Question

If $\Delta$ and $\nabla$ be the first forward difference and first backward difference operator respectively, then the expression $(1+\Delta)(1-\nabla)$ is equivalent to :

• A. 0
• B. 1
• C. $-1$
• D. 2

Hint:

Using the shift operator $E$ as a bridge is the most efficient way to prove or simplify any relationship involving $\Delta$, $\nabla$, or average operators.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses relations between the operator symbols in finite differences.
We use the shift operator ($E$), the forward difference operator ($\Delta$), and the backward difference operator ($\nabla$) to simplify the expression.
Key Formula or Approach:
The fundamental relationships between the operators are: \[ E = 1 + \Delta \] \[ E^{-1} = 1 - \nabla \] Where $E$ is the shift operator, defined as $E f(x) = f(x + h)$, and $E^{-1}$ is the inverse shift operator, defined as $E^{-1} f(x) = f(x - h)$.

Step 2: Detailed Explanation:
Let us substitute the shift operator relations into the given expression: \[ \text{Expression} = (1 + \Delta)(1 - \nabla) \] Substitute $1 + \Delta = E$: \[ \text{Expression} = E(1 - \nabla) \] Substitute $1 - \nabla = E^{-1}$: \[ \text{Expression} = E \cdot E^{-1} \] Since the shift operator and its inverse are mathematical inverses of each other, their product is the identity operator: \[ E \cdot E^{-1} = 1 \] Therefore, the expression $(1 + \Delta)(1 - \nabla)$ is equivalent to 1.

Step 3: Final Answer:
The expression is equivalent to 1, which corresponds to Option (B).