Graph of y=x^z: Exploring the Unknown

[DaveC426913]

I don't have a 3D graphing program, so I'm trying to figure out the graph of y=x^z on paper.

y=x^2 forms a parabola while y=x^3 forms a ... other thing. There's got to be some sensical values to y as the exponent climbs through the values between them.

 

[Werg22]

This function generates a surface rather than a curve. You can imagine it by considering different values of z, for example 0, 1, 2. At z = 0, you have x^0 and hence y = 1. At z = 1, you have y = x and at z = 2, y = x^2. Geometrically speaking, the so called surface will be so that it has the these functions of y as cross sections at the respective values of z. It's pretty easy to visualize on the positive side of z, for z < 1, and for x > 1, the surface concave in respect of X axis. For x < 1, the surface is convex. For z > 1, it's the inverse.

 

[christianjb]

I'm looking at a graph of z=x^y right now using 'Grapher' on the mac.

It's not an easy function to visualize by drawing. It's got a 1st order saddle point.

It's pretty cool in the region about the x-z axes intersect.

 

[DaveC426913]

I'm looking at a graph of z=x^y right now using 'Grapher' on the mac.

It's not an easy function to visualize by drawing. It's got a 1st order saddle point.

It's pretty cool in the region about the x-z axes intersect.

Mind sending me a screen grab?

 

[christianjb]

Mind sending me a screen grab?

What's the easiest way to send it?

 

[christianjb]

See if this works.

 

[christianjb]

Here's another view, and with contours.

You can see the saddle point quite clearly.

 

[Werg22]

The best way would be that Dave pm's you his email, and then you send it the picture to him as an attachment.

 

[Moo Of Doom]

Or upload it on Imageshack and post the link so we can all see it.

 

[christianjb]

http://img408.imageshack.us/my.php?image=graphxypy9.jpg

http://img511.imageshack.us/my.php?image=graph2dd6.jpg
Hmmm... that might work.

OK, that seems successful. Let me know if it works for you too.

 

[DaveC426913]

It does, though without labels I have little idea what the graph is showing. I'm presuming the x-axis runs top right to bottom left and the y-axis is vertical.

But if that graph were somewhere showing y=x^3, I would expect to see one of the cross sections symmetrical about (0,0) yet nowhere do I see any poiints below y=0

 

[christianjb]

It does, though without labels I have little idea what the graph is showing. I'm presuming the x-axis runs top right to bottom left and the y-axis is vertical.

But if that graph were somewhere showing y=x^3, I would expect to see one of the cross sections symmetrical about (0,0) yet nowhere do I see any poiints below y=0

The function is z=x^y The axis going out of the page (up) is the z axis. The function is not defined for -ve values of x.

Edit: This may be a little confusing, but e.g. (-1.4)^1.99 is not a real number, whereas (-1.4)^2 is. Thus the surface only exists for x>0

 

[DaveC426913]

Edit: This may be a little confusing, but e.g. (-1.4)^1.99 is not a real number,

So is it imaginary?


In the green graph: +x is upper left + y points lower left, right?

Shouldn't the slice through y at y=2 manifest as a parabola? I'm just not seeing it.

 

[DaveC426913]

Oh now I see it. At y=2, x<0 is not rendered on the graph.

So, when y is fractional does it create imaginary numbers?

 

[christianjb]

So is it imaginary?


In the green graph: +x is upper left + y points lower left, right?

Shouldn't the slice through y at y=2 manifest as a parabola? I'm just not seeing it.

No, you won't see it because x^y in the -ve x half is only real for integer values of y. It won't make a surface.

Again, for -ve values of x, x^2 exists, but x^1.999 doesn't (at least it's not real).

You will only see x^y for x>0.

 

[christianjb]

Oh now I see it. At y=2, x<0 is not rendered on the graph.

So, when y is fractional does it create imaginary numbers?

Exactly. Try doing -4 ^ 1.9 on a calculator. It will return an error.

And, when y is any non integer, x is -ve, x^y does not exist on the real number line.

 

[DaveC426913]

Exactly. Try doing -4 ^ 1.9 on a calculator. It will return an error.

And, when y is any non integer, x is -ve, x^y does not exist on the real number line.

Right. This is actually what I'm after. My suspicion is that, as y changes from 2 to 3, the curve actually jumps from +z to -z - but since it's a continuum, that curve is going somewhere, and I think where it's going is into the imaginary space - as if, conceptually, the imaginary number space were a sort of "fourth dimension".