By definition of the forward difference operator $\Delta f(x) = f(x + 1) - f(x)$ with differencing step $h = 1$:

$$\Delta\left(\frac{2^x}{x!}\right) = \frac{2^{x+1}}{(x + 1)!} - \frac{2^x}{x!}$$

Express over a common denominator $(x + 1)! = (x + 1)x!$:

$$= \frac{2 \cdot 2^x}{(x + 1)!} - \frac{2^x(x + 1)}{(x + 1)!}$$ $$= \frac{2^x [2 - (x + 1)]}{(x + 1)!} = \frac{2^x(1 - x)}{(x + 1)!}$$

Numerical integration using Simpson's 1/3rd Rule:
Interval $[0, 1]$ with step size $h = 0.5$ has $n = 2$ subintervals.
Node values for $f(x) = \frac{1}{1 + x}$:

• $x_0 = 0 \implies y_0 = \frac{1}{1 + 0} = 1.0000$
• $x_1 = 0.5 \implies y_1 = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667$
• $x_2 = 1.0 \implies y_2 = \frac{1}{1 + 1} = 0.5000$

Simpson's 1/3rd Rule formula:

$$I = \frac{h}{3}[y_0 + 4y_1 + y_2]$$ $$I = \frac{0.5}{3}\left[1.0 + 4\left(\frac{2}{3}\right) + 0.5\right] = \frac{1}{6}\left[1.5 + \frac{8}{3}\right] = \frac{1}{6}\left[\frac{12.5}{3}\right] = \frac{12.5}{18} \approx 0.69444 \approx 0.6945$$

In numerical analysis, the shift operator $E$ relates to finite difference operators as:
$E = 1 + \Delta$, and $E^{-1} = 1 - \nabla$.
Multiplying the two operators:
$(1 + \Delta)(1 - \nabla) = E \cdot E^{-1} = 1$.
Alternatively: $(1 + \Delta)(1 - \nabla) = 1 - \nabla + \Delta - \Delta\nabla$.
Since $\Delta - \nabla = \Delta\nabla$, the terms cancel, yielding 1.

Detailed analysis of options:

• 0 is incorrect.
• is the correct identity ($E \cdot E^{-1} = 1$).
• 1 is incorrect.
• is incorrect.

Numerical Integration (also called numerical quadrature) is the process of computing the value of a definite integral $\int_a^b f(x)\,dx$ when the function $f(x)$ is given as a table of discrete values $(x_i, y_i)$ rather than an explicit analytical form (e.g., Trapezoidal rule, Simpson's rules).

Detailed analysis of options:

• Numerical Integration is the standard scientific term for this process.
• Extrapolation is finding values outside the range of known data points.
• Finite Difference refers to differences between successive values.
• Interpolation is estimating the value of a function between known tabulated points.