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If the definition of the derivative is $$ f^\prime(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} $$ Would it make sense that the nth derivative would be (I know that the 'n' in delta x to the nth power is useless) $$ f^{(n)}(x)=\lim_{\Delta x \to 0} \sum_{k=0}^{n}(-1)^k{n \choose k}\dfrac{f(x+\Delta x(n-k))}{\Delta x^n} $$ I came to this conclusion using this method $$ f^\prime(x) = \lim_{\Delta x \to 0} \dfrac{f(x+\Delta x) - f(x)}{\Delta x} $$ (this is correct right?) $$ f^{\prime\prime}(x) = \lim_{\Delta x \to 0} \dfrac{f^\prime(x+\Delta x) - f^\prime(x)}{\Delta x}=$$
$$\lim_{\Delta x \to 0}\dfrac{\dfrac{f((x+\Delta x)+\Delta x)-f(x+\Delta x)}{\Delta x}-\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}{\Delta x}=$$
$$\lim_{\Delta x \to 0}\dfrac{f(x+2\Delta x)-2f(x+\Delta x)+f(x)}{\Delta x^2} $$ After following this method a couple of times(I think I used it to the 5th derivative) I noticed the pattern of $$(a-b)^n$$ And that is how i arrived at $$ f^{(n)}(x)=\lim_{\Delta x \to 0} \sum_{k=0}^{n}(-1)^k{n \choose k}\dfrac{f(x+\Delta x(n-k))}{\Delta x^n} $$ Have I made a fatal error somewhere or does this definition actually follow through?
Thanks for your time I really appreciate it.
P.S. Any input on using tags will be appreciated.

tiendbz
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  • looks good to me. Sometime folks prefer centered formulas. That is to make $x$ to be center of all points considered as opposed to having all other points after or before $x$. That gives better agreement in numerical work. – Maesumi Mar 06 '14 at 20:56
  • I'd be concerned that the $\Delta x$ from say the second derivative is a different object than the $\Delta x$ from the first derivative. You have a limit involving $\Delta x^n$ as $\Delta x\to0$, but the straightforward iterated interpretation would involve a limit $\Delta x_1\Delta x_2\cdots\Delta x_n$ as $\Delta x_n\to0$, $\Delta x_{n-1}\to0$, ..., $\Delta x_1\to0$. And I would guess that there are functions where the iterated (latter) limit is one thing, and the amalgamated limit (the former) is something else. – 2'5 9'2 Mar 06 '14 at 21:05
  • Ah, I see what you mean. Sense the end result does not change it was a bit hard to notice. Thanks alex.jordan – tiendbz Mar 06 '14 at 21:33
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    This method was used by Riemann (for 2nd order version only, I think) as a slightly weaker notion than ordinary 2nd order differentiability that he applied in the study of trigonometric series. The first few sentences of J. Marshall Ash's 1970 paper A characterization of the Peano derivative may be useful. For more than you'd ever possibly want to know about this topic, see Satya Mukhopadhyay's 2012 book Higher Order Derivatives (I was able to see, in "preview", the table of contents). – Dave L. Renfro Mar 06 '14 at 21:42
  • Thanks Dave L. Renfro. That book looks like it will answer most of my questions regarding this subject. – tiendbz Mar 07 '14 at 08:04
  • @tiendbz I'm still not sure, but I think it might be possible to find a function $f$ where $\lim\limits_{h\to0}\sum\limits_{k=0}^2(-1)^k\binom{2}{k}\frac{x+h(2-k)}{h^2}$ is not equal to $\lim\limits_{h_2\to0}\frac{\lim_{h_1\to0}\frac{f(x+h_2+h_1)-f(x+h_2)}{h_1}-\lim_{h_1\to0}\frac{f(x+h_1)-f(x)}{h_1}}{h_2}$. I could be wrong, and maybe these things are provably equal. But I can't help but feel that there could be a pathological $f$ where these are not equal. In general, the iterated limit as $h_2\to0,h_1\to0$ need not equal a corresponding limit as $h\to0$. – 2'5 9'2 Mar 07 '14 at 08:14
  • possible duplicate of Prove that $\lim_{\Delta x\rightarrow 0} \frac{\Delta ^{n}f(x)}{\Delta x^{n}} = f^{(n)}(x)$ – Paramanand Singh Mar 09 '14 at 07:45
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    The derivative is inherently dependent on the function in a very direct way. Thus $f'(a)$ depends on values of $f$ near (and on) $a$. And similarly $f''(a)$ depends on values of $f'$ near $a$ (which in turn depend on values of $f$ near $a$). But the dependence of $f''(a)$ on values of $f$ is much more indirect than its dependence on values of $f'$. It is better to defined n-th derivative as "derivative" of "(n - 1)-th derivative" rather than in terms of the original function. The limit formula you have mentioned is correct provided n-th derivative exists and not the other way round. – Paramanand Singh Mar 09 '14 at 07:50
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    It is possible to construct an example of a function where the limit formula gives a value but function is not differentiable n-times according to accepted definition. – Paramanand Singh Mar 09 '14 at 07:51
  • @Paramanand Sigh I see. If this is not the definition of the nth derivative can i still use it or is it flawed in the way that alex.jordan describes it? – tiendbz Mar 11 '14 at 09:13
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    @tiendbz: For most usual well behaved functions (i.e. function whose n-th derivatives exist) this can be used to calculate the n-th derivative directly by a single limit operation. But there can be cases where such a limit exists, but the function itself is not differentiable n times. – Paramanand Singh Mar 11 '14 at 10:32
  • Hi @ParamanandSingh... I write you beceause in this answer to a question I did, it is used the same procedure shown in the question above, and since you have made here some observations, I hope you could visit also the mentioned answer in order to check if its OK. Beforehand, thank you very much. – Joako Jul 09 '23 at 12:23

2 Answers2

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This is probably not a good definition of the $n$th derivative. To see this, consider the case $n = 2$: $$ f''(x) = \lim_{h \to 0} \frac{f(x + 2h) - 2f(x + h) + f(x)}{h^2} $$ Define $f: \mathbb{R} \to \mathbb{R}$ as follows. First, define $f(0) = 0$. Now define $f$ on the intervals $\left[-1, -\tfrac12\right)$ and $\left(\tfrac12, 1\right]$ to be your favorite unbounded function, for instance $\frac{1}{x^2 - 1/4}$ is a good choice. Now, for any $x$, let $k$ be the unique integer such that $2^k x$ is contained in one of these intervals, and define $f(x) = 2^{-k} f(2^k x)$.

This construction satisfies $f(2h) = 2f(h)$ for all $h \in \mathbb{R}$, so the derivative formula above gives $$ f''(0) = \lim_{h \to 0} \frac{f(2h) - 2f(h) + f(0)}{h^2} = \lim_{h \to 0} \frac{0}{h^2} = 0 $$ However, $f$ is wildly discontinuous at $0$, and is in fact unbounded in any neighborhood containing $0$.

Caleb Stanford
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  • Does this come from the fact that he should have used another variable for the second $\Delta x$ so as to not confuse it with the first? – DanielV May 01 '14 at 03:40
  • @DanielV Yes, probably if there were more than one variable it would work as a definition. See also my question here, which has not yet been answered. – Caleb Stanford May 01 '14 at 03:42
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when you did for f''(x) you are taking both limits as a single limit for delta x. what I did is take $h_1$ for first limit and $h_2$ for second limit, so that nth derivative is limit $(h_1, h_2, h_3,.... h_n)$ --> $(0,0,0,....,0)$ and some function.

Mc Cheng
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PremVijay
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