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Late answers - a mechanism for getting a fair share of upvotes?

It is a not unheard of phenomenon that answers don't get much upvotes in the following scenario:

First, a user asks a non-trivial question, which turns out to be non-trivial enough not to get an answer for a while.

Then, the question lands in the big heap of unanswered questions, where it does not receive much attention.

Finally, some user digs up the problem, devotes some of his time and energy to come up with the answer, and posts it. By now, it's probably only the OP who pays any attention (or maybe not even him).

The answer gets accepted, upvoted by the OP, and forgotten.

The net result is that much work is rewarded with some +20 rep, or maybe less. A situation in this spirit was discussed for instance here (but in a somewhat different context). This is in contrast with what happens with comparably easy problems where an immediate answer can earn more rep with much less effort.

I do realise that upvotes are not precisely the point of MSE, which is to generate good quality answers to mathematical questions. However, I am afraid that some users may be discouraged from posting late answers, because answering the easy ones "pays" better. This could result in there being a lot of answers to the easy problems that don't bring much new to the site, and with some interesting problems being left unanswered.

Is there some mechanism that could help such situations to occur less freqently? For instance, if there was an accessible list of questions where the question got much more upvotes than the accepted answer (perhaps with the extra condition that the answer was given relatively late), I would be happy to look through it and upvote under-appreciated answers if I happen to see any. But such query is rather too specific for the MSE search engine, isn't it?

I see that the "review" tool has the category "Late answers", but it apparently applies only to the ones by "new users", and seems not to be used much. (I, for one, have no possible reviews in this category.) Perhaps it could be considered to extend this feature? For instance, "Late answers" could also include late answers from experienced users, answers that got relatively little rep (my gut feeling is that if question deserves $x$ upvotes, then an accepted answer generically deserves at least $x/2$) or little attention, and so on.

(Disclaimer: I do not hide that I posted some late answers recently. It's not that I feel they are of particularly high quality or that they are not appreciated enough. It's rather that this served to direct my attention to the general possible problem.)

• "I am afraid that some users may be discouraged from posting late answers." - at least in my case, the fact that the problem took me more than a minute of thought was sufficient encouragement to keep working at it. Since my interests tend to not overlap very much with the interests of many users here (e.g. special functions), I've just gotten used to my answers, most especially late ones, to just not get that many eyes on them. Apr 22, 2013 at 12:33
• @J.M.: ... but some of that eyes were especially thankful for that late answers... :-) Sometimes I dream of a dedicated subset of users which are not so focused on "earning" something (may it be only electronic reputation or something else) but where I can get in resonance with the pure enthusiasm on interesting mathematical problems. Hmm. Perhaps that users should insert a "sign" in their user-name, like - hmm, what would be a good example... Apr 22, 2013 at 15:04
• @GottfriedHelms: I tend to think of reputation as (among other things) the indicator from the community that one is doing things right, and that there is some interest in it. So even in a case of an extremely dedicated and enthusiastic users, I would still feel it's a reasonable idea to give this sort of indication. (Also, posting an answer is not just about mathematics, but also the less inspiring task of writing things down). Apr 22, 2013 at 15:24
• @Feanor, I don't know, I think writing it down is in fact a way to force you to make what seemed crystal-clear in your head to be actually readable to at least one other person. That takes nontrivial amount of thinking, too. Apr 22, 2013 at 15:28
• @Gottfried, in contrast to the tendency elsewhere on the SE network to mindlessly strip out expressions of gratitude, I consider the expressions of thanks as the cherry on top. Anyway, I'm glad my (fashionably?) late answers to some of your questions were useful. Apr 22, 2013 at 15:29
• @Feanor: I hope I did not disgust you. I only feel a swift wind of change in MSE since I'm here. I didn't make a true statistic, but it seems to me, that the number, and possibly even the relative fraction, of meta-questions/problems focusing "reputation" in this or another way has increased. Or if not numerically increased: then possibly intensed. Even problems typical for bureaucracy - as if there were a bureaucracy here in action. Well, whatever - I'm getting slowly slowly tired of that impression and begin to wish a new SE-site where folks with less interest in reputation would crowd in... Apr 22, 2013 at 20:25
• @J.M. I do agree that writing answers down is non-trivial. My point is: it is often less exciting than thinking about the problem in more vague terms. When I think of a problem, I generally stop at the point where it becomes clear what would follow - after that, I would normally move on to something else. To write things down after I figure out the problem, and to post them publicly, I would need some additional incentive, except for the interest in the problem. Apr 23, 2013 at 8:09
• @GottfriedHelms: I don't think there is anything that could have disgusted me. Now that you mention it, I do feel slightly sorry to have contributed to the trend you describe. However, I think that the idea of SE relies, up to a point, on people pretending that reputation really matters. I think such delusion is part of what keeps people going in the short run (it reminds me of scores in computer games). But maybe that's just my impression. In any case, I haven't been involved enough to observe the trend directly. I absolutely agree caring about reputation should not be taken too far. Apr 23, 2013 at 8:17
• @Feanor: you're welcome! Apr 23, 2013 at 8:25
• My 4th a 5th highest-rated answers are on notation. I still find this odd, but I suppose everyone understand a question about unfamiliar notation... Apr 26, 2013 at 10:09
• May 1, 2013 at 20:14
• This happened to me here: math.stackexchange.com/questions/1143998/… Mar 25, 2015 at 10:27

It is a bit hard to find, but if you go to the 10k tools and click on the "stats" page.

At the bottom of the page, there's a list of links. One such link is the above:

• Thank you, it's good to know such a list exists. As a < 10k user, I did not know it's there. It makes me more confident in the voting system working smoothly. Apr 22, 2013 at 16:07
• Actually, I'm a 36k user and had not heard of that list yet. Where was I supposed to find it by myself (it is not among the "review" items, for example)? Apr 24, 2013 at 17:39
• @Hagen When in the Review page, you can click on "Tools" (see small red circle above) to get to the general 10K Tools page. Apr 24, 2013 at 18:00
• @Hagen: This is what I previously called "an elaborate measure of clicks".
– Asaf Karagila Mod
Apr 25, 2013 at 7:10
• For what reason the limit is 10k, in my opinion 1000 is enough. Oct 24, 2013 at 9:23
• @Sawarnik: I guess the reason is that folks should be introduced to the more "advcanced" tools little by little, so get more experience before using them. But some limits should be lower! Mar 19, 2015 at 12:14
• Too bad if the answer is to question which is old only say 20 days :)
– Sil
Sep 6, 2020 at 8:11

Obviously, this isn't a comment on existing functionality, but a suggestion of new functionality...

How about if the site were to add a new reward feature, whereby users with a reasonable number of points and the person asking the question are able to mark an answer as a 'major contribution'. Unlike the bounty, which is basically a transfer of points from one person to another, this would be a reward that is judged.

The idea is, the answer is marked as a 'major contribution', which then has to be reviewed by those with sufficient points, etc, and if that contribution is judged worthy, then the answer can be given an extra points boon, and perhaps featured on a separate list of questions with major contribution answers. On the flipside, there could be a small punishment (-10 points, or something) for users that try to suggest very minor contributions (like explaining a basic application of integration by parts, for instance), where the answer is clearly nowhere near being a major contribution, to keep people from spamming it (answers lying between the two extremes would simply be marked as "not major" and would continue as usual).

