There's a neat 2D counterpart to this problem. The area of an annulus (a circle with a concentric circle removed from the middle) can be determined just knowing the length of a chord of the outer circle which is also a tangent to the inner circle - this one falls right out of Pythagoras' theorem iirc.
Posting this here mostly for discussion.
What is the sum of all positive integers? The answer, if there is one, just might surprise you!
Well, some math can be used to come up with the answer -1/12, but the sensibility and meaning of it is up to discussion.
One such method is to use the Riemann zeta function with a value of -1. I'm no mathematician, but the meaning of said function for any complex parameter with the real part less than 1 is not clear-cut.
The Riemann zeta function is based on the infinite series
f(s) = sum(n=1->inf) 1/ns
This function has finite values only when the real part of s is larger than 1. The Riemann zeta function is the analytic continuation of that function for all complex values of s (so that all complex parameters will give a finite value, with the sole exception of s=1.)
While an analytic continuation of a complex function is well defined, it has some quirks to it. One of them is that if s=-1 then you get -1/12, but if you substitute s with -1 in the formula you get the infinite sum 1+2+3+4+...
Are those two things thus equal? I suppose it's up to definition. (Some argue that the equality is as real as eg. complex numbers themselves being "real". After all, it's up to definition whether the square root of -1 "really exists" or not.)
What is the sum of all positive integers? The answer, if there is one, just might surprise you!
If by the sum of all positive integers you mean 1+2+3+..., then we've had some discussion earlier this year about it. It starts here (Post #363689) and continues for eight posts. Relevant Wikipedia page.
Speaking of the Riemann zeta function, I have had for some time now an interest in understanding the Riemann hypothesis. Just out of curiosity, as a little "hobby project". I was wondering if someone could help me with it.
In order to understand the hypothesis, I first need to understand the Riemann zeta function. In order to understand said function, I need to understand analytic continuations. And the topic already becomes really complicated (not to talk about how complicated the zeta function itself is.)
Perhaps someone could guide me into the right direction. Baby steps. Not too much at once. Just a few essentials at a time.
(I understand this isn't a "math challenge" per se. Unless you consider trying to teach such a complicated topic to a layman a "math challenge" of sorts. But perhaps it could be an interesting topic not only for me, especially since there hasn't been a lot of traffic on this thread.)
The mechanistic idea behind analytic continuation appears to be to find a power-series representation at a point, then find *another* power-series representation inside the domain of convergence of the original power series, which itself has a domain of convergence outside the original one, and continue as much as necessary; the reason this works is that if two analytic functions share part of their domains (that is, if the interior of the intersection of their domains is not empty) and are equal on the shared part of their domains, then each function is the unique analytic continuation of the other into the non-shared portion of its domain: http://mathworld.wolfram.com/AnalyticContinuation.html
In general (for example, defining the natural logarithm), you need to deal with sheaf cohomology, or as this special case of it was once known, "multi-valued functions"; however, I think the Riemann zeta function has a single branch defined on the entire complex plane except for z=1, which is a simple pole (that is, every power-series representation of the zeta function based at the value a will have a radius of convergence equal to |a-1|, and no matter how many times you go around the point z=1, the value of the analytic continuation won't change).
So, we have the infinite series sum(n=1 → ∞) 1/ns. This is defined only when the real part of s is greater than 1.
Apparently a partial analytical continuation of that function can be written as:
(1-21-s)-1 sum(n=1 → ∞) -1n+1/ns
(This is the Dirichlet eta function divided by the factor (1-21-s).)
This new function is (apparently) defined for all values of s with a real part greater than zero, and equal to the first function for all values of s with a real part greater than 1.
Could someone help me with a concept proof of this (unless it's extremely complicated)?
Not to derail Warp's question, but I have a question of my own that concerns practical topology.
My girlfriend recently bought a wonderful looking top (the garment, not the kind you spin). The problem is there is a twist in the back. Let me do my best to describe it. There are five "holes" in this top: two for her arms, one for her neck, one for her torso, and then a fifth hole on the back. On a "normal" top, the fifth hole would not make the top unusual in any way and if you were to fill in the holes with fabric, you'd have something homeomorphic to a sphere's surface. The problem is that there is a full twist on the strip of fabric between the hole for the neck and the hole in the back. That particular strip of fabric is twisted 360 degrees before it is rejoined to the opposite side. (I'll draw a picture if necessary, but it will be time-consuming and I warn you that I'm not much of an artist.)
