1. Calculate Cubic Splines

X	0	1	2
Y	-5	-4	3

y(0.5)

Solution:

x	0	1	2
y	-5	-4	3

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=2`

`M_0=0,M_2=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+0=6/1*(-5-2*-4+3)`

`=>4M_1=36`

`=>M_1=9`

Substitute `i=1` in equation `(1)`, we get cubic spline in `1^(st)` interval `[x_0,x_1]=[0,1]`

`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)=(1-x)^3/6 *0 + (x-0)^3/6 *9 + ((1-x))/1 (-5-1/6 *0) + ((x-0))/1 (-4-1/6 *9)`

`f_1(x)=1/2(3x^3-x-10)`, for `0<=x<=1`

---

Substitute `i=2` in equation `(1)`, we get cubic spline in `2^(nd)` interval `[x_1,x_2]=[1,2]`

`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)=(2-x)^3/6 *9 + (x-1)^3/6 *0 + ((2-x))/1 (-4-1/6 *9) + ((x-1))/1 (3-1/6 *0)`

`f_2(x)=1/2(-3x^3+18x^2-19x-4)`, for `1<=x<=2`

---

For `y(0.5)`, `0.5 in [0,1]`, so substitute `x=0.5` in `f_1(x)`, we get

`f_1(0.5)=-5.0625`

---