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http://math.stackexchange.com/questions/216211/what-is-the-optimal-solution-for-covering-a-rectangle-with-circles
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# What is the optimal solution for covering a rectangle with circles?
Given a rectangle of area n*m, and identical circles with radiuses r. What is the optimal solution for covering this rectangle with minimum number of circles?
I found a relative solution here. Should I have to first partition this rectangle with a group of smaller rectangles, and then use circumcircles to cover these smaller rectangles?
Is there any optimal solutions?
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– MvG Oct 18 '12 at 8:54
Thanks for you answer. I have seen this article before. It uses only a few circles to cover the rectangle $B$. But if the circle is much smaller than the rectangle $B$, how should I solve the problem? – Kate YAN Oct 19 '12 at 14:45
– MvG Oct 19 '12 at 15:15
## 1 Answer
This question is in part based on answers to this question on Stack Overflow.
# No optimal solution
Is there any optimal solutions?
There is no known way to find optimal solutions.
Erich's Packing Center (by Erich Friedman) has a page on circles covering squares. This is related to your problem, and in fact your problem is at least as hard as that problem. Obviously, asking about squares only can be no harder than asking about all rectangles. Furthermore, assume you had an algorithm to find an optimal solution according to your question. Then you could use that algorithm in a binary search, increasing the size of the square (i.e. rectangle) by a bit and using the circle count from your algorithm to see if the number of circles is still sufficient. By this approach, you could approximate the maximal possible square size for a given circle count to arbitrary precision.
So even though your problem is at least as hard as the problem of finding the maximal size of a square that can be covered by a given number of unit circles, there appears to be no general solution to the latter. For some small counts, an solution has been proven to be optimal, denoted by the phrase “proven by …” on the page I referenced. Other configurations are termed “found by …”, indicating that they are the best configuration known to the author, but that he is not aware of any proof of optimality.
Therefore, you should not expect a really optimal solution in the strict sense, although of course a breakthrough discovery always is a possibility.
# Good solution
If you weaken your requirement of optimality, you are asking about “good” ways to cover your rectangle. As your comment indicates that your circles are much smaller than the rectangle, you may for the moment ignore the boundary, and instead examine infinite covers. It turns out that the infinite circle cover with least density is the hexagonal one. (I haven't found an article to that effect just now, so if you need a reliable source for this, feel free to post a reference-request.) So by cutting out a rectangular portion of such a hexagonal cover, you'll get the lowest possible density over the majority of your area, and perhaps slightly suboptimal solutions near the boundary.
Should I have to first partition this rectangle with a group of smaller rectangles, and then use circumcircles to cover these smaller rectangles?
As the hexagonal lattice will lead to smaller density than the square lattice, you should partition your rectangle not into rectangles but instead into regular triangles, inscribed into your disk of radius $r$. You may get partial triangles near the rim. Unless those parts are already completely covered by some other circle, you'll still need full circles for those partial triangles, which corresponds to the lack of optimality.
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Thank you for your detailed answer. I agree with the solution of partitioning the rectangle into hexagonal lattice. But I don't think that hexagonal lattice is always better than rectangles. It depends on the proportion of n, m and r. – Kate YAN Oct 22 '12 at 9:01
@KateYAN, don't claim that the hexagonal lattice will always be better. I'm just saying that it's better in the infinite case, and that according to your comment, $r ≪ m,n$ so it should be better for you. – MvG Oct 22 '12 at 14:39
Yes, you are right. Thanks. – Kate YAN Oct 23 '12 at 23:15
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http://mathoverflow.net/questions/25009?sort=oldest
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## Counting submanifolds of the plane
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After thinking about this question and reading this one I am led to ask for an uncountable collection of homeomorphism types of boundaryless connected path-connected submanifolds of the plane.
My guess is that is suffices to consider complements of Cantor sets. However, I do not know how to distinguish ends (up to homeomorphism) sufficiently well to ensure that this works. Are there other, easier, invariants?
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Given two perfect, totally disconnected subsets $A,B$ of $\mathbb R^2$, is there always a homeomorphism $\varphi$ of $\mathbb R^2$ such that $\varphi(A)=B$? If the answer is yes, then it is not enough to work with complements of Cantor sets. – Roland Bacher May 17 2010 at 14:29
## 1 Answer
See a theorem of Richards, which implies that homeomorphism types of planar surfaces are in 1-1 correspondence with homeo. types of compact subsets of the Cantor set. I think there should be uncountably many homeo. types of totally disconnected compactums, but I don't know a reference or an argument off the top of my head. I think this should be related to the ordinalities of the accumulation points, but I'm not sure which ordinals can occur.
Addendum: Googling, I found references to a result of Markiewicz-Sierpinski classifying countable compact metric spaces up to homeomorphism by their Cantor-Bendixson rank (see section 3 of this paper for a statement). The CB-rank must be a countable ordinal $\zeta$, and the space is homeomorphic to the ordinal $\omega^\zeta\cdot n+1$ with the order topology for some $n\in \mathbb{N}$. These may all be realized as compact subsets of the line. This gives uncountably many non-homeomorphic compacta, which by Richards' theorem implies that there are uncountably many planar surfaces.
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Very nice! I'll take a look at the paper you mention. – Sam Nead May 18 2010 at 10:59
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http://mathoverflow.net/questions/43240/what-is-the-l-function-version-of-quadratic-reciprocity/43243
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## What is the L-function version of quadratic reciprocity?
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Quadratic reciprocity theorems states that for two different odd prime p and q, we have (p/q)(q/p)=(-1)^(p-1)(q-1)/4.
What is the statement of this theorem in L-function?