It should help to give more attention to answers that require more work, that answer a difficult problem, and that might otherwise go unnoticed. It would both further encourage people to work on the hard problems, and draw more attention to the resulting answers - not just to have more chance of direct upvotes, but also to better showcase the use of the site itself.

Anyway, it's just a thought. Even if people don't like it, it might spur some alternate ideas.

• The idea sounds good to me. In fact, if there was just a list of such "major contributions" on which an interesting answer would land for a while (not necessarily with any reputation attached to it), this might already suffice to draw sufficient attention to it. Apr 25, 2013 at 6:28

Even new Answers which have few (or no) competing Answers on Questions of moderate difficulty may suffer in respect of getting less voting than more easily understood Q&A's. Certainly in my experience the Answers that I've toiled over are not the high vote count examples, but rather the short but perhaps surprising items, maybe because easily taken in at a glance.

I don't think there's anything to be done about this. At a fairly early met threshold the real reward for doing your best on Answers is not votes, other than perhaps the few garnered by the OP and others who try the Question for themselves, but the improvements to your own understanding and powers of exposition.

Also note the thrillingly named Necromancer Badge (silver), as a reward for netting as few as 5 votes on Questions over 60 days old. "This badge can be awarded multiple times."

• Don't forget "Revival"! Apr 30, 2013 at 14:36
• @J.M.: The Revival badge I've actually been able to snag a couple of times, but the Necromancer eludes me! Apr 30, 2013 at 14:54
• That requires patience in digging through old questions, as well as having the knack to find what hasn't been addressed by previous comments/answers... :) Apr 30, 2013 at 14:55


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Dixmier trace

In mathematics, the Dixmier trace, introduced by Jacques Dixmier (1966), is a non-normal trace on a space of linear operators on a Hilbert space larger than the space of trace class operators. Dixmier traces are examples of singular traces.

Some applications of Dixmier traces to noncommutative geometry are described in (Connes 1994).

Definition

If H is a Hilbert space, then L1,∞(H) is the space of compact linear operators T on H such that the norm

${\displaystyle \|T\|_{1,\infty }=\sup _{N}{\frac {\sum _{i=1}^{N}\mu _{i}(T)}{\log(N)}}}$

is finite, where the numbers μi(T) are the eigenvalues of |T| arranged in decreasing order. Let

${\displaystyle a_{N}={\frac {\sum _{i=1}^{N}\mu _{i}(T)}{\log(N)}}}$.

The Dixmier trace Trω(T) of T is defined for positive operators T of L1,∞(H) to be

${\displaystyle \operatorname {Tr} _{\omega }(T)=\lim _{\omega }a_{N}}$

where limω is a scale-invariant positive "extension" of the usual limit, to all bounded sequences. In other words, it has the following properties:

• limω(αn) ≥ 0 if all αn ≥ 0 (positivity)
• limω(αn) = lim(αn) whenever the ordinary limit exists
• limω(α1, α1, α2, α2, α3, ...) = limω(αn) (scale invariance)

There are many such extensions (such as a Banach limit of α1, α2, α4, α8,...) so there are many different Dixmier traces. As the Dixmier trace is linear, it extends by linearity to all operators of L1,∞(H). If the Dixmier trace of an operator is independent of the choice of limω then the operator is called measurable.

Properties

• Trω(T) is linear in T.
• If T ≥ 0 then Trω(T) ≥ 0
• If S is bounded then Trω(ST) = Trω(TS)
• Trω(T) does not depend on the choice of inner product on H.
• Trω(T) = 0 for all trace class operators T, but there are compact operators for which it is equal to 1.

A trace φ is called normal if φ(sup xα) = sup φ( xα) for every bounded increasing directed family of positive operators. Any normal trace on ${\displaystyle L^{1,\infty }(H)}$ is equal to the usual trace, so the Dixmier trace is an example of a non-normal trace.

Examples

A compact self-adjoint operator with eigenvalues 1, 1/2, 1/3, ... has Dixmier trace equal to 1.

If the eigenvalues μi of the positive operator T have the property that

${\displaystyle \zeta _{T}(s)=\operatorname {Tr} (T^{s})=\sum {\mu _{i}^{s}}}$

converges for Re(s)>1 and extends to a meromorphic function near s=1 with at most a simple pole at s=1, then the Dixmier trace of T is the residue at s=1 (and in particular is independent of the choice of ω).

Connes (1988) showed that Wodzicki's noncommutative residue (Wodzicki 1984) of a pseudodifferential operator on a manifold is equal to its Dixmier trace.

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Life in the Fourth Dimension

I would like to start off by introducing myself, my name is Jonathan Ploegman from Long Island, New York. I am 18 years old and I am new to the physics world, however I am very interested and eager to learn all that I can.

I was wondering what life in the "fourth" dimension would be life. I am not talking about the fourth dimension as time, but as the fourth dimension of space.

I read that we could walk through walls, and do all kinds of stuff nearly becoming God-like.

Anyway, I am just curious about the fourth dimensional world.

Have a great day everybody!

Funkee
Welcome, Jonathan.

Well, yes, higher dimensions offer incredible abilities that defy many of our laws of physics. But there is a flaw. Usually when we refer to the next higher spatial dimension, we use the 5th, accepting time as the 4th. Now, a theory that holds up to date to the superstring theory is that the universe began in 10 dimensions, but some disturbance caused all the matter to flow into and remain in 4 dimensions, space-time. The other 6, without any matter collapsed into a nearly impenetrable size, the Planck length. To probe it, it would require incredible amounts of energy we cannot harvest just yet.

This is from the book hyperspace, imagine a man that lives in 2D, and is somehow exposed to the 3rd dimension and is floating through it. His vision is limited to be able to conceive only 1 dimension. We, as 3 dimensional beings can observe only 2 dimensions at once without rotation. Now, he would be floating through space, and let’s say he encountered an apple. What he would see is a red line, that becomes progressively larger, then smaller, and then it turns brown and disappears. He would be seeing the apple in 1 dimensional cross-sections. Apply that to our 3 dimensions. We can only conceive 2 dimensions at a time, but supposedly a higher being, to visualize a hypercube for example; we should be able to conceive all 3 dimensions at once to be able to visualize higher dimensional movement. But we can’t do so without rotating the figure.

So let’s say we were floating through a higher dimension and we encountered a hyper-sphere or a hyper-apple to suppose. What we would see is a 2D view of the sphere, and as we progress through the cross-sections of the sphere, we would see the sphere appear out of nowhere, get bigger, then smaller, and then disappear.

What you’re speaking of, regarding walking through walls is often used in mystical sciences, to explain the movement of ghosts, spirits, etc. Take the same 2D person, and draw a box around him on a piece of paper. His vision is limited to the paper; he cannot visualize or move up and down as we have it. No matter which way he would move, he would not be able to escape. Now, take a prison cell for example, and place a higher dimensional being into it. Being able to move through a higher spatial dimension, he would be able to move ana or kata, the motions defined for the next spatial dimension, and go around the walls and appear back on the other side by reversing the motion.

