Why the negative number raised to X gives strange graph?
Y = (-1) ^ X
This is sine and cosine both functions in real and imaginary part.
How to explain this?
[Math] $(-1)^X?$
complex numbersgraphing-functionswolfram alpha
Related Solutions
Using this software you should get following graph for $\frac{1}{\ln x}$ ,when $x>0$
$\ln(x)$ is formally defined as the solution to the equation $e^y=x$.
If $x$ is positive, this equation has an unique real solution, anyhow if $x$ is negative this doesn't have a real solution. But it has complex roots.
Indeed, $\ln(x)= a+ib$ is equivalent to
$$x= e^{a+ib}= e^{a} (\cos(b)+i \sin (b)) \,.$$
If $x <0$ we need $e^{a}=|x|$, $\cos(b)=-1$ and $\sin(b)=0$.
Thus, $a= \ln(|x|)$ and $b=\frac{3\pi}{2}+2k\pi$....
Best Answer
If you heard about complex numbers, you should wonder how powers are evaluated in the complex.
A possible definition is with the polar representation
$$z=re^{i\theta}\implies z^x=r^xe^{i\theta x}$$ where $x$ is a real number.
Then
$$(-1)^x=(e^{i\pi})^x=e^{i\pi x}=\cos(\pi x)+i\sin(\pi x).$$
Beware that this is not the definition, because one also has
$$(-1)^x=(e^{i3\pi})^x=\cos(3\pi x)+i\sin(3\pi x)$$ and similar with other $2\pi$ increments.
More generally, a complex raised to a complex power can be defined by logarithms,
$$z^w=e^{w\log z}=e^{w(\log r+i\theta)}=e^{x\log r-y\theta+i(x\theta+y\log r)}=e^{x\log r-y\theta}(\cos(x\theta+y\log r)+i\sin(x\theta+y\log r)).$$