625
0.0625
0.0625
0.0625
m
0.5
0.25
0.75
0.125
0.375
0.625
0.875
0.0625
0.1875
0.3125
0.4375
0.5625
0.6875
0.8125
0.9375
(by symmetry)
hence a = h
(award same marks
for solution without
symmetry assumed)
hence  = h/sqrt(3)
m+h/sqrt3
0.788675
0.394338
0.894338
0.197169
0.447169
0.697169
0.947169
0.098584
0.223584
0.348584
0.473584
0.598584
0.723584
0.848584
0.973584
f(m+h/sqrt3)
1.646032
1.544084
1.667272
1.481855
1.55962
1.625438
1.676832
1.448594
1.490546
1.530218
1.567188
1.601085
1.631587
1.658417
1.681341
[A1]
[A1E1E1]
[subtotal 10]
integral
1.566278
diffs
ratios
[M1A1]
1.566321
4.29E-05
[M1A1]
1.566324
2.73E-06
0.06371
[A1]
1.566324
1.71E-07
0.06275
[A1]
Ratio of differences very close to the theoretical 0.0625 for fourth order
(iii)
[M1A1]
[M1A1]
[A1]
[A1]
[M1A1E1]
[subtotal 9]
e.g.:
k
1.22
k
I
h
0.5
0.25
0.25
0.125
0.125
0.125
0.125
0.0625
0.0625
0.0625
0.0625
0.0625
0.0625
0.0625
0.0625
m
0.5
0.25
0.75
0.125
0.375
0.625
0.875
0.0625
0.1875
0.3125
0.4375
0.5625
0.6875
0.8125
0.9375
m-h/sqrt3
0.211325
0.105662
0.605662
0.052831
0.302831
0.552831
0.802831
0.026416
0.151416
0.276416
0.401416
0.526416
0.651416
0.776416
0.901416
m+h/sqrt3
0.788675
0.394338
0.894338
0.197169
0.447169
0.697169
0.947169
0.098584
0.223584
0.348584
0.473584
0.598584
0.723584
0.848584
0.973584
f(m-h/sqrt3)
2.630873
2.480189
3.162099
2.404721
2.759934
3.095848
3.388775
2.367053
2.545548
2.72289
2.896046
3.061986
3.21775
3.360519
3.487672
1.2
2.948
1.3
3.229
1.21
2.975
1.23
3.030
1.22
3.002
2
f(m+h/sqrt3)
3.373721
2.886405
3.480932
2.610753
2.95778
3.271659
3.529836
2.470074
2.648271
2.823568
2.992924
3.153343
3.301937
3.435994
3.553039
integral
3.002297
3.002406
3.002413
3.002413
[M3A2]
[subtotal 5]
[TOTAL 24]
4777
3(i)
Method A
h
0.2
0.1
0.05
0.025
Method B
h
0.2
0.1
0.05
0.025
Mark Scheme
h
0.2
y(2)
3.669763
3.671640
3.672112
3.672231
h
0.2
y(2)
3.664805
3.670336
3.671778
3.672146
x
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
y
0
0.219089
0.475075
0.76573
1.089429
1.444924
1.831216
2.247484
2.693043
3.167305
3.669763
diffs
ratio
of diffs
0.001877
0.000473
0.000119
0.251926
0.250918
x
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
y
0
0.218322
0.473669
0.763764
1.086957
1.441983
1.827835
2.243686
2.688844
3.162721
3.664805
diffs
ratio
of diffs
0.005531
0.001442
0.000368
0.260722
0.255408
k1
0.2
0.238251
0.273867
0.307619
0.339966
0.37121
0.401558
0.431161
0.460132
0.488561
k2
0.219089
0.255986
0.290655
0.3237
0.355495
0.386292
0.416269
0.445559
0.474262
0.502457
June 2011
new y
0.219089
0.475075
0.76573
1.089429
1.444924
1.831216
2.247484
2.693043
3.167305
3.669763
