7. Example-6 (Fit 5 points)
2. Calculate Cubic Splines y(1.5)
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=4`
`M_0=0,M_4=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*2+5)`
`=>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+M_3=6/1*(2-2*5+11)`
`=>M_1+4M_2+M_3=18`
Substitute `i=3` in equation `(2)`
`M_2+4M_3+M_4=6/h^2(y_2-2y_3+y_4)`
`=>M_2+4M_3+0=6/1*(5-2*11+20)`
`=>M_2+4M_3=18`
Solving these 3 equations using elimination methodTotal Equations are `3` `4M_1+M_2+0M_3=12 -> (1)` `M_1+4M_2+M_3=18 -> (2)` `0M_1+M_2+4M_3=18 -> (3)`
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+M_3=18` | ` xx 4->` | | `` | `4M_1` | `+` | `16M_2` | `+` | `4M_3` | `=` | `72` | `` | | | |
| | | | | | `-` | `15M_2` | `-` | `4M_3` | `=` | `-60` | ` -> (4)` |
Select the equations `(3)` and `(4)`, and eliminate the variable `M_2`. | `M_2+4M_3=18` | ` xx 15->` | | | | `` | `15M_2` | `+` | `60M_3` | `=` | `270` | `` | | | + | | | `-15M_2-4M_3=-60` | ` xx 1->` | | | | `-` | `15M_2` | `-` | `4M_3` | `=` | `-60` | `` | | | |
| | | | | | | | `` | `56M_3` | `=` | `210` | ` -> (5)` |
Now use back substitution method From (5) `56M_3=210` `=>M_3=(210)/(56)=3.75` From (3) `M_2+4M_3=18` `=>M_2+4(3.75)=18` `=>M_2+15=18` `=>M_2=18-15=3` From (2) `M_1+4M_2+M_3=18` `=>M_1+4(3)+(3.75)=18` `=>M_1+15.75=18` `=>M_1=18-15.75=2.25` Solution using Elimination method. `M_1=2.25,M_2=3,M_3=3.75`
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 *2.25 + ((2-x))/1 (1-1/6 *0) + ((x-1))/1 (2-1/6 *2.25)`
`f_1(x)=0.375x^3-1.125x^2+1.75x`, 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 *2.25 + (x-2)^3/6 *3 + ((3-x))/1 (2-1/6 *2.25) + ((x-2))/1 (5-1/6 *3)`
`f_2(x)=0.125x^3+0.375x^2-1.25x+2`, 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 *3 + (x-3)^3/6 *3.75 + ((4-x))/1 (5-1/6 *3) + ((x-3))/1 (11-1/6 *3.75)`
`f_3(x)=0.125x^3+0.375x^2-1.25x+2`, for `3<=x<=4`
Substitute `i=4` in equation `(1)`, we get cubic spline in `4^(th)` interval `[x_3,x_4]=[4,5]`
`f_4(x)=(x_4-x)^3/(6h) M_3 + (x-x_3)^3/(6h) M_4 + ((x_4-x))/h (y_3-h^2/6 M_3)+((x-x_3))/h (y_4-h^2/6 M_4)`
`f_4(x)=(5-x)^3/6 *3.75 + (x-4)^3/6 *0 + ((5-x))/1 (11-1/6 *3.75) + ((x-4))/1 (20-1/6 *0)`
`f_4(x)=-0.625x^3+9.375x^2-37.25x+50`, for `4<=x<=5`
For `y(1.5)`, `1.5 in [1,2]`, so substitute `x=1.5` in `f_1(x)`, we get
`f_1(1.5)=1.3594`
This material is intended as a summary. Use your textbook for detail explanation.
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