Stirling's formula for equal interval.

Stirling's formula for equal interval:

Let u = (x-x₀)/h

Then Gauss's forward interpolation formula is

f(x) = y₀+u/1!∆y₀+{u(u-1)/2!}∆²y₋₁+{(u+1)(u)(u-1)/3!}∆³y₋₁+{(u+1)u(u-1)(u-2)/4!}∆⁴y₋₂ +.......    .....(1)

and Guass backward interpolation formula is

f(x) = y₀+u/1!∆y₋₁+{u(u+1)/2!}∆²y₋₁+{(u+1)(u)(u-1)/3!}∆³y₋₂+{(u+2)(u+1)u(u-1)/4!}∆⁴y₋₂ +......     .......(2)

Taking the mean of (1) and (2), we get

f(x) = = y₀+u/1![(∆y₀+∆y₋₁)/2] + (u²/2!) ∆²y₋₁ + u(u-1)(u+1)/3![(∆³y₋₁+∆³y₋₂)/2] + [u²(u²-1)/4!]∆⁴y₋₂ +..... .....(3)

Which is called Stirling formula.

It is used when u lies between -1/4 and 1/4.

Example: Employ Stirling formula to compute y₁₂.₂ from the following table (yₓ = 1+log₁₀sinx):

      x°:             10         11          12       13           14

10⁵yₓ:     23967  28060   31788  35209   38368

Solution: Rewriting

x°:        10         11           12           13          14

yₓ: .23967   .28060   .31788   .35209   .38368

Taking the origin at x₀=12°, h= 1, x=12.2,

then u = (x-x₀)/h = (12.2-12)/1 = 0.2,

which lies between -1/4 and 1/4.

Stirling formula is

f(x) = = y₀+u/1![(∆y₀+∆y₋₁)/2] + (u²/2!) ∆²y₋₁ + u(u-1)(u+1)/3![(∆³y₋₁+∆³y₋₂)/2] + [u²(u²-1)/4!]∆⁴y₋₂ +.....

Here y₀=0.31788, ∆y₀=0.03421, ∆y₋₁=0.03728, ∆²y₋₁= -0.00307, ∆³y₋₁= -0.00045, ∆³y₋₂=0.00058, ∆⁴y₋₂ = -0.00013

∴ y₁₂.₂ = 0.31788+(0.2)/1[ (.03421+.03728)/2] +(.2)²/2[-.00307] + {(.2)[(.2)²-1]/6} [(-.00045+.00058)/2] + {(.2)²[(.2)²-1]}/24(-0.00013)

= 0.31788+0.00715-0.00006-0.000002+0.0000002

=0.32497 Ans