The Unit Circle, a Complex One

My problem is this: I have an infinite number of (mostly, see later) random points on a unit circle, and I would like to know analytically what their sum would be (Sigma_x, Sigma_y).

I know Theta for each point; that is, what the arc length is for each point starting at the right-half of the x-axis.

There is a known relationship between Theta_n and Theta_n+1. Now, I wanted to use my knowledge of each point's Theta to help me find the sums mentioned earlier. I'm going to assume that this will involve the sum of my Theta's (Sigma_Theta).

So, the equation I'm after is Sigma_x + i * Sigma_y = f(Sigma_Theta). What I would like to know is, what the heck is the function f defined as?


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A few relationships:
Sigma_x = cos(Theta₁) + cos(Theta₂) + cos(Theta₃) ...
Sigma_y = sin(Theta₁) + sin(Theta₂) + sin(Theta₃) ...

A possibly useful equation:
n^bi = cos(b * ln(n)) + i * sin(b * ln(n))


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When visualizing this, I picture a unit circle with a bunch of little points on it. Then, when we want the sum, a simple average of the points shows up as yet another point inside the circle somewhere (the sum can be easily derived from this average). It seems simple enough to find a method of getting this average from each point's Theta, but I can't find a way to do it. It might be impossible, or ridiculously and self-defeatingly complicated. Oh well.

I'm going on a trip for a couple of days, so I won't be able to reply until I get back. Hope I've explained it well enough without any serious errors, and sorry for the puns in the title ;)

P.S. If you're curious, yeah, this is related to Riemann :)

There isn't a function like that...at least, i don't think so...
Because, if you add two random points on the unit circle, you're not likely to get another point on the unit circle.
However, if you multiply, then this is equivalent to adding thetas.

To reason what the f funciton would be, look at two vectors:
v1 = cis(t1), v2 = cis(t2)
=> v1 + v2 = (cos(t1) + cos(t2)) + i * (sin(t1) + sin(t2)
now, if v1 + v2 = v3 and v3 = r * cis(t3), then...
cos(t1) + cos(t2) = r * cos(t3)
sin(t1) + sin(t2) = r * sin(t3)
=> t3 = atan2(sin(t1) + sin(t2), cos(t1) + cos(t2)) ... not sure the exact formula for this though
=> r = (cos(t1) + cos(t2)) / cos(t3)

Hi,

One item that may be of use is that cos(x) + i sin(x) = e^(ix).

Incidentally, if you really have an infinite number of these points, I suspect you'll find that the average position of all the points is in the centre of the circle.
Think about it thus - suppose the points never overlap, i.e. you can never get back to where you started, then after an infinite number of points have been added you'll have put points all around the circumference and the average of this is clearly the centre of the circle.
Suppose then that at some point "n" you get back to your starting point - in that case your infinite points will be located at a finite number of positions. Since you must have gone around the circle a complete number of times to get back to the start, then you'll have a regular pattern of dots and you can construct a dividing line to split them in two horizontally or vertically (unproven, but I reckon one could prove it with some thought) - hence the average horizontally and vertically will again be 0 and the average at the centre of the circle.
This, of course, is dependent on your theta(n+1) to theta(n) relation - a set of random points could clearly be anywhere you want - and there being an infinite number of them so that they always balance out on each side of the circle.

Anyway, I reckon that that is where you'll probably end up - feel free to keep trying, but how you'll actually go about summing an infinite series number by number will be a poser. Unless you plan to do a sum to infinity of that function...

zaza

Turned out I got back early :)


sql_lall:
The idea of thinking about them as vectors is a cool one, but those formulas require you to know the sums of sets x and y before you can find those sums easily, lol. Thanks though. The extra thought is nice to have.

zaza:
Yeah, I can see your thinking on that, though I believe that an infinite number of possible positions on the unit circle occupied by a repeating pattern of otherwise random numbers would still produce results where the average is not zero. Of course, you thought all about this and have said so. This particular quandry--having some of this data sum to zero only in very special cases--I think is the guts of the difficulty of the Riemann Hypothesis. I believe your thinking about the symmetry of points may prove fruitful... maybe the line of symmetry need to pass through the center for the sum to be zero? I don't know yet.


I did get to thinking today that f(Sigma_Theta) would never be a general solution on random data. The reasoning behind this is that f(Sigma_Theta) could not be defined as a function in that it would have multiple values for each Sigma_Theta passed it. Take pi/2 + pi/2 + pi as Thetas. These added together make 2 * pi. This yields a sum of (-1, 2). Ok, now try 0, 2 * pi / 3, and 4 * pi / 3. This also adds to 2 * pi, but happens to yield (0, 0).

So, it can't be f(Sigma_Theta). You're intuition's exactly right then sql_lall.


