This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Approximation of Functions using Least Square Method”.
1. Fit the straight line to the following data.
a) y=x
b) y=x+1
c) y=2x
d) y=2x+1
Answer: a
Explanation: The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx2 + bΣx
Now,
x | y | x2 | xy |
1 | 1 | 1 | 1 |
2 | 2 | 4 | 4 |
3 | 3 | 9 | 9 |
4 | 4 | 16 | 16 |
5 | 5 | 25 | 25 |
Σx = 15 | Σy = 15 | Σx2 = 55 | Σxy = 55 |
Substituting in the equations,
15 = 15a + 4b and 55 = 55a + 15b
Solving these two equatons, we get a=1 and b=0,
Therefore the required straight line equation is y=x.
2. Fit the straight line to the following data.
a) y = 0.94x + 6.6
b) y = 6.6x + 0.94
c) y = 0.04x + 5.6
d) y = 5.6x + 0.04
Answer: a
Explanation: First drawing the table,
x | y | x2 | xy |
0 | 7 | 0 | 0 |
5 | 11 | 25 | 55 |
10 | 16 | 100 | 160 |
15 | 20 | 225 | 300 |
20 | 26 | 400 | 520 |
Σx = 50 | Σy = 26 | Σx2 = 750 | Σxy = 1035 |
The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx2 + bΣx.
Substituting the values, we get,
80 = 50a + 5b
1035 = 750 a + 50 b
Solving them, we get a = 0.94 and b = 6.6.
Therefore the straight line equation is y=0.94x + 6.6.
3. Fit the straight line curve to the following data.
x | 75 | 80 | 93 | 65 | 87 | 71 | 98 | 68 | 84 | 77 |
y | 82 | 78 | 86 | 72 | 91 | 80 | 95 | 72 | 89 | 74 |
a) y = 0.9288x + 7.78155
b) y = 7.78155x + 0.9288
c) y = 0.8288x + 6.78155
d) y = 6.78155x + 0.8288
Answer: a
Explanation: First drawing the table,
x | y | x2 | xy |
75 | 82 | 5625 | 6150 |
80 | 78 | 6400 | 6240 |
93 | 86 | 8349 | 7998 |
65 | 72 | 4225 | 4680 |
87 | 91 | 7569 | 7917 |
71 | 80 | 5041 | 5680 |
98 | 95 | 9605 | 9310 |
68 | 72 | 4624 | 4896 |
84 | 89 | 7056 | 7476 |
77 | 74 | 5929 | 5698 |
798 | 819 | 64422 | 66045 |
The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx2 + bΣx.
Substituting the values, we get,
819 = 798a + 10b
66045 = 64422a + 798b
Solving, we get
a = 0.9288 and b = 7.78155
Therefore, the straight line equation is :
y = 0.9288x + 7.78155.
4. Fit a second degree parabola to the following data.
a) y = -0.2673x2 + 3.5232x – 0.9286
b) y = 0.2673x2 + 3.5232x – 0.9286
c) y = 0.2673x2 + 3.5232x + 0.9286
d) y = -0.2673x2 + 3.5232x + 0.9286
Answer: a
Explanation: Here,
x | y | x2 | x3 | x4 | xy | x2y |
1 | 2 | 1 | 1 | 1 | 2 | 2 |
2 | 6 | 4 | 8 | 16 | 12 | 24 |
3 | 7 | 9 | 27 | 81 | 21 | 63 |
4 | 8 | 16 | 64 | 256 | 32 | 128 |
5 | 10 | 25 | 125 | 625 | 50 | 250 |
6 | 11 | 36 | 216 | 1296 | 66 | 396 |
7 | 11 | 49 | 343 | 2401 | 77 | 539 |
8 | 10 | 64 | 512 | 4096 | 80 | 640 |
9 | 9 | 81 | 729 | 6561 | 81 | 729 |
45 | 74 | 285 | 2025 | 15333 | 421 | 2771 |
The normal equations are:
Σy = aΣx2 + bΣx + nc
Σxy = aΣx3 + bΣx2 +cΣx
Σx2y = aΣx4 + bΣx3 + cΣx2
Substituting the values, we get
74 = 285a + 45b + 9c
421 = 2025 a + 285 b + 45 c
2771 = 15333a + 2025 b + 285 c
Solving them, we get the second order equation which is,
y = -0.2673x2 + 3.5232x – 0.9286.
