Question

o fit the best Curve fit among the following formulas for the given datain >C: 2 3 5 الدكتور سلمان حسین عباس 1.5 2 3.5 Y: 0.5 - loo 25 33 6.3 50 13.8 33.3 using the following methods. @ Lineur; second degree polynominal,c exponentials ② kit a straight line to the given data: 85 4.5 27 ② fit the formula Y= a + bx² to the given data: so 70 85 los 2700 2800 2920 3loo 9 Fit a power formula to the given datain 58 108 150 228 778 225 365 678 4332 CP 35 Y : 2620 Y: 88

Answer 1

(1)

	x	y	T	T	T	T	$x^2y$	$Y=ln(y)$	T
	1	0.5	0.5	1	1	1	0.5	-0.69315	-0.69315
	2	1	2	4	8	16	4	0	0
	3	1.5	4.5	9	27	81	13.5	0.405465	1.216395
	4	2	8	16	64	256	32	0.693147	2.772589
	5	3.5	17.5	25	125	625	87.5	1.252763	6.263815
total	15	8.5	32.5	55	225	979	137.5	1.658228	9.559652

To fit a linear curve. n=5

We need to fit the curve of the type:

$y=ax+b$

So the normal Equations are given as

$\\\sum y=a\sum x+nb \\\sum xy=a\sum x^2+b\sum x\\ \\i.e. \begin{bmatrix} \sum y\\ \sum xy \end{bmatrix}=\begin{bmatrix} \sum x &n \\ \sum x^2& \sum x \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix} \cdots (1)$

We have to solve for a and b by substituting the values

$\begin{bmatrix} \8.5\\ \32.5 \end{bmatrix}=\begin{bmatrix} \15 &5\\ \55& \15 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix} \\ R_1\rightarrow R_1 /5\, \& \,R_2\rightarrow R_2/5\\ \begin{bmatrix} \1.7\\ \6.5 \end{bmatrix}=\begin{bmatrix} \3 &1\\ \11& \3 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}$

On solving by Cramer's rule, we get

$a=\frac{\begin{vmatrix} 1.7& 1\\ 6.5&3 \end{vmatrix}}{\begin{vmatrix} 3 &1 \\ 11&3 \end{vmatrix}}=\frac{-1.4}{-2}=0.7$

and $b=\frac{\begin{vmatrix} 3& 1.7\\ 11&6.5 \end{vmatrix}}{\begin{vmatrix} 3 &1 \\ 11&3 \end{vmatrix}}=\frac{0.8}{-2}=-0.4$

Thus the linear curve is $y=0.7x-0.4$

Now to fit a second-degree polynomial

Now the normal equations are ,

$\\\sum y= a\sum x^2+b\sum x+nc \\ \sum xy=a\sum x^3+b\sum x^2+c\sum x\\ \sum x^2y=a\sum x^4 +b\sum x^3+c\sum x^2 \\ \\ i.e. \begin{bmatrix} \sum x^2&\sum x &n \\ \sum x^3& \sum x^2&\sum x \\ \sum x^4& \sum x^3 & \sum x^2 \end{bmatrix}\begin{bmatrix} a\\ b\\ c \end{bmatrix}=\begin{bmatrix} \sum y\\ \sum xy \\ \sum x^2y \end{bmatrix}$

We substitute the values and solve for a,b and c22

$\begin{bmatrix} \55&15 &5 \\ \225& 55&15\\ \979&225 &55 \end{bmatrix}\begin{bmatrix} a\\ b\\ c \end{bmatrix}=\begin{bmatrix} 8.5\\ 32.5 \\137.5\end{bmatrix}$

We write the systen as Augmented matrix and solve by gauss elimination method for a ,b and c.

55 15 5 8.5 225 55 15 32.5 979 225 55 | 137.5

Perform$R_1 \rightarrow R_1/5$

--: 1 3 17 11 11 110 225 55 13 32.5 979 225 55 | 137.5

Peform $R_2 \rightarrow R_2-225 R_1$ and $R_3 \rightarrow R_3-979 R_1$

3 1 17 1 110 25 11 11 70 60 0 11 11 0 -42 -34 11 -13.8

Now, Perform$R_2 \rightarrow \frac{-11}{70}R_2$

3 17. 1 1 11 11 110 6 5 0 1 14 7 0 -42 -34 | -13.8

Now perform

3 R1 + R1 - R2 and R3 R3 + 42R2 11

1 2 1 0 7 35 5 6 0 1 7 0 0 2 14 1.2

Perform$R_3 \rightarrow R_3 /2$

1 2 1 0 7 35 6 5 0 1 7 14 0 0 1 0.6

Perform$R_2 \rightarrow R_2-\frac{ 6}{7}R_3 \,\&\,R_1\rightarrow R_1 +\frac{1}{7}R_3$

