Why does the graph of x^x undefined for x is less than or equal to 0?

  • Thread starter tahayassen
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In summary: For positive x values:{x^x}\times \left({\cos(\pi x)+i\sin(\pi x)}\right) for positive x values.The function is defined as a real value only in the positive reals and the negative integers.This is not entirely true. The function is defined for all real values of x, however, it only has real values for positive reals and negative integers. For all other values, it has complex values. This is why Wolfram Alpha does not display the graph for x < 0, as it is not a real valued function for those values. In summary, f(x)=(x)^(x) is a function that is defined for all real values
  • #1
tahayassen
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1
Let f(x)=(x)^(x)

f(-1)=(-1)^(-1)
=1/-1
=-1

But according to wolframalpha, f(-1) does not exist.
 
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  • #2
The function works only in integer point. For each non integer point the function goes in Complex Set, and so isn't defined on real. Easilly, to avoid the calculator have problem on any other point lesser then 0, the programmer block each of that point with "does not exist" message.

Fruthermore I can rewrite the f(x) in this way:
f(x)=exp(x*ln(x)).
And ln(x<0) is undefined.
 
  • #3
http://www.wolframalpha.com/input/?i=f(x)+=+x^x

If you look at the graph of the function itself (without plugging anything in) you will see that it shows that at x=-1 it is imaginary.
 
  • #4
G.I. said:
Fruthermore I can rewrite the f(x) in this way:
f(x)=exp(x*ln(x)).
And ln(x<0) is undefined.

ln (-x) = ln(x) + [itex]i \pi[/itex]

tahayassen said:
Let f(x)=(x)^(x)
But according to wolframalpha, f(-1) does not exist.

http://www.wolframalpha.com/input/?i=(-1)^(-1)

It says "negative 1". :)
 
  • #5
Best Pokemon said:
http://www.wolframalpha.com/input/?i=f(x)+=+x^x

If you look at the graph of the function itself (without plugging anything in) you will see that it shows that at x=-1 it is imaginary.

That graph show that the imaginary part of f(-1) is 0 and that the real part is -1. That is, f(-1) = -1 + 0i.

It's as the first reply said, f(x)=x^x is not (in general) a real valued function for x<0, though it is real for the negative integers.
 
  • #6
So it's a matter of limitations of computers/programming that x<0 doesn't exist for the real-valued plot? Why does the real part for x<0 exist on the complex-valued plot?
 
  • #7
Remember x^x = (e^(log(x)))^x = e^(x*log(x))

log (-x) = log(x) + [itex]i\pi[/itex]

This means that you'll have the magnitude x^(-x), and the angle x*i*pi.

With the angle you'll have:

cos (x*pi) * x^(-x) for the real portion and
sin (x*pi) *x^(-x) for the imaginary portion.


Correct this if I made a mistake.. tired. :)
 
  • #8
Knowing that [itex]z=r\exp(i\theta + 2ki\pi)[/itex] where [itex]r=|z|=\sqrt{\Re^2(z)+\Im^2(z)}[/itex], [itex]\theta=\mathrm{arg}(z)=\mathrm{atan2}(\Im(z),\Re(z))[/itex] and k is an arbitrary integer, one has [itex]\log(z)=\log(r) + i\theta + 2ki\pi = \log|z| + i\mathrm{arg}(z) + 2ki\pi[/itex]. Now, we consider our function for an arbitrary negative integer, with the principal branch of the complex logarithm (k=0):

[tex](-x)^(-x) = \exp((-x)\log(-x)) = \frac{1}{\exp(x\log(-x))} = \frac{1}{\exp(\log(x)+xi\pi)}=
\frac{1}{x(\cos(x\pi)+i\sin(x\pi))}[/tex]

Now, note that if x was not an integer, we would be stuck here. However, we know for integer x, [itex]\cos(x\pi) = (-1)^x[/itex] and [itex]\sin(x\pi) = 0[/itex], which leaves us with our final answer:

[tex] = \frac{1}{x(-1)^x}[/tex]

So why did I go through this? I wanted to show why Wolfram does not display the graph for [itex]x \leq 0[/itex]. The function is defined as a real value only in the positive reals and the negative integers. For negative reals, we get the ugly-looking answer [itex]\displaystyle \frac{1}{x(\cos(x\pi)+i\sin(x\pi))} = \frac{\cos(\pi x)-i\sin(\pi x)}{x}[/itex], which can't be simplified further.
 
  • #9
tahayassen said:
Let f(x)=(x)^(x)

f(-1)=(-1)^(-1)
=1/-1
=-1

But according to wolframalpha, f(-1) does not exist.

this answer is correct
 
  • #10
Millennial said:
So why did I go through this? I wanted to show why Wolfram does not display the graph for [itex]x \leq 0[/itex].
I saw the graph on Wolfram Alpha. Really nice pretty one if you constrain the x variable:
http://www.wolframalpha.com/input/?...*Plot.plotlowerrange-.*Plot.plotupperrange---

Note... I am a fan of the Alpha. I also think the borderline ASD having character in Alphas is pretty awesome... and hilarious. Do you like strange tangents? :p

Millennial said:
The function is defined as a real value only in the positive reals and the negative integers. For negative reals, we get the ugly-looking answer [itex]\displaystyle \frac{1}{x(\cos(x\pi)+i\sin(x\pi))} = \frac{\cos(\pi x)-i\sin(\pi x)}{x}[/itex], which can't be simplified further.

It's actually (so you don't confuse people- look at the first post):
[tex]\dfrac{\cos(\pi x)-i\sin(\pi x)}{x^x}[/tex]
or (rewritten version of my post above your post):
[tex]{x^{-x}}\times \left({\cos(\pi x)-i\sin(\pi x)}\right)[/tex]
 
Last edited:

1. Why is the graph of x^x undefined for x is less than or equal to 0?

When x is a positive number, the exponent x means multiplication by itself x times. However, when x is a negative number, the exponent x does not have a clear meaning. Additionally, when x is equal to 0, any nonzero number raised to the power of 0 is equal to 1. Therefore, the graph of x^x is undefined for x is less than or equal to 0 because the exponent x does not have a clear meaning for negative numbers and 0.

2. Can I use imaginary numbers to define the graph of x^x for x is less than or equal to 0?

No, imaginary numbers cannot be used to define the graph of x^x for x is less than or equal to 0. Imaginary numbers are used to represent the square root of a negative number, but in this case, the issue is not with the square root, but with the exponent x. The concept of raising a number to a negative or 0 exponent is not defined for imaginary numbers.

3. Are there any real-world applications where the graph of x^x is undefined for x is less than or equal to 0?

Yes, there are real-world applications where the graph of x^x is undefined for x is less than or equal to 0. For example, in population growth models, the function P(t) = P0 * r^t is used, where P0 represents the initial population, r represents the growth rate, and t represents time. However, this model does not hold for negative values of t, as it would represent a negative time period.

4. Is there a way to extend the graph of x^x to include values of x less than or equal to 0?

Yes, there are ways to extend the graph of x^x to include values of x less than or equal to 0. One way is to use the concept of complex numbers, where the exponent x can be defined for negative and 0 values. Another way is to use the concept of limits, where the limit of x^x as x approaches 0 from the right is defined as 1, and the limit as x approaches 0 from the left is undefined.

5. Can the graph of x^x be defined for x is equal to 0?

No, the graph of x^x cannot be defined for x is equal to 0. This is because any nonzero number raised to the power of 0 is equal to 1, and therefore, the graph of x^x would be a constant line at y=1 for x=0, which does not fit the definition of a function.

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