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You can consider $\zeta(s)L(s,\chi)=\zeta_K(s)$ to be a reformulation of quadratic reciprocity, where $\chi$ is a primitive quadratic character of conductor $N$ and $K=\mathbb{Q}\left[\sqrt{\chi(−1) N} \right]$. This follows from the relation between Legendre symbols and splitting in quadratic extensions. – Sam Derbyshire Oct 22 2010 at 22:35
3
Maybe you find this beautiful article by T. Honda helpful. numdam.org:80/numdam-bin/… – BY Oct 22 2010 at 23:02
## 2 Answers
Say $K=\mathbf{Q}(\sqrt{p^{*}})$ is the unique quadratic field in which $p$ is the only ramified finite prime. That is, $p^*=(-1)^{(p-1)/2}p$. There are two L-functions one can concoct out of $K$, and quadratic reciprocity manifests in the fact that the L-functions coincide.
First, there is the L-function arising from the Galois character $\chi$ which cuts out $K$. This is a one-dimensional Artin L-function. Its Euler factor at a prime $q$ is $(1\pm p^{-s})^{-1}$, where you place $-$ whenever $q$ splits in $K$, and $+$ whenever $q$ is inert in $K$.
The other L-function is a Dirichlet L-function. Let $\varepsilon$ be the unique character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order exactly two. This extends to a Dirichlet character $\varepsilon$ on $\mathbf{Z}$, and the Dirichlet L-function is $\sum_{n\geq 1} \varepsilon(n)n^{-s}$.
Let us see that if the two L-functions are equal, then quadratic reciprocity holds. Indeed, look at the coefficient of $q^{-s}$. The coefficient in the first L-function is $\left(\frac{p^*}{q}\right)$. The coefficient in the second L-function is $\left(\frac{q}{p}\right)$. (This requires some explanation: $x\mapsto \left(\frac{x}{p}\right)$ is a character of $(\mathbf{Z}/p\mathbf{Z})^\times$ of order two, so it must equal $\varepsilon$ by uniqueness.) We find
$$\left(\frac{q}{p}\right)=\left(\frac{p^{*}}{q}\right)=\left(\frac{(-1)^{(p-1)/2}}{q}\right)\left(\frac{p}{q}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right).$$
Of course you didn't need L-functions to state any of this, but they are nonetheless very important. For instance, you can't prove that the Artin L-function has an analytic continuation and functional equation without knowing that it equals a Dirichlet L-function. The generalization of this coincidence to higher-degree Artin L-functions is still quite conjectural!
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What is the version of this for the $l$-tic reciprocity law, where $l$ is a prime $\neq2\;$? – Chandan Singh Dalawat Oct 23 2010 at 6:06
See for example Honda (Taira), Invariant differentials and L-functions. Reciprocity law for quadratic fields and elliptic curves over Q. Rend. Sem. Mat. Univ. Padova 49 (1973), 323–-335, available as numdam.org/item?id=RSMUP_1973__49__323_0 – Chandan Singh Dalawat May 12 2012 at 7:02
### 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.
Let me try to say a few things from my point of view concerning Chandan Singh Dalawat's question about higher reciprocity.
The "Galois character" of the quadratic extension $K$ ramified only at $p$ is given by the Kronecker symbol $\chi(q) = (p^*/q)$; the associated relation $\zeta(s) L(s,\chi) = \zeta_K(s)$ between zeta functions of the rationals, the Dirichlet L-function belonging to $\chi$, and the Dedekind zeta function of the field $K$ encodes the decomposition law of quadratic extensions (see the comment by Sam Derbyshire).
The quadratic reciprocity law in Euler's form states that this Kronecker symbol is a Dirichlet character with conductor $p$, i.e. that $\chi(q) = \chi(r)$ for positive primes $q \equiv r \bmod p$. Thus it must be one of the $p-1$ Dirichlet characters modulo $p$, and since $({\mathbb Z}/p{\mathbb Z})^\times$ is cyclic, is must be the quadratic Dirichlet character modulo $p$.
This has an interpretation in terms of L-functions for Dirichlet characters; the zeta relation $\prod L(s,\chi) = \zeta_L(s)$, where the product on the left hand side is over all Dirichlet characters mod $p$ (suitably interpreted at the bad prime $p$), and where $L$ is the field of $p$-th roots of unity, can be interpreted as the decomposition law for primes in cyclotomic extensions. The reason why there is a connection between these objects and the Kronecker character is, ultimately, the fact that $K \subset L$, something that can be proved directly e.g. with Gauss sums, another tool for proving reciprocity laws. In any case, the ultimate source of this proof is the realization of the Kummer extension $K/{\mathbb Q}$ as a class field (inside the ray class field $L/{\mathbb Q}$).
For getting the cubic reciprocity law, you can work with Kummer extensions over the field of cube roots of unity; comparing the decomposition laws for Kummer extensions and class fields gives you something like $(\pi/q)_3 = (q/\pi)_3$ for primary primes $\pi$ and rational primes $q$, which is strong enough to give you the full cubic reciprocity law with a couple of formal manipulations. Writing down the corresponding Artin and Dirichlet L-series is straightforward.
Something similar works for fourth powers, but then you run into trouble since you only get Eisenstein reciprocity out of this comparison of decomposition laws.
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Many thanks for your answer, Franz. – Chandan Singh Dalawat Oct 25 2010 at 3:33
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http://math.stackexchange.com/questions/36474/is-any-hamiltonian-system-with-just-one-degree-of-freedom-completely-integrable?answertab=oldest
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# is any hamiltonian system with just one degree of freedom completely integrable?
An hamiltonian system with $n$ degree of freedom is said to be completely integrable when there exists an system $f_1,\ldots,f_n$ of first integrals mutually Poisson-commuting, such that $df_1(x),\ldots,df_n(x)$ are linearly independent for any $x$ in a dense subset.
Consequently any hamiltonian system with one degree of freedom, whose non fixed points constitute a dense subset, is completely integrable, (infact in this case just its hamilton function gives a maximal set of Poisson-commuting first integrals which are independent on a dense subset).
