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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
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
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: Find the first term of the series whose second and subsequent terms are 8,3,0,-1,0.
Solution. Let y₀=?,y₁=8,y₂=3,y₃=0,y₄=-1,y₅=0.
Here number of known data = 5 values
∴∆⁵y=0
=> (E-1)⁵y = 0
=> (E⁵-5E⁴+10E³-10E²+5E-1)y=0 ....(1)
(i) Putting y = y₀ in (1), we get
 (E⁵-5E⁴+10E³-10E²+5E-1)y₀ = 0
=> y₅-5y₄+10y₃-10y₂+5y₁-y₀ = 0 [Eⁿy₀ = y₀₊ₙ]
=> 0-5(-1)+10(0)-10(3)+5(8)- y₀ =0
=> 5-30+40 =  y₀
=>  y₀ = 15 i.e, first term= 15 Ans.

Example: Find the missing values in the following table:
x: 0       5       10       15      20      25
y: 6      10        -        17       -        31
Solution. Let y₀, y₁, y₂, y₃, y₄, y₅ be given data in which y₂ and  y₄ are missing:
∴ Number of known data = 4 values, so that ∆⁴y = 0
=> (E-1)⁴y = 0 [∵ ∆=E-1]
=>  (E⁴-4E³+6E²-4E+1)y = 0 ......(1)
(i) Putting y =  y₀ in (1), we get
(E⁴-4E³+6E²-4E+1)y₀ = 0
=> y₄-4y₃+6y₂-4y₁+y₀ = 0 [Eⁿy₀ = y₀₊ₙ]
=>  y₄ -4(17)+6y₂-4(10)+6 = 0
=> 6y₂+y₄= 102 .......(2)
(ii) Putting y =  y₁ in (1), we get
 (E⁴-4E³+6E²-4E+1)y₁ = 0
=> y₅-4y₄+6y₃-4y₂+y₁ = 0
=> 31-4y₄+6(17)-4y₂+10 = 0
=> 4y₄+4y₂ = 143 .....(3)
Solving (2) and (3)
4(y₄+y₂) = 143 [From (3)]
y₄+y₂ = 143/4 = 35.75
y₄ = 35.75 - y₂
Put the value of y₄ in (2), we get
6y₂+35.75 - y₂ = 102
6y₂ - y₂ = 102-35.75
5y₂ = 66.25
y₂= 66.25/5 = 13.25
And
y₄ = 35.75-13.25 = 22.5
i.e f(10)= 13.25 and f(20) = 22.5 Ans.

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