Skip to main content

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.

Comments

Popular posts from this blog

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...

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 +...