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11. Hermite's Interpolation formula example ( Enter your problem )
  1. Formula & Example-1 (table data)
  2. Example-2 (table data)
  3. Example-3 (table data)
Other related methods
  1. Newton's Forward Difference Interpolation formula
  2. Newton's Backward Difference Interpolation formula
  3. Newton's Divided Difference Interpolation formula
  4. Lagrange's Interpolation formula
  5. Lagrange's Inverse Interpolation formula
  6. Gauss Forward Interpolation formula
  7. Gauss Backward Interpolation formula
  8. Stirling's Interpolation formula
  9. Bessel's Interpolation formula
  10. Everett's Interpolation formula
  11. Hermite's Interpolation formula
  12. Missing terms in interpolation table

2. Example-2 (table data)
(Previous example)
12. Missing terms in interpolation table
(Next method)

3. Example-3 (table data)





Find Solution using Hermite's formula
xf(x)
31.0986
3.51.2528
41.3863

x = 3.2
Finding f(2)


Solution:
The value of table for `x`, `f(x)` and `f'(x)`

xy=f(x)y'=f'(x)
31.09860.3333
3.51.25280.2857
41.38630.25

The Polynomials `I_i(x)` are

`I_0(x)=((x - x_1)(x - x_2))/((x_0 - x_1)(x_0 - x_2))=((x -3.5)(x -4))/((3 -3.5)(3 -4))=((x -3.5)(x -4))/((-0.5)(-1))=(x^2-7.5x+14)/(0.5)=2x^2-15x+28`

`I_1(x)=((x - x_0)(x - x_2))/((x_1 - x_0)(x_1 - x_2))=((x -3)(x -4))/((3.5 -3)(3.5 -4))=((x -3)(x -4))/((0.5)(-0.5))=(x^2-7x+12)/(-0.25)=-4x^2+28x-48`

`I_2(x)=((x - x_0)(x - x_1))/((x_2 - x_0)(x_2 - x_1))=((x -3)(x -3.5))/((4 -3)(4 -3.5))=((x -3)(x -3.5))/((1)(0.5))=(x^2-6.5x+10.5)/(0.5)=2x^2-13x+21`

`I_0'(x)=4x-15`

`I_1'(x)=-8x+28`

`I_2'(x)=4x-13`

`I_0'(x_0)=I_0'(3)=4 xx 3-15=-3`

`I_1'(x_1)=I_1'(3.5)=-8 xx 3.5+28=0`

`I_2'(x_2)=I_2'(4)=4 xx 4-13=3`

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-3) I_0'(3)][I_0(x)]^2`

`=>u_0(x)=[1-2(x-3) * (-3)][I_0(x)]^2`

`=>u_0(x)=[1 +6x-18][I_0(x)]^2`

`=>u_0(x)=(6x-17)(2x^2-15x+28)^2`

`v_0(x)=(x-x_0)[I_i(x)]^2`

`=>v_0(x)=(x-3)(2x^2-15x+28)^2`

`u_1(x)=[1-2(x-x_1) I_1'(x_1)][I_1(x)]^2`

`=>u_1(x)=[1-2(x-3.5) I_1'(3.5)][I_1(x)]^2`

`=>u_1(x)=[1-2(x-3.5) * (0)][I_1(x)]^2`

`=>u_1(x)=[1 +0][I_1(x)]^2`

`=>u_1(x)=(1)(-4x^2+28x-48)^2`

`v_1(x)=(x-x_1)[I_i(x)]^2`

`=>v_1(x)=(x-3.5)(-4x^2+28x-48)^2`

`u_2(x)=[1-2(x-x_2) I_2'(x_2)][I_2(x)]^2`

`=>u_2(x)=[1-2(x-4) I_2'(4)][I_2(x)]^2`

`=>u_2(x)=[1-2(x-4) * (3)][I_2(x)]^2`

`=>u_2(x)=[1 -6x+24][I_2(x)]^2`

`=>u_2(x)=(-6x+25)(2x^2-13x+21)^2`

`v_2(x)=(x-x_2)[I_i(x)]^2`

`=>v_2(x)=(x-4)(2x^2-13x+21)^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'+u_2(x)*y_2+v_2(x)*y_2'`

`H(x)=(6x-17)(2x^2-15x+28)^2 * (1.0986) + (x-3)(2x^2-15x+28)^2 * (0.3333)+(1)(-4x^2+28x-48)^2 * (1.2528) + (x-3.5)(-4x^2+28x-48)^2 * (0.2857)+(-6x+25)(2x^2-13x+21)^2 * (1.3863) + (x-4)(2x^2-13x+21)^2 * (0.25)`

Putting x=3.2 and simplifying, we obtain
`H(3.2)=1.1632`




This material is intended as a summary. Use your textbook for detail explanation.
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2. Example-2 (table data)
(Previous example)
12. Missing terms in interpolation table
(Next method)





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