≈ 0.25 so 2nd order
k1
0.2
0.238187
0.273764
0.307491
0.339821
0.371052
0.401389
0.430984
0.45995
0.488374
k2
0.236643
0.272507
0.306427
0.338896
0.370231
0.400651
0.430313
0.459333
0.487803
0.515794
new y
0.218322
0.473669
0.763764
1.086957
1.441983
1.827835
2.243686
2.688844
3.162721
3.664805
≈ 0.25 so 2nd order
Differences (and hence errors) in Method B about 3 times those in method A
setup:
[M2]
first run:
[A2]
further runs:
[A1A1A1]
diffs + ratios:
[M1A1]
[subtotal 9]
setup:
[M2]
first run:
[A2]
further runs:
[A1A1A1]
diffs + ratios:
[M1A1]
[M1E1]
[subtotal 11]
(ii)
h
0.3
y
k1
0.3
0.384428
0.461322
0.533436
0.602187
0.668416
0.732668
0.795322
0.856652
0.916864
Trial and error: y = 2x at x = 2.45 (accept 2.44 or 2.46)
k2
0.342053
0.422592
0.497069
0.567502
0.635002
0.700256
0.763723
0.825728
0.886511
0.946255
[G2]
[M1A1]
[subtotal 4]
[TOTAL 24]
3
4777
4(i)
Mark Scheme
-3
-2
-1
0
1
2
3
June 2011
-35.25
-8.01
2.51
-0.09
-4.07
-5.06
0.65
[G2]
Two tps: could be
cubic
[E1]
Almost passes
through origin
[E1]
[subtotal 4]
(ii)
(iii)
2
3 2
Q = Σ (y - ax - bx - cx )
∂Q/∂a = 0 gives
Σ xy =
other equations:
Σ x2y =
Σ x3y =
Σ x3 = 0
x
-3
-2
-1
0
1
2
3
Σ x5 = 0
y
-35.25
-8.01
2.51
-0.09
-4.07
-5.06
0.65
Normal equations:
x2
9
4
1
0
1
4
9
28
107.02 =
-365.24
=
986.32 =
a Σ x2 + b Σ x3 + c Σ x4
a Σ x3 + b Σ x4 + c Σ x5
a Σ x4 + b Σ x5 + c Σ x6
x4
81
16
1
0
1
16
81
196
x6
729
64
1
0
1
64
729
1588
xy
105.75
16.02
-2.51
0
-4.07
-10.12
1.95
107.02
28 a
+
196 c
196 a
196 b
+
1588 c
hence:
a=
-3.86425
b=
-1.86347
x
-3
-2
-1
0
1
2
3
y
-35.25
-8.01
2.51
-0.09
-4.07
-5.06
0.65
y-fitted
-34.826
-8.50983
0.902721
0
-4.62966
-6.39793
1.283537
res
0.424014
-0.49983
-1.60728
0.09
-0.55966
-1.33793
0.633537
-2.85714
res2
0.179788
0.24983
2.583345
0.0081
0.313219
1.790044
0.40137
5.525696
[M1]
[M1A1]
[B1]
[B1]
[subtotal 5]
as given
c=
x2 y
-317.25
-32.04
2.51
0
-4.07
-20.24
5.85
-365.24
x3 y
951.75
64.08
-2.51
0
-4.07
-40.48
17.55
986.32
1.098056
totals
[M1A3]
set up
& solve
[M1A3]
[M1A1A1A1]
[subtotal 12]
(iv)
[G2]
Comment on
goodness of fit
[E1]
[subtotal 3]
[TOTAL 24]
4
4777
Question
1 (i)
1 (ii)
Mark Scheme
Answer
2
Error in x2 is approximately k ε
x0 = a + ε, x1 = a + kε, x2 ≈ a + k2ε
Δx0 = (k  1) ε
Δx1 = k(k  1) ε
hence ∆2x0 = (k  1)2ε
Convincing algebra to given result
b
1
r
xr
1
0
1 0.540302
2 0.857553
3
0.65429
4
0.79348
----- -------30 0.739087
31 0.739084
32 0.739086 converging
33 0.739085 slowly
b
2
r
0
1
2
3
4
----30
31
32
33
xr
1
0.41615
0.673181
0.222554
0.902564
-------0.803826
0.03685
0.997286
0.4112 diverging
7
June 2012
Marks
E1
M1A1
A1
A1
A1
[6]
Guidance
Condone ‘=’ here
M1A1
setup
A1A1