Although it could be a fashion statement and she could do something like put a bow over the twist, my girlfriend and I both agreed that it looks kind of silly and we'd love to see how it looks without the twist. We both thought that it might be invertible such that the twist can be undone but after just a few minutes of playing with it, we quickly concluded that it just doesn't seem possible to get rid of the twist. My girlfriend's plan is to cut the top where the twist is and sew it back together.
But the topology question itself is enticing to me! I know next to nothing about topology, so I'll defer the following questions to the experts here:
• What is the top's topological class? (Is this the right term?)
• What is the topological class of the shape we're trying to turn the top into?
• If neither of the above is easy to define, can it still be elegantly shown that the twist cannot be undone?
• If it twist can be undone, how would we do so?
Like I said, I know next to nothing about topology, so the simpler the explanation, the better. A brief overview of the techniques used to characterize and/or solve the problem would be greatly appreciated. The only thing I thought to use was the Euler characteristic, but if I'm not mistaken, the Euler characteristic of the twisted and untwisted top are exactly the same (I calculated -3). Does this perhaps imply that the twist could be removed if the top were embedded in four dimensions? What techniques can and should I use?
BoboTheKing: showing that maths is relevant in the real world. I've never studied topology but I believe that the transformation cannot be made for the following reason: a jumper without a twist has two sides, an inside and an outside, and you cannot run your finger from one to the other without tracing over an edge (ie a neckhokle or similar). But as soon as you put the twist in you can trace your finger along the twist from the inside of the jumper to the outside, so it's now one-sided object. You can't change the number of sides a surface has just by deformation.
BoboTheKing: showing that maths is relevant in the real world. I've never studied topology but I believe that the transformation cannot be made for the following reason: a jumper without a twist has two sides, an inside and an outside, and you cannot run your finger from one to the other without tracing over an edge (ie a neckhokle or similar). But as soon as you put the twist in you can trace your finger along the twist from the inside of the jumper to the outside, so it's now one-sided object. You can't change the number of sides a surface has just by deformation.
That's only true for a half twist (a Mobius ring), for a full twist (which I think BoboTheKing is describing) the object still has two sides. You still can't remove the twist just by deformation in three dimensions, though.
Joined: 12/28/2007
Posts: 235
Location: Japan, Sapporo
The two tops belong to the same topological class, namely a compact orientable surface with five boundary components. Thus they are homeomorphic, but there's no transformation (in 3-space) which deforms one to the other. In other words, they are different realizations of that topological space in 3-space. It's tough to explain that, so I'll try making an easier example here.
First, the question reduces to the the following problem. Create two paper strips, and give a full twist (i.e. 360-degree twist ) to one of the two. (If you give it a half twist, it is a famous möbius strip.) Then can we deform the normal strip into the twisted strip? If you make three other holes on them, you see that this turns into the initial problem. Now the answer is no, though it's rather non-trivial. Let me explain further.
Maybe you know what is obtained by cutting the möbius strip along the center line. We do the same thing to the above two strips. The normal strip turns into a pair of strips, while the twisted strip turns into a pair of strips entangled with one another. If we were to turn the normal strip into the twisted one (in 3-space), then the two pairs of strips would also turn into one another, which is obviously impossible.
Untwisting in higher dimensions is also tough to describe. Here is a quick example (in lower dimensions) related to the problem. Consider a circle with a point inside on a plane. Obviously the point cannot escape out of the circle as long as they are forced on the plane, but via 3-space, it can easily escape. I'm not a topologist so I'm not sure how accurately this example accounts for the untwisting problem. (I think it's close enough, though.) Higher-dimensional topology is really confusing. :p
Retired because of that deletion event.
Projects (WIP RIP): VIP3 all-exits "almost capeless yoshiless", VIP2 all-exits, TSRP2 "normal run"
thatguy: the twist is 360°, not 180°, so it's still two-sided.
However, if you trace the borders of the neck and back openings, you have two rings. In the normal shirt, they're separate. In your shirt, they're intertwined. I believe that you cannot separate that link without cuts.
/edit: I need to type faster. ;)
I was thinking about Cantor's diagonal argument.
Suppose we have a list of (the decimal representation of) all rational numbers between 0 and 1. We take the first digit (after the decimal point) of the first number, the second digit of the second number, and so on. Then we change each of those digits. What we end up is a number between 0 and 1 that's not in the list (because it's different from every single number in the list.)