Some argue though that we do exist in all dimensions, but being that the dimensions are so microscopically small and looped, if we do move through them, our motion is limited past what we can observe. But that is another idea behind tunneling, where a particle can move from point to point through any barricade under a certain probability. Whether that fits under the Schrödinger wave function or that is up to the rest. But a higher dimensional being will not exist if there isn’t room for it in the shrunken higher dimensions, unless we are actually are those beings.

The fourth spatial dimension is a hard concept to visualize but u can infer its properties by comparing the other dimensions.....for instance take a square ABCD (A up-left, B down-left, C down-right, D up-right).......now how many vertical lines can u draw like AB, between the side AB and CD.....infinite...

The same is with a cube....how many vertical squares exist between the 2 faces of the cubes.....infinte.....so we can go on to say that there are infinte number of cubes between the two opposite faces of a hyper cube....thats another way of imagining...Funkee included all the cool ideas....

Welcome to the PFs, Ploegman . How do you pronounce "Ploegman"?

To help you visualize life in the fourth dimension, I suggest getting/borrowing a copy of "Journey Through the Tenth Dimension", by Michio Kaku. It's on audio tapes, and is a lecture that he gave in 1989.

An example of the illustrations that he uses would be this:

Imagine a two-dimensional person. She is absolutely flat. Her skin has formed "all-around" her, but, to us, it looks like a flat outline. You see, to other two-dimensional people, it appears that you would have to pierce through the skin, in order to see her insides. However, if you look at it from the third dimension (as we could) you would see both the outline, and the insides, at the same time! This is what it is like for someone that lives in the fourth dimension, even with regard to us (three-dimensional people). You see, you may think that you are covered "all around", but to a person in the fourth dimension, it is just an "outline", and they can see both it and the "insides".

There is much more in the lecture, and I highly recommend listening to it.

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RuroumiKenshin
That sounds like a wonderful lecture! I think I'll listen to it.

Anyway, in a 4D world, are there 4 spatial demensions? The 4th demension is perfectly active in our demension. So, in a 4D world, what would it be like, in the sense that 4D is time, which is defined as the increase of entropy? Would people themselves be increasing in entropy? I know how silly that sounds!! but I can't help but guess(note I didn't say infer!)

Originally posted by MajinVegeta
That sounds like a wonderful lecture! I think I'll listen to it.

Anyway, in a 4D world, are there 4 spatial demensions? The 4th demension is perfectly active in our demension. So, in a 4D world, what would it be like, in the sense that 4D is time, which is defined as the increase of entropy? Would people themselves be increasing in entropy? I know how silly that sounds!! but I can't help but guess(note I didn't say infer!)

Yes, it is a really fantastic lecture. If you are really interested in this whole 4th dimensional viewpoint, I also recommend the books, Flatland and Sphereland.

Yes, Majin, that's what they mean when they talk about life in the 4th dimension. Basically, when one theorizes about spacial dimensions, other than those that we are accustomed to (note: Prof. Kaku also briefly discusses why we are only used to three of the spacial dimensions), they "push 'time' up a notch" - so to speak. e.g. In ten-dimensional theory, time is the 11th dimension.

RuroumiKenshin
re: time is the 11th demension to 11D theory
How so?(do expand)

Funkee
I, too, seem to be a bit lost.

10 dimensional string theory explains that there is a total of 10 dimensions, due to quantum symmetry, and that even change (time) can alter symmetry, so time itself is within the 10. One popular idea supports that the universe split when all matter was confined to only 4 dimensions, while the 6 other collapsed to smallest measurement in each dimension that any matter can have.

For instance, take a sheet of paper. Draw a person on it. The 2D person is confined to the dimensions only when using the dimension to describe a part of an object. Not when you try to describe the object in the dimensions.

For you see, the man on the paper truly has a height(3rd dimension) when describing the man itself, and in that we do in fact use our 3, if not 4 dimensions. The point is, that any particle will have a measurement in higher dimensions, but it is supposedly the absolute minimal that any matter can have, which is the Planck Length.

If the object were to cease to have any properties in higher dimensions, no matter how infintessimaly small, the object would cease to have values in other dimensions and cease to exist. That means, that all matter does have a property and value in every higher dimension, including time, because even change can be symmetrical.

RuroumiKenshin
Originally posted by Funkee
I, too, seem to be a bit lost.

10 dimensional string theory explains that there is a total of 10 dimensions, due to quantum symmetry, and that even change (time) can alter symmetry, so time itself is within the 10. One popular idea supports that the universe split when all matter was confined to only 4 dimensions, while the 6 other collapsed to smallest measurement in each dimension that any matter can have.

Oh, you mean like branes?

For instance, take a sheet of paper. Draw a person on it. The 2D person is confined to the dimensions only when using the dimension to describe a part of an object. Not when you try to describe the object in the dimensions.
The person is confined to the demensions because it does not have demensions that exceed 2.

For you see, the man on the paper truly has a height(3rd dimension) when describing the man itself, and in that we do in fact use our 3, if not 4 dimensions. The point is, that any particle will have a measurement in higher dimensions, but it is supposedly the absolute minimal that any matter can have, which is the Planck Length.

Well, I agree that, since the piece of paper which the object is drawn is influenced by time, it can have a 4th demensional property.
But other than that, I don't see your point, since you seem to be missing a very important prospect: you need to know what a "higher" demension is in order to incorperate it to an object.

If the object were to cease to have any properties in higher dimensions, no matter how infintessimaly small, the object would cease to have values in other dimensions and cease to exist. That means, that all matter does have a property and value in every higher dimension, including time, because even change can be symmetrical.
So you believe that a demensional value is what dictates existence? Please explain.

I'm sorry for the confusion, I didn't expect that. Well, then, let me try to explain.

When Einstein was developing Relativity, he was only making reference to the three spacial dimensions that we commonly discuss/affect. These are represented by the x, y, and z coordinates. Time is supposed to be another dimension/coordinate, and is thus number four. Einstein thought that this was complete. However, hyperspacial theories dictate that there may be many more spacial dimension than Einstein thought. Now, the numbering doesn't really matter, but string theorists refer to time as the 11-th dimension (and the other ten are spacial). If it were discovered that there were in fact 11 spacial dimensions (instead of 10, like string theory postulates) then time would be the 12-th dimension. Does that make more sense?

Ploegman: Pronounced = Plue (like "glue") add a "g" sound, like in the beginning a glue, Plueg ... min, like mint. Pluegmin.

RuroumiKenshin
Originally posted by Mentat
I'm sorry for the confusion, I didn't expect that. Well, then, let me try to explain.

When Einstein was developing Relativity, he was only making reference to the three spacial dimensions that we commonly discuss/affect. These are represented by the x, y, and z coordinates. Time is supposed to be another dimension/coordinate, and is thus number four. Einstein thought that this was complete. However, hyperspacial theories dictate that there may be many more spacial dimension than Einstein thought. Now, the numbering doesn't really matter, but string theorists refer to time as the 11-th dimension (and the other ten are spacial). If it were discovered that there were in fact 11 spacial dimensions (instead of 10, like string theory postulates) then time would be the 12-th dimension. Does that make more sense?

