Question

44: The relationship between the operators E and D is

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Answer 1

1. Δ ≡ E −1

Proof:         From the definition of Δ we know that

Δ f (x) = f (x + h ) − f (x) and    

E[ f (x )] = f (x + h)

where h is the interval of difference.

Δ f (x) = f (x + h ) − f (x)

Δ f (x) = Ef (x ) − f (x)

⇒ Δ f (x) = (E −1) f (x)

Δ ≡ E −1

∴    E ≡ 1 + Δ

2.   Δ E       ≡       Δ E

Proof:

E(Δf(x)) = E[f(x+h)- f(x)]

= E f(x+h) - E f(x)

= f(x+2h) - f(x+h)

=Δf(x+h)

=ΔEf(x)

ΔE ≡ ΔE

3. ∇≡ E −1 / E

∇f (x) = f (x) - f (x-h)

= f (x) – E-1f (x)

= (1- E-1) f (x)

∇ ≡ (1- E-1)

∇ ≡ 1 – 1/E

Hence ∇ ≡ [E – 1]/E

Answer 2

Answer:

The relationship between the operators E and D is $E$ ≡$e^{hD}$

Step-by-step explanation:

• The proximity of the hues to one another increases with decreasing Delta E.
• There is no difference between the two hues when the delta E is 0.
• The farther the colours are separated and the greater colour difference is perceived, the higher the Delta E.

In the given question, we have been given operators E and D.

And we are supposed to find the relationship between operators E and D.

Lets see what operator E means,

The shifting operator, often known as the shift operator, E is defined as Ef(x) = f(x+h).

Where h in the above expression stands for step length

Now moving forward, lets find the relationship between the two given operators :

$f(x+h) = f(x) + hf1 (x) + \frac{h^2}{2!}f2(x) + \frac{h^3}{3!} f3(x) +$ …..

$Ef(x) = f(x) + hDf(x) + \frac{h^2}{2!}D^2f(x) + \frac{h^3}{3!}D^3f(x) +$….

$Ef(x) = [ 1+hD+\frac{h^2D^2}{2!} + \frac{h^3D^3}{3!} + ] f(x)$

$Ef(x) =e^h^Df(x)$

Therefore, $E$≡ $e^h^D$

#SPJ3