results
4777
Question
1 (ii)
cont
1 (iii)
Mark Scheme
Answer
b= 1
b= 2
2
∆xr
∆xr
∆ xr
est α
∆2xr
est α
r
xr
r
xr
0
1
0
1
1 0.540302 0.4597
1 0.41615 1.41615
2 0.857553 0.317251 0.776949 0.72801
2 0.673181 1.089328 2.505475 0.199564
0 0.72801
0 0.199564
1
0.7465 0.018489
1
0.9214 0.721836
2 0.73407 0.01243 0.03092 0.739067
2 0.26866 1.19006 1.9119 0.472093
0 0.739067
0 0.472093
1 0.739097 3.04E-05
1 0.586403
0.11431
2 0.739077
2E-05 5.1E-05 0.739085
2 0.387567 0.19884 0.31315 0.51382
0 0.739085
0
0.51382
1 0.739085 8.21E-11
1
0.51684
0.00302
2 0.739085 5.5E-11 1.4E-10 0.739085
2
0.51166 0.00518 0.0082 0.514932
makes divergent iteration convergent
eg
b
0
0.3
0.6
0.9
1.2
1.5
1.8
2.1
2.4
2.7
3
α
1
0.958907
0.867554
0.769576
0.682741
0.609904
0.549477
0.499139
0.456823
0.420881
0.39004
1.2
June 2012
Marks
Guidance
E1
M2A1
A1A1
[10]
Set up
Results
A1A1
G2
values
graph
M2
A2
trial
result
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
3
3.5
b
α
2.1
0.499139
2.0
0.514933
2.095
0.499907
2.094
0.500061
2.0944
0.499999
2.0943
0.500015
best is b = 2.0944
[8]
8
4777
Question
2 (i)
Mark Scheme
June 2012
Answer
Marks
M1A1
f(x) = 1
2h = 2a + b
(f(x) = x, x3 give 0 = 0, not reqd.)
f(x) = x2
2h3/3 = 2aa2
4
f(x) = x
2h5/5 = 2aa4
Convincing algebra to verify given results
Guidance
M1A1
M1A1
A1A1
[8]
2 (ii)
L
0
function values
weights
integral
R
1.570796
L
0
function values
weights
integral
L
0.785398
function values
weights
integral
R
0.785398
h = π/16 gives 2.160149
h = π/32 gives 2.160149
R
1.570796
m
0.785398
1.456475
0.698132
1.016812
h
0.785398
m
0.392699
1.493467
0.349066
0.521318
m
1.178097
1.29971
0.349066
0.453684
h
0.392699
h
0.392699
x1
0.177031
1.464531
0.436332
0.639022
x2
1.393765
1.156109
0.436332
0.504448
x1
0.088516
1.442418
0.218166
0.314687
x1
0.873914
1.431959
0.218166
0.312405
x2
0.696882
1.474966
0.218166
0.321788
x2
1.482281
1.082998
0.218166
0.236274
hence, given rate of convergence, correct to 6 sf
9
2.160281
M1A1
A1
A1
A1A1
1.157793
M1A1
1.002363
2.160156
A1
A1
A1A1
[12]
subdividing
4777
Question
2 (iii)
Mark Scheme
June 2012
Answer
Marks
Guidance
eg
k
0.381
L
0
function values
weights
integral
L
0.785398
function values
weights
integral
k
integral
0.380
1.999235
R
0.785398
R
1.570796
m
0.392699
1.357491
0.349066
0.473854
m
1.178097
1.221099
0.349066
0.426244
h
0.392699
h
0.392699
x1
0.088516
1.321988
0.218166
x2
0.696882
1.344658
0.218166
0.288413
x1
0.873914
1.314677
0.218166
0.286818
0.293359
x2
1.482281
1.062641
0.218166
0.231832
1.055626
0.944894
2.00052
0.381 (NB: these changes in the integral are greater than those
2.00052 arising from halving h again.)
