Newton-Gregory forward interpolation formula and Newton- Gregory backward interpolation formula.

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 + [u(u+1)]/2!×∇²yₙ+ .......+ {u(u+1)(u+2)[u+(n-1)]}/n!×∇ⁿyₙ

Example: The value of annuities are given for the following ages. Find the value of annuity at the age of 28.5?

Age:. 25 26 27 28 29

Annuity: 16.2 15.9 15.6 15.3 15

Solution: Here x=28.5, xₙ= 29 and h= 1.

Thus u = (x-xₙ)/h = (28.5-29)/1 = -0.5

Backward difference table.

Using Newton-Gregory backward interpolation formula

f(x) = yₙ + u∇yₙ + [u(u+1)]/2!×∇²yₙ+ .......+ {u(u+1)(u+2)[u+(n-1)]}/n!×∇ⁿyₙ

f(28.5) = yₙ + (-0.5)∇yₙ + [-0.5(-0.5+1)]/2!×∇²yₙ

f(28.5) = 15+(-0.5)(-0.3)+0

f(28.5) = 15+0.15

f(28.5) = 15.15 Ans.

Comments

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...

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) ...

Relation between the operators.

Relation between the operators: We can express each of Δ,∇,𝛿,μ and D in terms of E. 1) Δ = E - 1 Proof: By definition of Δ Δf(x) = f(x+h) - f(x) =Ef(x) - f(x) [∵Ef(x) = f(x+h)] Hence Δ = E -1 2) ∇ = 1 - E⁻¹ Proof: By definition of ∇ ∇f(x) = f(x) - f(x-h) = f(x) - E⁻¹f(x) [∵E⁻¹f(x) = f(x-h)] => ∇f(x) = (1 - E⁻¹)f(x) => ∇ = 1 - E⁻¹ 3) 𝛿 =E ¹/² - E⁻ ¹/² Proof: By definition of 𝛿 𝛿f(x) = f[x+(h/2)] - f[x-(h/2)] = E ¹/²f(x) - E⁻ ¹/²f(x) Hence 𝛿 =E ¹/² - E⁻ ¹/² 4) 𝝁 = 1/2 (E ¹/² + E⁻ ¹/²) Proof: By definition of 𝝁 𝝁f(x) = (1/2){f[x+(h/2)] + f[x-(h/2)]} = (1/2){E ¹/²f(x) + E⁻ ¹/²f(x)} =(1/2){E ¹/² + E⁻ ¹/²}f(x) Hence 𝝁 = 1/2 (E ¹/² +...

Numerical analysis

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