Alternate formula for second derivative of single-variable real valued function

Question

$\newcommand{\R}{\mathbb{R}}$ I am familiar with the following fact (second answer here): given a function $f: \R \to \R$ that is differentiable at $a \in \R$, it is true that $$ f'(a) = 2 \cdot \lim_{h \to 0} \frac{f(a + h) - f(a - h)}{h} .$$ This lead me to explore whether similar looking formula exist for higher derivatives (obviously we can replace $f$ above with $f''$, so more interesting ones). Assuming $f$ is twice differentiable at $a$, we also have \begin{align*} f''(a) &= \lim_{h \to 0} \frac{f'(a + h) - f'(a)}{h} \\ &\color{red}{=} \lim_{h \to 0} \frac{\lim_{h \to 0} \left( \frac{f'(a + 2h) - f'(a + h)}{h} \right) - \lim_{h \to 0} \left( \frac{f'(a + h) - f'(a)}{h} \right)}{h} .\\ \end{align*} At this point, if we could remove the inside limits, we would get the formula $$ f''(a) = \lim_{h \to 0} \frac{f(a + 2h) - 2f(a + h) + f(a)}{h^2} $$ which I think (?) I have seen before, but two things lead me to doubt this.

1. Can we remove the inside limits?
2. Even if we could, isn't the above mathematics abuse of notation? Really shouldn't we be using a variable different than $h$ for the inside limits? (Note this question is regarding the equality highlighted in red, whereas question #1 is concerning the step directly after it.) Maybe this just means that the statement is false, but I vaguely remember seeing a statement exactly like the one above, hence my hesitancy to reject this method of argument.

Answer 1

A little late, but here is the follow up on my comment.

Let start with a reminder on how to get the Taylor expansion with integral (FTC and integration by parts):

$\displaystyle f(x)-f(x_0)=\int_{x_0}^{x}f'(t)dt=\Big[(t-x)f'(t)\Big]_{x_0}^x- \int_{x_0}^x (t-x)f''(t)dt$

$$f(x)=f(x_0)+(x-x_0)f'(x_0)+\int_{x_0}^x (x-t)f''(t)dt$$

Applying in $\quad\begin{cases}x=a+2h\\x_0=a+h\end{cases}\quad$ and $\quad\begin{cases}x=a\\x_0=a+h\end{cases}\quad$ we get:

$\displaystyle f(a+2h)=f(a+h)+hf'(a+h)+\int_{a+h}^{a+2h}(a+2h-t)f''(t)dt$

$\displaystyle f(a)=f(a+h)-hf'(a+h)+\int_{a+h}^{a}(a-t)f''(t)dt$

After summation of these two formulas we obtain:

$\displaystyle f(a+2h)-2f(a+h)+f(a)=\int_{a+h}^{a+2h}(a+2h-t)f''(t)dt+\int_{a+h}^{a}(a-t)f''(t)dt$

Notice now that if we replace $f''(t)$ by a constant $K$ in these integrals, then they both evaluate to the same value $\frac 12h^2K$.

In particular:

• for $K=f''(a)$ we get $h^2f''(a)$ on the RHS
• for $K=\epsilon$ we get $h^2\epsilon$ on the RHS

Assuming the local continuity of $f''$ in $a$ we can bound $|f''(t)-f''(a)|<\epsilon$

Therefore subtracting $h^2f''(a)$ on both side, we make $f''(t)-f''(a)$ appear inside the integrals and:

$\displaystyle \Big|f(a+2h)-2f(a+h)+f(a)-h^2f''(a)\Big|<h^2\epsilon$

Which is the desired result (well we have to divide by $h^2$, select a delta such that $|t-a|<h$ gives the desired $\epsilon$, but you got the point).