10
M2
M1A1
[4]
Modification to
include k
4777
Question
3 (i)
3 (ii)
Mark Scheme
June 2012
Answer
Use central difference formulae for 2nd and 1st derivatives to obtain first given result
Hence obtain y1 (1  h) = h2 + 2  (1 + h) y–1
State or show that y1  y–1 = 2h or equivalent
Eliminate y–1; convincing algebra to given result for y1
h
0.1
x
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
y
1
0.895
0.782647
0.666837
0.552585
0.445715
0.352423
0.27885
0.230785
0.213615
0.232584
0.293408
0.403342
0.572952
0.819293
1.172342
1.689964
2.499465
3.945671
7.406389
30.95641
Marks
M1A1A1
M1A1
B1
M1A1
[8]
Guidance
35
M3
setup
A3
values
30
25
20
15
G2
10
5
0
1
1.5
2
2.5
3
[8]
11
4777
Mark Scheme
Question
3 (iii)
Answer
h
0.1
0.05
0.025
0.0125
0.00625
0.003125
June 2012
Marks
y(2)
diffs
ratios
y(3)
diffs
ratios
0.232584
30.95641
0.234604 0.00202
37.62817 6.671755
0.235106 0.000502 0.248632 39.83126 2.203093 0.330212
0.235232 0.000125 0.249662 40.42700 0.595738 0.27041
40.57898 0.151988 0.255125
0.235 to 3 sf
40.61718 0.038192 0.251283
40.6 to 3 sf
ratios about 0.25 (in each case), so second order method
M1A1A1
y(2)
M1A1A1
y(3)
M1E1
[8]
12
Guidance
4777
Question
4 (i)
4 (ii)
Mark Scheme
June 2012
Answer
A sufficient, but not necessary, condition for G-J and G-S to converge
is strict diagonal dominance: the diagonal elements in each row are, in magnitude,
greater than or equal to the sum of the magnitudes of the other elements,
and at least one inequality is strict.
k=1
x1
x2
x3
0
0
0
1
0
0
1
0.33333
0.66667
2.666667
0.33333
0.66667
2.666667
1.18519
1.92593
6.037037
1.18519
1.92593
6.037037
2.96708
4.50206
12.97119
2.96708
4.50206
12.97119
6.64289
9.80704
differences increasing so diverging
x4
0
0
0
0.444444
0.444444
1.432099
1.432099
3.478738
3.478738
k=3
x1
x2
x3
0
0
0
0.333333
0
0
0.333333
0.11111
0.22222
0.518519
0.11111
0.22222
0.518519
0.2716
0.39506
0.687243
0.2716
0.39506
0.687243
0.43759
0.56241
0.854138
0.43759
0.56241
0.854138
0.60418
0.72916
differences just increasing so diverging
Marks
E1
Guidance
E1
E1
[3]
x4
0
0
0
0.148148
0.148148
0.312757
0.312757
0.479195
0.479195
M3
Setup
A1E1
k=1
A1E1
k=3
k=5
x1
x2
0
0
0.2
0
0.2
0.06667
0.266667
0.06667
0.266667
0.14815
0.312593
0.14815
0.312593
0.21613
0.348971
0.21613
converges eventually:
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
x3
0
0
0.13333
0.13333
0.20741
0.20741
0.26436
0.26436
x4
0
0
0
0.088889
0.088889
0.167901
0.167901
0.232209
0.5
0.5
0.5
0.5
0.499999
0.499999
0.5
0.5
k=1 no diagonal dominance
and no convergence
k=3 no strict diagonal dominance
and no convergence
k=5 strict diagonal dominance
and convergence
E1
E1
E1
A1
A1
[12]
13
k=5
4777
Question
4 (iii)
Mark Scheme
Answer
June 2012
Marks
B1
T
First column of inverse matrix is (0.5, -0.5, -0.5, 0.5)
Modify routine to solve for the other three unit vectors
Obtain solutions
Write inverse matrix:
0.5
0.5
0.5
0.5
0.5 1.166667 0.666667
1
0.5 0.666667 0.916667
0.75
0.5
1
0.75
1.25
M1,1,1
A1,1,1
M1A1
[9]
14
Guidance
4777
Mark Scheme