Why does time have to be the last demension in any number of demensions? Is it because time is always the last demension relative to spacial demensions? if so why?
whats the difference between spacial demensions and non-special demension? i.e., what makes a non-spacial demension non-spacial? How in the world do you visualize this?

If you're really interested about learning all that stuff, i would recommend the book "Hyperspace" by Michio Kaku. I read it when i was 14 and since then i decided to orient my studies in physics. It really explains well, even without much mathematical concepts which is good if you are not familiar with advance math (and is not as fantastic when you actually know the math after and you wish he'd talk more about that). Anyway, it's a classic, and all the libraries i know have it so you don't even need to buy it to take a look. Hope you'll find it interesting.

RuroumiKenshin
You know, Max Born is an excellent author too. But a PhD physicist recommended his books; this must have been in the 70's...

So Michio Kaku's books are all basically on m-theory and string theory?

damgo
The ordering doesn't matter at all; it's just convention, ie what order you write stuff in. Actually time is usually written first, so an event's coordinates are given as (t, x, y, z).

A spatial dimension is just another one like the 3 we have now. A line is 1-D, a plane 2-D, and regular space 3-D. Higher spatial dimensions just add more independent directions. You can't really visualize more than three -- at least I haven't met anyone who could, except for maybe a topology professor or two -- so you have to just use the math.

A non-spatial dimension would be another time dimension, I suppose, though it's hard to imagine how there could be more than one.

Originally posted by pi-70679
If you're really interested about learning all that stuff, i would recommend the book "Hyperspace" by Michio Kaku. I read it when i was 14 and since then i decided to orient my studies in physics. It really explains well, even without much mathematical concepts which is good if you are not familiar with advance math (and is not as fantastic when you actually know the math after and you wish he'd talk more about that). Anyway, it's a classic, and all the libraries i know have it so you don't even need to buy it to take a look. Hope you'll find it interesting.

Michio Kaku is an excellent physicist, and an extraordinary author. He helps people visiualize/comprehend the more advanced concepts of string/M- theory, whether they have an advanced knowledge of math and science or not.

I second the recommendation of Hyperspace, but I would like to add Beyond Einstein, also by Prof. Kaku.

RuroumiKenshin
Originally posted by damgo
The ordering doesn't matter at all; it's just convention, ie what order you write stuff in. Actually time is usually written first, so an event's coordinates are given as (t, x, y, z).

Why x,y,z? Why not abc or something?

usually, the last half of the alphabet is reserved for variables and the first half for constants. It's just by convention.

hmm...

The 4th dimension is something i have never understood. Time, is it not just an invention of man? The speed of time is relative, is it not? I have often thought that if an object could reach the speed of light, time would stand still for that object. To me, there are only 2 limits in the universe. The speed of light, maximum kinetic energy, and absolute zero, minimum kinetic energy. I know we have come within a billionth of 0 kelvin, but i believe we are no closer to that than we are the speed of light, neither of which i believe man will ever reach. I probably sound like a complete idiot, but 10th grade doesnt teach much physics.

Locutus
The fourth spatial dimension gives one amazing perceptual abilities, among other things of course. He could read an entire book without turning a single page, escape from closed rooms, be able to see the sun at all times of day. Instead of painting is masterpieces on planar canvases, he would do so on and inside cubes. He may even be able to travel through space at alarming speeds...

Locutus

Originally posted by The_oMeGa
The 4th dimension is something i have never understood. Time, is it not just an invention of man? The speed of time is relative, is it not? I have often thought that if an object could reach the speed of light, time would stand still for that object. To me, there are only 2 limits in the universe. The speed of light, maximum kinetic energy, and absolute zero, minimum kinetic energy. I know we have come within a billionth of 0 kelvin, but i believe we are no closer to that than we are the speed of light, neither of which i believe man will ever reach. I probably sound like a complete idiot, but 10th grade doesnt teach much physics.

This post is about a 4th spacial dimension. However, I find it necessary to say that time is not just an invention of man. Time is a property of the universe as absolute (in existence) as something such as light. And you can't say man invented light, or for that matter, energy, space, gravity, etc.

BTW, i'm in 10th grade too and the problem with is it is that other kids just care more about getting drunk on prom night than learning things. Sadly, this deters us all.

RuroumiKenshin

Originally posted by The_oMeGa
The 4th dimension is something i have never understood. Time, is it not just an invention of man? The speed of time is relative, is it not? I have often thought that if an object could reach the speed of light, time would stand still for that object. To me, there are only 2 limits in the universe. The speed of light, maximum kinetic energy, and absolute zero, minimum kinetic energy. I know we have come within a billionth of 0 kelvin, but i believe we are no closer to that than we are the speed of light, neither of which i believe man will ever reach. I probably sound like a complete idiot, but 10th grade doesnt teach much physics.

What??!! I was looking forward to 10th grade (and highschool) 'cause I thought you'd learn physics more physics. so much for highschool.

Anyway, time is just the increase of entropy, and a coordinate. It tells when something existed. That's a simple definition. What part of time do you not understand?

dav2008
Gold Member
Nope...Most people only take like physics accel in high school and thats it...All our school has is regular physics and then AP B and AP C...theres no classes about 20th century physics or anything specific... u gotta take those in college...

Ok..well...the increase in entropy is an AFFECT of time, is it not? I mean...for our universe i guess it could define it..but yea..Its more complicated than that..Yea.. i dont know what im talking about

RuroumiKenshin
Lol, no, you've missed the point. Time is the increase of entropy. But to make it a little easier to understand, I'll describe it using coordinates:
Code:
     time l
l
l
l        x
l____________space
MV

The x, is lets say 12:00pm. MV is someone somehwere at 12:00pm. this is what I mean by 'time tells when something exists".

do you follow?

alright.....lets make the question clear......has the geometry of higher spatial dimensions been developed already???


text:
# HEP of subcomplex for product topology of CW-complexes

Suppose $X$ and $Y$ are CW-complexes and $A\subset X$ and $B\subset Y$ have properties such that the product topologies of $X\times B$ and $A\times Y$ are CW-complexes, such as when $A$ and $B$ are locally compact. So complexes $X\times B \cup A\times Y$ have the homotopy extension property in $(X\times Y)_c$, but they're also spaces that are subtopologies of $X\times Y$, so must they have the HEP in $X\times Y$?

In particular I'd like to show it for the case of $(X\times X, X\times \{e\} \cup \{e\}\times X)$ where $e$ is a $0$-cell in $X$. The case $(X\times X, X\times \{e\})$ is easy enough from extending the retract for the pair $(X,e)$, but beyond that I'm pretty stuck.


text:
## Saturday, November 26, 2011

### What do contractors in machine learning charge by the hour?

I saw this question on Quora this morning: "What do contractors in machine learning charge by the hour?"

Obviously the answer depends on the skill level, but based on the few machine learning contractors I've worked with on oDesk, I would guess about $30-$40/hour is average. However, I wanted to check with some real data. I have access to our full database, but a more useful answer would allow people to see how to do this for other skills by simply scraping our search results (we have an API, but I don't know how to use it yet and I wanted to try the BeautifulSoup package).

