y=x^x

[meee]

Heyyhey...just wondering, is the graph of y = x^x significant in anyway?

it looks kinda weird...?

[arildno]

Chinese cooks usually put the spaghetti threads in that particular shape on your plate.

Other than that, I don't know if that graph is "significant".

[benorin]

it is particularly weird for x<0, being that it takes complex values there...

[arildno]

it is particularly weird for x<0, being that it takes complex values there...
The Chinese have never liked the negatives.

[Dragonfall]

I can't get mathematica to plot this function for negative values. Anyone know how I can do it?

[arunbg]

I think someone has already answered your question.
it is particularly weird for x<0, being that it takes complex values there...

[Dragonfall]

I fail to see how I would be unable to plot it.

[arunbg]

Complex meaning they are imaginary.
Try x=-1/2

[heartless]

I can't get mathematica to plot this function for negative values. Anyone know how I can do it?

Yep, just do Plot[{y=x^2},{x,-10,10}] and you get all the values from -10 to 10 of this super significant function x^x.

[Dragonfall]

Complex meaning they are imaginary.
Try x=-1/2

Mathematica can plot complex functions, and this function in particular because it's R->C.

Yep, just do Plot[{y=x^2},{x,-10,10}] and you get all the values from -10 to 10 of this super significant function x^x.

How does plotting x^2 give me all the values of x^x?

[meee]

thnx cool guys... whats the derivative of y=x^x ?

[LeonhardEuler]

thnx cool guys... whats the derivative of y=x^x ?
y=x^x=e^{\ln{x^x}}=e^{x\ln{x}}
\frac{dy}{dx}=(1+\ln{x})e^{x\ln{x}}=(1+\ln{x})x^x

[Curious3141]

It is much more interesting and informative to plot the hyperpower function f(x) = x^{x^{x^{x^{...}}}}

Find the upper bound of x for which that function is defined and see if you can spot the relationship of that bound to a famous constant.

[benorin]

y=x^x=e^{\ln{x^x}}=e^{x\ln{x}}
\frac{dy}{dx}=(1+\ln{x})e^{x\ln{x}}=(1+\ln{x})x^x

\frac{dy}{dx}=x^x(1+\ln{x}) is not real when x is a real negative number, yet if x is negative and of the form x=\frac{p}{2q+1}, where p,q are positive or negative integers, then y is real. Curious, no? It has to do with the complex branch-cut structure of y=x^x=e^{x\ln{x}+2k\pi ix},k=0,\pm 1, \pm 2,\ldots.

Source: "A Course of Modern Analysis" by Whittaker & Watson, pg. 107.

[Curious3141]

\frac{dy}{dx}=x^x(1+\ln{x}) is not real when x is a real negative number, yet if x is negative and of the form x=\frac{p}{2q+1}, where p,q are positive or negative integers, then y is real. Curious, no? It has to do with the complex branch-cut structure of y=x^x=e^{x\ln{x}+2k\pi ix},k=0,\pm 1, \pm 2,\ldots.

Source: "A Course of Modern Analysis" by Whittaker & Watson, pg. 107.

I'm not sure about this, certainly for x in that domain and of that form, a real value of y exists if y is defined to be multivalued.

But if the principal value of ln(x) is used, which is Ln(x) + \pi i, then the value of y returned won't necessarily be real, right? :confused:

Sorry, I don't know that much about complex analysis, just the basics. I do know the principal branch for ln x, but not the one for x^x. I would've assumed it would be based on the branch cut of the log function, giving x^x = e^{x Ln(x)} = e^{x ln(|x|) + i\pi x} = \frac{\cos{(\pi |x|)} - i\sin{(\pi |x|)}}{|x|^{|x|}} (for negative real x) which would not necessarily return real values even for x of the form \frac{n}{2k+1}

[Vladimir]

f(x) = x^(x^(x^(x^x)))...

[BSMSMSTMSPHD]

I've always liked that the only critical point of f(x) = x^x is
x = \frac{1}{e}