# Rudin 5.11. Second order derivative

I got a slightly different result than suggested

\begin{align} f''(x)&=\lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x-h)}{2h} &(1)\\ &=\lim_{h\rightarrow 0}\frac{\frac{f(x+h)-f(x)}{h}-\frac{f(x)-f(x-h)}{h}}{2h}&(2)\\ &=\lim_{h\rightarrow 0}\frac{f(x+h)+f(x-h)-2f(x)}{2h^2}&(3) \end{align}

It seems that I got an extra factor of 2. I realized that such proof is not rigorous, because from (1) to (2), there are double limits and something mysterious could happen. Is it the reason why I got a wrong result?

• Try writing a Taylor approximation centered at $x$, evaluating it at $f(x+h)$ and also at $f(x-h)$. Work with the remainder term $R$ which satisfies $\lim_{h\to 0} R(h)/h^2=0$. – FakeAnalyst56 Jun 20 at 2:48

As you mentioned, the step from (1) to (2) is a little suspicious, especially since by definition, $$$$f'(x+h) = \lim_{k \to 0} \dfrac{f(x+h+k) - f(x+h)}{k}.$$$$

The most common, and the most easily generalizable approach to this question would be to use a Taylor expansion: \begin{align} f(x+h) = f(x) + h f'(x) + \dfrac{h^2}{2} f''(x) + \varphi(h), \tag{*} \end{align} for some function $$\varphi$$ (which is just LHS minus RHS) defined in a neighbourhood of $$x$$, such that $$\lim_{h \to 0} \dfrac{\varphi(h)}{h^2} = 0$$. The proof of the fact that $$\varphi$$ satisfies this limit condition can be found in Michael Spivak's Calculus, Chapter $$20$$, Theorem $$1$$ (or you can come up with your own proof; it's not that hard to prove... in the general case, just apply L'Hopital's rule several times, or you can give a proof by induction).

Similarly, we have \begin{align} f(x-h) = f(x) - h f'(x) + \dfrac{h^2}{2} f''(x) + \varphi(-h). \tag{*} \end{align}

Can you see how adding, rearranging terms of $$(*)$$ and $$(**)$$ yields the desired result?

• Yes. Thank you. I also see how 1/2 plays a role, as 1/2 appears in Taylor expansion to the second order. – Mou Jun 20 at 3:05

You made a slight mistake. Here, the corrected version.

\begin{align} f''(x)&=\lim_{h\rightarrow 0}\frac{f'(x+h)-f'(x-h)}{2h} &(1)\\ &=\lim_{h\rightarrow 0}\frac{\frac{f(x+2h)-f(x)}{2h}-\frac{f(x)-f(x-2h)}{2h}}{2h}&(2)\\ &=\lim_{h\rightarrow 0}\frac{f(x+2h)+f(x-2h)-2f(x)}{4h^2}&(3)\\ &=\lim_{h'\rightarrow 0}\frac{f(x+h')+f(x-h')-2f(x)}{(h')^2}&(4)\\ \end{align} where $$h'=\frac{h}{2}$$. Since, $$h$$ is small enough and it approaches $$0$$. Therefore, we can replace $$h'$$ by $$h$$ in $$(4)$$, and we get the desired formula.

• @ downvoter, can you please let me know my mistake? – Kumar Jun 20 at 3:02
• I did not downvote your post, but (2) does not immediately follow from (1). More precisely, it is not true that $f'(x+h) = \frac{f(x+2h) - f(x)}{2h}$. – Jesse Madnick Jun 20 at 3:03
• @JesseMadnick I see your point. I should be using some another variable in $(2)$ but as they both approach zero simultaneously, I made the abuse and used the same $h$. I have used the equation: $f'(x)=lim_{k\rightarrow 0}\frac{f(x+k)-f(x-k)}{2k}$. Maybe it makes some sense now. – Kumar Jun 20 at 3:08
• @jesse, assuming $f$ is twice differentiable, I think $\lim_{h\rightarrow0} f'(x+h)=f'(x) = \lim_{2h\rightarrow 0}\frac{f(x+2h)-f(x)}{2h}$. The main error in this proof, I believe, is still the use of double limits. We're now able to get arbitrary results if we use it wrongly. – Mou Jun 20 at 3:14