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Points on the outer edge of a circle

  1. Jul 27, 2011 #1
    My friend was telling me about something he read about involving points on a circle that seemed kind of cool.

    It went something like this:

    Have a circle of radius 1, and mark n equally spaced points on the outer perimeter.

    Choose one of the points, and connect all other points to it with a straight line running through the circle.

    The product of the lengths of each line is n.

    Sound familiar to anyone? I'd like to know more. That seems kind of neat.
     
  2. jcsd
  3. Jul 27, 2011 #2
    An easy way to write that is:

    [tex]
    \left| \prod_{k=1}^{n}\left(e^{\frac{2\pi i k}{n}}-1\right) \right|
    [/tex]

    Trying it out for a few values, it does indeed seem to work. It's not immediately obvious to me why from looking at a few of these expanded.
     
  4. Jul 27, 2011 #3

    micromass

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    Well, always, you'll need the product

    [tex]
    \left| \prod_{k=1}^{n-1}\left(e^{\frac{2\pi i k}{n}}-1\right) \right|
    [/tex]

    You can't take the root 1, can you? This would cause the product to be 0.

    Either way, the trick is to look at the polynomial

    [tex]
    \prod_{k=1}^{n}\left(z-e^{\frac{2\pi i k}{n}}\right)
    [/tex]

    This polynomial has n roots, and the roots are exactly the roots of unity. So, we have

    [tex]
    z^n-1\prod_{k=1}^n=\left(e^{z-\frac{2\pi i k}{n}}\right)
    [/tex]

    We want to eliminate the root 1 in the product, so we get

    [tex]
    \frac{z^n-1}{z-1}=\prod_{k=1}^{n-1}\left(e^{z-\frac{2\pi i k}{n}}\right)
    [/tex]

    This formula holds true in any number except z=1. But we must know the formula is z=1. So, what we must calculate is the limit

    [tex]
    \lim_{z\rightarrow 1}{\frac{z^n-1}{z-1}}
    [/tex]

    which can easily be done by l'Hopitals theorem, and which yields n.
     
  5. Jul 27, 2011 #4
    Ah, very nice. Thank you. I had not seen that before.

    You're right -- my error. I got it right in Mathematica when I was trying things out but typed it wrong here.
     
    Last edited: Jul 27, 2011
  6. Jul 28, 2011 #5
    I am trying to work an example with 8 points. I am being as accurate as possible, finding each length with trig, but my end product comes out to around 7.44.
     
  7. Jul 29, 2011 #6
    Can anyone show me the lengths of your segments for 8 points and what the product is?
     
  8. Jul 29, 2011 #7

    micromass

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    You mean like the actual numbers??

    If not, then I can easily say that the length of the segments for 8 points are [itex]|1-e^{2\pi i k/8}|[/itex] for 0<k<8. So the length is

    [tex]|(1-e^{2\pi i/8})(1-e^{4\pi i/8})(1-e^{6\pi i/8})(1-e^{8\pi i/8})(1-e^{10\pi i/8})(1-e^{12\pi i/8})(1-e^{14\pi i/8})|[/tex]

    Plugging those in in wolframalpha gets me the following lengths:

    0.765
    1.414
    1.847
    2
    1.847
    1.414
    0.765

    and the product is

    7.983

    So this seems to be correct modulo rounding errors.
     
  9. Jul 29, 2011 #8
    Can you explain that to me? Why i?
     
  10. Jul 29, 2011 #9

    micromass

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    Well, you want 8 points on the unit circle that are equidistant. The plane is of course the same as the complex numbers. So the 8 points on the unit circle are exactly (after a rotation):

    [tex]1, e^{2\pi i/8}, e^{4\pi i/8}, e^{8\pi i/8}, e^{10 \pi i/8}, e^{12 \pi i/8}, e^{14\pi i/8}[/tex]

    Do you follow this??
     
  11. Jul 30, 2011 #10
    [tex]
    \begin{array}{l}
    \sqrt{2-\sqrt{2}} \\
    \sqrt{2} \\
    \sqrt{2+\sqrt{2}} \\
    2 \\
    \sqrt{2+\sqrt{2}} \\
    \sqrt{2} \\
    \sqrt{2-\sqrt{2}}
    \end{array}
    [/tex]

    [tex]
    \begin{eqnarray*}
    \sqrt{2+\sqrt{2}} \times \sqrt{2-\sqrt{2}} & = & \sqrt{4-2} = \sqrt{2} \\
    \sqrt{2} \times \sqrt{2} \times \sqrt{2} \times \sqrt{2} & = & 4 \\
    2 \times 4 & = & 8 \\
    \end{eqnarray*}
    [/tex]
     
    Last edited: Jul 30, 2011
  12. Jul 31, 2011 #11
    I'm not sure I'm familiar with any of that, but I'd like to know more. A circle is the plane of complex numbers?
     
  13. Jul 31, 2011 #12
    No, the plane is the same as the complex numbers. That is any complex number x+yi (i=square root of -1) can be represented as a point (x,y) in the plane. This is called the Argand plane, or just the complex plane. Tha circle with radius 1 whose center is at 0 is called the unit circle. Every point on the unit circle can be written x+yi, with x2+y2=1. But it can also be written ei theta, where theta is the angle between the x axis and a line from 0 to the point. That's what we're doing here. This trick allows us to turn your geometric question into an algebraic question, which is much easier to solve.
     
