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In this post there is a faulty derivation of:

$$f''(x)=\lim_{h \to 0} \frac{f(x+2h)-2f(x+h)-f(x)}{h^2} \tag{1}$$

What I would like to do is derive it soundly. Although Taylor series and L'hopitals rule are great for conforming the limit, I don't think they are great for finding the formula from scratch.

I've always noted something about this limit and the other limits for the other derivatives. If we define $\Delta_h$ to map $f(x) \to f(x+h)-f(x)$. Then it seems as though $\lim_{h \to 0} \frac{\Delta_h^n}{h^n}=\frac{d^n}{dx^n}$. My question is why, and does this provide any insight as to how we may derive $(1)$ from scratch?

Ahmed S. Attaalla
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    They are not the same though. The limit expression is much more general and exists when the second derivative doesn't. Take the $\operatorname{sgn}$ function as an example. Are you asking for the derivation when we know the second derivative exists? – Jonathan Davidson Jul 21 '17 at 23:11
  • Yes, that is what I am asking @JonathanDavidson – Ahmed S. Attaalla Jul 21 '17 at 23:18
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    As a fun side note, if we defined have:$$f^{(a)}(x)\equiv\lim_{n\to\infty}n^a\sum_{k=0}^n(-1)^i\binom akf\left(x+\frac{a-k}n\right)$$then you can get silly things like $f^{(1/2)}(x)$ and, more importantly, $$f^{(-1)}(x)=\int_{x-1}^xf(t)~\mathrm dt$$i.e. negative derivatives translate into integrals. – Simply Beautiful Art Jul 21 '17 at 23:54
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    And as a last note of caution, I do believe there are necessary assumptions placed on $f(x)$ in order for these limits to equal the claimed derivatives. It shouldn't be too hard to find examples where $f''(a)$ doesn't exist but the symmetric limit does. (Just take any function that satisfies $f(2h)-2f(h)+f(0)=0$ for example) – Simply Beautiful Art Jul 21 '17 at 23:59

2 Answers2

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In fact Taylor's theorem itself can be stated in operator language as

$$e^{hD}=I+\Delta_h$$

where $D$ is the differentiation operator, $I$ is the identity operator, and $\Delta_h$ is what you called $\Delta$. Of course, strictly speaking this is not really Taylor's theorem; it is Taylor's theorem in the analytic case, and even then there is no "error term" included. Nevertheless, even in the non-analytic case, this tells you that "formally" (i.e. in the sense of asymptotic expansions),

$$D^n=\left ( \frac{\ln(I+\Delta_h)}{h} \right )^n = \frac{\ln(I+\Delta_h)^n}{h^n}.$$

Thus truncating a series expansion of $\ln(I+\Delta_h)$ and raising it to the $n$th power gives a consistent formula for approximating $D^n$. The leading order expansion is exactly $\Delta_h$ which gives your result.

Ian
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Directly, we see that:$$f''(x)=\lim_{h_2\to0}\lim_{h_1\to0}\frac{f(x+h_1+h_2)-f(x+h_1)-f(x+h_2)+f(x)}{h_1h_2}$$Now apply the mean value theorem, assuming $f'(x)$ is continuous and you will get$$f''(x)=\lim_{h\to0}\frac{\Delta^2_hf(x)}{h^2}$$Indeed, we may prove by induction + mean value theorem + continuity assumption that:$$f^{(n)}(x)=\lim_{h\to0}\frac{\Delta_h^nf(x)}{h^n}$$Which happens to be the Grünwald–Letnikov fractional derivative.

As a quick example, apply MVT to $\frac{f(x+h_1)-f(x)}{h_1}$ and $\frac{f(x+h_1+h_2)-f(x+h_2)}{h_1}$ to get

$$f''(x)=\lim_{h_2\to0}\lim_{h_1\to0}\frac{f'(c_1)-f'(c_2)}{h_2},c_1\in(x+h_2,x+h_2+h_1),c_2\in(x,x+h_2)$$

Apply the MVT again to get

$$f''(x)=\lim_{h_2\to0}\lim_{h_1\to0}f''(d),d\in(x,x+h_1+h_2)$$

whereupon we see we may set $h_1=h_2$ and we are basically done.

Simply Beautiful Art
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