Discussion
Let \(f(x) = e^{x}\), then \(\Delta^{n} e^{x} = \ldots\)
Answer: A
\(\Delta^{n} e^{x} = e^{x}(e^{h} - 1)^{n}\).
With step \(h\):
\[ \begin{aligned} \Delta e^{x} &= e^{x + h} - e^{x} = e^{x}(e^{h} - 1) \\ \Delta^{2} e^{x} &= (e^{h} - 1)\Delta e^{x} = e^{x}(e^{h} - 1)^{2} \end{aligned} \]
and by induction each application multiplies by the constant \(e^{h} - 1\). In operator form, \(\Delta = E - 1\) and \(E e^{x} = e^{h} e^{x}\).
The factor \((e^{h} + 1)\) confuses \(\Delta\) with the averaging operator, \(e^{x}\) alone would be the derivative rather than the difference, and \(e^{-x}\) has no source in the computation.
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