2. Fitting second degree parabola - Curve fitting Example-2

Calculate Fitting second degree parabola - Curve fitting using Least square method
X Y
1996 40
1997 50
1998 62
1999 58
2000 60

Solution:
The equation is

y=a+bx+cx2 $y = a + bx + cx^2$ and the normal equations are

∑y=an+b∑x+c∑x2 $sum y = an + b sum x + c sum x^2$

∑xy=a∑x+b∑x2+c∑x3 $sum xy = a sum x + b sum x^2 + c sum x^3$

∑x2y=a∑x2+b∑x3+c∑x4 $sum x^2y = a sum x^2 + b sum x^3 + c sum x^4$

X $X$	y $y$	x=X-1998 $x = X - 1998$	x2 $x^2$	x3 $x^3$	x4 $x^4$	x⋅y $x*y$	x2⋅y $x^2*y$
1996	40	-2	4	-8	16	-80	160
1997	50	-1	1	-1	1	-50	50
1998	62	0	0	0	0	0	0
1999	58	1	1	1	1	58	58
2000	60	2	4	8	16	120	240
---	---	---	---	---	---	---	---
9990	270	0	10	0	34	48	508

Substituting these values in the normal equations

270=5a+0b+10c $270=5a+0b+10c$

48=0a+10b+0c $48=0a+10b+0c$

508=10a+0b+34c $508=10a+0b+34c$

Solving these 3 equations using inverse matrix method,
Here

5a+10c=270 $5a+10c=270$

10b=48 $10b=48$

10a+34c=508 $10a+34c=508$

Now converting given equations into matrix form

[5010010010034][abc]=[27048508] $[[5,0,10],[0,10,0],[10,0,34]] [[a],[b],[c]]=[[270],[48],[508]]$

Now, A =

[5010010010034] $[[5,0,10],[0,10,0],[10,0,34]]$ , X =

[abc] $[[a],[b],[c]]$ and B =

[27048508] $[[270],[48],[508]]$

∴AX=B $:. AX = B$

∴X=A-1B $:. X = A^-1 B$

|A| $|A|$

5 $5$	0 $0$	10 $10$
0 $0$	10 $10$	0 $0$
10 $10$	0 $0$	34 $34$

5 $5$

	10 $10$	0 $0$
	0 $0$	34 $34$

+0 $+0$

	0 $0$	0 $0$
	10 $10$	34 $34$

+10 $+10$

	0 $0$	10 $10$
	10 $10$	0 $0$

=5×(10×34-0×0)+0×(0×34-0×10)+10×(0×0-10×10) $=5 xx (10 × 34 - 0 × 0) +0 xx (0 × 34 - 0 × 10) +10 xx (0 × 0 - 10 × 10)$

=5×(340+0)+0×(0+0)+10×(0-100) $=5 xx (340 +0) +0 xx (0 +0) +10 xx (0 -100)$

=5×(340)-+0×(0)+10×(-100) $=5 xx (340) - +0 xx (0) +10 xx (-100)$

=1700+0-1000 $= 1700 +0 -1000$

=700 $=700$

Here, |A|=700≠0 $"Here, " |A| = 700 != 0$

∴A-1 is possible. $:. A^(-1) " is possible."$

Adj(A) $Adj(A)$

Adj

5 $5$	0 $0$	10 $10$
0 $0$	10 $10$	0 $0$
10 $10$	0 $0$	34 $34$

	10 $10$	0 $0$
	0 $0$	34 $34$

	0 $0$	0 $0$
	10 $10$	34 $34$

	0 $0$	10 $10$
	10 $10$	0 $0$

	0 $0$	10 $10$
	0 $0$	34 $34$

	5 $5$	10 $10$
	10 $10$	34 $34$

	5 $5$	0 $0$
	10 $10$	0 $0$

	0 $0$	10 $10$
	10 $10$	0 $0$

	5 $5$	10 $10$
	0 $0$	0 $0$

	5 $5$	0 $0$
	0 $0$	10 $10$

+(10×34-0×0) $+(10 × 34 - 0 × 0)$	-(0×34-0×10) $-(0 × 34 - 0 × 10)$	+(0×0-10×10) $+(0 × 0 - 10 × 10)$
-(0×34-10×0) $-(0 × 34 - 10 × 0)$	+(5×34-10×10) $+(5 × 34 - 10 × 10)$	-(5×0-0×10) $-(5 × 0 - 0 × 10)$
+(0×0-10×10) $+(0 × 0 - 10 × 10)$	-(5×0-10×0) $-(5 × 0 - 10 × 0)$	+(5×10-0×0) $+(5 × 10 - 0 × 0)$

+(340+0) $+(340 +0)$	-(0+0) $-(0 +0)$	+(0-100) $+(0 -100)$
-(0+0) $-(0 +0)$	+(170-100) $+(170 -100)$	-(0+0) $-(0 +0)$
+(0-100) $+(0 -100)$	-(0+0) $-(0 +0)$	+(50+0) $+(50 +0)$

340 $340$	0 $0$	-100 $-100$
0 $0$	70 $70$	0 $0$
-100 $-100$	0 $0$	50 $50$

340 $340$	0 $0$	-100 $-100$
0 $0$	70 $70$	0 $0$
-100 $-100$	0 $0$	50 $50$

Now, A-1=1|A|×Adj(A) $"Now, "A^(-1)=1/|A| × Adj(A)$

Here, X=A-1×B $"Here, "X = A^(-1) × B$

∴X=1|A|×Adj(A)×B $:. X = 1/|A| × Adj(A) × B$

1700 $1/(700)$ ×

340 $340$	0 $0$	-100 $-100$
0 $0$	70 $70$	0 $0$
-100 $-100$	0 $0$	50 $50$

	270 $270$
	48 $48$
	508 $508$

1700 $1/(700)$ ×

	340×270+0×48-100×508 $340×270+0×48-100×508$
	0×270+70×48+0×508 $0×270+70×48+0×508$
	-100×270+0×48+50×508 $-100×270+0×48+50×508$

1700 $1/(700)$ ×

	41000 $41000$
	3360 $3360$
	-1600 $-1600$

	4107 $410/7$
	245 $24/5$
	-167 $-16/7$

∴[abc]=[4107245-167] $:.[[a],[b],[c]]=[[410/7],[24/5],[-16/7]]$

∴a=4107,b=245,c=-167 $:. a=410/7, b=24/5, c=-16/7$

Now substituting this values in the equation is

y=a+bx+cx2 $y = a + bx + cx^2$ , we get

y=4107+245x-167x2 $y = 410/7 +24/5x-16/7x^2$

y=4107+245(X-1998)-167(X-1998)2 $y = 410/7 +24/5(X-1998)-16/7(X-1998)^2$

This material is intended as a summary. Use your textbook for detail explanation.
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2. Example-2