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. |
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:
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)
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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