2. 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 + cx^2` and the normal equations are
`sum y = an + b sum x + c sum x^2`
`sum xy = a sum x + b sum x^2 + c sum x^3`
`sum x^2y = a sum x^2 + b sum x^3 + c sum x^4`
| `X` | `y` | `x = X - 1998` | `x^2` | `x^3` | `x^4` | `x*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`
`48=0a+10b+0c`
`508=10a+0b+34c`
Solving these 3 equations using inverse matrix method,
Here `5a+10c=270`
`10b=48`
`10a+34c=508`
Now converting given equations into matrix form
`[[5,0,10],[0,10,0],[10,0,34]] [[a],[b],[c]]=[[270],[48],[508]]`
Now, A = `[[5,0,10],[0,10,0],[10,0,34]]`, X = `[[a],[b],[c]]` and B = `[[270],[48],[508]]`
`:. AX = B`
`:. X = A^-1 B`
| `|A|` | = | | `5` | `0` | `10` | | | `0` | `10` | `0` | | | `10` | `0` | `34` | |
|
`=5 xx (10 × 34 - 0 × 0) +0 xx (0 × 34 - 0 × 10) +10 xx (0 × 0 - 10 × 10)`
`=5 xx (340 +0) +0 xx (0 +0) +10 xx (0 -100)`
`=5 xx (340) - +0 xx (0) +10 xx (-100)`
`= 1700 +0 -1000`
`=700`
`"Here, " |A| = 700 != 0`
`:. A^(-1) " is possible."`
| `Adj(A)` | = | | Adj | | `5` | `0` | `10` | | | `0` | `10` | `0` | | | `10` | `0` | `34` | |
|
|
| = | | `+(10 × 34 - 0 × 0)` | `-(0 × 34 - 0 × 10)` | `+(0 × 0 - 10 × 10)` | | | `-(0 × 34 - 10 × 0)` | `+(5 × 34 - 10 × 10)` | `-(5 × 0 - 0 × 10)` | | | `+(0 × 0 - 10 × 10)` | `-(5 × 0 - 10 × 0)` | `+(5 × 10 - 0 × 0)` | |
| T |
|
| = | | `+(340 +0)` | `-(0 +0)` | `+(0 -100)` | | | `-(0 +0)` | `+(170 -100)` | `-(0 +0)` | | | `+(0 -100)` | `-(0 +0)` | `+(50 +0)` | |
| T |
|
| = | | `340` | `0` | `-100` | | | `0` | `70` | `0` | | | `-100` | `0` | `50` | |
| T |
|
| = | | `340` | `0` | `-100` | | | `0` | `70` | `0` | | | `-100` | `0` | `50` | |
|
`"Now, "A^(-1)=1/|A| × Adj(A)`
`"Here, "X = A^(-1) × B`
`:. X = 1/|A| × Adj(A) × B`
| = | `1/(700)` × | | `340` | `0` | `-100` | | | `0` | `70` | `0` | | | `-100` | `0` | `50` | |
| × | |
| = | `1/(700)` × | | `340×270+0×48-100×508` | | | `0×270+70×48+0×508` | | | `-100×270+0×48+50×508` | |
|
`:.[[a],[b],[c]]=[[410/7],[24/5],[-16/7]]`
`:. a=410/7, b=24/5, c=-16/7`
Now substituting this values in the equation is `y = a + bx + cx^2`, we get
`y = 410/7 +24/5x-16/7x^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.
Any bug, improvement, feedback then
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