  14. Aug 2, 2011 #13
    When I put these values into wolfram, I get something times i, what are you guys doing to get rid of it?

    for e^(2i pi / 8) I get what looks like half of the square root of 2 plus half of the square root of 2 i.

    Sorry guys, this math is a bit beyond my level but I really am enthralled by it. Are you guys just getting rid of the i's when multiplying everything?
     
  15. Aug 2, 2011 #14
    When you understand how this works, you can look at an expression like [itex]e^{2\pi i * (\frac{1}{8})} [/itex]and see instantly that it's exactly what you got from Wolfram.

    That's because [itex]e^{2\pi i * (\frac{1}{8})}[/itex] corresponds to the point in the plane that's 1/8 of the way around the unit circle, starting from the point (1,0) and going counterclockwise.

    Going 1/8 around the circle is an angle of pi/4 radians, or 45 degrees. A line through the origin making an angle of pi/4 radians with the positive x-axis intersects the unit circle at the point [itex](\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})[/itex], because you have an isosceles right triangle with hypotenuse 1, so by Pythagoras, each leg must be [itex]\frac{\sqrt{2}}{2}[/itex].

    And by identifying the usual 2-dimensional plane with the complex numbers, we identify the point [itex](\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})[/itex] with the complex number [itex]\frac{\sqrt{2}}{2} + i * \frac{\sqrt{2}}{2}[/itex].

    In general, [itex]e^{2\pi i * t}[/itex] is the point in the plane you get when you go t radians around the unit circle. This is quite an amazing fact, actually, and was a great discovery in the history of mathematics. It took around 100 years, from 1700 to 1800, for mathematicians to get a handle on this surprising relationship between the complex exponential function and the angles around the unit circle. As usual one person, Euler, gets his name attached to the discovery; but the result was actually the work of a lot of people over a long period of time.

    Here's an article that explains all this and gives a bit of the history.

    http://en.wikipedia.org/wiki/Euler's_formula

    Now if you believe all that, then you can get eight points equidistant around the unit circle by letting t = 1/8, t = 2/8, t = 3/8, ..., t = 7/8, t = 8/8 = 1. That's where micromass and pmsrw3 are getting [itex]e^{\frac{2\pi ik}{n}}[/itex] from. You're just letting n = 8 and letting k = 1 through 8.
     
    Last edited: Aug 2, 2011
  16. Aug 2, 2011 #15
    We're taking the absolute value. A complex number x+yi is like a vector -- it has a length and a direction. The length is called the absolute value -- a term you've probably heard before, meaning, for real numbers, the number with a positive sign. For complex numbers [itex]\text{absolute value of }x+y i = \left|x+yi\right| = \sqrt{x^2+y^2}[/itex]. (Notice that if y = 0, then x+yi = x is real and this gives the familiar absolute value. So, the way we set up your problem is:

    (1) Have a circle of radius 1, and mark n equally spaced points on the outer perimeter.

    The circle is the unit circle, and the n points are [itex]e^{\frac{2\pi i k}{n}}[/itex], with k running from 0 to n-1.

    (2) Choose one of the points.

    It's convenient to choose k = 0. That point is just e0=1.

    (3) Connect all other points to it with a straight line running through the circle.

    To get the line from point k to point 0, we just subtract: [itex]e^{\frac{2\pi i k}{n}}-1[/itex]. (It doesn't matter in which direction you do the subtraction, since all we care about is the absolute value.)

    (4) Calculate the product of the lengths of each line

    Just multiply those n-1 lengths.

    [tex]
    \prod_{k=1}^{n-1}\left| e^{\frac{2\pi i k}{n}}-1 \right|
    [/tex]

    Since the absolute value of a product is the product of the absolute values, we can also write that as

    [tex]
    \left| \prod_{k=1}^{n-1} \left( e^{\frac{2\pi i k}{n}}-1 \right) \right|
    [/tex]
     
  17. Aug 3, 2011 #16
    Alright, I think I'm starting to understand.

    If we can name any point on the circle in the form of e^(i * angle), and our points are equidistant, and the difference with 1 is the length of the line segment, then our smallest (for example) line segment is e^(i2pi / n) - 1. Since n also represents each "fraction of a turn" on the circle between points, and those points are expressed e^iangle.


    Now, here's the big question. What about for a non-unit circle? We can no longer express points as e^iangle, but perhaps with a measurement or trig we can find values that are linked some way to those e^iangle points on the unit circle and still find some relation to the product?
     
  18. Aug 3, 2011 #17
    For a circle of radius r, equally spaced points are [itex]r e^{\frac{2\pi i k}{n}}[/itex]. The product of the lengths of the n-1 line segments joining one of those points to each of the others would be [itex]n r^{n-1}[/itex]. That is, it's just [itex]r^{n-1}[/itex] times the value for the unit circle, since each line is now longer by a factor r.
     
  19. Sep 22, 2011 #18
    not to dig this up again, but why does subtracting 1 give length?
     
  20. Sep 22, 2011 #19

    phinds

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    Isn't that just a scale factor? IOW if it works for the unit circle, it'll work for any circle. Am I oversimplifying?
     
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