What does this mean? Well, it means that the number we constructed this way is not a rational number. It's not in the list, therefore it's not rational.
Why doesn't the same logic work for real numbers? We take a list of all real numbers and do the same operations as above. We end up with a number that was not in the list. Therefore the number is not a real number?
I suppose a counter-argument would be that you can't build a(n indexed) list of all real numbers and thus the premise is impossible. However, to prove that you would need something else than Cantor's diagonal argument?
No, canthors argument is valid. Either the constructed number is not in the original set, or the list is incomplete. Possibly both. So in each case, you need to examine both possibilities. If you can exclude one, it must be the other.
With the rational numbers, we know that a complete list is possible (you can construct one if you like). But we shouldn't assume that the number we constructed is rational, since we didn't construct a fraction of integers, but an infinite decimal representation. A randomly constructed infinite decimal representation is almost surely not rational anyway; in this special case the construction guarantees that it's not.
With the reals, the constructed number is guaranteed to be a real number between 0 and 1. After all, we have its infinite decimal expansion and it starts with a zero before the decimal point. Thus, it's a real number between 0 and 1. You cannot conclude that it's "not a real number"; we know it has to be.
Since we know that the constructed number should have been in the list, but isn't, the list must be incomplete, thus the premise of enumerability must be false.
The rational numbers are known to be countable, proven by other means.
Cantor's argument is what showed that real numbers are uncountable. If you try to list the decimal expansions of all the real numbers between 0 and 1, you can still generate via the diagonal algorithm, an additional real number not in the initial set. This contradiction means that the original list is therefore not complete, and so |R is not countable.
Doing that on rational numbers |Q, does not lead to a contradiction, considering we know that these are in fact countable, so the list itself is not a problem. By using the diagonal algorithm you have in fact constructed a not in the set, but this would merely be an irrational numbers, which are known to exist. If trying to find an extra number not in |R, that is not possible considering |R is not countable, thus there's no list to perform the diagonal argument on in the first place.
Cantor's argument is what showed that real numbers are uncountable.
AFAIK CDA is not enough to show that real numbers are uncountable. The only thing it does is demonstrate that there exist sets that are uncountable. You need an additional step to demonstrate that there exists a one-to-one correspondence between the set of infinite digit strings and a subset of R.
But how about this: Let's take a list of all irrational numbers and construct a new number that's not in that list (using the diagonal algorithm). Since the new number was not in the list, it ought to mean it's not irrational. And by definition a number that's not irrational is rational.
(Therefore, I surmise, the CDA cannot be used to demonstrate that the set of irrational numbers is uncountable...)
AFAIK CDA is not enough to show that real numbers are uncountable. The only thing it does is demonstrate that there exist sets that are uncountable.
But these sets are contained in |R, and if you can't even count a subset of |R, then clearly you cannot count the whole thing.
Warp wrote:
You need an additional step to demonstrate that there exists a one-to-one correspondence between the set of infinite digit strings and a subset of R.
Digit representations are by definition just Cauchy sequences, of the form 3x10^0 + 1x10^-1 + 4x10^-2 + 1x10^-3 + 5x10^-4 + 9x10^-5+ etc. Real numbers are by definition the completion of |Q, IE the invented limits of these sequences. These digit representation are what allow real numbers to exist in the first place. If we have a given sequence of any digit strings, it is a Cauchy sequence which is automatically convergent in |R, and thus represents a real finite number. Every digit sequences is a real number, and every real number has a decimal expansion.
Warp wrote:
But how about this: Let's take a list of all irrational numbers...
...which cannot be done, thus any upcoming conclusion is not necessarily correct. Countable unions of countable sets are countable. We have |Q countable, so if irrational numbers were countable, their union |Q n (|R / |Q ) = |R would be countable too, which it isn't, thus irrational numbers are an uncountable set.
some numbers have two decimal expansions though (1.0000... = 0.99999...). Though the constructed number differs in at least one digit from any in the list, that alone doesn't suffice to claim that it's not equal to any.
That doesn't invalidate the proof, but it'd require at least an additional sentence to take care of, if you want to be rigorous.
Warp wrote:
(Therefore, I surmise, the CDA cannot be used to demonstrate that the set of irrational numbers is uncountable...)
Indeed it cannot. But as flip said, once the uncountability of the reals is established, there are simpler proofs for that.