Now in a paper I read that any hamiltonian system with one degree of freedom is completely integrable, because there should exist always a smooth function which is Poisson-commuting with the Hamilton function and whose regular points are every-where dense.
I have some difficulty in understanding this statetement, so is there someone who could give me an hint?
My guess is that, given an Hamilton function $H$ on $(M,\omega)$ whose regular points constitute an open subset $U$, if it is possible to prove that there exists a smooth function $\delta$ vanishing on $U$ and whose regular points are dense in $M\setminus\overline{U}$, then we could take $H+\delta$.
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1
Isn't this function the Hamiltonian itself? – Fabian May 2 '11 at 17:41
sure, but only when the non-fixed points of the hamiltonian system constitute a dense subset. – Giuseppe May 2 '11 at 19:28
My question is about the remaining case. – Giuseppe May 2 '11 at 19:28
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http://mathoverflow.net/questions/43514/how-do-eigenvectors-and-eigenvalues-change-when-we-remove-a-row-column-pair-of-a/43526
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## How do eigenvectors and eigenvalues change when we remove a row/column pair of a matrix?
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Let us have a symmetric matrix $C \in \mathbb{R}^{n\times n}$ having non-negative values. Suppose that we have the eigenvalue decomposition for this particular matrix such that
$$C e_i = \lambda_i e_i$$
where $e_i$ are the eigenvectors and $\lambda_i \geq 0$ are the corresponding eigenvalues. In matrix form,
$$CE = ES$$ where $E$ is the matrix involving eigenvectors as columns and $S$ is the diagonal matrix involving eigenvalues on the diagonal entries.
Now, we delete $k^\text{th}$ row and column of the matrix and form a new matrix $\tilde{C} \in \mathbb{R}^{(n-1) \times (n-1)}$ and we want to find its eigenvalues and eigenvectors such that
$$\tilde{C} \tilde{E} = \tilde{E} \tilde{S}$$
Instead of computing them from scratch, I wonder if there exists an analytical way to find the eigenvectors and eigenvalues iteratively using $E$ and $S$. In other words, is there a link between $E$, $S$ and $\tilde{E}$, $\tilde{S}$?
For example, for a 3-by-3 matrix $C$ where
$$C = \left[\begin{array}{ccc}a & b & c \\ b & d & e \\ c & e & f\end{array}\right]$$
if I delete the third row and column, then I get
$$C = \left[\begin{array}{cc}a & b \\ b & d \end{array}\right]$$
I know that the number of dimensions of the eigenvectors are one less and we have one less eigenvalues but may there be a projection of the others onto some bases?
P.S.(1) I've read the answers to the question "How does eigenvalues and eigenvectors change if the original matrix changes slightly", but I couldn't find a connection with this question. Sorry if I couldn't get a point and created a duplicate question.
P.S.(2) If you wonder why I'm asking this question, here it is: I'm computing the eigenvalues and eigenvectors of the covariance matrix of some samples. Let each sample have 3 dimensions and let $D \in \mathbb{R}^{k\times n}$ be the data matrix where each row is a sample and we have $k$ samples. Then the covariance matrix is $C = D^T D$. If we have 3 dimensions and the columns of $D$ are $d_1$, $d_2$ and $d_3$, then we have
$$C = \left[\begin{array}{ccc}d_1^T d_1 & d_1^T d_2 & d_1^T d_3 \\ d_2^T d_1 & d_2^T d_2 & d_2^T d_3 \\ d_3^T d_1 & d_3^T d_2 & d_3^T d_3\end{array}\right]$$
I'm interested in the decomposition when we have some dimensions missing in the data. If, as in the example, the third dimension is missing, then we have
$$\tilde{C} = \left[\begin{array}{cc}d_1^T d_1 & d_1^T d_2 \\ d_2^T d_1 & d_2^T d_2 \end{array}\right]$$
I wondered whether there's a way to compute the eigenvectors and eigenvalues of the missing data's covariance matrix using the ones that we computed from the full data in an iterative manner instead of computing it from scratch.
Kind regards and thanks for any ideas.
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## 3 Answers
Let $A$ be a symmetric matrix, with eigenvalues $\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n$. Let $B$ be the matrix obtained by deleting the $k$-th row and column from $A$, with eigenvalues $\mu_1 \leq \mu_2 \leq \cdots \leq \mu_{n-1}$. Then `$$\lambda_1 \leq \mu_1 \leq \lambda_2 \leq \mu_2 \leq \lambda_3 \leq \cdots \leq \lambda_{n-1} \leq \mu_{n-1} \leq \lambda_{n}.$$` This is a special case of Cauchy's interlacing theorem. The operator $P$ in the wikipedia article should be taken to be the projection on the coordinate vectors other than the $k$-th one.
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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.
The word you're looking for is downdating, and I cannot do better than to point out these two survey papers, and this article. I will also have to make the reminder that it makes better numerical sense to compute the singular values of $\mathbf{D}$ rather than the eigenvalues of $\mathbf{D}^T \mathbf{D}$.
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Removing a paired row/column from a symmetric matrix is kind of like setting the corresponding entries to zero. This reformulation would allow you to consider a perturbed matrix of the same dimensions.
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http://mathhelpforum.com/advanced-algebra/150692-component-vector-direction-another.html
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# Thread:
1. ## Component of a vector in direction of another
Practice problem asks for the component of vector A=(2,1,-4) in the direction of B=(1,2,3).
I've found the direction vector of B (by dividing B by its length of √14). Let's call this direction vector d.
Now, is the dot product (x.d) the answer?
I found -8/√14
2. Originally Posted by keysar7
Practice problem asks for the component of vector A=(2,1,-4) in the direction of B=(1,2,3).
I've found the direction vector of B (by dividing B by its length of √14). Let's call this direction vector d.
Now, is the dot product (x.d) the answer?
