Newton-Gregory forward interpolation formula: Let u = (x-x₀)/h then Newton-Gregory forward interpolation formula f(x) = y₀ + u∆y₀ + [u(u-1)]/2!×∆²y₀+ .......+ {u(u-1)(u-2)[u-(n-1)]}/n!×∆ⁿy₀ Example: Find f(15) by using Newton-Gregory forward interpolation formula? x: 10 20 30 40 50 y: 46 66 81 93 101 Solution. Here x₀=10, h = 10 and x= 15. Thus u = (x-x₀)/h = (15-10)/10 = 5/10 = 1/2 Forward difference table. By using Newton-Gregory forward interpolation formula f(x) = y₀ + u∆y₀ + [u(u-1)]/2!×∆²y₀+ .......+ {u(u-1)(u-2)[u-(n-1)]}/n!×∆ⁿy₀ f(15) = y₀ + 1/2∆y₀ + [1/2(1/2-1)]/2!×∆²y₀+ [1/2(1/2-1)(1/2-2)]/3!×∆³y₀ +[1/2(1/2-1)(1/2-2)(1/2-3)]/4!×∆⁴y₀ f(15) = 46 +(0.5)(20) + [(0.5)(-0.5)]/2×(-5) +[(0.5)(-0.5)(-1.5)]/6×(2) + [(0.5)(-0.5)(-1.5)(-2.5)]/24×(-3) f(15) = 46+10+0.625+0.125+0.1172 f(15) = 56.8672 Ans Newton-Gregory backward interpolation formula: Let u = (x-xₙ)/h then Newton-Gregory backward interpolation formula f(x) = yₙ + u∇yₙ
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 =