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What does the graph of the function $$ f(x) = (-1)^x $$ look like?

It seems as if on all even $x$, $y = 1$, and on all odd $x$, $y = -1$, but between these values (fractional indices) y's value is ambiguous.

insou
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    What would be the value on $x=1/2$? If you know complex numbers, then it can be extended to all $x$ in a reasonably straightforward manner. – TMM Mar 09 '17 at 02:29
  • @TMM $(-1)^{1/2}$ can be expressed as $(-1^2)^{1/4}$, which equals $1^{1/4}$, which is $\pm1$, but this is contradictory as $(-1)^{1/2}$ should obviously be $i$. – insou Mar 09 '17 at 02:35
  • This has to do with branches of the complex logarithm: you agree on a branch, and from then on the solution is uniquely defined. It's kind of comparable to $\sqrt{4}$ being $2$ and not $-2$, by convention. Treating the square root as the inverse of the square means it is not well-defined, but by restricting the range to be the positive numbers we obtain a well-defined function. – TMM Mar 09 '17 at 02:40
  • I'm afraid it doesn't really look like anything. On an $(x,y)$ plot, only integer $x$ makes sense. For it to make sense continuously, if you will, we need it to take complex inputs. But this gives a 4-D graph, which we can't visualize. Sad, I know. – The Count Mar 09 '17 at 02:49
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    1 actually has 4 distinct fourth roots. Only two are real, and two are indeed $i$ and $-i$. – The Count Mar 09 '17 at 02:50
  • @insou the expression $1^{\frac 14}$ only equals 1. – Radial Arm Saw Sep 29 '20 at 16:06
  • @insou $(-1^2)^{\frac 14}$ does not equal $1^{\frac 14}$. Did you mean to write $[(-1)^2]^{\frac 14}$ instead of $(-1^2)^{\frac 14}$? Either way, $(-1)^{\frac 12}$ cannot be written as either of those expressions because of the negative base. Also, as another comment mentioned, the number $1$ has four 4th roots: $1$, $-1$, $i$, and $-i$. – Radial Arm Saw Sep 29 '20 at 16:14

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