Unit - 4

Curve Fitting and Regression Analysis

Q1) Fit a straight line to   the following data regarding x as   the independent variables:

A1)

The   equation of straight line is

And the normal equations

We construct the data table:

X	Y	XY
0 1 2 3 4	1 1.8 3.3 4.5 6.3	0 1.8 6.6 13.5 25.2	0 1 4 9 16
Total	16.9

Here    (no. Of steps)

Substituting the values from table in normal equations:

16.9=5a+10b

47.1=10a+30b

On solving we getand

Therefore, the required equation of the straight line is .

Q2) Find the straight line that best fits of the following data by using method of least square.

A2)

Suppose the straight line

y = a + bx…….. (1)

Fits the best-

Then-

x	y	Xy
1	14	14	1
2	27	54	4
3	40	120	9
4	55	220	16
5	68	340	25
Sum = 15	204	748	55

Normal equations are-

Put the values from the table, we get two normal equations-

On solving the above equations, we get-

So that the best fit line will be- (on putting the values of a and b in equation (1))

Q3) Fit a second-degree parabola to the following data regarding x as an independent variable:

A3)

The equation of second-degree curve is

The normal equations are

We construct the data table:

X	Y
0 1 2 3 4	1 5 10 22 38	0 1 4 9 16	0 1 8 27 64	0 1 16 81 256	0 5 20 66 152	0 5 40 198 608
Total	= 76	30	100	354	243	851

Substituting these   values   from   the table in the above equations

On solving we get and

Therefore, equation of parabola is

Q4) Find the best values of a and b so that y = a + bx fits the data given in the table

A4)

y = a + bx

x	y	Xy
0	1.0	0	0
1	2.9	2.0	1
2	4.8	9.6	4
3	6.7	20.1	9
4	8.6	13.4	16
x = 10	y= 24.0	xy = 67.0

Normal equations, y= na+ bx (2)

On putting the values of

On solving (4) and (5) we get,

On substituting the values of a and b in (1) we get

Q5) Predict y at x = 3.75 by fitting a power curve  to the given data
 

A5)

Given equation is

Taking log on both sides,

Let    

Therefore, its normal equation is

We construct the data table:

				XY
1 2 3 4 5 6	2.98 4.26 5.21 6.10 6.80 7.50	0 0.301030 0.477121 0.602060 0.698970 0.778151	0.474216 0.629410 0.716838 0.785330 0.832509 0.875061	0 0.189471 0.342018 0.472816 0.581899 0.680930	0 0.090619 0.227644 0.362476 0.488559 0.605519
Total

Substituting the   values from the   table in the above equations

On solving we get

Hence the required equation is

So, 

Q6) Fit a second-degree parabola to the following data by least squares method.

	1929	1930	1931	1932	1933	1934	1935	1936	1937
	352	356	357	358	360	361	361	360	359

A6)

Taking

Taking 

The equation is transformed to

1929	-4	352	-5	20	16	-80	-64	256
1930	-3	360	-1	3	9	-9	-27	81
1931	-2	357	0	0	4	0	-8	16
1932	-1	358	1	-1	1	1	-1	1
1933	0	360	3	0	0	0	0	0
1934	1	361	4	4	1	4	1	1
1935	2	361	4	8	4	16	8	16
1936	3	360	3	9	9	27	27	81
1937	4	359	2	8	16	32	64	256
Total

Normal equations are

On solving these equations, we get

Q7) Find the least squares approximation of second degree for the discrete data

A7)

Let the equation of second-degree polynomial be

x	y	Xy
-2	15	-30	4	60	-8	16
-1	1	-1	1	1	-1	1
0	1	0	0	0	0	0
1	3	3	1	3	1	1
2	19	38	4	76	8	16
x=0	y=39	xy=10

Normal equations are

On putting the values of x, y, xy, have

On solving (5), (6), (7), we get,

The required polynomial of second degree is

Q8) Fit a second-degree parabola to the following data.

A8)

We shift the origin to (2.5, 0) antique 0.5 as the new unit. This amounts to changing the variable x to X, by the relation X = 2x – 5.