This last portion did bring up an interesting thing though, and it hits on what zaza was talking about: I think that the sum will only be (0, 0) when the Theta's are arranged around the circle in the shape of some polygon (or line). What the polygon/line is and what properties it has would then become the focus of further work.

If this is true, a world of possibilities are opened up for Riemann. The function degrades into a series of the same point inverted every other term for an infinite number of times, so the latter part of the series becomes inconsequential, and forms line after line which intersect the center of the unit circle, summing these ending points to zero. Using this straight line business... I just don't know. I'll have to digest it and think about it a bit. Still, I really like the geometric instead of algebraic approach to try to solve the Hypothesis.

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It strikes me that I should probably quick explain the actual Riemann Hypothesis; whoops, lol. Basically, it uses the complex equation, Sum (n = 1 to infinity) (-1)ⁿ⁺¹/n^a+bi. Then, it states that the only time the sum is equal to 0 + 0 * i is when a = 1/2 and b = an infinite number of really odd values, each of which has an undiscovered relationship to the others. The important part, though, is the a = 1/2 bit, and that's the part that needs a proof. Currently, it has been proven that 'a' must be between 0 and 1, though further progress has been... slow (over 100 years of nothing, in essence). Using computers, the hypothesis has been proven largely true, though there are sadly an infinite number of zeros to test, making a brute force effort in the end futile. Even a single point, z (z = a + bi in Riemann's notation), where a <> 1/2 but Riemann(z) = 0 + 0i would cause the Hypothesis to fall over on its head. This point could be the 120⁹⁴³²'th zero, meaning the computers wouldn't check it for many, many, many years to come.

On a happier and swifter note, the Hypothesis has been called the most urgent problem in modern mathematics, namely due to its relationship with the prime number function. Through some strange, twisted method, solving the one solves the other. (I've never looked into the strange, twisted method, by the way.)

Also, much amatuer interest in the function, including a large portion of my own to be honest, comes from the $1 million dollar reward for solving it, or the prestige of finally ending this deceptively simple curse (any time you try to simplify it, it becomes an even more ferocious monster, spawning into an ever-increasing Hydra, except geometrically increasing in complexity rather than arithmetically, and breathing fire upon all who touch its infinite number of variables added together in infinite sum of infinite sum of infinite sum). Well, I think it's time for me to get off for the night... I think I'm a bit too into this.

Thanks for the interest :)
I'm very curious as to whether there is some polygonal or linear relationship in the sum of these points and where they lie on the unit circle. It certainly would be cool, to say the least ;)

P.S. Excuse the ridiculously long post
P.P.S. Excuse the increasing length of the already long post
P.P.P.S. See P.P.S.
P.P.P.P.S. See P.P.P.S.
...
(P.)ⁿS. See (P.)^n-1S.

Well, I may just be rambling by now but...

I've thought about it, though I do not have any proofs to support my thinking yet. Given a set of points on the unit circle whose average is zero, I believe right now that there are two conditions that MUST be met for the average to ever be zero:
1. If the set has an even number of points: every point must have a corresponding point on the opposite side of the line. This sounds a bit far-fetched, but think about it (and then revisit it after the explanation after point 2). Say you have points that make a rectangle. It has this propertry, even if it is not a square, and the sum will always be 0. Then try a hexagon, and it doesn't have to be regular. If you try to move the points in any sort of way, and the sum is still zero, it will end up degenerating into three lines, each of which passes through the center of the circle. I know it seems odd, but I cannot find any counter-example to contradict this.

2. If the set has an odd number of points: the set must make a regular polygon to add up to zero. Again, it seems far-fetched that it cannot be any other way, though think about it for a minute. If you have three points, they must be arranged like an equilateral triangle to sum to zero. Now try five. It must be in the shape of a regular pentagon, if you take into account the counter-example in the next section.

Now, the big iffy part: the large set of data (in my case an infinite amount) can be separated into smaller sub-sets of data, and each part must then sum to 0.

What I mean by that is this: for example, take points at (0, 1) and then at (0, -1), and then at (1, 0), (-0.5, sqrt(3) / 2), and (-0.5, -sqrt(3) / 2). While this data itself does not meet the strict polygon restriction of point 2, it still sums to zero. Why? Because it is a line (the first two points) and a regular triangle (the last three points). The first two points, taken alone, satisfy requirement 1, and the last three points satisfy requirement 2.


How do you go about finding these sub-sets? I don't know yet. It could prove to be a fatal flaw, though I don't quite think so yet. I'm thinking that there is a unique way in which each point may be grouped with other points so that they meet either of those requirements. If there is duplicated data (the same point appearing multiple times), it would not matter with what sub-set it would be grouped, though it may add to the complexity of the problem. I don't really know if the data does ever repeat. Oh well, I'm running shorter on time finally, so enough rambling for now.