# Show that $\nabla\Delta = E$.

My assignment asks me to prove that
$$\nabla\Delta = E.$$ I was taught that \begin{align*} \nabla f(x) &= f(x) - f(x-h) \\ \Delta f(x) &= f(x+h) - f(x) \\ E f(x) &= f(x+h) \end{align*} I was taught to show equality assuming we have a function $f(x)$, so I essentially plugged the above part into the identity that has to be proven. I can't seem to do it though. How to prove this?

• Let $g(x)=\Delta f(x)=f(x+h)-f(x)$. Now apply $\nabla$ to $g(x)$, and then write out the result in terms of $f$. – Alex R. Sep 5 '16 at 18:40
• Wouldn't that end up as f(x+h) - 2f(x) + f(x-h) ? – Arctus Sep 5 '16 at 19:12

As you have found $\nabla\Delta\ne E$, since \begin{align*} \nabla\Delta f(x) &= \nabla(f(x+h)-f(x)) \\ &= \nabla f(x+h) - \nabla f(x) \\ &= (f(x+h)-f(x)) - (f(x)-f(x-h)) \\ &= f(x+h) - 2f(x) + f(x-h) \\ &\ne f(x+h). \end{align*} It is true that $\nabla\Delta = \Delta\nabla = \delta^2$, where $\delta f(x) = f(x+h/2)-f(x-h/2)$ is the central difference.
• Intuitively, for small $h$, $\nabla\Delta$, $\nabla^2$, and $\Delta^2$ are all proportional to the second derivative operator. See the discussion here, for example. – user26872 Sep 5 '16 at 23:30