How would you differentiate the function $(-1)^x$ w.r.t. $x$. Is the formula for $a^x$ valid here? Wolfram says that the derivative of $(-1)^x$ is $iπ(-1)^x$. So, does this mean than $ln(-1)=iπ$ ?
Another thing that intrigued me was that its domain and range are given to be $\emptyset$. Is it correct?(or does Wolfram Mathematica wants to say that a real domain and range do not exist?)
Update:
There arise 2 cases
1) $f_1 : \mathbb{R} \rightarrow \mathbb{R}$
$f_1(x)=(-1)^x$
2) $f_2 : \mathbb{C} \rightarrow \mathbb{C}$
$f_2(x)=(-1)^x$
Then as per the comments, $f_1$ is not differentiable and its domain and range cannot be specified, while $f_2$ is a differentiable function.(Right?)
$\endgroup$ 94 Answers
$\begingroup$Since it seems you are insisting in asking for the domain of $(-1)^x$ for $x \in \mathbb R$, then note that $$ \left( { - 1} \right)^{\,x} = \left( { - 1} \right)^{\,\left\lfloor x \right\rfloor + \left\{ x \right\}} = \left( { - 1} \right)^{\,\left\lfloor x \right\rfloor } \left( { - 1} \right)^{\,\left\{ x \right\}} $$ $(-1)$ elevated at the fractional part of $x$ is not defined in the reals, unless of course $\left\{ x \right\}$ be null $$ \left( { - 1} \right)^{\,\left\{ x \right\}} \text{n}\text{.}\,\text{d}\text{.}\quad \left| {\;0 < \left\{ x \right\}\; \wedge \;x \in R} \right. $$
so we can write $$ y = \left( { - 1} \right)^{\,x} \quad \left| {\;x \in R} \right.\quad = \left\{ {\begin{array}{*{20}c} {\left( { - 1} \right)^{\,x} = \left( { - 1} \right)^{\,\bmod \left( {x,2} \right)} } & { \Rightarrow \quad y \in \left\{ { - 1,1} \right\}} & {\left| {\;x \in Z\,} \right.} \\ {n.\,d.} & { \Rightarrow \quad y \in \emptyset } & {\left| {\;x \notin Z} \right.} \\ \end{array} } \right. $$
$\endgroup$ 0 $\begingroup$The function $f(x)=a^x$ is defined generally as
$$\begin{align} \bbox[5px,border:2px solid #C0A000]{f(x)=e^{x\log(|a|)+ix\arg(a)}} \end{align}$$
where $\arg (a)$ is the argument of $a$. Inasmuch as the argument of $a$ is multivalued (all values differing from one another by integer multiples of $2\pi$), the function $f$ is not unique.
Therefore, for $f(x)=(-1)^x$ we have
$$\begin{align} f(x)&=(-1)^x\\\\ &=e^{x\log(e^{i(2n+1)\pi})}\\\\ &=e^{i(2n+1)\pi x}\\\\ &=\cos((2n+1)\pi x)+i\sin((2n+1)\pi x) \end{align}$$
for integer values of $n$. Furthermore, the derivative $f'(x)$, of $f(x)$ is also multivalued and given by
$$\bbox[5px,border:2px solid #C0A000]{f'(x)=i(2n+1)\pi f(x)}$$
for integer values of $n$.
On the so-called Principal Branch, the complex logarithm is often denoted with a capital $L$ with
$$\text{Log}(z)=\log(|z|)+i\arctan2\left(\text{Im}(z),\text{Re}(z)\right)$$
where $\arctan2(y,x)$ is the arctangent function as described here. Therefore, on the principal branch, $-\pi<\text{Arg}(z)\le \pi$ and we find that
$$\begin{align} (-1)^x&=\cos(\pi x)+i\sin(\pi x) \end{align}$$
and
$\endgroup$ $\begingroup$$$\bbox[5px,border:2px solid #C0A000]{f'(x)=i\pi f(x)}$$
$log(y) = x log(-1)$
$d/dx(log(y)) = d/dx(x log(-1))$
$(1/y) (dy/dx) = log(-1)$
$dy/dx = log(-1) (-1)^x $
now using $-1 = e^{i\pi}$ - $log(-1) = log(e^{i\pi}) = i\pi$ - yes it does have this value, and also other values, since $-1 = e^{3i\pi}$ is also valid
$dy/dx = i\pi (-1)^x$
$\endgroup$ $\begingroup$$$f(z) = (-1)^{z} = \mathrm{e}^{z\ln(-1)}$$
\begin{align} \ln(-1) &= \ln|-1| + i\mathrm{Arg}(-1) + i2\pi n \\ &= 0 + i\pi +i2\pi n \\ &= i\pi (2n+1) \end{align} where
- $n \in \mathbb{Z}$
- We designate the principal branch of the logarithm as $-\pi \lt \mathrm{Arg}(z) \le \pi$
- On the principal branch, $n=0$ and $\ln(-1) = \mathrm{Ln}(-1) = i\pi$
Thus $$\frac{\mathrm{d}f}{\mathrm{d}z} = \ln(-1)\,(-1)^{z} = (-1)^{z}[i\pi (2n+1)]$$
On the principal branch $$\frac{\mathrm{d}f}{\mathrm{d}z} = i\pi (-1)^{z}$$
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