First step was to reverse-engineer our search syntax and the HTML for profile rates. A contractor search of "machine learning" gives this:

"https://www.odesk.com/contractors?nbs=1#q=machine+learning"

and if I click on the next 10 results, the url is:

"https://www.odesk.com/contractors?nbs=1#q=machine+learning&skip=10"

I then checked to see what happens if I set "skip=0"---it returns that same thing as the first search URL, so now know that I can just write one function that returns the query URL, with parameters of "q" and "skip."

Next, I needed to find out how the rates are stored. Using Chrome's awesome "Inspect element" feature, I found that the rates for the 10 returned results as listed as "rate_1", "rate_2", and so on:

The rest was pretty easy---I just wrote two loops to collect up wages. I saw that we had some clear false positives, which I filtered out. This actually brings up a big problem on oDesk that we've been working on---namely that until recently, we had no standardization of skills, which made it hard to match people or do really good, highly specific queries. We've now moved to a closed (but expandable) vocabulary of skills, ala StackOverflow which in the long run will make it much easier to do matching and recommendations (and little data projects like this). That's a topic for another blog post. So, returning to the original question, my answer based on a pretty tiny sample:

Min: 16.67
Max: 100.0
Mean: 39.6983333333

And here's my pretty crappy but for-the-moment-functional code on GitHub:


text:
# Revision history [back]

I suppose, we need "something like" a "multiplicative order" of $(x+1)$ in the ring. A first problem is that $(x+1)$ is not a unit. It is a zero divisor, since $X^n+1=X^n-1 = (X-1)(\dots)=(X+1)(\dots)$ in the polynomial ring in $X$ over the field with two elements. So we cannot use the word "order" (in the multiplicative sense). (Well, getting the additive order could not be the problem, so we try to give a sense to the question for the multiplication as considered operation...)

OK, let us define a "pseudomultiplicative order" by passing to an isomorphic ring and ignoring the factors where $(x+1)$ becomes nilpotent. Just as an example, to fix the framework, let us take $n=99$. Then $${\mathbb F}_2[X] /(X^99+1)\ \cong\ ({\mathbb F}_2[X]/(X+1))\times ({\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1))\ .$$ The image of $(x+1)$ in the cartesian product of the right side is $(0,x+1)$, so $(x+1)$ is not a unit. We will tacitly ignore the first cartesian factor and ask for the order of $(x+1)$ in $${\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1)$$ where $x$ is now the image of $X$ in the above quotient. This ring still splits corresponding to the factorization of the moded out polynomial:

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
R.<x> = A.quotient( MOD )
print "(X^%s+1) / (X+1) has the following factors over GF(2) :" % n
for f, _ in MOD.factor():
print f

And we get

(X^99+1) / (X+1) has the following factors over GF(2) :
X^2 + X + 1
X^6 + X^3 + 1
X^10 + X^7 + X^5 + X^3 + 1
X^10 + X^9 + X^5 + X + 1
X^10 + X^9 + X^8 + X^7 + X^6 + X^5 + X^4 + X^3 + X^2 + X + 1
X^30 + X^21 + X^15 + X^9 + 1
X^30 + X^27 + X^15 + X^3 + 1

Each factor is irreducible. We use the structure theorem (Chinese Reminder Theorem, CRT) $R/(fg)\cong (R/f)\times (R/g)$. (The map in one direction is canonical. The CRT insures it is a bijection.) We have more than two factors, so we need the inducitively extended version of CTR. So in our case, the ring ${\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1)$ is isomorphic to the cross product of the following fields:

${\mathbb F}_2[X]/(X^2 + X + 1)$, and

${\mathbb F}_2[X]/(X^6 + X^3 + 1 )$, and

${\mathbb F}_2[X]/(X^10 + X^7 + X^5 + X^3 + 1 )$, and

${\mathbb F}_2[X]/(X^10 + X^9 + X^5 + X + 1 )$, and

${\mathbb F}_2[X]/(X^10 + X^9 + X^8 + X^7 + X^6 + X^5 + X^4 + X^3 + X^2 + X + 1 )$, and

${\mathbb F}_2[X]/(X^30 + X^21 + X^15 + X^9 + 1 )$, and

${\mathbb F}_2[X]/(X^30 + X^27 + X^15 + X^3 + 1 )$.

The order of these fields are respectively $2^2$, $2^6$, $2^{10}$ three times, and $2^{30}$ twice.

The order of the respective subjacent (cyclic) multiplicative group of nonzero elements is $2^2-1$, $2^6-1$, $2^{10}-1$ three times, and $2^{30}-1$ twice.

The element $(x+1)$ has mapped images in all these fields. So it has an order dividing the lcm of all these numbers. Which is of course $M = 2^{30}-1$ in our case. It is still possible, that the order drops by a divisor of this number. Let us see first that $(x+1)^M$ is the one.

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
R.<x> = A.quotient( MOD )
M = lcm( [ 2^f.degree() - 1 for f, _ in MOD.factor() ] )
print "(x+1)^%s = %s" % ( M, (x+1)^M )

And we get:

(x+1)^1073741823 = 1

(Unfortunately we cannot obtain simply (x+1).multiplicative_order() although the method is there... Not no my machine with sage in version 'SageMath version 7.6, Release Date: 2017-03-25'. An error occurs.)

Let us get the order now:

divisors = M.divisors()
divisors . sort()    # i suspect they were already...
for d in divisors:
if (x+1)^d == R(1):
print "(x+1)^%s = %s" % ( d, (x+1)^d )
break

And we get:

(x+1)^3243933 = 1

Alternatively, we can start with the polynomial $f = (X^n+1)/(X+1)$, factor it, for each prime=irreducible factor $g$ of it we consider the field ${\mathbb F}_2[X]/(g)$, then the image of $x$, which is $X$ modulo $f$ is $y$ -- say -- , which is $X$ modulo $g$. We get its order. Then we consider the lcm of all these orders, when $g$ runs in the list of all factors of $f$. The code for this is:

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
ords  = []    # this is a list of orders, and we will append...
for g, gmultiplicity in MOD.factor():
R.<y> = GF( 2**g.degree(), modulus=g )
ords.append( (y+1).multiplicative_order() )

print "order(x+1) = lcm of %s = %s" % ( str(ords), lcm(ords) )

... getting:

order(x+1) = lcm of [3, 63, 1023, 1023, 341, 3243933, 3243933] = 3243933

So for a general $n$ the following function does the job in the sense above. (For an EVEN $n$ i will extract the right power of two to be used for making the program still work.)

def getOrder( n, verbose=False ):
"""get the order of (x+1) in the ring
R[ X ] / f
where f = (X^n+1) / (X+1)^(right power)
"""
powerof2 = Qp(2)(n).valuation()
q        = 2 ** powerof2
F        = GF(2)
A.<X>    = F[]
MOD      = A( (X^n + 1) / (X+1)^q )
ords     = []    # this is a list of orders, and we will append...
for g, gmultiplicity in MOD.factor():
R.<y> = GF( 2**g.degree(), modulus=g )
ords.append( (y+1).multiplicative_order() )

M = lcm( ords )
if verbose:
print "n = %s :: order(x+1) = lcm of %s = %s" % ( n, str(ords), M.factor() )
return M

Above, we fix an $n$. We write than $n= k\cdot q$, where $q=2^r$ is the maximal power of two dividing $n$. Then $$X^n+1 = (X^k+1)^q=(X+1)^q\cdot (X^{k-1}+\dots+1)^q\ .$$ And the last second factor does not have the root $1$, since $k$ is odd and we have $k$ terms, all of them evaluating to one when setting $X=1$.

