Question

If $\Delta$ and $\nabla$ be the first forward difference and first backward difference operator respectively of a function $f(x)$ then the expression $(\Delta - \nabla)$ is equivalent to

• A. $\frac{\Delta}{\nabla}$
• B. $\Delta^2$
• C. $\Delta \nabla$
• D. $\left( \frac{1}{\Delta} - \frac{1}{\nabla} \right)$

Hint:

Using the shift operator $E$ is the most robust way to prove any operator identity in finite differences. Always express $\Delta$ and $\nabla$ in terms of $E$ first.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Finite difference operators ($\Delta$, $\nabla$, and $E$) have established algebraic relationships used in numerical analysis.
Key Formula or Approach:
The relationships between the forward difference ($\Delta$), backward difference ($\nabla$), and shift operator ($E$) are: \[ \Delta = E - 1 \] \[ \nabla = 1 - E^{-1} \]

Step 2: Detailed Explanation:
Let us analyze the product of the two operators, $\Delta \nabla$: \[ \Delta \nabla = (E - 1)(1 - E^{-1}) \] Expand the terms: \[ \Delta \nabla = E(1) - E(E^{-1}) - 1(1) + 1(E^{-1}) \] \[ = E - 1 - 1 + E^{-1} \] \[ = E - 2 + E^{-1} \]
Now, let us analyze the difference between the two operators, $\Delta - \nabla$: \[ \Delta - \nabla = (E - 1) - (1 - E^{-1}) \] Simplify this expression: \[ = E - 1 - 1 + E^{-1} \] \[ = E - 2 + E^{-1} \]
Since both expressions simplify to the exact same form ($E - 2 + E^{-1}$), we establish the operator identity: \[ \Delta - \nabla = \Delta \nabla \]

Step 3: Final Answer:
The expression $\Delta - \nabla$ is equivalent to $\Delta \nabla$.