Advertisements
Advertisements
Question
Prove that Δ∇ = Δ – ∇
Sum
Advertisements
Solution
L.H.S = Δ∇
= (E – 1)(1 – E–1)
= E – EE–1 + E–1
= E – 1 – 1 – E–1
= E – 2 – E–1 .........(1)
R.H.S = Δ – ∇
= (E – 1) – (1 – E–1)
= E – 1 – 1 + E–1
= E – 2 + E–1 ........(2)
From (1) and (2)
L.H.S = R.H.S
Hence proved.
shaalaa.com
Finite Differences
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Evaluate Δ`[1/((x + 1)(x + 2))]` by taking ‘1’ as the interval of differencing
Find the missing entry in the following table
| x | 0 | 1 | 2 | 3 | 4 |
| yx | 1 | 3 | 9 | - | 81 |
Find the missing entries from the following.
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y = f(x) | 0 | - | 8 | 15 | - | 35 |
Choose the correct alternative:
Δf(x) =
Choose the correct alternative:
E ≡
Choose the correct alternative:
If c is a constant then Δc =
Choose the correct alternative:
If m and n are positive integers then Δm Δn f(x)=
Choose the correct alternative:
If ‘n’ is a positive integer Δn[Δ-n f(x)]
Choose the correct alternative:
∇f(a) =
Prove that EV = Δ = ∇E
