Fractional exponents of negative

  • #1
DaveC426913
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Main Question or Discussion Point

I've been toying around with stuff I probably shouldn't be. :biggrin:

I've been sketching a graph of y=x^n where n is a rational number (as opposed to an integer).

Of course, when I get into the fractional exponents, the negative portion of the curve ends up being imaginary (eg. x=-2,n=2.5 then y = squareroot(-32) or about 5.6i ).

Before I actually attempt to sketch this on paper, has it been done already?


It'll be a 3D graph where the z-axis is the imaginary axis. The positive x side will look normal, the negative side will flip between positive y and negative y (also normal), but it will pass through the imaginary space with each flip.
 
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Answers and Replies

  • #2
Office_Shredder
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I'm not sure what you mean by flip between positive y and negative y, but in terms of graphing the imaginary component of the output, it sounds like a pretty good idea. Go for it.

You're going to run into some issues of uniqueness. When you graph [tex]\sqrt{x}[/tex] we know to take the positive values because that's what we do. When you take [tex]\sqrt{-1}[/tex] what is the positive value? It's kind of "obvious" in this case that it's i, not -i. But what about when you're graphing [tex](-1)^{\frac{1}{8}}[/tex]? You have

0.923879533 + 0.382683432i
0.382683432 + 0.923879533i

(and some other numbers) raised to the eight power give -1. Which one do you take?
 
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  • #3
DaveC426913
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I'm not sure what you mean by flip between positive y and negative y
When graphing it looking at only the integer n's, x^2 has its -ive component in positive y. x^3 has its -ive component in negative y. So somehere in there it flips. Having not seen the graph using fractional exponents, I don't really know what form the flip takes (except that it's smooth and imaginary).
 

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