How do you graph y = x^x? What I mean is, when x = 0, y is undefined. How do you graph that? Then something weird happens when you try x= -0.3 and -0.4 and -0.5. Don't go too technical on me for the explanation. I'm just curious as to how you graph it.

0^0 is generally given as 1; and non-integer negative values will give imaginary (multiple of the square root of -1, i) numbers... generally you just want to graph positive values, which is easily accomplished.

when i put -0.2 into my calculator and try it, it works. but no other non integer negative number works. that's odd

like -0.4

 

(-0.4)^(-0.4) = 1/((-0.4)^(0.4)) = 1/((-0.4)^(2/5)) = 1/(fifthroot(.16)) = 1.4427

 

did i do that right?

 

how bout -0.5

 

(-0.5)^(-0.5) = 1/(-0.5)^(0.5)= 1/((-0.5)^(1/2)=1/squareroot(-0.5)

then this is imaginary. how do you graph that? or do you not graph it?

For non-negative values of x, the graphing problem is fairly trivial. It's when you extend the domain of the function to include negative numbers that you run into problems.

 

When x is a negative integer, then the function is defined, since you can have something like (-2)^(-2) = 1/4.

 

It's when you have x belonging to the set of rationals that you run into quite a few problems, since it'll work for some values and not for others. For instance, take x to be of the form (2p/q) where p is a negative integer and q is a positive integer (obviously with p and q not equal to zero). So you know that the numerator is always going to be an even number, so it'll involve making your negative x positive (effectively because you're squaring it p times).

 

But when you have x as some number (p/q) where p is odd and q is even (since you can cube/fifth/seventh.. root a negative number) you're going to be effectively square rooting a negative number, which is obviously not defined in the real line. There are a lot of problems when x is any negative real number.

 

Mathematically speaking, we can say that for all x in the open interval (-infinity, 0), x is discontinuous - it has lots of gaps in it, effectively making it impossible to graph properly. It also means it's not differentiable in this domain. But for the half-open interval [0, infinity), it's continuous - i.e. a smooth line.

 

Of course, you can get rid of the problem. Let's say the function f:N->R (N is the set of natural numbers, R is the set of reals, and f maps some number x from the naturals to the reals). Then f(x) = x^x is defined for all x in the domain.

 

If you continue with mathematics at university, you'll do a lot of this stuff. Hope this helps some.

fafalone said in post # :

0^0 is generally given as 1; and non-integer negative values will give imaginary (multiple of the square root of -1, i) numbers... generally you just want to graph positive values, which is easily accomplished.

 

0^0 is undefined, because the proof for 0 powers doesn't hold.

yeah it makes sense. thanks

MrL_JaKiri said in post # :

 

0^0 is undefined, because the proof for 0 powers doesn't hold.

 

How does that proof gO?

0^0 is undefined because

 

say that x is not equal to 0

0^0 = 0^(x) * 0^(-x) = 0^(x)/0^(x)

 

0^(x)=0

 

anything divided by 0 is undefined

0^0 is undefined because

 

say that x is not equal to 0

0^0 = 0^(x) * 0^(-x) = 0^(x)/0^(x)

 

0^(x)=0

 

anything divided by 0 is undefined

 

0^0 is undefined. But it is generally take to be equal to 1 in polynomial algebra.

 

for example if u have a series

 

k^k/(k+2) from 0 to infinity. then it would be taken as 0 + 1/3 + 1 + 9/5 + ...

 

well thats what i learned in my Analysis lectures anyway

cant seem to edit the last post

 

but

 

k^k/(k+2) from 0 to infinity. then it would be taken as 1 + 1/3 + 1 + 9/5 ..

 

the term x^0 is taken as 1 for all values of x

0^0 is an indeterminate form.

On the graph of f(x)=x^x at x=0, the graph is undefined:

lim x->0+ f(x) does not equal

lim x->0- f(x)

 

For more information:

http://www.faqs.org/faqs/sci-math-faq/specialnumbers/0to0/

Is it possible to incorperate an argand plane, plus your normal 2D axis?