Is the Graph of y = x^x Significant?

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In summary, the conversation discusses the graph of the function y=x^x and its significance, particularly for negative values. The speakers also mention the ability of Mathematica to plot complex functions and the derivative of y=x^x. The conversation ends with a discussion about the upper bound for which the function is defined and its relationship to a famous constant.
  • #1
meee
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Heyyhey...just wondering, is the graph of y = x^x significant in anyway?

it looks kinda weird...?
 
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  • #2
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".
 
  • #3
it is particularly weird for x<0, being that it takes complex values there...
 
  • #4
benorin said:
it is particularly weird for x<0, being that it takes complex values there...
The Chinese have never liked the negatives.
 
  • #5
I can't get mathematica to plot this function for negative values. Anyone know how I can do it?
 
  • #6
I think someone has already answered your question.
it is particularly weird for x<0, being that it takes complex values there...
 
  • #7
I fail to see how I would be unable to plot it.
 
  • #8
Complex meaning they are imaginary.
Try x=-1/2
 
  • #9
Dragonfall said:
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.
 
  • #10
arunbg said:
Complex meaning they are imaginary.
Try x=-1/2

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

heartless said:
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?
 
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  • #11
thnx cool guys... what's the derivative of y=x^x ?
 
  • #12
meee said:
thnx cool guys... what's the derivative of y=x^x ?
[tex]y=x^x=e^{\ln{x^x}}=e^{x\ln{x}}[/tex]
[tex]\frac{dy}{dx}=(1+\ln{x})e^{x\ln{x}}=(1+\ln{x})x^x[/tex]
 
  • #13
It is much more interesting and informative to plot the hyperpower function [tex]f(x) = x^{x^{x^{x^{...}}}}[/tex]

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.
 
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  • #14
LeonhardEuler said:
[tex]y=x^x=e^{\ln{x^x}}=e^{x\ln{x}}[/tex]
[tex]\frac{dy}{dx}=(1+\ln{x})e^{x\ln{x}}=(1+\ln{x})x^x[/tex]

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

Source: "A Course of Modern Analysis" by Whittaker & Watson, pg. 107.
 
  • #15
benorin said:
[tex]\frac{dy}{dx}=x^x(1+\ln{x})[/tex] is not real when x is a real negative number, yet if x is negative and of the form [tex]x=\frac{p}{2q+1}[/tex], 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 [tex]y=x^x=e^{x\ln{x}+2k\pi ix},k=0,\pm 1, \pm 2,\ldots[/tex].

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 [tex]Ln(x) + \pi i[/tex], 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 [tex]x^x = e^{x Ln(x)} = e^{x ln(|x|) + i\pi x} = \frac{\cos{(\pi |x|)} - i\sin{(\pi |x|)}}{|x|^{|x|}}[/tex] (for negative real x) which would not necessarily return real values even for x of the form [tex]\frac{n}{2k+1}[/tex]
 
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  • #16
f(x) = x^(x^(x^(x^x)))...
 
  • #17
I've always liked that the only critical point of [tex]f(x) = x^x[/tex] is
[tex]x = \frac{1}{e}[/tex]
 

FAQ: Is the Graph of y = x^x Significant?

What does the graph of y = x^x look like?

The graph of y = x^x looks like a curve that starts at the origin and gradually increases as x increases. It has a shape similar to the graph of y = x^2, but it is steeper and has a more exponential growth.

Is the graph of y = x^x symmetric?

No, the graph of y = x^x is not symmetric. It is a one-sided curve that only extends to the right side of the y-axis. This is because the exponent is always positive and thus, the output of the function will always be positive.

What is the domain and range of the graph of y = x^x?

The domain of y = x^x is all real numbers greater than 0, since the function is undefined for negative numbers. The range of the graph is all positive real numbers, as the output of the function will always be positive.

Does the graph of y = x^x have any asymptotes?

No, the graph of y = x^x does not have any asymptotes. The function is continuous and defined for all real numbers greater than 0, so there are no points where the graph approaches infinity or a specific value.

What is the significance of the graph of y = x^x?

The graph of y = x^x is significant because it is a function that involves a variable as both the base and the exponent. This makes it a useful tool in modeling various real-world situations, such as population growth, compound interest, and radioactive decay. It also has applications in calculus and other branches of mathematics.

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