dolg0514
dolg0514/alakkerekit
1. alakkerekit
multi
1.0 point 0.10 penalty Single Shuffle The normalized, regularly rounded representation of the number
5
36
in the floating point system
F = [a = 2, k− = −6, k+ = 6, t = 6]
is:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
2−2 · 0.100100 ✓
2−1 · 0.100100 (−100%)
2−3 · 0.100100 (−100%)
2−1 · 0.100011 (−100%)
2−2 · 0.100011 (−100%)
2−4 · 0.100011 (−100%)
20 · 0.100010 (−100%)
2−2 · 0.100010 (−100%)
2−1 · 0.100010 (−100%)
dolg0514/alaklevag
1. alaklevag
multi
1.0 point 0.10 penalty Single Shuffle 1
In the floating point number system
F = [a = 2, k− = −4, k+ = 3, t = 6]
the the normalized, choped representation of 0.490 is
(a)
(b)
(c)
(d)
(e)
(f)
2−1 · 0.111110
2−2 · 0.101111
2−1 · 0.111010
2−2 · 0.101110
2−1 · 0.111110
2−1 · 0.111111
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/alaklevag2
1. alaklevag2
1.0 point 0.10 penalty Single Shuffle multi
In the floating point number system
F = [a = 2, k− = −4, k+ = 3, t = 5]
the the normalized, choped representation of
(a)
(b)
(c)
(d)
(e)
(f)
2−1 · 0.11010
2−2 · 0.10000
2−2 · 0.11000
2−1 · 0.11011
2−1 · 0.11110
2−1 · 0.11010
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/chol1
1. chol1
2
16
39
is
multi
1.0 point 0.10 penalty Single Shuffle The Cholesky-decomposition of the matrix


9 6 9


A =  6 13 9 
9 9 19
is
(a)



3 0 0
3 2 3



LLT =  2 3 0   0 3 1 
3 1 3
0 0 3
✓
(b)



3 0 0
3 1 2



LLT =  1 3 0   0 3 −2 
2 −2 3
0 0 3
(−100%)
(c)



3
0 0
3 −1 1



LLT =  −1 3 0   0 3 −1 
1 −1 3
0 0
3
(−100%)
(d)


3 0 0
3 −1 0



LLT =  −1 3 0   0 3 2 
0 2 3
0 0 3

(−100%)
(e) There is no Cholesky-decomposition for the given matrix (−100%)
3
dolg0514/cond
1. cond
multi
1.0 point 0.10 penalty Single Shuffle If

−10 3 −6


A =  −7 0 −3 
−2 −8 3

then cond∞ (A) =
(a)
(b)
(c)
(d)
3097
15
3112
15
3188
15
3173
15
✓
(−100%)
(−100%)
(−100%)
dolg0514/cond2
1. cond2
multi
If
1.0 point 0.10 penalty Single Shuffle 

