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http://mathhelpforum.com/calculus/20921-implicit-differenciate.html
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# Thread:
1. ## implicit differenciate
I need help with these two problems! I keep working them out and getting wrong answers! BAH!
1.) Use implicit differentiation to find the equation of the tangent line to the curve at the point . The equation of this tangent line can be written in the form where is:
and where is:
AND
I keep got 0 for this one and its not right! :S
If and , find by implicit differentiation.
2. Differentiating implicitly:
Often, students forget the product rule when using implicit.
$3xy^{2}y'+y^{3}+xy'+y=0$
$y'=\frac{-y(y^{2}+1)}{x(3y^{2}+1)}$
Now, plug in your given points to find the slope.
Next, use your given points and the slope in y=mx+b to solve for b. You're done.
3. Thank you so much! I didn't differentiate correctly!
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http://math.stackexchange.com/questions/135541/range-terminology?answertab=oldest
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# Range Terminology
How should the following statement be interpreted: Let $f$ be a function with range in* $[0,1]$. (Here the word "in" has been deliberately included.)
In this context, the range is the same as the image.
I've always taken the statement to mean that the image of $f$ lies in $[0,1]$. Are there alternate interpretations of this statement? For example, can the above statement also mean that the image of $f$ is exactly $[0,1]$?
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## 1 Answer
The most common interpretation of that phrase would be that the range (image) of $f$ is a subset of $[0,1]$. That is, we may say $f:D \to [0,1]$ if the domain of $f$ is $D$. That is, the codomain of $f$ may be taken to be $[0,1]$.
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http://planetmath.org/closedoperator
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# closed operator
Let $B$ be a Banach space. A linear operator $A\colon\mathscr{D}(A)\subset B\to B$ is said to be closed if for every sequence $\{x_{n}\}_{{n\in\mathbb{N}}}$ in $\mathscr{D}(A)$ converging to $x\in B$ such that $Ax_{n}\xrightarrow[n\to\infty]{}y\in B$, it holds $x\in\mathscr{D}(A)$ and $Ax=y$. Equivalently, $A$ is closed if its graph is closed in $B\oplus B$.
Given an operator $A$, not necessarily closed, if the closure of its graph in $B\oplus B$ happens to be the graph of some operator, we call that operator the closure of $A$, and we say that $A$ is closable. We denote the closure of $A$ by $\overline{A}$. It follows easily that $A$ is the restriction of $\overline{A}$ to $\mathscr{D}(A)$.
A core of a closable operator is a subset $\mathscr{C}$ of $\mathscr{D}(A)$ such that the closure of the restriction of $A$ to $\mathscr{C}$ is $\overline{A}$.
The following properties are easily checked:
1. 1.
Any bounded linear operator defined on the whole space $B$ is closed;
2. 2.
If $A$ is closed then $A-\lambda I$ is closed;
3. 3.
If $A$ is closed and it has an inverse, then $A^{{-1}}$ is also closed;
4. 4.
An operator $A$ admits a closure if and only if for every pair of sequences $\{x_{n}\}$ and $\{y_{n}\}$ in $\mathscr{D}(A)$, both converging to $z\in B$, and such that both $\{Ax_{n}\}$ and $\{Ay_{n}\}$ converge, it holds $\lim_{n}Ax_{n}=\lim_{n}Ay_{n}$.
Type of Math Object:
Definition
Major Section:
Reference
## Mathematics Subject Classification
47A05 General (adjoints, conjugates, products, inverses, domains, ranges, etc.)
## Recent Activity
May 17
new image: sinx_approx.png by jeremyboden
new image: approximation_to_sinx by jeremyboden
new image: approximation_to_sinx by jeremyboden
new question: Solving the word problem for isomorphic groups by mairiwalker
new image: LineDiagrams.jpg by m759
new image: ProjPoints.jpg by m759
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new problem: Curve fitting using the Exchange Algorithm. by jeremyboden
new question: Undirected graphs and their Chromatic Number by Serchinnho
## Info
Owner: Koro
Added: 2003-07-28 - 16:29
Author(s): Koro
## Corrections
unclear by matte ✘
## Versions
(v9) by Koro 2013-03-22
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http://mathhelpforum.com/differential-equations/169184-1st-order-differential-equations-checking.html
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# Thread:
1. ## 1st Order Differential Equations: Checking
Is there an effective method of checking a solution to a differential equation once you have solved it (or thought you have)? Using a calculator or even MathCAD.
Both implicitly and explicitly and with the use of an arbitrary constant?
2. If you have a solution of the differential equation you can apply it to the equation and see if it's right.
3. I understand that, but how can you substitute (and consequently check) you solution to an equation such as:
$y+\frac{dy}{dx}tan(x)=7$
4. If you solve this equation for example you get $y=\frac{c}{sinx}+7$ where c i s a constant.
Now find that $y'=-c\frac{cosx}{(sinx)^2}$.
Your differential equation is $(tanx)y'+y=7$ and you know both y and y', apply them and you will see that it is true
5. Originally Posted by MaverickUK82
I understand that, but how can you substitute (and consequently check) you solution to an equation such as:
$y+\frac{dy}{dx}tan(x)=7$
That's a strange question. The answer- by doing exactly what it says! Differentiate whatever y you have (using implicit differentiation if necessary) and put y and its derivative into that equation. If the problem is that you cannot differentiate your y, what is your y?
6. I've had to remove previous post because I typed the wrong solution out. Sorry guys. I've wasted you time. Mods please remove this thread. I give up.
7. From the things we said before didn't you understand what to do?
8. I'm sorry for giving up. I'm getting really stressed out with the maths I'm trying to learn and and some aspects of this forum, but more on topic: I used the equation above as an example - the thing I wanted to check is this:
$\sqrt{t^2+9}\frac{dy}{dt}=y^2$
I thought i'd solved it with:
$y=\frac{1}{ln(t+\sqrt{t^2+9})+C}$
It's probably wrong - but I wondered whether there was an easy way of checking whether or not it was correct before trying to solve for C to find the value of the arbitrray constant for when y(0)=1
9. When integrating
$\frac{1}{\sqrt{t^2+9}}$
in math cad, it gives me $asinh(\frac{1}{3}t)$
But I can only see it to be:
$ln(x+\sqrt{x^2+3^2})+C$
Where am I going wrong?
10. The solution of mathcad is right.
It is in the form $\int\frac{1}{\sqrt{t^2+a^2}}dt$ so you set $t=|a|sinhz$ and you can find it.
How you worked to find your solution?
11. Hmmmm...
My course has given me a book of all the things I need to know when it comes to exam - it's like the bible - lol.
I have not came across sinh.
In the section showing the standard integrals, it gives me:
$\int\frac{1}{\sqrt{t^2+a^2}}dt=ln(x+\sqrt{x^2+a^2} )$
Could this be the same thing in a different format?
12. Ah ha!
I just found this on another website:
arcsinh x = ln [x + (x2 + 1)1/2]
All is not lost?
13. Originally Posted by MaverickUK82
I'm sorry for giving up. I'm getting really stressed out with the maths I'm trying to learn and and some aspects of this forum, but more on topic: I used the equation above as an example - the thing I wanted to check is this:
$\sqrt{t^2+9}\frac{dy}{dt}=y^2$
I thought i'd solved it with:
$y=\frac{1}{ln(t+\sqrt{t^2+9})+C}$
It's probably wrong - but I wondered whether there was an easy way of checking whether or not it was correct before trying to solve for C to find the value of the arbitrray constant for when y(0)=1
Originally Posted by MaverickUK82
When integrating
$\frac{1}{\sqrt{t^2+9}}$
in math cad, it gives me $asinh(\frac{1}{3}t)$
But I can only see it to be:
$ln(x+\sqrt{x^2+3^2})+C$
Where am I going wrong?
If only you had posted the real question (quoted above) in the first place, a lot of time (and frustration on your part) might have been avoided.
As you have discovered the hard way, both answers are correct. It is not uncommon for somethng like this to happen. You should research hyperbolic functions and inverse hyperbolic functions, just to fill out some useful mathematical background you haven't been formally taught.
14. Thank you Mr. Fantastic - I will research them this evening.
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http://mathhelpforum.com/advanced-algebra/86278-canonical-forms-linear-operators.html
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# Thread:
1. ## Canonical forms of linear operators
Im supposed to find the canonical form and corresponding change of coordinates for a few linear operators, heres two:
0 1 0
0 0 1
-1 -3 -3
2 1 1 -2
1 2 1 -2
1 1 2 3
3 3 3 5
I know im supposed to find the eigenvalues and vectors by finding detA = 0 for the characteristic equation then the roots, but I just am stuck when it comes to finding the equation from the matrixes.
Following from that im unsure of what to do after I plug back in lambda into the matrix...
2. You looking for the Jordan Canonical Form or the Rational Canonical Form?
Either way the first step is the same, find the characteristic polynomial. subtract x from the diagonal and calculate the determinant. It should factor into linear terms (if not, you cannot find the JCF unless you pass to the algebraic closure). The roots to this polynomial are the eigenvalues.
You will want to find the minimal polynomial, it must divide the characteristic polynomial (it will have the same roots, but possibly with less multiplicity). Now you either find the elementry divisors (JCF) or the invariant factors (RCF) respectively depending on which form you are looking for.
Once you find the eigenvalues you find the eigenvectors by solving $(A-I\lambda_i)v_i=0$.
3. Im not sure about either, but one of the lecture notes mentions the Jordan Block in one of the lectures about mapping to an n-dimensional space, so im guessing that.
I think my problem right now is more of getting the characteristic equation. I understand subtracting from the diagonal, but what is the best way going about calculating the determinant for a 4x4 matrix?
Once I can do this and get the roots he gives this tree:
Im still trying to understand the middle part and how to solve it to get the answer. Sorry for my ignorance, im literally working off of a few pages of his own lecture material since the guy doesnt teach from a book. Checking back in a few hours, thanks for the help so far!
4. ## JCF
Yeah, that is definitely the Jordan Canonical Form. As for calculating the determinant of a 4x4 matrix, pick your poison man. Subtract lambda from the diagonal and then take the determinant.
Method 1) You can row reduce to get it into an upper triangular matrix (if you do it right you can actually get it into the Smith Normal Form and that will tell you your JCF directly, but I think that is probably beyond the scope of your course) and multiply down the diagonal.
Method 2) The way I would suggest is to just do the cofactor expansion method though, it may take a while, but it is way less mistake prone. Just pick a particularly nice row or column to go down (one with as many 0s as possible) and get to it.
Neither way is particularly pleasant, but such is life. In the end you should be able to factor it completely. You may get some of the eigenvalues repeated with multiplicity. You will get n eigenvalues repeated with multiplicity, that is when you follow his tree.
Although I will be honest, I am not sure that last branch is true. It is certainly possible for you to have repeated eigenvalues and not necessarily get that Jordan block with a 1 in it, but maybe he has set the problems up ensuring that this is the case.
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http://math.stackexchange.com/questions/236292/how-can-i-evaluate-this-integral
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# how can I evaluate this integral?
how to evaluate this integral: $$l(y)=\int\limits_\beta^\infty \theta\exp(-y\theta)\alpha\exp(-\alpha\theta) \, d\theta$$ where $\alpha,\beta,\theta,y>0.$ Because I find it infinity! Can anyone help me to evaluate this integral? Thank you. I find this solution :$$\left(\left.\frac{-1}{(\alpha+y)^2}\exp(-(\alpha+y)\theta)\right)\right|_{\beta}^\infty-\left.\frac{\theta}{(\alpha+y)}\exp(-(\alpha+y)\theta)\right|_{\beta}^\infty.$$ in which in the second term, I obtain infinity value!
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Try substituting $t = exp(-\alpha \theta)$ . – Dilawar Nov 13 '12 at 8:14
4
Hint: Note that $\ell(y) = - \dfrac{\partial}{\partial y} \int_\beta^\infty \alpha \exp(-(y+\alpha) \theta)\ d\theta$ – Robert Israel Nov 13 '12 at 8:16
@Dilawar Or not. – Did Nov 13 '12 at 9:53
## 2 Answers
You're wrong, your 2nd term is not infinite ! Indeed exponential growth is much stronger than polynomial one, so $\lim_{x \to +\infty} x.e^{-x} = 0$.
Beware of the $\alpha$ factor inside the integral and signs, I found for my part: $$l(y) = \left( \frac\alpha{(y + \alpha)^2} - \frac{\alpha \beta}{y + \alpha} \right) e^{-(y + \alpha) \beta}$$
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The product $$\begin{align} x\exp(-x) &=x/\exp(x)\\ &=x/(1+x+x^2/2+x^3/6+\dots)\\ &=1/(1/x+1+x/2+x^2/6+\dots) \end{align}$$ $1/x\to0$ as $x\to\infty$ . Sum of other terms in the denominator $\to\infty$ as $x\to\infty$. So the product $x\exp(-x)\to0$ as $x\to\infty$.
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– robjohn♦ Jan 16 at 1:18
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http://mathoverflow.net/revisions/13045/list
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## Return to Answer
2 added 66 characters in body; added 3 characters in body
Formal can mean slightly different things in different contexts.
A commutative differential graded algebra (CDGA) is formal if it is quasi-isomorphic to it's homology. This is stronger than having all the higher Massey products equal to 0 (I think there are such examples in the Halperin-Stasheff paper).
To a space you can associate a CDGA (via Sullivan's $A_{pl}$ functor) which is basically the deRham complex when the space is a manifold. In nice cases this functor induces an equivalence from the rational homotopy category to the homotopy category of CDGA. Quasi-isormorphic CDGA correspond to (rationally) homotopy equivalent spaces. You can also tensor with the reals to get real CDGA.
If A is a CDGA which is quasi-isomorphic to $A_{pl}(X)$ for a space $X$ then A is often called a model of X. A space is formal if $A_{pl}$ of it is formal. So a formal space is modeled by its cohomology. In that sense its rational homotopy type is a formal consequence of its cohomology.
I think you have to be slightly careful with using $C^*$. This functor lands in differential graded algebra which are not commutative, so possibly the notion of formality could be different. In particular if you consider two CDGA there may be more maps strings of quasi-isomorphisms between them as DGAs then as CDGAs(. I don't know of any examples, but it's a point believe it is unknown if two CDGA that are quasi-isomorphic as DGA have to worry about)be quasi-isomorphic as CDGA.
1
Formal can mean slightly different things in different contexts.
A commutative differential graded algebra (CDGA) is formal if it is quasi-isomorphic to it's homology. This is stronger than having all the higher Massey products equal to 0 (I think there are such examples in the Halperin-Stasheff paper).
To a space you can associate a CDGA (via Sullivan's $A_{pl}$ functor) which is basically the deRham complex when the space is a manifold. In nice cases this functor induces an equivalence from the rational homotopy category to the homotopy category of CDGA. Quasi-isormorphic CDGA correspond to (rationally) homotopy equivalent spaces. You can also tensor with the reals to get real CDGA.
If A is a CDGA which is quasi-isomorphic to $A_{pl}(X)$ for a space $X$ then A is often called a model of X. A space is formal if $A_{pl}$ of it is formal. So a formal space is modeled by its cohomology. In that sense its rational homotopy type is a formal consequence of its cohomology.
I think you have to be slightly careful with using $C^*$. This functor lands in differential graded algebra which are not commutative, so possibly the notion of formality could be different. In particular if you consider two CDGA there may be more maps between them as DGAs then as CDGAs (I don't know of any examples, but it's a point to worry about).
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http://physics.stackexchange.com/questions/30134/intrinsic-structure-of-electron/30144
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# Intrinsic structure of electron
The electron contains finite negative charge.The same charges repel each other.What makes electron stable and why does it not burst? Is it a law of nature that the electron charge is the smallest possible charge that can exist independently? What is a charge after all? Is it like space and time or we can explain it in terms of some other physical quantities?
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Because the electron is a point, and it can't break. Why is this a mystery? You are imagining some classical structure to the electron that just isn't there. The role of physics is to answer the question "if you do such and such, what happens?" Charge is a primitive quantity in the description and does not resolve into something else more intuitive--- the charge is what builds up the intuitive things. – Ron Maimon Jun 16 '12 at 7:15
@RonMaimon I am not considering some classical structure to the electron and we know that electron behaves like a point and it can't break. My point is to explain the negative charge distribution inside the electron while considering it a material particle. However it would be equally interesting to think about the charge distribution when it behaves like a wave( or group of waves). – Kamran Jun 18 '12 at 9:47
@RonMaimon.. what do you think about Terry's answer? – Kamran Jun 18 '12 at 9:48
@Kamram: There is no "charge distribution" inside the QFT electron. The "wave" is not a physical wave, but an amplitude wave, and it doesn't push itself out, it is a charged fluid. I don't like the answers to the question because I don't think it's a sensible question. – Ron Maimon Jun 18 '12 at 20:08
## 4 Answers
That's a great question!
Unfortunately, the only honest answer is "that's what we see in nature, with great precision and complete reproducibility." There is no deep theoretical understanding.
The more exotic form of your question is phrased in terms the self-energy of an electron, and it's a question that plagued Nobel Laureate Richard Feynman his entire life. He first tried to imagine that the field that emanates from an electron simply wasn't seen by the electron itself, in somewhat the same fashion that a woman standing on a high plateau cannot "see" her own height, since from where she stands everything around her is flat.
It didn't work. It did accidentally lead him to some methods for dealing with the problem that won him that Nobel Prize, but his own conclusions later in life was that he and everyone else had pretty much flat-out failed, and that the self-energy of an electron was still a mystery. It continues to be.
So why is it this problem so hard?
You stated it pretty well in your question: If like charges repel, then why doesn't an electron simply blow itself apart? After all, if you take a hundred small negatively charged object and try to push them together, the energy need just keeps increasing as you push them closer together. For an infinitely small point object like an electron, that energy goes to infinity! So, point particles and charges just do not play well together... yet half the charges in our world are composed of just such particles! And even the protons have their own version of the problem because of the three point-like quarks that are bounding around within them.
Here's a somewhat different way to visualize the self-energy problem. I like it because when I was a kid I liked to put my thumb over the open end of a water hose to see how far and fast I could the water squirt out.
The analogy works like this: It turns out that you can model an electric field remarkably accurately by simply picturing a positive charge as the end of a hose that is spewing out a fixed number of liters of water per second into a pool of water (space). A negative charge then becomes the end of another type of hose that sucks water in from the pool at the same rate. James Clerk Maxwell, one of the most flat-out brilliant physicists in all of human history, was one of the first people to notice this analogy. That is why "lines of force" are also sometimes called "flux" (meaning "flow") lines. Maxwell made good use this analogy to help him derive his famous Maxwell's Equations, by which he was able to unify magnetism and static electricity into a single unified theory of forces. It was Maxwell who first realized what light really is, by applying his own theory.
Back to the self-energy problem: Have you ever put your thumb over a pipe end that wants to spew out water at a fixed rate no matter how small the opening is? What happens is that the water speeds up and becomes far more forceful. A hose that gently drops water out when is opening that is several centimeters across becomes a tiny but amazingly intense fire hose when most of that opening is blocked. The slow flow becomes a micro-torrent whose speed is so high it will cut into soft objects.
Picture the size of that hose end as the size of an electron. If it's a big opening, no problem. You get the full flow without ever reaching extreme velocities.
But what if you start making the end of the hose smaller and smaller? It still has to produce the same number of liters per second, so just like when you put your thumb over the end of a big opening and try to block it up, the flow of water speeds up. The smaller the opening, the more extreme the speed up, e.g. cutting the outlet size in half causes the speed of the water to double, just so it can keep up. The same is true for the electron, just with the "strength" of the field replacing the "speed" of the water, and with "field lines" replacing the path of flow of the water.
So, if you shrink the size of the "exit hole" (whether the end of a hose or the size of an electron) down to a point, what happens?
Easy: The speed goes to infinity... which of course can't happen! Infinite speeds aren't even possible for water, which is limited by the speed of light. They would require unlimited energy just to approach the speed of light.
The situation is no better for point-like charged particles, which similarly must acquire field densities (think speed) that approach infinity. That in a nutshell is another way to understand a bit more visually why the self-energy is so tough.
So, with all that said, is is possible that someday there will be a theory that truly explains such things -- deeply explains them, in that special way the gives people reading it that cool little "wow, now I finally get it!" feeling as everything finally clicks together?
Well, not right now. String theory doesn't really bother with such problems, and in fact arguably makes such issues worse by picturing everything as tiny strings whose self-energy is pretty astronomical. Harari-Shupe Rishon theory is more of a nice organizational observation than a theory (think of the Periodic Table), but all it does is make all the point charges one-third that of an electron. I don't know if Garrett Lisi has ever tried to address the self-energy issue in that remarkable E8 theory of his, but it would at least provide a new angle on the problem that might shed some fresh light on it.
So, again: Great question, but alas, there's no one who can answer it yet! But who knows, someone reading this answer may be the person to solve that one someday -- why not? Yes, it's a very hard problem... but then most folks have no idea what they are truly capable of if they have a knack for physics, are truly interested, and are willing to work hard at it.
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The electron is stable (from a QFT POV) because there is no particle with less energy (in the Standard Model) that it can decay into while also preserving all known conservation laws.
As for the "burst" part that question comes from a classical view of the electron which does not hold since we know from QM that electrons also behave like dual wave-particle entities. You could convert electron's mass in energy and calculate what would be the radius of a charged sphere (or spherical surface) with total charge $-|e|$ so that they would have the same energy but that would only give a classical radius. Near and inside that radius you would need QM or QFT to describe the electron. And from a QFT point of view we have not seen the electron behave like anything other than a point particle (when we observe it as a particle) no matter how close we "probe" it.
Regarding what charge is I d describe it as an observable, measurable and scalar quantity. (It is also conserved because of the U(1) symmetry of E/M.) Maybe not the best answer but that is how I approach that issue.
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It's a very good question. The electron is described by a wave field which resembles a charge distribution, so it is natural to wonder why it doesn't repel itself and spread out all over.
However, the wave is not a classical wave but is quantized, i.e. the energy in a given vibration mode has to come in discrete bundles. One can count how many excitations are present and the number is an integer N > 0, the number of "particles" present. Also, the different vibration modes are identified with momentum.
A single-electron state is a "superposition" of N=1 excitations. The main reason to superpose is that any given mode extends over all space, but by superposing modes of different momenta one forms a "wavepacket" which is peaked in the center and then falls off to zero at greater distances, because the modes of different frequencies interfere destructively. The wavepacket concentrates the electron's charge into a confined area, and it seems like it should repel itself and grow rapidly wider.
However, this doesn't happen because of quantization and momentum conservation. The electromagnetic field works by carrying energy from one mode to another. One can think of it in terms of resonance: the electron field vibrations excite a (non-resonant) mode of the electromagnetic field, which does not travel far due to lack of resonance, but can still transfer energy permanently to another field which is resonant.
If there are multiple excitations (N >= 2) then the electron's vibration frequency is the sum of the individual mode frequencies, and there are many different ways to sum two frequencies to get the same thing. In other words, a two (or multi) particle state has many non-trivial resonances with itself and it can exchange momentum amongst its modes, allowing the packet to split apart and separate.
However, for an N=1 state, the only resonance of a given mode is with itself. The electromagnetic field (or any other mechanism) cannot transfer any energy between the different modes, hence they all stay the same and the single-particle state is stable.
This doesn't mean that the single-particle state is unaffected! It still excites the electromagnetic field and puts some energy into it, hence the energy of the vibrating electron field is larger than it naively appears because it is carting around the associated electromagnetic vibrations everywhere. The extra energy depends mostly on the electron's amplitude (not frequency) and hence looks like a change to the electron's mass, (because mass in quantum field theory is a measure of how the energy depends on amplitude). Therefore the electromagnetic field renormalizes the electron's mass, but isn't able to split the electron into pieces.
Of course quantization, which is the crucial piece here, remains a black box in this explanation....but hopefully it's better than nothing!
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Can you give a link/reference to this model you are handwaving here? – anna v Dec 27 '12 at 5:07
Not really. It's just an attempt to put into words the difference, within quantum field theory, between a single-electron state, which is not affected by its own electric field (aside from mass renormalization), and a multi-electron state, which is affected, because the different electrons repel each other and separate. – William Nelson Dec 29 '12 at 17:06
It does not burst because of the force of attraction by the nucleus of the atom which is positively charged. Yes it can exist independently. A charge carried by an electron is 1.60217646 × 10^-19 coulombs. It is a negative charge as we all know. We can surely explain it in terms of physical quantities as I have told you the charge carried by the electron.
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This seems to have misunderstood the question. – dmckee♦ Jun 15 '12 at 14:11
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http://physics.stackexchange.com/questions/5217/what-are-dimensions/5221
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# what are dimensions?
First, discrete examples.
In a computer screen I can specify any "2D" point with just one single number (pixel order starting count from first at upper left, and going on, left2right and up2down like reading till last one at right bottom corner)
Then I don't need two numbers to specify position, just one.
Same can extend to other "dimensions" I can use same trick to describe a point inside a discrete cube, once have covered a screen like "area" descend a plane level
plane
````0 .....5
6 .....10
11.....15
````
then a cube (made of above planes)
````0 ____ 15
16 ____ 31
32 ____ 47
````
so a single number can map any point in a 3d discrete cube
same thing using differential dx instead of discrete points or planes, and that's all
(of course as differential is not well defined as discrete this won't be so easy)
Anyway my question is:
what is a good definition of dimention that avoid these tricks? there is one?
Finally another way to take 3d to 1D, is a reversible transformation like this
3d point = (x,y,z)
W = x + y * 2^20000 + z * 2^40000
the only "restriction" for this method is that equivalent 3d coords should be lesser than 2^20000, a very easy thing, at least in the known universe even in Plank's units
(20000 and 40000 could of course be changed for lesser numbers, but I like concrete examples)
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## 5 Answers
It's true that if you just have a set of points, with no additional mathematical structure, the notion of "dimension" is problematic as you say. But the spaces we deal with in physics usually have extra structures that make the notion well-defined. Often, the definition works by making precise a notion of different "directions" at a given point, and then finding a way of counting how many of those directions are independent of each other.
For instance, we often work with vector spaces (in which it makes sense to talk about adding vectors, etc.). The dimension of a vector space is well-defined: it's the maximum number of independent vectors you can find (such that no linear combination of them add to zero).
Also, we often talk about geometrical spaces such as differentiable manifolds, in which there's a notion of "smoothness" of some sort. Once again, these manifolds have a well-defined notion of dimension, and once again it's essentially the number of independent directions you can identify at any given point. In fact, for a smooth manifold one way to define the dimension is to note that you can smoothly map a subsection of the manifold onto a vector space, and then figure out the dimension of the vector space as before.
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Let me try to answer differently, in some sense more mathematically, and more directly to the point. Intuitively, the dimensionality is the number of independent numbers you need to identify a point. However, the redefinitions of the coordinates need to be continuous which prohibits your "labeling of the pixels"
By the way, one can easily define one-to-one maps from $R^d$ to $R$. For example, the point $$(0.147346,0.295002,0.139523)$$ may be identified with a single number $$0.121493759305402623.$$ I have simply taken the digits from all the three coordinates - from 1st; 2nd; 3rd; 1st; 2nd; 3rd, and so on. However, this map from $R^3$ to $R$ is not continuous so one is not allowed to do it when he computes the dimension. In the example above, only the 18 digits behind the decimal point(s) were rearranged but the same thing can clearly be done with the fully precise real numbers, too. In "set theory", an abstract branch of mathematics, they would interpret this construction as a proof that the sets $R$ and $R^3$ have the same cardinality (a generalized number of elements): the set theorists call the number of elements in either set simply "continuum".
In mathematics, the dimensionality - the number of dimensions - is defined by concepts such as Hausdorff dimension:
http://en.wikipedia.org/wiki/Hausdorff_dimension
You try to cover your manifold by a minimum number $N$ of balls of small radius $r$; a notion of a distance has to be available. A ball is a set of point with the distance from the center smaller or equal to $r$. How many balls you need? Well, the volume scales like $r^d$, so you will need $V/r^d$ balls or so. If you take the logarithm of the number of balls, it will be $\ln(V)-d\ln(r)$, and by sending $r\to 0$ i.e. $\ln(r)\to-\infty$, you may extract the coefficient $d$ by a limiting procedure.
Amusingly, this definition also works for fractals that can have fractional dimensions. The simplest ones have dimensionalities that are ratios of logarithms of simple numbers (e.g. integers).
In physics, we use the space to define theories and we differentiate the coordinates and fields with respect to time (and space, in the case of field theory). That's why the condition of "continuity" or "smoothness" is automatically required in physics. In particular, local quantum field theories must live in a spacetime with a well-defined number of spacetime dimensions. Relabeling the coordinates in your "pathological way" doesn't preserve the "differentiable structure" on the spacetime manifold, and because physics depends on derivatives (differentiation), such redefinitions would make physical laws meaningless. Also, quantum field theories are defined on a spacetime that has a well-defined metric tensor; so I can use the Hausdorff definition to count the dimension, too. (One should avoid the "balls" for the Minkowski case which has an indefinite metric.)
All these matters become ill-defined in quantum gravity, however.
In the Hausdorff dimension definition, I needed to consider arbitrarily small balls. But there are no objects smaller than the Planck length - the characteristic (tiny) distance scale of quantum gravity. Consequently, the limiting procedure cannot be performed in quantum gravity. It follows that the number of dimensions of spacetime isn't quite well-defined. Only the dimensions that are "large" - where the balls may still be much smaller than the size of these dimensions - have a physical meaning. However, the number of dimensions that are as small as the Planck length or so (or as curved as the Planck length curvature radius) can't be defined.
And indeed, there are often equivalent descriptions of a theory that actually disagree about the number of dimensions. For example, M-theory on a K3 manifold - which has 11 dimensions in total - is equivalent to heterotic strings on a 3-torus - which only have 10 spacetime dimensions. Also, the AdS/CFT correspondence shows that theories in spacetimes of different dimensions (by one) are equivalent: the radial, "holographic" dimension is invisible in the Lagrangian of CFT which is linked to its significant curvature.
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I don't agree physics nor reality need to preserve the "differentiable structure" mainly because I didn't say that those "pathological" (lol) coordinates would be used in a spacetime description, I've asked to get a deeper understanding of dimension concept, anyway it's a great answer, I liked your $\mathbb{R}^3$ to $\mathbb{R}$ transformation and I used to read your answers carefully, thanks – HDE Feb 16 '11 at 13:27
I'll just add to the others' answers by saying that I'm not sure the passage from discrete to continuous is so easy: even a finite continuous square could not be "numbered" (i.e.: a function (x,y) -> z defines such that each z is unique for a point) in the way you mentioned.
The "trick" in discrete space is that you can use the modulo to split the number in two distinct scalar quantities, while the same is not true for real (continuous) numbers.
Edit: Thanks to Ted Bunn, who has illuminated me to the existence of space-filling curves. The article talks about the existence of a bijective relation between a closed surface and a segment, which is all that is needed to extend the ideas in the question.
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HDE,
Ted Bunn has outlined the mathematical answer, but you have come at this question as a programmer with a programming example. Well in that case consider the programming construct of an unbounded 2 dimensional Array. Here there are two independent values for each pixel, and there is no limit to how large each array dimension value can become. In your model there is always a limit as it "wraps around". With such an Array (admittedly a slight idealisation for most real languages) we have two independent values for each pixel: hence two dimensions.
The reason dimensions are used is partly that there is no assumed limit for each direction/dimension in Physics. If we knew that the Universe had exactly N particles and M states (as some occasionally postulate), then the mathematics used to model it might be able to borrow from the simpler 1 dimensional computer models.
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I guess that's part of the story, but I really don't think it's the main thing. Suppose that we knew that the Universe was discretized in length and had finite extent, so that we knew the number of "pixels." I think it'd still be most natural to think of it as 3-dimensional, don't you? – Ted Bunn Feb 15 '11 at 19:06
1
@Ted Bunn, Yes I would be tempted to solve many physics problems using the best mathematical tools available which would include the 3 (N) dimensional representations we now have. However in this finite case we now have further information including the bounds on parameters (in principle at least). In turn this might mean that assuming independence (of parameters) was a local approximation, not valid when we consider global solutions.... – Roy Simpson Feb 15 '11 at 19:32
You can interpret the number of dimensions as the minimum amount of information required to uniquely specify an object in the "space" you are interested in. What you are doing, essentially, is providing an enumeration system with a specified counting scheme ("turning about the edges"). However when you subtract one piece of information, you are introducing another.
I know this is qualitative, and I do not know how to make it rigorous, but here is an example of what I mean to say:
Let us say I wish to describe the position of points lying on the circumference of a circle (2D plane). Then you can either
1. construct one of the coordinate systems. A point in such a system is described by a pair of numbers that is unique for every point.
Or, another way would be to 2. fix a point on the circumference and denote it your origin, and then say that every other point is labeled by just one _number_ which is the distance the point would have moved along that circle.
But in #2 you have introduced a constraint (additional information). The reduction of the number of degrees of freedom by the introduction of holonomic constraints in classical mechanics might serve as an analogy.
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Use the "edit" option on your post and you should get links and clues to Latex use on the RHS of your screen. – Roy Simpson Feb 15 '11 at 19:02
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You can use LaTeX (actually MathJaX) by surrounding it with dollar signs, e.g. `$\mathbb{R}^2$` gives $\mathbb{R}^2$. For a formula that constitutes a paragraph of its own (equivalent to the `equation*` environment), use double dollar signs instead, `$$a+b=c$$`. – David Zaslavsky♦ Feb 15 '11 at 19:03
Thank you, done. – Approximist Feb 15 '11 at 19:20
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http://math.stackexchange.com/questions/182747/prove-that-for-every-positive-x-in-mathbbq-there-exists-positive-y-in/182791
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# Prove that - for every positive $x \in \mathbb{Q}$, there exists positive $y \in \mathbb{Q}$ for which $y \lt x$
First my apologies if this question has been asked before.
Exposition
I'm new at learning how to prove theorems and among the given exercises from my reference material it is asked to prove the following:
The original question in words:
For every positive $x \in \mathbb{Q}$, there exists positive $y \in \mathbb{Q}$ for which $y \lt x$.
I translated it and got:
$\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$
Here is my attempted proof.
If $x \in \mathbb{Q}_{\gt0}$ then $\exists y \in \mathbb{Q}_{\gt0}$ such that $y \lt x$. Suppose $\forall y \in \mathbb{Q}_{\gt0}$, $y \geq x$.
So if $y \in \mathbb{Q}_{\gt0}$ then $y \geq x$. By contrapositive if $y \lt x$ then $y \notin \mathbb{Q}_{\gt0}$.
But this doesn't make sense. Hence we were wrong to assume that $\forall y \in \mathbb{Q}_{\gt0}$.
Question
I'm having trouble with the part starting from this doesn't make sense. I looked at the $y \notin \mathbb{Q}_{\gt0}%$ and made a somewhat educated guess regarding the fact that $y \notin \mathbb{Q}_{\gt0}%$ doesn't logically follow from the premise that $y \lt x$. This in the sense that the less than 'operator' can only be defined between two mathematical objects of the same kind.
Is there something i got wrong? Does this make sense? Is the proof complete anyway? What would be the correct proof?
In clear and concise terms, I'm trying to understand if my proof is correct.
Thanks
UPDATE
I re-read the question again from the material and $y$ is supposed to be a positive rational too. Yet i think given replies at the original time of this update still apply.
UPDATE 2
With regards to the answer provided by @crf i think i should provide the proof strategy. By this point if someone could see something wrong in the proof, i guess it came from me making something wrong in my strategy. So here is the proof strategy. All that follows of course is supposed to be part of draft work.
1. First i get the statement into symbolic form in order to 'safely' transform the expression:
$\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$
2. Transform the obtained statement into conditional form:
We know that $\forall x \in S, \ Q(x)$ is equivalent to $(x \in S) \Rightarrow Q(x)$.
Then we get:
$x \in \mathbb{Q}_{\gt0} \Rightarrow \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$
Reference: Book of proof by Richard Hammack, pp 54, Fact 2.2 available online here
3. Then we attempt a proof by contradiction for this conditional statement:
As such our hypotheses become:
$$x \in \mathbb{Q}_{\gt0} \\ \forall y \in \mathbb{Q}_{\gt0} \ (y \geq x)$$
And this is equivalent to :
$$x \in \mathbb{Q}_{\gt0} \\ y \in \mathbb{Q}_{\gt0} \Rightarrow y \geq x$$
and conclusion: will be a contradiction
4. Transform $\forall y \in \mathbb{Q}_{\gt0} \ (y \geq x)$ to get its contrapositive:
We get as new hypotheses:
$$x \in \mathbb{Q}_{\gt0} \\ y \lt x \Rightarrow y \notin \mathbb{Q}_{\gt0}$$
That where i got stuck and i started guessing: how does $y \notin \mathbb{Q}_{\gt0}$ follow from $y \lt x$? I couldn't see a rigorous contradiction between that and premises and got got stuck!
Thanks for bearing all this!
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What's wrong with writing $x = p/q$ and taking $y = p/(q+1)$, say? – t.b. Aug 15 '12 at 9:22
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@t.b.: Or simpler still: Take $y=x/2$. – Harald Hanche-Olsen Aug 15 '12 at 9:23
– t.b. Aug 15 '12 at 9:25
I understand there are various ways, and even simpler ways, of doing this but like i mentioned in the question i'm trying to understand if my proof is correct, not alternatives. While both the suggestions of @t.b. and Harald Hanche-Olsen are correct i'm trying to emphasize more on logical arguments directly from premises than computational 'intuition'. Thx – nt.bas Aug 15 '12 at 9:31
Hint: "but this doesn't make sense" is not a mathematical statement, and doesn't have a place in a proof. It is an invitation to duck the real issue at stake - mathematics is quite precise about such things and what is required is a mathematical statement of WHY it "doesn't make sense" in the mathematical context of the question. – Mark Bennet Aug 15 '12 at 9:43
show 4 more comments
## 6 Answers
What you've done so far actually has nothing to do with proving your theorem. If you do want such a symbolic lead-in, OK-it's not false, though it's worth noting that such intricate manipulations are ugly compared to a direct and brief argument. In any case, to finish, you must stop playing with symbols and say something mathematical, such as "But wait! $\frac{x}{2}\in Q_{>0}$ is less than $x,$ so by contradiction, my assumption $y<x \rightarrow y\notin Q_{>0}$ is false!" That last sentence is the entire mathematical content of the proof.
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My first observation is that you’re getting badly bogged down in symbols. For starters, there is absolutely no reason to replace the clear statement of the problem with the symbolic expression $\forall x \in \mathbb{Q}_{\gt0} \ \exists y \in \mathbb{Q}_{\gt0}, \ y \lt x$; that’s just introducing unnecessary obstacles for the reader. The same goes for your argument. Both it and its shortcomings would be much more easily read if you wrote it out in words, like this:
Suppose that that every positive rational $y$ is greater than or equal to $x$. Then if $y>0$ is rational, $y\ge x$. By taking the contrapositive it follows that if $y<x$, then $y$ is not a positive rational.
Without the fancy symbols to get in the way there’s a question that should almost leap out at you: what is $x$? Nowhere have you given any indication. And since you haven’t, what can it possibly mean to suppose that $y\ge x$ for all positive rationals $y$?
Back up now and think again about the actual statement: for each positive $x\in\Bbb Q$ there is a $y\in\Bbb Q$ such that $y<x$. Look at a few examples. If $x=7$, I can take $y=6$, for instance. If $x=6$, I can take $y=5$. If $x=3/2$, I can take $y=1/2$. In fact, no matter what positive rational $x$ may be, $x-1$ is a rational number less than $x$. I’m done: I’ve proved the statement by providing a recipe for finding a suitable $y$ given $x$.
And even then I’m working too hard. Is there any rational number that is less than all positive rational numbers? Sure: $0$, or for that matter any negative rational number. Now I’ve proved an even stronger statement: there is a $y\in\Bbb Q$ such that $y<x$ for each positive $x\in\Bbb Q$. If you insist on looking at quantifiers, this is $$\exists y\in\Bbb Q\forall x\in\Bbb Q_{>0}(y<x)\;.$$
Here’s an exercise for you to try: prove that for each positive $x\in\Bbb Q$ there is a positive $y\in\Bbb Q$ such that $y<x$. HINT: An idea something like my first argument works.
A mathematical proof is a piece of expository prose. Its purpose is to convince the reader that the theorem is true. Obviously it should be mathematically correct and logically sound, but it should also be clear and easy to follow. By all means use symbols when they’re appropriate: the quadratic formula is much easier to follow when expressed symbolically than when written out in words! But don’t fall into the trap of thinking that the more symbolism you use, the more professional your argument looks.
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Thanks.So if i understand your answer well, i should retain: 1/let your reader know what everything stands for in the proof and 2/avoid an unnecessary use of symbols. But i still don't understand why should $x$ stand for anything and this is because: 1/we know $x$ is any positive rational and 2/if we say (assuming this is correct and i understood you well) $y=x-1$, we know that that this proves that there is always something less than $x$ without giving $x$ any value. So what did you mean by what is x? Thanks – nt.bas Aug 15 '12 at 10:21
@nt.bas: Your argument began with (a symbolic version of) the statement Suppose that every positive rational number is greater than or equal to x. This is meaningful only if $x$ has already been defined in some way; otherwise $x$ might be Mars, for all the reader knows. If you really want to try a proof by contradiction, start by assuming that the result is false: Suppose that there is a rational number x such that every positive rational number is greater than or equal to x. Or, with a modest use of symbols, ... – Brian M. Scott Aug 15 '12 at 10:30
@nt.bas You've not specified in your proof what $x$ is. You should have had some kind of statement like, "let $x$ be a positive rational number" before even talking about whether $y\geq x$. – crf Aug 15 '12 at 10:31
... suppose that there is an $x\in\Bbb Q_{>0}$ such that $x\le y$ for every $y\in\Bbb Q$. That would be perfectly legitimate: now when you say that every positive rational is greater than or equal to $x$, you’re talking about a specific $x$, albeit one whose numerical identity isn’t known. It would also be very clear that this $x$ would be the smallest rational number, which would immediately point the way to the next step: get a contradiction by showing that there is no smallest rational number. – Brian M. Scott Aug 15 '12 at 10:31
I think my problem as rightfully mentioned by Brian M. Scott is i used to much symbols. The thing i gave was a raw translation of my draft work but if you look closely i did mention $x$ belonged to the rational positive integers, without the assuming part. Now suppose i had mentioned (even when i possibly didn't) and i didn't say this doesn't make sense, would the proof be correct?... – nt.bas Aug 15 '12 at 10:42
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So you want to prove that there always exists a rational $y$ in the interval $[0,x]$. We can do one better and prove the existence of rationals in any open interval $(a,b)$, for any real $a$ and $b$. This uses the Archimedean property of real numbers - that $\forall x,y \in R (x>0); \exists n \in N| nx > y$. So, suppose such a rational, $r$ exists. We have that: $$a<r<b$$ Let $p=[a]$ (the integral part of $a$). Since R is Archimedean, there exist natural $n$ and $m$ such that $n(b-a)>1$ and $m\left( \frac{1}{n} \right)>a-p$. Let us fix the $least$ such $m$, so that we have: $$\frac{m}{n}>a-p \quad but \quad \frac{m-1}{n} \le a-p$$ We therefore arrive at $$a<p+\frac{m}{n}\le a+\frac{1}{n} \le a+(b-a)=b$$
Therefore $p+\frac{m}{n}$ is our desired rational.
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Okay I see generally what your strategy is. You wanted to assume the negation of your symbolic translation $∀x∈ℚ_{>0} ∃y∈ℚ_{>0}, y<x$, and then derive a contradiction. One smaller issue—your translation into symbols is incorrect. The original statement says nothing about $y$ being positive. So your translation should have read $∀x∈ℚ_{>0} ∃y∈ℚ, y<x$. But there's a larger issue: you didn't quite get the negation right. Your assumption,
$∀y∈ℚ_{>0}, y≥x$
is not the negation of your sentence. The correct negation is
$\exists x\in \mathbb{Q}∀y∈ℚ_{>0}, y≥x$
Your hypothesis for the contradiction argument doesn't specify anything about $x$—do you mean for all $x$? For 2 values of $x$? Is $x$ a number? The existence quantifier is crucial—without it you don't really have an argument at all. But as Brian points out, my feeling is that all the symbolic manipulation obscures the logic and intuition, so I'm going to translate your argument back into English where it belongs, with to correct first step.
So what you really wanted to do was assume that there exists a positive rational number, which we'll call $x$, with the property that for every rational number $y$, $y\geq x$. This is the correct negation of the original statement, and now that you have a properly quantified sentence, I'm sure there are a number of explicit contradictions that you can derive. One nice thing is that, since every single $y\in\mathbb{Q}$ has to have this property under our hypothesis, all you have to do to find a contradiction is find just one counterexample. Can you find a rational number which is strictly less than every positive rational number?
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I have added the proof strategy i used because i believe it is the root of my troubles. And if you look at the first update, $y$ is to be positive. Thanks!!! – nt.bas Aug 15 '12 at 11:37
$x/2$ is rational and less than $x$, and it is positive if $x$ is positive.
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Maybe I am a bit sloppy, and forgive me if I am, but couldn't you just say the following:
I have two fractions, namely fraction $x$ and fraction $y$.
Suppose $x=\frac{1}{d}$.
Note that any fraction can be rewritten into something of this form. Any fraction of the form $\frac{a}{b}$ can be rewritten into $\frac{1}{b/a}$ such that $d=\frac{b}{a}$.
By picking $y=\frac{1}{d+1}$ I will always be able to have a fraction $y$ that is always smaller than $x$. Hence the statement is correct.
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Yes, you are right. but a young fella like me, new to all this beautiful theorem proving stuff, doesn't understand the importance of hanging on the fact than getting bogged down to unnecessary symbolic manipulations. If you look at the answers provided and comments, others have mentioned this notably Harald Hanche-Olsen. Otherwise thanks for helping me understand this even better! And of course, welcome to MSE!! :-) – nt.bas Aug 15 '12 at 15:19
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http://mathhelpforum.com/advanced-algebra/140698-linear-subspace.html
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# Thread:
1. ## Linear subspace
Hi everybody,
I don't know how to show that the following sets aren't linear subspaces:
$F_1$={ $(x;y;z)/x+y+z=1$}
$F_2$={ $(x;y;z)/xy=0$}
$F_3$={ $(x;y;z)/x^2=z^2$}
...
2. Originally Posted by lehder
Hi everybody,
I don't know how to show that the following sets aren't linear subspaces:
$F_1$={ $(x;y;z)/x+y+z=1$}
$F_2$={ $(x;y;z)/xy=0$}
$F_3$={ $(x;y;z)/x^2=z^2$}
...
To show that $U$ is a subspace of a vector space $V$ where $U \subset V$ you have two things to show,
that it is closed under scalar multiplication (for $u \in U$ you need to show that $\alpha u \in U$),
that it is closed under addition (for $u, v \in U$ you need to show that $u+v \in U$).
You need to show these two things as a subspace is a vector space contained in a bigger space, and this bigger space lends the smaller space the other properties. So all you need to prove is these two things.
So, for $F_1$ you need to show that $(x_1+x_2)+(y_1+y_2)+(z_1+z_2) = 1$ and that $\alpha x_1 + \alpha y_1 + \alpha z_1=1$ for $x_1+y_1+z_1=1$ and $x_2+y_2+z_2=1$.
Do you understand why this is what you need to show?
Now, does this hold?
The other two questions can be attacked similarly.
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http://mathhelpforum.com/advanced-algebra/211394-infinite-dimensional-cross-product.html
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3Thanks
• 1 Post By Deveno
• 1 Post By Deveno
• 1 Post By Deveno
# Thread:
1. ## Infinite Dimensional Cross Product
It has been proved that cross product of vectors just is satisfied in 3D space and also in 7D and 8D spaces as specific cases. Besides, there are some other multiplication schemes of vectors such as wedge product, Ersatz cross product ... that are not quite the cross product (they do not satisfy the required conditions as in 3D space). On the other hand, in Dirac bra-ket notation of quantum mechanical vectors, the operators are introduced as outer (cross) product of a pair of vector. But it is not clear that this definition of cross product is the same and familiar cross product of vectors with the known required conditions or it is a specific case such as the others. And in principle, is there exist a definition of "cross product" of two vectors in infinite dimensional space that satisfies the conditions as in 3D space? I will gratefully appreciate if anyone could guide me.
2. ## Re: Infinite Dimensional Cross Product
If you are looking at R^n, you would need the product of all vector lengths to be finite, or more correctly, the sum of all the logarithms of the lengths to also be finite.
For this you would need a stronger condition of little l squared (l^2) since you would need a finite amount of vectors to have length greater than 1 and an infinite amount to have length of 1 or less.
3. ## Re: Infinite Dimensional Cross Product
We know that inner product of vectors is satisfactorily defined in infinite dimensional Hilbert space in integral form. But what about "cross product". Is there exist a definition of "cross product" of two vectors in infinite dimensional space, particularly in integral form, that satisfies the conditions as in 3D space?
4. ## Re: Infinite Dimensional Cross Product
in general, the cross product is a product of n-1 vectors in n-dimensional space. this is in stark contrast to the inner product which is a product of 2 vectors in n-dimensional space that yields a scalar (increasing the dimensionality doesn't change the basic configuration). for this reason it is often maintained that the cross product can only be defined for n = 3 (at least as a binary operation).
it is possible to define an exterior product of vectors:
$\mathbf{u} \wedge \mathbf{v}$
which is a map from $V \times V \to \Lambda^2(V)$.
it turns out that $\dim(\Lambda^2(V)) = \binom{\dim(V)}{2}$
if dim(V) = n, this is (n-1)n/2. if n-1 = 2, the 2's cancel, and we get a vector space of the same dimension as V, that is: 3.
that is we can regard $\mathbf{u} \wedge \mathbf{v}$ as an element of F3, using the basis correspondence:
$\mathbf{e}_1 \leftrightarrow \mathbf{e}_2 \wedge \mathbf{e}_3$
$\mathbf{e}_2 \leftrightarrow \mathbf{e}_3 \wedge \mathbf{e}_1$
$\mathbf{e}_3 \leftrightarrow \mathbf{e}_1 \wedge \mathbf{e}_2$
and with this identification of $\Lambda^2(F^3) \cong F^3$ we get the usual cross-product.
for an infinite-dimensional space, it's not exactly clear how to generalize this, as the definition of the cross-product relies on unique arithmetical properties of 3.
if one relaxes the conditions for a cross-product somewhat, one can obtain a quasi-cross product by considering the non-real part of octonians (hence the 7-dimensional version), but the dimensionality of division rings over R is severely limited. one can also obtain a cross product of 3 vectors in 8 dimensions, which is a special case (and involves more than 2 factors).
in short, the possibilities for cross-products are rather limited.
5. ## Re: Infinite Dimensional Cross Product
Thanks for your excellent explanation. In quantum mechanical discussion of Hilbert space(J. J. Sakurai, Modern Quantum Mechanics, 2ed., p-15) and also some other mathemtical references, the operators, that for ket and bra of:
is defined as:
is mentioned as "outer product". It is not clear for me the relation of this definition with other familiar outer or cross product definitions in n-dimensional spaces.
6. ## Re: Infinite Dimensional Cross Product
regard u,v as nx1 real column vectors. then
<u,v> = uTv (this is a matrix product of size (1xn)x(nx1) = 1x1, a scalar).
we can also form the matrix product:
vTu which is an nxn matrix. this is the "outer product".
if we regard u,v as complex nx1 matrices, then we have to define the inner product as:
<u,v> = uHv, where uH is the conjugate-transpose. we then also have the outer product of uvH.
now in an arbitrary complex inner-product space (where we no longer have the ability to use matrices as representations of vectors and linear maps for the infinite-dimensional case, in general), we want some way to generalize this.
matrices naturally generalize to linear transformations, and linearity can be checked directly on the values, without producing a basis:
L(au+bv) = a(L(u)) + b(L(v)) for all u,v in V, is the linearity condition.
and nx1 column matrices can just be vectors, one does not need a basis to verify the axioms of a vector space.
the question arises, what corresponds to the "transpose" or "conjugate-transpose" of a vector now?
note that considered as a FUNCTION, a 1xn row-vector spits out a number when multiplied by an nx1 matrix.
since matrix multiplication is linear, we can regard the transpose (or conjugate-transpose in the complex case) of a vector as a function V--->F.
namely uT(v) = uTv (or uHv). in this case, we call uT (or uH) the DUAL vector to u, and denote it like so: u*.
we then (in the finite-dimensional case) get a vector space isomorphic to V, V*, the dual space of linear functionals on V.
sometimes you see this written like so:
u* = <u,__> which means that u*(v) = <u,v>.
given an inner product <_,_> it can be shown that in the finite-dimensional case, EVERY linear functional in V* arises in such a way.
so an inner product <u,v> can be seen as a combination of the linear functional u* = <_,v> and the vector v.
this gives us a way to regard the (conjugate-) transpose of a vector: as an element of the dual space.
in physics (such as in the study of quantum mechanics), instead of using the notation u* = <u,_>, they represent it like so: u* = <u|, or a "bra".
similarly, the vectors bras act upon are called "kets" and notated like so: v = |v>.
when a bra acts upon a ket, you get a scalar:
<u|v> = <u,v> (the inner product).
so in the infinite-dimensional case, instead of writing uH, we write <u|, which is the bra that corresponds to the ket |u>.
it turns out that for infinite-dimensional V, V* can be lots bigger than V, but the CONTINUOUS elements of V* are isomorphic to V (actually for complex vector spaces it's an anti-isomorphism, because scalar multiples are turned into conjugate scalar multiples, but that's not all that important).
now we're in a position to say what |u><v| means. this represents a linear operator: V-->V defined by:
(|>u<v|)(w) = |>u<v|w> = (<v|w>)|u> = <v,w>u
that is, the linear operator (transformation) that maps the ket (vector) |w> to the ket (vector) |u><v|w> (the vector u multiplied by the inner product of v and w).
that is the outer product is a mapping VxV*--->Hom(V,V), whereas the inner product is a mapping V*xV-->F
*********************
with regards to your earlier questions about integrals and inner products, the elements of the space of continuous (this is important, for technical reasons) linear functionals on the vector space of square-integrable functions all look like this:
$L_g(f) = \int_a^b f(x)g(x)\ dx$
in other words Lg is a "bra" (feed it a "ket", a function f in this case, you get a number).
7. ## Re: Infinite Dimensional Cross Product
I’m not sure whether my understanding is correct or not:
The outer product of |u><v| is defined (only?) by its operation on |w> as: |>u<v|w> and there is not any direct relationship between it and “cross product” of vectors.