1 10 0 7 11 010 70 001 0.6

Thus $a=\frac{1}{7} ; b=\frac{-11}{70}; c=0.6$

Thus the required second-degree polynomial is $y= \frac{x^2}{7}+\frac{-11x}{7}+0.6$

To fit an exponential curve :

It will be of the type$y=ab^{x}$

Consider

Now put

In(y) = Y &In(a) = A, In(b) = B

thus

= A + BC

Now the normal equations are

$\\\sum Y=B\sum x+nA \\\sum xY=B\sum x^2+A\sum x\\ \\i.e. \begin{bmatrix} \sum Y\\ \sum xY \end{bmatrix}=\begin{bmatrix} \sum x &n \\ \sum x^2& \sum x \end{bmatrix}\begin{bmatrix} B\\ A \end{bmatrix} \cdots (2)$

we put the values and solve for A and B

$\begin{bmatrix} \1.658228 \\ 9.559652 \end{bmatrix}=\begin{bmatrix} 15 &5 \\ 55&15 \end{bmatrix}\begin{bmatrix} B\\ A \end{bmatrix} \cdots (2)$

on solving we get

Perform

Rj/15

414557 1 3 3750000 55 15 9.559652

perform

R2 - 55 R1

414557 نم مانيا 3750000 652403 10 0 ملا 187500

perform

10 R2/ 3

1 414557 1 3 3750000 0 1 -1.0438448

perform

Ri R2 3

10| 0.4584968 0 1|-1.0438448

Thus

$B=0.4584968\, \&\, A=-1.04388448 \\ \Rightarrow ln(b)=0.4584968\, \&\, ln(a)=-1.04388448 \\ \Rightarrow b=e^{0.4584968}\, \&\, a=e^{-1.04388448} \\\Rightarrow b=1.581695 \, \&\, a=0.352084\\ y=(0.352084)(1.581695)^x$

2)

х у XX ху 112.5 25 4.5 33 6.3 625 1089 2500 50 85 207.9 690 2295 3330 6635.4 13.8 27 33.3 84.9 7225 10000 100 293 Total 21439

n=5

$\sum x ^{2}=21439$

to Fit a straight line means to fit at curve of the type $y=ax+b$

So the normal Equations are given as

$\\\sum y=a\sum x+nb \\\sum xy=a\sum x^2+b\sum x\\ \\i.e. \begin{bmatrix} \sum y\\ \sum xy \end{bmatrix}=\begin{bmatrix} \sum x &n \\ \sum x^2& \sum x \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix} \cdots (1)$

$\begin{bmatrix} \86.9\\ 6635.4 \end{bmatrix}=\begin{bmatrix} 293 &5 \\ 21439& 293 \end{bmatrix}\begin{bmatrix} a\\ b \end{bmatrix}$

Similarly solving by Cramer's rule we get

$a=\frac{77153}{213460}=0.36144$

$b =\frac{-811231}{213460}=-3.80038$

thus we get.

$y=0.36144x-3.80038$

3)

X У XX 35 2620 50 2700 70 XXX ху 1225 42875 91700 2500 125000 135000 4900 343000 196000 7225 614125 248200 11025 1157625 328650 26875 2282625 999550 2800 2920 3130 85 105 345 total 14170

Required equation is $y=a+bx^2$

$\\\sum y=b\sum x^2+na \\\sum xy=b\sum x^3+a\sum x\\ \\i.e. \begin{bmatrix} \sum y\\ \sum xy \end{bmatrix}=\begin{bmatrix} \sum x^2 &n \\ \sum x^3& \sum x \end{bmatrix}\begin{bmatrix} b\\ a \end{bmatrix} \cdots (2)$

$.\begin{bmatrix} 14170\\ 999550 \end{bmatrix}=\begin{bmatrix} 26875 &5 \\ 2282625& 345\end{bmatrix}\begin{bmatrix} b\\ a \end{bmatrix} \cdots (2)$

on solving by cramers rule

b=
2182 42825
= 0.05095 and a =
4385512 1713
=2560.13543

$y=+0.05095x^2+2560.13543$

4 problem can be done in same manner as exponential done in part 1