In the case of an extremely disconnected function such as ${\left(-2\right)}^{x}$. One definition requires that a derivative must be continous.

$(-2)^x$ has two paths that could make it discontinuous since

$$(-2)^x=\begin{cases} 2^x & n=\frac{\text{even integer}}{\text{odd integer}}\\ -\left(2^x\right) & n=\frac{\text{odd integer}}{\text{odd integer}}\\ \text{undefined} & n=\frac{\text{odd integer}}{\text{even integer}}\end{cases} $$

However, the derivative by the limit definition is possible at $\frac{\text{even integer}}{\text{odd integer}}$ as it approaches zero in... $$f'(x)=\lim_{x\to0}\frac{f(x+h)-f(x)}{h}$$

According to my answer in We can define the derivative of a function whose domain is a subset of rational numbers?

So does the derivative of ${\left(-2\right)}^{x}$ exist and if so is $\left(-2\right)^{x}$ continuous?

  • $\begingroup$ Ok then, I just needed some confirmation. $\endgroup$ – Arbuja Nov 16 '15 at 23:57

You can define the derivative of $f$ on any cluster point of its domain. This is not a new concept. For example, I phrased it like that because that is how Bartle describes it in The Elements of Real Analysis, though only briefly before restricting his attention to the usual sets.

Wherever a function $f$ is differentiable in its domain $D$, it is always continuous with respect to that domain: If $f'(x_0) = L$, then $$\lim_{D \ni x\to x_0}\frac{f(x) -f(x_0)}{x - x_0} = L.$$ Thus, letting $\epsilon = 1$, there is a $\delta > 0$ such that if $0 < |x - x_0| < \delta$ and $x \in D$, then $$\left | \frac{f(x) - f(x_0)}{x - x_0} - L \right | < 1$$ $$ -|x - x_0| < f(x) - f(x_0) - L(x - x_0) <|x - x_0|$$ $$|f(x) - f(x_0)| < (|L| + 1)|x - x_0|$$ From which it follows that $f(x) \to f(x_0)$ as $x \to x_0$ in $D$.

In this case, consider 3 possible domains for $f(x) = (-2)^x$ as you have defined it: $$D_0 = \left\{ {2n\over 2m+1}\ |\ n, m \in \Bbb Z\right\}$$ $$D_1 = \left\{ {2n+1\over 2m+1}\ |\ n, m \in \Bbb Z\right\}$$ $$D_2 = \left\{ {n\over 2m+1}\ |\ n, m \in \Bbb Z\right\} = D_0 \cup D_1$$

The differentiability of $f$ on $D_0$ follows from that of $2^x$ on $\Bbb R$, and the differentiability of $f$ on $D_1$ similarly follows from that of $-2^x$. But $f$ is obviously not continuous on $D_2$.

One final note: an additional loosening of the definition of derivative at $x_0$ would be the double limit:$$\lim_{x_1,x_2 \to x_0}\frac{f(x_1) - f(x_2)}{x_1 - x_2}$$ Under this definition $f$ no longer needs to be defined at $x_0$, so continuity fails. But by the same argument above, it must still satisfy a Cauchy condition there, so the discontinuity is removable.

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  • $\begingroup$ Is there other online sources where I can find this information? $\endgroup$ – Arbuja Nov 17 '15 at 13:48
  • $\begingroup$ Not sure what you are looking for. I don't know which Real Analysis texts other than Bartle might mentioned that derivatives are definable on general domains. He doesn't really say more about it than I've reproduced here. The proof of continuity is in any calculus/real analysis text. The discussion of your $f$ is obviously not going to be found elsewhere, and the final variant of the derivative is something I came up with so I can't tell you if anyone else discusses it. $\endgroup$ – Paul Sinclair Nov 17 '15 at 17:15
  • $\begingroup$ One last question: Must all the domains $D_0$, $D_1$ and $D_2$ be continuous for the derivative to exist for $\left({-2}\right)^{x}$? $\endgroup$ – Arbuja Nov 24 '15 at 19:51
  • $\begingroup$ "continuous" is a property of functions, not sets. Do you mean must $(-2)^x$ be continuous? That I answered already. A function is continuous at any point its derivative is defined. If you mean to aks if they have to be connected, then obviously not, as all three are completely disconnected (as are all sets of rationals), yet the derivative exists on $D_0$ and $D_1$. $\endgroup$ – Paul Sinclair Nov 24 '15 at 23:35

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