For $n$ among $124, 125, 126, 127, 128, 129$ we get the following orders:

for n in [ 124 .. 129 ]:
getOrder( n, verbose=True );

And we obtain:

n = 124 :: order(x+1) = lcm of [31, 31, 31, 31, 31, 31] = 31
31
n = 125 :: order(x+1) = lcm of [15, 25575, 140737488355327875] = 3 * 5^3 * 11 * 31 * 251 * 601 * 1801 * 4051
140737488355327875
n = 126 :: order(x+1) = lcm of [3, 7, 7, 21, 63, 63, 9, 63, 63, 63, 21, 63] = 3^2 * 7
63
n = 127 :: order(x+1) = lcm of [127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127] = 127
127
n = 128 :: order(x+1) = lcm of [] = 1
1
n = 129 :: order(x+1) = lcm of [3, 16383, 16383, 5461, 16383, 5461, 16383, 5461, 16383, 16383] = 3 * 43 * 127
16383

The posted question is answered by [ getOrder(n) for n in [1..300] ] and takes its seconds.

I suppose, we need "something like" a "multiplicative order" of $(x+1)$ in the ring. A first problem is that $(x+1)$ is not a unit. It is a zero divisor, since $X^n+1=X^n-1 = (X-1)(\dots)=(X+1)(\dots)$ in the polynomial ring in $X$ over the field with two elements. So we cannot use the word "order" (in the multiplicative sense). (Well, getting the additive order could not be the problem, so we try to give a sense to the question for the multiplication as considered operation...)

OK, let us define a "pseudomultiplicative order" by passing to an isomorphic ring and ignoring the factors where $(x+1)$ becomes nilpotent. Just as an example, to fix the framework, let us take $n=99$. Then $${\mathbb F}_2[X] /(X^99+1)\ /(X^{99}+1)\ \cong\ ({\mathbb F}_2[X]/(X+1))\times ({\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1))\ .$$ The image of $(x+1)$ in the cartesian product of the right side is $(0,x+1)$, so $(x+1)$ is not a unit. We will tacitly ignore the first cartesian factor and ask for the order of $(x+1)$ in $${\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1)$$ where $x$ is now the image of $X$ in the above quotient. This ring still splits corresponding to the factorization of the moded out polynomial:

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
R.<x> = A.quotient( MOD )
print "(X^%s+1) / (X+1) has the following factors over GF(2) :" % n
for f, _ in MOD.factor():
print f

And we get

(X^99+1) / (X+1) has the following factors over GF(2) :
X^2 + X + 1
X^6 + X^3 + 1
X^10 + X^7 + X^5 + X^3 + 1
X^10 + X^9 + X^5 + X + 1
X^10 + X^9 + X^8 + X^7 + X^6 + X^5 + X^4 + X^3 + X^2 + X + 1
X^30 + X^21 + X^15 + X^9 + 1
X^30 + X^27 + X^15 + X^3 + 1

Each factor is irreducible. We use the structure theorem (Chinese Reminder Theorem, CRT) $R/(fg)\cong (R/f)\times (R/g)$. (The map in one direction is canonical. The CRT insures it is a bijection.) We have more than two factors, so we need the inducitively extended version of CTR. So in our case, the ring ${\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1)$ is isomorphic to the cross product of the following fields:

${\mathbb F}_2[X]/(X^2 + X + 1)$, and

${\mathbb F}_2[X]/(X^6 + X^3 + 1 )$, and

${\mathbb F}_2[X]/(X^10 F}_2[X]/(X^{10} + X^7 + X^5 + X^3 + 1 )$, and

${\mathbb F}_2[X]/(X^10 F}_2[X]/(X^{10} + X^9 + X^5 + X + 1 )$, and

${\mathbb F}_2[X]/(X^10 F}_2[X]/(X^{10} + X^9 + X^8 + X^7 + X^6 + X^5 + X^4 + X^3 + X^2 + X + 1 )$, and

${\mathbb F}_2[X]/(X^30 + X^21 + X^15 F}_2[X]/(X^{30} + X^{21} + X^{15} + X^9 + 1 )$, and

${\mathbb F}_2[X]/(X^30 + X^27 + X^15 F}_2[X]/(X^{30} + X^{27} + X^{15} + X^3 + 1 )$.

The order of these fields are respectively $2^2$, $2^6$, $2^{10}$ three times, and $2^{30}$ twice.

The order of the respective subjacent (cyclic) multiplicative group of nonzero elements is $2^2-1$, $2^6-1$, $2^{10}-1$ three times, and $2^{30}-1$ twice.

The element $(x+1)$ has mapped images in all these fields. So it has an order dividing the lcm of all these numbers. Which is of course $M = 2^{30}-1$ in our case. It is still possible, that the order drops by a divisor of this number. Let us see first that $(x+1)^M$ is the one.

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
R.<x> = A.quotient( MOD )
M = lcm( [ 2^f.degree() - 1 for f, _ in MOD.factor() ] )
print "(x+1)^%s = %s" % ( M, (x+1)^M )

And we get:

(x+1)^1073741823 = 1

(Unfortunately we cannot obtain simply (x+1).multiplicative_order() although the method is there... Not no my machine with sage in version 'SageMath version 7.6, Release Date: 2017-03-25'. An error occurs.)

Let us get the order now:

divisors = M.divisors()
divisors . sort()    # i suspect they were already...
for d in divisors:
if (x+1)^d == R(1):
print "(x+1)^%s = %s" % ( d, (x+1)^d )
break

And we get:

(x+1)^3243933 = 1

Alternatively, we can start with the polynomial $f = (X^n+1)/(X+1)$, factor it, for each prime=irreducible factor $g$ of it we consider the field ${\mathbb F}_2[X]/(g)$, then the image of $x$, which is $X$ modulo $f$ is $y$ -- say -- , which is $X$ modulo $g$. We get its order. Then we consider the lcm of all these orders, when $g$ runs in the list of all factors of $f$. The code for this is:

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
ords  = []    # this is a list of orders, and we will append...
for g, gmultiplicity in MOD.factor():
R.<y> = GF( 2**g.degree(), modulus=g )
ords.append( (y+1).multiplicative_order() )

print "order(x+1) = lcm of %s = %s" % ( str(ords), lcm(ords) )

... getting:

order(x+1) = lcm of [3, 63, 1023, 1023, 341, 3243933, 3243933] = 3243933

So for a general $n$ the following function does the job in the sense above. (For an EVEN $n$ i will extract the right power of two to be used for making the program still work.)