Let the parabola of fit be y = a + bXThe values of X etc. Are calculated as below:

x	X	y	Xy
1.0	-3	1.1	-3.3	9	9.9	-27	81
1.5	-2	1.3	-2.6	4	5.2	-5	16
2.0	-1	1.6	-1.6	1	1.6	-1	1
2.5	0	2.0	0.0	0	0.0	0	0
3.0	1	2.7	2.7	1	2.7	1	1
3.5	2	3.4	6.8	4	13.6	8	16
4.0	3	4.1	12.3	9	36.9	27	81
Total	0	16.2	14.3	28	69.9	0	196

The normal equations are

7a + 28c =16.2; 28b =14.3; 28a +196c=69.9

Solving these as simultaneous equations we get

Replacing X bye 2x – 5 in the above equation we get

Which simplifies to y =

This is the required parabola of the best fit.

Q9) Create least   square regression line of set of points

A9)

Here is number of data points.

The simple linear regression is given by

Where

Here which is straight line.

Q10) Find the linear square regression  for the following data.   Also find y when

A10)

Here, number of data given. 

The simple linear regression is given by

Where

And

Hence the straight line is given by

Also 

Q11) What do you understand by linear multiple regression?

A11)

Let Y depends on two independent variables X1 and X2,

Then the linear multiple regression problem is to fit the regression plane given by Equation (1) to a

Given set of N triples.

To estimate the coefficient 0, 1, 2, apply the least squares method to minimize

This results in three normal equations given by

Here b0, b1, b2 are the least squares estimates of 0, 1, 2.

Q12) What is the Lagrange’s interpolation formula?

A12)

Let , be defined function we get

X				…..
f(x)				……

Where the interval is not necessarily equal. We assume f(x) is a polynomial of degree n.  Then Lagrange’s interpolation formula is given by

Q13) By means of Lagrange’s formula, prove that

A13)

We construct the table:

X	0	1	2	3	4	5	6
Y=f(x)

Here x = 3, f(x)=?

By Lagrange’s formula for   interpolation

Hence proved.

Q14) find the polynomial of fifth degree from the following data

A14)

Here

We get 

By Lagrange’s interpolation formula

Q15) What do you understand by interpolation?

A15)

Interpolation is a technique of estimating the value of a function for any intermediate value of the independent variable while the process of computing the value of the function outside the given range is called   extrapolation.

Let be a function of x.

The table given below gives corresponding values of y for different values of x.

X				….
y= f(x)				….

The process of finding the values of y corresponding to any value of x which lies between   is called interpolation.

If the given function is a polynomial, it is polynomial interpolation and given function is known as interpolating polynomial.

Conditions for Interpolation

1)     The function must be   a polynomial of independent variable.

2)     The function   should be either increasing or decreasing function.

3)     The value of   the function should be increase or decrease uniformly.

Q16) Using Newton’s forward difference formula, find the sum

A16)

Putting

It   follows that

Since is a fourth-degree polynomial in n.

Further,

By Newton Forward Difference Method

Q17) Given  find   , by using Newton forward interpolation method.

A17)

Let  , then

	0.7071	0.7660	-	0.8192	0.8660

The table of forward finite difference is given below:

45           50           55            60	0.7071            0.7660           0.8192            0.8660	0.0589          0.0532          0.0468	-0.0057        -0.0064	-0.0007

By Newton forward difference method

Here initial value = 45, difference of interval h = 5 and the value to be calculated at x=52.

By Formula

Q18) Find from the following table:

	0.20	0.22	0.24	0.26	0.28	0.30
	1.6596	1.6698	1.6804	1.6912	1.7024	1.7139

A18)

Consider the backward difference method

0.20         0.22         0.24         0.26         0.28         0.30	1.6596      1.6698      1.6804      1.6912     1.7024       1.7139	0.0102       0.0106      0.0108      0.0112     0.0115	0.0004    0.0002    0.0004    0.0003	-0.0002    0.0002   -0.0001	0.0004   -0.0003	-0.0007

Here

By Newton backward difference formula

Q19) The following are the marks obtained by 492 candidates in a certain examination

Find out the number of candidates:

a)     Who secured more than 48 but not more than 50 marks?

b)    Who secured less than 48 but not less than 45 marks?

A19)

Consider the forward difference table given below:

Marks upto x	No. Of candidates y
40       45       50       55       60       65	210   210+43=253   253+54=307   307+74=381   381+32=413   413+79= 492	43       54       74       32        79	11     20     -42      47	9     -62      89	-71        151	222

Here 

By Newton Forward Difference formula

f

a)     No. Of candidate secured more than 48 but not more than 50 marks

b)    No. Of candidate secured less than 48 but not less than 45 marks

Q20) Use the inverse interpolation to find value of x at for the following data:

A20)

Here , we have the data

The Lagrange’s inverse interpolation formula is given by

.