y=x^x

**meee** · May21-06, 04:49 AM

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

it looks kinda weird...?

**arildno** · May21-06, 06:29 AM

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** · May21-06, 10:32 AM

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

**arildno** · May21-06, 10:33 AM

Originally Posted by benorin

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

The Chinese have never liked the negatives.

**Dragonfall** · May22-06, 02:39 AM

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

**arunbg** · May22-06, 03:30 AM

I think someone has already answered your question.

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

**Dragonfall** · May22-06, 04:08 AM

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

**arunbg** · May22-06, 05:36 AM

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

**heartless** · May22-06, 12:51 PM

Originally Posted by Dragonfall

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** · May22-06, 01:05 PM

Originally Posted by arunbg

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

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

Originally Posted by heartless

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** · May23-06, 09:58 AM

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

**LeonhardEuler** · May23-06, 10:09 AM

Originally Posted by meee

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

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

**Curious3141** · May23-06, 10:40 AM

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

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** · May24-06, 06:58 PM

Originally Posted by LeonhardEuler

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

$LaTeX Code: \\frac{dy}{dx}=x^x(1+\\ln{x})$ is not real when x is a real negative number, yet if x is negative and of the form $LaTeX Code: x=\\frac{p}{2q+1}$ , 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 $LaTeX Code: y=x^x=e^{x\\ln{x}+2k\\pi ix},k=0,\\pm 1, \\pm 2,\\ldots$ .

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

**Curious3141** · May25-06, 08:11 AM

Originally Posted by benorin

$LaTeX Code: \\frac{dy}{dx}=x^x(1+\\ln{x})$ is not real when x is a real negative number, yet if x is negative and of the form $LaTeX Code: x=\\frac{p}{2q+1}$ , 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 $LaTeX Code: y=x^x=e^{x\\ln{x}+2k\\pi ix},k=0,\\pm 1, \\pm 2,\\ldots$ .

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 $LaTeX Code: Ln(x) + \\pi i$ , then the value of y returned won't necessarily be real, right?

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 $LaTeX Code: x^x = e^{x Ln(x)} = e^{x ln(|x|) + i\\pi x} = \\frac{\\cos{(\\pi |x|)} - i\\sin{(\\pi |x|)}}{|x|^{|x|}}$ (for negative real x) which would not necessarily return real values even for x of the form $LaTeX Code: \\frac{n}{2k+1}$

**Vladimir** · Jun6-06, 04:02 AM

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