I found -8/√14
You have found the dot product x.d correctly. But the answer to the question should be a vector, not a scalar. The component of A in the direction of B is the vector (x.d)d (its magnitude is x.d and its direction is given by the direction of d). So the answer should be $\frac{-8}{14}(1,2,3)$.
3. Opalg, many books use the opposite convention- that the "component" of a vector in a given direction is a scalar. For example, the "component" of <a, b, c> in the <1, 0, 0> direction is "a", not <a, 0, 0>. keysar7, better check your textbook or with your teacher to be sure which convention is being used.
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http://physics.stackexchange.com/questions/tagged/hilbert-space
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# Tagged Questions
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### Notational techniques for dealing with creation operators on Fock space
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### Show that for QM operator A: $\int_{-\infty}^{\infty}\psi A^{\dagger}A\psi dx = \int_{-\infty}^{\infty}(A\psi)^*(A\psi)dx$
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### Inner Product Spaces
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### What is a dual / cotangent space?
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### What is this state as a matrix?
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1answer
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### What really are superselection sectors and what are they used for?
When reading the term superselection sector, I always wrongly thought this must have something to do with supersymmetry ... DON'T laugh at me ... ;-) But now I have read in this answer, that for ...
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### Spontaneous symmetry breaking: How can the vacuum be infinitly degenerate?
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### State space of QFT, CCR and quantization, and the spectrum of a field operator?
In the canonical quantization of fields, CCR is postulated as (for scalar boson field ): $$[\phi(x),\pi(y)]=i\delta(x-y)\qquad\qquad(1)$$ in analogy with the ordinary QM commutation relation: ...
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### Can I prove boundedness of an operator without checking it for its whole domain?
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### Space of states in quantum mechanics
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### Differences between orthogonality and Kronecker delta function? [closed]
If $i$ and $j$ are two variables then Kronecker delta is written as \delta_{i,j}~:=~\begin{cases}1 \hspace{3mm} \mbox{if} \hspace{3mm} i=j,\\ 0 \hspace{3mm}\mbox{if} \hspace{3mm}i \neq ...
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### Wave function and Dirac bra-ket notation
Would anyone be able to explain the difference, technically, between wave function notation for quantum systems e.g. $\psi=\psi(x)$ and Dirac bra-ket vector notation? How do you get from one to the ...
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### Why must quantum logic gates be linear operators?
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### Intuitive meaning of Hilbert Space formalism
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### Banach Space representations of physical systems
I think most physicists mostly model physical systems as some kind of Hilbert space. Hilbert spaces are a strict subset of Banach spaces. Questions: Can physical systems really have non-compact ...
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### Don't understand the integral over the square of the Dirac delta function
In Griffiths' Introduction to Quantum Mechanics he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right).$$ ...
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### What's the physical significance of the inner product of two wave functions in quantum region?
I am a reading a book for beginners of the quantum mechanics. In one section, the author shows the inner product of two wave functions $\langle\alpha\vert\beta\rangle$. I am wondering what's the ...
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### The state of Indefinite metric in Quantum Electrodynamics
I faced difficulties to grasp why indefinite metric is introduced from no where in QED, after searching internet I found that this is a problem in QED, because one needs it to preserve theory's ...
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### Representing a polarization vector for light as a 'manifold of two state'
Explain me these projections please Context: I was reading a paper (Phys. Rev. A 68, 052307) which involved mesoscopic coherent states of light. There, in order to calculate the uncertainty of a ...
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### Existence of adjoint of an antilinear operator, time reversal
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### Scattering states of Hydrogen atom in non-relativistic perturbation theory
In doing second order time-independent perturbation theory in non-relativistic quantum mechanics one has to calculate the overlap between states E^{(2)}_n ~=~ \sum_{m \neq n}\frac{|\langle m | H' ...
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I need help understanding the role of representation theory in QM/QFT. My understanding of representation theory in this context is as follows: there are physical symmetries of the system we are ...
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### Rigged Hilbert space and QM
Are there any comprehensive texts that discuss QM using the notion of rigged Hilbert spaces? It would be nice if there were a text that went through the standard QM examples using this structure.
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### How much space to simulate a small Hilbert space?
I'm thinking about trying to do a numerical simulation of some very simple QM problems. How much space do I need? To simulate the Hilbert space? I'd like to eventually simulate the absorption or ...
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### Difficulties with bra-ket notation
I have started to study quantum mechanics. I know linear algebra,functional analysis, calculus, and so on, but at this moment I have a problem in Dirac bra-ket formalism. Namely, I have problem with ...
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### Kugo and Ojima's Canonical Formulation of Yang-Mills using BRST
I am trying to study the canonical formulation of Yang-Mills theories so that I have direct access to the $n$-particle of the theory (i.e. the Hilbert Space). To that end, I am following Kugo and ...
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156 views
### When and how do you represent a two body state as a tensor product?
I have read that in quantum mechanics, compound systems are constructed as tensor products. But on page 177 of Griffith, for example, a two body wavefunction is introduced as Psi ...
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### Why we use $L_2$ Space In QM?
I asked this question for many people/professors without getting a sufficient answer, why in QM Lebesgue spaces of second degree are assumed to be the one that corresponds to the Hilbert vector space ...
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### When we define the S-matrix, what are “in” and “out” states?
I have seen the scattering matrix defined using initial ("in") and final ("out") eigenstates of the free hamiltonian, with \left| \vec{p}_1 \cdots \vec{p}_n \; \text{out} \right\rangle = S^{-1} ...
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http://mathoverflow.net/questions/107508?sort=oldest
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## Is there an explicit formula for the modulus of an annulus given a parameterization of the inner and outer boundries?
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Every open set in the complex plane homeomorphic to an annulus is biholomorphic to exactly one annulus whose inner radius is 1 and whose out radius is $r>1$. Each value of $r$ gives a different complex manifold. This number r is called the "modulus" of the annulus. You could say that the set of real number $(0,\infty)$ give a moduli space of complex annuli.
