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The value of $(1+\Delta)(1-\nabla)$ is
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Verified Answer
The correct answer is:
1
We have, $(1+\Delta)(1-\nabla) f(x)$
$$
\begin{array}{l}
=(1+\Delta)\{(1-\nabla) f(x)\} \\
=(1+\Delta)\{f(x)-\nabla f(x)\} \\
=(1+\Delta)[f(x)-\{f(x)-f(x-h)\}] \\
=(1+\Delta) f(x-h)=E f(x-h) \\
\quad[\because(E=1+\Delta)] \\
=f(x)=1 \cdot f(x)
\end{array}
$$
Thus, $(1+\Delta)(1-\nabla) f(x)=1 \cdot f(x)$, for any function $f(x)$.
$$
\therefore \quad(1+\Delta)(1+\nabla)=1
$$
$$
\begin{array}{l}
=(1+\Delta)\{(1-\nabla) f(x)\} \\
=(1+\Delta)\{f(x)-\nabla f(x)\} \\
=(1+\Delta)[f(x)-\{f(x)-f(x-h)\}] \\
=(1+\Delta) f(x-h)=E f(x-h) \\
\quad[\because(E=1+\Delta)] \\
=f(x)=1 \cdot f(x)
\end{array}
$$
Thus, $(1+\Delta)(1-\nabla) f(x)=1 \cdot f(x)$, for any function $f(x)$.
$$
\therefore \quad(1+\Delta)(1+\nabla)=1
$$
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