Because you kindly regarded to my previous question, I allow myself to remind one of your phrases, that is not so clear for me:
“it is not true that all infinite-dimensional vectors are sums of the form:
some infinite-dimensional vector spaces are of uncountable dimension, and such a series "doesn't give enough terms". for example, the real numbers are a vector space over the rationals, but there is no countable set of (Q-linearly independent) real numbers that will serve to describe all reals as Q-linear combinations of them.”
Now, when we define, in general:
We in fact assumed that the set of vectors components is equipotent to natural number set N. So, the set of terms ukvk to be summed for calculating the inner product of the vectors is equipotent to N, too. But, when we introduce the inner product in integral form of:
he variable x is assumed to be continuously ranged from a to b thereby covering the uncountable real numbers in [a,b] range. Now, it seems that more terms are summed in the later expression than the former! (such an ambiguity can be found wherever an infinite-element summation is transformed to integral). Am I mistaken? or we need a more suitable notation for definition of the vectors emphasizing continuity property of their components? It is one of my old questions!
8. ## Re: Infinite Dimensional Cross Product
there is a certain "dynamic tension" you get when dealing with infinite-dimensional vector spaces.
on the one hand, every vector is a finite linear combination of basis elements. on the other, the set of basis elements itself may be uncountable. so you can use a "per-vector" index which is countable (indeed, even finite), but not a "per-space" index (if that makes sense to you).
but, i hesitate to regard an integral as a natural generalization of finite sums (in the sense of inner products). an inner product is just a bilinear function from VxV to F with certain properties. there are more possibilities than just sums or integrals.
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http://math.stackexchange.com/questions/195847/summation-of-modulo-sequences
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# Summation of Modulo Sequences
I came across through simulation that multiplying and adding certain modulo sequences yield equal results. Consider the following two sequences \begin{align} g_0[k] &= \sum_{n=0}^{N-1} \left< a n \right>_N \left< k n \right>_N \\ g_1[k] &= \sum_{n=0}^{N-1} \left< b n \right>_N \left< k n \right>_N \end{align} where $a, b \in \mathbb{Z}$, $\left<a\right>_N = a \pmod N$and $\text{GCD}(a,N) = \text{GCD}(b,N) = 1$. It appears that there exist $k_0, k_1 \in \mathbb{Z}$ such that \begin{align} g_0[k_1 k] &= g_1[k] \\ g_1[k_0 k] &= g_0[k] \end{align}
Outside of running several computer simulations that seem to indicate that such integers might always exist for arbitrary $a$ and $b$ with $\text{GCD}(a,N) = \text{GCD}(b,N)$, I cannot find out how to 1) determine under what conditions this holds and 2) figure out the value of $k_0$ and $k_1$ directly.
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## 1 Answer
Let $Z_N=\{0,1,\cdots,N-1\}$. Each integer $a$ creates a function on this set, given by multiplying by $a$ and subsequently reducing modulo $N$ (so the result lies in $Z_N$). Call such a map $\varphi_a$. Then $\varphi_a$ is a bijection if and only if $\gcd(a,N)=1$. By Bezout's identity, there are $x,y$ such that $ax+yn=1$ if $a$ is coprime to $N$, in which case $\varphi_a$ has an inverse map $\varphi_x$. And if $(a,N)>1$, then $\varphi_a$ sends both the elements $0\ne N/\gcd(a,N)\in Z_N$ and $0\in Z_N$ to $0$ and hence is not injective.
Therefore when $\gcd(a,N)=1=\gcd(b,N)$, and if $a'$ and $b'$ are inverses of $a,b$ mod $N$ resp., then
$$\sum_{n\in Z_N}\varphi_a(n)\varphi_k(n)=\sum_{m\in Z_N}m\varphi_k(\varphi_{a'}(m))=\sum_{\ell\in Z_N}\varphi_b(\ell)\varphi_k(\varphi_{a'}(\varphi_b(\ell)))=\sum_{\ell\in Z_N}\varphi_b(\ell)\varphi_{ka'b}(\ell).$$
We make the substitutions $m=\varphi_a(n)=\varphi_b(\ell)$. Thus $k_0=a'b$ and similarly $k_1=ab'$, which are only unique modulo $N$.
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Thanks for the answer. – dcdo Sep 15 '12 at 12:18
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http://physics.stackexchange.com/questions/17237/differential-equation-in-spherical-harmonics-derivation/17239
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# Differential Equation in Spherical Harmonics Derivation
I've been reviewing derivations of the spherical harmonics in quantum mechanics; mostly as review but also to make sure I understand where the concepts arise from.
However, every derivation I've seen makes use of the following differential equation/ identity that seemingly comes from nowhere. I've not yet been able to figure out where it comes from in looking through various texts. So, here I am to ask: where does this come from?
$$\frac{d}{d\theta}+l cot(\theta) \equiv \frac{1}{(sin(\theta))^l} \frac{d}{d\theta}(sin(\theta)^l)$$
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## 1 Answer
It comes from various differentiation rules. It is probably easier to grasp if one applies both sides of the identity on some function $f(\theta)$. Then the identity becomes
$$\frac{1}{\sin^\ell\theta} \frac{d}{d\theta}(\sin^\ell\theta~ f(\theta))~=~ f^\prime(\theta)+\ell \cot\theta ~f(\theta).$$
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Ah I see it now, thanks! – atomicpedals Nov 19 '11 at 22:23
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http://math.stackexchange.com/questions/173230/example-of-artinian-module-that-is-not-noetherian?answertab=votes
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Example of Artinian module that is not Noetherian
I've just learned the definitions of Artinian and Noetherian module and I'm now trying to think of examples. Can you tell me if the following example is correct:
An example of a $\mathbb Z$-module $M$ that is not Noetherian: Let $G_{\frac12}$ be the additive subgroup of $\mathbb Q$ generated by $\frac12$. Then $G_{\frac12} \subset G_{\frac14} \subset G_{\frac18} \subset \dots$ is a chain with no upper bound hence $M = G_{\frac12}$ as a $\mathbb Z$-module is not Noetherian.
But $M$ is Artinian: $G_{\frac{1}{2^n}}$ are the only subgroups of $G_{\frac12}$. So every decreasing chain of submodules $G_i$ is bounded from below by $G_{\frac{1}{2^{\min i}}}$.
Edit In Atiyah-MacDonald they give the following example:
Does one have to take the quotient $Q/Z$?
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2
– Dylan Moreland Jul 20 '12 at 11:56
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You might need to modify the example slightly. $G_{1/4}$ is not a submodule of $G_{1/2}$, so you haven't written down an increasing chain inside of your $M$. – Dylan Moreland Jul 20 '12 at 12:01
@DylanMoreland Thanks for pointing this out. So for rings I wouldn't be able to construct such an example. But is the example in my question about modules right? – Matt N. Jul 20 '12 at 12:02
1
– Bruno Stonek Jul 20 '12 at 12:24
1
@ClarkKent What you have is an ascending chain in $\mathbb{Q}$. – Joe Johnson 126 Jul 20 '12 at 14:24
show 6 more comments
2 Answers
Fix a prime $p$ and let $M_p={\Bbb Z}(\frac1p)/{\Bbb Z}$.
It is not difficult to see that the only submodules of $M_p$ are those generated by $\frac1{p^k}+{\Bbb Z}$ for $k\geq0$. From this it follows that $M_p$ is Artinian but not Noetherian.
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1
Thank you! So $\mathbb Z (\frac{1}{p}) = \{ \frac{k}{p^n} \mid n \in \mathbb N , k \in \mathbb Z \}$? – Matt N. Jul 20 '12 at 12:09
Right! I think that it can be described as the smallest divisible submodule of ${\Bbb Q}/{\Bbb Z}$ containing $1/p$. – Andrea Mori Jul 20 '12 at 12:16
Do we have to quotient by $\mathbb Z$? Does it not work with subgroups of $\mathbb Q$? – Matt N. Jul 20 '12 at 14:07
1
@ClarkKent: $A=\mathbb{Z}(\frac1p)$ is not artinian, as it contains the decreasing sequence of subgroups $A \supset qA \supset q^2A \supset q^3A \supset \cdots$ for any prime $q\neq p$. – Jack Schmidt Jul 20 '12 at 14:59
@JackSchmidt Right, thank you very much. I'm starting to understand it a bit better. – Matt N. Jul 21 '12 at 8:15
show 2 more comments
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1
Since links can expire, it is advisable to include a short summary to a link-answer. – Julian Kuelshammer Feb 22 at 14:52
1
This link has no new example, it is the famous example "$\mathbb{Z}_{P^{\infty}}$". – AmirHosein SadeghiManesh Feb 22 at 19:04
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It doesn't matter if it's new or not. – Julian Kuelshammer Feb 23 at 8:02
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http://mathoverflow.net/questions/tagged/invariant-theory
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## Tagged Questions
1answer
361 views
### What is the “fundamental theorem of invariant theory” ?
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### Generate a higher degree symmetric polynomial from an existing one
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### Can one pick generators for the ring of invariants of binary n-ic forms which have rational coefficients?
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### Ideal membership (concerning polynomial invariants of orthogonal groups)
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### Cyclically symmetric functions
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### Invariants and syzygies for 3x3 matrices
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### Quotient of affine space by cyclic permutation
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### Invariants of a set of real unit vectors in 3d space, under SO(3)
I have a set of $n$ real unit vectors, in 3-dimensional space. (It is a follow up of Sets of vectors related by a rotation.) Is there a construction providing a complete set of i …
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135 views
### Modules of invariants?
Let $G \subset SL(V, \mathbb{C})$ be a finite group and $R=(\operatorname{Sym}[V])^G$ is the ring of polynomial invariants, $W$ some irreducible complex representation of $G$. I wa …
0answers
126 views
### algebra of endomorphisms over the diagonal invariants
Let $k$ be a field of characteristic 0 (say $\mathbb{C}$). Consider the ring of polynomials $R = k[X_1,...,X_n]$ and its subring of invariant polynomials $S = R^{S_n}$. It is know …
0answers
45 views
### An almost permutation G-lattice
I've been trying to determine the rationality of certain fields of invariants coming from G-lattices. More precisely, letting $G$ be a finite group, $L=\mathbb{Z}^n$ a free abelian …
0answers
51 views
### A slightly odd (integral of Whittaker functions / sum of characters of $GL_n(\mathbb C)$ / sum of Schur functions)
Let $W$ be the normalized spherical Whittaker function attached to a spherical representation $\pi$ on $GL_n(k)$, where $k$ is a $p$-adic field and $n\ge 3$. I'm faced with the sl …
1answer
318 views
### A question on invariant theory of $GL_n(\mathbb{C})$.
Let $\rho$ denote the irreducible algebraic representation of $GL_n(\mathbb{C})$ with the highest weight $(2,2,\underset{n-2}{\underbrace{0,\dots,0}})$. Let $k\leq n/2$ be a non- …
2answers
166 views
### Intersection theory for $G$-varieties - an action on the chow ring?
Let $G$ be a reductive algebraic group. Let $X$ be a $G$-variety and consider any closed subvariety $Z$ of $X$. Since any $g\in G$ acts as an automorphism, we know that $g.Z$ is ag …
3answers
268 views
### Invariants of group action: SL_n acts simultaneously on m symmetric matrices
Let $\rm{SL}_n$ be the special linear group and let $\rm{Sym}_n$ be the set of all symmetric matrices of size n. $\rm{SL}_n$ acts on $(\rm{Sym}_n)^m$ by \$g(A_1, \ldots , A_m)=(gA_1 …
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http://mathhelpforum.com/math-topics/15753-quadratic-formula.html
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Thread:
1. Quadratic Formula
My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?
thank you!
2. Originally Posted by Fibonacci
My teacher told me this, but i just want to make sure for my exams on monday, what is the the quadratic formula used to solve x intercepts in an equation when it is in turning point form?
thank you!
Hello,
this is only a guess but...
If the vertex has the coordinates V(k, h) the the equation of the parabola is:
$y = a(x-k)^2 + h$ . Equation in turning point form.
To calculate the intercepts with the x-axis you use that y = 0:
$0 = a(x-k)^2 + h \Longleftrightarrow (x-k)^2 =\frac{-h}{a} \Longleftrightarrow x = k \pm \sqrt{\frac{-h}{a}}$
Maybe this helps a little bit.
3. Originally Posted by earboth
Hello,
this is only a guess but...
If the vertex has the coordinates V(k, h) the the equation of the parabola is:
$y = a(x-k)^2 + h$ . Equation in turning point form.
To calculate the intercepts with the x-axis you use that y = 0:
$0 = a(x-k)^2 + h \Longleftrightarrow (x-k)^2 =\frac{-h}{a} \Longleftrightarrow x = k \pm \sqrt{\frac{-h}{a}}$
Maybe this helps a little bit.
yah i think thats right but maybe a specific example if u could....
4. Originally Posted by Fibonacci
yah i think thats right but maybe a specific example if u could....
Hello,
$y = -\frac{1}{4} \cdot (x-3)^2 + 4$
The vertex is at V(3, 4) and according to the formula given in my previous post you'll get the zeros Z_1 and Z_2 at:
$x_{1,2} = 3 \pm \sqrt{\frac{-4}{-\frac{1}{4}}} = 3 \pm 4$
I've attached the graph of the parabola so you can check the answer.
Attached Thumbnails
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http://mathoverflow.net/questions/50006?sort=oldest
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## What is ample generator of a Picard group?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
First a note of caution: I am a physicist with a rudimentary knowledge of algebraic geometry picked up here and there. So don't assume I know anything besides basic properties of sheaves and try to give as simple answers as possible. Also, if my questions don't make sense for any reason, try to point me in the right direction.
So, my questions:
1. I'd like to hear the definition of the ample generator of a Picard group. I know what a Picard group is but I am having a hard time finding an actual definition of its ample generator.
2. Why is this notion important and where is it used primarily?
3. If you could provide some simple examples to illustrate the notion for some special $X$ and ${\rm Pic}(X)$ that would be welcome.
Motivation: recently, Scott Carnahan provided a very nice answer about conformal blocks in CFT over at physics.SE. My problem is that the punchline of his example involves the notion of an ample generator of a Picard group of certain moduli space of $SU(2)$ bundles on the Riemann surface (which is the arena for the CFT).
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1
Your difficulty is mostly semantical. There is no notion of "ample generator of a group". The Picard group is sometimes cyclic in which case it will have a generator, say g. Now -g is also a generator. Exactly one of these two will represent the class of an ample line bundle and this is the ample generator. – Felipe Voloch Dec 21 2010 at 1:36
@Felipe: yes, this is the picture I got from Henri's answer. But thanks for making that explicit. – Marek Dec 21 2010 at 9:01
## 2 Answers
First of all, the Picard group of a variety is not always monogen, so that the notion of "the ample generator" you are referring to surely concerns a restricted class of varieties.
Furthermore, an ample line bundle (or invertible sheaf) is a line bundle $L$ which satisfies any of the following properties :
1. For any coherent sheaf $\mathcal F$, the sheaf $\mathcal F \otimes L^{\otimes m}$ is generated by its global sections for every $m\geq m_0(\mathcal F)$.
2. The map $\Phi_{|L^{\otimes m}|} : X \to \mathbb P(H^0(X,L^{\otimes m})), x \mapsto [s_0(x): \ldots:s_N(x)]$ - where $(s_i)$ is a base of the space of global sections of $L^{\otimes m}$ - induces a embedding of $X$ in some projective space for every $m\geq m_0$.
3. If you are working over $\mathbb C$, which seems to be the case, this is also equivalent for $L$ to admit a smooth hermitian metric $h$ whose curvature $\Theta_h(L)$ is a (striclty) positive $(1,1)$-form.
Then I guess that the ample generator of some Picard group is in some case one generator which besides is ample.
For example, the most basic example consists in taking the ample line bundle $\mathcal O_{\mathbb P^n}(1)$ over $\mathbb P^n$, which generates $\mathrm{Pic}(\mathbb P^n)$.
Finally, if your question is : if the Picard group of a projective variety is monogen, then can we choose an ample generator? Then the answer is yes because for every $m\geq 1$, $L$ is ample iff $L^{\otimes m}$ is ample.
Edit : I forgot to mention that in this case, the unicity of "the" ample generator is clear : indeed, if $L$ and $L^{-1}$ admit non trivial sections (say $s$ and $t$) then $st$ is a non-zero section of $\mathcal O_X$ thus is constant, so that $s$ and $t$ are both non-vanishing sections, which implies that $L$ is trivial. You can apply this to $L^{\otimes m}$ to get the unicity property.
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Thank you Henri, this is very enlightening (unfortunately I can't yet vote up). I'll spend some more time with your answer to grok all of it and then will update my question to make it more precise. – Marek Dec 20 2010 at 23:16
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
In terms of why these might be important, having an ample line bundle is equivalent to your object of study being projective. This is since, as in Henri's answer, ample implies that there in an embedding into $\mathbb{P}H^0(X,L^{\otimes m})$ for some $m$, while conversely pulling back $\mathcal{O}(1)$ from an embedding into projective space yields an ample line bundle. Being projective gives it many nice properties.
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http://psychology.wikia.com/wiki/Markov_decision_process
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# Markov decision process
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Markov decision processes (MDPs) provide a mathematical framework for modeling decision-making in situations where outcomes are partly random and partly under the control of the decision maker. MDPs are useful for studying a wide range of optimization problems solved via dynamic programming and reinforcement learning. MDPs were known at least as early as in the fifties (cf. Bellman 1957). Much research in the area was spawned due to the book by Ronald A. Howard in 1960. Today they are used in a variety of areas, including robotics, automated control, economics and in manufacturing.
More precisely a Markov Decision Process is a discrete time stochastic control process characterized by a set of states; in each state there are several actions from which the decision maker must choose. For a state s and an action a, a state transition function $P_a(s)$ determines the transition probabilities to the next state. The decision maker earns a reward for each state visited. The states of an MDP possess the Markov property. This means that if the current state of the MDP at time $t$ is known, transitions to a new state at time $t+1$ are independent of all previous states.
## Definition
A Markov Decision Process is a tuple $(S,A,P_\cdot(\cdot,\cdot),R(\cdot))$, where
• $S$ is the State space,
• $A$ is the action space,
• $P_a(s,s') = \Pr(s_{t+1}=s' | s_t = s,\, a_t=a)$ is the probability that action $a$ in state $s$ at time $t$ will lead to state $s'$ at time $t+1$,
• $R(s)$ is the immediate reward received in state $s$.
The goal is to maximize some cumulative function of the rewards, typically the discounted sum under a discounting factor $\gamma$ (usually just under 1). This would look like:
• $\sum^{\infty}_{t=0}\gamma^t R(s_t)$
where $\gamma$ is the discount rate and satisfies $0 < \gamma \le 1$. The sum is to be maximised.
Markov decision processes are an extension of Markov chains; the difference is the addition of actions (allowing choice) and rewards (giving motivation). If there were only one action, or if the action to take were somehow fixed for each state, a Markov decision process would reduce to a Markov chain.
## Solution
The solution to a Markov Decision Process can be expressed as a policy $\pi$, which gives the action to take for a given state, regardless of prior history. Note that once a Markov decision process is combined with a policy in this way, this fixes the action for each state and the resulting combination behaves like a Markov chain.
The standard family of algorithms to calculate the policy requires storage for two arrays indexed by state: value $V$, which contains real values, and policy $\pi$ which contains actions. At the end of the algorithm, $\pi$ will contain the solution and $V(s_0)$ will contain the discounted sum of the rewards to be earned (on average) by following that solution.
The algorithm then has the following two kinds of steps, which are repeated in some order for all the states until no further changes take place.
$\ \pi(s) := \arg \max_a \sum_{s'} P_a(s,s') V(s')\$
$\ V(s) := R(s) + \gamma \sum_{s'} P_{\pi(s)}(s,s') V(s')\$
Their order depends on the variant of the algorithm; one can also do them for all states at once or state by state, and more often to some states than others. As long as no state is permanently excluded from either of the steps, the algorithm will eventually arrive at the correct solution.
### Notable variants
#### Value iteration
In value iteration (Bellman 1957), the $\pi$ array is not used; instead, the value of $\pi(s)$ is calculated whenever it is needed.
Substituting the calculation of $\pi(s)$ into the calculation of $V(s)$ gives the combined step:
$\ V(s) := R(s) + \gamma \max_a \sum_{s'} P_a(s,s') V(s')\$
#### Policy iteration
In policy iteration (Howard 1960), step one is performed once, and then step two is repeated until it converges. Then step one is again performed once and so on.
Instead of repeating step two to convergence, it may be formulated and solved as a set of linear equations.
This variant has the advantage that there is a definite stopping condition: when the array $\pi$ does not change in the course of applying step 1 to all states, the algorithm is completed.
#### Modified policy iteration
In modified policy iteration (Puterman and Shin 1978), step one is performed once, and then step two is repeated several times. Then step one is again performed once and so on.
#### Prioritized sweeping
In this variant, the steps are preferentially applied to states which are in some way important - whether based on the algorithm (there were large changes in $V$ or $\pi$ around those states recently) or based on use (those states are near the starting state, or otherwise of interest to the person or program using the algorithm).
## Extensions
### Partial observability
Main article: partially observable Markov decision process
The solution above assumes that the state $s$ is known when action is to be taken; otherwise $\pi(s)$ cannot be calculated. When this assumption is not true, the problem is called a partially observable Markov decision process or POMDP.
### Learning
If the probabilities are unknown, the problem is one of reinforcement learning.
For this purpose it is useful to define a further function, which corresponds to taking the action $a$ and then continuing optimally (or according to whatever policy one currently has):
$Q(s,a) = R(s) + \gamma \sum_{s'} P_a(s,s') V(s')$
While this function is also unknown, experience during learning is based on $(s, a)$ pairs (together with the outcome $s'$); that is, "I was in state $s$ and I tried doing $a$ (and $s'$ happened)". Thus, one has an array $Q$ and uses experience to update it directly. This is known as Q-learning.
### Minor extensions
These extensions are minor in that they complicate the notation, but make no real difference to the problem or its solution.
• The reward may be a function of the action as well as the state, $R(s,a)$.
• The reward may be a function of the resulting state as well as the action and state, $R(s,a,s')$.
• The action space may be different at each state, so that it is $A_s$ rather than $A$
## Alternative notations
The terminology and notation for MDPs are not entirely settled; there are two main streams — one using action, reward, value and $\gamma$, while the other uses control, cost, cost-to-go and $\alpha$. In addition, the notation for the transition probability varies.
action $a$ control $u$
reward $R$ cost $g$ $g$ is the negative of $R$
value $V$ cost-to-go $J$ $J$ is the negative of $V$
policy $\pi$ policy $\mu$
discounting factor $\gamma$ discounting factor $\alpha$
transition probability $P_a(s,s')$ transition probability $p_{ss'}(a)$
In addition, transition probability is sometimes written $Pr(s,a,s')$, $Pr(s'|s,a)$ or, rarely, $p_{s's}(a)$
## References
• R. Bellman. A Markovian Decision Process. Journal of Mathematics and Mechanics 6, 1957.
• R. E. Bellman. Dynamic Programming. Princeton University Press, Princeton, NJ.
• Ronald A. Howard Dynamic Programming and Markov Processes, The M.I.T. Press, 1960.
• M. L. Puterman. Markov Decision Processes. Wiley, 1994.
• H.C. Tijms. A First Course in Stochastic Models. Wiley, 2003.
• Sutton, R.S. On the significance of Markov decision processes . In W. Gerstner, A. Germond, M. Hasler, and J.-D. Nicoud (Eds.) Artificial Neural Networks -- ICANN'97, pp. 273-282. Springer.
• Sutton, R. S. and Barto A. G. Reinforcement Learning: An Introduction. The MIT Press, Cambridge, MA, 1998
## See also
• Partially observable Markov decision process
• Bellman equation for applications to economics.
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http://math.stackexchange.com/questions/56197/planar-curve-if-and-only-if-torsion
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# planar curve if and only if torsion
Again I have a question, it's about a proof, if the torsion of a curve is zero, we have that
$$B(s) = v_0,$$
a constant vector (where $B$ is the binormal), the proof ends concluding that the curve $$\alpha \left( t \right)$$ is such that $$\alpha(t)\cdot v_0 = k$$ and then the book says, "then the curve is contained in a plane orthogonal to $v_0$." It's a not so important detail but .... that angle might not be $0$, could be not perpendicular to it, anyway, geometrically I see it that $V_0$ "cuts" that plane with some angle.
My stupid question is why this constant $k$ must be $0$. Or just I can choose some $v_0$ to get that "$k$"?
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Any plane with normal vector $\vec{v}_0$ satisfies an equation of the form $\vec{x}\cdot\vec{v}_0 = k$. The quantity $k$ basically measures how far the plane is from the origin - it doesn't have to be $0$. – anon Aug 7 '11 at 21:47
Daniel, I attempted to edit your post so it was readable, but I had difficulty understanding your writing in some parts. You might want to check it - and make sure you communicate more clearly in the future. I'll leave figuring out a meaningful title up to you or someone else. – anon Aug 7 '11 at 21:55
## 1 Answer
The constant $k$ need not be $0$; that would be the case where $\alpha$ lies in a plane through the origin. You have $k=\alpha(0)\cdot v_0$, so for all $t$, $(\alpha(t)-\alpha(0))\cdot v_0=0$. This means that $\alpha(t)-\alpha(0)$ lies in the plane through the origin perpendicular to $v_0$, so $\alpha(t)$ lies in the plane through $\alpha(0)$ perpendicular to $v_0$. (If $0$ is not in the domain, then $0$ could be replaced with any point in the domain of $\alpha$.)
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Thanks Jonas!!! – Daniel Aug 7 '11 at 22:05
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http://mathoverflow.net/questions/121123?sort=newest
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## Distributive lattice embedding into a finite lattice.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose one has an inclusion $\iota : D \hookrightarrow S$ where $D$ is a finite distributive lattice and $S$ is a finite join-semilattice.
If $\iota$ preserves all meets and joins one can show that $|J(D)| \leq |J(S)|$ i.e. $D$ has no more join-irreducibles than $S$.
Does this also hold if $\iota$ only preserves finite joins?
I only know that it is true in the special cases where $S$ is a distributive lattice or $S \cong S^{op}$ is order-dual.
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If S had empty joins and i preserved meets then you get the inequality you want. I am skeptical joins work. – Benjamin Steinberg Feb 8 at 1:33
## 1 Answer
Well, I proved that it is true with a friend yesterday, so I'll include it here.
Recall that we have an inclusion $\iota : D \hookrightarrow S$ where $S$ is a finite join-semilattice and $D$ is a finite distributive lattice. More explicitly: $S$ has all finite joins, so it is a complete lattice with a top and bottom element and furthermore $D$ has a top and bottom too.
In the case where $\iota$ preserves all meets then the same function defines a meet-semilattice morphism $\iota : D^{op} \hookrightarrow S^{op}$, so by duality we have the surjection $\iota' : S \twoheadrightarrow D$, hence $|J(D)| \leq |J(S)|$ because join-irreducibles (we exclude zero) in join-semilattices form the minimal generating set.
But the question was whether the same is true when $\iota$ preserves all joins.
First recall that in any finite distributive lattice $D$, the cardinality $|J(D)|$ is the size of the maximal chain in $D$ (counting the edges). Since we have an embedding $\iota : D \hookrightarrow S$, it follows that $S$'s maximal chain has size more than or equal to $n = |J(D)|$. For a contradiction assume that $|J(S)| < n$. Then $S$ is generated by $n-1$ elements, so we have a join-semilattice morphism $f : 2^{n-1} \twoheadrightarrow S$. By duality there is an embedding $f' : S^{op} \hookrightarrow (2^{n-1})^{op}$, hence $S^{op}$'s maximal chain has size at most $n-1$, as does its order-dual $S$, this being a contradiction.
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Nice. Having the domain be distributive really helps. – Benjamin Steinberg Feb 9 at 15:59
@Benjamin Steinberg: Cheers. Perhaps you can see a pattern emerging regarding my other question that you answered i.e. I am searching for various conditions that imply a subalgebra has no more generators. By the way, the motivation is automata theory e.g. the above result can be used to identify a class of regular languages that have a canonical state-minimal nondeterministic acceptor. – Rob Myers Feb 9 at 16:56
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http://mathoverflow.net/questions/118662/cohomology-of-quotient-space/118809
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## Cohomology of quotient space
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let be $X$ a maximal torus in a Lie group $G$. I'd like to calculate the cohomology $H^{*}(G/N(T))$. I know that it is trivial in odd degree and the base-field is even but I haven't a basic method to do this. Is there a simple method of calculation when the gruop $G$ is the unitary gruop $U(n)$?
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## 2 Answers
If you take $G=U(n)$, then $G/T$ is the flag variety of $GL_n({\mathbb C})$. Hence its cohomology is given by Bruhat cells and as a representation of the Weyl group $W=N(T)/T)$ acting on the right on $G/T$, it is the regular representation (there are no fixed points for $W$!Use Lefschetz). Therefore, the space of invariant elements is one dimensional and is given by $H^0(G/T)$.
The same method applies to any compact connected LIe group $G$ and $T$ is a maximal torus.
For other groups, you need, I think, to know what the torus looks like (i.e. if it is split, or non-split etc).
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Have you some reference to study how Bruhat decomposition works? How can I use Lefschetz? However, do you know a complete reference in this topic?
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the space $G/T$ is the same as $G({\mathbb C}/B$ where $B$ is a Borel subgroup of the complexification $G({\mathbb C})$ containing $T$. One is doing Bruhat decomposition for this $B$. This is standard stuff in algebraic groups. The group $N(T)/T$ acts on the {\bf right} on $G/T$ and no element has a fixed point. Hence you can apply Lefschetz fixed point formula to say that this is the regular representation of $W$. Again, this is a well known fact (perhaps old papers of Bott or Borel). – Aakumadula Jan 13 at 14:23
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http://physics.stackexchange.com/questions/tagged/pressure?sort=active
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# Tagged Questions
Force per unit area.
3answers
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### Can a diver swim a short distance in great depths without being physically crushed by the pressure?
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I'm assuming the answer to this largely varies from person to person. Assuming you could instantly change the pressure around your head by amount $\Delta p$, what is the minimum $\Delta p$ for your ...
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### Thermodynamics, PV diagrams?
My teacher told me that the total amount of work done on or by a gas can be represented by the area enclosed in the process in a PV diagram. This is only valid for non isothermic processes, right?
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### Water draining from a height into the bottom of a reservoir
If I hang one 5 gal. Bucket directly above another, put a hole in the bottom of the upper bucket with a tube inserted just inside hanging all the way to the bottom of the lower bucket and fill the ...
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### Over-inflating weather balloons
Would it be safe to over-inflate a weather balloon if it isn't going to be used at high altitudes? I've been looking at several different balloons, but they are all measured by their burst diameter ...
1answer
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I want to measure the density in different layers of a suspension. To do this I want to place pressure sensors at different heights. Let's assume that the sensors are not by orders of magnitude more ...
2answers
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### Gas Circulation Using Pressure Difference
Dear all, see attached picture Please, is it possible to have the gas recirculated from the gas phase to the liquid as described in the diagram assuming the gas is not soluble in the water. These ...
0answers
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### Gas Circulation [duplicate]
Dear all, Please, is it possible to have the gas recirculated from the gas phase to the liquid as described in the diagram assuming the gas is not soluble in the water. These are the conditions. ...
0answers
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### What is the effect of an increase in pressure on latent heat of vaporization?
What is latent heat of vaporization ($L_v$) in the first place? Wikipedia seems to indicate that it is the energy used in overcoming intermolecular interactions, without taking into account at all any ...
1answer
54 views
### Calculating Air Density Lapse With Altitude (Specifically, pressures)
This might be a bit more of an engineering question, but I'm calculating air density drop-off with altitude, and I'm having some problems calculating the pressure (I'll run through my method). This ...
2answers
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### How is energy transferred from one incompressible fluid to another?
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1answer
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### How should holes in a Tesla turbine look like?
I think of building a Tesla turbine out of old hard drives. Now I wonder how to cut ventilation holes in the platters. On the internet there are a lot of different attempts on that matter. A lot of ...
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### Have negative pressures any physical meaning?
Some cubic thermodynamical equations of state predict negative pressures, have negative pressures any physical meaning? Could they be related to negative mass?
1answer
48 views
### Air pressure relative to a force on a bag?
Assume an airtight bag occupied by air such that the pressure inside the bag is equal to the atmospheric pressure. Assume the surface tension of the bag is negligible. What is the change in air ...
1answer
39 views
### Is this a correct interpretation of pressure?
So I am told that pressure = Force per Area --> F/A.. When considering the units of Force I find that force = kg * m/s^2 When considering the units of Area I find that area = m^2 Thus the units of ...
1answer
74 views
### What if the lid of a pressure cooker was suddenly released?
My dad and I have tried to calculate the strength of the explosion if the lid was suddenly freed. We took some measures: Lid mass: $0.7 \textrm{kg}$ Lid surface: $0.415 \textrm{m}^2$ Internal ...
3answers
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### Lethality of sounds and extreme “loudness”
In theory, could pure sound be lethal? How loud would it have to be? Also, which events are the loudest in the universe, and how loud are they? I'm confining attention to events which occur regularly, ...
0answers
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### Formula to calculate the density of satuared steam
I need to calculate (or look-up?) the density for saturated steam in a range of 8-12 bar. Is there a formula for this? It's ok if it's not perfectly exact.
1answer
112 views
### What happens to the vapor pressure when I expel the liquid phase from the vessel?
Suppose I put water in a closed vessel and heat it, so the vessel becomes pressurized above ambient pressure. Now, there should be liquid and gaseous water inside the vessel. Since it's under ...
1answer
41 views
### If a balloon is continuously filled with air and stays at a constant shape and size will there be any empty space in the balloon?
If a container like a balloon but with constant volume is filled, is it possible to pack air molecules so closely together that they don't have any empty space between them? If so, what would this ...
2answers
134 views
### Hydrostatic pressure on a teapot spout
The phenomenon where water flows on the outside side of a teapot spout is named "The teapot effect", and occurs due to a difference in pressure between water and the atmosphere. Consider the image of ...
1answer
43 views
### How does a pierced vacuum move?
If a can of compressed air is pierced on the right, air pushes out, and the can moves to the left. If a vacuum container is pierced on the right, which way does it move? Right? Left? Not at all? ...
0answers
51 views
### When do we have $p = -\frac{\partial F}{\partial V}$? [closed]
In what context can we say that : $$p = -\frac{\partial F}{\partial V}$$ ?
1answer
58 views
### How to calculate the amout of time for oil to get out from a pneumatic cylinder?
I have a system. You can assume it is just a 1" diameter steel pneumatic cylinder (1 foot long). Assume we ignore the bore size issue at this point. One chamber is filled up by oil (6" of volume), ...
2answers
116 views
### Physics behind the flow of gas coming out of a balloon
I'm working with stratospheric balloons (latex ones) and I want to put a valve on it so it can float for a longer time. I'm trying to define which valve I should use, which demands I estimate the flow ...
11answers
4k views
### How long a straw could Superman use?
To suck water through a straw, you create a partial vacuum in your lungs. Water rises through the straw until the pressure in the straw at the water level equals atmospheric pressure. This ...
1answer
86 views
### What is the pressure on an empty pneumatic tube?
At the residence I'm constructing a kitchen digester. The gas is to be collected into an old bicycle/scooter tube - initially as empty as may be after wringing it out.. Will the digester (total ...
1answer
61 views
### Does tapping at the side of a bottle prevent shaken soda from bubbling over?
Anecdotal evidence has it that a bottle of soda that was heavily shaken will not bubble over if tapped at the side multiple times. Yet I wonder: Has the tapping really any effect? Or could it be that ...
1answer
101 views
### Drinking juice through a straw
Why we are able to suck more drink through a larger diameter straw than a smaller diameter straw if $p_1 v_1 = p_2 v_2 = Q$ as per Bernoulli's Principle. The pressure difference I create in mouth ...
1answer
42 views
### Deflection of a membrane
I am currently working on a project which is described as the deflection of a circular membrane. What I am trying to model is the deflection of a piece of plastic film (E=200MPa,v=0.5) when placed ...
1answer
26 views
### How does a small radius connector affect gas flow?
Assume I have a 1m3 tank of air at 100 psi, they are going to release to a 1m3 tank which has air at 5 psi. There is a meter long pipe connect between two tanks. How to calculate the flow speed ratio ...
1answer
40 views
### How does the radius and length of a pipe affect vacuum performance?
Can I use Poiseuille's equation? Also If I have a vacuum pump which says that its performance is 30 cubic feet per minutes and I have two pipes (one is radius R1 and L1, the other one is 2*R1 and ...
2answers
146 views
### Why water in the sink follow a curved path?
When you fill the sink with water and then allow the water to be drained, the water forms a vortex.. And then it starts to follow a curved path downwards by effects of gravity.. Why this phenomena ...
3answers
113 views
### Is it easier to pump water up or down?
I'm trying to gauge the strength of aquarium air pumps, for use in a vertical hydroponic farm. A curious question arose: is it easier to pump water upwards vertically, or is it easier to pump it ...
2answers
136 views
### Freezing point of water with respect to pressure.
I know when the pressure is reduced, the boiling point of water is reduced as well. But how does the pressure affect the freezing point of water ? In a low pressure environment is the freezing point ...
2answers
77 views
### Pascal's law: pressure of fluid at different locations
I know that's stupid question, but I'm really confused what my teachers says, so I need to check that theory. Here are just two ordinary connected containers, which are full of water. On grounds ...
1answer
83 views
### A water balloon in vacuum: does it boil?
If I put water in a vacuum it will boil. But what if I put this water inside a balloon ? I searched for answers and fount this video: http://www.youtube.com/watch?v=9q8F3ClUuV0 It seams that the ...
1answer
71 views
### Hydrostatic pressure at lateral directions
I just read that, with respect to a stationary tiny cube, suspended in a fluid, that has a negligible weight and dimensions: Pressure is the same in every direction in a fluid at a given depth, ...
1answer
66 views
### Real world usage; pump flow vs pressure vs pump head, what is more useful in this situation?
I am currently making a water cooling system for my computer and decided to see if I could get some better help here than what people think they know on the computer forums. So for my cooling system ...
1answer
74 views
### What pressure or tension occurs in a glass jar when I pour boiling water inside?
Sometimes it happens that when you pour a boiling water into a glass jar, it cracks. Since glass is very hard material and resilient to pressure, the tension must be very high. Is it possible to ...
2answers
76 views
### Temperature change inside pressure chamber
Let's say there is a pressure chamber with some sort of sample / specimen (e.g. protein crystal) in it. Now I apply a certain amount of gas pressure, e.g. 10 or 20 atm. Let's say I use xenon as a gas. ...
0answers
43 views
### How to solve state parameters using these givens for an ideal gas?
In a thermodynamic turbine using air as an ideal gas, given that you have a known inlet temperature value $T_i$, a known exit pressure value $P_e$, a known inlet and exit velocity $V_i$ and $V_e$, a ...
3answers
3k views
### Explanation for different boiling points of water on different altitudes
I understand water boils at different temperatures depending on altitude. I am seeking to get an illustrative explanation for this, including a diagram if possible.
0answers
29 views
### pressure required for displacing a single electron off a crystal
I need to know this for my project- "power generation using the pressure applied on a keypad of a mobile electronic device". How much pressure does it take to displace a single electron off its ...
0answers
129 views
### pressure loss in a syringe
I'm currently working on a problem which is really giving me some issues. The problem concerns the force required to expel water from a syringe. We have a 20ml syringe (which is $2\times10^{-5}$ ...
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http://physics.stackexchange.com/questions/23295/why-cant-we-know-the-speed-vecvt-and-position-vecrt-of-an-ele/23299
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# Why can't we know the speed, $\vec{v}(t)$, and position, $\vec{r}(t)$, of an electron (the two) at the same time $t$?
I've read something about this and I conclude that it happens because of the uncertainty principle. But I don't understand very well the meaning of that.
I mean, it's very abstract that the speed, $\vec{v}(t)$, and position, $\vec{r}(t)$, of a particle can't be known at the same time. I don't understand that statement.
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1
Read an introductory book to quantum mechanics, there's no way to explain it in the scope of a SE question. – leftaroundabout Apr 5 '12 at 12:53
## 3 Answers
You can think of it in this way:
To find position of any object we use reflected light from that object. For day-to-day life objects there is no problem.
But for subatomic particle it means that we are giving them considerable amount of momentum and energy through photons. Thus the very moment we measure their position we are also changing their momentum.
Thus both cannot be known with absolute certainty at the same time.
Hope this helps.
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Welcome to Physics StackExchange! – Manishearth♦ Apr 5 '12 at 14:06
Thanks Manishearth – anuragsn7 Apr 5 '12 at 14:28
This is a very basic property of quantum mechanics which is summarized by the Heisenberg Uncertainty Principle. Essentially, the fact that the speed and position cannot be simultaneously known follows from the fact that "particles" are really waves. Thus, this just boils down to a very simple property of waves and fourier transforms. So imagine that you have a particle localize at x=0. In terms of its wavefunction, you would have something like a guassian (bell curve) centered at x=0. You could imagine the particle precisely at x=0 but this would just be the limiting case of a bell curve so it doesn't add anything new to the discussion. Now, this bell curve may be written as a sum of simple sine and cosine waves via fourier transforms. Each one of these sine/cosine waves represents a particle moving with a particular momentum. (Actually, sine waves are a superposition of left moving and right moving waves.) As with any wave phenomenon, there is a dispersion relation which tells you how the frequency relates to the wavelength, and hence you can get the velocity as v=(frequency)*(wavelength). Thus, the "velocity" of the bell curve representing position is just a superposition of all the velocities of the waves making up the bell curve. It turns out that the sharper the bell curve is (the more localized) then the more waves of larger velocity (momentum) are needed to represent it using fourier transforms. Thus, the uncertainty in the velocity is larger.
This may sound complicated because one is not normally used to thinking about particles in terms of waves. However, as far as fourier transforms go, this can be understood very easily in the context of sound. There is an analogous relationship between frequency and time for a sound wave which says that you can't know the frequency and duration of a sound precisely at the same time. For example, middle C is around 261 Hz. The uncertainty principle says that I can't determine the duration that middle C is played to an accuracy greater than 1/(261 Hz), or about .004s. Why? Simply because .004s is the period of middle C and if I'm playing a note for less time than it takes for the sound vibrations to oscillate once then there's no sense in saying that you're even playing that note! It's hard to develop an intuition for how this applies to a particle, but the mathematics is the same.
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As @leftaroundabout said, it can take a bit to explain--it has to do with what 'speed' and 'position' mean in quantum and wave mechanics.
These books http://www.lightandmatter.com/ are free and I have found them to provide very clear conceptual explanations for many physics topics. Read them and see if you understand then.
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http://math.stackexchange.com/questions/249862/fastest-primality-test-using-n-1-factorization
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# Fastest Primality test using $N-1$ Factorization?
If $N-1$ could be factored easily with several small prime factors, then what is the fastest way to check $N$ for primality?
Updated
I'm aware of Pocklington primility test which is not good for small factors. I'm looking for a reduction in modular exponentitation when $N-1$ has several small factors.
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– Gerry Myerson Dec 3 '12 at 9:59
It is not fast, since finally it requires modular exponentiation with $N-1$ in exponent, which requires the same time of Fermat Little Theorem PRP test. – Mohsen Afshin Dec 3 '12 at 10:15
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The pocklington test relies on the fact that $N-1$ is divisible by some relatively large prime, which doesn't seem to be the case. – Arthur Dec 3 '12 at 11:05
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– TonyK Dec 3 '12 at 11:11
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@Arthur: You should do these exponentiation in parallel, to save even more time. I don't mean using multiple processors, I mean exponentiating them all at once by incorporating the successive square only in those results that have the appropriate bit set. (If you are familiar with square-and-multiply exponentiation, this might make some sense!) – TonyK Dec 3 '12 at 11:15
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http://openwetware.org/wiki/User:TheLarry/Notebook/Larrys_Notebook/2009/11/21
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# User:TheLarry/Notebook/Larrys Notebook/2009/11/21
### From OpenWetWare
Generating Data Main project page
Previous entry Next entry
## Creating a Trajectory
I have a pretty good way to make a trajectory now. The user can put his trajectory together piece wise. They can add a horizontal line, vertical line, or circle in any order they want. And make each one of those as long as they want. I can add more geometries to create more intricate trajectories like a slanted line, but for right now this can do what i want. This .vi can also calculate the distances between adjacent points as well. It does this through
$s=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$
So i have everything i need now to make any trajectory and then make a microtubule of a certain length move along it. I still have to worry about the width of the microtubule, but for right now i'll go with it being a line and get this working. Then i'll add width.
This works decent, but i have a hard time transitioning from the horizontal line to the circle. I think it is the abrupt change that does it. The points around the change from line to circle have a shorter length than all the other points. This might be ok to work with. i can take these points our or since i know the actual answers this can sill be used to show if the tracking software is working alright. So i'll just go ahead with this.
Here is a movie of the trajectory i made. I think everything is working well. I am happy with this. Now i have to take this image and send it through my convolution sub.vi and i think i'd have a good looking simulation of the movie we get.
oh wait i gotta get this width right first. Alright let me see how badly i can screw this up. hmm maybe i don't have to worry about screwing this up since the width of the microtubule is so small if keep the high res image pixel higher than the width of the microtubule i should be golden.
Maybe the next thing i gotta do is automate the convolution .vi. But i am getting tired so maybe i'll do that tomorrow.
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http://math.stackexchange.com/questions/180133/showing-that-217-1-is-prime/180137
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# Showing that $2^{17} - 1$ is prime.
The question is:
Show that $131071=2^{17} - 1$ is prime.
So in this section we have the corollary to a theorem that clearly applies to this question.
Corollary. Any Divisor of $2^p -1$ is of the form $2kp + 1$.
How would you approach this problem?
WHAT I HAVE:
Let $q=2kp +1$ be an odd prime.
Then we have $2^{17} \equiv 1$ (mod q).
Using the a theorem that states that is the order of 2 (mod q) is $t$ then $2^n \equiv 1$ (mod q) iff $t|n$
this implies that $t=1$ or $t=17$
if $t=1$ we have that $2 \equiv 1$(mod q) so $q=1$ obviously impossible since we said $q$ was an odd prime.
So, $t=17$.
Using this I get that $q=131071$ is an odd prime.
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In your case, the corollary indicates that any possible prime divisor should be of the form $17\times 2k+1=34k+1$. – J. M. Aug 8 '12 at 0:42
@anon The theorem states that if $p$ and $q$ are odd primes and $q|a^p -1$, then either $q|a-1$ or $q=2kp + 1$ for some integer $k$. – HowardRoark Aug 8 '12 at 0:54
1
Oops I misread. – anon Aug 8 '12 at 1:05
## 1 Answer
As J.M. noted, any divisor must be of the form $34k+1$. We only have to check potential divisors up to $\sqrt{131071} \approx 362$, so only about 10 numbers. You can do this manually relatively quickly. Certain cases can be ruled out quickly like anything ending in a five (35, 205), etc.
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1
There are 10 numbers of that form in this range: 35, 69, 103, 137, 171, 205, 239, 273, 307, 341. 35 and 205 are divisible by 5, while 69, 171, 273 are divisible by 3, 341 is divisible by 11, which leaves 103, 137, 239 and 307. I believe they are all prime. That's still pretty hard to check by hand... – tomasz Aug 8 '12 at 0:55
I know that :). But I was thinking of approaching the problem a little more elegantly then by trying each divisor. – HowardRoark Aug 8 '12 at 0:56
1
@tomasz: You don't have to verify the primality of the divisors; you just have to check to see if they divide $131071$. – tskuzzy Aug 8 '12 at 0:57
@tomasz, 341 is obviously divisible by 11. – Rick Decker Aug 8 '12 at 0:57
2
Howard, there's no elegant reason why $2^{17}-1$ is prime (and, say, $2^{23}-1$ is not); there's no way to avoid doing a certain amount of trial division. Well, there is something called the Lucas-Lehmer test, but that's probably more trouble than it's worth for this particular problem. – Gerry Myerson Aug 8 '12 at 1:17
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http://mathoverflow.net/questions/43962/local-to-global-principle-for-reductive-groups/43969
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## Local to Global principle for reductive groups
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Let $G$ be a reductive group over an algebraic number field $k$. Denote with $k_v$ a local field and with $A$ the ring of its adeles, let $G_k$, $G_{k_v}$ resp. $G_A$ be the group of its $k$- resp. $A$- points. What are necessary and sufficient conditions for a local representation $\pi_v$ of $G_{k_v}$ to appear as $\otimes_v$ $\pi_v$ in the right regular representation of $G_k \backslash G_A$? What is a good reference to study this local to global process?
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The special case of this for $\mathrm{GL}_2 / \mathbf{Q}$ was asked a little while ago, here: mathoverflow.net/questions/42728/… – David Loeffler Oct 28 2010 at 11:08
## 3 Answers
If $\pi_v$ is supercuspidal, then (after making a twist if necessary) it should be possible to find an automorphic $\pi$ with $\pi_v$ as the local factor at $v$. This kind of result is usually proved (although their are sometimes other possibilities) by an application of the simple trace formula. This method will also give good (although perhaps not complete) control of the ramification at other places. If you look at the paper of Deligne, Kazhdan, and Vigneras on Jacquet--Langlands for $GL_n$, I think you will find an expose of the technique.
If $\pi_v$ is not supercuspidal (or if one is not willing to make a twist), then this is not generally possible, just for cardinality reasons. The link that David Loeffler provides to the discussion of the $GL_2$ case is somewhat indicative of the situation.
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I think you mean Deligne-Kazhdan-Vigneras. Rogawski also came up with an independent proof, and personally I find Rogawski's paper easier to read. – Kimball Oct 28 2010 at 16:34
Dear Kimball, Thanks! – Emerton Oct 28 2010 at 23:57
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
There is also a theorem of I think Hakim (my apologies if this is not Jeff's theorem), generalized by Prasad and Schulze-Pillot, that allows you to globalize representations distinguished with respect to a subgroup. This one uses a simple relative trace formula.
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Also, if $\pi_v$ is square-integrable and $G=Sp$ or $SO$, Arthur proves this in his upcoming book. See my reply to the question http://mathoverflow.net/questions/72354/embedding-of-local-representation-into-automorphic-representation/81552#81552
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http://nrich.maths.org/8586
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### Knight's Swap
Swap the stars with the moons, using only knights' moves (as on a chess board). What is the smallest number of moves possible?
### Beads and Bags
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### How Do You See It?
Here are some short problems for you to try. Talk to your friends about how you work them out.
# Dice in a Corner
##### Stage: 2 Challenge Level:
There are three dice sitting in the corner with the simple rule that where two dice meet there must be the same numbers facing each other.
So, in the first picture above there are $3$'s at the bottom of the red dice and on the top of the middle green and there are $4$'s on the bottom of the green dice and the top of the white dice. The numbers on the seven faces that can be seen are then added and make $21$.
In the second picture above there are $4$'s at the left of the red dice and on the right of the green dice and there are $3$'s on the left of the green dice and the right of the white dice. The numbers on the seven faces that can be seen are then added and make $23$.
#### Your challenge is to arrange dice (using at least $2$ and up to as many as you like) in a line from the corner, so as the faces you can see add up to $18$ (instead of the $21$ and $23$ above), in as many ways as possible.
Each line of dice must be along or up a wall (or two walls). A line going up is counted the same as a line going along. Remember the dice must touch face to face and have the same numbers touching. The lines of dice must be of a single thickness, so this one below is not allowed;
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://physics.stackexchange.com/questions/2196/anti-gravity-in-an-infinite-lattice-of-point-masses/14745
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# Anti-gravity in an infinite lattice of point masses
Another interesting infinite lattice problem I found while watching a physics documentary.
Imagine an infinite square lattice of point masses, subject to gravity. The masses involved are all $m$ and the length of each square of the lattice is $l$.
Due to the symmetries of the problem the system should be in (unstable) balance.
What happens if a mass is removed to the system? Intuition says that the other masses would be repelled by the hole in a sort of "anti-gravity".
• Is my intuition correct?
• Is it possible to derive analytically a formula for this apparent repulsion force?
• If so, is the "anti-gravity" force expressed by $F=-\frac{Gm^2}{r^2}$, where $r$ is the radial distance of a point mass from the hole?
Edit:
Video here (start at 7min): http://www.disclose.tv/action/viewvideo/45729/Stephen_Hawking__The_Story_of_Everything_pt_2_9/
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Interesting question! :) – Noldorin Dec 23 '10 at 16:17
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I think it should be correct because of superposition principle, but the sign of force might be positive. – hwlau Dec 23 '10 at 16:28
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The question is if the vector sum of all forces converge to 0 or diverge, if it diverges the question is meaningless. – Holowitz Dec 23 '10 at 16:35
@kalle43: the force on a point goes to zero as $o(r^{-2})$ as r->infinity, so the sum should converge – Sklivvz♦ Dec 23 '10 at 16:41
@Sklivvz: That's not considering it rigourously enough. This is quite a mathematically challenging question, and involves the Euler-Maclaurin formula for sums. – Noldorin Dec 23 '10 at 16:46
show 6 more comments
## 4 Answers
I think that your initial intutiion is right--before the point particle is removed, you had (an infinite set of) two $\frac{G\,m\,m}{r^{2}}$ forces balancing each other, and then you remove one of them in one element of the set. So initially, every point particle will feel a force of $\frac{G\,m\,m}{r^{2}}$ away from the hole, where $r$ is the distance to the hole. An instant after that, however, all of the particles will move, and in fact, will move in such a way that the particles closest to the hole will be closer together than the particles farther from the hole. The consequence is that the particles would start to clump in a complicated way (that I would expect to depend on the initial spacing, since that determines how much initial potential energy density there is in the system)
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This is a qualitative answer, but really just states the obvious. Doesn't help a huge amount. – Noldorin Dec 23 '10 at 16:36
@Noldorin: Well, yes, but time-evolving this problem isn't going to be simple at all, and most likely won't be analytically solveable. The many-body problem is highly nonlinear. A simple analysis of the particles with a pairwise consideration of the particles shows that you instantaneously have a hole in the system. After the initial timestep, however, that is gone, and the system will behave in an unpredictable manner. – Jerry Schirmer Dec 25 '10 at 19:20
On second look, this problem is rather intractable analytically, so your intuitive answer is actually growing on me. KennyTM provides a good reason why the quantities blow up, but I suppose there's still more to the story. – Noldorin Dec 26 '10 at 14:55
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This is incorrect--- but Newton made the same mistake, so no downvote. The constant density over all space Newtonian gravitating system is not allowed, except in an expanding or contracting Newtonian cosmology, and there is no antigravity for the missing particle. The symmetry arguments are misleading, because the limit is subtle. – Ron Maimon Sep 16 '11 at 18:55
@Jerry You followed your intuition in the last part of your answer and you are expecting a "clump in a complicated way". But the clumping will be along the outer shell of a growing 'void' as I show in my answer. Either analytically either by simulation it is what really happens. – Helder Velez Sep 19 '11 at 7:53
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This is not correct, but Newton believed this. The infinite system limit of a finite mass density leads to an ill defined problem in Newtonian gravity because $1/r^2$ falloff is balanced by density contributions of size $r^2\rho$, and there is no well defined infinite constant-mass-density system. The reason is that there is no equilibrium of infinite masses in Newtonian gravity--- you need an expanding/contracting Newtonian big-bang.
