We can define the derivative of a function whose domain is a subset of rational numbers?

Question

Usually the derivative is defined for a function $f:A\to \mathbb{R}$ where $A \subset \mathbb{R}$, and the usual definition of the derivative at a point $a$ require the existence of an open neighborhood of $a$ where the function is defined.

So, if $A\subset \mathbb{Q}$ it seems that we cannot define a derivative, since $A$ is totally disconnected.

But the definition

$$ f'(a)=\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h} $$

require only the existence of the limit that, with the $\epsilon -\delta$ definition, can be found using only rational values of $h$.

So it seams that a ''derivative'' can be defined. Or there is something that does not works?

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This question is suggested by Clarification if a disconnected function has a derivative at defined points. , where the OP asks for the derivability of the function $$ f:\{x=\dfrac{n}{2k+1} | n,k \in \mathbb{Z}\} \to \mathbb{R} \quad;\quad f(x)=(-2)^x $$

Answer 1

There's no reason that a derivative couldn't be defined on such a set like $\mathbb Q$. As you note, the limit $$\lim_{h\rightarrow 0}\frac{f(a+h)-f(h)}{h}$$ may still be calculated even if $a+h$ is restricted to only rational values.

I think it's worth noting that the fact that $\mathbb Q$ is totally disconnected might give the wrong impression. The above definition only fails when fed isolated points - that is points with no other points in the domain in an open neighborhood. In the language of topology, we could say points such that $\{x\}$ is open in the domain of the function. Whether or not the domain is connected it somewhat irrelevant. So, you can't use this definition of for a function defined only on $\mathbb Z$ where every point is isolated. But it works fine on any dense subset of $\mathbb R$, like $\mathbb Q$. Another similar thing to think about is that not all function $f:\mathbb Q\rightarrow\mathbb R$ are continuous, despite $\mathbb Q$ being totally disconnected.

We might avoid using such a derivative as one might notice that a lot of theorems (e.g. the fundamental theorem of calculus) really do need conditions like "$f:\mathbb R\rightarrow\mathbb R$ is everywhere differentiable" which can't be replaced by $f:A\rightarrow\mathbb R$ being differentiable everywhere in its domain.

Answer 2

This is not an answer but this could lead to an answer

Taking the definition of a limit.... $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

I found that there can be a derivative of $(-2)^x$ but only if $h\to0$ in a way that the numerator is even and the denominator is odd (even/odd). Such as $h\to\frac{2}{5}\to\frac{2}{101}\to\frac{2}{1000001}\to0$.

This is because using odd numerators and denominators (odd/odd) will result in the derivative diverging to $\infty$.

This is because for $(-2)^x$ the output is negative when x is (odd/odd) or positive when x is (even/odd).

For example $(-2)^{-1/15}=-.954841$ and $(-2)^{12/71}=1.124289$, which means. $$(-2)^x=\begin{cases} 2^x & s=\left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\left(2^x\right) & s=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\ \\ \text{undefined} & s=\left\{ {2n+1\over 2m}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{even integer}} \end{cases} $$

So then when we apply the formal definition of $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ where $f(x)=\left({-2}\right)^{x}$ $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}$$

First if $h=\text{odd}/\text{even}$ such as $1/2$ and we take $x$ as $odd/odd$ or $x=1/3$ then $f(x)$ is already undefined. This means the entire limit is undefined. Now to get $f(x+h)$ we must get $x+h$ which is $5/6$ which means $f(x+h)$ is also undefined whenver $h$=odd/even. Thus...

$$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}-2^x}{h}$$ $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{\text{undefined}-\text{undefined}}{h}$$ $$\text{Limit does not exist}$$

Now if $h=\text{odd}/\text{odd}$ for example $h=1/3$ and let $x$ be (even/odd) fraction like $x=2/3$. So $f(x)=\left({2^x}\right)$; however $x+h=1$. This shows under these conditions $x+h$ will always be (odd/odd) and thus $f(x+h)=-\left(2^{x+h}\right)$

So then when x is odd/odd the limit will be $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}-2^x}{h}$$
$$\lim_{h\to0}\frac{\left(\left({-{2}^{h}-1}\right)*{{2}^{x}}\right)}{h}\approx$$ $$\lim_{h\to0}\frac{-2\left(2\right)^{x}}{h}=$$ $$\lim_{h\to{0}}\frac{1}{h}*{-2\left(2\right)^{x}}$$ Now since$-2({2})^x$ is always postive we have to analyze $\lim_{h\to0}\frac{1}{h}$. Since that part of the limit does not exist there is no derivative.

$$\text{Limit Does Not Exist}$$

If $h=\text{odd}/\text{odd}$ for example $h=1/3$ but $x$ is (odd/odd) fraction like $x=1/5$. So $f(x)=-\left(2^x\right)$; however, $x+h=\frac{8}{15}$. This shows that under these conditions $x+h$ is always (even/odd) so $f(x+h)=2^{x+h}$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{2^{x+h}+2^x}{h}$$ $$\lim_{h\to0}\frac{\left(2^{h}+1\right)\left({2^x}\right)}{h}\approx$$ $$\lim_{h\to0}\frac{2\left({2}^{x}\right)}{h}=$$ $$\lim_{h\to0}\frac{1}{h}*2\left({2}^{x}\right)$$

Since $2\left({2}\right)^{x}$ is always positive but also has $\lim_{h\to0}\frac{1}{h}$ with a limit that does not exist there no derivative.

$$\text{Limit Does Not Exist}$$

But if $h=\text{even}/\text{odd}=2/3$ and $x=\text{odd}/{\text{odd}}=1/3$ then $f(x)=-\left(2^x\right)$. Since $x+h=3/3$, under these conditions, $x+h$ is always (odd/odd) and so $f(x+h)=-\left(2^{x+h}\right)$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{-2^{x+h}+2^{x}}{h}=-\ln{(2)}{2^{x}}\to\text{For x=odd/odd}$$

And if $h=\text{even}/\text{odd}=2/3$ and $x=\text{even}/{\text{odd}}=2/3$ then $f(x)=2^x$. Since $x+h=4/3$, under these conditions, $x+h$ is always (even/odd) and thus $f(x+h)=2^{x+h}$. $$\lim_{h\to0}\frac{(-2)^{x+h}-(-2)^x}{h}=\lim_{h\to0}\frac{2^{x+h}-2^x}{h}=\ln{(2)}{2^x}\to\text{For x=even/odd}$$

$$\frac{d}{dx}(-2)^x=\begin{cases} \ln(2)2^x & s=\left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{even integer}}{\text{odd integer}}\\ -\ln(2)2^x & s=\left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}\frac{\text{odd integer}}{\text{odd integer}}\end{cases}=\text{Limit does not exist}$$

According to what I have heard from mathematicians it is possible to define a derivative at cluster points. However for the defined cluster points of each group must have the same limit.

Take $\lim_{x\to\infty}\left({-1}^{x}\right)$ for example. It ossilates between $-1,1,-1,1$ and cannot exist. If I choose (even/odd) for $x\to\infty$ ($2/3\to2000/3\to200000/3$) then the only get $1$ but if I choose (odd/odd) for $x\to\infty$ ($1/3\to10/3\to10000/3$) then I only get $-1$. Thus all sequences for defined intervals of values that approaches some value must have the same limit.

Thus for $\left({-2}\right)^{x}$ the derivative must not exist. However if we instead took $|\left({-2}\right)^{x}|$ then the derivative would be $\ln{\left(2\right)}|\left(-2\right)^{x}|$. Infact when you put the absolute value the derivative satisfies the mean value theorem and rolles theorem.