If I give you a parameterization of the inner boundary and the outer boundary of a topological annulus, is there some explicit formula for the modulus of the annulus it is conformally equivalent to? I am guessing there is some kind of integral formula. I looked through a few papers, and have not found such a formula, but it must certainly be classical.
My question is very similar to
http://mathoverflow.net/questions/11239/conformal-maps-of-doubly-connected-regions-to-annuli
but that question has an accepted answer which does not answer my question. The advice on Meta was just to ask a new question, but explain the link to the old one.
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Are the boundaries just Jordan curves? – Igor Rivin Sep 18 at 22:48
The open set $\mathbb{C}-{0}$ is homeomorphic to the annulus but is not biholomorphic to any annulus whose inner radius is 1 and whose outer radius is $r>1$. – Lee Mosher Sep 18 at 23:18
True. I should have been more precise. – Steven Gubkin Sep 18 at 23:21
## 3 Answers
I think that there is no formula. The best one can do is to estimate. Here is a simpler problem of the same sort: suppose you have a parametrization of the boundary of a simply connected region, and suppose that 0 is inside. Consider the Riemann mapping f of this region sending 0 to 0. The problem is to find |f'(0)|. There is no formula in any reasonable case.
Of course this is not a theorem, because one cannot define what a "formula" is. Both quantities, the modulus of a ring, and |f'(0)| in the simplified problem are solutions of certain extremal problems. So one can write a "formula" involving sup over some class of functions.
Added on 9.19: I don't know why the question about a "formula" is important. There are reasonably good converging algorithms for finding moduli of rings, of course. The closest thing to a "formula" for a conformal map of a simply connected region that I know is described in the papers of Wiegmann and Zabrodin, for example, MR1785428. Perhaps this can be modified to make a formula for the modulus of a ring.
Added on the same day: Here is a "formula". Let $\mu$ and $\nu$ be two probability measures, one sitting on each boundary component. Let $\rho=\mu-\nu$. Then $$\log r=-2\pi\sup\int\int\log|z-w|d\rho(z)d\rho(w),$$ where the $\sup$ is taken over all such measures. If your boundaries are smooth, the measures are also smooth, and can be described by smooth densities.
Explanation. Think of the boundaries as bases of metal cyinders, and put unit charges on them, one positive another negative. Then allow the charges to flow according to Coulomb Law. they will occupy the equilibrium position (minimizing the energy). This minimal energy is $\log r/(2\pi)$ and it is conformally invariant. It is the so-called capacity of a condenser.
This was given as an example of what I meant by a formula containing a sup over a set of functions.
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It would be interesting to formulate a reasonable class of "possible formulas", and show that none of them work to answer the question. – Steven Gubkin Sep 18 at 22:37
Though I haven't studied it yet, I wonder if Greg Hjorth's paper The complexity of the classification of Riemann surfaces and complex manifolds says anything about the possibility of a "formula" at least in analogous situations. math.ucla.edu/~greg/research.html – David Feldman Sep 19 at 1:17
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You can read all about numerical approximation schemes to compute this in this Diplomarbeit. link text
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This looks beautiful! – Steven Gubkin Sep 18 at 23:25
I also think there is no "explicit formula" for the conformal modulus, even in simple cases. However, as mentioned in Igor Rivin's answer, there are methods for approximating the conformal map and the conformal modulus. These methods are generelizations to doubly-connected domains of the well-known Bergman Kernel Method for simply connected domains.
For an introduction to these methods, I suggest you take a look at this document, sections 1.3 and 2.5.
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http://www.kurims.kyoto-u.ac.jp/~nakajima/Talks/200306_rims.html
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# Introduction to moduli spaces of instantons on $\mathbf R^4$
In this introductory lecture, I will explain basic properties of moduli spaces of $SU(r)$-instantons on $\mathbf R^4$. These spaces can be interpreted as framed moduli spaces of locally free sheaves on the projective plane, and hence have a natural smooth (partial) compactification (by adding non locally free sheaves). These give resolutions of singularities of Uhlenbeck (partial) compactifications. If I have time, I will also introduce quiver varieties as subvarieties of (partial) compactifications.
nakajima@kusm.kyoto-u.ac.jp
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http://mathoverflow.net/questions/46636/sasaki-but-not-einstein
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## Sasaki but not Einstein
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Hi, I search for an example of a Sasaki-manifold which is not Einstein. Can you give one?
Thank you and best regards!
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Since you are having difficulty getting hold of the book mentioned in my answer, feel free to get in touch with me by email (my address should be easy to find from my user page). – José Figueroa-O'Farrill Nov 21 2010 at 2:24
## 3 Answers
Chapter 11 in Boyer-Galicki's Sasakian Geometry discusses two obstructions to the existence of Sasaki-Einstein structures on Sasakian manifolds. They go on to discuss many such examples, obtained as links of conical singularities of projective varieties in weighted projective spaces.
They are not hard to find! The cone of a Sasakian manifold is Kähler, whereas the cone of a Sasaki-Einstein manifold is Calabi-Yau. Hence your question is the analogue of asking for a Kähler manifold which is not Calabi-Yau. This is the generic situation.
As in my answer to your first question on Sasakian Geometry, there is no excuse for these questions given the wealth of information and the clarity of style of the book by Boyer and Galicki.
Go read the book! It really is good.
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Thanks for the answer! I want to read the book, but the problem is that I am not able to get the book.... – Differentialgeometer Nov 20 2010 at 8:38
### 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.
This is not an answer to your question, but rather an extension of José Figueroa-O'Farrill's answer. Roughly a year ago, a quick Google search led me to a preprint of Boyer & Galicki's wonderful book; it can be found here
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I am not very sure of everything here but I wonder if the example given in Appendix A on Page 21 of this paper by one of my professors meets your criteria.