Question
1
(i)
Answer
June 2013
Marks
x1    k ( x0   )
x2    k ( x1   )
x2  x1
x1  x0
kx  x1
Solve first equation for    0
k 1
Subtract and rearrange to k 
Guidance
M1
Setup
M1
Convincing algebra
M1
To given result
[3]
1
(ii)
Eg m = 1.5
3
2.5
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
And/or
Gradient of line > 1
Gradient of curve = 1 at x = 0
Exponential grows faster than linear
So root
B3
[3]
5
For convincing
argument
4777
1
Question
(iii)
Mark Scheme
Answer
June 2013
Marks
(*)
0.5
0.559616
0.609452
0.649289
0.680028
0.703118
0.720119
0.732454
Differences getting smaller so converging, but slowly
(**)
0.5
0.432481
0.360717
0.289572
0.223904
0.1673
0.121406
0.086056
Converging to root at zero
M1
A1
E1
M1
A1
E1
[6]
6
Guidance
4777
Question
1
(iv)
Mark Scheme
Answer
Marks
(v)
Guidance
(*)
0.5
k
α
0.559616 0.059616
0.609452 0.049837 0.835962 0.863426
0.863426
0.830793 -0.03263
0.809236 -0.02156 0.660625 0.767271
0.767271
0.765889 -0.00138
0.764925 -0.00096 0.697717
0.7627
0.7627
0.762697 -3.6E-06
0.762694 -2.5E-06 0.699611 0.762689
(*) converges more quickly
(**)
0.7
k
α
0.675835 -0.02416
0.643783 -0.03205 1.326411 0.774032
0.774032
0.778995 0.004963
0.786187 0.007192 1.449254 0.762986
0.762986
0.763113 0.000128
0.763296 0.000182 1.429871 0.762689
0.762689
0.762689
9E-08
0.762689 1.29E-07 1.429356 0.762689
(**) converges provided a good enough starting point is used
1
June 2013
M2
A2
E1
M1
A1
E1,E1
[9]
M1
A1
A1
[3]
Value is 1.482 to 3dp
7
T&e
Answer
4777
2
Question
(i)
Mark Scheme
June 2013
Answer
The data are not evenly spaced so (ordinary) differences will not work
Divided differences, however, can accommodate variable spacing
Lagrange's method is not well suited to increasing the degree of the approximating
polynomial because it requires complete recalculation
Marks
E1
E1
E1
Guidance
[3]
2
(ii)
45.00
35.00
25.00
G2
15.00
5.00
-5.00
0.0
2.0
4.0
6.0
8.0
10.0
[2]
2
(iii)
x
2.0
4.0
1.0
0.5
6.0
8.0
y(2.9) =
1DD
y
1.04
1.22 0.09000
1.43 -0.07000
2.23 -1.60000
6.96 0.86000
40.53 16.78500
1.040
2DD
3DD
0.16000
0.43714
0.49200
2.12333
+
0.081
+
-0.1584
+ 0.347537
+ -0.23948
4DD
-0.18476
0.02743 0.05305
0.23305 0.05140
=
=
=
=
1.121
0.963
1.310
1.071
5DD
Re-order values
M1
A1
DD table
-0.00027
linear
quadratic
cubic
quartic
No convergence of estimates, but 4DDs are nearly equal, so quartic a good fit
Answer 1.07 (only 2dp in data)
8
M1
M1A1
M1A1
M1A1
M1A1
E1 E1
A1
[14]
4777
Mark Scheme
Question
2 (iv)
Answer
1DD
x
y
4.0
1.22
6.0
6.96
2.87000
2.0
1.04
1.48000
1.0
1.43 -0.39000
0.5
2.23 -1.60000
8.0 40.53
5.10667
x=
4.79 gives
1.220
+
+
+
+
2DD
0.69500
0.37400
0.80667
0.95810
2.2673
-0.66435
-0.28536
-0.53619
June 2013
Marks
3DD
4DD
5DD
M1
A1
0.10700
-0.07867
0.02524
=
=
=
=
0.05305
0.05195
3.487
2.823
2.538
2.001
Guidance
rearrange
-0.00027
linear
quadratic
cubic
quartic
M1
A1
A1
[5]
9
T&e
answer
4777
Mark Scheme
Question
3 (i)
June 2013
Answer
h
0.5
0.5
0.5
0.5
0.5
x
0
0.5
1
1.5
2
Cf y  1  0.25 x 4
h
0.5
0.5
0.5
0.5
0.5
Cf y  1  0.2 x 5
y
1
1.015625