This is subtle, because symmetry leads you to believe that it is possible. This is not so, because any way you take the limit, the result does not stay put. This was only understood in Newtonian Gravity after the much more intricate General Relativistic cosmology was worked out.
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but the elapsed time, since the beginning of matter, is finite. – Helder Velez Sep 16 '11 at 19:22
but the question is asking about Newton's model, where gravity is instantaneous. – Ron Maimon Sep 17 '11 at 19:12
Why "infinite mass density" ? In my answer I made clear that your statement "there is no equilibrium of infinite masses in Newtonian gravity" is simply wrong. – Helder Velez Sep 19 '11 at 7:19
@Helder: I was using shorthand for "infinite system/finite density". There is no equilibrium of infinite masses in Newtonian gravity. They always collapse. – Ron Maimon Sep 22 '11 at 15:34
I assume by square lattice you mean a 3D cubic lattice because there's no translational symmetry along the $z$-axis for a 2D square lattice.
Suppose the masses are located at $(n_x, n_y, n_z)$ where $n_{x,y,z}\in\mathbb Z$. Let's also define the unit of mass and length so that $m=l=1$.
Consider the total force acted on the mass point at (0, 0, 0) just due to the 1st octant $(x>0,y>0,z>0)$: $$\begin{aligned} \mathbf F_{+++} &= -G \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x \hat{\mathbf x} + n_y \hat{\mathbf y} + n_z \hat{\mathbf z}}{(n_x^2+n_y^2+n_z^2)^{3/2}} \\ &= -G \left( \hat{\mathbf x} \sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} + \dotsb \right), \end{aligned}$$ however, the sum actually diverges, since, $$\begin{aligned}\sum_{n_x=1}^\infty \sum_{n_y=1}^\infty \sum_{n_z=1}^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} &\ge \int_1^\infty \int_1^\infty \int_1^\infty \frac{n_x}{(n_x^2+n_y^2+n_z^2)^{3/2}} dn_x dn_y dn_z \\ &= \int_1^\infty \int_1^\infty \frac1{\sqrt{1+n_y^2+n_z^2}} dn_y dn_z \\ &= \infty, \end{aligned}$$ so while symmetry suggests that the force at center is 0, mathematically it is not well defined.
Of course, if we assume the net force can be well-defined as 0 (e.g. the gravity actually decays faster than $1/r^3$!), then Points 1 and 3 are correct. When we remove a particle from the lattice, the contribution $-\frac{GMm\hat{\mathbf r}}{r^\alpha}$ will be subtracted from it, so it is as if there is a particle of negative mass $-m$ put to that point. This is because force is additive and gravity is proportional to mass.
(Yeah this is stating the obvious.)
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Let me play devil's advocate here: one can easily fix the problem mathematically by considering faster decaying force: $F \sim r^{\alpha}$ for alpha suitably small. Then the nature of the problem doesn't change but the math starts to converge (and would give the same answer as the intuitive one). So in this case it's obvious that computing integrals is useless math masturbation; pure physical intuition is all this problem requires. Actually this is very similar to regularizations in QFT and why mathematicians argue that QFT doesn't work. Needless to say, experiments prove them wrong :-) – Marek Dec 23 '10 at 18:31
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@Marek: Yep. But this is mainly a respond to OP's comment that "the sum should converge". – KennyTM Dec 23 '10 at 18:57
@Marek: I agree with you, and I have nothing against QFT, but (just to be fair) experiments can't prove them wrong because experiments can't prove QFT works. Experiments can only prove a theory is wrong or doesn't work. For all we know, all of the successful predictions could be a series of bizarre coincidences. ;-) – Bruce Connor Dec 23 '10 at 21:13
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Are you saying he won't come? =( – Bruce Connor Dec 23 '10 at 21:32
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@KennyTM: but you can regularize that divergence by considering the particles pairwise. Start with the central mass, and add two opposing partices at a time. The Newtonian force only depends on pairwise interactions, so it doesn't cause a problem doing it this way. Once you have built the lattice, then it clearly has translational symmetry, so any point is the same as any other point, and the solution is static. – Jerry Schirmer Dec 25 '10 at 19:17
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Your intuition is OK. This apparent «anti-gravity» is just the usual attractive gravity. The problem is treated analytically, in a light way (*), starting in the post O nascimento de uma Bolha (The birth of a Void). For convenience I grabbed a few images from the blog (CC license) and I adapted some legends.
Image A-top - The gravitational field is null everywhere in an uniform and isotropic distribution.
Image A-middle - due to fluctuations particles acquire motion, and a local defect of density, here represented by a removed particle, originates a field that is symetric of the field of the missing particle.
Image A-bottom - The acceleration of particles surrounding the point of depletion leads them to exceed the following ones; Now, every particle of its boundary will be subject to a field corresponding to all matter that is missing inside. The "Void" grows rapidly because the field grows as the "Void" grows.
Image B -- void + sphere = isodense
Image C -- The field of a void
Image D -- The field of an homogeneous sphere, mass M and radius R.
( graphs C+D = 0 )
Image E -- The fields considering the outer shell
Image F -- The acceleration of the Void.
I already did a simulation of a lattice respecting this conditions (tricky because it is an infinite universe ;-) and the result is the same as I show in this answer.
(*) a complete mathematical treatment is known to me since 1992/Oct and it will be the subject of a paper by my friend Alfredo, as the author states in the last page of his recent paper A self-similar model of the Universe unveils the nature of dark energy (no peer reviewed). It will be shown that the evolution of the initial homogeneous and uniform universe, with T=0, will match the observed large-scale structure of the universe (without DM).
## The birth of a Void:
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@all . My answer goes against all others and also against some conceptions that are deep inside our believes. If you feel that my answer is wrong I kindly ask you to leave a few words to explain your viewpoint. I'm used to be down-voted and it's your privilege. I know that I have controversial opinions that are not shared by the community but and when I saw unpleasant words like 'crackpot' and not a single argument left is offensive. I'm used to be responsible and justify my sayings and I expect the same from others. – Helder Velez Sep 19 '11 at 8:27
Hey. -1 because of first sentence. The two other answers are right. – Holowitz Sep 26 '11 at 15:02
@solomon - Thanks for the explanation. I see your point. Can you read the 2nd sentence? – Helder Velez Sep 26 '11 at 17:38
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http://mathoverflow.net/questions/46877/natural-examples-of-sequences-of-adjoint-functors/89817
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## Natural examples of sequences of adjoint functors
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I am looking for examples of sequences of adjoint functors. That are (possibly bounded) sequences $$(...,F_{-1}, F_{0}, F_1, F_2,...)$$ such that each $F_n$ is left adjoint to $F_{n+1}$. We call such a sequence cyclic of order $k$ if for one $n$ (and hence for all) we have $F_{n} \cong F_{n+k}$. It is relatively easy to prove that cyclic sequences of all orders and non-cyclic sequences of all possible length exist. This can e.g. be done using posets, see http://www.springerlink.com/content/pmj5074147116273/.
I am looing for more "natural" examples of such sequences that are as long as possible. By natural I mean that they grow out of "usual functors" (sorry for this vague statement...)
Let my give two short examples: 1) Let $U: Top \to Set$ be the forgetful functor from locally connected topological spaces to sets. This induces a sequence of length 4: $$(\pi_0 , Dis , U , CoDis)$$ where $Dis$ and $CoDis$ are the functor that equip a set with the discrete and indiscrete topology. Then the sequence stops. Tons of examples of this type are induced by pullback functors in algebraic geometry.
2) a cyclic sequence of order 2: the Diagaonal functor $\Delta: A \to A \times A$ for any abelian category $A$ is left and right adjoint to the direct sum $$( ...,\Delta,\oplus,\Delta,\oplus,...)$$
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Dorette Pronk gave a talk once on this topic. I'll see if I can dig up my notes. There is an abstract for what I think is the same talk here: cms.math.ca/Events/ete06/abs/ct.html (search for 'Pronk') – David Roberts Nov 21 2010 at 23:43
You don't actually get pi_0 in your first example. I tried this once and was corrected here: qchu.wordpress.com/2009/12/26/… . As DT says, the functor which equips a set with the discrete topology doesn't preserve infinite products, so it can't have a left adjoint. – Qiaochu Yuan Nov 21 2010 at 23:48
Oh, thats right. Thank you. I have changed it to "locally connected topological spaces" now. That should fix it. – Thomas Nikolaus Nov 21 2010 at 23:56
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Related question: mathoverflow.net/questions/9849/… I mention this just for cross-referencing purposes, since I remembered such a question but had to look for it. – José Figueroa-O'Farrill Nov 21 2010 at 23:56
I had seen that question too, but I could not give the reference, since I was somehow not allowed to place a second link in my question... – Thomas Nikolaus Nov 22 2010 at 9:43
## 8 Answers
For any category $B$ with small hom-sets one can form the yoneda embedding $y:B\to[B^{op},Set]$ (although if $B$ is not small, $[B^{op},Set]$ may not itself have small hom-sets).
Rosebrugh and Wood showed link text here that if $B$ is itself the category of of sets then there is an adjoint string $u\dashv v\dashv w\dashv x\dashv y$, and that this characterizes $Set$.
The adjunctions $\pi_0\dashv Dis\dashv U\dashv CoDis$s also work if we replace Top by any of the categories SSet of simplicial sets, Cat of categories, Gpd of groupoids, or Preord of preorders.
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That is very nice, although it causes me some trouble to reproduce each of the five functors ;) – Thomas Nikolaus Nov 22 2010 at 0:15
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The functor from the category of abelian groups to the category of arrows of abelian groups that sends an object to its identity morphism has three adjoints to the left and three to the right, for a chain of seven functors. The extreme adjoints are the functors that assign to an arrow its kernel or cokernel, as an object.
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Pardon my ignorance, but what are the morphisms in the category of arrows? – Qiaochu Yuan Nov 22 2010 at 16:04
This is really a striking-simple example. I wonder that I have not come across it so far. Morphisms in the arrow-category are commuting squares. – Thomas Nikolaus Nov 22 2010 at 16:43
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This isn't specific to Abelian groups, it works in any pointed category (i.e. a category with an object $0$ which both initial and final) having pullbacks and pushouts (where the kernel of a morphism $A \to B$ is the pullback of $A\to B \leftarrow 0$, and dually for the cokernel) – Omar Antolín-Camarena May 8 2012 at 2:40
If $f:X \to Y$ is a proper morphism of algebraic varieties, and $D(X)$, $D(Y)$ are the (unbounded) derived categories of quasicoherent sheaves then `$(f^*,f_*,f^!)$` is such a sequence of adjoint functors. If moreover, $f$ has finite Tor-dimension then $f^!(F) \cong f^* (F)\otimes f^!(O_Y)$. If moreover the relative dualizing complex $f^!(O_Y)$ is an invertible sheaf then the functor $T$ of tensoring with $f^!(O_Y)$ is an autoequivalence, hence we have an infinite sequence of adjoint functors ```$$
(\dots,T^{-1}\circ f^*,f_*\circ T,f^*,f_*,T\circ f^*,f_*\circ T^{-1},T^2\circ f^*,f_*\circ T^{-2},\dots).
$$``` The same happens for arbitrary pair of adjoint functors between categories which have Serre functors.
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A nice one from the representation theory of $p$-adic reductive groups:
Let $k$ be a finite extension of $Q_p$. Let $G$ be the $k$-points of a connected reductive group over $k$. We do not distinguish between algebraic groups over $k$ and their $k$-points in what follows. Let $P$ be a parabolic $k$-subgroup of $G$. Let $M$ be a Levi subgroup of $P$, and $N$ the unipotent radical of $P$, so $P = MN$. Let $Q$ be the opposite parabolic to $P$, so that $Q \cap P = M$. Let $U$ be the unipotent radical of $Q$, so $Q = MU$.
Let $Rep(G)$ and $Rep(M)$ denote the categories of smooth representations of $G$ and $M$, respectively.
Let $R_P^G$ (respectively $R_Q^G$) denote Jacquet's restriction from $Rep(G)$ to $Rep(M)$, taking a representation $V$ of $G$ to its $N$-coinvariants $V_N$ (resp. $U$-coinvariants $V_U$), viewed as a representation of $M$.
Let $I_P^G$ (respectively $I_Q^G$) denote Jacquet's induction, from $Rep(M)$ to $Rep(G)$, defined by extending a representation from $M$ to $P$ (resp. $Q$), then inducing smoothly.
Then there are the following adjointnesses, for $V$ a smooth rep of $G$ and $W$ a smooth rep of $M$: $$Hom_M(R_P^G(V), W) \cong Hom_G(V, I_P^G(W)).$$ $$Hom_G(I_P^G(W), V) \cong Hom_M(W, R_Q^G(V)).$$ $$Hom_M(R_Q^G(V), W) \cong Hom_G(V, I_Q^G(W)).$$ $$Hom_G(I_Q^G(W), V) \cong Hom_M(W, R_P^G(V)).$$
The cyclic sequence of functors is: $$R_P^G, I_P^G, R_Q^G, I_Q^G.$$ The adjointnesses in the second and fourth come from Bernstein's "Second Adjointness Theorem" -- a highly nontrivial result!
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Is the distinction between $P$ and $Q$ not rather trivial, since they are conjugated. But nice example! – Marc Palm Feb 28 2012 at 18:48
@Marc: No, for example the (2,1) parabolic in GL(3) is not conjugate to its opposite. – fherzig Apr 3 2012 at 16:03
There are some really nice answers here. Here's another contribution.
Let [n] denote the totally ordered (n+1)-element set, regarded as a category. For each positive integer n, we have the usual n+1 order-preserving injections from [n-1] to [n], and the usual n order-preserving surjections from [n] to [n-1]. (I mean the ones used all the time for simplicial anything.) When you regard them as functors, these injections and surjections interleave to form an adjoint chain of length 2n.
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I hadn't seen this question before, but this is almost a canonical answer. Just as $\Delta$ is the "generic monoid", you could call the 2-category $\Delta$ the "generic Kock-Zoeberlein monad", where the multiplication is left adjoint to a unit. – Todd Trimble Feb 29 2012 at 2:59
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Yeah, now that I think about it, this is probably one of the examples used in the paper Thomas cites. (I only just noticed that he talked about it being "done using posets".) But what's noticeable, I think, is that the functors involved come up naturally in non-categorical mathematics, e.g. the definition of singular homology. – Tom Leinster Feb 29 2012 at 3:08
@Tom: Thats exactly how they do it in the paper (of course with some modifications). – Thomas Nikolaus Mar 5 2012 at 2:27
Similar to Ben Wieland's and Sasha's answer: Let $\mathcal{C}$ be the category of complexes in an abelian category. Let $\underline{\mathcal{C}(\Delta_0)}$ be the homotopy category of $\mathcal{C}$. Let $\underline{\mathcal{C}(\Delta_1)}$ be the category of arrows with values in $\mathcal{C}$, with pointwise homotopy equivalences formally inverted (not to be confused with $\underline{\mathcal{C}(\Delta_0)}(\Delta_1)$). The functor $\underline{\mathcal{C}(\Delta_0)}\to\underline{\mathcal{C}(\Delta_1)}$ that sends an object of $\mathcal{C}$ to the identity arrow on this object, morphisms accordingly, gives rise to an infinite chain of adjoint functors ("walking along a distinguished triangle"). [Which functors, such as $\Delta_1\to\Delta_0$ considered here, have this property?]
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A simple version of your first example is to look at the forgetful functor $U : \text{Graph} \to \text{Set}$, where $\text{Graph}$ is, say, the category of simple undirected graphs. (To be explicit, the morphisms in this category are maps of vertices which respect the edge relation, and in the edge relation a vertex is considered related to itself.) This functor has a left adjoint $E : \text{Set} \to \text{Graph}$ which sends a set to the empty graph on that set and a right adjoint $K : \text{Set} \to \text{Graph}$ which sends a set to the complete graph on that set. $K$ doesn't preserve coproducts, so it doesn't have a right adjoint. $E$ has a left adjoint $\pi_0 : \text{Graph} \to \text{Set}$ which sends a graph to its set of connected components (no topological difficulties here). I am not sure whether $\pi_0$ has a left adjoint in this case.
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$\pi_0$ does not preserve the equalizer of the two maps from $K_1$ to $K_2$. – Steve Lack Nov 22 2010 at 0:09
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so it doesn't have a left adjoint – Steve Lack Nov 22 2010 at 0:10
(Sorry for the bump, everyone, but I only just saw this question.)
Here's an example similar in feel to the $\operatorname{Dis} \dashv U \dashv \operatorname{Codiss}$ example - so perhaps not of the sort that you were really after.
Let ${\bf Op}_1$ be the category whose objects are (complete) operator spaces (or "quantum/quantized Banach spaces" according to some authors) and whose morphisms are the completely contractive maps. Let ${\bf Ban}_1$ be the category of Banach spaces and contractive (a.k.a. short) linear maps. Then if $U:{\bf Op}_1\to{\bf Ban}_1$ is the forgetful functor, we have adjunctions $\operatorname{MAX} \dashv U \dashv \operatorname{MIN}$.
The left and right adjoints to $U$ are sometimes called the maximal and minimal quantizations, respectively, of a Banach space. (One also sees the terminology of "maximal and miimal operator space structures", but then we wouldn't be able to have the magic word quantum and its important-sounding derivatives.)
See Prop 3.3 of this article by Pestov for a brief mention of left adjoints and MAX.
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http://mathoverflow.net/questions/106544?sort=oldest
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## Partitions of $\mathbb{R}^d$ by implicit polynomial equations
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Given a polynomial $p(x_1,x_2,\ldots,x_d)$ in $d$ variables, with maximum degree $k$, what is the maximum number of components of $\mathbb{R}^d$ minus $p(\ldots)=0$? In other words, into how many pieces can an implicit polynomial equation partition $\mathbb{R}^d$?
For example, the following three equations partition $\mathbb{R}^2$ or $\mathbb{R}^3$ into $3$, $4$, and $2$ pieces respectively (I think!):
$$x^3 y^2+x^3 -3 x^2 y -y^2 +4 x y+x=0$$
$$x^6 y^8+x^3+4 x y-y=0$$
$$x^4+3 \left(x^2+y^4+z\right)- \left(x^2+y^2+z^2\right)^2+y^3+z^5 + 2 xy=3$$
Of course the answer is $k+1$ in $\mathbb{R}^1$. I suspect this is well known for $\mathbb{R}^d$; if so, I would appreciate a pointer. Thanks!
Update. Greg Martin's idea (from the comments), using the 5th Chebyshev polynomial of the first kind:
As Aaron Meyerowitz points out, here the degree $k=10$, and the plane is partitioned into $28$ pieces. But using Pietro Majer's line-arrangement idea leads to (now corrected:) $56$ pieces for a degree $10$ polynomial.
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An observation, which can probably be generalized: let $T_n$ be the $n$th Chebyshev polynomial of the first kind. For $n$ odd, consider the polynomial $(y-T_n(x))(x-T_n(y))$. The graph of the zero locus of this degree-$2n$ polynomial splits the plane into $n^2+3$ pieces (4 unbounded pieces and $n^2-1$ bounded pieces). The bounded pieces come up because the graph of $y-T_n(x)=0$ oscillates up and down $n$ times inside the unit box $[-1,1]^2$, while $x-T_n(x)$ oscillates $n$ times side to side. See: en.wikipedia.org/wiki/… – Greg Martin Sep 6 at 23:15
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If $p$ is a product of $k$ affine functions in $\mathbb{R}^d$, its zero-set is the union of $k$ affine hyperplanes. If these are in generic position, the number of components of the co-set is the well-known $\sum_{j=0}^d\binom kj$. I'd conjecture this is not worse than any polynomial of degree $k$. Note that $2n$ lines disconnect the plane into $2n^2+n+1\ge n^2+3$ components. – Pietro Majer Sep 7 at 0:42
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I'm pretty sure it is $56$ and not $58$ regions. I may have heard once that this is optimal. You lose something in not having curved borders which can multiply intersect but gain more in having lots of linear borders heading off in many directions. Look at the quintic factor $y-T_5(x).$ Wouldn't you rather straighten the rounded turns and replace them with $5$ lines going off to infinity? The $\pm \epsilon$ adjustment gives a prettier picture but then the number of regions is $n^2+O(n).$ – Aaron Meyerowitz Sep 8 at 3:17
See the edit to my response for a sharper (in fact, I am pretty sure it is sharp) bound. – Igor Rivin Sep 8 at 15:40
## 4 Answers
In some contexts (for example, in the study of spherical harmonics), the connected components of the complement of the zero set of a polynomial are called nodal domains.
The maximum number of nodal domains in the real projective plane of a polynomial of degree $d$ (i.e. a homogeneous polynomial in $\mathbb{R}^3$) is bounded above by $d(d-1)+2.$ A nice exposition of this result can be found Leydold's paper On the number of nodal domains of spherical harmonics.
A related result is Harnack's curve theorem. It says that the number of connected components of the zero set of a polynomial in the real projective plane is bounded by $(d-1)(d-2)/2+1$.
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@Matthew: Thanks! I didn't know this theorem. But I am confused because the Chebyshev example I posted has $d=6$, and so should have at most $11$ regions, but I count $24$ bounded regions... – Joseph O'Rourke Sep 7 at 1:28
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@Joseph: Oops! I confused components of the curve and components of the complement of the curve. I've updated the answer to fix this. – Matthew Badger Sep 7 at 3:32
@Matthew: isn't your latter bound shifted by one? For a 2-variables real polynomial of degree $d$, I'd say at least $d(d-1)/2+1$ components in the real projective plane (and $d(d+1)/2+1$ in the affine plane), which is reached if the polynomial is a product of linear terms (see my comment above). – Pietro Majer Sep 7 at 8:01
(And, as Aaron pointed out, $d=10$, not $6$ as I claimed above.) – Joseph O'Rourke Sep 7 at 11:52
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You are asking for the number of components of the semialgebraic set defined by $P\geq 0.$ It is a classical result that the number of connected components is of order $O(k^d).$ In the below paper, the author gives all the relevant references, and extends the results to estimating higher Betti numbers.
Basu, Saugata(1-GAIT-CC) Different bounds on the different Betti numbers of semi-algebraic sets. (English summary) ACM Symposium on Computational Geometry (Medford, MA, 2001).
EDIT In fact, the result is much stronger than a Big-O result: Milnor (in his paper MR0161339 (28 #4547) Reviewed Milnor, J. On the Betti numbers of real varieties) Shows that in this situation the upper bound is $\frac12(2+k)(1+k)^{d-1}.$
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I think Joseph is asking for the semi algebraic set that is the union of $P>0$ and $P<0$. Nonetheless, I was also wondering if work of Basu would be the place to look. – Patricia Hersh Sep 7 at 3:08
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@Patricia: I am sure he is, but once you have $O(),$ what's a factor of two among friends... – Igor Rivin Sep 7 at 13:28
@Igor: good point. My real concern was that throwing in $P=0$ as your statement does might reduce the number of connected components. I also wondered if the fact that Joseph is actually considering a Zariski open set might mean that there are also relevant results in algebraic geometry. – Patricia Hersh Sep 7 at 13:57
@Patricia: No, I don't think this reduces the number of connected components, which should be equal to the number of components of $P\geq 0$ plus the number for $P\leq 0$ but I am not 100% certain -- it is conceivable that one has two components meeting at a point, or some lower dimensional stratum -- presumably one can bound the number of such occurrences but a lower order term than the main term in the bound... – Igor Rivin Sep 7 at 14:36
@Igor: Very useful to know the asymptotic growth rate. (I should have thought of Basu's work myself.) Thanks! – Joseph O'Rourke Sep 7 at 15:42
show 1 more comment
Your example of $f(x,y)=\left(y-T_5(x)\right)\left(x-T_5(y)\right)$ is ```$$ 256\,{x}^{5}{y}^{5}-320\,{x}^{5}{y}^{3}-320\,{x}^{3}{y}^{5}-16\,{x}^{6
}+80\,{x}^{5}y+400\,{x}^{3}{y}^{3}+80\, x{y}^{5}\\-16\,{y}^{6}+20\,{x}^{4
}-100\,{x}^{3}y-100\,x{y}^{3}+20\,{y}^{4}-5\,{x}^{2}+26\,xy-5\,{y}^{2}
$$```
If I count correctly, it has $28$ regions. Would that qualify as maximum degree $10?$ If so, then as Pietro points out, $g(x,y)=\prod_{i=1}^{10}\left(a_ix+b_iy+c_i\right)$ will have $56$ regions (if $a_i,b_i,c_i$ are such that the $10$ lines are in general position: no two parallel and no three meeting at a common point). Similar things (as he says) can be done with hyperplane arrangements in higher dimension. You can color the regions $g \gt 0$ white and $g \lt 0$ black so that each region is bounded by regions of the opposite color.
If you want the curve itself to have many disjoint connected components then $g(x,y)+\epsilon$ and $g(x,y)-\epsilon$ are nice to look at. Then all the regions of one color fuse together but each of the others becomes a nicely bordered region. I think that (in the two variable case with a projective viewpoint, at any rate) these achieve that bound given by Harnack's theorem.
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A slightly more general result -- namely on the number of connected components of the non-zeros of a polynomial, restricted to the zeros of another polynomial -- where the degrees of the two polynomials could be different, can be deduced from the main theorem in the paper titled
"Refined Bounds on the Number of Connected Components of Sign Conditions on a Variety" by Barone and Basu, DISCRETE & COMPUTATIONAL GEOMETRY Volume 47, Number 3 (2012), 577-597, DOI: 10.1007/s00454-011-9391-3
Note that this is a bound only on the number of connected components and not on he higher Betti numbers. I don't know if the added generality is helpful to the original poster.
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Welcome to MO! Perhaps you can address @Patricia's question: The Oleinik/Milnor/Thom bound is for the number of connected components of $P\geq 0.$ Does anything particularly unpleasant happen if the inequalities are strict? – Igor Rivin Sep 9 at 2:00
Or is the "semi-algebraically connected component" in fact the open set? – Igor Rivin Sep 9 at 2:07
The number of connected components of $P> 0$ is the same as the number for $P\geq \epsilon$ for $\epsilon$ sufficiently small. This doesn't work as well for two-sided bounds but if it's one-sided I don't think that's a big problem – Will Sawin Sep 9 at 4:52
It could happen of course that $P>0$ is empty, even though $P\geq 0$ is not. However, the reduction mentioned above works to give the same bound. If one is interested in bounding the number of connected components of the basic semi-algebraic defined by $P_1 > 0,\ldots,P_s >0$, then a slightly better bound (better dependence on $s$) is available in the case where the inequalities are strict. Need to thank Patricia Hersh for pointing me to this discussion since I don't follow this forum regularly. – Saugata Basu Sep 9 at 12:04
In fact, looking at the Milnor paper, that seems to be exactly what he does (jiggle the $\epsilon$ a bit...) – Igor Rivin Sep 10 at 1:32
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http://nrich.maths.org/1836/index
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### Doodles
A 'doodle' is a closed intersecting curve drawn without taking pencil from paper. Only two lines cross at each intersection or vertex (never 3), that is the vertex points must be 'double points' not 'triple points'. Number the vertex points in any order. Starting at any point on the doodle, trace it until you get back to where you started. Write down the numbers of the vertices as you pass through them. So you have a [not necessarily unique] list of numbers for each doodle. Prove that 1)each vertex number in a list occurs twice. [easy!] 2)between each pair of vertex numbers in a list there are an even number of other numbers [hard!]
### Russian Cubes
How many different cubes can be painted with three blue faces and three red faces? A boy (using blue) and a girl (using red) paint the faces of a cube in turn so that the six faces are painted in order 'blue then red then blue then red then blue then red'. Having finished one cube, they begin to paint the next one. Prove that the girl can choose the faces she paints so as to make the second cube the same as the first.
### N000ughty Thoughts
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!
# Counting Binary Ops
##### Stage: 4 Challenge Level:
This question is about the number of ways of combining an ordered list of terms by repeating a single binary operation.
For example with three terms $a$, $b$ and $c$ there are just two ways $((a\oplus b)\oplus c)$ and $(a\oplus (b\oplus c))$. Suppose the binary operation $\oplus$ is just ordinary subtraction and $a=12, \; b=7, \; c=5$ then $((a\oplus b)\oplus c)= 5 - 5 =0$ and $(a\oplus (b\oplus c))= 12 - 2 = 10$.
We are not concerned in this question with doing the 'arithmetic' or with whether the answers are the same or different. We just want to find out how many ways there are of combining the terms, or if you like of putting brackets into the expression. Note that we need the brackets because the answers may be different as in the subtraction example.
These two tree diagrams show the two cases for combining 3 terms."
Show that for four terms, (three binary operations) there are five cases and find the number of cases for five terms and six terms.
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http://mathoverflow.net/questions/100991?sort=oldest
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Are there any neat algorithm to factor a homogenous polynomial, given we know this polynomial factors into linear forms?
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Are there any neat algorithm to factor a multivariable (more than 2 variables) homogenous polynomial, given we know this polynomial factors into linear forms?
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3 Answers
As Denis Serre points out, this problem subsumes the problem of factoring single variable polynomials, so I'm going to assume that you already have software on hand to do that (probably numerically). In that case, here's what I'd do. For simplicity, presented first in the case of a homogenous polynomial $F(x,y,z)$ in three variables, so a curve in the projective plane.
Pick a line $L$ in the plane and find out where it meets your curve. This amounts to solving a single variable polynomial equation $F(x_0+at, y_0+bt, z_0+ct)=0$ for some constants $(x_0, y_0, z_0$ and $(a,b,c)$. If any of the roots are extremely close to each other, this suggests that $L$ is passing very close to where two of your lines cross; discard this data and try again with another $L$. So now we have $n$ points $z_1$, $z_2$, ..., $z_n$ in $\mathbb{P}^2$, one on each of your lines.
Repeat with a second line $M$ to get points $w_1$, $w_2$, ..., $w_n$. Then, for each $(z_i, w_j)$, see whether your polynomial vanishes on the line through $z_i$ and $w_j$. This is just a matter of plugging $z_i + t w_j$ into $F$. Remember that, when you work with floating point numbers, you should never check whether something equals zero, but rather whether it is very small.
Because we chose the $z_i$ and $w_i$ well away from the intersections of your lines, the restriction of $F$ to $z_i + t w_j$ should not be particularly small if $z_i$ and $w_j$ do not lie on the same line. So you can figure out which $z_i$ lies on a factor of $F$ with which $w_j$, and then recover that factor by linear algebra.
For a polynomial in $d+1$ variables, use $d$ lines to get $n d$ points. Computing the restriction of $F$ to $(d-1) n^2$ lines will then tell you which of your points to group together.
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Why are you working with floating point numbers? – Igor Rivin Jun 30 at 23:53
In my experience, working with numeric solutions to polynomials is significantly faster than passing to the number field where they have roots, once the degree gets past $4$ or $5$. – David Speyer Jul 1 at 1:17
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
No. Take a one-variable polynomial $P(X)$. Make it homogenous: $P(X)=Q(X,1)$. Pretend that $Q(X,Y)$ is a polynomial in $3$ variables $X,Y,Z$. If the field $k$ is algebraically closed, you know that $Q$ splits. If you have that algorithm, then you find the factors and this gives you the roots of $P$. This contradicts Abel's theorem that the general polynomial of degree $\ge5$ is not solvable.
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This doesn't make sense, as the algorithm doesn't need to output the roots in radical form. This does raise the question however, of what form of input and output does the OP expect. – Felipe Voloch Jun 30 at 13:07
I think this answer is valuable; it is unclear what the OP wants and this certainly answers one interpretation. – David Speyer Jun 30 at 15:50
There are a number of ways to factor multivariate polynomials, and the problem is greatly muddled by the fact that a general multivariate polynomial in $n$ variables of degree $d$ has something like $d^n$ terms, so any algorithm designed for general polynomials is likely to be useless in practice, so one needs to keep track of the sparsity of the intermediate results, etc. Most algorithms I am aware of use Hensel lifting in some form, see this paper for example (Inaba, Factorizations of Multivariate polynomials, 2005). Whether this method is "neat" is in the eye of the beholder.
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http://www.nag.com/numeric/cl/nagdoc_cl23/html/G08/g08ecc.html
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# NAG Library Function Documentnag_triplets_test (g08ecc)
## 1 Purpose
nag_triplets_test (g08ecc) performs the triplets test on a sequence of observations from the interval $\left[0,1\right]$.
## 2 Specification
#include <nag.h>
#include <nagg08.h>
void nag_triplets_test (Integer n, const double x[], Integer max_count, double *chi, double *df, double *prob, NagError *fail)
## 3 Description
nag_triplets_test (g08ecc) computes the statistics for performing a triplets test which may be used to investigate deviations from randomness in a sequence of $\left[0,1\right]$ observations.
An $m$ by $m$ matrix, $C$, of counts is formed as follows. The element ${c}_{jkl}$ of $C$ is the number of triplets (x$\left(\mathit{i}\right)$, x$\left(\mathit{i}+1\right)$, x$\left(\mathit{i}+2\right)$), for $\mathit{i}=1,4,\dots ,n-2$, such that
$j-1m≤Xi<jm$
$k-1m≤Xi+1<km$
$l-1m≤Xi+2<lm.$
Note that all triplets formed are non-overlapping and are thus independent under the assumption of randomness.
Under the assumption that the sequence is random, the expected number of triplets for each class (i.e., each element of the count matrix) is the same, that is the triplets should be uniformly distributed over the unit cube ${\left[0,1\right]}^{3}$. Thus the expected number of triplets for each class is just the total number of triplets, ${\sum }_{j,k,l=1}^{m}{c}_{jkl}$, divided by the number of classes, ${m}^{3}$.
The ${\chi }^{2}$ test statistic used to test the hypothesis of randomness is defined as:
$X 2 = ∑ j , k , l = 1 m c jkl - e 2 e$
where $e={\sum }_{j,k,l=1}^{m}{c}_{jkl}/{m}^{3}=\text{}$ expected number of triplets in each class.
The use of the ${\chi }^{2}$ distribution as an approximation to the exact distribution of the test statistic, ${X}^{2}$, improves as the length of the sequence relative to $m$ increases, hence the expected value, $e$, increases.
## 4 References
Dagpunar J (1988) Principles of Random Variate Generation Oxford University Press
Knuth D E (1981) The Art of Computer Programming (Volume 2) (2nd Edition) Addison–Wesley
Morgan B J T (1984) Elements of Simulation Chapman and Hall
Ripley B D (1987) Stochastic Simulation Wiley
## 5 Arguments
1: n – IntegerInput
On entry: $n$, the number of observations.
Constraint: ${\mathbf{n}}\ge 3$.
2: x[n] – const doubleInput
On entry: the sequence of observations.
Constraint: $0.0\le {\mathbf{x}}\left[\mathit{i}-1\right]\le 1.0$, for $\mathit{i}=1,2,\dots ,n$.
3: max_count – IntegerInput
On entry: the size of the count matrix to be formed, $m$.
Constraint: ${\mathbf{max_count}}\ge 2$.
4: chi – double *Output
On exit: contains the ${\chi }^{2}$ test statistic, ${X}^{2}$, for testing the null hypothesis of randomness.
5: df – double *Output
On exit: contains the degrees of freedom for the ${\chi }^{2}$ statistic.
6: prob – double *Output
On exit: contains the upper tail probability associated with the ${\chi }^{2}$ test statistic, i.e., the significance level.
7: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
NE_G08EC_CELL
The expected value for the counts in each element of the count matrix is less than or equal to 5.0. This implies that the ${\chi }^{2}$ distribution may not be a very good approximation to the test statistic.
NE_G08EC_TRIPLETS
No triplets were found because less than 3 observations were provided in total.
NE_INT_ARG_LE
On entry, max_count must not be less than or equal to 1: ${\mathbf{max_count}}=〈\mathit{\text{value}}〉$.
NE_INT_ARG_LT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 3$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_REAL_ARRAY_CONS
On entry, ${\mathbf{x}}\left[〈\mathit{\text{value}}〉\right]=〈\mathit{\text{value}}〉$.
Constraint: $0<{\mathbf{x}}\left[\mathit{i}\right]<1.0$, for $\mathit{i}=0,1,\dots ,n-1$.
## 7 Accuracy
The computations are believed to be stable. The computations of prob given the values of chi and df will obtain a relative accuracy of five significant figures for most cases.
## 8 Further Comments
The time taken by nag_triplets_test (g08ecc) increases with the number of observations, $n$.
## 9 Example
The following program performs the pairs test on 10000 pseudorandom numbers from a uniform distribution $U\left(0,1\right)$ generated by nag_rand_basic (g05sac). nag_triplets_test (g08ecc) is called with max_count set to 5.
### 9.1 Program Text
Program Text (g08ecce.c)
None.
### 9.3 Program Results
Program Results (g08ecce.r)
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http://math.stackexchange.com/questions/73457/the-convergence-of-fourier-series
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# the convergence of Fourier series
Assume now we have $f(x)\in L^1([0,1])$, then we don't necessarily have the convergence of the partial sum of the Fourier series, moreover, by the theorem of kolmogorov, we can even have a.e. divergence of the partial sum.
Now my question is, for $f(x)\in L^1([0,1])$, denote the Fourier Transform as $\{a_n\}_{n=-\infty}^{\infty}$, and assume that the partial sum $S_n(x)=\sum_{k=-n}^{n}a_ke^{ikx}$ converges pointwise almost everywhere in $[0,1]$, then can we expect that the partial sum will converge back to the original function $f(x)$?
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A Fourier transform is different from a Fourier sequence. – Thomas Andrews Oct 17 '11 at 21:57
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@Thomas Andrews: but I think in some general sense, there's no problem to call it like this, or? – bonnnnn2010 Oct 17 '11 at 22:00
No, in the "general sense," the Fourier transform of a function is different from the sequence of coefficients of the Fourier series of that function. In mathematics, words are precise things. You can argue, perhaps, that the Fourier series of a function is somehow akin to the Fourier transform, but then the Fourier series is "a (generalized) Fourier transform" of the function, not "the Fourier transform" of that function. – Thomas Andrews Oct 18 '11 at 17:00
## 1 Answer
The Cesaro means $\sigma_n(x)$ of the Fourier series of an $L^1$ function $f(x)$ converge almost everywhere to $f(x)$. At any point where the partial sums $S_n(s)$ converge, the Cesaro means $\sigma_n(x)$ converge to the same value. So if the partial sums converge pointwise almost everywhere, the limit must almost everywhere be $f(x)$. Of course, as user16892 noted, the limit might not be $f(x)$ everywhere (but that's obvious, because you can change an $L^1$ function on a set of measure 0 and not affect the Fourier series).
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nice observation, thanks a lot, but pity that I'm not able to vote up – bonnnnn2010 Oct 17 '11 at 22:02
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http://mathoverflow.net/questions/63490/maximum-of-gaussian-random-variables/69705
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## Maximum of Gaussian Random Variables
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Let $x_1,x_2,…,x_n$ be zero mean Gaussian random variables with covariance matrix $\Sigma=(\sigma_{ij})_{1\leq i,j\leq n}$.
Let $m$ be the maximum of the random variables $x_{i}$
m=\max\{x_i:i=1,2,\ldots,n\}
What can one say about $m$? Can we at least compute its mean and variance?
More specifically the problem that I'm interested is the following. Consider a triangular array of random variables where the $n$-th row looks like $$x_{1}^{(n)},x_{2}^{(n)},\ldots,x_{n}^{(n)}$$ and all the random variables are zero mean and Gaussian. Moreover, $$\mathbb{Var}(x_{i}^{(n)})=1 \quad \text{for all $1\leq i\leq n$}$$ and $$\mathbb{Var}(x_{i}^{(n)}x_{j}^{(n)})=\sigma_{ij}(n)\to 0\quad \text{as $n$ increases for $i\neq j$.}$$
Is there anything that can be said about the behavior of $m$ asymptotically?
Thanks!
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## 3 Answers
If the correlations decay fast enough $\sigma_{ij}(n) = o(1/\log n)$, then the asymptotic distribution of the maximum is the same as if the variables were independent (i.e. the standard Gumbel distribution) - see:
Limit Theorems for the Maximum Term in Stationary Sequences, S.M. Berman (Ann. Math. Statist. 1964) http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.aoms/1177703551
and also: On the asymptotic joint distribution of the sum and maximum of stationary normal random variables H.C. Ho and T. Hsing (Journal of applied probability, 1996). http://www.jstor.org/pss/3215271
For the general case (correlations decay slower or not at all) I don't know of exact results for the limit, but there is a work showing how to compute bounds on the expectation for finite $n$:
Useful Bounds on the Expected Maximum of Correlated Normal Variables, A.M. Ross (2003) http://people.emich.edu/aross15/q/papers/bounds_Emax.pdf
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Thanks for the references Or! I will take a careful look at these papers. – ght May 1 2011 at 12:33
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
See: On the distribution of the maximum of random variables, by J. Galambos (Annals of Math. Stat, 1972). For your convenience, the pdf is here.
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@Igor: Thanks for the paper! However, I'm not so sure that this helps with the triangular array problem though. What do you think? It might be possible that I'm missing something. – ght Apr 30 2011 at 13:48
I did not think about the triangular array thing, but would think that it could be gotten out the reference with some work. Maybe @Or Zuk'z Ross reference does it? I did not check yet... – Igor Rivin Apr 30 2011 at 15:55
C.E.Clark's paper on Maximum of a finite set of random variables provides a reasonable closed form approximation. You can always write max(x1,x2,x3) as max(x1,max(x2,x3)). Clark's paper basically uses this fact and tries to create a chain for finite number of variables
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http://stats.stackexchange.com/questions/38218/low-classification-accuracy-what-to-do-next
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# Low classification accuracy, what to do next?
So, I'm a newbie in ML field and I try to do some classification. My goal is to predict the outcome of a sport event. I've gathered some historical data and now try to train a classifier. I got around 1200 samples, 0.2 of them I split off for test purposes, others I put into grid search (cross-validation included) with different classifiers. I've tried SVM with linear, rbf and polynominal kernels and Random Forests to the moment. Unfortunately, I can not get accuracy significantly larger than 0.5 (the same as random choice of class). Does it mean I just can't predict outcome of such a complex event? Or I can get at least 0.7-0.8 accuracy? If it's feasible, then what should I look into next?
• Get more data? (I can enlarge dataset up to 5 times)
• Try different classifiers? (Logistic regression, kNN, etc)
• Reevaluate my feature set? Are there any ML-tools to analyze, which features make sense and which don't? Maybe, I should reduce my feature set (currently I have 12 features)?
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What is your training accuracy? And how many samples you have in each class? – Leo Sep 28 '12 at 23:24
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What sport is this and what do you consider a "correct" classification? If you're simply trying to predict a win/loss outcome in virtually any major sport it's almost inconceivable that even the simplest of classifiers wouldn't predict better than 0.5. If you are, say, trying to predict win/loss against a spread or some other handicapped outcome, then much better than 0.5 may be difficult. – cardinal Sep 29 '12 at 14:31
@Leo Training accuracy is around 0.5. Classes are evenly distributed, I have classes 0 and 1. – fspirit Sep 29 '12 at 15:36
@cardinal Yes, I try to predict win/loss outcome, no handicaps. Is it feasible to reach, say, 0.8 accuracy on test set? – fspirit Sep 29 '12 at 16:00
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@fspirit: That depends on the sport and the inequity in ability between the participants, for one thing. Just knowing who is participating in each contest can often be a strong predictor. Here and here are a couple of related posts. – cardinal Sep 29 '12 at 16:13
show 2 more comments
## 4 Answers
First of all, if your classifier doesn't do better than a random choice, there is a risk that there simply is no connection between features and class. A good question to ask yourself in such a position, is weather you or a domain expert could infer the class (with an accuracy greater than a random classifier) based on given features. If no, then getting more data rows or changing the classifier won't help. What you need to do is get more data using different features.
IF on the other hand you think the information needed to infer the class is already in the labels, you should check whether your classifier suffers from a high bias or high variance problem.
To do this, graph the validation error and training set error, as a function of training examples.
If the lines seem to converge to the same value and are close at the end, then your classifier has high bias and adding more data won't help. A good idea in this case is to either change the classifier for a one that has higher variance, or simply lower the regularization parameter of your current one.
If on the other hand the lines are quite far apart, and you have a low training set error but high validation error, then your classifier has too high variance. In this case getting more data is very likely to help. If after getting more data the variance will still be too high, you can increase the regularization parameter.
This are the general rules I would use when faced with a problem like yours.
Cheers.
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Did you mean validation set error and TEST set error? Otherwise, I'm confused. I dont even know train set error, cause I use validation set error to choose model and them check selected model on test set. – fspirit Sep 29 '12 at 15:40
– sjm.majewski Sep 29 '12 at 19:45
I've competed the course, when it was ran for the first time. As for the training set error, I now output it too, for SVM its quite high - 0.5, but for random forests its 0. – fspirit Oct 2 '12 at 7:35
It's good that you separated your data into the training data and test data.
Did your training error go down when you trained? If not, then you may have a bug in your training algorithm. You expect the error on your test set to be greater than the error on your training set, so if you have an unacceptably high error on your training set there is little hope of success.
Getting rid of features can avoid some types of overfitting. However, it should not improve the error on your training set. A low error on your training set and a high error on your test set might be an indication that you overfit using an overly flexible feature set. However, it is safer to check this through cross-validation than on your test set. Once you select your feature set based on your test set, it is no longer valid as a test set.
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I use separate train, validation and test sets. I select hyper-params based on validation set error and then apply selected model to the test set. I doubt there is a bug in training algorithm, because I use off-the-shelf lib. – fspirit Sep 29 '12 at 15:45
You had to connect that library to your data somehow. Always check that you are training correctly. If you are getting a training error rate of $50\%$, this either means your features are terrible, or else you are not training correctly. – Douglas Zare Sep 29 '12 at 20:00
In the "features are terrible" possibility, I include the case that there is no solution possible. However, I doubt that very much. There is no sport I know where there aren't ways to see that one competitor is a favorite over another. It is even possible in rock-paper-scissors. – Douglas Zare Sep 29 '12 at 20:31
Why not follow the principle "look at plots of the data first". One thing you can do is a 2 D scatterplot of the two class conditional densities for two covariates. If you look at these and see practically no separation that could indicate lack of predictability and you can do this with all the covariates. That gives you some ideas about the ability to use these covariates to predict. If you see some hope that these variables can separate a little then start thinking about linear discriminants, quadratic discriminants, kernel discrimination, regularization, tree classification, SVM etc.
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Sorry, um, is covariate == feature? – fspirit Sep 29 '12 at 15:50
I would suggest taking a step back and doing some exploratory data analysis prior to attempting classification. It is worth examining your features on an individual basis to see if there is any relationship with the outcome of interest - it may that the features you have do not have any association with the class labels. How do you know if the features you have will be any use?
You could start with doing hypothesis testing or correlation analysis to test for relationships. Generating class specific histograms for features (i.e. plotting histograms of the data for each class, for a given feature on the same axis) can also be a good way to show if a feature discriminates well between the two classes.
It is important to remember though not to let the results of your exploratory analysis influence your choices for classification. Choosing features for classification based on a prior exploratory analysis on the same data, can lead to overfitting and biased performance estimates (see discussion here) but an exploratory analysis will at least give you an idea of whether the task you are trying to do is even possible.
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I'll try to draw the histograms and see what they will look like. – fspirit Sep 29 '12 at 15:58
@BGreene - your third paragraph is a tough one for me. If exploratory analysis shows predictor x1 to be highly correlated with the outcome, wouldn't it defeat the purpose of checking that correlation if one didn't use x1 as at least a candidate predictor in a multivariate model? – rolando2 Sep 29 '12 at 21:06
@rolando2 - I'm not suggesting that you don't include the feature as a candidate as part of a feature selection routine but you should not choose features based on such an exploratory analysis as this will overfit. However for the purposes of evaluating the generalized performance of a classifier model, feature selection should be done within the model selection routine (i.e. within each fold of cross validation). What I am suggesting is that exploratory analysis and classification should be treated as separate activities - each tells you different things about your data – BGreene Sep 29 '12 at 22:47
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http://physics.stackexchange.com/questions/27176/significance-of-the-hyperfinite-iii-1-factor-for-axiomatic-quantum-field-theor?answertab=active
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# Significance of the hyperfinite $III_1$ factor for axiomatic quantum field theory
Using a form of the Haag-Kastler axioms for quantum field theory (see AQFT on the nLab for more details), it is possible in quite general contexts to prove that all local algebras are isomorphic to the hyperfinite $III_1$ factor or to the tensor product of the $III_1$ factor with the center of the given local algebra.
(A local algebra is the algebra of observables that is associated to an open bounded subset of Minkowski space. The term $III_1$ factor refers to the Murray-von Neumann classification of factors of von Neumann algebras).
Also see this question on math overflow for more details.
So one could say that quantum mechanics has the $I_n$ and $I_{\infty}$ factors as playground, while QFT has the hyperfinite $III_1$ factor as playground.
My questions has two parts:
1) I would like to know about a concrete physical system where it is possible to show that the local algebras are hyperfinite $III_1$ factors, if there is one where this is possible.
2) Is there an interpretation in physical terms of the presence of the hyperfinite $III_1$ factor in QFT?
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## 2 Answers
Yngvason, J. (2005). The role of type III factors in quantum field theory. Reports on Mathematical Physics, 55(1), 135–147. (arxiv)
The Type III property says something about statistical independence. Let $\mathcal{O}$ be a double cone, and let $\mathfrak{A}(\mathcal{O})$ be the associated algebra of observables. Assuming Haag duality, we have $\mathfrak{A}(\mathcal{O}')'' = \mathfrak{A}(\mathcal{O})$. If $\mathfrak{A}(\mathcal{O})$ is not of Type I, the Hilbert space $\mathcal{H}$ of the system does not decompose as $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_2$ in such a way that $\mathfrak{A}(\mathcal{O})$ acts on the first tensor factor, and $\mathfrak{A}(\mathcal{O}')$ on the second. This implies that one cannot prepare the system in a certain state when restricted to measurements in $\mathcal{O}$ regardless of the state in the causal complement. It should be noted that if the split property holds, that is there is a Type I factor $\mathfrak{N}$ such that $\mathfrak{A}(\mathcal{O}) \subset \mathfrak{N} \subset \mathfrak{A}(\widehat{\mathcal{O}})$ for some region $\mathcal{O} \subset \widehat{\mathcal{O}}$, a slightly weaker property is available: a state can be prepared in $\mathcal{O}$ irregardless of the state in $\widehat{\mathcal{O}}'$. An illustration of the consequences can be found in the article above.
Another consequence is that the Borchers property B automatically holds: if $P$ is some projection in $\mathfrak{A}(\mathcal{O})$, then there is some isometry $W$ in the same algebra such that $W^*W = I$ and $W W^* = P$. This implies that we can modify the state locally to be an eigenstate of $P$, by doing the modification $\omega(A) \to \omega_W(A) = \omega(W^*AW)$. Note that $\omega_W(P) = 1$ and $\omega_W(A) = \omega(A)$ for $A$ localised in the causal complement of $\mathcal{O}$. Type III$_1$ implies something slightly stronger, see the article cited for more details.
As to the first question, one can prove that the local algebras of free field theories are Type III. This was done by Araki in the 1960's. You can find references in the article mentioned above. In general, the Type III condition follows from natural assumptions on the observable algebras. Non-trivial examples probably have to be found in conformal field theory, but I do not know any references on the top of my head.
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Ok, I somehow missed that paper, although Yngvason is on my watch list :-) – Tim van Beek Sep 16 '11 at 7:28
Feel free to expand you answer later, even though I will accept it already. – Tim van Beek Sep 16 '11 at 7:29
Yes, I will. Have to read the paper again myself first, I don't recall the details... – Pieter Sep 16 '11 at 11:47
I updated the question, feel free to ask for more details :) – Pieter Sep 17 '11 at 0:21
Regarding the first question. As Pieter already said for a conformal net the $III_1$ property holds (if it is not $\mathbb C$). Further $e^{-\beta L_0}$ being trace class for all $\beta>0$ with $L_0$ the generator of the rotations implies the split property, which implies $\mathcal A(I)$ to be the hyperfinite $III_1$-factor.
edit The property $III_1$ and trace class implies split can be found in - D'Antoni,Longo,Radulescu. Conformal Nets, Maximal Temperature and Models from Free Probability [arXiv:math/9810003v1]
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http://mathoverflow.net/questions/106256?sort=newest
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## About the strength of representation-theoretic obstructions for orbit closure problems
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Let `$G$` be a reductive, affine, algebraic group over `$\newcommand{\C}{\mathbb C}\C$`. Let `$X$` be a `$G$`-variety. For `$x\in X$`, we write $$G_x:=\{ g\in G\mid g.x=x\}$$ for its stabilizer and for any subgroup `$H\subseteq G$`, we write $$X^H:=\{x\in X\mid H.x=x\}$$ for the `$H$`-invariants of `$X$`. We say that `$x\in X$` is characterized by its stabilizer if `$X^{G_x}=\{x\}$`. Let `$\{V_\lambda\mid \lambda\in\Lambda\}$` be the irreducible `$G$`-modules.
Given two points `$x,y\in X$`, then `$x\in\overline{G.y}$` implies `$\overline{G.x}\subseteq\overline{G.y}$`. Hence, `$\C[\overline{G.y}]\twoheadrightarrow\C[\overline{G.x}]$` and thus,
$$\DeclareMathOperator{\mult}{mult}\forall \lambda\in\Lambda:\quad \mult\nolimits_\lambda(\C[\overline{G.x}])\le\mult\nolimits_\lambda(\C[\overline{G.y}])$$
Finding `$\lambda\in\Lambda$` violating the above is therefore an "obstruction" for the inclusion of orbit closures.
My question now is the following: If `$x$` and `$y$` are characterized by their respective stabilizers, does the converse hold? I.e., does the above inequality imply that `$x\in\overline{G.y}$`? I have been trying to come up with a counterexample, but without success so far.
Intuition: If `$G$` acts on a variety `$Y$` and `$y\in Y$` is characterized by its stabilizer, then you can very easily find counterexamples if you give up the condition that both points are characterized by their respective stabilizers: Consider `$X:=Y\times\{z_1,z_2\}$` with `$G$` acting trivially on `$Z=\{z_1,z_2\}$`. Now, the points `$x_i:=(y,z_i)$` satisfy `$x_1\notin\overline{G.x_2}$` and `$\C[\overline{G.x_1}]\cong\C[\overline{G.x_2}]$`. In the cases of interest to me, however, both points are characterized by their stabilizer and the question arises whether there are counterexamples under this additional condition.
