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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.

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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 6 '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. – alex.jordan Mar 6 '14 at 21:05
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 9 '14 at 7:50
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 9 '14 at 7:51
@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

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 $0$.

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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 1 '14 at 3:40
@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. – 6005 May 1 '14 at 3:42

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