−6 −10 −10


A =  −6 4
3 
−9 10
5
then cond∞ (A), cond2 (A) and cond1 (A) are respectively
(a)
(b)
(c)
(d)
24.844, 15.945, 28.444
24.844, 28.444, 15.945
15.945, 28.444, 24.844
15.945, 24.844, 28.444
✓
(−100%)
(−100%)
(−100%)
4
dolg0514/egyenes
1. egyenes
multi
1.0 point 0.10 penalty Single Shuffle If L is the linear function that interpolates the points (10, −10) and
(−6, 1), then L(7) is:
(a)
(b)
(c)
(d)
(e)
(f)
− 127
✓
16
− 33
(−100%)
4
67
− 8 (−100%)
− 121
(−100%)
16
31
− 4 (−100%)
− 17
(−100%)
2
dolg0514/eig
1. eig
multi
1.0 point 0.10 penalty Multiple Shuffle Consider the matrix
[
9
3
−16 −10
]
Choose the true statements about its eigenvalues and eigenvectors.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
It has two different eigenvalues (20%)
It has infinitely many eigenvectors (20%)
It has at least two different eigenvectors (20%)
The two eigenvalues are: 6 and −7 (20%)
The sum of its eigenvalues is: −1 (20%)
It has exactly two eigenvectors (−20%)
It has exactly one eigenvector (−20%)
It has exactly one eigenvalue (−20%)
5
(i)
(j)
(k)
(l)
(m)
The sum of its eigenvalues is: 3 (−20%)
The two eigenvalues are: 9 and −5 (−20%)
The only eigenvalue is: −3 (−20%)
It has infinitely many eigenvalues. (−20%)
It has zero eigenvalues. (−20%)
dolg0514/fixapprox
1. fixapprox
1.0 point 0.10 penalty Single Shuffle multi
In order to approximate a root of the function
x3 − 13x + 3
we apply fixed-point iteration for the function
3
1 3
x +
13
13
with the starting point x0 =
(a)
(b)
(c)
(d)
(e)
(f)
(g)
0.231727
0.347591
0.139036
0.115864
0.162209
0.324418
Pass.
7
15
and perform 3 iterations. Then x3 =
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/foeh3
1. foeh3
multi
1.0 point 0.10 penalty Single Shuffle 6
If L is the minimal degree polynomial that interpolates the points
(− 74 , 139
), ( 12 , − 35
) and ( 54 , − 269
), then its leading coefficient is:
32
8
32
(a)
(b)
(c)
(d)
(e)
(f)
− 12 ✓
−5 (−100%)
− 78 (−100%)
4 (−100%)
− 92 (−100%)
1
(−100%)
2
dolg0514/foeh4
1. foeh4
multi
1.0 point 0.10 penalty Single Shuffle If L is the minimal degree polynomial that fits the points (−2, −76),
(−5, −1132), (1, 8) and (3, 164), then its leading coefficient is:
(a)
(b)
(c)
(d)
(e)
(f)
8✓
12 (−100%)
1 (−100%)
13 (−100%)
2 (−100%)
6 (−100%)
dolg0514/fokszam
1. fokszam
multi
1.0 point 0.10 penalty Single Shuffle If L is the minimal degree polynomial that fits the data
t
3
−2
−5 −5
f
7
then its degree is:
(a)
(b)
(c)
(d)
0
4
1
3
✓
(−100%)
(−100%)
(−100%)
dolg0514/integral1
1. integral1
multi
1.0 point 0.10 penalty Single Shuffle The approximate value of the definite integral
∫
−16
cos(2x) sin(2x2 − π)dx
−22
computed by the function ’integral’ is:
(a)
(b)
(c)
(d)
(e)
(f)
0.023503 ✓
0.042306 (−100%)
0.025854 (−100%)
0.0 (−100%)
0.007051 (−100%)
0.014102 (−100%)
dolg0514/integral2
1. integral2
multi
1.0 point
0.10 penalty
Single
Shuffle
The approximate value of the integral
∫
4
−5
∫
1
5
x2
1
2
2
e−x −y dy dx
+1
8
computed by the function ’integral2’ is:
(a)
(b)
(c)
(d)
(e)
(f)
0.187259
0.149807
0.056178
0.299614
0.262162
0.112355
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/integral2inf
1. integral2inf
multi
1.0 point 0.10 penalty Single Shuffle The approximate value of the integral
∫
−2
−∞
∫
−2
2y 3 e−x
2 −|y|
−∞
(1 + |x − y|)dy dx
computed by the function ’integral2’ is:
(a)
(b)
(c)
(d)
(e)
(f)
−0.137745
−0.165294
−0.055098
−0.179068
−0.261715
−0.234166
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/integral2var
1. integral2var
multi
1.0 point 0.10 penalty Single Shuffle 9
The approximate value of the integral
∫
9
∫
4
x2
−x2
sin(2x) −0.1x2 −0.2y2
dy dx
e
y4 + 1
computed by the function ’integral2’ is:
(a)
(b)
(c)
(d)
(e)
(f)
0.050075
0.030045
0.035053
0.025038
0.060091
0.015023
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/integralinf
1. integralinf
multi
1.0 point 0.10 penalty Single Shuffle The approximate value of the definite (improper) integral
∫
∞
sin(2x)
4.0
2
dx
1 + e|x|
computed by the function ’integral’ is:
(a)
(b)
(c)
(d)
(e)
(f)
0.004977 ✓
0.005475 (−100%)
0.000498 (−100%)
0.009955 (−100%)
0.007466 (−100%)
0.00647 (−100%)
10
dolg0514/integralsing
1. integralsing
multi
1.0 point 0.10 penalty Single Shuffle The approximate value of the definite integral
∫
2
−2
sin(πx) √
1
dx
2−x
computed by the function ’integral’ is:
(a)
(b)
(c)
(d)
(e)
(f)
−0.548671
−0.603538
−0.877874
−0.823007
−0.109734
−0.054867
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/intsimple
1. intsimple
1 point 0.10 penalty If we approximate the integral
∫
1
−3
x3 − 2x + 2 dx
using the simple midpoint(tangent), Simpson and trapesoidal methods, we get respectively:
multi
1 point Single Shuffle • 12, −4, −36 ✓
• 10, 1, −40 (−100%)
11
•
•
•
•
•
5, 1, −44 (−100%)
17, −2, −42 (−100%)
16, −9, −42 (−100%)
14, −1, −44 (−100%)
Pass.
The exact value of the integral above is:
numerical
1 point -4.0 ± 5e-3 ✓
dolg0514/leslie1
1. leslie1
multi
1.0 point 0.10 penalty Single Shuffle In an age-classified population we have the following transition matrix:


