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8. Stirling's 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
Stirling's formula
`p = (x - x_0)/h`
`y_p=y_0+p*(Delta y_0+Delta y_(-1))/2 + (p^2)/(2!) * Delta^2y_(-1) + (p(p^2 - 1^2))/(3!) * (Delta^3y_(-1)+Delta^3y_(-2))/2 + (p^2(p^2 - 1^2))/(4!) * Delta^4y_(-2) + ...`

Examples
1. Find Solution using Stirling's formula
xf(x)
2049225
2548316
3047236
3545926
4044306

x = 28


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

x2025303540
y4922548316472364592644306

Stirling's method to find solution

`h=25-20=5`

Taking `x_0=30` then `p=(x-x_0)/h=(x-30)/5`

The difference table is
`x``p=(x-30)/5``y``Deltay``Delta^2y``Delta^3y``Delta^4y`
20-249225
-909
25-148316-171
-1080-59
30047236-230-21
-1310-80
35145926-310
-1620
40244306


`x = 28`

`p = (x - x_0)/h = (28 - 30)/5 = -0.4`

`y_0=47236, Delta y_0=-1310,Delta^2y_(-1)=-230,Delta^3y_(-1)=-80,Delta^4y_(-2)=-21`

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

`y_(-0.4) = 47236 + (-0.4)*((-1310-1080))/2 + ((0.16))/(2)*(-230) + ((-0.4)(0.16 - 1))/(6)*((-80-59))/2 + ((0.16)(0.16 - 1))/(24)*(-21)`

`y_(-0.4)=47236+478 -18.4 -3.892 +0.1176`

`y_(-0.4)=47691.8256`


Solution of Stirling's interpolation is `y(28) = 47691.8256`




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