def getOrder( n, verbose=False ):
"""get the order of (x+1) in the ring
R[ X ] / f
where f = (X^n+1) / (X+1)^(right power)
"""
powerof2 = Qp(2)(n).valuation()
q        = 2 ** powerof2
F        = GF(2)
A.<X>    = F[]
MOD      = A( (X^n + 1) / (X+1)^q )
ords     = []    # this is a list of orders, and we will append...
for g, gmultiplicity in MOD.factor():
R.<y> = GF( 2**g.degree(), modulus=g )
ords.append( (y+1).multiplicative_order() )

M = lcm( ords )
if verbose:
print "n = %s :: order(x+1) = lcm of %s = %s" % ( n, str(ords), M.factor() )
return M

Above, we fix an $n$. We write than $n= k\cdot q$, where $q=2^r$ is the maximal power of two dividing $n$. Then $$X^n+1 = (X^k+1)^q=(X+1)^q\cdot (X^{k-1}+\dots+1)^q\ .$$ And the last second factor does not have the root $1$, since $k$ is odd and we have $k$ terms, all of them evaluating to one when setting $X=1$.

For $n$ among $124, 125, 126, 127, 128, 129$ we get the following orders:

for n in [ 124 .. 129 ]:
getOrder( n, verbose=True );

And we obtain:

n = 124 :: order(x+1) = lcm of [31, 31, 31, 31, 31, 31] = 31
31
n = 125 :: order(x+1) = lcm of [15, 25575, 140737488355327875] = 3 * 5^3 * 11 * 31 * 251 * 601 * 1801 * 4051
140737488355327875
n = 126 :: order(x+1) = lcm of [3, 7, 7, 21, 63, 63, 9, 63, 63, 63, 21, 63] = 3^2 * 7
63
n = 127 :: order(x+1) = lcm of [127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127] = 127
127
n = 128 :: order(x+1) = lcm of [] = 1
1
n = 129 :: order(x+1) = lcm of [3, 16383, 16383, 5461, 16383, 5461, 16383, 5461, 16383, 16383] = 3 * 43 * 127
16383

The posted question is answered by [ getOrder(n) for n in [1..300] ] and takes its seconds.

I suppose, we need "something like" a "multiplicative order" of $(x+1)$ in the ring. A first problem is that $(x+1)$ is not a unit. It is a zero divisor, since $X^n+1=X^n-1 = (X-1)(\dots)=(X+1)(\dots)$ in the polynomial ring in $X$ over the field with two elements. So we cannot use the word "order" (in the multiplicative sense). (Well, getting the additive order could not be the problem, so we try to give a sense to the question for the multiplication as considered operation...)

OK, let us define a "pseudomultiplicative order" by passing to an isomorphic ring and ignoring the factors where $(x+1)$ becomes nilpotent. Just as an example, to fix the framework, let us take $n=99$. Then $${\mathbb F}_2[X] /(X^{99}+1)\ \cong\ ({\mathbb F}_2[X]/(X+1))\times ({\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1))\ .$$ The image of $(x+1)$ in the cartesian product of the right side is $(0,x+1)$, so $(x+1)$ is not a unit. We will tacitly ignore the first cartesian factor and ask for the order of $(x+1)$ in $${\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1)$$ where $x$ is now the image of $X$ in the above quotient. This ring still splits corresponding to the factorization of the moded out polynomial:

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
R.<x> = A.quotient( MOD )
print "(X^%s+1) / (X+1) has the following factors over GF(2) :" % n
for f, _ in MOD.factor():
print f

And we get

(X^99+1) / (X+1) has the following factors over GF(2) :
X^2 + X + 1
X^6 + X^3 + 1
X^10 + X^7 + X^5 + X^3 + 1
X^10 + X^9 + X^5 + X + 1
X^10 + X^9 + X^8 + X^7 + X^6 + X^5 + X^4 + X^3 + X^2 + X + 1
X^30 + X^21 + X^15 + X^9 + 1
X^30 + X^27 + X^15 + X^3 + 1

Each factor is irreducible. We use the structure theorem (Chinese Reminder Theorem, CRT) $R/(fg)\cong (R/f)\times (R/g)$. (The map in one direction is canonical. The CRT insures it is a bijection.) We have more than two factors, so we need the inducitively extended version of CTR. So in our case, the ring ${\mathbb F}_2[X]/(X^{98}+\dots+X^2+X+1)$ is isomorphic to the cross product of the following fields:

${\mathbb F}_2[X]/(X^2 + X + 1)$, and

${\mathbb F}_2[X]/(X^6 + X^3 + 1 )$, and

${\mathbb F}_2[X]/(X^{10} + X^7 + X^5 + X^3 + 1 )$, and

${\mathbb F}_2[X]/(X^{10} + X^9 + X^5 + X + 1 )$, and

${\mathbb F}_2[X]/(X^{10} + X^9 + X^8 + X^7 + X^6 + X^5 + X^4 + X^3 + X^2 + X + 1 )$, and

${\mathbb F}_2[X]/(X^{30} + X^{21} + X^{15} + X^9 + 1 )$, and

${\mathbb F}_2[X]/(X^{30} + X^{27} + X^{15} + X^3 + 1 )$.

The order of these fields are respectively $2^2$, $2^6$, $2^{10}$ three times, and $2^{30}$ twice.

The order of the respective subjacent (cyclic) multiplicative group of nonzero elements is $2^2-1$, $2^6-1$, $2^{10}-1$ three times, and $2^{30}-1$ twice.

The element $(x+1)$ has mapped images in all these fields. So it has an order dividing the lcm of all these numbers. Which is of course $M = 2^{30}-1$ in our case. It is still possible, that the order drops by a divisor of this number. Let us see first that $(x+1)^M$ is the one.

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
R.<x> = A.quotient( MOD )
M = lcm( [ 2^f.degree() - 1 for f, _ in MOD.factor() ] )
print "(x+1)^%s = %s" % ( M, (x+1)^M )

And we get:

(x+1)^1073741823 = 1

(Unfortunately we cannot obtain simply (x+1).multiplicative_order() although the method is there... Not no my machine with sage in version 'SageMath version 7.6, Release Date: 2017-03-25'. An error occurs.)

Let us get the order now:

divisors = M.divisors()
divisors . sort()    # i suspect they were already...
for d in divisors:
if (x+1)^d == R(1):
print "(x+1)^%s = %s" % ( d, (x+1)^d )
break

And we get:

(x+1)^3243933 = 1

Alternatively, we can start with the polynomial $f = (X^n+1)/(X+1)$, factor it, for each prime=irreducible factor $g$ of it we consider the field ${\mathbb F}_2[X]/(g)$, then the image of $x$, which is $X$ modulo $f$ $f$, is $y$ -- say -- , which is $X$ modulo $g$.

We get its order. the multiplicative order of $y+1$ in this field.