5. The normal equations for a straight line y = ax + b are:
a) Σy = aΣx + nb and Σxy = aΣx2 + bΣx
b) Σxy = aΣx + nb and Σy = aΣx2 + bΣx
c) Σy = aΣx + nb and Σxy = aΣx2 + bΣxy
d) Σy = aΣx + nb and Σx2y = aΣx2 + bΣx
Answer: a
Explanation: Let the sum of residues be E.
E = Σdi2 = Σ(yi – (axi+b))2
Here (frac{∂E}{∂a})=0 and (frac{∂E}{∂b})=0
Solving these two equations, we get the normal equations as
Σy = aΣx + nb and Σxy = aΣx2 + bΣx.
6.The normal equations for a second degree parabola y = ax2 + bx + c are Σy = aΣx2 + bΣx + nc, Σxy = aΣx3 + bΣx2 + cΣx and Σx2y = aΣx4 + bΣx3 + cΣx2.. Is it true or false?
a) True
b) False
Answer: a
Explanation: The second order parabola is given by
y = ax2 + bx +c.
Let the sum of residues be E.
E = Σdi2 = Σ(yi – (axi2 + bxi +c))2.
Here (frac{∂E}{∂a})= 0, (frac{∂E}{∂b})= 0 and (frac{∂E}{∂c})= 0.
Solving these three equations, we get the normal equations as
Σy = aΣx2 + bΣx + nc, Σxy = aΣx3 + bΣx2 + cΣx and Σx2y = aΣx4 + bΣx3 +cΣx2.
7. If the equation y = aebx can be written in linear form Y=A + BX, what are Y, X, A, B?
a) Y = logy, A = loga, B=b and X=x
b) Y = y, A = a, B=b and X=x
c) Y = y, A = a, B=logb and X=logx
d) Y = logy, A = a, B=logb and X=x
Answer: a
Explanation: The equation is
y = aebx.
Taking log to the base e on both sides,
we get logy = loga + bx.
Which can be replaced as Y=A+BX,
where Y = logy, A = loga, B = b and X = x.
8. If the equation y=abx can be written in linear form Y=A+BX, what are Y, X, A, B?
a) Y=logy, X=x, A=loga and B=logb
b) Y=y, A=a, B=b and X=x
c) Y=y, A=a, B=logb and X=logx
d) Y=logy, A=a, B=logb and X=x
Answer: a
Explanation: The given curve is y=abx.
Taking log on bothe the sides, we get,
logy = loga + x logb.
This can be written in the format of Y=A+BX
where
Y = logy, X = x, A = loga and B = logb.
9. If the equation y=axb can be written in the linear form Y=A+BX, what are Y, X, A, B?
a) Y=logy, A=loga, B=b and X=logx
b) Y=y, A=a, B=b and X=x
c) Y=y, A=a, B=logb and X=logx
d) Y=logy, A=a, B=logb and X=x
Answer: a
Explanation: The given curve is y=axb.
Taking log on bothe the sides, we get,
logy = loga + blogx.
This can be written as Y=A+BX,
where
Y=logy, A=loga, B=b and X=logx.
10. The parameter E which we use for least square method is called as ____________
a) Sum of residues
b) Residues
c) Error
d) Sum of errors
Answer: a
Explanation: E is given by
E = Σdi2 = Σ(yi – f(xi))2.
Where the term inside the summation is called as residues and the sum is said to be a sum of residues.
Therefore, E is said to be the sum of residues.