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I do not fully grasp the role of the condition "$x\in X$ is characterized by its stabilizer", but I want to point out that it fails for any linear action of $G$ on a non-zero vector space $V$, because the stabilizer of any non-zero vector fixes all its scalar multiples. This difficulty persists for all $G$-invariant cones in $V$. Of course, it could be avoided by passing to the projectivization $\mathbb{P}(V).$ – Victor Protsak Sep 4 at 21:10
Inclusions of the closures of nilpotent orbits have been studied in various contexts. If my memory serves, for quiver representations, Bobinski and Zwara proved a complete representation-theoretic characterization of the inclusion of the closures of nilpotent orbits, but it involved finer invariants than just the G-multiplicities. – Victor Protsak Sep 4 at 21:21
Concerning your first comment, yes, this is usually dealt with by passing to the projectivization, or by defining "characterized by its stabilizer" different, i.e. requiring $X^{G_x}=\mathbb{C}x$. I have added a paragraph to somewhat explain the origin behind this condition. Thanks also for the reference, I will look into it – Jesko Hüttenhain Sep 5 at 8:04
With the condition $X^{G_x}=\mathbb{C}x$ imposed on $X$ my counterexample below works perfectly for $g=sl_2(\mathbb{C})$ and the adjoint action of $G=SL_2(\mathbb{C})$. In this case $l=1$ and $f_1$ is just the determinant of a $2\times 2$ matrix. One cannot pass to projectivisations either as the regular functons on the orbit closures would then reduce to scalars (for $G$ connected) and all multiplicities would be zero. – Alexander Premet Sep 5 at 11:45
So you mean $X=\mathfrak{sl}_2(\C)$ with $G=\mathrm{SL}_2(\C)$ acting on $X$ by conjugation (i.e. $g.x = gxg^{-1}$) and the points are the semisimple element $h=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and the nilpotent $x=\begin{pmatrix}0&0\\1&0\end{pmatrix}$. Is that correct? I am confused because my calculations yield $\overline{G.x}=X$, while $G.h=\C h$. – Jesko Hüttenhain Sep 5 at 15:23
## 1 Answer
Take the nilpotent cone $\mathcal N$ in $g={\rm Lie}(G)$ and the $G$-orbit of a regular semisimple element, $h$ say. The categorical quotient $g\rightarrow g//G\cong\mathbb{A}^l$, $l={\rm rk}(G),$ is equidimensional and each of its fibres is an trreducible complete intersection and contains a unique open $G$-orbit. Since the orbit $Gh$ is closed it coincides with one of the fibres and the algebra of regular functions $\mathbb{C}[Gh]$ is just a filtered deformation of the graded algebra $\mathbb{C}[\mathcal{N}]=\mathbb{C}[g]/(f_1,\ldots,f_l)$ whose defining ideal is generated by $f_1-\lambda_1,\ldots, f_l-\lambda_l$ for some $\lambda_i\in\mathbb{C}$ (here $f_1,\ldots, f_l$is a set of free homogeneous generators for $\mathbb{C}[g]^G$). Since $G$ is reductive, we are in characteristic $0$ and the action of $G$ on $\mathbb{C}[Gh]$ is rational, we have that $\mathbb{C}[\mathcal{N}]\cong \mathbb{C}[Gh]$ as $G$-modules. So all multiplicities will be the same in both cases. However, $Gh$ is not contained in $\mathcal N$ (and vice versa).
However this example does not answer the question as the stabilisers $G_x$ of regular elements $x\in g$ are not self-normalising (I have completely overlooked the extra condition on $x$ in the first reading, which implies that $N_G(G_x)=G_x$).
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Side remark: mathoverflow.net/questions/85560/… - "Are nilpotent orbits degenerations of semisimple ones" – Alexander Chervov Sep 4 at 12:17
Thanks a lot already, but I am having some difficulties understanding the counterexample, mostly for lacking terminology: I assume $G$ acts on $g$ by $\mathrm{Ad}$. What is a filtered deformation? Why does it follow from $Gh$ being closed that $\mathbb C[Gh]$ is a filtered deformation of $\mathbb C[\mathcal N]$? It would probably take a while to clear up all these questions (and their possible follow-ups). Maybe you could, instead, apply your (very general) counterexample to the case $G=Gl_n(\mathbb C)$ and explain what $g$, $\mathcal N$, etc. are, in this case? – Jesko Hüttenhain Sep 4 at 15:45
It is quite problematic that I do not know the definition of a filtered deformation. I really think that making this more concrete would help immensely. – Jesko Hüttenhain Sep 4 at 17:46
I confess that I do not fully grasp the role of the condition "$x$ is characterized by its stabilizer", but it's not fulfilled in the present case. The stabilizer of a regular semisimple element $h$ is its centralizer $H=Z(h)$, which is a Cartan subgroup, so it is commutative and stabilizes each element of $H.$ Similarly, the stabilizer of a regular nilpotent element $e$ is its centralizer $Z(e)$, which is commutative and stabilizes each of its elements. Thus in both cases, $X^{G_x}$ has dimension $\ell.$ – Victor Protsak Sep 4 at 20:58
Hm. In this case, one can construct easier counterexamples, see my edit. The crux is for both points to be characterized by their respective stabilizers. – Jesko Hüttenhain Sep 5 at 7:58
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http://unapologetic.wordpress.com/2008/07/29/polynomials-as-functions/?like=1&_wpnonce=84af3d0b4c
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# The Unapologetic Mathematician
## Polynomials as Functions
When I set up the algebra of polynomials I was careful to specify that the element $X$ is not a “variable”, as in high school algebra. Why do I have to do that? What is the “variable” thing that we were all taught, then?
We’ve got the algebra of polynomials $\mathbb{F}[X]$ over the base field $\mathbb{F}$. Now I’m going to define a function $\mathrm{ev}:\mathbb{F}[X]\times\mathbb{F}\rightarrow\mathbb{F}$ called the “evaluation map”. We define $\mathrm{ev}(p,x)$ by first writing out $p$ in terms of the standard basis
$p=c_0+c_1X+c_2X^2+...+c_nX^n$
Remember here that the sum must terminate after a finite number of basis elements. Then we just stick the field element $x$ in for $X$ to get an expression written out in the field $\mathbb{F}$ itself:
$\mathrm{ev}(p,x)=c_0+c_1x+c_2x^2+...+c_nx^n$
Now the superscripts on each $x$ must be read as exponents. This defines a particular element of the field. If we keep the polynomial $p$ fixed and let $x$ range over $\mathbb{F}$ we get a function from $\mathbb{F}$ to itself, which we can abuse notation to write as $p(x)$. This is the notion of polynomial-as-function we were taught in high school.
But it’s actually more interesting to see what happens as we fix $x$ and let $p$ vary over all polynomials. The map $p\mapsto p(x)$ turns out to be a homomorphism of $\mathbb{F}$-algebras! Indeed, given polynomials
$p=c_0+c_1X+c_2X^2+...+c_nX^n$
$q=d_0+d_1X+d_2X^2+...+d_nX^n$
(the top coefficients here may be zero, and all higher coefficients definitely are) and a field element $k$ we find
$\begin{aligned}\left[p+q\right](x)=(c_0+d_0)+(c_1+d_1)x+(c_2+d_2)x^2+...+(c_n+d_n)x^n\\=c_0+d_0+c_1x+d_1x+c_2x^2+d_2x^2+...+c_nx^n+d_nx^n\\=c_0+c_1x+c_2x^2+...+c_nx^n+d_0+d_1x+d_2x^2+...+d_nx^n\\=p(x)+q(x)\end{aligned}$
$\begin{aligned}\left[kp\right](x)=(kc_0)+(kc_1)x+(kc_2)x^2+...+(kc_n)x^n\\=kc_0+kc_1x+kc_2x^2+...+kc_nx^n\\=k(c_0+c_1x+c_2x^2+...+c_nx^n)\\=kp(x)\end{aligned}$
I’ll let you write out the verification that it also preserves multiplication.
In practice this “evaluation homomorphism” provides a nice way of extracting information about polynomials. And considering polynomials as functions provides another valuable slice of information. But we must still keep in mind the difference between the abstract polynomial
$p=c_0+c_1X+c_2X^2+...+c_nX^n$
and the field element
$p(x)=c_0+c_1x+c_2x^2+...+c_nx^n$
### Like this:
Posted by John Armstrong | Algebra, Ring theory
## 7 Comments »
1. [...] of Polynomials I When we consider a polynomial as a function, we’re particularly interested in those field elements so that . We call such an a [...]
Pingback by | July 30, 2008 | Reply
2. The finer points of the distinction between polynomials as
indeterminates and polynomials as functions do not seem to be well enough known, despite the fact that it is closely
related to some interesting mathematics. See for instance
the discussion in Section 2.1 (pages 3-7) of
http://math.uga.edu/~pete/4400ChevalleyWarning.pdf
This discussion is adapted (and expanded) from a similar
discussion in Ireland and Rosen’s text.
Comment by Pete L. Clark | July 30, 2008 | Reply
3. Pete, that’s a great reference, thanks. I’m planning on mentioning some of the stuff in that section, but not all of it. It’d be worth an interested reader’s time to check it out.
Comment by | July 30, 2008 | Reply
4. [...] . But there’s one big difference here: we have a relation that must satisfy. When we use the evaluation map we must find . And, of course, any polynomial which includes as a factor must evaluate to as [...]
Pingback by | August 7, 2008 | Reply
5. [...] going to want to talk about “evaluating” a power series like we did when we considered polynomials as functions. But when we try to map into our base field, we get a sequence of values and we ask questions about [...]
Pingback by | August 26, 2008 | Reply
6. [...] evaluation of power series is specified by two conditions: it should agree with evaluation of polynomials when we’ve got a power series that cuts off after a finite number of terms, and it should be [...]
Pingback by | August 27, 2008 | Reply
7. [...] it as a function in its own right. In a way, this is just extending what we did when we considered polynomials as functions and we can do everything algebraically with abstract “variables” as we can with [...]
Pingback by | October 2, 2009 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://unapologetic.wordpress.com/2007/10/02/the-snake-lemma/?like=1&_wpnonce=81a70c627a
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# The Unapologetic Mathematician
## The Snake Lemma
And now for what has to be the most popular theorem in all of homological algebra: the snake lemma.
This shows up in every first course in algebraic topology, and always with the same diagram-chasing proof (though not with the same semantic sugar as we’re using). And some grad student always raises the same objection halfway through (I did it in my class). And the follow-up is always the same. Weibel’s An Introduction to Homological Algebra points out that the proof is best seen visually, but I don’t have a video camera like The Catsters. Weibel also points out that a good visual presentation appears in the 1980 movie It’s My Turn. I looked around everywhere but couldn’t find a clip online. So, if you have access to the movie it’s old enough that I’m sure the producers wouldn’t mind a teensy clip posted for The Good of Mathematics. Call it “fair use”.
[UPDATE]: Down in the comments, Graham points out that someone at Harvard has posted the clip here. Tough to see, though.
Now, down to business. We consider the following diagram:
We start with two short exact sequences and a morphism $(f,g,h)$ from one to the other. We construct the kernel and cokernel of each arrow, and then use their universal properties to construct arrows between the kernels and cokernels. It is straightforward to show that the given rows are exact — in particular, $m_0$ is monic and $e_1$ is epic.
However, the upper and lower rows are not necessarily short exact sequences. That is, $e_0$ may fail to be epic and $m_1$ may fail to be monic. Amazingly, though, these two fail to happen in exactly the same way at exactly the same time! It turns out that there is a morphism $\delta:\mathrm{Ker}(h)\rightarrow\mathrm{Cok}(f)$ so that the sequence
$\mathbf{0}\rightarrow\mathrm{Ker}(f)\rightarrow\mathrm{Ker}(g)\rightarrow\mathrm{Ker}(h)\rightarrow\mathrm{Cok}(f)\rightarrow\mathrm{Cok}(g)\rightarrow\mathrm{Cok}(h)\rightarrow\mathbf{0}$
is exact.
Now, if you really want to see the chase you can get someone to walk you through it, try to construct it yourself, or beg Netflix to stock more early 1980s romances between Jill Clayburgh and Michael Douglas. Here I’ll do a much more insightful proof that gets to the heart of what the connecting morphism really is. The proof depends on this diagram:
We start with the same middle rows as before and construct $\mathrm{Ker}(h)$ and $\mathrm{Cok}(f)$. Now we use one of our lemmas (and its dual) to construct the pullback $D$, the pushout $D'$, and the two dashed arrows. Notice that here we’re using the (unshown) facts that $m$ is monic and $e'$ is epic. We have an arrow $\delta_0=p'\circ g\circ k'$, which satisfies $s'\circ\delta_0=h\circ k\circ u$ and $\delta_0\circ s=u'\circ p\circ f$. And then since $u'$ is the kernel of $s'$ and $u$ is the cokernel of $s$, we see that $\delta_0$ factors as $u'\circ\delta\circ u$ for some unique $\delta:\mathrm{Ker}(h)\rightarrow\mathrm{Cok}(f)$.
What does this arrow do to a member $x\in\mathrm{Ker}(h)$? Well first it gets sent to $k\circ x$. Then because $e$ is epic there must be a $y\in_mB$ with $e\circ y\equiv k\circ x$. We calculate that $e'\circ g\circ y=h\circ e\circ y\equiv h\circ k\circ x\equiv0$. But then by exactness there must be $z\in_mA'$ with $m'\circ z\equiv g\circ y$. This is the usual diagram chase given to construct the connecting morphism. Now I claim that $\delta\circ x\equiv p\circ z$.
This happens because $D$ is a pullback, so there is a member $x_0\in_mD$ with $u\circ x_0\equiv x$ and $k'\circ x_0\equiv y$. Then
$u'\circ\delta\circ x\equiv u'\circ\delta\circ u\circ x_0\equiv\delta_0\circ x_0\equiv p'\circ g\circ y\equiv u'\circ p\circ z$
Since $u'$ is monic we have $\delta\circ x\equiv p\circ z$, as desired. This also shows that the equivalence class of $p\circ z$ is independent of the choices made in the construction of the last paragraph, which is the standard objection I mentioned earlier, now handled in a very slick (but not diagram-chasing) way.
So now we need to verify the exactness of the long sequence. We’ll show the exactness at $\mathrm{Ker}(h)$, and that at $\mathrm{Cok}(f)$ will follow by duality.
First of all we need to see that $\delta\circ e_0=0$. For this, we show that $\delta\circ e_0\circ w=0$ for any $w\in_m\mathrm{Ker}(g)$. Now when we use the diagram chase our first step is to choose a $y\in_mB$ with $e\circ y\equiv k\circ e_0\circ w=e\circ j\circ w$, where $j:\mathrm{Ker}(g)\rightarrow B$ is the kernel map. This shows we can choose $y=j\circ w$. But then $g\circ y=g\circ j\circ w\equiv0$, and so we choose $z=0$ and find $\delta\circ e_0\circ w=p\circ z=0$.
Now take $x\in_m\mathrm{Ker}(h)$ with $\delta\circ x\equiv0$, meaning that we construct $p\circ z\equiv0$. By the definition of the cokernel there will be $w\in_mA$ with $f\circ w\equiv z$, so $g\circ m\circ w=m'\circ z\equiv g\circ y$. The “subtraction” rule gives us a $y_0\in_mB$ with $e\circ y_0\equiv e\circ y=k\circ x$ and $g\circ y_0$. But now by the definition of $\mathrm{Ker}(g)$ we have $x_0\in_m\mathrm{Ker}(g)$ with $j\circ x_0\equiv y_0$. Then $k\circ e_0\circ x_0=e\circ y_0\equiv k\circ x$, and so $x\equiv e_0\circ x_0$ because $k$ is monic. And thus the sequence is exact at $\mathrm{Ker}(h)$.
Incidentally, what the setup of the snake lemma tells us is that the category $\mathbf{Ses}(\mathcal{C})$, while always enriched over $\mathbf{Ab}$, is not generally abelian itself. If it were an abelian category the upper and lower rows of the first diagram would have to be short exact sequences themselves. That doesn’t quite happen, but we can exactly capture the obstruction to existence of kernels and cokernels in a single connecting morphism. If this morphism $\delta$ is the zero morphism, then we can insert $\mathbf{0}$ into the long exact sequence and then break it into the two short exact sequences, and thus we have a kernel and a cokernel for the triple $(f,g,h)$.
Notice here that $\delta\in\hom_\mathcal{C}(\mathrm{Ker}(h),\mathrm{Cok}(f)$, so the obstruction lives in this abelian group, which is determined by $f$ and $h$. Thus in a sense a morphism of short exact sequences has or fails to have a kernel and a cokernel on $g$ alone. I’ll leave this to you to ponder: if we’ve given $f$ and $h$, must there exist a $g$ so that $(f,g,h)$ has a kernel and a cokernel — so that $\delta=0$? And if so, how can we find it?
### Like this:
Posted by John Armstrong | Category theory
## 6 Comments »
1. I might be missing something here, but your $\delta_0$ isn’t an arrow…you can’t do that composition with the diagram as written.
Comment by | October 2, 2007 | Reply
2. Oops, I mistyped… editing…
There. $\delta_0$ should be the vertical arrow straight down the middle of the second diagram.
Comment by | October 2, 2007 | Reply
3. Mathematics in Movies, sixth clip.
Comment by | October 2, 2007 | Reply
4. You lost me in the middle. Why is u the cokernel of s?
Comment by Homologically Confused | October 3, 2007 | Reply
5. HC: I’m composing this on the fly while I wait for my plane, so I’m hoping I don’t make a mistake here. If I recall correctly, that’s something that follows from the lemma about pullbacks we proved earlier. It should be in one of the posts linked at the top of this post.
If I’m wrong, point out where you lose it and I’ll give it another look when I’m back at a real computer instead of my iPhone.
Comment by | October 3, 2007 | Reply
6. HC: I think that follows from the axioms of abelian categories: every epic is cokernel of its kernel. Since e is epic, so is u with kernel s by the lemma. Then u is cokernel of its kernel s.
Comment by Dohyeong Kim | September 25, 2008 | Reply
« Previous | Next »
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
|
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http://mathoverflow.net/questions/57197?sort=oldest
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## fundamental groups of curves
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I saw the following statement made without proof in a paper of Bogomolov and Tschinkel:
If $X$ is an algebraic surface, and $C$ is an ample smooth curve in $X,$ then the fundamental group of $C$ surjects onto that of $X.$
I was wondering if someone could supply a reference, and perhaps some idea of what the most general version of this statement was...
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3
This statement goes back to Zariski and Lefschetz, but it is usually proved by Morse theory now days. (If it wasn't so late I'd say more.) – Donu Arapura Mar 3 2011 at 3:33
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It will be earlier tomorrow :) – Igor Rivin Mar 3 2011 at 3:36
You're right, it is, and you have answers too. Regarding, your comment below. Ample means that a multiple is equivalent (in the appropriate sense) to a hyperplane section, or if you prefer that the associated line bundle is positive (by Kodaira). – Donu Arapura Mar 3 2011 at 21:32
## 2 Answers
There is a general Lefschetz hyperplane theorem for the homotopy groups of an ample divisor $D$ on a smooth complex variety $X$. Basically, this theorem says that the relative homotopy groups $\pi_i(X,D)$ are zero for all $i$ less than $\dim X$. In particular, the map $\pi_1(D)\to\pi_1(X)$ is an isomorphism for $\dim X\ge 3$ and surjective for $\dim X= 2$. There is a very nice proof of this theorem using Morse theory which can be found in Lazarsfeld's book 'Positivity in algebraic geometry'.
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That is a a nice theorem! (I fixed a small typo in your LaTeX, hope you don't mind) – David Roberts Mar 3 2011 at 3:40
Ah, the right reference seems to be the following paper of Raoul Bott's: R. Bott On a theorem of Lefschetz Michigan Math Journal, 1959 (Bott attributes the result to Thom). – Igor Rivin Mar 3 2011 at 3:43
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I did not notice @J.C.'s (or @David's) edit -- the version I saw missed the last three lines. I will check out Lazarsfeld's book as well (which can, hopefully, also shed light on when ampleness can/should be expected -- as a non-algebraic-geometer I have little feeling for what it means...) – Igor Rivin Mar 3 2011 at 4:00
1
another standard reference for this classical result of Lefschetz is Milnor, Morse theory, p. 41-42. – roy smith Mar 3 2011 at 4:09
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
There is also an algebraic proof which works over fields of positive characteristic.
See, e.g., Lemma 5.1 of the paper Zariski's conjecture and related problems by Madhav Nori (Annales scientifiques de l'École Normale Supérieure, Sér. 4, 16 no. 2 (1983), p. 305-344).
The proof of that Lemma is more general than what you need. It goes as follows. Let $X$ be a proper smooth algebraic surface and let $A\subset X$ be an ample curve. That $\pi_1(A)$ surjects onto $\pi_1(X)$ means the following: any finite étale cover $f:Y\to X$ which is split over $A$ is split. So if $Y$ is connected, $\deg(f)=1$.
Let us prove that if $A$ is ample; in fact, we only need $A$ big and nef. If $f$ has a section on $A$, one can write $f^{-1}(A)=B+R$ where $B\to A$ is an isomorphism, and $R$ is disjoint from $B$. The Hodge index theorem says that the intersection form restricted to the space generated by $B$ and $R$ has at most one +-sign. Since $(B+R)^2=(f^*A)^2=\deg(f) A^2>0$, it has exactly one +-sign, and the determinant $$\begin{vmatrix} (B+R)^2 & (B+R)\cdot B \cr B\cdot (B+R) & B^2\end{vmatrix}$$ is nonpositive. By the projection formula, one has $$(B+R)\cdot B=f^*A\cdot B=A\cdot f_*B=A^2.$$ Since $B$ and $R$ are disjoint, we obtain $B^2=A^2$. Then, $$\deg(f)A^2 = (B+R)^2=B^2+R^2,$$ so R^2=(\deg(f)-1) A^2$. The above determinant is equal to$ (\deg(f)-1)A^2$. Since$A^2>0$,$\deg(f)\leq 1\$.
This proof generalizes to the so-called Ramanujam lemma according to which an effective divisor on a surface which is big and nef is numerically connected (doesn't decompose as the sum of two nonzero effective divisors with 0-intersection), hence connected. In our case, the existence of the section implies that $f^*A$ is not numerically connected; it is however big and nef because this property is stable under finite pull-back.
See also the paper of J-B. Bost, Potential theory and Lefschetz theorems for arithmetic surfaces (Annales scientifiques de l'École Normale Supérieure, Sér. 4, 32 no. 2 (1999), p. 241-312) where this argument is explained for surfaces and adapted for arithmetic surfaces.
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This is very nice! – J.C. Ottem Mar 3 2011 at 17:26
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http://mathoverflow.net/questions/141?sort=newest
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## Model category structures on categories of complexes in abelian categories
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Section 2.3 of Hovey's Model Categories book defines a model category structure on Ch(R-Mod), the category of chain complexes of R-modules, where R is a ring. Lemma 2.3.6 then essentially states (I think) that taking projective resolutions of a module corresponds to taking cofibrant replacements of the module, at least in nice cases (e.g. when the projective resolution is bounded below). There is of course also a "dual" model category structure which gives the "dual" result for injective resolutions and fibrant replacements (Theorem 2.3.13).
1. I think the results in Hovey are proven for not-necessarily-commutative rings. Do things become nicer if we restrict our attention to commutative rings only?
2. Do these results generalize? For example, is there an analogous model category structure and an analogous result for Ch(OX-Mod), the category of chain complexes of OX-modules, where X is a scheme? More generally, how about for Ch(A), where A is an abelian category?
If the answers to these questions are known, then I assume they would be "standard", but I don't know a reference.
I've re-asked my question in a different form here.
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I assume so as well, though I likewise can't provide a reference. – Ben Webster♦ Oct 6 2009 at 18:25
The link to the reask of the question is broken (... well, it redirects to a question about numerical methods for calculating digits of pi ...). – cdouglas Oct 27 2011 at 11:53
## 2 Answers
I don't think the existence of the dual "injective" model structure merits an "of course," since its generators are much less obvious to construct. However, it turns out that injective model structures actually exist in more generality than projective ones, for instance they exist for most categories of sheaves. I believe this was originally proven by Joyal, but it was put in an abstract context by Hovey and Gillespie.
The basic idea is that model structures on Ch(A) correspond to well-behaved "cotorsion pairs" on A itself. The projective model structure comes from the (projective objects, all objects) cotorsion pair (which is well-behaved for R-modules, but not for sheaves), and the injective one comes from (all objects, injective objects). There is also e.g. a flat model structure coming from (flat objects, cotorsion objects) which is monoidal and thus useful for deriving tensor products. A good introduction, which I believe has references to most of the literature, is Hovey's paper Cotorsion pairs and model categories.
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Awesome, thanks for the reference! – Kevin Lin Oct 13 2009 at 21:30
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I would also add the following references: J.D. Christensen and M. Hovey, Quillen model structures for relative homological algebra, Math. Proc. Camb. Phil. Soc. 133 (2002), no. 2, 231-293. The following paper should answer your questions 1) and 2) quite precisely. intlpress.com/HHA/v11/n1/a11 – Denis-Charles Cisinski Oct 25 2009 at 21:36
The paper of Hovey that I linked to includes a reference to the Christensen-Hovey paper, and to a number of others; I didn't want to give a huge list of references here. Thanks for the other link. – Mike Shulman Oct 27 2009 at 0:32
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I don't know the "standard" answer, but the exact same construction should work for any abelian category with a small projective generator, where "small" means that any map into a sufficiently large well-ordered colimit factors through some stage. This is exactly what is needed to make the small object argument work. Just replace the ring with your small projective generator in the "sphere" and "disk" objects.
For the dual (injective) model structure, cosmall injectives don't tend to exist in practice (for example, they don't exist in abelian groups), so you have to use a more complicated set of maps that I don't understand well and don't know how to generalize. In particular, in the case of quasicoherent sheaves one would need to generalize the injective model structure, and I don't know anything about that.
I don't know of any reason to expect commutative rings to give nicer results.
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This answer also fits into the framework of Gillespie's work. Asking for a small projective generator is related to asking for the cotorsion pair $(A,B)$ of Mike's answer to be cogenerated by a one-element set (cogenerated by a set means there is a set $S\subset A$ such that $b\in B$ iff $Ext^1(S,b)=0$). In a Grothendieck category with enough projectives, being cogenerated by a set forces $(A,B)$ to be a complete cotorsion pair. You'll also get the dual complete cotorsion pair, and thence the Hovey model structure described by Mike. – David White Aug 8 at 21:41
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http://www.math.uah.edu/stat/applets/PokerExperiment.html
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### Poker Experiment
Card 1 Card 2 Card 3 Card 4 Card 5
Distribution graph
#### Description
The poker experiment consists of dealing 5 cards at random from a standard deck of 52 cards. The value of the hand is defined as follows:
1. no value
2. one pair
3. two pair
4. three of a kind
5. straight
6. flush
7. full house
8. four of a kind
9. straight flush
Random variable $$V$$ gives the value of the hand and is displayed on each update in the data table. The probability density function of $$V$$ is shown in blue in the distribution graph and recorded in the distribution table. On each update, the empirical density function of $$V$$ is shown in red in the distribution graph and recorded in the distribution table. The record table also records the card hand $$(X_1, X_2, X_3, X_4, X_5)$$ The simulation can be set to stop automatically when $$V$$ is a particular value.
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http://mathoverflow.net/questions/17731?sort=oldest
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## Properties of collections (functions) that make them proper classes (uncomputable)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
There are collections too big to be a set, e.g. the collection of all sets (in ZFC), and there are collections that cannot be sets for "pure" logical reasons, e.g. the collection of sets that do not contain themselves.
There are functions growing too fast to be computable, e.g. the busy beaver function, and there are functions that cannot be computed for "pure" logical reasons, e.g. the halting function.
In the final end it is of course shown by logical means that being too big (growing too fast) prohibit a collection to be a set (a function to be computable), but those properties are not "purely" logical (they are about sizes and growth rates), opposed to the "pure" logical reasons mentioned above.
Is there a simple lesson to be learned from these observations? Are there other not "purely" logical properties of collections (functions) that prohibit them to be sets (computable)?
Edit: It's from the very definition of a collection -- $\lbrace x | x = x \rbrace$ or $\lbrace x | x \not\in x \rbrace$ -- that it is shown by logical means, that it cannot be a set. And not, firstly, from a hard to define meta-property of "defining too big a collection" or "being intrinsically inconsistent".
Then it's at least somehow astonishing, that some of those definitions cohere with the meta-property of "defining too big a collection".
Cannot - after all - the meta-property of "defining too big a collection" be rigorously defined such that it can be shown that no definition of a collection with this meta-property defines a set?
(The same should go - mutatis mutandis - for functions and computability.)
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## 1 Answer
One plausible necessary condition for a collection c to satisfy the meta-property in question is tthe following: the existence of c must not be a logical consequence of either ZFC or any large cardinal axioms expressible in the language of ZF. However, it is unclear what to say about collections that follow from large cardinal axioms expressible only in some larger language that includes the language of ZF (e.g., a language obtained by adding a truth-predicate to the language of ZF). It's not obvious a priori whether to call such collections "sets" or not; it would seem to depend on the specific nature of the relevant large cardinal axiom(s). It is tempting to say that the collections in question should be regarded as sets if they exist within a cumulative hierarchy satisfying the axioms of ZFC. Such a position is not clearly correct, however; for it is possible, using a certain language containing a truth-predicate, to distinguish "sets" from "proper classes" within this hierarchy itself, using omega-inconsistency as a criterion: see "Omega-inconsistency and the universe of sets," in IeCCS 2007 (International e-Conference on Computer Science), ed. T.E. Simos and G. Psihoyios. Thus, the attempt to rigorously define your meta-property raises some difficult questions and issues that make me skeptical about the possibility of such a definition.
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http://gowers.wordpress.com/2007/09/
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# Gowers's Weblog
Mathematics related discussions
## Archive for September, 2007
### One way of looking at Cauchy’s theorem
September 19, 2007
Cauchy’s theorem is the assertion that the path integral of a complex-differentiable function around a closed curve is zero (as long as there aren’t any holes inside the curve where the function has singularities or isn’t defined). This theorem, which is fundamental to complex analysis, can be vastly generalized and seen from many different points of view. This post is about a little idea that occurred to me once when I was teaching complex analysis. I meant to include it as a Mathematical Discussion on my web page but didn’t get round to it. Now, while the novelty of having a blog still hasn’t worn off, I find I have the energy to put it here.
Let’s start with a simpler fact: that if $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$ and the derivative of $f$ is everywhere zero, then $f$ is constant. What is the natural generalization of this fact to functions defined on the plane? (more…)
Posted in Demystifying proofs | 19 Comments »
### About this blog
September 15, 2007
As I mentioned in my first post, I am very busy at the moment, so my rate of posting will soon go right down, probably until about February. However, I wanted to have at least one post that had mathematical content (as opposed to remarks about the presentation of mathematics) since I hope that in the end that will be the main focus of the blog. So the post about cubics is supposed to be more representative of what this blog is about. Thanks to all who have contributed so far. Although I had read about the power of blogs it nevertheless came as a surprise to see from the stats how many more visits a blog gets than a homepage, and also to discover how easy it is to get useful advice. I’ve even had a serious offer of technical help with setting up new wiki-style websites, so there’s a good chance that some of those ideas will actually be realized.
Posted in General | 14 Comments »
### Discovering a formula for the cubic
September 15, 2007
In this post I want to revisit a topic that I first discussed on my web page here. My aim was to present a way in which one might discover a solution to the cubic without just being told it. However, the solution that arose was not very nice, and at the end I made the comment that I did not know a way of removing the rabbit-out-of-a-hat feeling that the usual much neater formula for the cubic (together with its derivation) left me with.
A couple of years ago, I put that situation right by stumbling on a very simple idea about quadratic equations that generalizes easily to cubics. More to the point, the stumble wasn’t completely random, so the entire approach can be justified as the result of standard and easy research strategies. I am no historian, but I would imagine that this idea is pretty similar to the idea (in some equivalent form) that first led to this solution. (more…)
Posted in Demystifying proofs | 20 Comments »
### Alternative maths reviews
September 15, 2007
Here’s another idea for a wiki-style website, one that might bring closer the day when mathematicians ceased to bother about print journals. It’s a site where people can post reviews of mathematical papers. Such a site, if it did what I have in mind, would have one disadvantage and two advantages over Math Reviews. The disadvantage, which is also one of the advantages actually, is that by no means every paper would be included. If you want a list of all published papers in mathematics, then Math Reviews (or Zentralblatt) does the job very well. However, it’s not really a site where one would browse for fun, and part of the reason is that all papers are given equal status, so if one is looking for an interesting paper one has to look amongst a whole lot of uninteresting ones. With a bit of skill and prior knowledge one can find interesting things of course, but that’s not really what I’d call browsing, in the sense of just having a look at what’s there and finding all sorts of gems. (more…)
Posted in Mathematics on the internet | 16 Comments »
### How should vector spaces be introduced?
September 14, 2007
This question arose in the discussion of my previous post, but deserves a place to itself because (in my opinion, which I shall try to justify) it involves different issues. For example, how does one explain the point of the abstract notion of finite-dimensional vector spaces when, unlike with groups, you don’t seem to have an interesting collection of different spaces? Why not just use $\mathbb{R}^n$? I addressed this point on my home page here, so won’t discuss it further on this post. But another point, which was raised in the previous discussion, concerns the relationship between theory and computation. I think it’s pretty uncontroversial to say that if you don’t know how to invert a matrix, or extend a linearly independent subset of $\mathbb{R}^n$ to a basis, then you don’t truly understand linear algebra, even if you can state and prove conditions for a linear map to be invertible and can prove that every linearly independent set extends to a basis. Equally, as was pointed out, if you can multiply matrices but don’t understand their connection with linear maps, then you don’t truly understand matrix multiplication. (For example, it won’t be obvious to you that it is associative.) But how does one get people to understand the theory, be able to carry out computations, and see the links between the two? This is another situation where my own experience was not completely satisfactory: I’d be taught the theory in lectures and given computational questions to do, as though once I knew the theory I’d immediately see its relevance. But in fact I found the computational questions pretty hard, and some of the links to the theory were things I didn’t appreciate until years later and I found myself having to explain the subject to others.
Posted in Mathematical pedagogy | 34 Comments »
### How should logarithms be taught?
September 13, 2007
Having a blog gives me a chance to defend myself against a number of people who took issue with a passage in Mathematics, A Very Short Introduction, where I made the tentative suggestion that an abstract approach to mathematics could sometimes be better, pedagogically speaking, than a concrete one — even at school level. This was part of a general discussion about why many people come to hate mathematics.
The example I chose was logarithms and exponentials. The traditional method of teaching them, I would suggest, is to explain what they mean and then derive their properties from this basic meaning. So, for example, to justify the rule that xa+b=xaxb one would say something like that if you have a xs followed by b xs and you multiply them all together then you are multiplying a+b xs all together. (more…)
Posted in Mathematical pedagogy | 39 Comments »
### What might an expository mathematical wiki be like?
September 11, 2007
This post has its origins in a discussion that arose as a result of a very interesting post of Terence Tao. Both the post and the discussion can be found here . The post outlines a rather general idea, or trick, that can be used in many mathematical situations. With such tricks, it is usually difficult, and in any case not desirable, to formalize them as lemmas: if you try to do so then almost certainly your formal lemma will not apply in all the situations where the trick does. This has the unfortunate consequence that they are relegated to something like “folklore,” transmitted orally (to a lucky few) or rediscovered over and over again (the more usual experience). (more…)
Posted in Mathematics on the internet | 59 Comments »
### The Princeton Companion to Mathematics
September 6, 2007
I have decided to follow the excellent example of Terence Tao and start up a blog. For the moment I am too busy to do this properly, because, with the help of June Barrow-Green and Imre Leader, I am editing a book called The Princeton Companion to Mathematics. However, it is partly for that very reason that I want to set up the blog. It is somewhat hard to explain what the book is, but if you want to get quite a good idea, there is a substantial (though out of date) description of it, with several sample articles available here. You can get into this site with userid Guest and password PCM. Comments welcome. A sufficiently sensible comment could even influence what goes into the book, but I should warn that, because we are at a rather late stage of the editing process, I no longer have much room for manoeuvre. So I may end up having to say, “Yes, great point, but unfortunately it’s too late to do anything about it.”
Actually, I hope that the PCM blog will really come into its own after the book comes out. In particular, if you feel that there are unfortunate gaps (as there undoubtedly will be) then maybe it will be possible to do something about it online — I might even start up a wiki consisting of PCM supplements. (The distinction between that and the regular mathematics articles in Wikipedia would be some kind of certification that an article had reached PCM levels of comprehensibility. I’d probably be unwilling to put in the sort of editorial efforts I’ve been putting in over the last few years, but would try to distribute that task by using the blog medium. If you are reading this, maybe you will have a suggestion about how to go about it — in particular, I don’t yet know anything about the technicalities of this kind of thing.)
Posted in Princeton Companion To Mathematics | 37 Comments »
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http://nrich.maths.org/1926/clue
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### Route to Root
A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?
### Rain or Shine
Predict future weather using the probability that tomorrow is wet given today is wet and the probability that tomorrow is wet given that today is dry.
### Climbing Powers
$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?
# Dalmatians
##### Stage: 4 and 5 Challenge Level:
Start with different numbers and write down the sequences. What patterns do you notice? Can you explain what happens?
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://physics.stackexchange.com/questions/15025/am-i-doing-this-conservation-of-angular-momentum-problem-correctly?answertab=active
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# Am I doing this conservation of angular momentum problem correctly? [closed]
I am doing some calculations for a practical project of mine. Basically I have a levitating sphere with gravity countered and close to 0 friction when the sphere rotates(due to airdrag). Inside the sphere is a servo spinning at $.12sec/60 degrees$, or $2\pi /.72sec$(8.73 rad/s). The spinning servo on the inside has masses attached to both arms to allow it to have enough moment to make the globe spin. The arm's radius is 1 or .0254m.
I have the following equation:
````JA=sum of torques on globe due to servo
````
J is the moment of inertia for the sphere and many other things inside, I did a rough calculation and got .5kg·m²
A is the angular acceleration, for now i have it be $\pi /6$ radians, I assume that this acceleration is good enough for the globe to turn at a reasonable rate.
Finally to find the torques, there are two torque motion on the servo, since its servo arm has two arms. All I need to do is calculate the torque on one arm and multiply by 2. $\omega$ is the angular velocity of the servo, r is the radius of the arm. m is the mass attached to the end of the servo arm, and v is the velocity. Assume the servo arm itself is massless.
$F=ma$
$T=Fr$
$a=\omega v$
$velocity=\omega r$
$F=m\omega \omega r$
$T=m\omega \omega rr$
Now to solve:
$T=m\omega ^2r^2$
$2T=m\omega ^2r^2$
$2T=JA$
$m\omega ^2r^2=JA$
$m=JA/(\omega ^2r^2)=.5(\pi /6)/(8.73^2*.0254^2)=5.324grams$
Thus, to make my globe rotate(at $\pi /6$ angular acceleration) in reaction to the servo on the inside spinning, the servo must hold $5.3 grams$ of weight on both ends.
Does this calculation make sense?
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Is there a reason you think it doesn't make sense? It'll help us give a more focused answer if you explain why you're wondering about your result. – David Zaslavsky♦ Sep 24 '11 at 4:19
1
@DavidZaslavsky its not that I don't think it doesn't make any sense. But I'm unsure if I'm taking the right steps and setting up the right equations. I just needed some help to verify my approach – mugetsu Sep 24 '11 at 8:31
OK, well then, how else do you think you should be doing it? I mean, clearly there's something you're not sure of, otherwise you wouldn't be asking the question. I'm trying to get you to ask about the specific concept you're unsure about, since it'll make a much more useful question. This site is really meant for questions about physical concepts, not general "check my math" questions. – David Zaslavsky♦ Sep 24 '11 at 19:25
@DavidZaslavsky I'm not sure if my concept of sum of torque from the arms=moment of inertia of body * angular acceleration of body is applicable here. As you can see I did a lot of substitution, playing around with omega. The specific concept I'm not understanding would have to be how conservation of angular momentum would work in a system that essentially applies torque to itself. – mugetsu Sep 27 '11 at 0:33
## closed as too localized by David Zaslavsky♦Dec 20 '12 at 1:26
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.
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http://mathhelpforum.com/differential-geometry/183630-finding-limits-w-definition-not-shortcuts.html
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# Thread:
1. ## Finding limits w/definition not shortcuts
My question is how to get the limit of lim ((3n+1)/(n+2))=3.
I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.
However, I have to use "for each epsilon greater than 0 there exists a real number N such that for all n elements of the natural numbers n > N implies that absolute value of the sequence-s < epsilon.
Any suggestions?
2. ## Re: Finding limits w/definition not shortcuts
Originally Posted by CountingPenguins
My question is how to get the limit of lim ((3n+1)/(n+2))=3.
I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.
I'm sort of confused by your post. So are you suppose to prove that $\lim\frac{3n+1}{n+2}=3$?
If yes, then start with the inequality $|\frac{3n+1}{n+2}-3|=|\frac{3n+1}{n+2}-\frac{3(n+2)}{n+2}|<|\frac{-5}{n}|$.
3. ## Re: Finding limits w/definition not shortcuts
Originally Posted by CountingPenguins
Any suggestions?
Some more details: $\left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$ now, for every $\epsilon>0$ choose $N>\dfrac{5}{ \epsilon}-2$ .
4. ## Re: Finding limits w/definition not shortcuts
[QUOTE=FernandoRevilla;662680]Some more details: $\left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$
Thank you for your response, but can you tell me how you got from $\left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$ it isn't obvious to me. Those are the details I am missing.
Thanks,
CP
5. ## Re: Finding limits w/definition not shortcuts
$\left |\dfrac{3n+1}{n+2} -3\right |< \epsilon \Leftrightarrow \left |\dfrac{3n+1-3n-6}{n+2}\right |< \epsilon\Leftrightarrow$
$\left |\dfrac{-5}{n+2}\right |< \epsilon\Leftrightarrow \dfrac{5}{n+2}< \epsilon\Leftrightarrow\ldots \Leftrightarrow n>\dfrac{5}{ \epsilon}-2$
6. ## Re: Finding limits w/definition not shortcuts
Originally Posted by CountingPenguins
My question is how to get the limit of lim ((3n+1)/(n+2))=3.
I used the trick where you divide every term by the variable with the largest exponent which is n in this case and came up with 3/1=3 which gives the lim = 0. I also checked it on it's graph.
However, I have to use "for each epsilon greater than 0 there exists a real number N such that for all n elements of the natural numbers n > N implies that absolute value of the sequence-s < epsilon.
Any suggestions?
To prove $\displaystyle \lim_{x \to \infty}f(x) = L$, you need to show that $\displaystyle x > N \implies |f(x) - L| < \epsilon$ for $\displaystyle \epsilon > 0$. So in this case, to prove $\displaystyle \lim_{n \to \infty}\frac{3n + 1}{n + 2} = 3$, you need to show $\displaystyle n > N \implies \left|\frac{3n + 1}{n + 2} - 3\right| < \epsilon$.
$\displaystyle \begin{align*} \left|\frac{3n + 1}{n + 2} - 3\right| &< \epsilon \\ \left|\frac{3n + 1 - 3(n + 2)}{n + 2}\right| &< \epsilon \\ \left|-\frac{5}{n+2}\right| &< \epsilon \\ \frac{5}{|n + 2|} &< \epsilon \\ \frac{|n + 2|}{5} &> \frac{1}{\epsilon} \\ |n + 2| &> \frac{5}{\epsilon} \\ n + 2 &> \frac{5}{\epsilon}\textrm{ since }\epsilon > 0\textrm{ and }n > 0 \\ n &> \frac{5}{\epsilon} - 2\end{align*}$
So by letting $\displaystyle N = \frac{5}{\epsilon} - 2$, the proof will follow.
7. ## Re: Finding limits w/definition not shortcuts
Thank you very much for filling in all the blanks for me. I don't want the answer so much as how and why it works and you showed me how. Thanks.
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http://physics.stackexchange.com/questions/14968/superluminal-neutrinos/14989
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# Superluminal neutrinos
I was quite surprised to read this all over the news today:
Elusive, nearly massive subatomic particles called neutrinos appear to travel just faster than light, a team of physicists in Europe reports. If so, the observation would wreck Einstein's theory of special relativity, which demands that nothing can travel faster than light.
—source
Apparently a CERN/Gran Sasso team measured a faster-than-light speed for neutrinos.
• Is this even remotely possible?
• If so, would it be a real violation of Lorentz invariance or an "almost, but not quite" effect?
The paper is on arXiv; a webcast is/was planned here.
News conference video here
Edit febr 23 '12 .Since this is controversial still, there is a report that a possible error has been identified in the GPS measurements.
Two separate issues were identified with the GPS system that was used to time the arrival of neutrinos at an underground lab in Italy, James Gillies, spokesman for the European Organization for Nuclear Research, or CERN, said Wednesday.
One could have caused the speed to be overestimated, the other could have caused it to be underestimated, he said.
"The bottom line is that we will not know until more measurements are done later this year," Gillies told The Associated Press.
More in another report
the 60 nanoseconds discrepancy appears to come from a bad connection between a fiber optic cable that connects to the GPS receiver used to correct the timing of the neutrinos' flight and an electronic card in a computer. After tightening the connection and then measuring the time it takes data to travel the length of the fiber, researchers found that the data arrive 60 nanoseconds earlier than assumed. Since this time is subtracted from the overall time of flight, it appears to explain the early arrival of the neutrinos. New data, however, will be needed to confirm this hypothesis.
And the CERN press release
UPDATE 23 February 2012
The OPERA collaboration has informed its funding agencies and host laboratories that it has identified two possible effects that could have an influence on its neutrino timing measurement. These both require further tests with a short pulsed beam. If confirmed, one would increase the size of the measured effect, the other would diminish it. The first possible effect concerns an oscillator used to provide the time stamps for GPS synchronizations. It could have led to an overestimate of the neutrino's time of flight. The second concerns the optical fibre connector that brings the external GPS signal to the OPERA master clock, which may not have been functioning correctly when the measurements were taken. If this is the case, it could have led to an underestimate of the time of flight of the neutrinos. The potential extent of these two effects is being studied by the OPERA collaboration. New measurements with short pulsed beams are scheduled for May.
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I was just thinking of asking this. One way or the other, this is an epic media meltdown. I'm interested to hear what the real story is. – AlanSE Sep 22 '11 at 19:24
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I think I'll wait for T2K and Minos to have a look at their data before I get all excited about this, but if I were them I'd check my beam characteristics and site-to-site timing agreement pretty carefully. – dmckee♦ Sep 22 '11 at 19:32
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– Z.Sun Sep 22 '11 at 21:56
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– Jim Graber Sep 23 '11 at 0:53
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Interesting paragraph, taking directly from the paper : "Despite the large significance of the measurement reported here and the stability of the analysis, the potentially great impact of the result motivates the continuation of our studies in order to investigate possible still unknown systematic effects that could explain the observed anomaly. We deliberately do not attempt any theoretical or phenomenological interpretation of the results." – Frédéric Grosshans Sep 23 '11 at 12:40
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## 13 Answers
Last (?) Edit: The "problem" is solved: it was mainly a problem in the timing chain, due to a badly screwed optical fibre. A high level description of the problem is given here and a more detailed explanation of the investigation is here.
# List of possible systematic biases
I thought it might be a good idea to list the possible systematic biases which could lead xkcd's character to win his bet. As many physicists (including, I guess, many people from the OPERA collaboration), I think it will end like the Pioneer anomaly. Of course, the current list only contains biases which are unlikely, but less unlikely than a causality violation.
## Location errors and clocks drifts
The arXiv paper studied them, and seem to exclude it. The distance seems to be known within 20 cm and the synchronisation seems to be within 15 ns (6.9 statistical and 7.4 systematic). If this would however end up to be the explanation, it would be quite boring.
Update: Rumors seems to tell that the boring explanation is the good one.
## Not the same neutrinos detected
The neutrinos are emitted on a 10.5 µs window, 175 times longer than the observed effect. It might be possible that the neutrino emitted early are not exactly the same as the one emitted late. Neutrino oscillation might, for example, then make early neutrino more detectable by the distant detector.
However, the detectors were built to measure the oscillation, so I guess that the OPERA collaboration thought about it, and rejected it for whatever reason. I suppose an explanation along these lines would mean interesting new particle physics.
Update: This possibility excluded by a new experiment with 3 ns pulses.
## Errors in the statistical timing analysis
The timing itself is based on a quite elaborate statistical analysis. Furthermore, the pulses are quite long (10 μs), so an error in this analysis could easily be of the good order of magnitude.
Update: This possibility excluded by a new experiment with 3 ns pulses.
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Concerning your #2: they purport to have dealt with this using the shape-shape fitting between the proton current monitor and the timing of the detection. Several of my colleague suspect there may be a subtle effect hiding here, but it is not as if they didn't think of it. – dmckee♦ Sep 23 '11 at 18:11
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Thanks for making a community wiki reply. This is good because otherwise the voting process could drown out important updates that are otherwise ignored in the media. The researchers who released this data themselves will be one of the most likely sources for resolution of the paradox. – AlanSE Sep 23 '11 at 19:53
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The new setup (3 ns pulses, 20 times shorter than the observed effect) has eliminated the last two points. The actual timing and positioning hasn't changed, so point one still stands. (However, that's been perhaps the most scruntinized of all explanations). – MSalters Nov 22 '11 at 8:35
@MSalters: I agree. But the time and distance measurements have been verified by multiple methods, and the methods are ones that are standard and reliable. E.g., the delay in the 8.3-km optical fiber has been measured both by two-way timing and using a portable clock, and it's been measured repeatedly over time so that one can rule out changes in optical properties due to aging of the plastic. These are simple measurements that could be checked in an afternoon by a competent 2nd-year grad student. I really have a hard time imagining a plausible "goof" explanation at this point. – Ben Crowell Nov 23 '11 at 21:17
• Is this even remotely possible?
Well... "possible," yes, but kind of like how tunneling through a brick wall is "possible": while you can't definitively prove it impossible, you'd feel pretty safe saying "this will never happen." Relativity is really well-tested, and it's really hard to conceive of a way that neutrinos could travel faster than light without it having other consequences that we would have discovered by now. That being said, I don't know the field inside-out and I'm sure some theorist has come up with some wacky idea that allows it. I asked another question that might come up with something.
• If so, would it be a real violation of Lorentz invariance or an "almost, but not quite" effect?
If the results from OPERA are accurate, this effect would be a full-blown real Lorentz violation, not just an apparent effect like Cerenkov radiation or astronomical superluminal motion. That's why everyone is so excited about it. (Unless the neutrinos are tachyons; in that case, I guess Lorentz invariance is technically still intact, but the observation of a tachyon would be equally big news.)
The setup of CERN and OPERA is conceptually very simple, basically just two observers located a known distance apart with synchronized clocks. It's a direct measurement of average velocity. There's no complicated theoretical analysis that needs to be done to determine whether the speed of light was exceeded. Either they are wrong about either the distance (mismeasurement, or there is a spacetime "rift" within the Earth :-P) or the time (clock synchronization error or drift), or they have actually discovered superluminal neutrinos.
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Are the observers using exactly identical detectors? Can't the "timing offset" of detection depend on some build parameters that are different, or is the measured excess velocity simply too large for being caused by something like that? – Bjorn Wesen Sep 23 '11 at 9:59
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No, the detectors are not identical, but the offset they're measuring is not just what they read off their clocks. They account for the time it takes to process the signal and work backwards from their measurements to determine the time at which the neutrino actually interacted with the detector. You can see their analysis in section 6 of the paper. There is some uncertainty in exactly how much time this takes, but it's much smaller than the time difference they detected. So it would appear that differences between the detectors are not the cause of the time difference. – David Zaslavsky♦ Sep 23 '11 at 10:13
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They are not actually using a near detector at all in the usual sense, they are measuring the beam current directly after the pick off magnet, and then correcting for beam TOF down to the target. This is a place that people are examining for subtle effects. I found that odd given that they do have a downstream muon detector system, but they may be concerned about backgrounds. – dmckee♦ Sep 23 '11 at 18:13
UPDATE 2011-10-15
This phenomena may have been explained. The crux of the problem had to do with differing reference frames - the distance traveled according to the satellites which measured the time was different from the distance traveled according to us on earth. If you're going to measure speed (distance / time), you have to get the distance and time both from the same reference frame. We were getting distance from our reference frame and time from the (very fast) satellite's reference time.
To understand how relativity altered the neutrino experiment, it helps to pretend that we're hanging out on one of those GPS satellites, watching the Earth go by underneath you. Remember, from the reference frame of someone on the satellite, we're not moving, but the Earth is. As the neutrino experiment goes by, we start timing one of the neutrinos as it exits the source in Switzerland. Meanwhile, the detector in Italy is moving just as fast as the rest of the Earth, and from our perspective it's moving towards the source. This means that the neutrino will have a slightly shorter distance to travel than it would if the experiment were stationary. We stop timing the neutrino when it arrives in Italy, and calculate that it moves at a speed that's comfortably below the speed of light.
"That makes sense," we say, and send the start time and the stop time down to our colleagues on Earth, who take one look at our numbers and freak out. "That doesn't make sense," they say. "There's no way that a neutrino could have covered the distance we're measuring down here in the time you measured up there without going faster than light!"
And they're totally, 100% correct, because the distance that the neutrinos had to travel in their reference frame is longer than the distance that the neutrinos had to travel in our reference frame, because in our reference frame, the detector was moving towards the source. In other words, the GPS clock is bang on the nose, but since the clock is in a different reference frame, you have to compensate for relativity if you're going to use it to make highly accurate measurements.
The original paper publishing these findings is here: Times of Flight between a Source and a Detector observed from a GPS satelite.
Original Post
Sources: [1] (Associated Press), [2] (Guardian.co.uk), [3] (Original Publication - Cornell University)
Scientists around the world reacted with cautious shock on Friday to results from an Italian laboratory that seemed to show that certain subatomic particles can travel faster than light.
The journey would take a beam of light around 2.4 milliseconds to complete, but after running the Opera experiment for three years and timing the arrival of 15,000 neutrinos, the scientists have calculated that the particles arrived at Gran Sasso 60 billionths of a second earlier, with an error margin of plus or minus 10 billionths of a second. The speed of light in a vacuum is 299,792,458 metres per second, so the neutrinos were apparently travelling at 299,798,454 metres per second.
Ignoring the boilerplate media hype about the possibilities of time travel and alternate dimensions - I'm looking for academic sources that might suggest how this could be true, or alternatively, how this discrepancy could be accounted for.