Then we consider the lcm of all these orders, when $g$ runs in the list of all factors of $f$. The code for this is:

n     = 99
F     = GF(2)
A.<X> = PolynomialRing( F )
MOD   = A( (X^n + 1) / (X+1) )
ords  = []    # this is a list of orders, and we will append...
for g, gmultiplicity in MOD.factor():
R.<y> = GF( 2**g.degree(), modulus=g )
ords.append( (y+1).multiplicative_order() )

print "order(x+1) = lcm of %s = %s" % ( str(ords), lcm(ords) )

... getting:

order(x+1) = lcm of [3, 63, 1023, 1023, 341, 3243933, 3243933] = 3243933

So for a general $n$ the following function does the job in the sense above. (For an EVEN $n$ i will extract the right power of two to be used for making the program still work.)

def getOrder( n, verbose=False ):
"""get the order of (x+1) in the ring
R[ X ] / f
where f = (X^n+1) / (X+1)^(right power)
"""
powerof2 = Qp(2)(n).valuation()
q        = 2 ** powerof2
F        = GF(2)
A.<X>    = F[]
MOD      = A( (X^n + 1) / (X+1)^q )
ords     = []    # this is a list of orders, and we will append...
for g, gmultiplicity in MOD.factor():
R.<y> = GF( 2**g.degree(), modulus=g )
ords.append( (y+1).multiplicative_order() )

M = lcm( ords )
if verbose:
print "n = %s :: order(x+1) = lcm of %s = %s" % ( n, str(ords), M.factor() )
return M

Above, we fix an $n$. We write than $n= k\cdot q$, where $q=2^r$ is the maximal power of two dividing $n$. Then $$X^n+1 = (X^k+1)^q=(X+1)^q\cdot (X^{k-1}+\dots+1)^q\ .$$ And the last second factor does not have the root $1$, since $k$ is odd and we have $k$ terms, all of them evaluating to one when setting $X=1$.

For $n$ among $124, 125, 126, 127, 128, 129$ we get the following orders:

for n in [ 124 .. 129 ]:
getOrder( n, verbose=True );

And we obtain:

n = 124 :: order(x+1) = lcm of [31, 31, 31, 31, 31, 31] = 31
31
n = 125 :: order(x+1) = lcm of [15, 25575, 140737488355327875] = 3 * 5^3 * 11 * 31 * 251 * 601 * 1801 * 4051
140737488355327875
n = 126 :: order(x+1) = lcm of [3, 7, 7, 21, 63, 63, 9, 63, 63, 63, 21, 63] = 3^2 * 7
63
n = 127 :: order(x+1) = lcm of [127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127, 127] = 127
127
n = 128 :: order(x+1) = lcm of [] = 1
1
n = 129 :: order(x+1) = lcm of [3, 16383, 16383, 5461, 16383, 5461, 16383, 5461, 16383, 16383] = 3 * 43 * 127
16383

The posted question is answered by [ getOrder(n) for n in [1..300] ] and takes its seconds.


text:
Please, try EDU on Codeforces! New educational section with videos, subtitles, texts, and problems. ×

epsilon_573's blog

By epsilon_573, history, 4 weeks ago, ,

I have tried 20-30 times , using floor() and type casting but I can't seem to get AC. What is the problem here? Thanks for the help.

// Author : Epsilon573
// If it works, don't touch it.

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

int main()
{
ios_base :: sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);

ll a; ld b;
cin >> a >> b;

ll c = b*100;

cout << a*c/100LL;

return 0;
}

• 0

 » 4 weeks ago, # |   0 Floating point types have a precision, so the result of calculations is often not exact. If you want to cast to an integer type this can lead to wrong result.In this case you can do something like ll c=(b+0.001)*100; to force kind of round-up. Or read the number as a string, and parse by hand ignoring the dot.


text:
## gjhfdfg Group Title Use a graphing calculator to solve the equation -3 cos t = 1 in the interval from 0 to 2p. Round to the nearest hundredth. one year ago one year ago

1. gjhfdfg Group Title

I typed it in on my calculator but did not do a thingg!

2. swissgirl Group Title

$$\cos(t)=\large{1 \over- 3}$$ $$\cos^{-1}(\large{1 \over- 3} )=t$$

3. gjhfdfg Group Title

How did you get that?

4. swissgirl Group Title

First I divided -3 from both sides

5. swissgirl Group Title

$$-3\cos(t)=1$$ $$\large {-3\cos(t) \over -3}={1 \over -3}$$ $$\cos(t)=\large{1 \over -3}$$ $$\cos^{-1}(\cos(t))=Cos^{-1}(\large{1 \over -3})$$ $$t=Cos^{-1}(\large{1 \over -3})$$

6. gjhfdfg Group Title

How would you put this in a decimal form.?

7. swissgirl Group Title

lol you gotta put this into your calc first and then your answer will be a decimal

8. gjhfdfg Group Title

Oh wow that was dumb of me lol. I was thinking this had to be put in a decimal form.! before solving it

9. gjhfdfg Group Title

I got 109.47

10. swissgirl Group Title

Are you supossed to be working in radians or degrees?

11. gjhfdfg Group Title

Ah, I had my calc set in degrees, I'm pretty sure I'm suppose to be using radians.! Gah I keep forgetting to change the mode

12. gjhfdfg Group Title

Ok now I got 1.91.!

13. swissgirl Group Title

YUP THAT IS WHAT I GOT :)

14. gjhfdfg Group Title

Thanks for the help.! (-:


text:
# Post-quantum cryptography from supersingular isogeny problems?

Christophe Petit
University of Birmingham
https://www.cs.bham.ac.uk/~petitcz/

Date(s) : 17/05/2018   iCal
11 h 00 min - 12 h 00 min

We review existing cryptographic schemes based on the hardness of computing isogenies between supersingular isogenies, and present some attacks against them. In particular, we present techniques to accelerate the resolution of isogeny problems when the action of the isogeny on a large torsion subgroup is known, and we discuss the impact of these techniques on the supersingular key exchange protocol of Jao-de Feo.

Catégories


text:
## Tsukuba Journal of Mathematics

### A product formula defined by the Beta function and Gauss's hypergeometric function

#### Abstract

Let $c$ be a constant in ${\bf R}^t$. For a plane algebraic curve $r^{2m-n} = 2c^n \cos n\theta$, which depends on $m$ and $n$ in ${\bf N}$, we show that the whole length of the curve are given by a value of a product formula defined by the Beta function and Gauss's hypergeometric function depending $m$ and $n$ in ${\bf N}$. Besides, we point out the fact to be a similar model and an expansion for the complete elliptic integral of the second kind. Last, we give a background for the fact explaining the special case $m = n$.

#### Article information

Source
Tsukuba J. Math., Volume 34, Number 1 (2010), 13-30.

Dates
First available in Project Euclid: 8 September 2010

Permanent link to this document
https://projecteuclid.org/euclid.tkbjm/1283967405

Digital Object Identifier
doi:10.21099/tkbjm/1283967405

Mathematical Reviews number (MathSciNet)
MR2723721

Zentralblatt MATH identifier
1223.33029

#### Citation

Ogawa, Takuma; Kamata, Yasuo. A product formula defined by the Beta function and Gauss's hypergeometric function. Tsukuba J. Math. 34 (2010), no. 1, 13--30. doi:10.21099/tkbjm/1283967405. https://projecteuclid.org/euclid.tkbjm/1283967405