0 2 3


L =  12 0 0 
0 12 0
What is the stable size distribution w of the age categories?
[
]
T
(a) w = 0.64397 0.25503 0.101 ✓
[
]
(b) wT = 0.48953 0.15456 0.35591 (−100%)
[
]
(c) wT = 0.41608 0.06414 0.51977 (−100%)
[
]
T
(d) w = 0.35687 0.24052 0.4026 (−100%)
(e) There is no stable distribution for this problem. (−100%)
dolg0514/linreg
1. linreg
12
multi
1.0 point 0.10 penalty Single Shuffle Determine the best approximating (in the sense of least squares) line
for the data below.
t
2 −6 0
f
3 −4 7 −10 −1
4
−8 10
5
−8
7
(a) 17 t − 1 ✓
(b) 17 t − 12
(−100%)
5
89
2
(c) 35 t − 5 (−100%)
61
(d) 35
t + 14
(−100%)
5
114
7
(e) − 35 t + 5 (−100%)
(−100%)
(f) − 67 t + 13
5
dolg0514/linregVal
1. linregVal
numerical
1.0 point 0.10 penalty of the best approximating (in the sense of
Determine the value at 19
2
least squares) line for the data below.
t
3
f
10
−2 −3 3
1
10
dolg0514/lu1
1. lu1
1.0 point 0.10 penalty Single Shuffle 13
3
3 −1 −8
• -5.153061224 ± 1e-4 ✓
multi
3
The LU decomposition of the matrix


−2 3 1


A =  −6 12 6 
−6 15 12
is
(a)


1 0 0
−2 3 1



LU =  3 1 0   0 3 3 
0 0 3
3 2 1

✓
(b)



−2 3 1
1 0 0



LU =  5 1 0   0 0 0 
5 −1 1
0 0 0
(−100%)
(c)



1 0 0
−2 3 1



LU =  6 1 0   0 1 0 
6 4 1
0 0 1
(−100%)
(d)



1 0 0
−2 3 1



LU =  6 1 0   0 4 1 
4 4 1
0 0 5
(−100%)
(e) There is no LU decomposition for the given matrix (−100%)
14
dolg0514/lu2
1. lu2
1.0 point 0.10 penalty Single Shuffle multi
In the LU decomposition of the matrix

−1 3 2


A =  −1 5 1 
−2 0 5

the elements l32 and u33
(a)
(b)
(c)
(d)
(e)
(f)
-3 and -2 ✓
-4 and -5 (−100%)
2 and -9 (−100%)
-2 and 4 (−100%)
-1 and 5 (−100%)
There is no LU decomposition for the given matrix (−100%)
dolg0514/lu3
1. lu3
1 point 0.10 penalty For the matrix A we know partially its LU decomposition:

 


−3 4 −4
1 0 0
−3 4 −4

 


A =  −9 13 −13  =  3 1 0   0 1 −1 
3 −3 4
−1 X 1
0 0 Y
Then the value of X is:
numerical
1 point 1 ✓
15
the value of Y is:
numerical
1 point 1 ✓
the value of det(A) is:
numerical
1 point -3 ✓
dolg0514/lu4
1. lu4
1 point 0.10 penalty Once we have the A = LU decomposition, the usual way to solve
the Ax = b equation is to solve Ly = b for y and U x = y for x.
Triangular systems can be solved by the back-substitution. For a given
decomposition



1
0 0
−1 1 −4



A = LU =  2
1 0  0 3 1 
−1 −1 1
0 0 4
and a constant vector (right hand side)
bT =
[
−9 −3 6
]
apply the method of back-substitution. Then
the value of x2 is:
numerical
1 point 4 ✓
16
and the value of y3 is:
numerical
1 point 12 ✓
dolg0514/newtonApprox
1. newtonApprox
multi
1.0 point 0.10 penalty Single Shuffle In order to approximate a root of the function
10 2 220
80
x +
x+
3
9
3
we apply Newton method with the starting pont x0 = −11 and perform
3 iterations. Then x3 =
(a)
(b)
(c)
(d)
(e)
(f)
−6.024032
−6.570907
−6.055282
−6.211532
−6.633407
−6.320907
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/newtonGyokA
1. newtonGyokA
multi
1.0 point 0.10 penalty Single Shuffle Approximate the root of the function f (x) = x2 − 13 with Newtonmethod, starting from x0 = 13. Then x3 =:
(a)
(b)
799
217
114
31
✓
(−100%)
17
(c)
(d)
(e)
(f)
786
217
796
217
789
217
113
31
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/newtonGyokX0
1. newtonGyokX0
multi
1.0 point
0.10 penalty
Single
Shuffle
In order to approximate a root of the function
−
5 2
35
15
x −
x−
16
128
512
we apply Newton method with the starting pont x0 =− 34 and perform
6 iterations. Then x6 =
(a)
(b)
(c)
(d)
(e)
(f)
−0.75 ✓
−1.453125 (−100%)
−1.734375 (−100%)
0.015625 (−100%)
−0.796875 (−100%)
−1.28125 (−100%)
dolg0514/newtonPeriod2
1. newtonPeriod2
1 point 0.10 penalty Approximate the root of the function
1 2
1
1
x + x+
14
14
2
18
using the Newton method with initial point x0 = −2. Choose the
correct value of x2 from below.
multi
•
•
•
•
•
•
•
1 point Single Shuffle −2 ✓
− 14 (−100%)
9
(−100%)
2
9 (−100%)
− 32 (−100%)
10 (−100%)
Pass.
The value of x17 is
numerical
1 point 1.0 ± 5e-4 ✓
dolg0514/nodom
1. nodom
multi
1.0 point 0.10 penalty Single Shuffle Find an eigenvector v for the dominant eigenvalue of



A=


27 0 −9 15
0 27 15 9
−9 15 14 0
15 9 0 14






(a) There[is no dominant eigenvalue for the given matrix. ✓ ]
(b) v T = −0.489480711 −0.293688426 0.0 0.821069877 (−100%)
[
]
T
(c) v = −0.704061611 −0.422436966 0.0 −0.570827695 (−100%)
19
[
]
−1.408123221 −0.844873933 0.0 −1.141655391 (−100%)
]
(e) v T = 0.422436966 −0.704061611 −0.570827695 0.0 (−100%)
[
]
T
(f) v = 1.267310899 −2.112184832 −1.712483086 0.0 (−100%)
(d) v T =
[
dolg0514/normakm
1. normakm
multi
1.0 point 0.10 penalty Single Shuffle 
If


