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Graphing y=x^z

  1. May 19, 2007 #1

    DaveC426913

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    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.
     
    Last edited: May 19, 2007
  2. jcsd
  3. May 19, 2007 #2
    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.
     
  4. May 19, 2007 #3
    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.
     
  5. May 19, 2007 #4

    DaveC426913

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    Mind sending me a screen grab?
     
  6. May 19, 2007 #5
    What's the easiest way to send it?
     
  7. May 19, 2007 #6
    See if this works.
     

    Attached Files:

  8. May 19, 2007 #7
    Here's another view, and with contours.

    You can see the saddle point quite clearly.
     

    Attached Files:

    Last edited: May 19, 2007
  9. May 19, 2007 #8
    The best way would be that Dave pm's you his email, and then you send it the picture to him as an attachment.
     
  10. May 19, 2007 #9
    Or upload it on Imageshack and post the link so we can all see it.
     
  11. May 19, 2007 #10
    Last edited: May 20, 2007
  12. May 20, 2007 #11

    DaveC426913

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    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
     
    Last edited: May 20, 2007
  13. May 20, 2007 #12
    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
     
    Last edited: May 20, 2007
  14. May 20, 2007 #13

    DaveC426913

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    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.
     
    Last edited: May 20, 2007
  15. May 20, 2007 #14

    DaveC426913

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    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?
     

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  16. May 20, 2007 #15
    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.
     
  17. May 20, 2007 #16
    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.
     
  18. May 20, 2007 #17

    DaveC426913

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    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".
     
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