Now for a question of my own: the cardinality of the reals is supposedly equal to the cardinality of the power set of all integers, or
| |R | = 2^{ | |N | }
Is there an actual, intuitive bijection between the set of reals and the power set of the integers? Wikipedia lists proofs that the cardinalities must be equal, but nothing as nice as a bijection.
The beautiful thing about CDA is that it is, to my knowledge, the last mathematical discovery that can be understood with no background learning required. Maths has gotten so obscure since then...
Is there an actual, intuitive bijection between the set of reals and the power set of the integers?
Yes. One way to uniquely represent a real would be to construct a bit string starting with some number of 1's followed by a 0. If there were no 1's, the integer portion is 0. Otherwise, as many bits as there were 1's follow, with the first being a sign bit and the others constituting the integer portion, with an implied 1 at the beginning. Subsequently, the fractional bits follow. If the number terminates, the first fractional bit is omitted, a 0 is appended at the end and finally the first fractional bit is repeated an infinite number of times.
Edit: I think I spoke too soon. This does not account for the case when there is just an infinite string of 1's for the fractional part.
Now for a question of my own: the cardinality of the reals is supposedly equal to the cardinality of the power set of all integers, or
| |R | = 2^{ | |N | }
Is there an actual, intuitive bijection between the set of reals and the power set of the integers? Wikipedia lists proofs that the cardinalities must be equal, but nothing as nice as a bijection.
There is a relatively simple bijection. It's not that intuitive, but you can get the idea behind it.
It's easy to see that an element of the power set of N is in bijection to a set of infinite strings that can contain only 0's and 1's. So, as a first idea, we try to map a string S of ones and zeros to the real number with binary representation 0.S
There are two problems with this. First is that the image is the interval [0,1], not the entire real line. Second is that not every real number admits a unique binary representation, so 0.01011111... and 0.01100000.... would map to the same value.
We try to get rid of the first problem by using a function that maps an interval to the entire real line, like tan(x). If we scale and shift tan(x), we get tan(pi*x-pi/2), which maps (0,1) to R, which is still not quite right, since we were working with the closed interval [0,1], not the open one.
Using a small trick we can get rid of all these annoying problems. I won't prove anything I'll say below, but it's more or less simple to see it's correct. List all real numbers of (0,1) that are problematic (have two binary representations). These numbers are all irreducible fractions whose denominator is a power of 2 (1/2, 1/4, 3/4, 1/8, ...). Since this is a subset of the rationals, it's countable.
Now get the set of all problematic strings. The problematic strings are those that are eventually constant, like 01101010000000... and 10101111111..., this set is also countable.
The idea is that you can use the normal mapping for all elements that aren't problematic, and for the problematic ones you use a completely different bijection, which is guaranteed to exist because both problematic sets are countable.
So, we can make the algorithm:
1) Initialize the sequence of problematic strings sorted like this (it's annoying to specify the ordering in full detail, but you can get the idea):
A = (1111..., 0000..., 0111..., 1000..., 00111..., 01000...)
2) Initialize the sequence of problematic numbers sorted like this:
B = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, ...)
3) Pick a subset of N (call it S)
4) Map S to an infinite string of 0's and 1's. If n is in S, the n-th character in the string is 1, and 0 otherwise. Call the string u
5) If u is not problematic, construct the number x in the usual way, by interpreting the string as the sequence of binary digits.
6) If u is problematic, find its position in the sequence A, and pick the number x which occupies the same position in sequence B
7) Return tan(pi*x-pi/2)
So, that should be it. The function is basically the shifted and scaled tangent, except for a countable set of points, which are mapped to unrelated numbers.
The trick I used here is very useful when you have to construct bijections and the most obvious ones fail because of some corner cases.
Here's something really simple that I came up with yesterday after thinking about it, but I was too tired to post:
An infinite string of 1's represents the number 0.
Otherwise, it starts with some number of 1's followed by a 0, then a sign bit which is applied at the very end. As many bits as there were 1's at the start then encode an integer, with an implied 1 prepended. Then, consider what follows as the fractional part:
- if it terminates in an infinite string of 0's, we shift it right 1 place and subtract it from the integer
- if it terminates in an infinite string of 1's, we shift it right 1 place and add it to the integer, and subtract 1
- otherwise, we subtract it from the integer as is
Finally, apply the sign bit.
