Points on the outer edge of a circle

[1MileCrash]

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.

 

[pmsrw3]

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.

 

[micromass]

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.

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.

 

[pmsrw3]

Ah, very nice. Thank you. I had not seen that before.

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.

You're right -- my error. I got it right in Mathematica when I was trying things out but typed it wrong here.

 

[1MileCrash]

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.

 

[1MileCrash]

Can anyone show me the lengths of your segments for 8 points and what the product is?

 

[micromass]

Can anyone show me the lengths of your segments for 8 points and what the product is?

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.

 

[1MileCrash]

Can you explain that to me? Why i?

 

[micromass]

Can you explain that to me? Why i?

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??

 

[pmsrw3]

Can anyone show me the lengths of your segments for 8 points and what the product is?

[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]

 

[1MileCrash]

The plane is of course the same as the complex numbers.

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?

 

[pmsrw3]

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?

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 e^{i 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.

 

[1MileCrash]

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?

 

[SteveL27]

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?

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.

 

[pmsrw3]

When I put these values into wolfram, I get something times i, what are you guys doing to get rid of it?

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]

 

[1MileCrash]

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?

 

[pmsrw3]

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.

 

[1MileCrash]

not to dig this up again, but why does subtracting 1 give length?

 

[phinds]

Now, here's the big question. What about for a non-unit circle?

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