Sketching x^x (1 Viewer)

swagswagyoloswag

Member
How would i sketch x^x? Also, why is f(x) not defined for x<0 ? When (-2)^(-2) = 4?

klee98

Member
u could implicitly differentiate to find the stat pts, or you can just plot them like x = 0/ y =1 and stuff.
(-2)^(-2) isn't 4

Carrotsticks

Retired
How would i sketch x^x? Also, why is f(x) not defined for x<0 ? When (-2)^(-2) = 4?
Try x=-1/2 to see why negative x values can get nasty.

Also, to obtain a rough sketch for positive x, realise that you are sketching $\bg_white y=e^{\ln \left (x^x \right )}=e^{x \ln x}$

swagswagyoloswag

Member
Try x=-1/2 to see why negative x values can get nasty.

Also, to obtain a rough sketch for positive x, realise that you are sketching $\bg_white y=e^{\ln \left (x^x \right )}=e^{x \ln x}$
So would it only work for negative integer values?

InteGrand

Well-Known Member
So would it only work for negative integer values?
It also works for reciprocals of odd roots (like cube roots, fifth roots, etc.), e.g. $\bg_white \left(-\frac{1}{5} \right)^{-\frac{1}{5}}$ is real.

swagswagyoloswag

Member
It also works for reciprocals of odd roots (like cube roots, fifth roots, etc.), e.g. $\bg_white \left(-\frac{1}{5} \right)^{-\frac{1}{5}}$ is real.
So it wouldn't be worth it to graph the negative x values as only those work?

InteGrand

Well-Known Member
So it wouldn't be worth it to graph the negative x values as only those work?
It'd be quite dodgy to graph the negative x (if you're only graphing on the real plane). The function is going to be well-defined if you allow complex values.

InteGrand

Well-Known Member
So it wouldn't be worth it to graph the negative x values as only those work?
If you were interested in plotting the function for negative x ($\bg_white f: (-\infty ,0)\to \mathbb{C}), f(x)=x^x)$ using complex values, you'd be able to produce a visualisable 3D graph with negative x on the negative x-axis, and you could use the other two axes to represent $\bg_white \Re (f(x))$ and $\bg_white \Im (f(x))$.

Using certain results like Euler's Formula etc., one may show that $\bg_white f(x)=x^x = |x|^x \cos(\pi x)+i\cdot |x|^x \sin (\pi x)$, so on one axis, you could plot $\bg_white \Re (f(x)) = |x|^x \cos(\pi x)$ and on the other plot $\bg_white \Im (f(x))=|x|^x \sin (\pi x)$, for all real x < 0.

Or, you could plot the real and imaginary parts on the one graph, and just have a 2D plot: Wolfram does this for you here -> http://www.wolframalpha.com/input/?i=plot+x^x . (Note that there is a graph for the real part, and one for the imaginary part. You can compare these to the graphs of $\bg_white |x|^x \cos(\pi x)$ and $\bg_white |x|^x \sin (\pi x)$ here to see that they match (only of interest to us are the parts of the graphs for x < 0): http://www.wolframalpha.com/input/?i=plot+|x|^x+cos+(pi*x),+|x|^x+*sin(pi*x))

Edit: those formulas for the Real and Imaginary parts of f give the principal roots. So they won't give the real-valued cube root for $\bg_white \left( -\frac{1}{3} \right)^{-\frac{1}{3}}$, say, but a complex one.

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Kaido

be.
How would i sketch x^x? Also, why is f(x) not defined for x<0 ? When (-2)^(-2) = 4?
I'm loling so hard rite now