Discussion
Let E be the shift operator and \(\mu\) be the averaging operator. Then :
Answer: B
The averaging operator is μ = (E1/2 + E−1/2)/2.
By definition
\[ \begin{aligned} \mu f(x) &= \tfrac{1}{2}\left[f\left(x + \tfrac{h}{2}\right) \right. \\ &\quad \left. + f\left(x - \tfrac{h}{2}\right)\right] \end{aligned} \]
Since \(E^{1/2}f(x) = f(x + h/2)\) and \(E^{-1/2}f(x) = f(x - h/2)\),
\[ \mu = \frac{1}{2}\left(E^{1/2} + E^{-1/2}\right) \]
\(E^{1/2} - E^{-1/2}\) is the central difference operator \(\delta\), not \(\mu\); half of it is not a standard operator, and dividing by 3 does not average two values.
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