Fitting cubic equation - Curve fitting calculator

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Solution

Solution provided by AtoZmath.com

1. Fit a straight line y = a + bx using the following data

x	5	4	3	2	1
y	1	2	3	4	5

2. Fit a straight line to the following data on production.

Year	1996	1997	1998	1999	2000
Production	40	50	62	58	60

3. Fit second degree parabola equation `y = a + bx + cx^2` using the following data.

x	12	10	8	6	4	2
y	6	5	4	3	2	1

4. Fit cubic equation `y = a + bx + cx^2 + dx^3` using the following data.

x	12	10	8	6	4	2
y	6	5	4	3	2	1

Example

1. Calculate Fitting cubic equation - 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 + cx^2 + dx^3` and the normal equations are

`sum y = an + b sum x + c sum x^2 + d sum x^3`

`sum xy = a sum x + b sum x^2 + c sum x^3 + d sum x^4`

`sum x^2y = a sum x^2 + b sum x^3 + c sum x^4 + d sum x^5`

`sum x^3y = a sum x^3 + b sum x^4 + c sum x^5 + d sum x^6`

`X`	`y`	`x = X - 1998`	`x^2`	`x^3`	`x^4`	`x^5`	`x^6`	`x*y`	`x^2*y`	`x^3*y`
1996	40	-2	4	-8	16	-32	64	-80	160	-320
1997	50	-1	1	-1	1	-1	1	-50	50	-50
1998	62	0	0	0	0	0	0	0	0	0
1999	58	1	1	1	1	1	1	58	58	58
2000	60	2	4	8	16	32	64	120	240	480
---	---	---	---	---	---	---	---	---	---	---
9990	270	0	10	0	34	0	130	48	508	168

Substituting these values in the normal equations
`270=5a+0b+10c+0d`

`48=0a+10b+0c+34d`

`508=10a+0b+34c+0d`

`168=0a+34b+0c+130d`

Solving these 4 equations using inverse matrix method,
Here `5a+10c=270`
`10b+34d=48`
`10a+34c=508`
`34b+130d=168`

Now converting given equations into matrix form
`[[5,0,10,0],[0,10,0,34],[10,0,34,0],[0,34,0,130]] [[a],[b],[c],[d]]=[[270],[48],[508],[168]]`

Now, A = `[[5,0,10,0],[0,10,0,34],[10,0,34,0],[0,34,0,130]]`, X = `[[a],[b],[c],[d]]` and B = `[[270],[48],[508],[168]]`