A=


−8 −2 −4 8
8 −7 −5 −4
6 −6 −1 7
−9 7
2 −7






then ||A||1 , ||A||2 and ||A||∞ are respectively:
(a)
(b)
(c)
(d)
31.0, 19.295, 25.0
31.0, 25.0, 19.295
19.295, 31.0, 25.0
25.0, 31.0, 19.295
✓
(−100%)
(−100%)
(−100%)
dolg0514/normakv
1. normakv
multi
1.0 point 0.10 penalty Single Shuffle If
[
x=
0 −1 9 −1 3 −1
then ||x||∞ , ||x||1 and ||x||2 is respectively:
(a) 9.0, 15.0, 9.644 ✓
(b) 9.644, 9.0, 15.0 (−100%)
20
]
(c) 9.644, 15.0, 9.0 (−100%)
(d) 15.0, 9.644, 9.0 (−100%)
dolg0514/normakv2
1. normakv2
multi
1.0 point 0.10 penalty Single Shuffle For the vectors
[
]
x = 9 0 6 −8 0
[
]
y = −6 −7 9 −2 4 3
[
]
z = 0 6 8 1 −3 −7
choose the correct ordering:
(a)
(b)
(c)
(d)
||z||9 ≤ ||y||5 ≤ ||x||2
||y||5 ≤ ||z||9 ≤ ||x||2
||x||2 ≤ ||z||9 ≤ ||y||5
||x||2 ≤ ||y||5 ≤ ||z||9
✓
(−100%)
(−100%)
(−100%)
dolg0514/optim1var
1. optim1var
multi
1.0 point 0.10 penalty Single Shuffle Among the locations at which the function
(
f (x) = cos
)
9 2
x + sin (3x)
4
has a local minimum the closest to zero is:
(a) −1.043144 ✓
21
(b)
(c)
(d)
(e)
(f)
−1.061144
−1.168144
−0.965144
−1.177144
−0.891144
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/optim2var
1. optim2var
multi
1.0 point 0.10 penalty Single Shuffle Among the points in [−2, 2] × [−2, 2] at which the function
(
f (x, y) =
)2 (
)2
5
25 2 5
25 2
x − y − 11 +
x+ y −7
4
2
2
4
has a local minimum, the smallest in absolute value is:
(a)
(b)
(c)
(d)
(e)
(f)
(1.2, −0.8) ✓
(1.131, −0.71) (−100%)
(1.311, −0.975) (−100%)
(1.095, −0.691) (−100%)
(1.151, −0.603) (−100%)
(1.387, −0.781) (−100%)
dolg0514/parabola
1. parabola
multi
1.0 point 0.10 penalty Single Shuffle If L is the quadratic function that interpolates the points (−5, 121), (9, 177)
and (−4, 86), then its value at L(4) is:
(a) 22 ✓
22
(b)
(c)
(d)
(e)
(f)
16
24
12
18
17
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/power
1. power
multi
1.0 point 0.10 penalty Single Shuffle For the given A and v0 , compute v6 with power iteration.


