2. Example-1 (Fit 4 points)
Formula
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Cubic spline formula is
`f(x)=(x_i-x)^3/(6h) M_(i-1) + (x-x_(i-1))^3/(6h) M_i + ((x_i-x))/h (y_(i-1)-h^2/6 M_(i-1)) + ((x-x_(i-1)))/h (y_i-h^2/6 M_i) ->(1)`
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`M_(i-1)+4M_(i)+M_(i+1)=6/h^2(y_(i-1)-2y_(i)+y_(i+1))->(2)`
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Examples
1. Calculate Cubic Splines y(1.5), y'(2)
Solution:
Cubic spline formula is
`f(x)=(x_i-x)^3/(6h) M_(i-1) + (x-x_(i-1))^3/(6h) M_i + ((x_i-x))/h (y_(i-1)-h^2/6 M_(i-1)) + ((x-x_(i-1)))/h (y_i-h^2/6 M_i) ->(1)`
We have, `M_(i-1)+4M_(i)+M_(i+1)=6/h^2(y_(i-1)-2y_(i)+y_(i+1))->(2)`
Here `h=1,n=3`
`M_0=0,M_3=0`
Substitute `i=1` in equation `(2)`
`M_0+4M_1+M_2=6/h^2(y_0-2y_1+y_2)`
`=>0+4M_1+M_2=6/1*(1-2*5+11)`
`=>4M_1+M_2=12`
Substitute `i=2` in equation `(2)`
`M_1+4M_2+M_3=6/h^2(y_1-2y_2+y_3)`
`=>M_1+4M_2+0=6/1*(5-2*11+8)`
`=>M_1+4M_2=-54`
Solving these 2 equations using elimination methodTotal Equations are `2` `4M_1+M_2=12 -> (1)` `M_1+4M_2=-54 -> (2)`
Select the equations `(1)` and `(2)`, and eliminate the variable `M_1`. | `4M_1+M_2=12` | ` xx 1->` | | `` | `4M_1` | `+` | `M_2` | `=` | `12` | `` | | | − | | | `M_1+4M_2=-54` | ` xx 4->` | | `` | `4M_1` | `+` | `16M_2` | `=` | `-216` | `` | | | |
| | | | | | `-` | `15M_2` | `=` | `228` | ` -> (3)` |
Now use back substitution method From (3) `-15M_2=228` `=>M_2=(228)/(-15)=-15.2` From (2) `M_1+4M_2=-54` `=>M_1+4(-15.2)=-54` `=>M_1-60.8=-54` `=>M_1=-54+60.8=6.8` Solution using Elimination method. `M_1=6.8,M_2=-15.2`
Substitute `i=1` in equation `(1)`, we get cubic spline in `1^(st)` interval `[x_0,x_1]=[1,2]`
`f_1(x)=(x_1-x)^3/(6h) M_0 + (x-x_0)^3/(6h) M_1 + ((x_1-x))/h (y_0-h^2/6 M_0)+((x-x_0))/h (y_1-h^2/6 M_1)`
`f_1(x)=(2-x)^3/6 *0 + (x-1)^3/6 *6.8 + ((2-x))/1 (1-1/6 *0) + ((x-1))/1 (5-1/6 *6.8)`
`f_1(x)=1.1333x^3-3.4x^2+6.2667x-3`, for `1<=x<=2`
Substitute `i=2` in equation `(1)`, we get cubic spline in `2^(nd)` interval `[x_1,x_2]=[2,3]`
`f_2(x)=(x_2-x)^3/(6h) M_1 + (x-x_1)^3/(6h) M_2 + ((x_2-x))/h (y_1-h^2/6 M_1)+((x-x_1))/h (y_2-h^2/6 M_2)`
`f_2(x)=(3-x)^3/6 *6.8 + (x-2)^3/6 *-15.2 + ((3-x))/1 (5-1/6 *6.8) + ((x-2))/1 (11-1/6 *-15.2)`
`f_2(x)=-3.6667x^3+25.4x^2-51.3333x+35.4`, for `2<=x<=3`
Substitute `i=3` in equation `(1)`, we get cubic spline in `3^(rd)` interval `[x_2,x_3]=[3,4]`
`f_3(x)=(x_3-x)^3/(6h) M_2 + (x-x_2)^3/(6h) M_3 + ((x_3-x))/h (y_2-h^2/6 M_2)+((x-x_2))/h (y_3-h^2/6 M_3)`
`f_3(x)=(4-x)^3/6 *-15.2 + (x-3)^3/6 *0 + ((4-x))/1 (11-1/6 *-15.2) + ((x-3))/1 (8-1/6 *0)`
`f_3(x)=2.5333x^3-30.4x^2+116.0667x-132`, for `3<=x<=4`
For `y(1.5)`, `1.5 in [1,2]`, so substitute `x=1.5` in `f_1(x)`, we get
`f_1(1.5)=2.575`
For `y'(2)`, `2 in [1,2]`, so find `f_1^(')(x)`
`f_1^(')(x)=3.4x^2-6.8x+6.2667`
Now substitute `x=2` in `f_1^(')(x)`, we get
`f_(1)^(')(2)=6.2667`
This material is intended as a summary. Use your textbook for detail explanation.
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