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Are you sure that is the right paper? Can you give me the construction of the manifold? Best regards! – Differentialgeometer Nov 19 2010 at 17:39
1
The geometries in that paper are orbifolds $\mathbb{C}^2/\mathbb{Z}_n$. The $\mathbb{Z}_n$ lies inside an $\mathrm{SU}(2)$ subgroup and hence the orbifold will be Calabi-Yau, whence the link of the conical singularity at the origin will be Sasaki-Einstein. – José Figueroa-O'Farrill Nov 19 2010 at 18:37
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http://mathoverflow.net/questions/72569/harmonic-function-on-surface/72655
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## harmonic function on surface
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Dear all,
I am looking for reference books about real-valued harmonic functions on complete Riemannian surfaces, do you have any reference in your mind about this? I found some books about harmonic functions on the plane (but mainly discussing on conformal maps) or about general harmonic maps (which is too general to possess nice properties). They do not fit my need...
Precisely, I would like to understand the zero set (for example, is it a one dimensional manifold?), the existence and linearity of such functions.
Thanks for your help in advance!
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The zero set will will have prong singularities, like the zeroes of $Re(z^n)$ or $Im(z^n)$, at least. Meeks used to carry around a book by Rick Schoen that had lots of information about harmonic maps on Riemann surfaces. I tried to find it but its been too long ago to remember. Lots of theorems about minimal surfaces are proved via arguments with harmonic mappings as the coordinate functions are harmonic in the conformal structure underlying the pull back Riemannina metric. – Charlie Frohman Aug 10 2011 at 13:46
Thank you Charlie for the examples. (I will try to find the book of R. Schoen.) – Chih-Wei Chen Aug 12 2011 at 14:36
## 1 Answer
Two remarks:
1. the notion of harmonic function on a surface is conformally invariant, so if your question is local the it is about harmonic function on $C$.
2. on a surface, a harmonic function is the same as the real part of a holomorphic function, see wikipedia. So you can reformulate your question on zeros as a question on the inverse image of the real line by a holomorphic function defined on (a subset of) $C$.
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Thank you Jean-Marc, I had found these two remarks but I cannot figure out more "explicit" properties from them in general. However I think they will be helpful when certain particular cases are discussed. Thanks again. – Chih-Wei Chen Aug 12 2011 at 14:34
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http://mathhelpforum.com/discrete-math/181027-proper-use-universal-instantiation.html
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# Thread:
1. ## Proper use of Universal Instantiation
Is this a proper use of universal instantiation? If not, how could this proof be corrected?
Suppose:
$\forall x P(x) \rightarrow \exists x Q(x)$
Prove:
$\exists x (P(x) \rightarrow Q(x))$
Proof:
$y$ is arbitray, so $P(y)$
$P(y)$, so choose a $z_{0}$ such that $Q(z_{0})$
By universal instantiation let $y = z_{0}$.
Therefore $P(z_{0}) \rightarrow Q(z_{0})$.
$\square$
2. Originally Posted by Ontolog
Is this a proper use of universal instantiation?
No, it's not.
$y$ is arbitray, so $P(y)$
It seems that you start with ∀x P(x) and instantiate x with y. However, ∀x P(x) is not given: it is an assumption of an assumption and thus has to be proved. It is also not clear why you need to instantiate x since you can immediately derive ∃x Q(x) by MP and from there derive the conclusion.
If, on the other hand, you assume P(y), then you can conclude ∀x P(x) because the assumption contains y free. Also, it is not clear where you close the assumption P(y).
To say something more precise, you have to say what formalism you are using, even informally (e.g., natural deduction or sequent calculus).
This conclusion can't be proved without the law of excluded middle. Suppose that ∀x P(x) is true. Then ∃x Q(x) from the assumption, and so ∃x (P(x) -> Q(x)). If ∀x P(x) is false, then there exists an x0 such that P(x0) is false. Therefore, P(x0) -> Q(x0) is true.
3. Originally Posted by emakarov
No, it's not.
It seems that you start with ∀x P(x) and instantiate x with y. However, ∀x P(x) is not given: it is an assumption of an assumption and thus has to be proved.
I can see this, yes I should have added another assumption that $\forall x P(X)$.
Originally Posted by emakarov
It is also not clear why you need to instantiate x since you can immediately derive ∃x Q(x) by MP and from there derive the conclusion.
If, on the other hand, you assume P(y), then you can conclude ∀x P(x) because the assumption contains y free. Also, it is not clear where you close the assumption P(y).
Yes, I can see that by assuming $\forall x P(x)$ I wouldn't need to instantiate x to derive $\exists x Q(x)$. However, wouldn't I need to instantiate x to show that $\exists x(P(x) \rightarrow Q(x))$, since in this expression we are saying that the x for which P(x) is the same as the x for which Q(x).
Originally Posted by emakarov
To say something more precise, you have to say what formalism you are using, even informally (e.g., natural deduction or sequent calculus).
This is something I have been unclear on. I am studying from a book "How To Prove It" (Vellemen 2006) and I am (attempting to) use the formalism presented there. I believe it is "first-order logic."
Originally Posted by emakarov
This conclusion can't be proved without the law of excluded middle. Suppose that ∀x P(x) is true. Then ∃x Q(x) from the assumption, and so ∃x (P(x) -> Q(x)).
I don't follow this argument for the same reason I mentioned earlier. To show that $\exists x(P(x) \rightarrow Q(x))$ is true, don't we need to demonstrate that the x for which P(x) is the same as the x for which Q(x)?
Originally Posted by emakarov
If ∀x P(x) is false, then there exists an x0 such that P(x0) is false. Therefore, P(x0) -> Q(x0) is true.
Again I can't follow. Where is it shown that $\lnot Q(x0)$?
$\forall A x P(x) \rightarrow \exists x Q(x)$
Therefore, by modus tollens:
$\lnot \exists x Q(x) \rightarrow \lnot \forall A x P(x)$
But the converse is not necessarily true:
$\lnot \forall A x P(x) \rightarrow \lnot \exists x Q(x) ?$
I realize that there must be something blindingly obvious that I'm missing here so please bear with me!
