Formula
Examples
Find Solution using Hermite's formula
x = 0.1
Finding f(2)
Solution:
The value of table for `x`, `f(x)` and `f'(x)`
The Polynomials `I_i(x)` are
`I_0(x)=((x - x_1))/((x_0 - x_1))=((x -1))/((0 -1))=((x -1))/((-1))=(x-1)/(-1)=-x+1`
`I_1(x)=((x - x_0))/((x_1 - x_0))=((x -0))/((1 -0))=((x -0))/((1))=(x)/(1)=x`
`I_0'(x)=-1`
`I_1'(x)=1`
`I_0'(x_0)=I_0'(0)=-1=-1`
`I_1'(x_1)=I_1'(1)=1=1`
Hermite Interpolation Formula is
`H(x)=sum u_i(x)*y_i + sum v_i(x)*y_i'`
where `u_i(x)=[1-2(x-x_i) I_i'(x_i)][I_i(x)]^2` and `v_i(x)=(x-x_i)[I_i(x)]^2`
`u_0(x)=[1-2(x-x_0) I_0'(x_0)][I_0(x)]^2`
`=>u_0(x)=[1-2(x-0) I_0'(0)][I_0(x)]^2`
`=>u_0(x)=[1-2(x-0) * (-1)][I_0(x)]^2`
`=>u_0(x)=[1 +2x][I_0(x)]^2`
`=>u_0(x)=(2x+1)(-x+1)^2`
`v_0(x)=(x-x_0)[I_i(x)]^2`
`=>v_0(x)=(x)(-x+1)^2`
`u_1(x)=[1-2(x-x_1) I_1'(x_1)][I_1(x)]^2`
`=>u_1(x)=[1-2(x-1) I_1'(1)][I_1(x)]^2`
`=>u_1(x)=[1-2(x-1) * (1)][I_1(x)]^2`
`=>u_1(x)=[1 -2x+2][I_1(x)]^2`
`=>u_1(x)=(-2x+3)(x)^2`
`v_1(x)=(x-x_1)[I_i(x)]^2`
`=>v_1(x)=(x-1)(x)^2`
Hermite Interpolation formula is
`H(x)=u_0(x)*y_0+v_0(x)*y_0'+u_1(x)*y_1+v_1(x)*y_1'`
`H(x)=(2x+1)(-x+1)^2 * (0) + (x)(-x+1)^2 * (0)+(-2x+3)(x)^2 * (1) + (x-1)(x)^2 * (1)`
Putting x=0.1 and simplifying, we obtain
`H(0.1)=0.019`
This material is intended as a summary. Use your textbook for detail explanation.
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