Question

(∆-∇) = (∆∆) solve and prove that

Answer 1

down vote
accepted
As you have found ∇Δ≠
∇
Δ
≠
E
, since
∇Δ()=∇((+ℎ)−())=∇(+ℎ)−∇()=((+ℎ)−())−(()−(−ℎ))=(+ℎ)−2()+(−ℎ)≠(+ℎ).
∇Δf(x) =∇(f(x+h)−f(x)) =∇f(x+h)−∇f(x) =(f(x+h)−f(x))−(f(x)−f(x−h)) =f(x+h)−2f(x)+f(x−h) ≠f(x+h).
It is true that ∇Δ=Δ∇=2
∇
Δ
=
Δ
∇
=
δ
2
, where ()=(+ℎ/2)−(−ℎ/2)
δ
f
(
x
)
=
f
(
x
+
h
/
2
)
−
f
(
x
−
h
/
2
)
is the central difference.