I read the published article, Measurement of the neutrino velocity with the OPERA detector in the CNGS beam, with their findings. It looks like they took an insane amount of care with their measurement of distance and time.
One of the most common skepticism of people who no nothing about the experiment is stuff like:
You might worry about[...] have they correctly accounted for the time delay of actually reading out the signals? Whatever you are using as a timing signal, that has to travel down the cables to your computer and when you are talking about nanoseconds, you have to know exactly how quickly the current travels, and it is not instantaneous. [2]
This experiment doesn't use that sort of 'stopwatch' timing mechanism though. There is no 'T=0', and no single firing of neutrinos. What is detected is watermark patterns in the steady stream of particles. The streams at the input and output are time stamped using the same satellites and any position along each stream has a precise time associated with it. By identifying identical patterns at input and output streams, they can identify how long it took particles to travel between the points. [1]
As for distance, they use GPS readings to get the east, north, and altitude position along the path travelled to great precision. So much so that they even detect slow earth crust migration and millimetres of changes in distance between source and destination when something like an earthquake occurs. When your particles are travelling on the scale (730534.61 ± 0.20) metres, this is more than enough precision:
It's going to take a lot more than grassroots skepticism to think of what could have caused this discrepancy. I've seen suggestions such as the gravity of the Earth being different along the path of the neutrinos, which warps space/time unevenly. The neutrino might not actually be travelling as far as they think if space/time is contracted at one or more points along the path where gravity varies.
Anyway, I'll be interested in seeing how it pans out. Like most scientists, my guess is an unaccounted for systematic error (because they definitely have statistical significance and precision on their side) that has yet to be pointed out, but it probably won't take too long with all the theoretical physicists that will be pouring through this experiment.
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One possibility is that the widespread use of GPS for measurments of earth has redefined the meter. The meter is defined as a specific fraction of the speed of light in vacuum. The GPS is not working in vacuum but its electromagnetic pulses go through the atmosphere and ionosphere and are corrected for that. If a systematic error enters there though, the fact of the precision of measurement with GPS, not disputed, would be a demonstration of the difference between accuracy and precision. The neutrinos are little affected by matter and seem to be covering more "meters" than vacuum meters. – anna v Sep 25 '11 at 12:55
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The problem with the GPS position measurements (I think that the time measurements are accurate) is that the relative position is not subject to the same systematics as the aboslute position. The different rotational velocity at Geneva vs. Central Italy gives diurnal abberation which must be corrected for to get an accurate absolute distance. You must convince yourself that the absolute measurements have the same error bars as the relative measurements, and I did not see that in the arxiv paper. – Ron Maimon Sep 25 '11 at 20:32
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– dmckee♦ Oct 22 '11 at 21:11
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There have been plenty of papers (well, preprints) have been put forward offering various explanations of the OPERA results, but none of them has been widely accepted yet as far as I know so it's rather premature to say the results have been explained. It's a nice answer other than that, though. – David Zaslavsky♦ Nov 4 '11 at 3:16
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@dmckee: The "partial apology and retraction" is not an apology or a retraction. The explanation for the error provided is cogent, clear, and almost certainly correct. The author is only clarifying that the GPS community doesn't need to read his paper, because it has no impact GPS best-practices, since the issue of precise time-of-flight is not relevant for most GPS uses. – Ron Maimon Nov 4 '11 at 3:45
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According to Dr. Phil Plait, there's a rumour that it's been a faulty connection. In summary: nothing is wrong with the calculation, the theoretical assumptions, rotation of the Earth, etc... A hardware problem caused the 60 ns time gap.
It's still gossip, so take this with abundance of caution, but here's what he had to say:
According to sources familiar with the experiment, the 60 nanoseconds discrepancy appears to come from a bad connection between a fiber optic cable that connects to the GPS receiver used to correct the timing of the neutrinos’ flight and an electronic card in a computer. After tightening the connection and then measuring the time it takes data to travel the length of the fiber, researchers found that the data arrive 60 nanoseconds earlier than assumed. Since this time is subtracted from the overall time of flight, it appears to explain the early arrival of the neutrinos. New data, however, will be needed to confirm this hypothesis.
—source
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This doesn't seem right--- could a hardware problem actually do this? Can you plausibly make a 60ns delay by a loose cable? Usually, you just lose some pulses travelling down the cable. I find it hard to believe its hardware. If I were conspiratorially minded, I would say they are covering up an uncorrected relativistic effect with a bogus story of a hardware error. How could a hardware error cause a systematic bias through two different runs of the same size. They have an incentive to lie, and they are incompetent, and incompetent people lie. – Ron Maimon Feb 23 '12 at 13:52
Suppose this is real, that the neutrinos arrive very slightly faster than light would through the vacuum. Wouldn't that point to there being a slightly higher c which actually limits speeds, and some slight slow-down for light from this maximum due to interactions of the electromagnetic field with other particles, including virtual particles?
After all, you can move an electron faster than a photon in glass, and we don't call it the end of relativity, we call it Cherenkov radiation.
So the definition of refractive index might need adjusting, but effectively the vacuum has a non-zero refractive index, or rather the vacuum is not entirely empty. Which we know.
It makes sense that a neutrino is not subject to the same interactions, given its famed reluctance to interact with anything. Perhaps it is just an indication that the particles in a vacuum are more likely to be electromagnetic-interacting than weak-interacting.
Or am I labouring under a false premise? Is the speed of light in a vacuum already adjusted for virtual particle interactions?
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You would still need to explain why a massive particle (the neutrino) moves faster than a massless particle (the photon). – Sklivvz♦ Sep 23 '11 at 12:37
@Sklivvz The mass of the neutrino is so small that it is irrelevant in the argument, if the refraction is of the order of magnitude of the measurement. We end up with statistical errors. I think what is true is that the group velocity of light as assumed by the experimenters is shown to be smaller than the group velocity of the neutrinos as measured by them. "Assumed" because there is no discussion of the effect of the collective refraction index due to the atmosphere, ionosphere, magnetic field (and maybe etc) of the earth in the measure of time they use. – anna v Sep 23 '11 at 13:21
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@Sklivvz: a massive particle moving faster than massless photons is what also happens in Cherenkov radiation. What one would need to explain is why hadrons and non-neutrino leptons experience exactly the same "braking" effekt as photons do. – leftaroundabout Sep 23 '11 at 13:33
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This is not supported by the supernova data. – dmckee♦ Sep 23 '11 at 18:15
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@leftaroundabout: we can only measure the speed of light in a vacuum through a vacuum. So given a constant density of vacuum particles, the speed of light through the vacuum would always be constant. Only with a different particle (e.g. a neutrino) would we be able to measure a higher speed. Inevitably, if this turned out to be the case, the real upper limit is slightly higher again, since neutrinos are massive and thus move below the maximum speed. – Phil H Sep 26 '11 at 11:47
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•Is this even remotely possible?
Well yes, of course it's possible in the same way that it's possible that invisible neutrino fairies are messing around with the neutrinos underground and hence causing havoc with the mental health of physicists around the world. It's just... unlikely, very unlikely, just as the 4-sigma evidence for new CP violation in like-sign dimuons was possible, only to fall flat on its face when ATLAS and CMS failed to see the same thing. But, it's still possible! Even so, let's focus on what's more likely: There are no neutrino fairies, and the conflict between data and special relativity lies with >> 6-sigma likelyhood of it being an error with the experiment.
•If so, would it be a real violation of Lorentz invariance or an "almost, but not quite" effect?
It would be one hell of a kick-up-Einsten's-backside violation of Lorentz invariance. All particles show the same speed limit as light, yet neutrinos with a rest mass greater than light possess a larger speed limit?
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No one has forgotten this. I can assure you that the OPERA people are acutely and painfully aware of the long history of highly "significant" bumps just going away. // also in Big Physics (tm) – dmckee♦ Sep 22 '11 at 21:01
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– anna v Sep 23 '11 at 5:01
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Given how big this question is, maybe it would be best to delete this answer? – Jonathan. Sep 25 '11 at 20:16
@jonathan I'll delete my answer if neutrinos travelling faster than c is confirmed, big question or not ;) – Physiks lover Sep 25 '11 at 23:19
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@jonathan light travels at a velocity below c in fibre optic cable. – Physiks lover Sep 28 '11 at 20:50
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There are strong reasons for disbelieving this result.
[This paragraph is disproved by the Nov. 17 result.] It uses an experimental design that was never intended for this purpose, and that is inherently poorly suited to it; the beam pulses were 10,000 ns wide, and the shift they claim to have measured is only 60 ns. This means that the shift can only be detected statistically, and it makes the result extremely vulnerable to unanticipated systematic errors, e.g., correlations between the time of emission of the neutrinos and their energy (which strongly affects the efficiency of detection) or the direction of emission. They did another run at the end of October, with beam pulses 1-2 ns wide. That's the correct design if you want to measure the speed of the neutrinos reliably. They should have simply waited until after they had those data before announcing their results. (In fact, five senior members of the collaboration did not put their names on the paper.) I have a bet running with a colleage, for a six-pack of Fat Tire, that the new run will show that the original result was bogus. The result may be announced as soon as November or December. [The result was announced Nov. 17, and I lost my six-pack.]
Another reason to disbelieve it is that there are strong and fairly model-independent reasons to believe that it cannot be correct. A superluminal neutrino beam would have lost a lot of its energy via radiation, but a measurement by another detector shows that this was not the case: http://arxiv.org/abs/1110.3763 Superluminal motion for neutrinos would also cause superluminal motion for electrons, which is contrary to observation http://arxiv.org/abs/1109.5682 , and it would also have caused a suppression of pion decay, so that the beam could never have been produced in the first place http://arxiv.org/abs/1109.6630 . All of this holds regardless of the details of the model. E.g., it holds both for tachyonic neutrinos without a preferred frame and for models in which neutrinos are not tachyonic and there is a preferred frame.
Yet another reason for disbelief is that the velocity of propagation of neutrinos has been measured to much higher precision by other techniques, so if you want to believe the OPERA result, you have to posit a very strange energy-dependence of the velocity.
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is this the result of the experiment you're talking about? arxiv.org/abs/1109.4897v2 ... science20.com/quantum_diaries_survivor/… – Hrant Khachatrian Nov 18 '11 at 14:02
@Hrant Khachatrian: Yes. It shows that the effect was not a statistical artifact as I proposed above. IMO what really needs to happen now is two things: (1) Other groups will try to reproduce the anomaly. (2) OPERA should try to verify that the anomaly has an energy dependence. (If the result is wrong, then it should be independent of the energy.) – Ben Crowell Nov 23 '11 at 21:05
@BenCrowell, if this were verified, what energy-dependent effects would we see in Nature? By analogy, if Einstein relativised the classical picture, how would this result "relativize" Einstein's theory of gravity? – bwkaplan Nov 23 '11 at 22:30
May be the case that this problem has to do with the «one-way» light speed and the referential that is used. Afaik the only known measures of the c are done in a «two-way» version (mean value in a closed path). When a photon is released in space it starts its journey at c speed independently of the source and of the receiver. The CMB referential clearly is the only referential to «observe» the light as isotropic. As the Earth moves we observe a dipole, and in different directions we measure different wavelengths for the same physical object (photon).
This paper (Cosmological Principle and Relativity - Part I) analyses the anisotropy of light speed for a moving observer.
Fig.3 and eq. 18, pag 14
The one-way light speed is : $c_{A}^{r}=\frac{c_{0}}{1+V/c_{0}\cdot\cos\phi_{A}}$
Using $c_0=299792.458$ Km/s is two-way light speed, $V\;$ is the speed of the lab in relation to the CMB: $V=V_{SS}+V_E$=369$\pm$30 km/s (data from here)
gives the max value of $\frac{\left|c_{V\pm\delta V}-c_{V}\right|}{c_{V}}\cdot10^{5}$=10.2.
All experimental measures of |v-c|/c are within this limit.
Anyway Einstein is correct.
I suspect that the syncronization used in the GPS is in the same as in the above paper and not as Einstein did.
In the pic Sat A must be synchronized with C at the same time thru the shortest red path and thru the longest blue path. At the same time B is in sync with C thru other paths with different lengths. IMO this is only possible if they are synchronised as in the above paper (instant observer) and not in the Einstein way that only considers one path between the observer and any other point (Synchronisation around the circumference of a rotating disk gives a non vanishing time difference that depends on the direction used).
Anyway Einstein is correct, and the neutrinos are not superluminal.
A careless reading of the paper might make you think that it is contrary to Einstein, but it is not.
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I will bet all my beans into the idea that they didn't estimate the spacetime curvature inside the earth well and over the beam trajectory, and what they actually discovered is a great way to measure space-time inside the Earth.
EDIT it seems this effect is settled to be a missing correction due to sattelite-speed terms: http://arxiv.org/abs/1110.2685. Until i hear or read any counter-claims to that paper, i'll consider this to be a settled matter
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Maybe a control would be to send photons along the same trajectory and measure THEIR speed? – Lagerbaer Sep 23 '11 at 17:38
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General relativistic effects near the surface of the Earth are of order $(9\text{ mm})/(6400\text{ km}) \approx 10^{-9}$. It is less important that the rotation of the Earth. – dmckee♦ Sep 23 '11 at 18:08
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@Lagerbaer I think the trajectory is all underground... it starts in a deep tunnel at CERN and ends under a mountain at Gran Sasso :-) – Sklivvz♦ Sep 23 '11 at 20:40
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@nominator: Any relativistic effect cannot make the speed superluminal. The only explanation is systematic errors in GPS position, GPS time, or bunching statistics. – Ron Maimon Sep 25 '11 at 20:34
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@Ron, any (general) relativistic effect cannot make the speed superluminal, but it can make your length measurement based on GPS incorrect. Read again what i wrote – lurscher Sep 26 '11 at 19:36
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Did anybody catch this?
Speedy neutrino result may be due to instrument glitch
http://www.newscientist.com/blogs/shortsharpscience/2012/02/speedy-neutrino-result-may-be.html
Loose Cable Explains Faulty 'Faster-than-light' Neutrino Result
http://www.space.com/14654-error-faster-light-neutrinos.html?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+spaceheadlines+%28SPACE.com+Headline+Feed%29
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there are updates to the question – anna v Feb 23 '12 at 13:54
There was a very reliable report of finding a monopole in 1980s by Caberera(?). There was no other explanation of the glitch in the arrangement of the SQUID, but a capture of one monopole. That never repeated. Never confirmed. Never rejected as being a fake effect. I believe this question needs a couple of years more investigation. A question of physics should be repeatedly be confirmed before a postulate or an inference can be derived. Even after that derivation a sensitive experiment should be perceived to break it through further.
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This probably should be a comment. Even so, this very experiment was a repeat of a MINOS experiment, which found the same effect at much lower levels of confidence, and this time it involved 15.000+ neutrino detections (which, however, could not be individually labelled faster or slower than light) – MSalters Oct 3 '11 at 14:05
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In addition, this paper was signed by a large collaboration. – Carl Brannen Nov 6 '11 at 3:21
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@Carl: and this is supposed to make one trust their report more? Large groups are vastly stupider than competent individuals. – Ron Maimon Nov 21 '11 at 16:52
This is not a true answer... none is knowing the explanation, so far. However, I will post this "consideration" anyway... Now, November 21, 2011, with 3ns pulses, the new value for the "missing time" is 62.1ns +/-3.7 (only 20 events). My answer is only a "would-be" consideration that, if read by the experimenters, could give them some "debug" clues.
First of all my assumptions:
it is unlikely that the neutrinos go superluminal or SR is not holding true anymore
• it is unlikely that the distance is measured incorreclty
• it is unlikely that the GPS setup/usage is incorrect
• it is also unlikely that the light speed has been measured incorreclty so far.
Then two hypotesis:
• the "missing time" is 62.5ns (compatible with 62.1 +/-3.7ns)
• the electronics involved in the time measurement has some clock domain running at 16MHz.
Of course the conclusion would be to investigate if there is one circuit running on one clock pulse less than expected by design / testing.
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I do not agree with the superluminal neutrinos news for very simple reasons. The difference they found with respect to the speed of light is very small, so some errors in the calulations must have been made. Neutrino is not faster than light. The Special Theory of Relativity (STR) of Einstein, through the principle of the speed limit, makes the magnetic force come from the electric one and the magnetic force is an electric force, as physicists know; an easy demonstration of that can be found in chapter 3 of my file at the following link (also English inside):
http://www.fisicamente.net/FISICA_2/UNIFICAZIONE_GRAVITA_ELETTROMAGNETISMO.pdf
If you get rid of the speed limit principle, the magnetic field cannot exist anymore.
Moreover, as c=1/square root of(epsilon x µ), if you change c with a c'>c, then you have to accept a µ'<µ, so you have to accept different intensities of magnetic fields from a given electric current, so you have to get rid of the electromagnetism, but it's describing so well the currents, the fields, the real world etc. Therefore, there's a mistake in the computation of the speed of neutrinos, in the calculations on the run lenght, in the interaction time calculations, during the generation and also the detection of those evanescent particles!
(another interesting file, also related to this subject):
http://www.mednat.org/new_scienza/strani_legami_numerici_universo.pdf
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## protected by David Zaslavsky♦Nov 4 '11 at 3:27
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://mathhelpforum.com/geometry/1884-solved-geometry-shapes.html
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Thread:
1. [SOLVED] geometry shapes
i need to find the area of a octagon with a perimeter of 40 ft and also find the same thing for a triangle and trapezoid. can you please help me fast?
2. Originally Posted by britt12790
i need to find the area of a octagon with a perimeter of 40 ft and also find the same thing for a triangle and trapezoid. can you please help me fast?
Assuming it is a regular octagon meaning that angles and sides are equal then using the formula,
$A=\frac{1}{2}s\sin\frac{180}{n}$
Where $n$ is the number of sides and,
$s$ is the length of the sides.
But since it perimeter is 40 and there are 8 sides then each sides is 5. Thus,
$A=\frac{1}{2}(5)\sin\frac{45}{2}$
But $\sin\frac{45}{2}=\frac{\sqrt{2-\sqrt{2}}}{2}$.
Thus,
$A=\frac{5}{4}\sqrt{2-\sqrt{2}}$
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http://mathoverflow.net/questions/73098/negative-values-of-riemann-zeta-function-on-the-critical-line/73143
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## Negative values of Riemann zeta function on the critical line.
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From parametric plots of $\zeta \left( \frac{1}{2} + it \right)$ it seems to be the case that:
(1) except for $\zeta(\frac{1}{2})$ the Riemann zeta function does not attain any negative real value on the critical line.
(2) the curve $(t, \zeta(1/2+it))$ is dense in the complex plane.
Are these statements known to be false, if not, is there any proof affirming them?
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1
(2) is an open problem and is conjectured to be true. This conjecture is usually attributed to Ramachandra. – Micah Milinovich Aug 17 2011 at 20:19
1
For a recent paper supporting this conjecture see math.ethz.ch/~kowalski/zeta-density.pdf – quid Aug 17 2011 at 20:27
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The first conjecture seems to be disproved by the graph at fredrik-j.blogspot.com/2010/01/… It doesn't say which of the curves shows the real part and which the imaginary, but either way there's a $t \in (10^{10}+53,10^{10}+54)$ where one of them crosses the $x$-axis and the other is negative. – Noam D. Elkies Aug 17 2011 at 21:17
If the dotted line is the imaginary part, instead of the dashed line, then there is no point in this graph where the imaginary curve crosses 0 and meanwhile the dashed curve is negative. But I do get the point. – Eren Mehmet Kiral Aug 17 2011 at 22:21
True — I somehow misread the graph. But see below for an actual counterexample. – Noam D. Elkies Aug 17 2011 at 23:58
## 5 Answers
A numerical counterexample to the first conjecture is $$t = 282.4547208234621746108397940690599354\ldots$$ where both gp and Wolfram Alpha agree that $\zeta(\frac12 + it)$ has negative real part $\simeq -0.02763$ and negligible imaginary part, so the actual zero of ${\rm Im}(\zeta(\frac12+it))$ near $t=295.5839\ldots$ yields a negative value of $\zeta(\frac12+it)$.
This was found by approximating $\zeta'(\frac12+it)$ at each of the first "few" zeros of $\zeta$ tabulated by Odlyzko in http://www.dtc.umn.edu/~odlyzko/zeta_tables/zeros1 and looking near the first zero (the 127th overall) at which $\zeta'$ has negative imaginary part. There are $22$ such zeros of the $649$ zeros whose imaginary part lies in $[0,1000]$; there's probably a counterexample near each of those, e.g. looking around the second such zero (#136) yields $$t = 295.583906974228176092587915204356841\ldots$$ with $\zeta(\frac12+it) \simeq -0.0169004$.
EDIT 1) Henry Cohn (in a comment below) provides gp code that looks for solutions in an interval by dividing it into segments $(t_0, t_0 + 0.01)$, testing whether ${\rm Im}(\zeta(\frac12+it))$ changes sign between the endpoints, and if so whether the real part is negative at the crossing. Extending his computation to $0 \leq t \leq 1000$ finds the expected $22$ solutions; in particular $282.45472+$ seems to be the first.
2) Once one has calculated an answer one can ask Google for its previous appearances. Google recognizes $282.45472$ from J.Arias-de-Reyna's paper "X-Ray of Riemann zeta function" (http://arxiv.org/abs/math/0309433) where it appears (to within $10^{-5}$) as the first counterexample to "Gram's law" — see the plot on page 26 (thick and thin curves show where $\zeta(s)$ is real and imaginary respectively).
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7
Incidentally, these seem to be the first two counterexamples. Here's some pari code: {for(i=1,300*100, if(imag(zeta(1/2+I*i/100))*imag(zeta(1/2+I*(i+1)/100))<0, x = solve(y=i/100,(i+1)/100,imag(zeta(1/2+I*y))); if(real(zeta(1/2+I*x))<-10^(-10),print(x))));} – Henry Cohn Aug 18 2011 at 0:47
That would have looked a little nicer if markdown worked in the comments, but at least the code runs. – Henry Cohn Aug 18 2011 at 0:49
@Henry: Thanks, that was what I guessed too. Extending the computation to imaginary part $10^3$, which takes a bit under 10 minutes here, finds 22 solutions, again corroborating the suggestion that they are near zeta zeros $\frac12 + i \gamma$ with negative ${\rm Im}(\zeta'(\frac12+i\gamma))$. [But eventually there are bound to be exceptions when the image of the critical line under $\zeta$ gets more complicated, or simply when some $\zeta'(\frac12+i\gamma)$ is very close to the real axis.] – Noam D. Elkies Aug 18 2011 at 1:34
I do not get why a zero of $\zeta$ with $\Im(\zeta')$ negative at that point would imply that a counterexample to (1) is near, even heuristically. – Eren Mehmet Kiral Aug 18 2011 at 8:19
@E.M.Kiral – to see why sign of ${\rm Im}(\zeta'(\frac12+i\gamma))$ might be relevant: look at a plot (such as the first plot in G.Helms' post) of the image under $\zeta$ of an initial segment of the critical line, and imagine what it would take for one of the arcs joining consecutive zeros to pass through the negative real axis. – Noam D. Elkies Aug 18 2011 at 14:30
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The reason (1) 'appears' to be true for small $t$ is related to Gram's Law for the zeros of $\zeta(s)$. Edwards' book Riemann's Zeta Function (Dover) has a good explanation starting on p.125. The short version is that the Euler Maclaurin formula for $\zeta(1/2+i t)$ starts with a $+1$, and,
"as long as it is not necessary to use too large a value of $N$, it will be unusual for the smaller terms which follow to combine to overwhelm this advantage on the plus side. As Gram puts it, equilibrium between plus and minus values of Re$\zeta$ will be achieved only very slowly as $t$ increases."
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Is it known whether the apparent bias in the value distribution of $\mathrm{Re}\zeta(1/2+it)$ dissolves as $t \to \infty$? – David Hansen Aug 18 2011 at 0:11
Paul, the zeta function is not real-valued on the critical line. – David Hansen Aug 18 2011 at 0:12
Here is the graph of real and imaginary parts along the critical line, from wikipedia: en.wikipedia.org/wiki/… with $t$ (they call it $x$) between -30 to 30 – Will Jagy Aug 18 2011 at 1:20
I agree that Gram's Law is the thing that is relevant. – Junkie Aug 18 2011 at 4:47
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Sorry, my earlier comment that zeta itself is real-valued on the criticial line is of course false, but the normalized version from the functional equation is. The gamma factor is non-vanishing, etc. – paul garrett Aug 18 2011 at 12:41
The zeta function is real on the critical line only at the zeros and at Gram points, this is because zeta(1/2+it)=exp(-ivartheta(t)) Z(t).
At the Gram point g_k we have by definition vartheta(g_k)=pi k. so that zeta(1/2+ig_k) =(-1)^k Z(g_k).
Now a Gram point g_k is said a good Gram point if (-1)^k Z(g_k) >0. In other case it is said a bad Gram point.
Since it appear improbable a zero just at a Gram point. You are asking if there exists bad Gram points, there are plenty. The first few bad Gram points are g_126, g_134, g_195, g_211, ...
g_126 = 282.45472082346217461077
In fact it is proved there are infinite bad Gram points.
Also we may easily obtain large negative values. For example using data of T. Kotnik "Computational estimation of the order of zeta(1/2+it) Math of Comp. (2003) we easily locate the point t = grampoint(2601005843707) were we have
zeta(0.5+i t) = -119.6304321077241661374
This is easily confirmed in mpmath (or Mathematica) ( grampoint(2601005843707) = 669980906189.53552206792 ).
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Update, some recent information on (1):
Kalpokas, Korolev, Steuding recently released a preprint showing that $\zeta(1/2 + it)$ takes aribtrarily large positive and negative (real) values; and also show analog statements for the other lines through the origin, that is positive and negative (real) values of arbitary says of $e^{-i \phi} \zeta(1/2 + it)$ for any $\phi$. The paper contains also more quantitative results along these lines (cf. in particular Corollary 3 and the preceeding discussion).
Since (1) already received several answers, I expand and upgrade the comments on (2):
Yes, indeed it is conjectured, but unproved, that $\zeta(1/2 + i t)$ for $t \in \mathbb{R}$ is dense in the complex plane. [Side note: It is well-known that this is so for the lines $\sigma +it$ with $1/2 < \sigma < 1$.]
It seems that this conjecture was first formulated by Ramachandra (Durham, 1979), however only appeared in print in the second edition of Titchmarsh's book (note's by Heath-Brown), see the articles below for details.
There is very recent work on this problem due to Delbaen, Kowalski, and Nikeghbali. See in particular this preprint by the latter two and this by all three. Among others: in the former, they show how this result would follow "from a suitable version of the Keating--Snaith moment conjectures"; in the latter, they propose a refinement of the density conjecture, a quantitative version (see Conj. 1, in Sec. 3.9).
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@ Q1: After the counterexample of Noam Elkies I used Pari/GP to draw that parametric plot to get more visual impression;
[update] The visual impression in the 1:1000 zoomed picture had artifacts; I deleted the picture and provide a more precise one and corrected in my original answer [/update]
Plot 1 shows the known curve in the complex plane, when t increases from 0 to 100:
``` \\ Pari/GP:
ri_zeta(t)=local(tmp);tmp=zeta(1/2+I*t);return([real(tmp),imag(tmp)])
ploth(x=0,100,ri_zeta(x),1)
```
From the drawing one cannot discern, whether there is some crossing of the negative real axis. Here is a rescaling; the values of the zeta-function are scaled by the tanh-function:
``` \\ Pari/GP:
ri_zeta(t)=local(tmp);tmp=10*zeta(1/2+I*t);return([tanh(real(tmp)),tanh(imag(tmp))])
ploth(x=0,100,ri_zeta(x),1)
```
and then a strong scaling factor of 1:1000 applied. [update] To remove artifacts, there is an option "recursive" in the plot-routine to scatter the coordinates more regularly; the strong zoom separated the dots of the plot too much so that artifacts are likely to occur. With an improvement of the sampling no crossings of the negative real axis can be seen [/update]
``` \\ Pari/GP:
ri_zeta(t)=local(tmp);tmp=1000*zeta(1/2+I*t);return([tanh(real(tmp)),tanh(imag(tmp))])
```
I used internal precision of 200 dec digits, [update] so I think the computation of the single points do not introduce artefacts, but the connection by lines may do due to the strong scaling required. This type of plotting seems to require much resources; I'll see whether it can verify the crossing in the near of t=282 visually; I'll update then this answer again.
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This is suspicious because it would mean that Henry's computation missed some solutions with $t<100$. That is possible but unlikely: there would have to be at least two solutions in an interval $(t,t+0.01)$. Even more suspicious are angles in what purports to be the graph of a smooth function ;-) It must be that only the vertices, not the line segments between them, come from actual data points accurate to about $10^{-200}$. I'd expect that each of the crossings of the real axis, whether positive or negative in the above picture, is exactly at zero in reality. – Noam D. Elkies Aug 18 2011 at 2:39
[that is, the crossings might not be artifacts, but their position relative to the origin is.] – Noam D. Elkies Aug 18 2011 at 3:26
@Noam: you'right, that were artifacts. When I plotted without joining lines then I could see, that the single dots were far apart of each other so the joining lines crossed the axis at artificial points. An improvement of the plot routine showed then no more such crossings. Unfortunately the applied method is still too rough to show the actual crossing to which you refer in the even improved picture for the range t=200 to 300. So that all has only been "a good idea"... – Gottfried Helms Aug 18 2011 at 7:26
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http://physics.stackexchange.com/questions/41738/can-auxiliary-fields-be-thought-of-as-lagrange-multipliers
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# Can auxiliary fields be thought of as Lagrange multipliers?
In the BRST formalism of gauge theories, the Lautrup-Nakanishi field $B^a(x)$ appears as an auxiliary variable $$\mathcal{L}_\text{BRST}=-\frac{1}{4}F_{\mu\nu}^a F^{a\,\mu\nu}+\frac{1}{2}\xi B^a B^a + B^a\partial_\mu A^{a\,\mu}+\partial_\mu\bar\eta^a(D^\mu\eta)^a,$$ and in the superfield formalism of SUSY, the field $F(x)$ also appears as an auxiliary variable: $$\mathcal{L}_\text{SUSY}=\partial_\mu \phi\partial^\mu\phi+i\bar\psi^\dagger\bar\sigma^\mu\partial_\mu\psi+F^*F+\ldots\,.$$
It is very tempting to view $B^a(x)$ and $F$ as Lagrange multipliers since their equations of motion leads to constraints. But, these variables do not enter into the Lagrangian linearly, like a conventional Lagrange multiplier. Rather, they enter into the Lagrangian quadratically.
However, in Kugo and Ojima's paper Manifestly Covariant Canonical Formulation of the Yang-Mills Field Theores (1978), they refer to the $B^a(x)$ fields as the 'Lagrange Multiplier' fields (p.1882).
So my question is: Can these auxiliary fields be viewed as Lagrange multipliers? and in what ways do they behave differently/similar to the conventional Lagrange multipliers that enter into the function linearly?
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## 2 Answers
By definition, Lagrange multipliers are only coefficients that enter the extremized quantity (action) etc. linearly – and that multiply constraints. In some exceptional cases, an auxiliary field could enter in this way. However, they typically appear in a more complicated way and bilinear terms in the auxiliary fields are a rule rather than an exception. So strictly speaking, they're not Lagrange multipliers. But they are very similar. If no derivatives of these objects appear in the action, they're also "non-dynamical" (not involving time derivatives) and the variation with respect to them implies "non-dynamical" i.e. algebraic equations of motion.
Note that in the normal treatment of extremization, we introduce Lagrange multipliers because we want to extremize the quantity given the assumption that another quantity or other quantities are kept fixed. "Kept fixed" is translated as "conservation laws" into the physics jargon. However, in physics, we rarely consider conserved quantities that are conserved because the conservation law is explicitly written down as an independent constraint. Instead, in physics we usually discover conservation laws nontrivially – the conserved quantity has to be determined by a somewhat non-trivial procedure due to Emmy Noether out of a symmetry. In almost all physical theories, conservation laws are non-trivial consequences of some other, "more elementary" equations of physics.
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Nice clarification, I like this :-) – Dilaton Oct 26 '12 at 10:57
1) OP writes (v1):
Can auxiliary fields be thought of as Lagrange multipliers?
No, not necessarily. Auxiliary fields usually mean non-propagating fields, and there may be other non-propagating fields, e.g. ghost fields and antighost fields. In case of so-called reducible gauge symmetries, one also has e.g. ghosts-for-ghost fields. Moreover, if one works in the Hamiltonian formalism, one has non-propagating momentum fields for all the above mentioned auxiliary fields. In fact, the converse statement is true: Lagrange multipliers are examples of auxiliary fields.
2) Strictly speaking according to the original definition, it is true that the Lagrange multipliers $\lambda^a$ should enter linearly (as opposed to e.g. quadratically) in the action,
$$S~=~\int \!d^4x ~{\cal L}, \qquad {\cal L}~=~ \ldots + \lambda^a \chi_a+ \ldots,$$
where $\chi_a\approx 0$ are the conditions that we impose via the Lagrange multiplier method.
In Lagrangian gauge theories, the conditions $\chi_a$ are typically gauge-fixing conditions, and it turns out that when treated consistently$^1$, the gauge-invariant physical observables do not depend on the choice of gauge-fixing conditions $\chi_a$.
This independence of the gauge-fixing conditions $\chi_a$ extends to situations where the gauge-fixing conditions $\chi_a$ themselves depend e.g. linearly on the auxiliary $\lambda$ fields, so that the action depends quadratically on the $\lambda$'s.
Many aspects of a gauge theory can be discussed before choosing particular gauge-fixing conditions, and in practice the auxiliary $\lambda$ fields are referred to as Lagrange multipliers anyway whether or not the gauge-fixing conditions $\chi_a$ themselves depend on $\lambda^a$. See also e.g. this Phys.SE answer.
For example, the gauge-fixed Yang-Mills action that OP mentions(v1) can precisely be viewed as a situation where the gauge-fixing conditions $\chi_a$ themselves depend linearly on the Lautrup-Nakanishi fields $\lambda$.
--
$^1$ Note that the Faddeev-Popov determinantal term also depends on the gauge-fixing condition $\chi_a$. For general gauge theories, a consistent treatment is given by the Batalin-Vilkovisky (BV) recipe, cf. e.g. this Phys.SE answer.
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This (and the answer linked to) is again an example of a very nice and pedagogical answer, explaining patiently and step by step how things work. Thanks for taking time to write so many of these extremely helpful posts here on physics SE, I always like them a lot :-) – Dilaton Oct 27 '12 at 10:59
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http://physics.stackexchange.com/questions/13695/a-priori-validity-of-w-int-fdx-in-relativity?answertab=oldest
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# a priori validity of $W=\int Fdx$ in relativity?
There are lots of different ways of arriving at the relativistic relations involving mass, energy, and momentum such as $E=mc^2$ and $m^2=E^2-p^2$ (the latter with $c=1$). One that I've seen in some textbooks is to start with the equation $W=\int F dx$ for mechanical work. Here is a particularly careful and rigorous version, which makes it explicit that this is a nontrivial assumption, along with $F=dp/dt$: http://www.physicsforums.com/showthread.php?p=2416765 Historically, Einstein made use of $W=\int F dx$ in section 10 of his 1905 paper on SR http://www.fourmilab.ch/etexts/einstein/specrel/www/ . This was before the "Does the inertia of a body..." paper, in which he derived $E=mc^2$ and gave it its full relativistic interpretation. Einstein later decided that force wasn't a very useful concept in relativity and stopped appealing to it.
What I have never seen in any of these treatments (neither the careful ones above nor the sloppy ones in some freshman physics texts) is any argument as to why the nonrelativistic relation $W=\int F dx$ should be expected a priori to hold without modification in SR. If one has already established the form and conservation of the energy-momentum four-vector, then I don't think it's particularly difficult to show that $W=\int F dx$. But what justification is there for assuming $W=\int F dx$ before any of that has been established? Is there any coherent justification for it?
(A secondary problem with some of these treatments based on $W=\int F dx$ is that they need to establish the constant of integration.)
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I do not see where the problem is. $W=\int F dx$ follows from the relativistic formula $F=\frac{dp}{dt}$ which has been proved experimentally. – Martin Gales Aug 18 '11 at 7:44
## 1 Answer
There is no reason for the formula to obviously work, as you say, just because there are at least two different generalizations of force, the proper-time relativistic curvature of the trajectory ($d^2x\over d\tau^2$ not the right notion) and the derivative of the relativistic energy times the velocity with respect to time (${d\over dt} (m{dx\over d\tau})$ the one that works). Which one is the proper generalization?
Einstein came from a thermodynamic background, and he always accorded the laws of thermodynamics an absolute status, being deduced as they were so inexorably from such elementary postulates. It was the deductive form of thermodynamics that led him to write the special relativity in such a deductive style, building up from simple axioms.
Potential energies, and energies in general, have a thermodynamic meaning mostly independent of their mechanical meaning, which is unchanged by relativity (if you like to think of this statistically, the Boltzmann distribution is still $\exp(-\beta E)$ in either theory). The electrostatic potential is the energy per unit charge in a region of space. The electrostatic energy of a charged ball moving with velocity v in an electrostatic potential $\phi$ can't depend on the velocity, because then I could transfer the charge from a fast moving object to a stationary object as the fast object comes whooshing by, and make a perpetual motion machine. So the potential energy of the charge is still $q\phi$ even in relativity, for thermodynamic reasons.
The electric field is the gradient of the static potential, and this is true in Maxwell's equations, which are already relativistic. The integral of the gradient of the static potential along the path, times the charge, then gives the work done along the path. So the correct notion of force to use is whatever mechanical thing is equal to $qE$. It is physically all but certain that different forces, no matter what their origin, do the same work if they have the same magnitude, so you can conclude that the work formula is generally true for this notion of force, which is the notion of force in the equation of motion. For Einstein this is so clear, he doesn't bother to explain.
This assumption seems unnatural to many people nowadays, since we are comfortable with symmetries much more than thermodynamics. But Einstein's exposition, survives in textbooks to this day.
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You're a gold mine on these historical topics! It seems that you're saying that the four-force $ma^j$, where $a^j$ is the acceleration four-vector and $m$ the invariant mass, is "not the right notion" of force. Why not? If I'm generalizing the three-force, I'd really dearly love to have the generalization be a valid four-vector. The quantity $d(E v^j)/dt$, where $E$ is the mass-energy, is not a four vector, so I react to it with horror. Why do you consider it "the one that works?" – Ben Crowell Aug 18 '11 at 1:53
I don't think that the four-force is "wrong", just wrong for generalizing the horribly non-covariant expression $\int F dx$. I personally am more comfortable with four-force, which is why I asked myself the exact same question "why does Einstein consider this formulat obvious?" many years ago. – Ron Maimon Aug 18 '11 at 2:01
The relationship between the second half of your answer and the first half confuses me. The first half says we have multiple possibilities for generalizing the notion of force to SR. The second half argues that $W=\int F dx$ holds in SR, without ever having to invoke either of the two definitions of F. Surely its validity should depend on which definition one uses...? – Ben Crowell Aug 18 '11 at 2:02
What I am saying is that the notion of force which can be seen to have the property that $W = \int F dx$ is whatever mechanical quantity that appears on the left hand side of "F=qE" in relativity (where the right hand side is just the spatial gradient of the electrostatic potential, without any velocity factors). Einstein figures out this quantity in his special relativity paper by transforming from a frame moving along with the particle. – Ron Maimon Aug 18 '11 at 2:11
Also, in skimming over the paper, he does say that this quantity must be the work done, because it must equal the gain in electrostatic potential energy along the integration path. This is the thermodynamic a-priori argument I was talking about. – Ron Maimon Aug 18 '11 at 2:12
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http://math.stackexchange.com/questions/116009/choices-for-integrationg-by-parts/116016
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# Choices for Integrationg by Parts
I am working on an integration by parts problem, and, it looks like I got something incorrect somewhere. I have been told to follow LIATE (L ogarithmic, I nverse trigonometric, A lgebraic, T rigonometric, E xponential) when making a choice for selecting the `u` term in integration by parts. So, here is where I start:
Find $\displaystyle \int \frac{x \cdot e^{2x}}{{\left(2x + 1\right)}^{2}}$
$u = {(2x + 1)}^{2} = 4x^2 + 4x + 1$
$du = 16x + 4 \ dx$
$v = \frac{x}{2} \cdot e^{2x}$
$dv = x \cdot e^{2x} \ dx$
... and the rest of the problem is solved in this manner.
In the end I come up with:
${\left(2x + 1\right)}^{2} \cdot \frac{x}{2} \cdot e^{2x} - \left(4x^2 + x\right)\left(e^{2x}\right) + \left(16x + 1\right)\left(\frac{1}{2} \cdot e^{2x}\right) - {8e}^{2x} \ dx$
Comparing my answer with that from the student solutions handbook, here is how they start:
Find $\displaystyle \int \frac{x \cdot e^{2x}}{{\left(2x + 1\right)}^{2}}$
$u = x \cdot e^{2x}$
$du = x \cdot 2e^{2x} + e^{2x} \ dx = e^{2x} \left(2x + 1\right) \ dx$
$v = -\frac{1}{2\left(2x + 1\right)}$
$dv = \frac{1}{{\left(2x + 1\right)}^{2}} \ dx$
... and the rest of the problem is solved in this manner.
In the end, they come up with:
$$\frac{e^{2x}}{4\left(2x + 1\right)} + C$$
The answer they came up with appears to be the case since they chose their `u` term differently than I did. I purposefully didn't make the choice they did, since it was an exponential function, which I was taught should be the last choice for `u`. I chose the algebraic function.
There must be something that I am missing. Could some one please show me as to where I went wrong, and perhaps some rules that would help me make better choices for my `u` term in the future?
-
In general, try to pick $u$ to be a function that has a simple derivative and $dv$ to be a function that has a simple integral. – Austin Mohr Mar 3 '12 at 17:09
Your calculation has errors. It should be $du=(8x+4)dx$. You had $16x+4$. You chose $dv=xe^{2x}$. But then $v$ is not what you say it is. You can check that by differentiating. The "rules" are not universal. Because of the very specific choice of $e^{2x}$, the derivative of $xe^{2x}$ happens to be $(2x+1)e^{2x}$, so there is miraculous cancellation. General rules can't take care of all very contrived special cases. – André Nicolas Mar 3 '12 at 18:15
## 2 Answers
The formula for integration by parts is $\int udv=uv-\int vdu.$ You have $\int\frac{dv}u,$ which won't work.
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Thank Mike, but I did use `v` and `du` in the integral. Since I used integration by parts twice, maybe that is why it looks as though I had done that. – spryno724 Mar 3 '12 at 17:16
@spryno724 What do you mean it looks like you did that? You selected a u that was in the denominator. Look at your choice of u and dv. $\int udv$ would be $\int(2x+1)^2xe^{2x}dx$, which was not what you started with... – Mike Mar 3 '12 at 17:21
Errummm.... that was silly mistake. Ok, that was my problem! Thank you for pointing that out Mike! How could I have missed that? – spryno724 Mar 3 '12 at 17:23
There's no hard-and-fast rule for integration by parts, much like the rest of integration. You have to (unfortunately =P ) apply your mental facilities here as well.
Over here, choice of $u$ eliminated a part of the denominator. A definite plus. Again, there's no rule for this, you've got to learn to perceive the hidden things like this.
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If that is the case, then shouldn't the answer come out the be the same, regardless of what (within reason) I pick? – spryno724 Mar 3 '12 at 17:17
@spryno724 Integration is not like algebra. There are multiple methods of simplification, but only a few will work for a given problem. You must pass through certain 'gates' to reach it – Manishearth Mar 3 '12 at 17:26
Wait, is your first answer the final answer or the resultant integrand? I sort of assumed the latter as it has a dx. If not, you've made a mistake somewhere. I thought you were asking why you get something easy to integrate with one substitution, and hard with the other – Manishearth Mar 3 '12 at 17:29
1
Your mistake is that your u was in the denominator. $\frac{x}{(2x+1)^2}$ makes more sense as u by LIATE. – Manishearth Mar 3 '12 at 17:35
1
@spryno724 possibly. No guarantee that you'll reach an answer, though. – Manishearth Mar 3 '12 at 18:15
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http://mathhelpforum.com/differential-geometry/107971-compute-curve-given-implicitly-print.html
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# Compute Curve Given Implicitly
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• October 14th 2009, 12:35 AM
bumcheekcity
Compute Curve Given Implicitly
So I have a curve given implicitly by the two equasions:
$\frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 = 1$
$a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$
So I've defined functions such that:
$F_{1}(x,y,z) = \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 - 1$
$F_{2}(x,y,z) = a\sqrt{b^2 - c^2}z - c\sqrt{a^2 - b^2}$
For the NEW curve (which I'll call C for argument's sake, and I appreciate here a sketch might be useful, but I don't know how to do one online), I'll let $p_{0}$ be the "centre" of the curve (which a sketch will show you is an ellipse/circle) and will define the vectors $e_{1}, e_{2}$ to be an orthonormal basis in this new curve.
I'm doing this to make it easier to paramatrise, as we will effectively take it down to 2-Dimensions, in this new basis.
$p_{0} = (0,0,0)$
$e_{1} = (a\sqrt{b^2 - c^2}, 0, c\sqrt{a^2 - b^2})$
$e_{2} = (0,1,0)$
Now for the bit I'm stuck on. I'd LIKE to substitute $p_{0} + \lambda e_{1} + \mu e_{2}$ into $F_{1}$ or $F_{2}$, but it just doesn't work. I don't get the equasion of a circle (our lecturer told us it would be a circle, because we have to arc length parametrise it later.
I've tried "swapping round" the values in $e_{2}$, but I'm sure they're right, and that gives me something even worse than what I get the other way. Any help?
• October 16th 2009, 04:29 AM
HallsofIvy
Quote:
Originally Posted by bumcheekcity
So I have a curve given implicitly by the two equasions:
$\frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 = 1$
$a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$
So z is a constant?
Quote:
So I've defined functions such that:
$F_{1}(x,y,z) = \frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 - 1$
$F_{2}(x,y,z) = a\sqrt{b^2 - c^2}z - c\sqrt{a^2 - b^2}$
For the NEW curve (which I'll call C for argument's sake, and I appreciate here a sketch might be useful, but I don't know how to do one online), I'll let $p_{0}$ be the "centre" of the curve (which a sketch will show you is an ellipse/circle) and will define the vectors $e_{1}, e_{2}$ to be an orthonormal basis in this new curve.
I'm doing this to make it easier to paramatrise, as we will effectively take it down to 2-Dimensions, in this new basis.
$p_{0} = (0,0,0)$
$e_{1} = (a\sqrt{b^2 - c^2}, 0, c\sqrt{a^2 - b^2})$
$e_{2} = (0,1,0)$
Now for the bit I'm stuck on. I'd LIKE to substitute $p_{0} + \lambda e_{1} + \mu e_{2}$ into $F_{1}$ or $F_{2}$, but it just doesn't work. I don't get the equasion of a circle (our lecturer told us it would be a circle, because we have to arc length parametrise it later.
I've tried "swapping round" the values in $e_{2}$, but I'm sure they're right, and that gives me something even worse than what I get the other way. Any help?
The original surface is a sphere so taking z= constant gives a circle in the z= constant plane.
From $a \sqrt{b^2 - c^2}z = c\sqrt{a^2 - b^2}$[/quote]
$a^2(b^2- c^2)z^2= c^2(a^2- b^2)$ so $\frac{z^2}{c^2}= \frac{a^2- b^2}{a^2(b^2- c^2)}$
$\frac{x}{a}^2 + \frac{y}{b}^2 + \frac{z}{c}^2 = 1$
$\frac{x}{a}^2 + \frac{y}{b}^2 = 1- \frac{z}{c}^2$
$\frac{x}{a}^2 + \frac{y}{b}^2 = 1- \frac{a^2- b^2}{a^2(b^2- c^2)}$
That's a circle with center on the z-axis at $(0, 0, z_0)= (0, 0, \frac{c\sqrt{a^2- b^2}}{a\sqrt{b^2- c^2}})$ and radius $r= \sqrt{1- \frac{a^2- b^2}{a^2(b^2- c^2)}}$ so a good parameterization would be
$x= r cos(\theta)$, $y= r sin(\theta)$, $z= z_0$, with $\theta$ going from 0 to $2\pi$.
I don't see any good simplification of $z_0= \frac{c\sqrt{a^2- b^2}}{a\sqrt{b^2- c^2}}$ or $r= \sqrt{1- \frac{a^2- b^2}{a^2(b^2- c^2)}}$.
• October 17th 2009, 11:24 AM
bumcheekcity
Bah, of course. All I had to do was sub in the new value of z in terms of x. Thanks, you're a lifesaver :D:D:D
All times are GMT -8. The time now is 04:36 AM.
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http://www.physicsforums.com/showpost.php?p=2626707&postcount=5
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Mentor OK, I think I see what I did wrong. $(P\land Q)\Rightarrow R$ is equivalent to $\lnot R\Rightarrow (\lnot P\land Q)\lor(P\land\lnot Q)$, and you can't just drop one of the terms on the right in the last expression, which is essentially what I did. So let's go back to the implication that I want to rewrite: $$(I\subset\tau)\ \land\ \bigg(\bigcup_{i\in I}i\supset K\bigg)\Rightarrow\exists I_0(I_0\subset I,\ |I_0|<\infty,\ \bigcup_{i\in I_0}i\supset K)$$ It implies several different things, one of which is $$\bigg(\forall I_0(I_0\subset I,\ |I_0|<\infty) \bigcap_{i\in I_0}F_i\cup K\neq\emptyset\bigg)$$ $$\Rightarrow \bigg((I\not\subset\tau)\land\bigg(\bigcup_{i\in I}i\supset K\bigg)\bigg)\lor\bigg((I\subset\tau)\land\bigg(\bigcap_{i\in I}F_i\cup K\neq\emptyset\bigg)\bigg)$$ And this implies $$(I\subset\tau)\land\bigg(\forall I_0(I_0\subset I,\ |I_0|<\infty) \bigcap_{i\in I_0}F_i\cup K\neq\emptyset\bigg) \Rightarrow \bigcap_{i\in I}F_i\cup K\neq\emptyset$$ Alternatively: $$\forall I\bigg((I\subset\tau)\land\bigg(\forall I_0(I_0\subset I,\ |I_0|<\infty) \bigcap_{i\in I_0}F_i\cup K\neq\emptyset\bigg)\bigg)\quad \bigcap_{i\in I}F_i\cup K\neq\emptyset$$ This is the expression I want. I thought it would be possible to obtain it in a way that corresponds to $A\Rightarrow B$ if and only if $\lnot B\Rightarrow\lnot A$, but it was more difficult than that.
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http://mathoverflow.net/questions/21122?sort=votes
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## Intuition behind existence of moduli space of stable curves
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I'm not entirely sure that the title is what I'm looking for. What I'm really asking is for intuition as to why $\bar{\mathcal{M}_g}$ is the compactification of $\mathcal{M}_g$. I'm sure this is covered in the more classic papers (like Deligne and Mumford), but I still find those hard to penetrate.
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2
Can I answer with the anthropic principle? If it were not, then it would not be called `$\bar{\mathcal{M}}_g$`. – Theo Johnson-Freyd Apr 12 2010 at 19:10
Actually, Deligne is highly readable and "easy to penetrate". – Anton Fonarev Nov 15 2011 at 9:25
## 4 Answers
You can test "compactness" of $\bar M_g$ by the valuative criterion of properness, and this in turn comes down to the stable reduction theorem. In more down to earth terms, given a map of a punctured disk D* into the above space, one wants it to extend uniquely to a map from the whole disk. The essential case is given by as a family of smooth projective curves over D*. The stable reduction theorem says that it extends to family of stable curves over D provided one passes to a ramified cover. Thus one gets the desired extension.
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2
This part is really the miracle of the construction for me, that the condition that makes a pointed nodal curve have finitely many automorphisms is also exactly the condition that makes the valuative criterion work. (e.g. if we allow rational componenents with two markings we can get more than one extension to D by blowing up, etc.) – Dan Petersen Apr 12 2010 at 19:40
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I think a good first step to understanding how the Deligne-Mumford compactification works is to understand the moduli space of genus $0$ curves with marked points. This can be worked out very concretely. I recommend reading the first couple of chapters of "An Invitation to Quantum Cohomology" by Kock and Vainsencher, which has an inspired discussion of this.
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One perspective is geometric. Non-singular closed Riemann surfaces have a unique complete hyperbolic metric by uniformization, with a decomposition into thin and thick regions, where the thin regions are collar neighborhoods of short geodesics, and the thick regions are compact, with each component having uniformly bounded diameter (depending only on the genus). In the case of a non-compact Riemann surface of finite hyperbolic areaa, one also has cusp components of the thin part. One also sees that the geometry of a Riemann surface is uniquely determined by the geometry of the thick part, together with which pairs of boundary components bound collar neighborhoods. One may take a limit of a sequence of surfaces with geodesics whose length approaches zero, and the thick parts converge geometrically to a finite area hyperbolic surface, with a pair of cusps for each geodesic whose length has gone to zero. These cusp pairs are naturally identified with noded Riemann surfaces, giving the Deligne-Mumford compactification. The fact that it is compact follows from the compactness of the thick regions in the Gromov-Hausdorff topology.
It is an interesting result of Masur that the compactification is also given by the metric completion of the Weil-Petersson metric on moduli space.
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As Donu already pointed out, the key result in this sense is the stable reduction theorem. It says that any punctured curve in $\mathscr M_g$ has a unique limit in $\overline{\mathscr M}_g$, so the latter is compact by the valuative criterion of properness. Let me add that this includes also that it is separated!
As far as understanding what stable reduction means and why it should even hold it might be helpful to think about the more general case of compactifying moduli spaces of canonically polarized varieties in arbitrary dimension. This still works for curves, but in higher dimension you need to say different words and it is more clear what's happening on account of our lesser knowledge in general. (By this last sentence I mean that since we know less we ought to talk about it the "right" way whereas in the case of curves there are in general several ways to get the same result.)
From the higher dimensional perspective a stable reduction is a relative canonical model. We need the base change to get rid of multiple fibers and then we need to find a well-defined relative model. The canonical model is the model to use if we want unique limits (which means separatedness of the moduli space). It turns out that if the total space is a canonical model then the fibers are stable.
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http://stats.stackexchange.com/questions/19173/generating-random-matrices-with-specific-equality-constraints
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# Generating random matrices with specific equality constraints
Suppose I want to generate a nonnegative $n \times n$ matrix $\mathbf A$ for an odd $n$ (say, $n=5$ for a good enough example), such that
• the individual elements are drawn from a uniform distribution
but with the equality constraints
• the sum of all the elements in the rows above the mid row = the sum of all the elements in the rows below the mid row
• the sum of all the elements in the columns left of the middle column = the sum of all the elements in the columns right of the middle column
• the sum of all elements = 1.0
That is, in effect I want to generate matrices "balanced" in a specific way (I intend to use them as convolution matrices and I don't care about translation).
