Test: Approximation of Functions using Least Square Method - Civil Engineering (CE) MCQ

First drawing the table,

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.

The normal equation are:
Σy = aΣx + nb
and
Σxy = aΣx2 + bΣx
Now,

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.

Here,

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.

First drawing the table,

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.

The equation is
y = ae^bx.
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.

Let the sum of residues be E.

Solving these two equations, we get the normal equations as
Σy = aΣx + nb and Σxy = aΣx² + bΣx.

The given curve is y=ax^b.
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.

The given curve is y = ab^x.
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.

The second order parabola is given by
y = ax² + bx +c.
Let the sum of residues be E.
E = Σd_i² = Σ(y_i – (ax_i² + bxi +c))².
Here 
Solving these three equations, we get the normal equations as
Σy = aΣx² + bΣx + nc, Σxy = aΣx³ + bΣx² + cΣx and Σx²y = aΣx⁴ + bΣx³ +cΣx².

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.