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

Comments

Gauss's central difference formula for equal intervals.

Gauss's central difference formula for equal intervals: We shall develop central difference formulae which are best suitable for interpolation near the middle of the tabulated set (table). x: x₋₂ x₋₁ x₀ x₁ x₂ y=f(x): y₋₂ y₋₁ y₀ y₁ y₂ Difference table. Gauss's forward interpolation formula for equal intervals: 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 +......., Where u= (x-x₀)/h Remark: This formula is applicable when u lies between 0 and 1 i.e (0<u<1). Example: Using Gauss's forward formula to evaluate y₃₀ given that y₂₁=18.4708, y₂₅=17.8144, y₂₉=17.1070, y₃₃=16.3432 and y₃₇=15.5154. Solution. The difference table is Forward difference table. To find y=f(x) at x=30, i.e f(30): Taking x₀ = 29, h=4, x=30, then u= (x-x₀)/h = (30-29)/4 = 0.25 Using Gauss's forward difference formula 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 +...... => f(30)

Different operators

Different operators We will study the following operators: 1) The Shifting Operator (E): Ef(x) = f(x+h) i.e Ey₀= y₁ E²f(x) = f(x+2h) i.e E²y₀ = y₂ Eⁿf(x) = f(x+nh) i.e Eⁿy₀ = yₙ Here n takes integral or fractional, positive or negative values. For example: E⁻¹f(x) = f(x-h) i.e E⁻¹ y₂ = y₁ E¹/²f(x) = f[x+(1/2)h] i.e E¹/² y₁ = y₃ₗ₂ Properties of Operator E 1) Operator E is distributive. 2) Operator E is commutative with respect to constant. 3) Operator E obeys laws of indices. 2) Forward difference operator (∆): If x₀, x₁,x₂,......., xₙ are equally spaced with interval of differencing h and if y = f(x), then ∆f(x) = f(x+h) - f(x) i.e ∆yᵢ = yᵢ₊₁ - yᵢ for i = 0,1,2,3,..... n-1, The symbol ∆ is called forward difference operator and ∆yᵢ is called first forward difference. Similarly, the second forward differences are ∆²yᵢ =∆yᵢ₊₁ - ∆yᵢ For example: ∆²y₀ =∆y₁ - ∆y₀ = (y₂ - y₁)-(y₁-y₀) = y₂ -2y₁+y₀ Clearly any higher order differences can easily be expresse

Questions (Forward difference operator, Backward difference operator and Shifting operator)

Questions Example: Given that y₅=4, y₆=3, y₇=4, y₈=10, y₉=24, find the value of ∆⁴y₅: (i) By using the difference table and (ii) Without using the difference table. Solution: (i) Forward difference table. From the difference table ∆⁴y₅=0. Ans. (ii) We have ∆⁴y₅= (E-1)⁴y₅ [∵ ∆=E-1] = (E⁴-4E³+6E²-4E+1)y₅ = E⁴y₅-4E³y₅+6E²y₅-4Ey₅+1y₅ = y₉ -4y₈+6y₇-4y₆+y₅ [Eⁿyₓ=yₓ₊ₙ] = 24-4×10+6×4-4×3+4 = 24-40-24-12+4 =0Ans. Example: Given x: 1 2 3 4 5 y: 2 5 10 17 26 Find the value of ∇²y₅. Solution. (i) Backward difference table. From the difference table, we get ∇²y₅=2 Ans. (ii) Without using the difference table to find ∇²y₅: We have ∇²y₅ = (1-E⁻¹)²y₅ [∵ ∇=1-E⁻¹] = (1+E⁻² -2E⁻¹)y₅ = y₅ + y₃ -2y₄ [E⁻ⁿyₓ=yₓ₋ₙ] = 26+10-2×17 = 36-34 = 2 Ans. Example: F

Numerical analysis

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