For my current purposes I think I have a good enough ad hoc (not entirely correctly distributed) solution that generates a uniform matrix and then enforces the equality constraints by correcting the individual elements using a least squares approach.
However I'm curious, mainly for the learning experience, if there would be some easy-ish way to sample the resulting conditional distribution. So far I've had a quick peek on Wikipedia into the Bates distribution (mean of uniform random variables) and into Monte Carlo Markov Chain models, but I'm not sure if they help here or if there would be an easier solution.
I'm not necessarily looking for a complete solution; pointers to insightful or helpful material are also appreciated.
Edit: I'm looking for the conditional probability distribution, i.e. the elements do not need to be uniformly distributed; rather, I'm looking for a distribution where each sample that satisfies the equality constraints has the same probability (hence conditional uniform distribution given the equalities).
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Your two conditions (1) individual elements uniformly distributed and (2) your equality constraints are mutually incompatible. So, it's not entirely clear what you're aiming for. – cardinal Nov 30 '11 at 17:42
The Sinkhorn-Knopp algorithm is somewhat similar in that the goal there is to generate a matrix of nonnegative elements that satisfy individual row and column sum constraints. Quite a lot is known about its convergence properties. – cardinal Nov 30 '11 at 17:43
1
– cardinal Nov 30 '11 at 17:51
1
What I'm thinking about is, if I'm not mistaken about the terminology, sampling the conditional probability distribution of uniformly distributed values conditioned with the given equations hold. Sure, the conditioned distribution is not necessarily uniform, but that's not a problem. Or in even other words, is there an efficient algorithm to replace the following: 1. Generate such a uniformly random matrix 2. If the equations do hold to within a difference of $\epsilon$, return the matrix; else goto 1 What I'm looking for is essentially this with $\lim \epsilon$ -> 0. – SLi Nov 30 '11 at 20:20
2
Ah, then @Cardinal is (as usual) correct: the marginals will no longer be uniform and you're not requiring that they be so. In fact, they can be far from uniform. (But for medium to large $n$ they ought to be pretty close, because the constraints are so mild.) What you're asking is this: the constraints along with the bounds of the uniform distributions describe a polytope of $n^2-3$ dimensions; you would like to sample this polytope uniformly. – whuber♦ Nov 30 '11 at 23:21
show 1 more comment
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http://math.stackexchange.com/questions/3231/how-to-convert-a-hexadecimal-number-to-an-octal-number
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# How to convert a hexadecimal number to an octal number?
How can I convert a hexadecimal number, for example `0x1A03` to its octal value?
I know that one way is to convert it to decimal and then convert it to octal
````0x1A03 = 6659 = 0o15003
````
• Is there a simple way to do it without the middle step (conversion to decimal or conversion to binary)?
• Why do we tend to convert it to Base10 every time?
-
Does this really fall under "number theory"? I thought number theory was all about the values of numbers, not the notation. Maybe I'm being nitpicky, or maybe I'm just totally wrong. – MatrixFrog Aug 25 '10 at 8:02
## 5 Answers
A simpler way is to go through binary (base 2) instead of base 10.
0x1A03 = 0001 1010 0000 0011
Now group the bits in bunches of 3 starting from the right
0 001 101 000 000 011
This gives
0 1 5 0 0 3
Which is your octal representation.
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I think the easiest way is to go through binary. A hex digit corresponds to 4 bits, an octal digit to 3.
0x1A03 in binary is 0001 1010 0000 0011 (grouped into 4-bit nibbles). If I regroup into 3-bit groups (from the right), I have 001 101 000 000 011. That's octal 0o15003.
-
When you're converting among bases $2^n$, you can often do so more quickly with a comprehensive conversion table. For example, between binary and octal, each block of 3 binary digits will convert to one octal digit: $$\begin{matrix} 000_2=0_8 & 001_2=1_8 & 010_2=2_8 & 011_2=3_8 \\ 100_2=4_8 & 101_2=5_8 & 110_2=6_8 & 111_2=7_8 \end{matrix}$$ Similarly, when converting between binary and hexadecimal, each block of 4 binary digits converts to a single hexadecimal digit. Because of these two facts, each block of 12 (least common multiple of 3 and 4) binary digits is 4 octal digits and 3 hexadecimal digits, so each block of 3 hexadecimal digits can be converted to a block of 4 octal digits. The table for doing hexadecimal<->octal directly is quite large, though, so it's usually simpler to convert to binary as an intermediate form.
It is also possible to do the conversion directly by performing division in octal or hexadecimal, though this can be tricky to get used to.
I think we tend to use decimal as an intermediate form because we are most familiar and comfortable with base 10 (since we generally have 10 fingers and use it most often).
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There is a trick for hex to/from octal because they relate to binary but lets ignore that and consider the general idea (which works for any bases).
Since we want to convert to octal lets start using octal as our number system - as if were always were. So we count like this 1,2,3,4,5,6,7,10,11,12,13,14,15,16,17,20,21,22,...
And our multiplication table is:
$$\begin{array}{|c|c|c|c|c|c|c|} \hline \times & \mathbf 1 & \mathbf 2 & \mathbf 3 & \mathbf 4 & \mathbf 5 & \mathbf 6 & \mathbf 7\\ \hline \mathbf 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \mathbf 2 & 2 & 4 & 6 & 10 & 12 & 14 & 16 \\ \mathbf 3 & 3 & 6 & 11 & 14 & 17 & 22 & 25 \\ \mathbf 4 & 4 & 10 & 14 & 20 & 24 & 30 & 34 \\ \mathbf 5 & 5 & 12 & 17 & 24 & 31 & 36 & 43 \\ \mathbf 6 & 6 & 14 & 22 & 30 & 36 & 44 & 52\\ \mathbf 7 & 7 & 16 & 25 & 34 & 43 & 52 & 61 \\ \hline \end{array}$$
So the hexadecimal digits are $0\text x1 = 1$, $0\text x2 = 2$, $0\text x3 = 3$, $0\text x4 = 4$, $0\text x5 = 5$, $0\text x6 = 6$, $0\text x7 = 7$, $0\text x8 = 10$, $0\text x9 = 11$, $0\text {xA} = 12$, $0\text xB = 13$, $0\text xC = 14$, $0\text xD = 15$, $0\text xE = 16$, $0\text xF = 17$.
Anyway, we want to re-interpret the number `0x1A03` which is equal to $20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3$. Now we just "do the math", multiply everything out and add it up.
$$\begin{align} & 20^3 \cdot 1 + 20^2 \cdot 12 + 20 \cdot 0 + 3 \\ =& 10000 + 400 \cdot 12 + 3 \\ =& 10000 + 5000 + 3 \\ =& 10000 + 5000 + 3 \\ =& 15003 \end{align}$$
Now the fact that we have all these zeros suggests there would be a quicker method for these particular bases (normally we would not get so many zeros in our calculations).
-
1
Please check your result. Should be 0 1 5 0 0 3. – Américo Tavares Aug 24 '10 at 20:52
Thanks for pointing out that arithmetic mistake, I've fixed it now. – anon Aug 24 '10 at 21:11
1
You should use `$20_8^3$` for example to indicate that those are base 8 numbers: $20_8^3 \cdot 1 + 20_8^2 \cdot 12 + 20_8^1 \cdot 0 + 20_8^0 \cdot 3$ – Dennis Williamson Aug 24 '10 at 22:36
Dennis Williamson, no, I intentionally did not do that. The whole point here is to show that if we use octal as our natural base the 'conversion' calculation is just evaluation and there is nothing special about base 10. – anon Aug 25 '10 at 6:16
1
@muad: thanks a lot! – Lazer Aug 25 '10 at 8:05
There's a fast procedure involving no intermediate representation. You need only four lookup tables totaling 544 entries.
Earlier answers have established that the conversion can be done in blocks of 3 (working right to left). Consider the rightmost block in your example:
0xA03 = 1010 0000 0011 B
You need to break this binary string into four groups of three bits, which I will number 1, 2, 3, 4 from the right to the left:
1: 011 B
2: 000 B
3: 000 B
4: 101 B
The first depends only on the rightmost hex digit 0x3. The second depends only on the two rightmost digits 0x03 (it gets its rightmost bit from the 0x3 and its first two bits from 0x0). The third depends only on the second and third digits 0xA0. The fourth depends only on the third digit 0xA. Whence, you only need to perform the following conversions, each of which can be stored in its own static lookup table:
1: 0x3 --> 03 [16 entries cover all possibilities]
2: (0x0, 0x3) --> 00 [16 * 16 entries]
3: (0xA, 0x0) --> 00 [16 * 16 entries]
4: 0xA --> 5 [16 entries].
Now you repeat with the next block, padding with zeros on the left as needed. Continuing the example, the next block is 0x001:
1: 0x1 --> 01
2: (0x0, 0x1) --> 00.
You can stop here because all the original input has been consumed. The output, working backwards, is 0015003.
This directly answers both parts of the original question: it provides a simple conversion without the middle step and it avoids base 10 (which is rarely used for computer conversion anyway: usually the job is done with bit shifting and masking, essentially a binary operation). In case this procedure still looks too much like the other proposed solutions, please note that it performs absolutely no arithmetic (apart from decrementing pointers to input and output as it proceeds): it takes a string representation (hex) as its input, uses the characters to index its tables, and outputs an octal string.
-
Indeed this is great as a computer algorithm; I however got the impression that he was doing these conversions manually, so I suppose an intermediate conversion to binary is still a nicer approach. – J. M. Aug 26 '10 at 0:08
Good point: "simple" depends on your computational model (e.g., Turing machine vs. human being). It's easily done with mental arithmetic by means of the binary conversion. – whuber Aug 26 '10 at 0:58
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http://mathhelpforum.com/advanced-algebra/133228-annihilators-subspaces.html
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# Thread:
1. ## Annihilators and subspaces
Hi,
Problem:
If M and N are subspaces of a finite-dimensional vector space, then $(M \cap N)^0=M^0+N^0$ and $(M+N)^0=M^0 \cap N^0$.
( $M^0$ and $N^0$ are the annihilators of M and N, respectively)
Attempt:
Not much I can do here, but I know that if $M \subset N$, then $N^0 \subset M^0$.
I also know that $M^{00}=(=(M^0)^0)=M$. I think the author wants me to use these two facts to prove the given statement.
Any hints are greatly appreciated.
Thanks.
2. dont think that you need to use those facts that you mentioned, basicly what you have is two sets, so you need to show that one set is a subset of other and opposite.
3. Originally Posted by Mollier
Hi,
Problem:
If M and N are subspaces of a finite-dimensional vector space, then $(M \cap N)^0=M^0+N^0$ and $(M+N)^0=M^0 \cap N^0$.
( $M^0$ and $N^0$ are the annihilators of M and N, respectively)
Attempt:
Not much I can do here, but I know that if $M \subset N$, then $N^0 \subset M^0$.
I also know that $M^{00}=(=(M^0)^0)=M$. I think the author wants me to use these two facts to prove the given statement.
Any hints are greatly appreciated.
Thanks.
Hints (easy): (a) $U\subset W\Longrightarrow W^0\subset U^0$ , (b) $A\subset A^{00}\,,\,\,\forall$ subset $A\subset V$ , (c) $U=U^{00}$ if $U$ is a subspace of $V$, and thus we get:
(1) $M^0+N^0\subset (M\cap N)^0$ (why?);
(2) $(M+N)^0\subset M^0\cap N^0$ (why?)
Take now the annihilator in both sides in (1) and use (a) and (c) above:
(3) $M\cap N=(M\cap N)^{00}\subset (M^0+N^0)^0$ , and now use (2) above (with $M^0\,,\,N^0$ instead of $M\,,\,N$), and you get
(4) $(M^0+N^0)^0\subset M^{00}\cap N^{00}= M\cap N$ , so from (3)-(4) you get
(5) $M\cap N\subset (M^0+N^0)^0\subset M\cap N$ , and this means the inclusions (3)-(4) must be equalities ; but this means that the annihilators of both sides in (1) are equal , and thus we get the wanted equality because of the
Hint (easy, too) (d) If $U\,,\,W$ subspaces of $V\,,\,\,s.t.\,\,\,U\neq W$ , then $U^0\neq W^0$
Try now to complete the proof of the second equality by yourself.
Tonio
Pd. I don't know any proof of these equalities that is less messy than the above, which is not hard but requires concentration in details and lots of careful "connecting the dots"
4. Hi, apologies for the late reply. The last few days have been a mess so no time for math
Originally Posted by tonio
(1) $M^0+N^0\subset (M\cap N)^0$ (why?);
(2) $(M+N)^0\subset M^0\cap N^0$ (why?)
The only thing I can come up with is that since $M \cap N \subset M+N$, then $(M+N)^0 \subset (M \cap N)^0$, but that doesn't really prove either (1) nor (2)..
Originally Posted by tonio
Try now to complete the proof of the second equality by yourself.
I figure that I need to show that $M^0 \cap N^0 \subset (M^0+N^0)$. If I assume that (2) is true, I should be able to take the annihilator of both sides of (2) and get to the wanted result.
$((M+N)^0)^0 \subset (M^0 \cap N^0)^0$
Then by (a) I have that
$(M^0 \cap N^0) \subset (M+N)^0$
Ah, I find this to be really hard!
Thanks!
5. Originally Posted by Mollier
Hi, apologies for the late reply. The last few days have been a mess so no time for math
The only thing I can come up with is that since $M \cap N \subset M+N$, then $(M+N)^0 \subset (M \cap N)^0$, but that doesn't really prove either (1) nor (2)..
Indeed, that doesn't...but $M\cap N\subset M\Longrightarrow M^0\subset (M\cap N)^0\,,\,\,and\,\,\,also\,\,\,M\cap N\subset N\Longrightarrow N^0\subset (M\cap N)^0$ and now you get what you wanted...
I figure that I need to show that $M^0 \cap N^0 \subset (M^0+N^0)$. If I assume that (2) is true, I should be able to take the annihilator of both sides of (2) and get to the wanted result.
$((M+N)^0)^0 \subset (M^0 \cap N^0)^0$
Then by (a) I have that
$(M^0 \cap N^0) \subset (M+N)^0$
Ah, I find this to be really hard!
Yes, it really requires concentration. Try again, since the last days you did no much math (or at all).
Tonio
Thanks!
.
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http://physics.stackexchange.com/questions/36157/why-is-the-critical-density-of-the-universe-non-zero/36158
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# Why is the critical density of the Universe non-zero?
I'm missing something and I wondered if someone could enlighten me. As I understand it, Einstein's equations say, among other things, that matter curves spacetime. I also understand, the Universe has a 'critical density', defined to be the mass (or energy) density required for space to have zero curvature (found by solving Friedmann's equations).
Why does the Universe need a non-zero density (the critical density) for zero curvature?
Many thanks.
-
## 2 Answers
First of all it is good to differentiate what can be curved. One way to deal with Einsteins equations is to choose a space-time foliation (global when possible). In the Friedmann models one has a set isometries which define the homogeneus and isotropic spatial hypersurfaces, it can be shown that these hypersurfaces have a normal vector field $n^\mu$ which is geodesic $(n^\nu\nabla_\nu{}n^\mu=0)$ where $\nabla_\nu$ is the connection compatible with the Friedmann metric $g_{\mu\nu}$ (signature $(-1,1,1,1)$), i.e., $\nabla_\alpha{}g_{\mu\nu}=0$.
Given this choice of hypersurfaces we have the projector $\gamma_{\mu\nu}\equiv{}g_{\mu\nu}+n_\mu{}n_\nu$ which acts as a metric in the spatial sections. This metric we have a unique covariant derivative (in the hypersurfaces) $D_\alpha\gamma_{\mu\nu} = 0$, and this derivative in turn defines a Riemann tensor on the spatial section $$(D_\mu{}D_\nu-D_\nu{}D_\mu)v_\alpha = \mathcal{R}_{\mu\nu\alpha}{}^\beta{}v_\alpha,$$ where $\gamma_\alpha{}^\beta{}v_\beta=v_\alpha$. For the Friedmann metric we have that $$\nabla_\mu{}n_\nu = \Theta_{\mu\nu} = \frac{\Theta}{3}\gamma_{\mu\nu},$$ where $\Theta_{\mu\nu}$ is the extrinsec curvature and $\Theta$ its trace.
In short, the notion of a foliation induces a concept of curvature and a extrinsec curvature in the hypersurfaces in a way that the four dimensional Riemann tensor $$R_{\mu\nu\alpha}{}^\beta{}v_\beta = (\nabla_\mu\nabla_\nu-\nabla_\nu\nabla_\mu)v_\alpha,$$ is split in $\mathcal{R}_{\mu\nu\alpha}{}^\beta{}v_\alpha$ plus combinations of $\Theta_{\mu\nu}$ and $D_\mu$, i.e., $$R_{\mu\nu\alpha\beta} = \mathcal{R}_{\mu\nu\alpha\beta} + 2\Theta_{\mu[\alpha}\Theta_{\beta]\nu} - 4(D_{[\mu}\Theta_{\nu][\alpha}n_{\beta]} + D_{[\alpha}\Theta_{\beta][\mu}n_{\nu]}) + 4(n^\sigma\nabla_\sigma(n_{\mu}\Theta_{\nu][\alpha}n_{\beta]}) + n_{[\mu}\Theta_{\nu]}{}^\sigma\Theta_{\sigma[\alpha}n_{\beta]}),$$ where $T_{[\mu\nu]} = (T_{\mu\nu}-T_{\nu\mu})/2$.
All this said, we can see that it is possible to have a zero spatial curvature (Ricci tensor) $\mathcal{R}_{\mu\alpha} = 0$, but a non-zero four dimensional $R_{\mu\alpha}$. For a Friedmann metric we have $D_\mu\Theta=0$, $\mathcal{R}_{\mu\nu} = 2K\gamma_{\mu\nu}$ and $D_\mu K=0$. Contracting the indexes in the expression for the Riemann tensor in terms of the foliation objects and using the Friedmann simplifications we obtain $$R_{\mu\alpha} = \mathcal{R}_{\mu\alpha} + \frac{\dot{\Theta}+\Theta^2}{3}\gamma_{\mu\nu} - \left(\dot{\Theta}+\frac{\Theta^2}{3}\right)n_\mu{}n_\alpha,$$ where $\dot{\Theta} \equiv n^\sigma\nabla_\sigma\Theta$. Thus, the Einsteins tensor is simply $$G_{\mu\nu} = \mathcal{G}_{\mu\nu} + \frac{\mathcal{R}n_\mu{}n_\nu}{2} -\left(\frac{2\dot{\Theta} + \Theta^2}{3}\right)\gamma_{\mu\nu} + \frac{\Theta^2}{3}n_\mu{}n_\nu ,$$ where $G_{\mu\nu} = R_{\mu\nu}-g_{\mu\nu}R/2$ and $\mathcal{G}_{\mu\nu} = \mathcal{R}_{\mu\nu}-\gamma_{\mu\nu}\mathcal{R}/2$ and consequently the Einsteins equation $G_{\mu\nu}=\kappa T_{\mu\nu}$ ($\kappa = 8\pi G$) have the time-time component (the so called Friedmann equation) $$\frac{\mathcal{R}}{2}+\frac{\Theta^2}{3} = \kappa\rho,\quad\Rightarrow\quad H^2 = \frac{\kappa\rho}{3}-K,$$ where $\rho = n_\mu{}n_\nu{}T^{\mu\nu}$, $\Theta = 3H$. Therefore, we have that $H^2 = {\kappa\rho}/{3}$ if and only if $K=0$, which defines the critical density. Then, finally the if the universe has energy density equal to critical density then the curvature of the spatial hypersurfaces of homogeneity and isotropy is null.
Note, however, that the spatial curvature depends on the choice of hypersurfaces and even if $\mathcal{R}_{\mu\nu}$ is null in the usual foliation we can always find another foliation in which this is not true. In Friedmann this is not a issue since the foliation is induced by the symmetries. However, not that, in de Sitter metric any other choice of foliation gives also homogeneous and isotropic hypersurfaces, in this case different choices of foliation can leads to flat, hyperbolic and spherical spatial sections.
Additionally, the Einsteins equations relates the energy momentum density with the Ricci tensor. Therefore, if the energy density is zero the Einsteins equations leads to zero Ricci tensor, however, one can still have non zero Riemann tensor $R_{\mu\nu\alpha\beta}$.
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The critical density is the density required to make the universe spatially flat. It will still be curved in the time dimension. A universe containing no matter (and no radiation) would be COMPLETELY flat.
If you want some insight into how you can have curvature in other dimensions, imagine a sphere--it is curved in two dimensions, while a cylinder will have one ''flat'' dimension and one ''curved'' dimension.* Much in this way, you can have spacetimes where the spatial dimensions are flat, but the time dimension has curvature.
*For the technical minded, and the pedantic: this example is slightly cheating, since a cylinder has no intrinsic curvature--you can turn a sheet of paper into a cylinder without distorting the surface. But I wanted an example that was immediately obvious to someone with no math background using common shapes.z
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"The critical density is the density required to make the universe spatially flat." OK, I will change "spacetime" to "space" in my question (thanks for pointing that out btw). However, I am still none the wiser on the main point: why does the Universe need a non-zero density (the critical density) for zero curvature? – user12345 Sep 11 '12 at 15:02
Also, be as technical and/or as mathematical as you wish. In fact, I encourage it! – user12345 Sep 11 '12 at 15:05
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http://mathhelpforum.com/differential-equations/71815-differential-equation-verification.html
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# Thread:
1. ## differential equation verification
(dy/dt) + y = cos(e^t) i got the integrating factor as e^t and the answer i got is y = (1/(e^t))*(sin (e^t)) + C is this correct???
2. Originally Posted by crafty
(dy/dt) + y = cos(e^t) i got the integrating factor as e^t and the answer i got is y = (1/(e^t))*(sin (e^t)) + C is this correct???
Very close, when you integrate, you get
$e^t y = \sin e^t + c$
Dividing by $e^t$ you must keep the c where it is,
$y = \frac{\sin e^t + c}{e^t}$ a little different from your answer.
3. but isnt c just an arbitrary constant and assuming my c outside is a different constant would it still be correct or is there something special about this case, if there is can you explain why ... thanks in advance
4. Originally Posted by crafty
but isnt c just an arbitrary constant and assuming my c outside is a different constant would it still be correct or is there something special about this case, if there is can you explain why ... thanks in advance
Your right c is an arbitrary constant but when multiplied by a function of t then it will change with respect to t (it's no longer fixed). This is really seen if you were given an initial condition were the value of c would be set.
5. oh right i get it now and can you please reply to my other thread called its called "how do is solve this differential eqn???" ive been waiting a while for that one and i still don't get it
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http://mathoverflow.net/revisions/32389/list
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## Return to Answer
2 added 138 characters in body
I should state first that this reply has only to do with the above mentioned ideas in the category of models of a first order theory.
John Goodrick's work is referenced in Joel's post above, and I have heard John Goodrick speak about this at least once. Specifically, John mentioned the following (and a lot more that I didn't write down):
Fix some countable, complete first-order theory, $T$. Suppose $T$ has the following property: Whenever we are given two models $\mathcal M_1$ and $\mathcal M_2$ of $T$ which have elementary embeddings into each other, then $\mathcal M_1 \cong \mathcal M_2.$
Then $T$ is superstable and nonmultidimensional (and I know if John replies to this, he can mention many other things, but I don't remember now). In the case that $T$ is actually $\omega -$stable, nonmultidimensional implies the bi-embedding property stated in the above paragraph.
1
John Goodrick's work is referenced in Joel's post above, and I have heard John Goodrick speak about this at least once. Specifically, John mentioned the following (and a lot more that I didn't write down):
Fix some countable, complete first-order theory, $T$. Suppose $T$ has the following property: Whenever we are given two models $\mathcal M_1$ and $\mathcal M_2$ of $T$ which have elementary embeddings into each other, then $\mathcal M_1 \cong \mathcal M_2.$
Then $T$ is superstable and nonmultidimensional (and I know if John replies to this, he can mention many other things, but I don't remember now). In the case that $T$ is actually $\omega -$stable, nonmultidimensional implies the bi-embedding property stated in the above paragraph.
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http://mathoverflow.net/questions/28222?sort=oldest
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## Planar graph drawing [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
how i determine a face of a planar graph is convex polygon or not..............
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I can't make sense of this as it stands. A planar graph is one which can be embedded into the plane. Your question only makes sense for an actual emebedding in the plane, and I cannot see any sensible answer save that a face is a convex polygon if it's a convex polygon. Perhaps a more precise question is lurking? – Robin Chapman Jun 15 2010 at 6:37
With a protractor, I suppose. You may want to expand on your question to make it easier to know what exactly you are asking. – Gerry Myerson Jun 15 2010 at 6:38
I'm going to take a wild guess and say the question is asking whether one can tell if a face of a planar graph is a convex polygon before the graph is drawn with edges as straight line segments. That is, in the beginning, you are given a random drawing where the edges could be curved. Still, I believe the question will be closed unless the OP makes it clear what is being asked. – Gjergji Zaimi Jun 15 2010 at 7:32
Perhaps you are asking for the result of Steinitz's Theorem: Which planar graphs are $1-$skeletons of convex polytopes. Answer: The underlying abstract graph has to be $3-$edge connected. Such graphs have essentially (up to isotopies and an orientation reversing homeomorphism) only one planar embedding coming from a realization as the $1-$skeleton of a convex polytope. – Roland Bacher Jun 15 2010 at 8:37
2
The question has been closed: we simply don't know what you're asking. If you edit it to provide enough information, precision and context so that we can understand it (and, for bonus points, use proper punctuation and grammar, to the best of your ability), it can be reopened. – Pete L. Clark Jun 15 2010 at 15:14
show 1 more comment
## 1 Answer
this is not a well defined question. Perhaps you mean that you want to know if a face of a planar embedding of a planar graph is convex ? Note that it's always possible to draw a planar graph so all faces (with the exception of the outer face) are convex
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3
(I think that this should have been left as a comment, not an answer) – Andrew Stacey Jun 15 2010 at 7:54
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http://mathoverflow.net/revisions/114972/list
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## Return to Question
4 retagged
3 added 295 characters in body
Let $\{x_1,\dots,x_n\}$ be pairwise distinct complex numbers and $l_1+l_2+\dots+l_n=N$. The $N\times N$ confluent Vandermonde matrix is defined as $$V= \begin{bmatrix} v_{1,0}&v_{2,0}&\dots&v_{n,0}\\ v_{1,1}&v_{2,1}&\dots&v_{n,1}\\ \vdots\\ v_{1,N-1}&v_{2,N-1}&\dots&v_{n,N-1} \end{bmatrix}$$ where $v_{j,k}=\begin{bmatrix}x_j^k,&kx_j^{k-1},&\dots&k(k-1)\times\dots\times (k-l_j+1) x_j^{k-l_j+1}\end{bmatrix}$. Let $\|\cdot\|$ denote the row sum matrix norm. In some applications (e.g. interpolation, signal processing) one would like to estimate the quantity $\|V^{-1}\|$.
Gautschi [1] has shown that for $l_1=\dots=l_n=2$ one has $$\|V^{-1}\| \leq \max_{1\leq \lambda\leq n} \beta_{\lambda} \prod_{\nu=1,\nu\neq\lambda}^n \biggl(\frac{1+|x_{\lambda}|}{|x_{\nu}-x_{\lambda}|}\biggr)^2$$ where $\beta_{\lambda}=\max\biggl(1+|x_{\lambda}|,1+2(1+|x_{\lambda}|)\sum_{\nu\neq\lambda}{1\over |x_\nu-x_\lambda|}\biggr)$.
I am interested in a somewhat cruder estimates, as follows: if $|x_j|\leq 1$ and $|x_i-x_j|\geq \delta$, then for the above case we have $$\|V^{-1}\| \leq C n 2^N\delta^{-N+1}\qquad (*)$$ for some absolute constant $C$.
Is it true that something like $(*)$ holds for the general configuration $\{l_1,\dots,l_n\}$?
EDIT: using [2], this seems to boil down to the following. Let $$h_j(x)=\prod_{i \neq j}(x-x_i)^{-l_i}.$$ For $t=0,1,\dots,l_j$ evaluate $h_j^{(t)}(x_j).$
[1] W.Gautschi, "On Inverses of Vandermonde and Confluent Vandermonde matrices II", Numerische Mathematik 5, 425-430, 1963.
[2] R.Schapelle, "The Inverse of the Confluent Vandermonde Matrix", IEEE Trans. on Automatic Control, October 1972, pp.724-725.
2 cleanup of formula
Let $\{x_1,\dots,x_n\}$ be pairwise distinct complex numbers and $l_1+l_2+\dots+l_n=N$. The $N\times N$ confluent Vandermonde matrix is defined as $$V= \begin{bmatrix} v_{1,0}&v_{2,0}&\dots&v_{n,0}\\ v_{1,1}&v_{2,1}&\dots&v_{n,1}\\ \vdots\\ v_{1,N-1}&v_{2,N-1}&\dots&v_{n,N-1} \end{bmatrix}$$ where $v_{j,k}=\begin{bmatrix}x_j^k,&kx_j^{k-1},&\dots&k(k-1)\times\dots\times (k-l_j+1) x_j^{k-l_j+1}\end{bmatrix}$. Let $\|\cdot\|$ denote the row sum matrix norm. In some applications (e.g. interpolation, signal processing) one would like to estimate the quantity $\|V^{-1}\|$.
Gautschi [1] has shown that for $l_1=\dots=l_n=2$ one has $$\|V^{-1}\| \leq \max_{1\leq \lambda\leq n} \beta_{\lambda} \prod_{\nu=1,\nu\neq\lambda}^n \biggl(\frac{(1+|x_{\lambda}|)}{|x_{\nu}-x_{\lambda}|}\biggr)^2$$ biggl(\frac{1+|x_{\lambda}|}{|x_{\nu}-x_{\lambda}|}\biggr)^2 where $\beta_{\lambda}=\max\biggl(1+|x_{\lambda}|,1+2(1+|x_{\lambda}|)\sum_{\nu\neq\lambda}{1\over |x_\nu-x_\lambda|}\biggr)$.
I am interested in a somewhat cruder estimates, as follows: if $|x_j|\leq 1$ and $|x_i-x_j|\geq \delta$, then for the above case we have $$\|V^{-1}\| \leq C n 2^N\delta^{-N+1}\qquad (*)$$ for some absolute constant $C$.
Is it true that something like $(*)$ holds for the general configuration $\{l_1,\dots,l_n\}$?
[1] W.Gautschi, "On Inverses of Vandermonde and Confluent Vandermonde matrices II", Numerische Mathematik 5, 425-430, 1963.
1
# Norm of inverse confluent Vandermonde matrix
Let $\{x_1,\dots,x_n\}$ be pairwise distinct complex numbers and $l_1+l_2+\dots+l_n=N$. The $N\times N$ confluent Vandermonde matrix is defined as $$V= \begin{bmatrix} v_{1,0}&v_{2,0}&\dots&v_{n,0}\\ v_{1,1}&v_{2,1}&\dots&v_{n,1}\\ \vdots\\ v_{1,N-1}&v_{2,N-1}&\dots&v_{n,N-1} \end{bmatrix}$$ where $v_{j,k}=\begin{bmatrix}x_j^k,&kx_j^{k-1},&\dots&k(k-1)\times\dots\times (k-l_j+1) x_j^{k-l_j+1}\end{bmatrix}$. Let $\|\cdot\|$ denote the row sum matrix norm. In some applications (e.g. interpolation, signal processing) one would like to estimate the quantity $\|V^{-1}\|$.
Gautschi [1] has shown that for $l_1=\dots=l_n=2$ one has $$\|V^{-1}\| \leq \max_{1\leq \lambda\leq n} \beta_{\lambda} \prod_{\nu=1,\nu\neq\lambda}^n \biggl(\frac{(1+|x_{\lambda}|)}{|x_{\nu}-x_{\lambda}|}\biggr)^2$$ where $\beta_{\lambda}=\max\biggl(1+|x_{\lambda}|,1+2(1+|x_{\lambda}|)\sum_{\nu\neq\lambda}{1\over |x_\nu-x_\lambda|}\biggr)$.
I am interested in a somewhat cruder estimates, as follows: if $|x_j|\leq 1$ and $|x_i-x_j|\geq \delta$, then for the above case we have $$\|V^{-1}\| \leq C n 2^N\delta^{-N+1}\qquad (*)$$ for some absolute constant $C$.
Is it true that something like $(*)$ holds for the general configuration $\{l_1,\dots,l_n\}$?
[1] W.Gautschi, "On Inverses of Vandermonde and Confluent Vandermonde matrices II", Numerische Mathematik 5, 425-430, 1963.
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http://nrich.maths.org/44/note?nomenu=1
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## Newspapers
I don't know whether in your homes, at the weekend, there are lots of newspapers lying about. Maybe you travel by train a lot and when you get into the train there are lots of bits of newspapers on the seats and tables. These things happen because they're not stapled together like magazines usually are [like the ones that tell you what's on the T.V.].
So what we're thinking about is newspaper that has had the pages separated and they're lying about instead of being like they were when they were bought at the shop. We'll think about a very small newspaper in that it only has eight pages. It would be a good idea to have two sheets of paper [the same size] ready now so that we can talk this through more easily. You could use two sheets of paper from a real newspaper [ask the adult who bought the newspaper first!] or just two pieces of A4 folded down the middle. They should look like this:-
Now that you've got these you will need to put one inside the other, fold the two together and then number the pages 1, 2, 3, 4, 5, 6, 7 and 8.
To help you to see my pictures better I've done the numbers in a special way.
When the pages are facing you I put the numbers in a yellow box.
I've pretended that you can see through the thin paper and so you can see the number that's written on the side not facing you. That number I've put into a pink box.
You should write the numbers the correct way round from 1 to 8.
So the middle sheet of the newspaper looks like:-
If I remove that sheet the outside sheet will look like:-
Now you have all that you need so let's have a look at the challenge.
When pages get separated at home we have to try to sort them out and get things in the correct order so that we can follow the stories and pictures.
Let's suppose that in our eight-page newspaper there is just one complete story on each page and so it does not matter what order the pages are in, as long as the writing is not upside down!
How many ways can we arrange the pages [no tearing!] so that the numbering may be different?
Here is a start I've made:
I N N E R - S H E E T - - - - - - - - O U T E R - S H E E T
A)
B)
C)
In A) I've turned the inside sheet over so now the pages go 1, 2, 5, 6, 3, 4, 7, 8.
In B) I've got the inside sheet the correct way round but have turned over the first sheet, so now the pages go 7, 8, 3, 4, 5, 6, 1, 2.
In C) I've got both sheets turned over so now the pages go 7, 8, 5, 6, 3, 4, 1, 2.
Now of course there are other things we can do, like we can change the order of the sheets so that one arrangement could be like this.
I N N E R - S H E E T - - - - - - - - - - O U T E R - S H E E T
The page numbers now go 3, 4, 1, 2, 7, 8, 5, 6.
Well have a go yourself, be as inventive as you like BUT the writing must be the correct way up.
It would be good to find a way of recording all the page number orders, and look at them all together.
Good luck.
I usually ask you to provide a question like "I wonder what would happen if ...?''. Well I'm going to give you one this time:
What would happen if we had three sheets of paper and so we had twelve pages to number?
If you found a way of doing the first challenge you may be able to use that same system for getting ALL the ways of arranging these three sheet and looking at the numberings that occur. I guess it would be good to then explore these sets of numbers that show the page numberings. [That's a funny word!]
### Why do this problem?
Use this activity to present youngsters with a problem that challenges them to think 'outside the box'. It can help to develop general problem solving strategies and find suitable ways of recoding their results.
### Possible approach
This presents a good opportunity to hand over to the pupils the decision as to how to approach this challenge.
This activity allows the youngsters to create good ways of recording what they have done and communicate their findings to a whole class by producing a display. It may end up as an assessment for patience and perseverence!
### Key questions
When pupils have made decisions then ask, "Are you allowed to do that?"
Do you think there are any more possibilities?
### Possible extension
Some children could be asked what they could do about changing one of the rules, but beware of going further than using three sheets of paper i.e. 12 pages! You can also challenge them to look at methods of progressing so as to make sure that they get all the possible combinations.
### For the exceptionally mathematically able
For those very able pupils we would expect them to pursue the three sheet of the paper making $12$ pages. They might be encouraged to explore methods of ensuring that all the combinations are found. This could lead to a method for n numbers of sheets.
### Possible support
For those who struggle with the whole idea it would be good to have the sheets of paper ready so that they can rearrange them physically.
------------------------------------------------------
I found that a group of children who were 8 and 9 years old coped with this very enthusiastically. They got a lot out of it and were very creative with their ideas.
Even if all the combinations are not found by your class we'd be pleased to hear how it went and what aspects of the activity caught their interest and which did not.
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http://mathoverflow.net/questions/42276
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## Noether’s normalization lemma over a ring A
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given a field $k$ and a finitely generated $k$-algebra $R$ without zero divisors, one knows that there exist $x_1, \ldots, x_n$ algebraically independent such that $R$ is integral over $k[x_1, \ldots, x_n]$.
Does one have a similar statement, under good assumptions, if $k$ is not a field but a ring ? In this discussion, I am also interested by geometric explanations.
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1
I am not sure this is really what you are looking for, but on the geometric side, let me mention the following result of Barry Green (MR1458302): let A be a Dedeking ring whose fraction field is a local or global field, then every normal projective curve over Spec(A) has a finite morphism to $\mathbf{P}^1_A$. See also arxiv.org/abs/0902.2039. – Jérôme Poineau Oct 19 2010 at 11:02
## 3 Answers
http://www.math.lsa.umich.edu/~hochster/615W10/supNoeth.pdf
Supplementary notes from Mel Hochster's commutative algebra class. They discuss, in particular, a generalization of Noether normalization to integral domains.
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I have not read carefully, but it seems to be the same proof? – Martin Brandenburg Oct 17 2010 at 15:42
@Martin: Yes, essentially. – Harry Gindi Oct 17 2010 at 21:09
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The geometric interpretation of Noether's normalization lemma is that any affine algebraic variety has a finite surjective morphism to the affine space $\mathbb A^d_k$ of dimension $d=\dim X$. When $k$ is an integral domain, and $X$ dominates $\mathrm{Spec}(k)$, the finite surjective morphism of the generic fiber of $X$ to $\mathbb A^d_{K}$, where $K=\mathrm{Frac}(k)$ extends to a finite surjective morphism $X_V\to V$ for some dense open subset $V$ of $\mathrm{Spec}(k)$. This is the geometric interpretation of the statement in M. Hochster's note in Harry Gindi's post.
In general, such a morphism can not exist because it would imply that the fibers of $X\to \mathrm{Spec}(k)$ all have the same dimension.
Suppose that this condition is satisfied: the fibers of $X\to\mathrm{Spec}(k)$ all have the same dimension $d$. Then I think that a reasonable statement (Noether's normalization lemma over a ring $k$) would be: there exists a quasi-finite and surjective morphism $X\to \mathbb{A}^d_k$. If $k$ is noetherian, then by Zariski's Main Theorem, this implies that $X$ is an open subscheme of scheme which is finite surjective over $\mathbb{A}^d_k$. In general one can not expect better result than quasi-finite (consider the case $d=0$).
The above "reasonable statement" should be easy to prove when $k$ is a local ring.
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I recently came to want this generalization of Noether normalization for my own commutative algebra course and notes. So I just wanted to report that I found what seems to me to be the optimally efficient and clear treatment of this result, at the beginning of Chapter 8 of these commutative algebra notes of K.M. Sampath. All in all I highly recommend Sampath's notes: they are excellent.
I am starting to find it surprising that this simple and useful generalization of Noether Normalization is not the standard version: it has some important applications, e.g. finiteness of integral closure of domains which are finitely generated over $\mathbb{Z}$. Does anyone know who first came up with this version (Hochster, perhaps)?
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The Noether normalization lemma in these notes seem to be a copy of the treatment in Eisenbud's book on commutative algebra. He mentions that the case with one ideal is due to Nagata. – Jakob Dec 7 at 10:01
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http://crypto.stackexchange.com/questions/2820/why-do-block-ciphers-need-a-non-linear-component-like-an-s-box
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# Why do block ciphers need a non-linear component (like an S-box)?
Why is there a requirement of "Non-Linear functions" as a component of many popular block ciphers (e.g. the S-box in DES or 3DES)?
How does it make the cipher more secure?
The only intuition I have is a non linear function can have many roots (solutions). E.g. for a non linear $y = F(x)$, if attacker knows $y$ to find out the $x$, there might be many choices which satisfy $y= F(x)$ hence it adds up complexity at attacker's end when he tries to make his way back (trace back) from cipher text to plain text.
Any references or detailed material are appreciated.
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– duskwuff Jun 6 '12 at 4:48
Welcome to Cryptography Stack Exchange. Your question was migrated here because of being not directly related to IT security (the topic of IT Security Stack Exchange), and being fully on-topic here. Please register your account here, too, to be able to comment and accept an answer. – Paŭlo Ebermann♦ Jun 6 '12 at 19:34
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It's important to note that combining two operations which are linear in different fields, can be non linear. For example addition in GF(2^32) aka 32 bit integer addition and addition in GF(2) aka XOR is a common choice. – CodesInChaos Jun 7 '12 at 1:51
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@David: If you combine functions linear of the same field you get again a function linear over that field, and linear equations are easy to solve. So you do need something non-linear to make solving difficult. Non-linear does not necessarily mean that there are several solutions (looks like you think of polynomials over the complex numbers). For the AES the S-Box is a permutation, so there exists always exactly one solution $x$ for $y = S(x)$. – jug Jun 7 '12 at 10:33
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So, we use addition in different rings. The ring $\mathbb Z/2^{32} \mathbb Z$ (integers modulo $2^{32}$) is not a field, although it has the same number of elements as $GF(2^{32})$. – Paŭlo Ebermann♦ Jun 7 '12 at 18:52
show 3 more comments
## 2 Answers
Here's the cryptography theory perspective.
We want block ciphers to resemble pseudo-random permutations (PRPs). PRPs are a desirable modeling goal because a block cipher under a given key is a permutation on the input, and a PRP is simply a random collection of permutations. The block cipher's key can never be better at creating permutations than an actual random sampling of them, but we want it to be as close as possible. Detectable deviation from PRP-like behavior is considered a weakness in a block cipher.
To compare a block cipher to a PRP we use the CPA model, where an attacker queries a black box with plaintext and receives the corresponding permuted output. They try to determine if the black box is choosing the ciphertext output by apply a random permutation or the given block cipher with an unknown key to the plaintext. If they guess correctly they win the game. If they can win the game with probability greater than 50% then they've broken the block cipher. (Look at these notes, pages 1 - 8, specifically 7, for a picture and more precise definition of this model.)
An attacker can distinguish a block cipher from a PRP if said block cipher has linear properties. They can learn those linear properties then query the black box with plaintexts that should produce certain properties in the ciphertext. If the black box replies back with ciphertext that matches those expected properties for more than 50% of the input queries, then the attacker guesses that the black box houses the block cipher because a random permutation would honor those linear properties with only 50% probability. (Otherwise, they guess that the black box is using a random permutation.) Then the attacker wins the distinguishing game.
Of course, if you're looking for a practical reason, this raises the question of why we care about PRP models to begin with.
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Hi B-Con thanks for the elaboration on the Block Cipher and PRP. i agree with the model you have presented but my basic query is how in the construction of these block ciphers non linearity play a role. – David Jun 9 '12 at 6:09
If a block cipher is linear with respect to some field, then, given a few known plaintext-ciphertext pairs, it is possible to recover the key using a simple Gaussian elimination. This clearly contradicts the security properties one expects from a secure block cipher.
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http://www.conservapedia.com/Quantitative_Introduction_to_General_Relativity
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# Quantitative Introduction to General Relativity
### From Conservapedia
For a general overview of the theory, see General Relativity
$\pi_1(S^1)=?\,$ This article/section deals with mathematical concepts appropriate for a student in late university or graduate level.
General Relativity is a mathematical extension of Special Relativity. GR views space-time as a 4-dimensional manifold, which looks locally like Minkowski space, and which acquires curvature due to the presence of massive bodies. Thus, near massive bodies, the geometry of space-time differs to a large degree from Euclidean geometry: for example, the sum of the angles in a triangle is not exactly 180 degrees. Just as in classical physics, objects travel along geodesics in the absence of external forces. Importantly though, near a massive body, geodesics are no longer straight lines. It is this phenomenon of objects traveling along geodesics in a curved spacetime that accounts for gravity.
The mathematical expression of the theory of general relativity takes the form of the Einstein field equations, a set of ten nonlinear partial differential equations. While solving these equations is quite difficult, examining them provides valuable insight into the structure and meaning of the theory.
In their general form, the Einstein field equations are written as a single tensor equation in abstract index notation relating the curvature of spacetime to sources of curvature such as energy density and momentum.
$G_{\mu\nu} = 8\,\pi\,G\,T_{\mu\nu}$
In this form, Gμν represents the Einstein tensor, G is the same gravitational constant that appears in the law of universal gravitation, and Tμν is the stress-energy tensor (sometimes referred to as the energy-momentum tensor). The indices μ and ν range from zero to three, representing the time coordinate and the three space coordinates in a manner consistent with special relativity.
The left side of the equation — the Einstein tensor — describes the curvature of spacetime in the region under examination. The right side of the equation describes everything in that region that affects the curvature of spacetime.
As we can clearly see even in this simplified form, the Einstein field equations can be solved "in either direction." Given a description of the gravitating matter, energy, momentum and fields in a region of spacetime, we can calculate the curvature of spacetime surrounding that region. On the other hand, given a description of the curvature of a region spacetime, we can calculate the motion of a test particle anywhere within that region.
Even at this level of examination, the fundamental thesis of the general theory of relativity is obvious: motion is determined by the curvature of spacetime, and the curvature of spacetime is determined by the matter, energy, momentum and fields within it.
## Contents
### The right side of the equation: the stress-energy tensor
In the Newtonian approximation, the gravitational vector field is directly proportional to mass. In general relativity, mass is just one of several sources of spacetime curvature. The stress-energy tensor, Tμν, includes all of these sources. Put simply, the stress-energy tensor quantifies all the stuff that contributes to spacetime curvature, and thus to the gravitational field.
First we will define the stress-energy tensor technically, then we'll examine what that definition means. In technical terms, the stress energy tensor represents the flux of the μ component of 4-momentum across a surface of constant coordinate xν.
Fine. But what does that mean?
In classical mechanics, it's customary to refer to coordinates in space as x, y and z. In general relativity, the convention is to talk instead about coordinates x0, x1, x2, and x3, where x0 is the time coordinate otherwise called t, and the other three are just the x, y and z coordinates. So "a surface of constant coordinate "xν" simply means a 3-plane perpendicular to the xν axis.
The flux of a quantity can be visualized as the magnitude of the current in a river: the flux of water is the amount of water that passes through a cross-section of the river in a given interval of time. So more generally, the flux of a quantity across a surface is the amount of that quantity that passes through that surface.
Four-momentum is the special relativity analogue of the familiar momentum from classical mechanics, with the property that the time coordinate $\mathbf{P}^0$ of a particle's four-momentum is simply the energy of the particle; the other three components of four-momentum are the same as in classical momentum.
So putting that all together, the stress-energy tensor is the flux of 4-momentum across a surface of constant coordinate. In other words, the stress-energy tensor describes the density of energy and momentum, and the flux of energy and momentum in a region. Since under the mass-energy equivalence principle we can convert mass units to energy units and vice-versa, this means that the stress-energy tensor describes all the mass and energy in a given region of spacetime.
Put even more simply, the stress-energy tensor represents everything that gravitates.
The stress-energy tensor, being a tensor of rank two in four-dimensional spacetime, has sixteen components that can be written as a 4 × 4 matrix.
$T_{\mu \nu }=\begin{pmatrix} \color{Purple}T_{00} & \color{Blue}T_{01} & \color{Blue}T_{02} & \color{Blue}T_{03} \\ \color{Red}T_{10} & \color{Orange}T_{11} & \color{Green}T_{12} & \color{Green}T_{13} \\ \color{Red}T_{20} & \color{Green}T_{21} & \color{Orange}T_{22} & \color{Green}T_{23} \\ \color{Red}T_{30} & \color{Green}T_{31} & \color{Green}T_{32} & \color{Orange}T_{33} \end{pmatrix}$
Here the components have been color-coded to help clarify their physical interpretations.
$\color{Purple}T_{00}$
energy density, which is equivalent to mass-energy density; this component includes the mass contribution
$\color{Red}T_{10}$, $\color{Red}T_{20}$, $\color{Red}T_{30}$
the components of momentum density
$\color{Blue}T_{01}$, $\color{Blue}T_{02}$, $\color{Blue}T_{03}$
the components of energy flux
The space-space components of the stress-energy tensor are simply the stress tensor from classic mechanics. Those components can be interpreted as:
$\color{Green}T_{12}$, $\color{Green}T_{13}$, $\color{Green}T_{23}$, $\color{Green}T_{21}$, $\color{Green}T_{31}$, $\color{Green}T_{32}$
the components of shear stress, or stress applied tangential to the region
$\color{Orange}T_{11}$, $\color{Orange}T_{22}$, $\color{Orange}T_{33}$
the components of normal stress, or stress applied perpendicular to the region; normal stress is another term for pressure.
Pay particular attention to the first column of the above matrix: the components $\color{Purple}T_{00}$, $\color{Red}T_{10}$, $\color{Red}T_{20}$ and $\color{Red}T_{30}$, are interpreted as densities. A density is what you get when you measure the flux of 4-momentum across a 3-surface of constant time. Put another way, the instantaneous value of 4-momentum flux is density.
Similarly, the diagonal space components of the stress-energy tensor — $\color{Orange}T_{11}$, $\color{Orange}T_{22}$ and $\color{Orange}T_{33}$ — represent normal stress, or pressure. Not some weird, relativistic pressure, but plain old ordinary pressure, like what keeps a balloon inflated. Pressure also contributes to gravitation, which raises a very interesting observation.
Imagine a box of air, a rigid box that won't flex. Let's say that the pressure of the air inside the box is the same as the pressure of the air outside the box. If we heat the box — assuming of course that the box is airtight — then the temperature of the gas inside will rise. In turn, as predicted by the ideal gas law, the pressure within the box will increase.
The box is now heavier than it was.
More precisely, increasing the pressure inside the box raised the value of the pressure contribution to the stress-energy tensor, which will increase the curvature of spacetime around the box. What's more, merely increasing the temperature alone caused spacetime around the box to curve more, because the kinetic energy of the gas molecules inside the box also contributes to the stress-energy tensor, via the time-time component $\color{Purple}T_{00}$. All of these things contribute to the curvature of spacetime around the box, and thus to the gravitational field created by the box.
Of course, in practice, the contributions of increased pressure and kinetic energy would be miniscule compared to the mass contribution, so it would be extremely difficult to measure the gravitational effect of heating the box. But on larger scales, such as the sun, pressure and temperature contribute significantly to the gravitational field.
In this way, we can see that the stress-energy tensor neatly quantifies all static and dynamic properties of a region of spacetime, from mass to momentum to electric charge to temperature to pressure to shear stress. Thus, the stress-energy tensor is all we need on the right-hand side of the equation in order to relate matter, energy and, well, stuff to curvature, and thus to the gravitational field.
#### Example 1: Stress-energy tensor for a vacuum
The simplest possible stress-energy tensor is, of course, one in which all the values are zero.
$T_{\mu \nu }\ =\ 0$
This tensor represents a region of space in which there is no matter, energy or fields, not just at a given instant, but over the entire period of time in which we're interested in the region. Nothing exists in this region, and nothing happens in this region.
So one might assume that in a region where the stress-energy tensor is zero, the gravitational field must also necessarily be zero. There's nothing there to gravitate, so it follows naturally that there can be no gravitation.
In fact, it's not that simple. We'll discuss this in greater detail in the next section, but even a cursory qualitative examination can tell us there's more going on than that. Consider the gravitational field of an isolated body. A test particle placed somewhere near but outside of the body will move in a geodesic in spacetime, freely falling inward toward the central mass. A test particle with some constant linear velocity component perpendicular to the interval between the particle and the mass will move in a conic section. This is true even though the stress-energy tensor in that region is exactly zero. This much is obvious from our intuitive understanding of gravity: gravity affects things at a distance. But exactly how and why this happens, in the model of the Einstein field equations, is an interesting question which will be explored in the next section.
#### Example 2: Stress-energy tensor for an ideal dust
Imagine a time-dependent distribution of identical, massive, non-interacting, electrically neutral particles. In general relativity, such a distribution is called a dust. Let's break down what this means.
time-dependent
The distribution of particles in our dust is not a constant; that is to say, the particles may be motion. The overall configuration you see when you look at the dust depends on the time at which you look at it, so the dust is said to be time-dependent.
identical
The particles that make up our dust are all exactly the same; they don't differ from each other in any way.
massive
Each particle in our dust has some rest mass. Because the particles are all identical, their rest masses must also be identical. We'll call the rest mass of an individual particle m0.
non-interacting
The particles don't interact with each other in any way: they don't collide, and they don't attract or repel each other. This is, of course, an idealization; since the particles are said to have mass m0, they must at least interact with each other gravitationally, if not in other ways. But we're constructing our model in such a way that gravitational effects between the individual particles are so small as to be be negligible. Either the individual particles are very tiny, or the average distance between them is very large. This same assumption neatly cancels out any other possible interactions, as long as we assume that the particles are far enough apart.
electrically neutral
In addition to the obvious electrostatic effect of two charged particles either attracting or repelling each other — thus violating our "non-interacting" assumption — allowing the particles to be both charged and in motion would introduce electrodynamic effects that would have to be factored into the stress-energy tensor. We would greatly prefer to ignore these effects for the sake of simplicity, so by definition, the particles in our dust are all electrically neutral.
The easiest way to visualize an ideal dust is to imagine, well, dust. Dust particles sometimes catch the light of the sun and can be seen if you look closely enough. Each particle is moving in apparent ignorance of the rest, its velocity at any given moment dependent only on the motion of the air around it. If we take away the air, each particle of dust will continue moving in a straight line at a constant velocity, whatever its velocity happened to be at the time. This is a good visualization of an ideal dust.
We're now going to zoom out slightly from our model, such that we lose sight of the individual particles that make up our dust and can consider instead the dust as a whole. We can fully describe our dust at any event P — where event is defined as a point in space at an instant in time — by measuring the density ρ and the 4-velocity u at P. If we have those two pieces of information about the dust at every point within it at every moment in time, then there's literally nothing else to say about the dust: it's been fully described.