4. Originally Posted by Ontolog
Suppose:
$\forall x P(x) \rightarrow \exists x Q(x)$
Prove:
$\exists x (P(x) \rightarrow Q(x))$
Is the above the exact statement of the exercise as it appears in the mentioned textbook?
From read his solution, it does not appear to be the complete quote.
5. Originally Posted by Plato
Is the above the exact statement of the exercise as it appears in the mentioned textbook?
From read his solution, it does not appear to be the complete quote.
OK, some context for how this problem came up. In the book there is a straightforward exercise:
Prove that if $\exists x (P(x) \rightarrow Q(x))$ is true, then $\forall x P(x) \rightarrow \exists x Q(x)$ is true.
Then there is a note that "The other direction of the equivalence is quite a bit harder to prove. See exercise 29 of Section 3.5", where we find the problem:
Prove that if $\forall x P(x) \rightarrow \exists x Q(x)$ then $\exists x (P(x) \rightarrow Q(x))$.
And just as I look up the original problem text I find that the author gives a hint: "Remember that $P \rightarrow Q$ is equivalent to $\lnot P \lor Q$." Which is something I'll have to think about...
6. The hint suggests considering the case ~AxPx and the case ExQx. Show that in either case we have Ex(Px -> Qx).
7. Originally Posted by Ontolog
I don't follow this argument for the same reason I mentioned earlier. To show that $\exists x(P(x) \rightarrow Q(x))$ is true, don't we need to demonstrate that the x for which P(x) is the same as the x for which Q(x)?
Recall that an implication is true iff the conclusion is true (regardless of the premise) or the premise is false (regardless of the conclusion).
Suppose ∀x P(x) is true, which also means that ∃x Q(x) is true. Fix x0 that makes Q true. Then P(x0) -> Q(x0) is true regardless of whether P(x0) is true or not. Similarly, if ∀x P(x) is false, suppose P(x0) is false. Then P(x0) -> Q(x0) is true regardless of whether Q(x0) is true.
8. Originally Posted by emakarov
Recall that an implication is true iff the conclusion is true (regardless of the premise) or the premise is false (regardless of the conclusion).
Suppose ∀x P(x) is true, which also means that ∃x Q(x) is true. Fix x0 that makes Q true. Then P(x0) -> Q(x0) is true regardless of whether P(x0) is true or not. Similarly, if ∀x P(x) is false, suppose P(x0) is false. Then P(x0) -> Q(x0) is true regardless of whether Q(x0) is true.
I understand now. I suppose there are some alternative systems of logic that treat implications differently? Because intuitively, an implication is defined by a premise and a consequence which follows from that premise. In fact, I wonder how many typical proofs in mathematics make use of the fact that if Q(x) then $P(x) \rightarrow Q(x)$ for any P and x. In other words, is this merely an idiosyncrasy of first-order logic or does it have any practical implications? (No pun intended )
9. Originally Posted by Ontolog
I wonder how many typical proofs in mathematics make use of the fact that if Q(x) then $P(x) \rightarrow Q(x)$ for any P and x. In other words, is this merely an idiosyncrasy of first-order logic or does it have any practical implications?
The statement that $A\subseteq B$ means that $\left( {\forall x} \right)\left[ {x \in A \to x \in B} \right]$.
We use the to say in the class of sets $\left( {\forall C} \right)\left[ {\emptyset \subseteq C} \right]$.
10. Yes, without the rule that true conclusion makes the whole implication true we would not be able to say general facts whose converse is false, both in mathematics and in real life. An example from Wikipedia:
Being at least 30 years old is necessary of serving in the U.S... That is, if you are a senator, it follows that you are at least 30 years old.
The converse of the intermediate value theorem is false: roughly speaking, if for all a, b and every u between f(a) and f(b) there exists a c such that f(c) = u, it does not follow that f is continuous.
Still another example: If f : A -> B and g : B -> C are injections, then so is the composition g o f, but not conversely (if g o f is injective, then so is f, but not necessarily g).
Other concepts of implications have been considered in philosophy and logic. See this Wikipedia section, for example. You can probably find a lot of information in the Stanford Encyclopedia of Philosophy.
11. Originally Posted by Plato
The statement that $A\subseteq B$ means that $\left( {\forall x} \right)\left[ {x \in A \to x \in B} \right]$.
We use the to say in the class of sets $\left( {\forall C} \right)\left[ {\emptyset \subseteq C} \right]$.
That's very interesting! I worked it out and have seen that it is true, so I think I'm starting to absorb this concept of "rigorously defined" logical implication. Thanks!
12. Originally Posted by emakarov
Yes, without the rule that true conclusion makes the whole implication true we would not be able to say general facts whose converse is false, both in mathematics and in real life. An example from Wikipedia:
The converse of the intermediate value theorem is false: roughly speaking, if for all a, b and every u between f(a) and f(b) there exists a c such that f(c) = u, it does not follow that f is continuous.
Still another example: If f : A -> B and g : B -> C are injections, then so is the composition g o f, but not conversely (if g o f is injective, then so is f, but not necessarily g).
Other concepts of implications have been considered in philosophy and logic. See this Wikipedia section, for example. You can probably find a lot of information in the Stanford Encyclopedia of Philosophy.
Fascinating Thanks for the link to that Wikipedia section, it basically laid out precisely the concerns I had with logical implication. I think my problem is that I get too philosophical when I should be more "hyper-literal" when dealing with mathematics. I'm slowly starting to get it. Thanks for all your help so far!
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http://math.stackexchange.com/questions/217166/use-undetermined-coefficients-to-find-a-particular-solution-for-y8y-8x-3
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# Use undetermined coefficients to find a particular solution for $y'''+8y'=-8x-3$
Guess is $y = Ax+B$.
