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6. Gauss Forward Interpolation formula example ( Enter your problem )
  1. Formula & Example-1 (table data)
  2. Example-2 (table data)
  3. Example-3 (table data)

1. Formula & Example-1 (table data)





Formula
Gauss Forward formula
`p = (x - x_0)/h`
`y_p=y_0+p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_(-1) + ((p + 1)p(p - 1))/(3!) * Delta^3y_(-1) + ((p + 1)p(p - 1)(p - 2))/(4!) * Delta^4y_(-2) + ...`

Examples
1. Find Solution using Gauss Forward formula
xf(x)
11
2-1
31
4-1
51

x = 3.5


Solution:
The value of table for `x` and `y`

x12345
y1-11-11

Gauss's forward method to find solution

`h=2-1=1`

Taking `x_0=3` then `p=(x-x_0)/h=(x-3)/1`

Now the central difference table is
`x``p=(x-3)/1``y``Deltay``Delta^2y``Delta^3y``Delta^4y`
1-21
-2
2-1-14
2-8
301-416
-28
41-14
2
521


`x = 3.5`

`p = (x - x_0)/h = (3.5 - 3)/1 = 0.5`

`y_0=1, Delta y_0=-2,Delta^2y_(-1)=-4,Delta^3y_(-1)=8,Delta^4y_(-2)=16`

Gauss's forward interpolation formula is
`y_p=y_0+p Delta y_0 + (p(p - 1))/(2!) * Delta^2y_(-1) + ((p + 1)p(p - 1))/(3!) * Delta^3y_(-1) + ((p + 1)p(p - 1)(p - 2))/(4!) * Delta^4y_(-2)`

`y_(0.5) = 1 + (0.5)(-2) + ((0.5)(0.5 - 1))/(2) * (-4) + ((0.5 + 1)(0.5)(0.5 - 1))/(6) * (8) + ((0.5 + 1)(0.5)(0.5 - 1)(0.5 - 2))/(24) * (16)`

`y_(0.5)=1 -1 +0.5 -0.5 +0.375`

`y_(0.5)=0.375`


Solution of Gauss's forward interpolation is `y(3.5) = 0.375`




This material is intended as a summary. Use your textbook for detail explanation.
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