##### Density
Let's start by figuring out the density of dust at a the event P, as measured from the perspective of an observer moving along with the flow of dust at P. The density ρ is calculated very simply:
$\rho = m_0 n\,$
where m0 is the mass of each particle and n is the number of particles in a cubical volume one unit of length on a side centered on P. This quantity is called proper density, meaning the density of the dust as measured within the dust's own reference frame. In other words, if we could somehow imagine the dust to measure its own density, the proper density is the number it would get.
Clearly proper density is a function of position, since it varies from point to point within the dust; the dust might be more "crowded" over here, less "crowded" over there. But it's also a function of time, because the configuration of the dust itself is time-dependent. If you measure the proper density at some point in space at one instant of time, then measure it at the same point in space at a different instant of time, you may get a different measurement. By convention, when dealing with a quantity that depends both on position in space and on time, physicists simply say that the quantity is a function of position, with the understanding that they're referring to a "position" in four-dimensional spacetime.
##### 4-velocity
The other quantity we need is 4-velocity. Four-velocity is an extension of three-dimensional velocity (or 3-velocity). In three dimensional space, 3-velocity is a vector with three components. Likewise, in four-dimensional spacetime, 4-velocity is a vector with four components.
Directly measuring 4-velocity is an inherently tricky business, since one of its components describes motion along a "direction" that we cannot see with our eyes: motion through time. The math of special relativity lets us calculate the 4-velocity of a moving particle given only its 3-velocity v (with components vi where i = 1,2,3) and the speed of light. The time component of 4-velocity is given by:
$u^0 = \gamma c\,$
and the space components u1, u2 and u3 by:
$u^i = \gamma v^i\,$
where γ is the boost, or Lorentz factor:
$\gamma = \frac{1}{\sqrt{1-\frac{\|v\|^2}{c^2}}}\,$
and where $\|v\|^2$, in turn, is the square of the Euclidean magnitude of the 3-velocity vector v:
$\|v\|^2 = (v^1)^2+(v^2)^2+(v^3)^2\,$
Therefore, if we know the 3-velocity of the dust at event P, then we can calculate its 4-velocity. (For more details on the how and why of 4-velocity, refer to the article on special relativity.)
Just as proper density is a function of position in spacetime, 4-velocity also depends on position. The 4-velocity of our dust at a given point in space won't necessarily be the same as the 4-velocity of the dust at another point in space. Likewise, the 4-velocity at a given point at a given time may not be the same as the 4-velocity of the dust at the same point at a different time. It helps to think of 4-velocity as the velocity of the dust through a point in both space and time.
##### Assembling the stress-energy tensor
Since the density and the 4-velocity fully describe our dust, we have everything we need to calculate the stress-energy tensor.
$T(x) = \rho (x) u(x)\otimes u(x)\,$
where the symbol $\otimes$ indicates a tensor product. The tensor product of two vectors is a tensor of rank two, so the stress-energy tensor must be a tensor of rank two. In an arbitrary coordinate frame xμ, the contravariant components of the stress-energy tensor for an ideal dust are given by:
$T^{\mu\nu} = \rho u^{\mu} u^{\nu}\,$
From this equation, we can now calculate the contravariant components of the stress-energy tensor for an ideal dust.
###### Time-time component
$T^{00} = \rho u^0 u^0 = \rho (\gamma c) (\gamma c) = \rho \gamma^2 c^2 \,$
If we rearrange the terms in this equation slightly, something important becomes apparent:
$T^{00} = \gamma^2 (\rho c^2)\,$
Recall that ρ is a density quantity, in mass per unit volume. By the mass-energy equivalence principle, we know that E = mc2. So we can interpret this component of the stress-energy tensor, which is written here in terms of mass-energy, to be equivalent to an energy density.[1]
###### Off-diagonal components
The off-diagonal components of the tensor — Tμν where μ and ν are not equal — are calculated this way:
$T^{10} = \rho u^1 u^0 = \rho (\gamma v^1) (\gamma c) = \gamma^2 c \rho v^1\,$
Again, recall that ρ is a quantity of mass per unit volume. Multiplying a mass times a velocity gives momentum, so we can interpret ρv1 as the density of momentum along the x1 direction, multiplied by constants c and γ2. Momentum density is an extremely difficult quantity to visualize, but it's a quantity that comes up over and over in general relativity. If nothing else, one can take comfort in the fact that momentum density is mathematically equivalent to the product of mass density and velocity, both of which are much more intuitive quantities.
Note that the off-diagonal components of the tensor are equal to each other:
$T^{10} = \rho u^1 u^0 = \rho (\gamma v^1) (\gamma c) = \rho \gamma^2 c v^1\,$
$T^{01} = \rho u^0 u^1 = \rho (\gamma c) (\gamma v^1) = \rho c \gamma^2 v^1\,$
In other words, in the case of an ideal dust, the stress-energy tensor is said to be symmetric. A rank two symmetric tensor is said to be symmetric if Tab = Tba.
###### Diagonal space components
The diagonal space components of the stress-energy tensor are calculated this way:
$T^{11} = \rho u^1 u^1 = \rho (\gamma v^1) (\gamma v^1) = \gamma^2 \rho (v^1)^2\,$
In this case, we're multiplying a four-dimensional mass density, ρ, by the square of a component of 4-velocity. By dimensional analysis, we can see:
$\frac{\mbox{kg}}{\mbox{m}^4} \cdot \frac{\mbox{m}^2}{\mbox{s}^2} = \frac{\mbox{kg} \cdot \mbox{m}^2}{\mbox{m}^4 \cdot \mbox{s}^2} = \frac{\mbox{kg}}{\mbox{m}^2 \cdot \mbox{s}^2}$
Recall that the force has units:
$\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}^2}$
If we divide the units of the diagonal space component by the units of force, we get:
$\frac{\mbox{kg}}{\mbox{m}^2 \cdot \mbox{s}^2} \cdot \frac{\mbox{s}^2}{\mbox{kg} \cdot \mbox{m}} = \frac{1}{\mbox{m}^3}$
So the diagonal space components of the stress-energy tensor come are expressed in terms of force per unit volume. Force per unit area are, of course, the traditional units of pressure in three-dimensional mechanics. So we can interpret the diagonal space components of the stress-energy tensor as the components of "4-pressure"[2] in spacetime.
###### The big picture
We now know everything we know to assemble the entire stress-energy tensor, all sixteen components, and look at it as a whole.[3]
$T_{\mu \nu }=\begin{pmatrix} \gamma^2 \rho c^2 & \gamma^2 \rho v^1 c & \gamma^2 \rho v^2 c & \gamma^2 \rho v^3 c \\ \gamma^2 \rho c v^1 & \gamma^2 \rho (v^1)^2 & \gamma^2 \rho v^2 v^1 & \gamma^2 \rho v^3 v^1 \\ \gamma^2 \rho c v^2 & \gamma^2 \rho v^1 v^2 & \gamma^2 \rho (v^2)^2 & \gamma^2 \rho v^3 v^2 \\ \gamma^2 \rho c v^3 & \gamma^2 \rho v^1 v^3 & \gamma^2 \rho v^2 v^3 & \gamma^2 \rho (v^3)^2 \end{pmatrix}$
The large-scale structure of the tensor now becomes apparent. This is the stress-energy tensor of an ideal dust. The tensor is composed entirely out of the proper density and the components of 4-velocity. When velocities are low, the coefficient γ2, even though it's a squared value, remains extremely close to one.
The time-time component includes a mass multiplied by the square of the speed of light, so it has to do with energy. The rest of the top row and left column all include the speed of light as a coefficient, as well as density and velocity; in the case of an ideal dust which is made up of non-interacting particles, the energy flux along any basis direction is the same as the momentum density along that direction. This is not the case in other, less simple models, but it's true here.
The diagonal space components of the tensor represent pressure. For example, the T11 component represents the pressure that would be exerted on a plane perpendicular to the x1 direction.
The off-diagonal space components represent shear stress. The T12 component, for instance, represents the pressure that would be exerted in the x2 direction on a plane perpendicular to the x1 axis.
The overall process for calculating the stress-energy tensor for any system is fairly similar to the example given here. It involves taking into account all the matter and energy in the system, describing how the system evolves over time, and breaking that evolution down into components which represent individual densities and fluxes along different directions relative to a chosen coordinate basis.
As can easily be imagined, the task of constructing a stress-energy tensor for a system of arbitrary complexity can be a very daunting one. Fortunately, gravity is an extremely weak interaction, as interactions go, so on the scales where gravity is interesting, much of the complexity of a system can be approximated. For instance, there is absolutely nothing in the entire universe that behaves exactly like the ideal dust described here; every massive particle interacts, in one way or another, with other massive particles. No matter what, a real system is going to be very much more complex than this approximation. Yet, the ideal dust solution remains a much-used approximation in theoretical physics specifically because gravity is such a weak interaction. On the scales where gravity is worth studying, many distributions of matter, including interstellar nebulae, clusters of galaxies, even the whole universe really do behave very much like an ideal dust.
### The left side of the equation: the Einstein curvature tensor
We will recall that the Einstein field equations can be written as a single tensor equation:
$G_{\mu\nu} = 8\,\pi\,G\,T_{\mu\nu}$
The right side of the equation consists of some constants and the stress-energy tensor, described in significant detail in the previous section. The right side of the equation is the "matter" side. All matter and energy in a region of space is described by the right side of the equation.
The left side of the equation, then, is the "space" side. Matter tells space how to curve, and space tells matter how to move. So the left side of the Einstein field equation must necessarily describe the curvature of spacetime in the presence of matter and energy.
#### Some assumptions about the universe
Before we proceed into a discussion of what curvature is and how the Einstein equation describes it, we must first pause to state some fundamental assumptions about the universe.[4]
The first assumption we're going to make is that spacetime is continuous. In essence, this means that for any event P in spacetime — that is, any point in space and moment in time — there exists some local neighborhood of P where the intrinsic properties of spacetime differ from those at P by only an infinitesimal amount.
The second assumption we're going to make is that spacetime is differentiable everywhere. In other words, the geometry of spacetime doesn't have any sharp creases in it.
If we hold these two assumptions to be true, then a convenient property of spacetime emerges: Given any event P, there exists a local neighborhood where spacetime can be treated as flat, that is, having zero curvature. It is not necessarily true that all of spacetime be flat — in fact, it most definitely is not — but given any event in spacetime, there exists some neighborhood around it that is flat. This neighborhood may be arbitrarily small in both time and space, but it is guaranteed to exist as long as our two assumptions remain valid.
With these two assumptions and this convenient property in hand, we will now examine what it means to say that spacetime is curved.
#### Flatness versus curvature
Let's start by considering the simplest possible geometry[5]: the Euclidean plane.
The Euclidean plane is an infinite, flat, two-dimensional surface. A sheet of paper is a good approximation of the Euclidean plane. Onto this plane, we can project a set of Cartesian coordinates. By "Cartesian," we mean that the coordinate axes are straight lines, that they are perpendicular, and that the unit lengths of the axes are equal. A fancier term for a Cartesian coordinate system is an orthonormal basis.
Note carefully the distinction between the Euclidean plane and Cartesian coordinates. The plane exists as a thing in and of itself, just as a blank piece of paper does. It has certain properties, which we'll get into below. Those properties are intrinsic to the plane. That is, the properties don't have anything to do with the coordinates we project onto the plane. The plane is a geometric object, and the coordinates are the method by which we measure the plane. (The emphasis on the word measure there is not accidental; please keep this idea in the foreground of your mind as we continue.)
Cartesian coordinates are not the only coordinates we can use in the Euclidean plane. For example, instead of having axes that are perpendicular to each other, we could choose axes that are straight lines, but that meet at some non-perpendicular angle. These types of coordinates are called oblique.
For that matter, we're not bound to use straight-line coordinates at all. We could instead choose polar coordinates, wherein every point on the plane is described by a distance from a fixed but arbitrary point and an angle from a fixed but arbitrary direction. Polar coordinates are often more convenient than Cartesian coordinates. For example, when navigating a ship on the ocean, the location of a fixed point is usually described in terms of a bearing and a distance, where the distance is the straight-line distance from the ship to the point, and the bearing is the clockwise angle relative to the direction in which the ship is sailing. Polar coordinates in two and three dimensions are often used in physics for similar reasons.
But there's a fundamental problem with polar coordinates that is not present with Cartesian coordinates. In Cartesian coordinates, every point on the Euclidean plane is identified by exactly one set of real numbers: there is precisely one set of x and y coordinates for every point, and every point corresponds to precisely one set of coordinates.
This is not true in polar coordinates. What are the unique polar coordinates for the origin? The radial distance is obviously zero, but what is the angle? In actuality, if the radial distance is zero, any angle can be used, and the coordinates will identify the same point. The one-to-one correspondence between points in the plane and pairs of coordinates breaks down at the origin.
In mathematical terms, polar coordinates in the Euclidean plane have a coordinate singularity at the origin. A coordinate singularity is a point in space where ambiguities are introduced, not because of some intrinsic property of space, but because of the coordinate basis you chose.
So clearly there may exist a reason to choose one coordinate system over another when measuring — there's that word again — the Euclidean plane. Polar coordinates have a singularity at the origin — in this case, a point of undefined angle — while Cartesian coordinates have no such singularities anywhere. So there may be good reason to choose Cartesian coordinates over polar coordinates when measuring the Euclidean plane.
Fortunately, this is always possible. The Euclidean plane can always be measured by Cartesian coordinates; that is, coordinates wherein the axes are straight and perpendicular at their intersection, and where lines of constant coordinate — picture the grid on a sheet of graph paper — are always a constant distance apart no matter where you measure them.
Imagine taking a piece of graph paper, which is printed in a pattern that lets us easily visualize the Cartesian coordinate system, and rolling it into a cylinder. Do any creases appear in the paper? No, it remains smooth all over. Do the lines printed on the paper remain a constant distance apart everywhere? Yes, they do. In technical mathematical terms, then, the surface of a cylinder is flat. That is, it can be measured by an orthonormal basis, and there is everywhere a one-to-one correspondence between sets of coordinates and points on the surface. It's possible not to use an orthonormal basis to measure the surface; one might reasonably choose polar coordinates, or some other arbitrary coordinate system, if it's more convenient. But whichever basis is actually used, it's always possible to switch to an orthonormal basis instead.
Now imagine wrapping a sheet of graph paper around a basketball. Does the paper remain smooth? No, if we press it down, creases appear. Do the lines on the paper remain parallel? No, they have to bend in order to conform the paper to the shape of the ball. In the same technical mathematical terms, the surface of a sphere is not flat. It's curved. That is, it is not possible to measure the surface all over using an orthonormal basis.
But what if we focus our attention only on a part of the sphere? What if instead of measuring a basketball, we want to measure the whole Earth? The Earth is a sphere, and therefore its surface is curved and can't be measured all over with Cartesian coordinates. But if we look only at a small section of the surface — a square mile on a side, for instance — then we can project a set of Cartesian coordinates that work just fine. If we choose our region of interest to be sufficiently small, then Cartesian coordinates will fit on the surface to within the limits of our ability to measure the difference.
The surface of a sphere, then, is globally curved, but locally flat. In physicist jargon, the surface of a sphere can be flattened over a sufficiently small region. Not the whole sphere all at once, nor half of it, nor a quarter of it. But a sufficiently small region can be dealt with as if it were a Euclidean plane.
But this brings up an important point. The entire surface of the sphere is curved, and thus can't be approximated with Cartesian coordinates. But a sufficiently small patch of the surface can be approximated with Cartesian coordinates. This implies, then, that "curvedness" isn't an either-or property. Somewhere between the locally flat region of the surface and the entire surface, the amount of curvature goes from none to some value. Curvature, then, must be something we can measure.
#### The metric tensor
It is a fundamental property of the Euclidean plane that, when Cartesian coordinates are used, the distance s between any two points A and B is given by the following equation:
$s^2 = \Delta x^2 + \Delta y^2\,$
where Δx and Δy are the distance between A and B in the x and y directions, respectively. This is essentially a restatement of the universally known Pythagorean theorem, and in the context of general relativity, it is called the metric equation. Metric, of course, comes from the same linguistic root as the word measure, and since this is the equation we use to measure distances, it makes sense to call it the metric equation.
But this particular metric equation only works on the Euclidean plane with Cartesian coordinates. If we use polar coordinates, this equation won't work.[6] If we're on a curved surface instead of a plane, this equation won't work. This metric equation is only valid on a flat surface with Cartesian coordinates.
Which makes it pretty useless, since so much of physics revolves around curved spacetime and spherical coordinates.
What we need is a generalized metric equation, some way of measuring the interval of any two points regardless of what coordinate system we're using or whether our local geometry is flat or curved.
The metric tensor equation provides this generalization.
If v is any vector having components vμ, the length of v is given by the following equation:
$s^2 = g_{\mu\nu} v^{\mu} v^{\nu}\,$
where gμν is the metric tensor, and μ and ν range over the number of dimensions. Recall that Einstein summation notation means that this is actually a sum over indices μ and ν. If we assume that we're in the two-dimensional Euclidean plane, the metric tensor equation expands to:
$s^2 = g_{11} v^{1} v^{1} + g_{12} v^{1} v^{2} + g_{21} v^{2} v^{1} + g_{22} v^{2} v^{2}\,$
The terms of the metric tensor, then, must be numerical coefficients in the metric equation. We already know what these equations need to be to make the metric equation work in the Euclidean plane with Cartesian coordinates:
$s^2 = (1) v^{1} v^{1} + (0) v^{1} v^{2} + (0) v^{2} v^{1} + (1) v^{2} v^{2}\,$
Now we can write the metric tensor for the Euclidean plane in Cartesian coordinates in the form of a 2 × 2 matrix:
$g_{\mu\nu} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
So in the case of the Euclidean plane with Cartesian coordinates, the metric tensor is the Kronecker delta:
$\delta^i_j = \begin{cases} 1, & \mbox{if }i = j \\ 0, & \mbox{if }i \ne j \end{cases}$
Of course, the same concepts apply if we expand our interest from the plane to three-dimensional Euclidean space with Cartesian coordinates. We just have to let the indices of the Kronecker delta run from 1 to 3.
$g_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$
Which gives us the following metric equation for the length of a vector v (omitting terms with zero coefficient):
$s^2 = (v^1)^2 + (v^2)^2 + (v^3)^2\,$
Which precisely agrees with the Pythagorean theorem in three dimensions. So given a metric tensor gμν for any space and coordinate basis, we can calculate the distance between any two points. The metric tensor, therefore, is what allows us to measure curved space. In a very real sense, the metric tensor describes the shape of both the underlying geometry and the chosen coordinate basis.
But relativity is concerned not with geometrically abstract space; we're interested in very real spacetime, and that requires a slightly different kind of metric.
#### The cosmological constant
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http://math.stackexchange.com/questions/73316/do-we-need-the-digits-of-pi/73320
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# Do We Need the Digits of $\pi$?
I was reading today that someone found $\pi$ to the ten trillionth digit. Whenever I read that $\pi$ has been calculated to more digits, I ask myself whether this is useful. I know that there are conjectures out there about distributions of numbers in $\pi$ and such. So, I supposed knowing more digits helps us test conjectures. But, are there more reasons that we would want to know the digits? Anything really cool I'm ignoring or forgetting?
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This should probably be community. Do I have the ability to do that? – Joe Johnson 126 Oct 17 '11 at 10:58
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No you can't do it yourself. You need to flag for moderator attention and ask the moderators to convert the question into CW mode. However, I don't really see why it should be a community wiki. – t.b. Oct 17 '11 at 11:00
@t.b.: I figured CW because I'm not looking for one particular answer. – Joe Johnson 126 Oct 17 '11 at 11:07
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– J. M. Oct 17 '11 at 11:16
Thanks for all the great replies. I'll accept the highest answer, as I feel is customary with questions of this nature. – Joe Johnson 126 Oct 21 '11 at 20:24
show 1 more comment
## 6 Answers
not for any calculations from the wiki
Practically, one needs only 39 digits of π to make a circle the size of the observable universe accurate to the size of a hydrogen atom.
however it is useful to test supercomputers for accuracy
Today the high precision calculation of $\pi$ finds practical use in testing the "global integrity" of a supercomputer. "A large scale calculation of pi is entirely unforgiving; it soaks into all parts of the machine and a single bit awry leaves detectable consequences.
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Something went wrong with your first link. I tried to fix it by linking to the relevant part of the Wikipedia page on $\pi$. Please check that it's the link you intended. – t.b. Oct 17 '11 at 11:18
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"any calculations" is a bit of a stretch; as mentioned in the answers to a related question, PSLQ and other integer-relation algorithms do require a whole pile of digits to assist in finding new relations between constants, as well as new formulae (and thus methods) to compute them... – J. M. Oct 17 '11 at 11:52
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@J.M.: Although those algorithms would not need a trillion digits. I would guess at most a couple hundred. – Grumpy Parsnip Oct 17 '11 at 15:48
A less mathematical reason for calculating more and more decimal places is because we know they are there. Man is inherently curious and always wants to see what's over the next hill, round the next bend etc.
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More to the point, we are curious about something which we already know is very important. I could roll dice until I die to find out what will happen next, but that doesn't mean that I would find this very enlightening. The difference is that the digits of $\pi$ are more likely to have at least some interesting properties, compared to any typical set of dice-rolls. (Also, for what it's worth, I've met a number of Women who are also very curious.) – Niel de Beaudrap Oct 17 '11 at 11:50
Knowing more and more digits of $\pi$ has absolutely no value to anybody. Digits of $\pi$, beyond the twentieth or so, are completely worthless. The only value in this enterprise, if any, lies in the process by which the digits are generated, not the digits themselves.
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"While such an exercise might seem frivolous, the fact is we learned a lot from the continual refinement of our algorithms to work efficiently at ultrahigh precision. The reward is a deeper understanding of the theory, and often a better algorithm for low-precision cases." – J. M. Oct 18 '11 at 0:54
@J.M.: That seems to back me up (if you ignore my "if any", that is). Who said it? – TonyK Oct 18 '11 at 7:36
– J. M. Oct 18 '11 at 9:32
Another reason is for its properties of randomness.
This can be used to test software for analyzing random sequences. It can also be used as a teaching aide. Example: if the digits of pi exhibit statistical randomness (it is believed they do), then at some point in pi's expansion there will be a sequence of one million consecutive 0's.
This surprises a lot of students.
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It's not a good test for randomness tests because we lack a proof that π has the properties we want to test. – sigfpe Oct 17 '11 at 18:03
@user80: That's irrelevent, because we know those properties hold for all the digits we use to test (when used to test software, we won't use anywhere near the first trillion digits). Similarly, Goldbach's Conjecture is often used in proofs of algorithmic complexity: we don't know that it's true, but we know that it's true for much, much larger numbers than we'll ever use in practice, so it's still useful. – BlueRaja - Danny Pflughoeft Oct 17 '11 at 19:50
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Whenever we calculate the first $n$ digits of an irrational number, we simply estimate it and there is a small error in our approximation.
If we use that for practical applications, we should always be aware of the error and check if it is resonable or not for the practical application. If not we need more digits, so theoretically there is always a need for extra digits.
The above answers covered well why 1 billion digits are probably more than enough for $\pi$, anyhow in general it is hard to say how many digits we need to know to cover any possible application. A resonable number of digits shouldn't suffice, we always need an unresonable number.
Not for $\pi$, but here is a known example where 26 digits (well not exactly, they were 26 binary digits) were not enough for a practical application, and unfortunatelly some people died because the error in that application was too big:
http://ta.twi.tudelft.nl/users/vuik/wi211/disasters.html#patriot
I find this example interesting, because most people would think that 5-6 digits should suffice in all cases, and it is easy to understand why in this case the estimation wasn't good (of course people should had thought about it before it happend)...
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http://physics.stackexchange.com/tags/lagrangian-formalism/new
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Tag Info
New answers tagged lagrangian-formalism
3
Retrieving Maxwell's equations from the minimum action principle
Author gives a clue on the transition: Let us assume that $\delta\vec{A}$ vanishes at infinity and integrate (formula (1)) by parts... This is the usual step in the Lagrangian theory of field (actually, of anything). At first, we have the variation of action written in an awkward form: \delta S=\int_{\substack{\text{domain of least}\\\text{action ...
1
Is there a Lagrangian formulation of statistical mechanics?
I am not sure if this is what you are up to (it is related to what Xiao-Qi Sun said) to but I'll give it a try too ... At the beginning of Chapter V.2 of his QFT Nutshell, Anthony Zee explains how classical statistical mechanics (characterized by the corresponding partition function involving the Hamilton function) in $d$- dimensional space is related to ...
0
Determinant for a coupled fluctuation Lagrangian
I asked my professor and in a discussion we came up with the following. The process of establishing the effective action for a fluctuation Lagrangian to consist of the functional determinant of the initial differential operator involved, relies on the equality: $$\det(A)=e^{Tr(\log(A))}$$ for a matrix $A$, which is only true for diagonalizable ...
3
Why does Lagrangian of free particle depend on the square of the velocity ?
The Lagrangian should not only be independent of the direction of $\vec{v}$ but it should also change correctly under a Galilean transformation. For instance, if $K$ and $K'$ are two frames of reference with a relative velocity $\vec{V}$ then the two Lagrangians $L$ and $L'$ should differ only by a total time derivative. If $L$ is a function of fourth power ...
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Independent systems and Lagrangians
Well, you are not specifying the kind of probabilities you are talking about. Therefore, I refer to the tags and assume that we are talking about quantum (not statistical) probabilities. (See the edit at the end) Then, I have to note that your probabilistic definition of independece does not make much sense. Edit Well, according to the comments, I should ...
4
Independent systems and Lagrangians
How about path integrals? The probability that a system evolves between state $|\phi_1\rangle$ and $|\phi_2\rangle$ is $$\langle \phi_2|\phi_1 \rangle =\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)$$ where the measure $\mathcal{D}\phi$ is suitably defined and the action $S(\phi)$ is the integral of the Lagrangian (over ...
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Lagrangian density for a Piano String
Let's look for solutions of the PDEs in a form $\eta(x, t)= \eta(z)$ and $\xi(x, t)= \xi(z)$, where $z= x-c_T t$. If we substitute these solution to the PDEs, they are reduced to ODEs. But to answer you question, we only need the second equation, which has the form: c_T^2 \eta_{zz} = {\lambda \over \rho_0} [\eta_{zz} ({\tau_0 \over \lambda} + \xi_z + ...
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A Type of Pendulum
Your equation is wrong. Intuitively as either $r$ or $\omega$ vanishes you should recover the equation for a simple pendulum which is not true about your equation. In the Lagrangian formalism we equate $\frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i}$ for each generalised coordinates to zero, not to each other. Usually ...
1
Total energy is extremal for the static solutions of equation of motions
I) Here is at least a partial answer. Assume the following set-up. Let there be given a classical Lagrangian field theory in $d+1$ spacetime dimensions, with dynamical field variables $\phi^{\alpha}(x,t)$, and with no explicit time dependence. Action $S[\phi]:=\int \! dt~ L[\phi(t,\cdot)]$. Lagrangian functional $L:=T-V$. Energy functional $E=T+V$. ...
1
What is the mathematical justification for the quadratic approximation to the energy of a spring in a one-dimensional lattice?
Your answer is indeed what's going on in the lecture, but it doesn't explain what was wrong with your initial argument: you'd expect a model with $l_0>0$ to be a closer representation of reality than one with $l_0=0$, wouldn't you? Actually, your initial reasoning was correct: transverse displacements of springs under zero tension do indeed result in ...
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What is the mathematical justification for the quadratic approximation to the energy of a spring in a one-dimensional lattice?
Interestingly, the quadratic potential can be justified by considering that the springs have natural length zero, i.e. $l_{0}=0$ and not $1$ (this means $l=1$ for $x=0$ just the same, but the spring is not relaxed and the potential energy will have a different dependency). I understand that this is silently assumed behind the explanation in the 1h15min mark. ...
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What is the mathematical justification for the quadratic approximation to the energy of a spring in a one-dimensional lattice?
Consider the first three terms of the Taylor expansion of the potential. The first term is a constant, which will not affect the motion. The next term is the linear term, which will only change the equilibrium point. So the first interesting term is the quadratic term.
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http://mathhelpforum.com/algebra/185813-completing-square-quadratics-form-x-2-bx-c.html
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# Thread:
1. ## Completing the square in quadratics of the form x^2 + bx + c
I am trying to understand where I am getting these types of questions wrong, it seems algebra is not my best subject at the moment.
x^2 - 8x - 5
(x - 4)^2 - 16 + 5 = (x - 4)^2 - 11
Check it;
(x - 4)^2 - 11 = x^2 - 8x + 5 + 11 = x^2 - 8x + 16
I think the problem is in the (x - 4) area as this does not work back out to the original question. I have also tried (x - 2)^2 but this also results in x^2 + 4x + 4, which is not the same.
My course book talks a lot about halfing the coefficients of x, but I am missing something somewhere?
2. ## Re: Completing the square in quadratics of the form x^2 + bx + c
$(x-4)^2=x^2-8x+16$.
So, $(x-4)^2-11=x^2-8x+16-11\dots$
3. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Why did you change the original -5 to 5 in the second line?
4. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by SammyS
$(x-4)^2=x^2-8x+16$.
So, $(x-4)^2-11=x^2-8x+16-11\dots$
So if I am reading your example correctly, you are saying that it is correct to carryout a subtraction from the LHS and Add - 11 to the RHS.
(x - 4)^2 - 11 - 11 = x^2 - 8x + 16 - 11 = x^2 - 8x - 5
I tried another method of algebra;
x2 – 8x – 5
(x – 4)2 – 16 + 5
(x – 4)2 = – 11
Check;
(x – 4)2– 11 = x2 – 8x - 5 – 11
= x2 – 8x - 16
(x – 4)2 – 11 = x2 – 8x - 16 + 11
= x2 – 8x – 5
5. ## Re: Completing the square in quadratics of the form x^2 + bx + c
16-11=5, not -5.
I would write:
$x^2-8x-5=(x-4)^2-21$
Because:
$(x-4)^2-21=x^2-8x+16-21=x^2-8x-5$
Or does it has to be $x^2-8x+5$? In that case: $(x-4)^2-11$ is correct.
6. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by Siron
16-11=5, not -5.
I would write:
$x^2-8x-5=(x-4)^2-21$
Because:
$(x-4)^2-21=x^2-8x+16-21=x^2-8x-5$
If (x - 4)^2 gives - 16 and I add 5 then this = - 11.
Or does it has to be $x^2-8x+5$? In that case: $(x-4)^2-11$ is correct.
The original equation was as you say; x^2 - 8x - 5. I understand the half term (x - 4)^2 but -4^2 = - 16 which is why I changed the sign convention from + 5 to - 5.
So in your example above because you are saying x^2 - 8x + 5, you are saying that (x - 4)^2 - 11 is correct, but I changed the - 5 to a + 5 from - 16 to get - 11.
when I add -11 to both sides I get x^2 - 8x - 5, which is the original equation, so are you saying that I am wrong?
Thanks
David
7. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Notice: $(x-4)^2=x^2-8x+16$, you're right about $-4^2=-16$ but in this case it's $(-4)^2$ which is 16.
8. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by David Green
The original equation was as you say; x^2 - 8x - 5. I understand the half term (x - 4)^2 but -4^2 = - 16 which is why I changed the sign convention from + 5 to - 5.
-16 is the right thing to put, but it looks like you got it for the wrong reason. When you expand (x-4)^2, you should do (-4)^2, which equals 16. But that has nothing to do with the 5.
What completing the square does is to represent algebraically what it would mean to literally fill in an incomplete square. You're on the right track with (x-4)^2, but this gives x^2-8x+16. But you can't just go adding 16 to one side of an equation, so you have to add it to the other side or subtract it again from the same side, which is why you're right by having -16 there.
This has nothing to do with the -5 though, and I'm not sure why you tried to change it.
9. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by Siron
Notice: $(x-4)^2=x^2-8x+16$, you're right about $-4^2=-16$ but in this case it's $(-4)^2$ which is 16.
So would I be right along these lines;
x^2 - 8x - 5
(x - 4)^2 16 - 5
(x - 4)^2 = 11
(x - 4)^2 16 = x^2 - 8x + 11
(x - 4)^2 16 - 16 = x^2 - 8x + 11 - 16
= x^2 - 8x - 5
Thanks
David
10. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by David Green
So would I be right along these lines;
(x - 4)^2 16 - 5
What does $(x-4)^2 16$ means?
11. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by Siron
What does $(x-4)^2 16$ means?
(x - 4)^2 = 16 as you said previously, it's just the way I have written it out.
12. ## Re: Completing the square in quadratics of the form x^2 + bx + c
You've to be careful with notations like that, it can be very confusing.
$(x-4)^2=x^2-8x+16$
Well, you want to have $x^2-8x-5$ so you have to substract a term -21 from each side, because $16-21=-5$ and this what you want, so:
$(x-4)^2-21=x^2-8x+16-21 \Leftrightarrow (x-4)^2-21=x^2-8x-5$
13. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by Siron
You've to be careful with notations like that, it can be very confusing.
$(x-4)^2=x^2-8x+16$
Well, you want to have $x^2-8x-5$ so you have to substract a term -21 from each side, because $16-21=-5$ and this what you want, so:
$(x-4)^2-21=x^2-8x+16-21 \Leftrightarrow (x-4)^2-21=x^2-8x-5$
Very much appreciated for all your help.
So writing the quadratic equation x^2 - 8x - 5 in completed square form is;
(x - 4)^2 = 16
14. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Originally Posted by Siron
Notice: $(x-4)^2=x^2-8x+16$, you're right about $-4^2=-16$ but in this case it's $(-4)^2$ which is 16.
Hi Siron, I have been on a learning curve with this, but now I understand the method used to find the square. We had different ideas regarding the value 16?
This is what I have learned;
x^2 - 8x - 5
(x - 4)^2 - 16 - 5
(x - 4)^2 - 21
(x - 4)^2 - 21 = x^2 - 8x + 16 - 21
= x^2 - 8x - 5
However all though all the above has been a learning curve, I did not find the solution in completed square form?
(x - 4)^2 - 16
David
15. ## Re: Completing the square in quadratics of the form x^2 + bx + c
Can you be more clear about your question? We've found the complete square of $x^2-8x-5$ which is $(x-4)^2-21$.
Where are you stuck? What do you not understand? ...
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http://mathhelpforum.com/calculus/209035-just-me-evaluating-integrals-not-easist-thing-do-please-help.html
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2Thanks
• 1 Post By Prove It
• 1 Post By HallsofIvy
# Thread:
1. ## [SOLVED] Is it just me, or evaluating integrals is not the easist thing to do?
This is my first time evaluating integrals on my own and, unfortunately, I just can't seem to get it. As easy as they may seem, integrals are giving me such a hard time, even the simplest ones are proving to be a challenge.
I need help evaluating the integral below:
I ended up with 15ln(-1)-15ln(-3)=DNE, but I'm almost certain I got it wrong.
2. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
You should have a list of standard integrals, if you haven't got one then get one. Later, it helps if you can commit them to memory.
The one you need here is $\int \frac{1}{x}\, dx = \log_{e}x + C.$
3. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Actually it's $\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}$
4. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by Prove It
Actually it's $\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}$
That's what I was about to say. Where do I take it from there though? I know that one would be $15ln{|x|} + C$ but the numbers, -1 and -3, on that symbol are confusing the hell out of me!
If I knew what to do, I wouldn't be asking...
5. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
The absolute value turns anything negative positive.
6. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by Prove It
The absolute value turns anything negative positive.
Oh hold on a sec! Would the answer be:
$15ln|-1|-15ln|-3|=-16$ (approximated)
?
7. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by Prove It
Actually it's $\displaystyle \begin{align*} \int{\frac{1}{x}\,dx} = \ln{|x|} + C \end{align*}$
Agreed, lazy of me.
8. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by theunforgiven
Oh hold on a sec! Would the answer be:
$15ln|-1|-15ln|-3|=-16$ (approximated)
?
No. Like I said, absolute values turn anything negative positive. So $\displaystyle \begin{align*} 15\ln{|-1|} - 15\ln{|-3|} = 15\ln{(1)} - 15\ln{(3)} = 0 - 15\ln{(3)} = -15\ln{(3)} \end{align*}$
9. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by Prove It
No. Like I said, absolute values turn anything negative positive. So $\displaystyle \begin{align*} 15\ln{|-1|} - 15\ln{|-3|} = 15\ln{(1)} - 15\ln{(3)} = 0 - 15\ln{(3)} = -15\ln{(3)} \end{align*}$
And that's exactly what I did. I just didn't show that extra step where you put the positive values between parentheses. I did, however, turn them to positive and put them between parentheses when I plugged in the equation into my calculator. But yeah, thank you so much!
10. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by theunforgiven
Oh hold on a sec! Would the answer be:
$15ln|-1|-15ln|-3|=-16$ (approximated)
?
Yes, although -16.5 would be a better approximation.
One of the first things you should have learned about "definite integrals" (actually, the definition) is that if F is an anti-derivative of f (that is, if dF/dx= f) then $\int_a^b f(x)dx= F(b)- F(a)$.
Actually, yes, finding anti-derivatives is, typically harder than differentiating- although evaluating definite integrals should not be more difficult than evaluating any functions. There is a general distinction in mathematics between "direct problems", where we are given a specific definition or formula, and "inverse problems" where we are to "reverse" some direct problem. For example, if I told you that $y= x^5- 3x^4+ 4x^2- 7x+ 9$ and ask you to find y when x= 3, that would be easy- just do the arithmetic in that formula. But if I told you that y= 3 and asked you to find x,, that would be a much more difficult problem- and there might be many solutions or none.
We learn, in introductory Calculus, a formula for the derivative of a function so finding a derivative can always go back to using that formula (though we can also use the many "theorems" for special cases derived from that formula). That's a direct problem. But an anti-derivative of f, say, is only defined as "a function whose derivative is f". We don't have a direct formula to use, we have to try to "remember" a function whose derivative is the given function. That's the "inverse" problem.
11. ## Re: Is it just me, or evaluating integrals is not the easist thing to do? Please help
Originally Posted by HallsofIvy
Yes, although -16.5 would be a better approximation.
One of the first things you should have learned about "definite integrals" (actually, the definition) is that if F is an anti-derivative of f (that is, if dF/dx= f) then $\int_a^b f(x)dx= F(b)- F(a)$.
Actually, yes, finding anti-derivatives is, typically harder than differentiating- although evaluating definite integrals should not be more difficult than evaluating any functions. There is a general distinction in mathematics between "direct problems", where we are given a specific definition or formula, and "inverse problems" where we are to "reverse" some direct problem. For example, if I told you that $y= x^5- 3x^4+ 4x^2- 7x+ 9$ and ask you to find y when x= 3, that would be easy- just do the arithmetic in that formula. But if I told you that y= 3 and asked you to find x,, that would be a much more difficult problem- and there might be many solutions or none.
We learn, in introductory Calculus, a formula for the derivative of a function so finding a derivative can always go back to using that formula (though we can also use the many "theorems" for special cases derived from that formula). That's a direct problem. But an anti-derivative of f, say, is only defined as "a function whose derivative is f". We don't have a direct formula to use, we have to try to "remember" a function whose derivative is the given function. That's the "inverse" problem.
That helped immensely! Thank you
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http://mathhelpforum.com/discrete-math/16236-direct-proof-contraposition-print.html
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# direct proof or contraposition?
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• June 24th 2007, 12:25 PM
Discrete
direct proof or contraposition?
I am asked to proof this conjecture:
If x is rational and x is not equal to 0, then 1/x is rational. How do I proof this? Should I use direct or contraposition proof? I just learned about this stuff and I don't know where to start at.
• June 24th 2007, 12:40 PM
Plato
The answer depends upon the axiom set you are basing the proof upon.
What are you given about rational numbers?
• June 24th 2007, 12:52 PM
Discrete
Quote:
Originally Posted by Plato
The answer depends upon the axiom set you are basing the proof upon.
What are you given about rational numbers?
What do you mean what I am given about rational numbers?
• June 24th 2007, 01:02 PM
Plato
Quote:
Originally Posted by Discrete
What do you mean what I am given about rational numbers?
What is the set of axioms you have been given?
You can prove nothing without axioms.
• June 24th 2007, 01:07 PM
Soroban
Hello, Discrete!
Quote:
$\text{Prove: If }x\text{ is rational and }x \neq 0\text{, then }\frac{1}{x}\text{ is rational.}$
You must have been given a defintion of a rational number.
Maybe somthing like:
. . $x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
That's a good place to start . . .
• June 24th 2007, 01:12 PM
Discrete
Quote:
Originally Posted by Soroban
Hello, Discrete!
You must have been given a defintion of a rational number.
Maybe somthing like:
. . $x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
That's a good place to start . . .
yes soroban that's what I was actually given.
$x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
but then I don't know the next step after that.
• June 24th 2007, 01:15 PM
Jhevon
Quote:
Originally Posted by Discrete
yes soroban that's what I was actually given.
$x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$.
but then I don't know the next step after that.
Let $x$ be a rational number, then $x = \frac {a}{b}$ for $a,b \in \mathbb {Z}, b \neq 0$. Since $x \neq 0$ it means that $a \neq 0$
So $\frac {1}{x} = \frac {1}{ \frac {a}{b}} = \frac {b}{a}$
Since $a,b \in \mathbb {Z}$ and $a \neq 0$, the fraction above represents a rational number
QED
• June 24th 2007, 01:20 PM
Plato
Quote:
Originally Posted by Jhevon
So $\frac {1}{x} = \frac {1}{ \frac {a}{b}} = \frac {b}{a}$
How do you know that? What axiom are you to use?
You are using properties of rational numbers that the student is expected to prove.
• June 24th 2007, 01:43 PM
Ilaggoodly
not true, his argument is based off of the given assumption that x is a rational number, rewriting 1/x as b/a is nothing more than perforrming simple algebra, and then reusing the given definition of rationals
• June 24th 2007, 01:52 PM
ThePerfectHacker
Quote:
Originally Posted by Ilaggoodly
not true, his argument is based off of the given assumption that x is a rational number, rewriting 1/x as b/a is nothing more than perforrming simple algebra, and then reusing the given definition of rationals
I think, Plato is asking how do you define 1/x to be.
The more elegant way to define a rational number is by using a "field of quotients of an integral domain"
In that case we can define Q=F(Z) where F(Z) represents the field of quotients of Z (integral domain). In that case it is valid to flip the fraction.
• June 24th 2007, 01:53 PM
Plato
Quote:
Originally Posted by Ilaggoodly
rewriting 1/x as b/a is nothing more than perforrming simple algebra, and then reusing the given definition of rationals
Have you ever done a rigorous construction of the real numbers? Let’s say in a mathematical logic or set theory course.
• June 24th 2007, 02:00 PM
Ilaggoodly
yes as a matter of fact i have taken a logic and set theory course, and im wel aware that you can make this problem MUCH more complicated than it is likely intended to be, i believe the way it has been shown by soroban and jhevon would satisfy the poster... in which case it was my mistake in saying that you were not correct, my apologies
• June 24th 2007, 02:06 PM
Plato
Quote:
Originally Posted by Ilaggoodly
i believe the way it has been shown by soroban and jhevon would satisfy the poster...
But the point is: we don't know that!
The student should have given more of what sort of proof is expected.
• June 24th 2007, 02:11 PM
Ilaggoodly
true enough, however considering some of his other posts...
S.O.S. Mathematics CyberBoard :: View topic - discrete math basic's
it would seem as if he/she is taking his or her first course in introduction to advanced mathematics...anyway, whatever
• June 24th 2007, 02:25 PM
Jhevon
Yes, Plato is right. It is up to the poster to tell us how rigorous a proof is needed. My proof is not all that rigorous, it is simple algebra which assumes the reader knows the axioms behind it. If a more rigorous proof is needed, my post should only serve as an outline of what to do
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http://mathoverflow.net/questions/43680/example-in-dimension-theory/43688
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## Example in dimension theory
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Could you give me an example of a complete metric space wiht covering dimension $> n$ all of which compact subsets have covering dimension $\le n$?
-
## 1 Answer
A guess
In $l^2$ Hilbert space, consider the set $E$ of points with all coordinates rational. Erdös (reference) showed that $E$ has topological dimension $1$. (In separable metric space, all notions of topological dimension coincide.)
Does this $E$ have the property that every compact subset is zero-dimensional? This space (and thus any subset of it) is totally disconnected, and isn't it the case that for compact (metric) spaces, this implies zero-dimensinal?
-
@Gerald: I can't quite figure out how compactness can be used to rule out the element on the boundary that Erdos constructed. Help please? – Willie Wong Oct 26 2010 at 17:44
Thank you, I add "completeness" but Erdös constructs also a complete set ($R_1$). It seems to work, but I need to think a bit. – ε-δ Oct 26 2010 at 17:48
1
@Willie, If your set $K$ is compact then there is projection $\pi:K\to\mathbb R^n$ to a coordinate n-plane such that preimage of a set of small diameter has small diameter --- from this you get that $\dim K=0$ – ε-δ Oct 26 2010 at 17:58
1
Yes, complete Erdös-space (all sequences with only irrational coordinates) is complete, separable, dim= 1, and totally disconnected, so this example will do for the complete case. – Henno Brandsma Nov 7 2010 at 14:49
Why does the Erdös space or its irrational variation admit a complete metrics? – Wlodzimierz Holsztynski Mar 1 at 4:29
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http://mathhelpforum.com/advanced-statistics/189837-can-i-improve-accuracy-method.html
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Thread:
1. Can I improve the accuracy of this method?
I hope it is OK to post here - I am a total non-expert with only basic maths ability, but with what I think is a tricky problem to solve. I hope someone here may be able to help.
I need to estimate the value of X (unknown) when I know only the value for Y. To do this I have a list of other occurrences of X and Y, and I have to determining any pattern in the correlation of X to Y. Although X and Y are not totally random, there is no definite or consistent correlation other than there is a tendency for the value of X to be larger if Y is larger. The best that can be hoped for is as accurate a guess as possible. Here is an example:
1. X = 380 Y = 320
2. X = 840 Y = 5500
3. X = 5176 Y = 499
4. X = 5315 Y = 1074
5. X = 6484 Y = 140
6. X = 7244 Y = 5002
7. X = 10950 Y = 1104
8. X = 11040 Y = 279
9. X = 16791 Y = 3001
10. X = 17096 Y = 2788
11. X = 20700 Y = 9194
12. X = 169901 Y = 20886
13. X = 178322 Y = 94437
14. X = 184739 Y = 1198
15. X = 525257 Y = 29129
X = ? if Y = 1198
X = ? if Y = 8062
X = ? if Y = 31419
As a non-expert, at the moment my best guess at the Xs involves the following process: I divide X by Y for all known values, in the example above this gives results ranging from 0.15 (No.2) to 154.21 (No. 14). I then find the median value across the sample set, in the example above this is 6.13 (represented by No. 10). I then apply this value to Y in the case of unknown X, so in this case I would estimate the 3 unknown Xs to be 7346 and 49420 and 192599.
I have two questions.
a) My simple method gets within +-50% in 60% of cases (the actual data sets are larger than the small example above, with anything from 50 to 500 known X and Ys). Is there a more advanced approach that would give more accurate estimates?
b) If I have to use a formula for estimating 100 unknown X's, and my goal is to get, say, 75% of my estimates to within, say, +-50%. As all values are positive, the most my result can be in error on the negative side is close to -100%, but my method can easily give errors on the positive side in excess of +100%, +200%, even 1,000%. So the scope for positive error is unlimited while the scope for negative error is limited to -100%. Can this fact be harnessed to improve the statistical accuracy and get more results within my target tolerance range?
For practical purposes any suggestions or solutions ideally need to be executable on Excel.
Thank you in advance for your patience with this problem and any help.
2. Re: Can I improve the accuracy of this method?
Originally Posted by davidwhite
I hope it is OK to post here - I am a total non-expert with only basic maths ability, but with what I think is a tricky problem to solve. I hope someone here may be able to help.
I need to estimate the value of X (unknown) when I know only the value for Y. To do this I have a list of other occurrences of X and Y, and I have to determining any pattern in the correlation of X to Y. Although X and Y are not totally random, there is no definite or consistent correlation other than there is a tendency for the value of X to be larger if Y is larger. The best that can be hoped for is as accurate a guess as possible. Here is an example:
1. X = 380 Y = 320
2. X = 840 Y = 5500
3. X = 5176 Y = 499
4. X = 5315 Y = 1074
5. X = 6484 Y = 140
6. X = 7244 Y = 5002
7. X = 10950 Y = 1104
8. X = 11040 Y = 279
9. X = 16791 Y = 3001
10. X = 17096 Y = 2788
11. X = 20700 Y = 9194
12. X = 169901 Y = 20886
13. X = 178322 Y = 94437
14. X = 184739 Y = 1198
15. X = 525257 Y = 29129
X = ? if Y = 1198
X = ? if Y = 8062
X = ? if Y = 31419
As a non-expert, at the moment my best guess at the Xs involves the following process: I divide X by Y for all known values, in the example above this gives results ranging from 0.15 (No.2) to 154.21 (No. 14). I then find the median value across the sample set, in the example above this is 6.13 (represented by No. 10). I then apply this value to Y in the case of unknown X, so in this case I would estimate the 3 unknown Xs to be 7346 and 49420 and 192599.
I have two questions.
a) My simple method gets within +-50% in 60% of cases (the actual data sets are larger than the small example above, with anything from 50 to 500 known X and Ys). Is there a more advanced approach that would give more accurate estimates?
b) If I have to use a formula for estimating 100 unknown X's, and my goal is to get, say, 75% of my estimates to within, say, +-50%. As all values are positive, the most my result can be in error on the negative side is close to -100%, but my method can easily give errors on the positive side in excess of +100%, +200%, even 1,000%. So the scope for positive error is unlimited while the scope for negative error is limited to -100%. Can this fact be harnessed to improve the statistical accuracy and get more results within my target tolerance range?
For practical purposes any suggestions or solutions ideally need to be executable on Excel.
Thank you in advance for your patience with this problem and any help.
We would need to know more about the process producing these numbers, and would need to see a lot more data to offer any meaningful opinion.
CB
3. Re: Can I improve the accuracy of this method?
The X and Y values are asset types, and I can have data sets where, for example, I have 500 records in a data set. For 200 records both X and Y values may be known, for the remaining 300 records only Y is known. I'm trying to establish the most accurate method of estimating X in these cases.
I know there is no consistent relationship between the two values, but there is a tendency / probability (I use these terms as a layperson) for both to increase in size. I notice that X does not always increase with Y, so I suspect part of the problem may involve quantifying the probability for X and Y to both increase, as well as calculating the most probable proportion of one value to the other. Or maybe the approach required is totally different?
I have attached a larger data set of several hundred records as requested.
Attached Files
• X and Y Relationship.xls (135.5 KB, 8 views)
4. Re: Can I improve the accuracy of this method?
Originally Posted by davidwhite
The X and Y values are asset types, and I can have data sets where, for example, I have 500 records in a data set. For 200 records both X and Y values may be known, for the remaining 300 records only Y is known. I'm trying to establish the most accurate method of estimating X in these cases.
I know there is no consistent relationship between the two values, but there is a tendency / probability (I use these terms as a layperson) for both to increase in size. I notice that X does not always increase with Y, so I suspect part of the problem may involve quantifying the probability for X and Y to both increase, as well as calculating the most probable proportion of one value to the other. Or maybe the approach required is totally different?
I have attached a larger data set of several hundred records as requested.
Look at a log-log plot of this data.
CB
5. Re: Can I improve the accuracy of this method?
I've produced a log-log of the example data set where X and Y values are both known. I have attached the log-log graph generated in Excel using base 10 (Excel default, it can be changed). It produces two Y axis, which is no doubt correct. I'm sure this is the approach and I can see that there is a clustering which appears to hint at potential order. But I have no idea how to use this data to produce an Excel formula which I can then apply to future known Y values to estimate unknown X values.
I understand there is a degree of 'figure it out yourself' appropriate for students of maths but I'm not - not even a CSE in maths 30 years ago! If this forum isn't the right place for non-mathematicians to get help can you point me in the right direction? Thanks.
Attached Thumbnails
6. Re: Can I improve the accuracy of this method?
Originally Posted by davidwhite
I've produced a log-log of the example data set where X and Y values are both known. I have attached the log-log graph generated in Excel using base 10 (Excel default, it can be changed). It produces two Y axis, which is no doubt correct. I'm sure this is the approach and I can see that there is a clustering which appears to hint at potential order. But I have no idea how to use this data to produce an Excel formula which I can then apply to future known Y values to estimate unknown X values.
I understand there is a degree of 'figure it out yourself' appropriate for students of maths but I'm not - not even a CSE in maths 30 years ago! If this forum isn't the right place for non-mathematicians to get help can you point me in the right direction? Thanks.
The problem with a non-educational problem that may be used commercially is the MHF cannot do anything that might result in us being liable in any way for losses financial or otherwise of the advice.
But a linear realationship in a log-log plot implies a relationship of the form:
$Y=k X^{\alpha}$
where $k$ and $\alpha$ are constants.
CB
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http://physics.stackexchange.com/questions/38352/am-i-making-the-right-assumption-about-a-jump-discontinuity-in-the-acceleration?answertab=active
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# Am I making the right assumption about a jump discontinuity in the acceleration?
A train heads from Station A to Station B, 4 km away. If the train begins at rest and ends at rest, and its maximum acceleration is 1.5 m/s^2 and maximum deceleration is -6 m/s^2, what's the least time required to complete the journey?
My attempt at a solution: I assumed the acceleration function, $a(t)$, has a jump discontinuity at some time, say $t_1$, where it jumps from 1.5 to -6. Under this assumption, it's easy to use the velocity boundary conditions to get the total time $t_f$ is $\frac{5}{4}t_1$, because the velocity $v(t)$ for $t>t_1$ is just $1.5t_1-6(t-t_1)=7.5t_1-6t$. Setting this equal to 0 at $t_f$ gives this result.
The position $x(t)$ for $t\leq t_1$ is just $\frac{1}{2}(1.5)t^2$ and for $t>t_1$ it's just $\frac{1}{2}(1.5)t_1^2+(1.5t_1)(t-t_1)-\frac{1}{2}(6)(t-t_1)^2$. Using the fact that $x(t_f)=4000$, and substituting $t_f=\frac{5}{4}t_1$, I find $t_f\approx 86.06$ seconds.
But my teacher's solution says the minimum time is $81.65$ seconds. Who's wrong? Is my assumption about the jump discontinuity of $a(t)$ what's wrong?