1.25
2.265625
5
y (2)  5
k1
0
0.0625
0.5
1.6875
k2
0.007813
0.210938
0.976563
2.679688
k3
0.007813
0.210938
0.976563
2.679688
k4
0.0625
0.5
1.6875
4
new y
1.015625
1.25
2.265625
5
solution to y  x 3 is exact
k1
x
y
0
0
1
0.5 1.00651 0.03125
0.5
1 1.200521
1.5 2.519531 2.53125
2 7.401042
y (2)  7.4
Marks
k2
0.001953
0.158203
1.220703
4.689453
Guidance
M2
setup
A2
M1
A1
Result
Exact
E1
k3
k4
new y
0.001953 0.03125 1.00651
0.158203
0.5 1.200521
1.220703 2.53125 2.519531
4.689453
8 7.401042
error in solution to y  x 4
M1
Rerun
A1
Result
B1
Exact
E1
[11]
10
4777
Question
3 (ii)
Mark Scheme
June 2013
Answer
h
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
x
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
k1
0.168294
0.09783
0.020509
-0.03163
-0.04498
-0.02279
0.029779
0.11
0.204615
0.268871
y
1
1.134994
1.193243
1.18463
1.143096
1.10653
1.107613
1.175408
1.332919
1.575176
1.848193
k2
0.136333
0.056918
-0.01171
-0.044
-0.03754
0.001266
0.069232
0.1621
0.251627
0.278311
Marks
k3
0.135593
0.058692
-0.00856
-0.04242
-0.03854
-0.00191
0.063581
0.151587
0.23499
0.272474
k4
0.097818
0.020444
-0.03164
-0.04474
-0.02226
0.030575
0.111362
0.207694
0.275691
0.267657
new y
1.134994
1.193243
1.18463
1.143096
1.10653
1.107613
1.175408
1.332919
1.575176
1.848193
Guidance
M3
Setup
A3
result
G2
graph
B1
M1
A1
First
Further
B1
B1
[13]
answer
2
1.5
1
0.5
0
0
0.5
1
1.5
2
Local maximum, eg
h
0.1
0.01
0.001
0.0001
x
y
0.5 1.19552
0.46 1.196441
0.464 1.196451
0.4638 1.196451
(0.464, 1.196)
11
4777
Mark Scheme
Question
4 (i)
June 2013
Answer
pivots in boxes:
4
2
2
4
3
4
6
1
2.5
4.5
-2
5
3
2
9
0.5
-0.5
4
0.777778
3.777778
1
4
5
5
3.5
4.5
4
1
6
-0.23529
Marks
solutions
1
6.5625 = x1
2
3
4
1.5
2.5
-3 = x2
3
0.111111
4.111111
-3.875 = x3
-0.73529
3.125 = x4
magnitude of determinant (product of moduli of pivots):
check:
1
2
3
4
M3
Setup
A2
1st elimination
A2
2nd elimination
A1
3rd elimination
M2
Back substitution
A2,A2
M1
A1
[16]
16
12
Guidance
Solution and check
determinant
4777
Mark Scheme
Question
4 (ii)
June 2013
Answer
4.01
2
2
4
3
4
6
1
2.503741
4.503741
-1.99252
5
3
2
9
0.506234
-0.49377
4.012469
0.780731
3.79402
1
4
5
5
3.501247
4.501247
4.002494
0.998893
5.993909
-0.23453
1
2
3
4
1.501247
2.501247
3.002494
0.110742
4.109081
-0.73482
magnitude of determinant (product of moduli of pivots):
4
2.01
2
4
3
4
6
1
2.4925
4.5
-2
5
3
2
9
0.4875
-0.5
4
0.764444
3.777778
1
4
5
5
3.4975
4.5
4
1.005
6
-0.20912
Marks
check:
solutions % change
6.533914
-0.43559
1
2
3
4
-3
2.44E-13
-3.86683
3.133167
-0.21077
0.261357
16.07
M1
A1
Re-run
M1
A1
% change
M1
Re-run
A1
% change
0.4375
solutions % change
1 7.383966
12.51758
2
3
4
1.4975
2.5
-3.3692
12.30661
3
0.112778
4.111111 -4.37342
12.86239
-0.71912 3.438819
10.04219
magnitude of determinant (product of moduli of pivots): 14.22
Guidance
check:
1
2
3
4
-11.125
A small change in a produces small changes in the solutions and the determinant
A small change in b produces large changes in the solutions and the determinant
13
E1
E1
[8]