# If $f \in C^\infty$, then $f''(x) = \lim_{t \to 0} \frac{f(x + t) + f(x - t) - 2f(x)}{t^2}$ [duplicate]

I would like help to show the following:

If $$f \in C^\infty$$, then $$f''(x) = \lim_{t \to 0} \frac{f(x + t) + f(x - t) - 2f(x)}{t^2}$$

This is the definition of second order derivative by differential coefficients. The first order case is just the definition of derivative, but I'm having trouble with the second order case.

I'm really confused, don't even know where to start.

• I would say, start with Taylor's theorem. See if your textbook has a section on it. – GEdgar Feb 18 at 21:31

Apply Taylor's theorem to get

\begin{align} f(x+t) &=f(x)+f'(x)t+\frac{f''(x)}{2}t^2+O(t^3)\\ f(x-t) &=f(x)-f'(x)t+\frac{f''(x)}{2}t^2+O(t^3) \end{align}

Thus, you get that

$$f(x+t)+f(x-t)-2f(x)=f''(x)t^2+O(t^3)$$ which should allow you to readily finish the proof.

Hint.

The assumption that $$f\in C^\infty$$ is very strong.

Simply applying the L'Hopital rule (and the chain rule) once to the RHS, you have $$\lim_{t\to 0}\frac{f'(x+t)-f'(x-t)}{2t}$$

You can keep using L'Hotital (because of the luxury that $$f\in C^\infty$$) or observe that $$\frac{f'(x+t)-f'(x-t)}{2t}=\frac12\left(\frac{f'(x+t)-f'(x)+f'(x)-f'(x-t)}{t}\right)$$ where you can use the definition of $$f''(x)$$.