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Hi symplectomorphic, and welcome to Physics Stack Exchange! I changed the formatting of your question a bit to make it easier to read. – David Zaslavsky♦ Sep 26 '12 at 3:38
I think your change of title is misleading -- it's a much less interesting question. The question is a brachiostone-style time minimization problem: the point of putting "time minimization" in the title was to bring out this connection. – symplectomorphic Sep 26 '12 at 3:39
brachistochrone* – symplectomorphic Sep 26 '12 at 3:41
Ah, well, that didn't seem clear to me. Feel free to change the title (or more of the post) to reflect what you wanted it to ask. (It's highly recommended, but not required, to make the title a question that corresponds to the main thing you are asking.) – David Zaslavsky♦ Sep 26 '12 at 3:52
## 1 Answer
I agree with you that the distance travelled up to the changeover point is:
$$x = \frac{1}{2}a_1\left(\frac{4t_f}{5}\right)^2$$
but what I would do next is start from the other end and point out that:
$$4000 - x = \frac{1}{2}a_2\left(\frac{t_f}{5}\right)^2$$
adding these together and rearranging gives:
$$t_f = \sqrt{\frac{50 \times 4000}{16a_1 + a_2}} \approx 81.65$$
so your teacher is indeed correct (it had to happen one day :-). Your equation for the total distance:
$$4000 = \frac{1}{2}(1.5)t_1^2+(1.5t_1)(t-t_1)-\frac{1}{2}(6)(t-t_1)^2$$
looks fine to me, so I'd guess you just made a mistake in the algebra.
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1
Found my algebra mistake after reading through my work for about the fiftieth time. Thanks for confirming I wasn't losing my mind about the general approach. – symplectomorphic Sep 26 '12 at 6:42
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http://mathhelpforum.com/peer-math-review/145515-pdf-quadratic-equations.html
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# Thread:
1. ## PDF on quadratic equations
Hello,
I've been writing a paper going through quadratic equations (algebra and graphing), and I intent to ask a moderator to review and sticky it if appropriate in the Pre-Algebra section of MHF. I feel it might be useful to a whole bunch of people, probably infinitely many xD. But first, I am looking for peer review, so I provide the first version of my paper for you all to read. The paper is attached to the thread.
The PDF does not expect you to be a layman and does require some basic skills at algebra and some notions of graphs (nothing very difficult though). I will accept any constructive critics and tips on how to present ideas fluently, or format the paper so as to make it as clear and enjoyable to read as possible. There is (hopefully) going to be a reviewed version based on your opinion, maybe adding some more stuff into it and correcting some flaws that I might have missed.
Thank you all
Attached Files
• Quadratics.pdf (212.9 KB, 203 views)
2. Originally Posted by Bacterius
Hello,
I've been writing a paper going through quadratic equations (algebra and graphing), and I intent to ask a moderator to review and sticky it if appropriate in the Pre-Algebra section of MHF. I feel it might be useful to a whole bunch of people, probably infinitely many xD. But first, I am looking for peer review, so I provide the first version of my paper for you all to read. The paper is attached to the thread.
The PDF does not expect you to be a layman and does require some basic skills at algebra and some notions of graphs (nothing very difficult though). I will accept any constructive critics and tips on how to present ideas fluently, or format the paper so as to make it as clear and enjoyable to read as possible. There is (hopefully) going to be a reviewed version based on your opinion, maybe adding some more stuff into it and correcting some flaws that I might have missed.
Thank you all
In section 2.2 in the list of methods of solving quadratics you should have completing the square as this is often taugt befor the quadratic formula.
In the bit about factorization there is no mention of the rational roots theorem.
In the derivation of the quadratic formula you provide no words telling the reader what you are doing.
CB
3. Originally Posted by CaptainBlack
In section 2.2 in the list of methods of solving quadratics you should have completing the square as this is often taugt befor the quadratic formula.
In the bit about factorization there is no mention of the rational roots theorem.
In the derivation of the quadratic formula you provide no words telling the reader what you are doing.
CB
Thank you Captain Black for your comments,
I will take them into account for the rewriting of the paper (well not exactly rewriting, "update" would suit better )
4. At the bottom of page 4, where you have
$ax^2 + bx + c = \left(x - \frac{-b - \sqrt{b^2 - 4ac}}{2a}\right)\left(x + \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) = 0$
should actually read
$ax^2 + bx + c = a\left(x - \frac{-b - \sqrt{b^2 - 4ac}}{2a}\right)\left(x + \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) = 0$.
You don't factorise equations, you factorise expressions which happen to be on one side of the equation.
You should give more examples and exercises for everything you are trying to teach.
You haven't explained how to factorise non-monic quadratics - the method where you break the middle term and factorise by grouping is the best.
Except for one sentence in the introduction, you have not given any real world applications.
5. Originally Posted by Prove It
Except for one sentence in the introduction, you have not given any real world applications.
I wonder: are there really any honest and direct real-world applications of the quadratic formula? Many math textbooks give improbable and contrived "word problems" as their application, and in my opinion, it does more damage than good.
Maybe it is better to leave it out, if nothing sensible can be found.
6. Of course ! The only example I actually gave in my PDF (at this point, it's currently being updated) about projectile motion is fundamental.
But indeed, usually, math textbooks do give stupid problems that lead to thinking that the QF is useless. But it's not !
Ooh I need to get some communication skills asap xD
7. Apologies for the update being so sluggish, I really can only work on it on week-ends
Hopefully it will be finished by Sunday.
8. This looks very neat. Thanks for sharing. May i ask what type of utility you used to create the graphs?
On a more general note, anybody recommend a good resource/tutorial for authoring with LaTeX for beginners?
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http://mathoverflow.net/questions/5670/maximal-compact-subgroup-as-fixed-points-of-some-involution-on-p-adic-group
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## maximal compact subgroup as fixed points of some involution on p-adic group?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
As is well known, maximal compact subgroup of real Lie group is just the fixed points of Cartan involution.
Now the question is what's the possible p-adic analog?
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## 2 Answers
There are some complications in the p-adic case. For GL_n, every maximal compact subgroup is conjugate to GL_n of the ring of integers, but for more general groups (eg PGL_2, SU_3) there can be more than one conjugacy class of maximal compact subgroups. One thing that you can say is that every maximal compact subgroup is the stabaliser of a point on the Bruhat-Tits building.
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That doesn't really answer the question: is GL_n of Z_p the fixed points of some natural involution? – Ben Webster♦ Nov 16 2009 at 4:36
I realise it doesn't answer the question, but provides some information on maximal compact subgroups that might be useful. I'll see if I can go away and give any sort of a better answer. – Peter McNamara Nov 16 2009 at 4:42
Fair enough. It's wise in a situation like that to say something like "I realize this isn't quite your question, but..." just so people don't think you misread. – Ben Webster♦ Nov 16 2009 at 16:26
@Ben, I like your interpretation of the question, but I disagree with your implicit claim that the question was even well-defined as stated. – S. Carnahan♦ Nov 16 2009 at 18:40
I think that Peter's answer is an insightful one to the precise question asked: "what's the possible p-adic analog[ue]?" – Pete L. Clark Mar 9 2010 at 4:15
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Here is how the real and p-adic situations are the same.
Let $G$ be a connected reductive algebraic group defined over a field $F$ not of characteristic two. Let $\theta$ be an involution of $G$ defined over $F$. Then the group $G^\theta$ of fixed points is a reductive algebraic subgroup of $G$.
Here are two ways in which they are different.
In the real case, one can always choose $\theta$ so that the group of rational points of $G^\theta$ is compact. In the p-adic case, compact reductive groups are quite rare, and so in most cases there is no analogous way to choose $\theta$.
Second, compact subgroups do not play the same roles in the real and p-adic cases. Think of the fields themselves. In the p-adic case, the maximal compact subring is the ring of integers. In the real case, there are no nontrivial compact subrings. There is a ring of integers, but it is not compact. Moreover, since $G^\theta$ has smaller dimension than $G$, it cannot be an open subgroup, and maximal compact subgroups are always open in the p-adic case. Thus, even in the rare cases where $G^\theta$ is compact, it is not maximal.
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http://math.stackexchange.com/questions/128666/matrix-multiplication-proof
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Matrix multiplication proof
I have the following details about the matrices $A,B$:
$$A_{n\times n},B_{n\times n}$$ $$A^2 = B^2 = 0$$ $$AB=BA$$
I need to find $x \in \mathbb{N}$ so $(A+B)^x=0$
What implications can I make from the given details?
1. $A,B$ doesn't have to equal $0$ so $A^2 = B^2 = 0$
2. Can I imply from the 2nd and 3rd given details that $A=B$?
3. Any other leads?
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2
– bgins Apr 6 '12 at 11:42
Concering 2.: $B$ might be a scaled version of $A=\begin{pmatrix}0& 1\\0 &0\end{pmatrix}$, so no. – draks ... Apr 6 '12 at 11:49
1 Answer
For all $x \ge 3$, The equation $(A+B)^x=0$ is true because for $x =3$, it is true.
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http://math.stackexchange.com/questions/20820/algebraically-integral-expansions-of-differential-forms
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# Algebraically integral expansions of differential forms?
Let C be a smooth, genus g curve defined over a number field K'. Suppose we have a K'-valued point P on C. We can view C as a Riemann surface; then the space of holomorphic differential forms has dimension g over C.
It seems to me that we can choose a uniformizer x at P and a basis of differential forms $\{ \omega_i \}$ such that the expansion of each $\omega_i$ in terms of x has algebraically integral coefficients; that is, there is some finite extension K of K', with ring of integers A, such that $\omega_i = \sum_{n=0}^\infty a_{i,n} x^n dx$ with all $a_{i,n} \in A$.
I would really appreciate critiques, help, or references for my reasoning. We can start with dx, which is a meromorphic differential form on X. We just need to multiply dx by rational functions with zeros in the right places to get holomorphic differential forms, but we want to choose rational functions with integral expansions in x.
First, we make an initial choice of uniformizer x. If P=(C0:C1: ... :Cn) then we know each monomial CiXj-CjXi has a zero of some order at P; we can just choose some convenient quotient of powers of these monomials to get a zero of order one and call that expression x (for now).
The poles of dx will be at various points of C with coordinates in some finite extension of K' (as will the zeros of dx). It seems to me that we can build enough rational functions from monomials involving these coordinates to get what we need. (Similar to how we chose the uniformizer x.) We can then expand these rational functions in terms of x. We may get some denominators (say powers of N) in the expansions of our functions, but then we can replace x with x/N to absorb them.
What do you think? Many thanks for the assistance!
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The result you want is true. It seems to me that you may be making too much of it: by definition, a smooth projective genus $g$ curve over a number field $K$ has $g$ $K$-linearly independent global regular differential forms. These will look like $\sum_i f_i dg_i$. If you have any denominators in the coefficients of the rational functions appearing in your differential, multiplying by a suitable positive integer $N$ will get rid of them. Aren't we done? – Pete L. Clark Feb 7 '11 at 17:11
@Pete: I think it depends crucially on the chosen uniformizer at P. We can't just multiply the differential form by N: Consider 1/(1-x/N), which has unbounded denominators when expanded in x, but we could just use x/N instead of x. Also, the expansion may be integral for a uniformizer x, but not integral for y=ln(x+1), which is also a local uniformizer at P. It seems like there is some "natural" family of uniformizers that makes this work (probably rational functions), and I'm trying to nail it down precisely. – Hypocycloid Feb 7 '11 at 18:20
for one thing, you seem to be identifying a differential with its power series expansion, which is valid but certainly not necessary. (And I don't think it's helping you out here.) Second: $y = \ln x+1$ is not a rational function on any compact Riemann surface I know... – Pete L. Clark Feb 7 '11 at 18:24
@Pete: Thanks for the comments. You're right; it is clear to me now that rational functions are what I want to work with. By uniformizer at P, I mean any analytic function on a neighborhood of P with a simple zero at P. – Hypocycloid Feb 7 '11 at 19:21
@Pete: Here's my motivation, which I'll post as another question: I'm thinking of modular curves and the equivalence of weight 2 modular forms and differential forms. We study the q-expansions of cusp forms f, where q=e^{2pi iz/m} is a uniformizer at i\infty on the upper half-plane. Is q a rational function on the modular curve? Or at least an integral power series in a rational function uniformizer? If we expand f in (z-a) at the point z=a, then we are using something like the shape z=ln(q+1); which I want to avoid. – Hypocycloid Feb 7 '11 at 19:26
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http://math.stackexchange.com/questions/226664/coloring-graph-problem
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# Coloring Graph Problem
If G is a graph containing no loops or multiple edges, then the edge-chromatic number $X_e(C)$ of G is defined to be the least number of colours needed to colour the edges of G in such a way that no two adjacent edges have the same colour.
How to find edge-chromatic number :
1)Complete Graph($K_n$)
[I know $k_n$ is (n-1)regular and if n is even then edge-chromatic number of $k_n$ is n-1 and if n is odd then edge-chromatic number of $k_n$ is n,but how to prove it??]
2)Wheel($W_n$)
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3
– joriki Nov 1 '12 at 9:22
1
Also, it seems unlikely that you don't have at least some rudimentary thoughts of your own on at least part 4) of the problem -- answers will be much more useful if you point out specifically what you already understand and what's causing you trouble. – joriki Nov 1 '12 at 9:26
## 1 Answer
For a complete graph with an even number of vertices, this amounts to finding a 1-factorisation, such as the following:
A general construction is formed by rotating a "starter", such as the following:
Each rotation describes where to place the edges of a single colour. Picture source:
Mendelsohn and Rosa, One-factorizations of the complete graph—A survey, Journal of Graph Theory 9 (1985) 43–65. (link)
See also the above reference for further details on how to construct 1-factorisations of $K_{2n}$ and near 1-factorisations of $K_{2n-1}$. Hence
Observation: The edge-chromatic number of $K_{2n}$ is $2n-1$ (its maximum degree) for $n \geq 1$.
For $K_{2n-1}$, we simply start with a 1-factorisation of $K_{2n}$ and delete a vertex, which results in $2n-1$ distinct edge colours. This is the best possible, since each edge colour can occur at most $n-1$ times (without having adjacent monochromatic edges), so we need at least $$\frac{{2n-1} \choose 2}{n-1}=2n-1$$ colours.
Observation: The edge-chromatic number of $K_{2n-1}$ is $2n-1$ for $n \geq 2$.
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http://www.physicsforums.com/showthread.php?t=409044
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Physics Forums
## solving inequalities with three variables
Can you help me solve this inequality for x?
[tex]
\frac{1+(\gamma+x(r-\alpha)-1)t}{1+\frac{\gamma+x(r-\alpha)-1}{2}}>0
[/tex]
where $$\gamma>1, 0<t<1, 0<r<3\alpha, \alpha>0$$
I really don't know where to start ...
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Mentor I would multiply both sides of the inequality by what's in the denominator, and then make two cases: one in which the denominator is assumed to be positive, and the other in which the denominator is assumed to be negative. The reason for the two cases is that if you multiply both sides of an inequality by a variable quantity, the direction of the inequality symbol changes if what you multiplied by is negative. If the quantity you multiply by is positive, the inequality symbol doesn't change direction. For example, 2 < 3, and 2(2) < 2(3), but -1(2) > -1(3).
Well, my way of thinking of solving such inequalities is similiar to the one presented by Mark44, I just get to it in a different way. Since it's fairly obvious, that $$\frac{x}{y} >0 \iff xy>0$$, you can simply write: $$(1+(\gamma+x(r-\alpha)-1)t)(1+\frac{\gamma+x(r-\alpha)-1}{2})>0$$ insted of the fraction. And the product of 2 numbers is >0 only if both of them are of the same sign. Just choose whatever interpretation you prefer.
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| | Introductory Physics Homework | 2 |
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http://mathoverflow.net/revisions/663/list
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## Return to Answer
2 deleted 6 characters in body; added 2 characters in body
A while ago I worked on the question of what we can say if d^n=0, $d^n=0$, but I got distracted by more concrete problems. A few people have certainly thought about this question. One place to start looking is "$d\sp N=0$: $d^N=0$: generalized homology" or "Generalized homologies for $d\sp N=0$ d^N=0$and graded$q\$-differential algebras" both by Michel Dubois-Violette.
(Sorry for the lack of links; I'm off-campus so I can't actually get to the MathSciNet entries right now.)
1
A while ago I worked on the question of what we can say if d^n=0, but I got distracted by more concrete problems. A few people have certainly thought about this question. One place to start looking is "$d\sp N=0$: generalized homology" or "Generalized homologies for $d\sp N=0$ and graded $q$-differential algebras" both by Michel Dubois-Violette.
(Sorry for the lack of links; I'm off-campus so I can't actually get to the MathSciNet entries right now.)
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http://physics.stackexchange.com/questions/52637/what-is-the-curvature-of-the-universe/52639
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# What is the curvature of the universe?
What is currently the most plausible model of the universe regarding curvature, positive, negative or flat?
(I'm sorry if the answer is already out there, but I just can't seem to find it...)
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## 2 Answers
The primary authority on this is the WMAP project (though there are other observational cosmology projects springing up all the time). They analyze the CMB observations every two years, and the nine-year data was just published. The paper can be found here.
Basically, everyone assumes flatness because we know it is a very good approximation, because it simplifies the equations, and also because aesthetically being exactly flat is more appealing than being just close to flat. Thus most of the analysis is done assuming the curvature parameter $\Omega_k$ vanishes. Toward the middle of the paper, in Table 9, they show the results of fits that allow curvature to vary. They report $\Omega_k = -0.037^{+0.044}_{-0.042}$. When combined with results from other projects and surveys, they find $\Omega_k = -0.0027^{+0.0039}_{-0.0038}$.
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According to NASA the universe is flat to within 0.4%.
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http://math.stackexchange.com/questions/231966/proof-that-a-hermitian-matrix-has-orthogonal-eigenvectors-real-eigenvalues/232002
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# Proof that a Hermitian Matrix has orthogonal eigenvectors, real eigenvalues
Here's my feeble attempt:
A = A* is the condition for a Hermitian matrix. So I try expanding both sides in terms of their SVD:
A = USV* and A* = (USV*)* = ... = VSU*
So equating I get:
VSU* = USV*
Now maybe through this I can conclude something about how V and U relate to each other but I don't know how...
Thanks for any help.
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It will help to conclude that $U=V$. What happens if you multiply $VSU^*$ by $USV^*$? Note that order will matter, so consider both. – Alex Nov 7 '12 at 6:47
## 2 Answers
A more efficient route is to consider the Schur decomposition, $A=QTQ^*$, where $T$ is upper triangular and $Q$ is unitary. Note that the diagonal of $T$ consists of the eigenvalues of $A$. The equality $A^*=A$ is $QTQ^*=QT^*Q^*$, from where we deduce $T=T^*$. As $T$ is upper triangular and $T^*$ is lower triangular, we get that $T$ is diagonal. Moreover, as the diagonal of $T^*$ is the conjugate of that of $T$ and they are equal, we have that the eigenvalues of $A$ are real.
The eigenspaces of $T$ are orthogonal to each other (easy to see since $T$ is diagonal). But $Q$ takes the eigenspaces of $T$ to the eigenspaces of $A$, so these have to be pairwise orthogonal too.
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The process is fairly straightforward given the fact that any square matrix has at least one eigenvalue and eigenvector.
Suppose $Ax = \lambda x$. Then $\langle x , Ax \rangle = \langle Ax , x \rangle = \overline{\langle x , Ax \rangle} = \lambda \|x\|^2$, hence $\lambda \in \mathbb{R}$.
The point about being Hermitian is that if $x$ is an eigenvector of $A$, then both $\text{sp} \{x\}$ and the subspace $\{x\}^\bot$ are invariant under $A$. To see the latter, suppose $v \bot x$, then $\langle x , Av \rangle = \langle Ax , v \rangle = \lambda \langle x , v \rangle = 0$, hence $A v \bot x$. Now let $v_k$ be a basis for $\{x\}^\bot$, then in the basis $x, v_1,...,v_k$, $A$ must have a block diagonal form: $$A \sim \begin{bmatrix} \lambda & \\ & \tilde{A}\end{bmatrix}$$ (By $\sim$ I mean that the two matrices are similar.) Now find an eigenvalue of $\tilde{A}$ (which must also be Hermitian) and repeat the process until $A$ is diagonalized.
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Nice, you made that seem easier than previous proofs I've read. – littleO Nov 7 '12 at 8:11
Thanks! Well, I skipped some details like writing out the similarity transformation that reduces $A$ to block diagonal form. However, I personally find the details more straightforward when I have the 'big picture'. – copper.hat Nov 7 '12 at 8:19
Yeah, you made a nice choice of which details to skip, in order to show how simple the proof really is. – littleO Nov 7 '12 at 8:31
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http://math.stackexchange.com/questions/210584/create-a-periodic-function-with-given-minima
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Create a periodic function with given minima
For a simulation, I need to create a function $r(\theta,\phi)$ that has minima at specific pairs of $(\theta,\phi)$, and that is periodic, i.e. $r(\theta,\phi) = r(\theta+2\pi,\phi)$ and $r(\theta,\phi) = r(\theta,\phi+2\pi)$ etc. At least the first and second derivatives of the function should be continuous.
How can I construct such a function? I imagine that when drawn as a surface, it will look a bit like a bowling ball, but without the sharp edges.
EDIT I can imagine several ways of creating a function with a single minimum at $(\theta_0,\phi_0)$, but I am stuck creating a function that has a second (or third) minumum at e.g. $(\theta_1,\phi_1)$, but not at $(\theta_0,\phi_1)$ or $(\theta_1,\phi_0)$
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2 Answers
$$r(\theta,\phi) = (1-\cos(\theta-\theta_0))\cdot(1-\cos(\phi-\phi_0))$$
looks like a good candidate :)
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Thank you for your suggestion! This is what I originally started with. How can I adapt it to multiple minima? – Jonas Oct 10 '12 at 18:09
The function $r(\theta,\phi)=\sin(\theta)+\cos(\phi)$ has the periodicity wanted, and is minimized at point $(\frac{3\pi}{2}+2k\pi,\pi+2k\pi), k\in \mathbb{Z}$.
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I apologize for not being clear. I can manage single minima, but how would I do multiples? – Jonas Oct 10 '12 at 18:09
What do you mean multiples minima, this one as an infinity of them – Jean-Sébastien Oct 10 '12 at 18:10
1
I believe he would like to have a function that has a given number of minima in $(0, 2\pi)$ – peterm Oct 10 '12 at 18:15
@Jean-Sébastien: Yes, user44010 is correct. – Jonas Oct 10 '12 at 19:23
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http://mathoverflow.net/revisions/92744/list
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## Return to Answer
2 added 40 characters in body
In general you can not expect that $f$ will be flat on $X-U$, but you have a locally finite stratification of $X$ for which $U$ is the dense stratum, and such that $f$ is flat over each strata. This notion is called "platification" (in French) and it was dealed with great details in the paper of Gruson and Raynaud "Techniques de platification d'un module".
1
In general you can not expect that $f$ will be flat on $X-U$, but you have a locally finite stratification of $X$ such that $f$ is flat over each strata. This notion is called "platification" (in French) and it was dealed with great details in the paper of Gruson and Raynaud "Techniques de platification d'un module".
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http://mathhelpforum.com/advanced-algebra/206517-ideals-euclidean-domains.html
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# Thread:
1. ## Ideals, Euclidean Domains
$J$ is an Euclidean Domain with $v:J-[0]\rightarrow\mathbb{Z}\cup[0]$. Let $1$ be the multiplicative identity.Let
$I=[a\in{J}|v(a)>v(1)]$
Is $I$ an ideal?
The only thing confusing me about this definition is whether the elements $0$ and $1$ are in $I$, as if they aren't then $I$ isn't a subring, so it's not an ideal.
$0$ isn't in the domain of $v$, so I'm not sure if it's in $I$ and $v(1)=v(1)$, so $1$ should not be in $I$.
2. ## Re: Ideals, Euclidean Domains
I think you've a little confusion about ideals.
You're correct that $1 \notin I$, so that I isn't a ring. But not containing 1 doesn't mean that I isn't an ideal. The "usual" situation is for an ideal not to contain 1. The only ideal in ring that contains 1 is the entire ring itself.
----------------------------
$\text{Proof: Let }I \text{ be an ideal in a ring } R \text{ such that }1 \in I.$
$\text{Then for all }r \in R, \text{ since } 1 \in I, \text{ have that } r1 = r \in I.$
$\text{Thus if }r \in R, \text{ then } r \in I, \text{ and so } R \subset I.$
$\text{But also }I \subset R, \text{ and therefore } I = R.$
----------------------------
Being an ideal means being a subgroup of the addition group of the ring, and being closed under multiplication by *any* element of the ring.
Think of the ideal E of even numbers in the integers. E = (2) in (Z, +, *, 0, 1). 0 is even, and addition and subtraction of even numbers is again an even number, so E is a subgroup of the additive group of the integers. Also, an even times *any* integer (even or odd) is again even. Thus E is an ideal. But it's *not* a ring, since it doesn't contain 1.
Note that there's nothing special about 2 for that. The ideal I = (n) in (Z, +, *, 0, 1) behaves the exact same way: sums and differences of integers that are both multiples of n are again multiples of n, 0 is a multiple of n (0 = n0), so (n) is a subgroup of (Z, +, 0). If m is *any* integer, then m times a multiple of n is again a multiple of n, so m times anything in (n) is also in (n). Thus (n) is an ideal.
So ideals differ from rings in that they don't necessarily contain a 1, and that the defined multiplication in the ideal is BIGGER than the ideal itself. We define multiplication in the ideal of even integers as mutliplication of an even integer times ANY integer, even or not.
----------------------------
As to your problem, I always get confused about things like whether 0 is in I. in fact, I'd have to look it up, and write it down carefully since, as you note, 0 is not in the domain of v. Thus whether 0 is in I, or not, is a matter of the truth of the vacuous proposition and the technical definition of that set. This is something that I won't say one way or another off the top of my head (my "guess" is that 0 is in I, but don't quote me.) You're right that 0 be in I *is* necessary for I to be an ideal (since as a subgroup of the additive group of J, I must contain the identity of the additive group, 0.)
----------------------------
However, you can see that that's not an ideal w/o answering whether 0 in I or not. Counter examples are easy.
Consider J = the usual Euclidean Domain of the integers. J = (Z, +, *, 0, 1), where v(x) = |x|. Then I is the set of integers whose absolute value is greater than the absolute value of 1 (i.e. I is the set of integers whose absolute value is greater than or equal to 2.) Is 5 in I? Is -4 in I? Is (5) + (-4) in I? Is I a subgroup of the additive group of the ring (i.e. of (Z, +, 0))? Is I an ideal of the Euclidean domain that's the ring of integers?
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http://physics.stackexchange.com/questions/tagged/statistics?sort=active&pagesize=30
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# Tagged Questions
The statistics tag has no wiki summary.
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### Randomly sampling a “well-mixed” solution of Brownian particles
I place $N$ Brownian particles in $V$ liters of solution, shake until I assume that the particles are "well-mixed", and sample and randomly sample an $S$ liter volume. What is the probability ...
2answers
121 views
### How to decide convergence of the ensemble average in a Monte Carlo process
I am simulating electromagnetic scattering off a rough surface. The usual process is to do a Monte Carlo simulation, which is briefly described as follows. Generate a randomly rough surface, and ...
2answers
159 views
### Average Neighbouring Impurity Separation in a Random 1D chain [closed]
I have a finite and discrete 1D chain (edit: linear chain, i.e. a straight line) of atoms, with unit separation, with a set number of impurities randomly distributed in the place of these atoms in ...
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### Computing an average escape distance for a particle
Somewhere in a two dimensional convex bulk of particles (pic related) on a random position a reaction takes place and a particle is sent out in a random direction with a constant velocity $v$. What ...
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286 views
### Use of Monte-Carlo simulation in High-energy Physics
I've been doing some research into the analysis used in particle physics when determining the significance of a finding (e.g. the recent Higgs candidate was announced as a boson in the 125-126 ...
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### What is the minimal set of expectation values I need in a statistical model?
At least if $\vec v$ is really only a one dimensional parameter, measuring all the moments $\langle v^n \rangle_f$ seems to give me all the information to compute $\langle A \rangle_f$ with $A(v)$ ...
2answers
513 views
### Combining multiple theories with 5 $\sigma$ confidence level
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### polynomial fit and uncertainty [closed]
I am trying to fit a polynomial function by least square method and I want to know the uncertainty on my fitting parameters. In which book can I find a clear answer?
3answers
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### why the difference between $\langle \hat p^2 \rangle_{\psi}$ and $\langle \hat p \rangle_{\psi}^2$ is NOT zero?
Well, the difference between the two expressions $\langle \hat p^2 \rangle_{\psi}$ and $\langle \hat p \rangle_{\psi}^2$ is exactly $\Delta p^2$ , i.e. the squared uncertainty (variance) of the ...
1answer
108 views
### Lacking of scale and distribution moments
Given a physical random variable x, $E(x)$ and $E((x-<x>)^2)$ defines mean and variance. From a statistical point of view variance represents the statistic error (isn't it?). If variance is not ...
0answers
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### Fitting of data to a model
Imagina that I have some observable value predicted with a theory for some process to be: 1+a x + b x^2 and observed value is 1.3 with an error 0.2; a and b are some numerical constants. I also have ...
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http://www.physicsforums.com/showthread.php?t=67527
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Physics Forums
## text book recommendation
I need a recommendation on a good second level college physics textbook. Pretty much any price range is fine. Also, i would like a recommendation for a good first semester Calculus book. Thanks to all who help out!!
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions: Gold Member Science Advisor Second level, what text was used for the first and we can make suggestions.
This is the book from my first level class: Physics: Algebra/Trig by Eugene Hecht ISBN: 0534261000 for the calculus, i just need a good introductory level book. Thanks!
Recognitions:
Gold Member
Science Advisor
## text book recommendation
Halliday, Resnik and Krane or Serway or Sears and Zemansky are all fairly decent calculus based books to work from.
Calculus i use Thomas' 9th Edition. Its ok, also Hughes-hallet. Those are the books I have used, they seem pretty decent. For Physics, I use Knight. It is a calculus based approach, but the calculus is just basic derivatives and integrals. Aside from mastering physics, it is a pretty good book. There are at least 4 volumes though, it is a really long book. I think volumes 3-4 are the second level ones though.
For calculus I recommend an older edition of one of Stewarts books along with a solutions manual for the book. Also stewart has a few books, I'd get one that covers single and multivariable calculus, that way you can learn more. I recommend looking on ebay or amazon. I have read and worked through all 1168 pages of Stewart's Calculus Early Transcendentals 5th Edition including the appendix and I think it does a very good job. It really helps to have the solutions manual if you are doing self study also. When I took multivariable Calculus I had a horrible teacher and I was essentially learning on my own. Stewart's book helped me through the course with it's excellent pictures and explanations. The only other book I have read on Calculus is Michael Spivak's Calculus. I don't think that it is very good for self study if don't know anything about Calculus and want to learn it quickly. The examples are really interesting but the material is too difficult I think if you know nothing about calculus. It would take a very long time to read it front to back without knowing any calculus I think. I also own a copy of Calculus by Ostebee and Zorn, I got it for a few dollars off ebay. I have only read the section on Limits and it seems extremely easy to read. However, I already know Calculus so I can't comment on how good it is for learning from scratch. Maybe a good idea is to get stewarts book with a solutions manual along with another book from ebay. Books are very cheap there. If not try amazon. It's always a good idea to have more than one reference when you get stuck. Also there is the internet too to help as a reference when you get stuck.
Thanks Eratosthenes, i looked into both of those books (Stewart's Calculus Early Transcendentals 5th Edition, & Michael Spivak's Calculus). I'm kind of leaning towards Spivak's book after reading reviews on amazon.com. I plan on using it for self study, so i'm a little hesitant since you said that may not be a good idea with this book. Does anyone else have any suggestions or comments about either of these books? or any others? Thanks!!
Ohanian is also a pretty good calculus based Physics book. University Physics from Hugh Young is exhaustive covering in depth almost all topics. I think this was mentioned earlier (Sears and Zemanswki).
Quote by gr3gg0r Thanks Eratosthenes, i looked into both of those books (Stewart's Calculus Early Transcendentals 5th Edition, & Michael Spivak's Calculus). I'm kind of leaning towards Spivak's book after reading reviews on amazon.com. I plan on using it for self study, so i'm a little hesitant since you said that may not be a good idea with this book. Does anyone else have any suggestions or comments about either of these books? or any others? Thanks!!
Np.
Well Michael Spivak's book is worth buying regardless. I bought it off amazon along with the answer book, it's called "Answer book for calculus" and cost $35. The total cost for both books was$105 plus shipping but it was well worth it. If you are serious about mathematics it is a good investment.
The problem with Spivaks book is that it takes alot of time and effort and you might get discouraged.
If you know nothing about Calculus I think you will learn more concepts reading stewarts book than spivak's simply because you will be able to work through stewarts quicker. I think Stewarts is the most widely used book in the US for Calculus 1, 2, and 3 at universities. I think that's a good indication that it at least does a decent job of teaching the basic concepts and ideas.
Maybe the solution is to get both books. After reading spivaks book for an hour reading stewart will seem like a joke:)
BTW I think Stewarts 5th Edition is more expensive than the 4th edition. You might be able to get the 4th edition for alot less along with a solutions manual. The only reason I have the 5th edition is because I needed it for three courses I took. If I was purchasing something for self study I would get the 4th edition or older as long as I could get a solutions manual with whatever I got.
As for physics books I used Serway and Jewett's Physics for scientist and engineers for 2 semesters of physics. I like you was learning Calculus at the same time I was learning Physics, and the physics book was much harder to read than stewarts book. I had a really hard time with physics but I think it's because I was learning Calculus at the same time. I actually read part of chapter 7 of Serway and Jewett's tonight because I was helping a friend and it is much more readable to me now than it was when I first used it.
For the calculs books, it had also crossed my mind to pick them both up. Perhaps that is what i will do, i'll have to see how much money i have in the next few days. (i would probably just get the 5th edition because i like new stuff :) ) Of course i'm still open to suggestions. For the physics book, that would also be self study. I'll look into University Physics, and the others that were mentionsed as well. I'm not actually enrolled in school right now because in the next month or so i should be leaving my area for about 2 years, but i took physics B ap in high school, and i really enjoy that class, and i've always enjoyed math. I figured i would get a head start so when i get back i'll be able to jump right back on where i left off.
You can also get spivak's book directly from the publisher, they give a 10% discount and shipping is FREE. I think the book w/ answer book comes to \$94.50... or, you can always get one used for much less. Warning... his book assumes you know some of the logic behind proofs already (ie: contrapositive etc.. ) It is highly recommended though... some other good ones for calc are apostol and courant.
thanks mathstudent, i'll look into the publisher and see if i can find it there
Quote by ramollari Ohanian is also a pretty good calculus based Physics book. University Physics from Hugh Young is exhaustive covering in depth almost all topics. I think this was mentioned earlier (Sears and Zemanswki).
^^
Probably the best book, I'd recommend that also..
There's also a companion book that goes with it..
Quote by gr3gg0r I need a recommendation on a good second level college physics textbook. Pretty much any price range is fine. Also, i would like a recommendation for a good first semester Calculus book. Thanks to all who help out!!
If you're cheap, like me, go out and buy Schaum's Outline of Physics for Scientists and Engineers and Schaum's Outline of Engineering Mechanics.
Forgot to mention, the publisher's website for the spivak book is www.mathpop.com
Recognitions: Homework Help Science Advisor i recommend the 2nd edition of stewart. it went downhill after that. spivak is a thousand times better. on a different level entirely. stewart is a joke compared to spivak. i.e. stewart (2nd edition) is a good non honors book. spivak is an excellent super honors book,(not just regular honors).
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http://mathematica.stackexchange.com/questions/5933/solving-a-linear-equation-in-mathematica?answertab=oldest
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# Solving a linear equation in Mathematica
This should be easy but I can't seem to find the right way to do it.
I have an equation of the form $a x + b x + c y + a z + d z = 0$, and I'd like to solve for relations between the parameters $a,b,c,d$, etc. such that the equation holds for all values of $x,y,z$, etc.
This is a very simplified example of the kind of equation I'm dealing with; there are actually 308 variables $x,y,\ldots$ and 42 parameters $a,b,\ldots$, making this nontrivial. As in the short sample equation, different variables could have the same parameter and could appear multiple times with different parameters.
So far my best attempt is:
````Solve[ForAll[{x, y, ...}, a*x + b*y + ... == 0], {a, b, ...}]
````
This yields the correct solution on a smaller equation (about 20 variables and 9 parameters) but is far too slow for this one. Is there a more specific technique I could use that's more optimized for this kind of problem specifically? Leaving out `ForAll` yields solutions in terms of the x-variables, which is kind of useless.
EDIT: nikie mentioned a much better way to do it (in the comments)! Thanks!
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2
Isn't this equivalent to solving the equation system `a+b==0 && c==0 && a+d==0`, without `x,y,z` and `ForAll`? – nikie May 22 '12 at 20:26
Huh. Yes, I think so, now that you point it out. Thanks! I'll see if I can rework the larger equation into that form, shouldn't be too hard. – Abilinglortly May 22 '12 at 20:40
## 2 Answers
````Coefficient[a x + b x + c y + a z + d z, #] == 0 & /@ {x, y, z} // Reduce[#, {a, b, c, d}] &
b == -a && c == 0 && d == -a
````
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You can use `SolveAlways` for the type of problems you describe. I haven't looked if it's fast enough for a very large system though.
````In[1]:= SolveAlways[a x + b x + c y + a z + d z == 0, {x, y, z}]
Out[1]= {{a -> -d, b -> d, c -> 0}}
````
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lang-mma
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http://mathhelpforum.com/calculus/161718-uniform-convergence.html
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# Thread:
1. ## Uniform convergence
I am asked to show that $f_n (x) = x^n$ is uniformly convergent in $(0, 1)$ but not in $[0, 1]$.
I proceed by computing the pointwise convergence to $f(x) = 0$ for $x \in [0, 1)$ and $f(x) = 1$ for $x = 1$.
Then I compute for $x \in [0, 1)$:
$\vert f_n(x) - f(x) \vert < \epsilon \Leftrightarrow \vert x^n \vert < \epsilon \Leftrightarrow x < (\epsilon)^{1/n}$
This is not fullfilled for all $x \in (0, 1)$ (since I can always find an $x : (\epsilon)^{1/n} < x < 1$) and therefore I have shown that $f_n (x) = x^n$ is not uniformly convergent in $(0, 1)$.
I am confused since I was asked to show that it is unformly convergent in $(0, 1)$. What is wrong?
2. Originally Posted by tholan
I am asked to show that $f_n (x) = x^n$ is uniformly convergent in $(0, 1)$ but not in $[0, 1]$.
I proceed by computing the pointwise convergence to $f(x) = 0$ for $x \in [0, 1)$ and $f(x) = 1$ for $x = 1$.
Then I compute for $x \in [0, 1)$:
$\vert f_n(x) - f(x) \vert < \epsilon \Leftrightarrow \vert x^n \vert < \epsilon \Leftrightarrow x < (\epsilon)^{1/n}$
This is not fullfilled for all $x \in (0, 1)$ (since I can always find an $x : (\epsilon)^{1/n} < x < 1$) and therefore I have shown that $f_n (x) = x^n$ is not uniformly convergent in $(0, 1)$.
I am confused since I was asked to show that it is unformly convergent in $(0, 1)$. What is wrong?
The question is wrong. The sequence of functions $f_n (x) = x^n$ is not uniformly convergent in $(0, 1)$, as the above argument correctly shows.
3. Thank you.
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http://mathoverflow.net/questions/121028/rank-of-a-matrix-with-missing-entries
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## Rank of a matrix with missing entries
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$ be a $2^n \times 2^n$ matrix over real number field, where the rows and columns are indexed by subsets of $[n] := {1,2,\ldots,n}$, and defined as follows, $M_{A, B} = 1$ if $A \subseteq B$; $M_{A, B} = -1$ if $B \subsetneq A$; $M_{A, B}$ can take arbitrary value over $\mathbb{R}$.
In words, $M$ is a matrix with some missing entries. Is there some result lower bounds the rank of matrix $M$, for example, $rk(M) \ge n^{\Omega(\log n)}$.
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What does "undefined" mean? Do you need the lower bound for all matrices with some prescribed values? – Ilya Bogdanov Feb 7 at 15:16
"undefined" = could take any value. Lower bound means the lower bound for all such matrices – jsliyuan Feb 8 at 0:43
Do you have constructions that allow reducing the rank so much?? – S. Sra Feb 8 at 3:57
(it seems to be easy to reduce the rank $2^{n/2}$, but one needs something more clever to reduce it even further...) – S. Sra Feb 8 at 4:00
I think I have constructions achieving this bound. But I don't know how to prove the lower bound. – jsliyuan Feb 8 at 7:26
## 1 Answer
Amazingly (to me) the rank can be as low as $n$. Simply define $M_{A,B}=1$ when $|A| \le |B|$ and $M_{A,B}=-1$ when $|A| \gt |B|.$ This is consistent with the previous requirements and makes the $A$ row depend only on $|A|.$
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http://stats.stackexchange.com/tags/suppressor/hot
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# Tag Info
## Hot answers tagged suppressor
13
### How can adding a 2nd IV make the 1st IV significant?
Although collinearity (of predictor variables) is a possible explanation, I would like to suggest it is not an illuminating explanation because we know collinearity is related to "common information" among the predictors, so there is nothing mysterious or counter-intuitive about the side effect of introducing a second correlated predictor into the model. ...
6
### How can adding a 2nd IV make the 1st IV significant?
I think this issue has been discussed before on this site fairly thoroughly, if you just knew where to look. So I will probably add a comment later with some links to other questions, or may edit this to provide a fuller explanation if I can't find any. There are two basic possibilities: First, the other IV may absorb some of the residual variability ...
3
### Differences in coefficients
"How much" matters a great deal! The adjustment is unlikely to be zero, after all; this would only happen if z were totally uncorrelated with x or y. By common convention one would test the statistical significance of the relationship between z and y as a way of deciding whether it is necessary to use z to adjust x's coefficient. That said, significance ...
2
### How to interpret a predictor with a positive structure coefficient and a negative standardised coefficient in discriminant function analysis?
The main difference between the coefficients and the correlations (elements of structure matrix) is not that these are less stable than those. Coefficient shows partial (i.e. unique) contribution of the variable to the discriminant function score, it is like regression coefficient. Correlation shows omnibus (i.e. unique + shared with other variables) ...
1
### Interpreting Omitted Variables
I think this hinges on what you means by "not changes much". The estimated parameters could change and so could their standard errors. That's two separate effects. Let's focus on the variance part for now. Suppose the true DGP is really $y=x\beta+z\alpha+v$, but you leave out the relevant variable $z$ from your model, so you actually estimate ...
1
### How to interpret a predictor with a positive structure coefficient and a negative standardised coefficient in discriminant function analysis?
I agree with @ttnphns. There are good reasons to interpret both standardised and structure coefficient tables. In some senses they contribute two complementary perspectives on what it means to say that a variable is an important predictor: (a) prediction after controlling for other predictors (i.e., standardised coefficients); (b) prediction without ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://nrich.maths.org/6493
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### Real(ly) Numbers
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?
### How Many Solutions?
Find all the solutions to the this equation.
### After Thought
Which is larger cos(sin x) or sin(cos x) ? Does this depend on x ?
# Curve Match
##### Stage: 5 Challenge Level:
The following five functions have been plotted accurately on the chart below
$$y=\sqrt{x}\quad\quad y=x^2\quad\quad y=x\quad\quad y=(x^2+x)/2\quad\quad y=2\sqrt{x}-x$$
Which is the $x$-axis and which is the $y$-axis? Which curves correspond to which functions?
Imagine that you wish to plot a route between $(0, 0)$ and $(1, 1)$. Can you find a function which does this without intersecting any of the existing curves, except at the endpoints?
What other curves can you plot which only intersect the existing curves at the end points?
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://scicomp.stackexchange.com/questions/tagged/iterative-method
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# Tagged Questions
A method which produces a sequence of numerical approximations which converges (provided technical conditions are satisfied) to the solution of a problem, generally through repeated applications of some procedure. Examples include Newton's method for root finding, and Jacobi iteration for ...
1answer
40 views
### Using fixed point iteration to decouple a system of pde's
Suppose I had a boundary value problem: $$\frac{d^2u}{dx^2} + \frac{dv}{dx}=f \text{ in } \Omega$$ $$\frac{du}{dx} +\frac{d^2v}{dx^2} =g \text{ in } \Omega$$ $$u=h \text{ in } \partial\Omega$$ My ...
1answer
57 views
### Is it possible to ensure global convergence of a fixed point iteration?
Suppose I have a fixed point iteration of the form $$x_{n+1}=f(x_n).$$ Suppose further that after some initial testing, I found that it does not converge to an a priori known fixed point $x^*$. I ...
1answer
63 views
### Convergence of GMRES
From what I understand the GMRES method is (using Arnoldi Iterations/Modified Gram-Schmidt): The first vector of the Krylov subspace span of A is the normalized vector \$\frac{\vec b - A\vec x_0} {|| ...
1answer
91 views
### Jacobi Iteration diverges?
I'm working through a problem in a textbook as follows: "Consider the $d \times d$ Toeplitz matrix A = \left[ \begin{array}{ccccc} 2 & 1 & 0 & \cdots & 0 \\ -1 &2 ...
0answers
52 views
### What are Implications of Commutative Diagrams?
This question may be too broad. But I really want to know some concrete explanations. I often find various commutations appear here and there, which concerns the application order of two operators. ...
2answers
151 views
### Looking for an algorithm that allocates climbing hold colors to wall sectors
I posted this question earlier on stackoverflow, where it was closed as off-topic. I hope it survives here. I our climbing gym, the routes need to be re-set from time to time. The following rules ...
2answers
54 views
### Levenberg optimizer halts quickly when given more variables, or fewer constraints
I'm using the g2o C++ optimization library to refine a GPS trajectory using accelerometer data. The program uses a Levenberg-Marquardt optimizer over data points representing the position and ...
3answers
156 views
### Fastest polynomial root finder for a given accuracy
I am looking for a very fast polynomial root finder, hopefully with a matlab code. I don't need very accurate results, 2-3 decimal places would be fine. Also the method should be able to optionally ...
3answers
256 views
### Understanding the “rate of convergence” for iterative methods
According to Wikipedia the rate of convergence is expressed as a specific ratio of vector norms. I'm trying to understand the difference between "linear" and "quadratic" rates, at different points of ...
2answers
102 views
### Checking for error in conjugate gradient algorithm
What is a good way to check if the any numerical error is occured in conjugate gradient algorithm. Additionally why is it not suggested to check error by checking A-orthogonality of search direction ...
2answers
107 views
### Krylov subspace iterative methods in floating point arithmetic
Is there any work that considers Krylov subspace iterative methods in floating point arithmetic? I'm especially interested in how rounding errors influence the convergence and the accuracy of the ...
2answers
261 views
### How to remove Rigid Body Motions in Linear Elasticity?
I want to solve $K u = b$ where $K$ is my stiffness matrix. However some constraints may be missing an therefore some rigid body motion may be still present in the system (due to eigenvalue zero). ...
2answers
118 views
### Does right-hand side influence convergence rate of a Krlylov supspace method?
Consider general system $Ax=b$. Does convergence of the Krylov subspace methods depend on actual vector $b$ assuming initial guess is zero? I mean such factors as locality of the source (with the ...
1answer
130 views
### What is the current state of polynomial preconditioners?
I wonder what has happened to polynomial preconditioners. I am interested in them, because they appear to be comparatively elegant from a mathematical perspective, but as far as I have read in surveys ...
3answers
296 views
### Eigenvectors with the Power Iteration
To compute the eigenvector corresponding to dominant eigenvalue of a symmetric matrix $A\in\mathbb{R}^{n\times n}$, one used Power Iteration, i.e., given some random initialization, ...
4answers
406 views
### Simulated Annealing proof of convergence
I implemented downhill simplex simulated annealing algorithm. Algorithm is very hard to tune, w.r.t. parameters including cooling schedule, starting temperature... My first question is about ...
3answers
153 views
### Convergence of the gradient descent and linear vs non-linear fixed point iteration
Suppose a system $$Ax=b$$ is given, with $A\in\mathbb{R}^{n\times n}$ being a symmetric positive-definite matrix, and some non-zero $b\in\mathbb{R}^n$. The gradient method with optimum step length can ...
2answers
648 views
### How to choose a method for solving linear equations
To my knowledge, there are 4 ways to solving a system of linear equations (correct me if there are more): If the system matrix is a full-rank square matrix, you can use Cramer’s Rule; Compute the ...
5answers
275 views
### Iterative solution to a nonlinear equation
I appologize in advance if this question is silly. I need to compute the root of \begin{equation} u -f(u) =0 \end{equation} Where $u$ is a real vector and $f(u)$ is a real-vector valued function. ...
2answers
146 views
### Basin of attraction for Newton's method
Newton's method for solving nonlinear equations is known to converge quadratically when the starting guess is "sufficiently close" to the solution. What is "sufficiently close"? Is there literature ...
1answer
154 views
### Is the sparsity pattern of a linear system important for iterative (KSP) solvers?
Pretty much the question. Given a general sparse, non-symmetric (both numerically and structurally) matrix, how important is the sparsity pattern (i.e. row/column permutation of matrix/vector) for ...
1answer
130 views
### Solving a system with a small rank diagonal update
Suppose I have the original large, sparse linear system: $A\textbf{x}_0=\textbf{b}_0$. Now, I do not have $A^{-1}$ as A is too large to factor or any sort of decomposition of $A$, but assume that I ...
2answers
222 views
### What is a robust, iterative solver for large 3-d linear-elastic problems?
I'm diving into the fascinating world of finite element analysis and would like to solve a large thermo-mechanical problem (only thermal $\rightarrow$ mechanical, no feedback). For the mechanical ...
1answer
270 views
### Projecting out the null-space of $A$ from $b$ in $Ax=b$
Given the system $$Ax=b,$$ where $A\in\mathbb{R}^{n\times n}$, I read that, in case Jacobi iteration is used as a solver, the method will not converge if $b$ has a non-zero component in the null-space ...
2answers
159 views
### How to establish that an iterative method can be applied to large matrices whose size may reach 10^3?
I have an iterative method for computing the Moore-Penrose generalized inverse of matrices, that is $$X_{k+1} = ((I-\beta X_{k}A)^t) + X_{k}$$ with initial approximation: $$X_{0} = \beta AA^t$$ ...
1answer
113 views
### Gradient descent to stationary, or accumulation point
I recently came across the notion of an accumulation point as a result of a certain gradient descent variation. The following definition was found: An accumulation point $P$ is such that there are an ...
3answers
183 views
### Unique coordinates (solutions) in a single Gauss-Seidel iteration
I managed to reduce certain computational problem to the Gauss-Seidel solution of the following linear system: $$Ax=Ly,$$ where $A, L\in\mathbb{R}^{n\times n}$ are weighted Laplacian matrices ...
1answer
237 views
### A Comparison between GMRES, QMR and LU for Dense Matrices
As I see it, there are 3 ways to solve unstructured dense system of equations: GMRES, QMR and LU. Has anyone done a comparison for these three? As far as I know, LU is the preferred choice and it is ...
2answers
598 views
### Safe application of iterative methods on diagonally dominant matrices
Suppose the following linear system is given $$Lx=c,\tag1$$ where $L$ is the weighted Laplacian known to be positive $semi-$definite with a one dimensional null space spanned by ...
3answers
226 views
### On Vanilla Preconditioners for solving dense $Ax=b$ iteratively
I am looking for preconditioners which don't assume anything about the matrix or its origins. I basically want to be able to type in the following in MATLAB and have quick solving time: ...
4answers
179 views
### Determining the algorithmic complexity
A few of the iterative matrix algorithms (CG,GMRES etc.) I have authored are acting rather funny. They converge to the right answers but take abnormally long time to run. I am in the process of ...
1answer
264 views
### Which iterative linear solvers converge for positive semidefinite matrices?
I want to know which of the classic linear solvers (e.g Gauss-Seidel, Jacobi, SOR) are guaranteed to converge for the problem $Ax=b$ where $A$ is positive semi definite and of course $b \in im(A)$ ...
1answer
366 views
### Jacobi iteration to reduce the quadratic function
Given certain function $f(X)$ which is quadratic in $X\in\mathbb{R}^{n\times d}$, $$\frac{1}{2}tr(X^TAX) - tr(Y^TBX)$$ for positive definite weighted Laplacian matrices \$A, B\in\mathbb{R}^{n\times ...
1answer
96 views
### Question about the smoothing operators in multigrid methods for nonlinear PDEs
Suppose we are dealing with a nonlinear problem, say $$A u := L u + G(u) = f$$ the nonlinearity of the operator $A$ is the polynomial type, ie, $L$ is a linear operator, and $G(u) = u^k$, or ...
1answer
196 views
### How to establish that an iterative method for large linear systems is convergent in practice?
In computational science we often encounter large linear systems which we are required to solve by some (efficient) means, e.g. by either direct or iterative methods. If we focus on the latter, how ...
3answers
275 views
### Why does iteratively solving the Hartree-Fock equations result in convergence?
In the Hartree-Fock self-consistent field method of solving the time-independent electronic Schroedinger equation, we seek to minimize the ground state energy, $E_{0}$, of a system of electrons in an ...
3answers
354 views
### What guidelines should I use when searching for good preconditioning methods for a specific problem?
For the solution of large linear systems $Ax=b$ using iterative methods, it is often of interest to introduce preconditioning, e.g. solve instead $M^{-1}(Ax=b)$, where $M$ is here used for ...
1answer
775 views
### Are there any heuristics for optimizing the successive over-relaxation (SOR) method?
As I understand it, successive over relaxation works by choosing a parameter $0\leq\omega\leq2$ and using a linear combination of a (quasi) Gauss-Seidel iteration and the value at the previous ...
1answer
251 views
### Can a Krylov subspace method be used as a smoother for multigrid?
As far as I am aware, multigrid solvers use iterative smoothers such as Jacobi, Gauss-Seidel, and SOR to dampen the error at various frequencies. Could a Krylov subspace method (like conjugate ...
3answers
270 views
### Iterative methods for indefinite systems without block structure
Indefinite systems of matrices appear for example in the discretization of saddle point problems by mixed finite elements. The system matrix can then be put in the form \begin{pmatrix} A & B^t ...
3answers
450 views
### Krylov Subspace Methods for Dense Systems
I am currently researching on the viability of using KS methods for solving large dense systems. What I wish to prove (or disprove) is that methods like CG, BiCG and QMR are as good (if not better) ...
1answer
329 views
### What is a good stop criterion when using an iterative method to find eigenvalues?
I read this answer, and realized I have been using the difference between sucessive iterates to define a stop criterion for an iterative method of finding eigenvalues/vectors. What are good stop ...
2answers
1k views
### Why is my iterative linear solver not converging?
What can go wrong when using preconditoned Krylov methods from KSP (PETSc's linear solver package) to solve a sparse linear system such as those obtained by discretizing and linearizing partial ...
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