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Describe graph of y=[-1] exp x

  1. Feb 18, 2010 #1
    I am talking about the three dimensional graph in y real; y imaginary; and x real. I understand it is a cylinder centered on x axis, but can you say more? In particular, is it possible that it is an infinitly dense spiral centered around x axis?
    -harry wertmuller
  2. jcsd
  3. Feb 18, 2010 #2
    I am afraid there is something wrong in your question. Namely, if x is real, then y=-exp(x) is always real too.
    Perhaps the function you are interested in is:


    where i id the imaginary unit, and x is a real number.

    Plotting the graph in the coordinate-system you mentioned gives you a spiral with its axis in x. But it is not an "infinitely dense" cylinder.
    To increase/decrease what you call "density" (and what physicists and engineers call frequency), you can introduce a real constant k and write:


    These are essentially the complex sinusoids used in the Fourier transform.
    Hope it helps.
  4. Feb 18, 2010 #3
    No, not y=-exp(x). Y=(-1) exp x. I believe some posters use ^ for exponential so you might write
    y=(-1)^x. When x is one-half, y is definitely not real. For infinitesimal increments of x we obtain values of y real and y imaginary at various points centered around the x axis. I am wondering if there is any order to this chaos.
    -harry wertmuller
  5. Feb 18, 2010 #4
    Ah! Now I understand! You meant [tex]y=-1^x[/tex].
    If I got your question right, you can re-write it as:

    [tex]y=(-1)^x = e^{i\pi x}[/tex]

    At this point you can go back and re-read my previous post and everything should make sense.
  6. Feb 18, 2010 #5
    Not infinitely dense.
    Would be correct to say one full loop for every two units on X-axis?
  7. Feb 19, 2010 #6


    Staff: Mentor

    I think the OP meant (-1)x, which is very different from -1x. The latter function is identically equal to -1.
  8. Feb 19, 2010 #7
    Yes, that function has period = 2, in fact for any arbitrary x you have:

    [tex]y(x+2)=e^{i\pi (x+2)}=e^{i\pi x}e^{i2\pi}=e^{i\pi x}=y(x)[/tex]

    and there is no other number k (where [itex]0<|k|<2[/itex]) such that you get [tex]y(x+k)=y(k), \forall x[/tex]

    And btw, you are definitely dealing with complex sinusoids. You might want want to familiriaze more with that subject, and why not, also play around with http://www.math.uu.nl/people/beukers/voorlichting/CompVis.htm" [Broken].

    Yes, sorry...I forgot the parentheses, I guess the rest was correct, though.
    Last edited by a moderator: May 4, 2017
  9. Feb 23, 2010 #8
    Hey mnb96 - So period for my dumb equation is 2 but how did you know that? I found out just from computing some intermediate results. And from looking at the nifty Java applet you listed at Utrecht University. You removed the '2' from the exponent as if it were clear 'by inspection'. Me fail inspect.
  10. Feb 23, 2010 #9


    Staff: Mentor

    For a function f, if f(x + p) = f(x), the function is periodic with period p. mnb96 showed you that y(x + 2) = y(x), so the period of this function is 2.
  11. Feb 23, 2010 #10
    Well, it isn't "shown" is my point. If someone would say "We know by DeMoivre's Theorem e^(2pi)=1" then I could review DeMoivre. Otherwise left stumbling around the Internet. Thanks, though. Definitely answered my original question.
  12. Feb 23, 2010 #11


    Staff: Mentor

    mnb96 used Euler's formula: eix = cos(x) + isin(x). If x = pi, you have ei pi = cos(pi) + i sin(pi) = -1 + 0i = -1.

    He replaced -1 by ei pi
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