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Question
Prove that Δ∇ = Δ – ∇
Sum
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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.
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Finite Differences
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