2 −3 −3 3

−3 −3 −2 −3 

3
3 −1 1 

3 −1 3
3


A=


[
v0 =
−3 1 1 −2
]
Here, by power iteration we mean the following simple procedure:
v0
is the initial vector
vk+1 =
(a)
v6T
[
=
[
Avk
||Avk ||2
k≥0
−0.59035 0.41142 −0.14316 −0.6795
]
]
✓
0.45258 0.45453 −0.15816 −0.75071 (−100%)
]
T
(c) v6 = 0.26535 0.8528 −0.09273 −0.44014 (−100%)
[
]
(d) v6T = 0.19698 0.63309 0.67352 −0.32675 (−100%)
(b) v6T =
[
23
dolg0514/quadreg
1. quadreg
multi
1.0 point 0.10 penalty Single Shuffle Determine the best approximating (in the sense of least squares) second
order polynomial for the data below.
t
f
0
9 −9 −9 9 −3 3
−4 4
2
8
8 −6 7
671 2
(a) 7542
t + 61 t − 684
✓
419
(b) 63691
t2 − 37
t − 5096
(−100%)
37710
30
2095
11729 2
31
5934
(c) − 37710 t − 30 t − 2095 (−100%)
(d) 63691
t2 + 29
t − 6353
(−100%)
37710
30
2095
1
154
64523 2
(e) − 37710 t − 30 t + 419 (−100%)
(f) − 41897
t2 + 41
t − 3839
(−100%)
37710
30
2095
dolg0514/quadregVal
1. quadregVal
numerical
1.0 point 0.10 penalty Evaluate at 11
the best approximating (in the sense of least squares)
2
second order polynomial for the data below.
t
7
7
6
f
7 −4 −4 −1 7 5 −2
• 2.239964957 ± 1e-4 ✓
dolg0514/ray
1. ray
24
0
5 6
2
multi
1.0 point 0.10 penalty Single Shuffle 
For the matrix


A=



0 −1 1 −3

1 −3 2 −3 

3
3
3 −3 

−1 −3 −3 0
and vector
T
v =
[
3 −1 0 3
]
find the λ for which
||Av − λv||2
is as small as possible.
(a)
(b)
(c)
(d)
(e)
(f)
λ = −1.10526 ✓
λ = −1.06376 (−100%)
λ = −1.05602 (−100%)
λ = −1.08152 (−100%)
λ = −1.09572 (−100%)
There is no such λ. (−100%)
dolg0514/recipreg
1. recipreg
multi
1.0 point 0.10 penalty Single Shuffle Determine the parameters a, b of the best approximating function of
type
a
+b
t
for the data below.
t
4 5 −2 −2 −4 2
f
2 6 −6
25
1
−4 4
(a)
(b)
(c)
(d)
(e)
(f)
a=
a=
a=
a=
a=
a=
49
, b = 109
✓
6
120
167
37
, b = 120 (−100%)
30
281
83
, b = − 120
(−100%)
30
349
299
, b = 120 (−100%)
30
263
13
, b = 120
(−100%)
30
137
373
, b = 120 (−100%)
30
dolg0514/recipregVal
1. recipregVal
numerical
1.0 point 0.10 penalty Evaluate at 5 the best approximating function of type
a
+b
t
for the data below.
t
f
−9 −8
0
3
4
1 −4
10 8
9
• 6.15 ± 1e-4 ✓
dolg0514/rendszer
1. rendszer
multi
1.0 point 0.10 penalty Single Shuffle In the floating point number system
F = [a = 2, k− = −4, k+ = 3, t = 4]
the quantities 1+ M∞ ε1 are:
26
(a)
(b)
(c)
(d)
(e)
(f)
9
8
9
16
9
4
9
8
9
16
9
8
15
2
15
2
15
4
15
4
15
15
2
1
8
1
4
1
8
1
8
1
8
1
4
✓
(−100%)
(−100%)
(−100%)
(−100%)
(−100%)
dolg0514/trigreg
1. trigreg
multi
1.0 point 0.10 penalty Single Shuffle Determine the best approximating (in the sense of least squares) f
function of type
f (t) = a cos(πt) + b
for the data below.
t
13
2
−12
3
2
3
f
2
4
− 12
5
2
(a) 34 cos(πt) + 25
✓
14
7
141
(b) 4 cos(πt) − 70 (−100%)
1
(c) − 41
cos(πt) − 70
(−100%)
20
3
47
(d) 20 cos(πt) − 14 (−100%)
(e) 51
cos(πt) − 113
(−100%)
20
70
31
377
(f) 20 cos(πt) + 70 (−100%)
dolg0514/trigregVal
1. trigregVal
numerical
1.0 point 0.10 penalty 27
3
2
7
2
9
2
− 12
−2
3
Evaluate at 9 the best approximating (in the sense of least squares) f
function of type
f (t) = a cos(πt) + b
for the data below.
t
0 1
f
3 1
13
2
− 52
• -1.333333333 ± 1e-4 ✓
28
1
2
1
2
13
2
− 19
2
−3
−1