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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 expressed in terms of ordinates.
Forward difference table
Forward difference table.

Here the first entry y₀ is called leading term and ∆y₀, ∆²y₀,.... are called leading differences.
Properties of ∆:
1) ∆[f(x)±𝝓(x)] = ∆f(x)± ∆𝝓(x)
2) ∆[cf(x)] = c∆f(x); c is constant.
3) ∆ᵐ∆ⁿf(x) = ∆ᵐ⁺ⁿf(x), m,n being positive integers.
4) ∆c = 0.
5) If given n observation (xᵢ ,yᵢ), then ∆ⁿy=0.

Example : Evaluate ∆²x³, when h= 1.
Solution. ∆²x³ = ∆[(x+1)³-x³] = ∆[x³+3x²+3x+1-x³] = ∆(3x²+3x+1) = 3∆x²+3∆x+∆1 = [3(x+1)²-3x²+3(x+1)-3x+0]
= 3x²+6x+3-3x²+3x+3-3x = 6x+6 = 6(x+1)Ans.

3) Backward 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) - f(x-h)
i.e ∇yᵢ = yᵢ - yᵢ₋₁, for i = 1,2,3,.....,n
The symbol  ∇ is called the backward difference operator and ∇yᵢ is called first backward difference.
Also the second backward differences are 
∇²yᵢ = ∇yᵢ - ∇yᵢ₋₁, i = 2,3,4,.....,n
For example
 ∇²y₂ = ∇(∇y₂) =∇( y₂- y₁) =  ∇y₂- ∇ y₁ = (y₂- y₁) -( y₁-y₀) = y₂ -2y₁+y₀
Similarly, the higher order backward differences are defined as follows:
∇ᵏyᵢ = ∇ᵏ⁻¹yᵢ - ∇ᵏ⁻¹yᵢ₋₁
Backward difference table
Backward difference table.


Example: Find the first backward difference of x²+2x,
Solution. We know that
 ∇f(x) = f(x) - f(x-h)
∇(x²+2x) = [x²+2x] - [(x-h)²+2(x-h)]
                =  [x²+2x] - [(x²+h²-2xh+2x-2h]
                = 2hx +2h-h² Ans.

4) Central difference operator (𝜹):

If x₀, x₁,x₂,......., xₙ are equally spaced with interval of differencing h and if y = f(x), then the first order central difference of y is defined as:
𝜹f(x+h/2) = f(x+h) - f(x) or
𝜹f(x) = f(x+h/2) - f (x-h/2)
The symbol 𝜹 is called the central difference operator.
Central difference table
Central difference table.

5) Averaging operator (𝝁):

𝝁f(x) = [f(x+h/2) + f(x-h/2)]/2

6) Differential operator (D):

Df(x) = f'(x)
D²f(x) = f"(x) and so on.

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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)

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