`:. AX = B`

`:. X = A^-1 B`

`|A|`

`5`	`0`	`10`	`0`
`0`	`10`	`0`	`34`
`10`	`0`	`34`	`0`
`0`	`34`	`0`	`130`

`5`

`10`	`0`	`34`
`0`	`34`	`0`
`34`	`0`	`130`

`+0`

`0`	`0`	`34`
`10`	`34`	`0`
`0`	`0`	`130`

`+10`

`0`	`10`	`34`
`10`	`0`	`0`
`0`	`34`	`130`

`+0`

`0`	`10`	`0`
`10`	`0`	`34`
`0`	`34`	`0`

`=5 × (4896)+0 × (0)+10 × (-1440)+0 × (0)`

`=24480+0-14400+0`

`=10080`

`"Here, " |A| = 10080 != 0`

`:. A^(-1) " is possible."`

`Adj(A)`

Adj

`5`	`0`	`10`	`0`
`0`	`10`	`0`	`34`
`10`	`0`	`34`	`0`
`0`	`34`	`0`	`130`

`10`	`0`	`34`
`0`	`34`	`0`
`34`	`0`	`130`

`0`	`0`	`34`
`10`	`34`	`0`
`0`	`0`	`130`

`0`	`10`	`34`
`10`	`0`	`0`
`0`	`34`	`130`

`0`	`10`	`0`
`10`	`0`	`34`
`0`	`34`	`0`

`0`	`10`	`0`
`0`	`34`	`0`
`34`	`0`	`130`

`5`	`10`	`0`
`10`	`34`	`0`
`0`	`0`	`130`

`5`	`0`	`0`
`10`	`0`	`0`
`0`	`34`	`130`

`5`	`0`	`10`
`10`	`0`	`34`
`0`	`34`	`0`

`0`	`10`	`0`
`10`	`0`	`34`
`34`	`0`	`130`

`5`	`10`	`0`
`0`	`0`	`34`
`0`	`0`	`130`

`5`	`0`	`0`
`0`	`10`	`34`
`0`	`34`	`130`

`5`	`0`	`10`
`0`	`10`	`0`
`0`	`34`	`0`

`0`	`10`	`0`
`10`	`0`	`34`
`0`	`34`	`0`

`5`	`10`	`0`
`0`	`0`	`34`
`10`	`34`	`0`

`5`	`0`	`0`
`0`	`10`	`34`
`10`	`0`	`0`

`5`	`0`	`10`
`0`	`10`	`0`
`10`	`0`	`34`

`+(10 xx (34 × 130 - 0 × 0) +0 xx (0 × 130 - 0 × 34) +34 xx (0 × 0 - 34 × 34))`	`-(0 xx (34 × 130 - 0 × 0) +0 xx (10 × 130 - 0 × 0) +34 xx (10 × 0 - 34 × 0))`	`+(0 xx (0 × 130 - 0 × 34) -10 xx (10 × 130 - 0 × 0) +34 xx (10 × 34 - 0 × 0))`	`-(0 xx (0 × 0 - 34 × 34) -10 xx (10 × 0 - 34 × 0) +0 xx (10 × 34 - 0 × 0))`
`-(0 xx (34 × 130 - 0 × 0) -10 xx (0 × 130 - 0 × 34) +0 xx (0 × 0 - 34 × 34))`	`+(5 xx (34 × 130 - 0 × 0) -10 xx (10 × 130 - 0 × 0) +0 xx (10 × 0 - 34 × 0))`	`-(5 xx (0 × 130 - 0 × 34) +0 xx (10 × 130 - 0 × 0) +0 xx (10 × 34 - 0 × 0))`	`+(5 xx (0 × 0 - 34 × 34) +0 xx (10 × 0 - 34 × 0) +10 xx (10 × 34 - 0 × 0))`
`+(0 xx (0 × 130 - 34 × 0) -10 xx (10 × 130 - 34 × 34) +0 xx (10 × 0 - 0 × 34))`	`-(5 xx (0 × 130 - 34 × 0) -10 xx (0 × 130 - 34 × 0) +0 xx (0 × 0 - 0 × 0))`	`+(5 xx (10 × 130 - 34 × 34) +0 xx (0 × 130 - 34 × 0) +0 xx (0 × 34 - 10 × 0))`	`-(5 xx (10 × 0 - 0 × 34) +0 xx (0 × 0 - 0 × 0) +10 xx (0 × 34 - 10 × 0))`
`-(0 xx (0 × 0 - 34 × 34) -10 xx (10 × 0 - 34 × 0) +0 xx (10 × 34 - 0 × 0))`	`+(5 xx (0 × 0 - 34 × 34) -10 xx (0 × 0 - 34 × 10) +0 xx (0 × 34 - 0 × 10))`	`-(5 xx (10 × 0 - 34 × 0) +0 xx (0 × 0 - 34 × 10) +0 xx (0 × 0 - 10 × 10))`	`+(5 xx (10 × 34 - 0 × 0) +0 xx (0 × 34 - 0 × 10) +10 xx (0 × 0 - 10 × 10))`

`+(10 xx (4420 +0) +0 xx (0 +0) +34 xx (0 -1156))`	`-(0 xx (4420 +0) +0 xx (1300 +0) +34 xx (0 +0))`	`+(0 xx (0 +0) -10 xx (1300 +0) +34 xx (340 +0))`	`-(0 xx (0 -1156) -10 xx (0 +0) +0 xx (340 +0))`
`-(0 xx (4420 +0) -10 xx (0 +0) +0 xx (0 -1156))`	`+(5 xx (4420 +0) -10 xx (1300 +0) +0 xx (0 +0))`	`-(5 xx (0 +0) +0 xx (1300 +0) +0 xx (340 +0))`	`+(5 xx (0 -1156) +0 xx (0 +0) +10 xx (340 +0))`
`+(0 xx (0 +0) -10 xx (1300 -1156) +0 xx (0 +0))`	`-(5 xx (0 +0) -10 xx (0 +0) +0 xx (0 +0))`	`+(5 xx (1300 -1156) +0 xx (0 +0) +0 xx (0 +0))`	`-(5 xx (0 +0) +0 xx (0 +0) +10 xx (0 +0))`
`-(0 xx (0 -1156) -10 xx (0 +0) +0 xx (340 +0))`	`+(5 xx (0 -1156) -10 xx (0 -340) +0 xx (0 +0))`	`-(5 xx (0 +0) +0 xx (0 -340) +0 xx (0 -100))`	`+(5 xx (340 +0) +0 xx (0 +0) +10 xx (0 -100))`

`4896`	`0`	`-1440`	`0`
`0`	`9100`	`0`	`-2380`
`-1440`	`0`	`720`	`0`
`0`	`-2380`	`0`	`700`

`4896`	`0`	`-1440`	`0`
`0`	`9100`	`0`	`-2380`
`-1440`	`0`	`720`	`0`
`0`	`-2380`	`0`	`700`

`"Now, "A^(-1)=1/|A| × Adj(A)`

`"Here, "X = A^(-1) × B`

`:. X = 1/|A| × Adj(A) × B`

`1/(10080)` ×

`4896`	`0`	`-1440`	`0`
`0`	`9100`	`0`	`-2380`
`-1440`	`0`	`720`	`0`
`0`	`-2380`	`0`	`700`

	`270`
	`48`
	`508`
	`168`

`1/(10080)` ×

	`4896×270+0×48-1440×508+0×168`
	`0×270+9100×48+0×508-2380×168`
	`-1440×270+0×48+720×508+0×168`
	`0×270-2380×48+0×508+700×168`

`1/(10080)` ×

	`590400`
	`36960`
	`-23040`
	`3360`

	`410/7`
	`11/3`
	`-16/7`
	`1/3`

`:.[[a],[b],[c],[d]]=[[410/7],[11/3],[-16/7],[1/3]]`

`:. a=410/7, b=11/3, c=-16/7, d=1/3`

Now substituting this values in the equation is `y = a + bx + cx^2 + dx^3`, we get

`y = 410/7 +11/3x-16/7x^2+1/3x^3`

`y = 410/7 +11/3(X-1998)-16/7(X-1998)^2+1/3(X-1998)^3`