$y''' = 0$
$y' = A$
Thus, the differential equation becomes:
$0 + 8(A) = -8x-3$
Where can I go from here? I can't find an explicit solution for A, and my work doesn't even involve the variable B. Any help?
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Well, you started with a guess and got stuck. Which kind of suggests that a different guess might be a way out.... Hint: You're going to need at least some $x$-term on the left-hand side, so maybe a polynomial with degree $\geq 2$ wouldn't be a bad idea. – fgp Oct 20 '12 at 0:49
## 1 Answer
You'll need $y_p = (Ax+B)x = Ax^2 + Bx$, since the characteristic eqn. of your ODE has $0$ as a root.
$$y'_p = 2Ax + B$$
$$y'''_p = 0$$
So we have:
$$0 + 8(2Ax + B) = -8x - 3$$
And you should be able to take it from there.
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http://mathoverflow.net/questions/15673?sort=votes
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## An unfamiliar (to me) form of Hensel’s Lemma
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In his very nice article
Peter Roquette, History of valuation theory. I. (English summary) Valuation theory and its applications, Vol. I (Saskatoon, SK, 1999), 291--355, Fields Inst. Commun., 32, Amer. Math. Soc., Providence, RI, 2002
Roquette states the following result, which he attributes to Kurschak:
Hensel-Kurschak Lemma: Let $(K,|\ |)$ be a complete, non-Archimedean normed field. Let $f(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \in K[x]$ be a polynomial. Assume (i) $f(x)$ is irreducible and (ii) $|a_0| \leq 1$. Then $|a_i| \leq 1$ for all $0 < i < n$.
He says that this result is today called Hensel's Lemma and that Hensel's standard proof applies.
This is an interesting result: Roquette explains how it can be used to give a very simple proof of the fact that, with $K$ as above, if $L/K$ is an algebraic field extension, there exists a unique norm on $L$ extending $| \ |$ on $K$. This is in fact the argument I gave in a course on local fields that I am currently teaching.
It was my initial thought that the Hensel-Kurschak Lemma would follow easily from one of the more standard forms of Hensel's Lemma. Indeed, in class last week I claimed that it would follow from
Hensel's Lemma, version 1: Let $(K,| \ |)$ be a complete non-Archimedean normed field with valuation ring $R$, and let $f(x) \in R[x]$ be a polynomial. If there exists $\alpha \in R$ such that $|f(\alpha)| < 1$ and $|f'(\alpha)| = 0$, then there exists $\beta \in R$ with $f(\beta) = 0$ and $|\alpha - \beta| < 1$.
Then in yesterday's class I went back and tried to prove this...without success. (I was not at my sharpest that day, and I don't at all mean to claim that it is not possible to deduce Hensel-Kurschak from HLv1; only that I tried the obvious thing -- rescale $f$ to make it a primitive polynomial -- and that after 5-10 minutes, neither I nor any of the students saw how to proceed.) I am now wondering if maybe I should be trying to deduce it from a different version of Hensel's Lemma (e.g. one of the versions which speaks explicitly about factorizations modulo the maximal ideal).
This brings me to a second question. There are of course many results which go by the name Hensel's Lemma. Nowadays we have the notion of a Henselian normed field, i.e., a non-Archimedean normed field in which the exended norm in any finite dimensional extension is unique. (There are many other equivalent conditions; that's rather the point.) Therefore, whenever I state a result -- let us restrict attention to results about univariate polynomials, to fix ideas -- as "Hensel's Lemma", I feel honorbound to inquire as to whether this result holds in a non-Archimedean normed field if and only if the field is Henselian, i.e., that it is equivalent to all the standard Hensel's Lemmata.
Is it true that the conclusion of the Hensel-Kurschak Lemma holds in a non-Archimedean valued field iff the field is Henselian?
More generally, what is a good, reasonably comprehensive reference for the various Hensel's Lemmata and their equivalence in the above sense?
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@Harry: if $(R,\mathfrak{m})$ is a valuation ring, it is $\mathfrak{m}$-adically separated iff it is Noetherian iff it is a DVR. So your first claim is not correct. It is true that Henselianity can be characterized in terms of etale morphisms (Milne, Etale Cohomology, Section 1.4), but what does this have to do with the Hensel-Kurschak Lemma? – Pete L. Clark Feb 18 2010 at 6:33
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@Harry Gindi: Sorry, I'm not going to play the disappearing comments game with you tonight. Please give an answer if you have one. – Pete L. Clark Feb 18 2010 at 6:45
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@Everyone: in case it was not apparent, this was a rather embarrassing question to ask: I should have worked this out myself, but I wanted a quick answer for my students' sake. In fact I got, in addition to exactly the information I wanted/needed, lots of other interesting stuff. The moral of the story is....what is it, exactly? You can use MO to help teach your graduate courses? :) – Pete L. Clark Feb 18 2010 at 13:53
## 3 Answers
What you say at the beginning of your post is right: Hensel-Kurschak's lemma may be deduced from some refined version of Hensel's lemma. Actually, it's what Neukirch does in Algebraic Number Theory (see chapter II, corollary 4.7). His proof relies on the following (see 4.6)
Hensel's lemma: Let $(K,|.|)$ be a complete valued field with valuation ring $R$, maximal ideal $\mathfrak{m}$. Let $f(x) \in R[x]$ be a primitive polynomial (ie $f\ne 0$ mod $\mathfrak{m}$). Suppose $f=\bar{g}\bar{h}$ mod $\mathfrak{m}$, with $\bar{g}$ and $\bar{h}$ relatively prime. Then you can lift $\bar{g}$ and $\bar{h}$ to polynomials $g$ and $h$ in $R[x]$ such that $\textrm{deg}(g)=\textrm{deg}(\bar{g})$ and $f=gh$.
Neukirch goes on with proving that the valuation on $K$ extends uniquely to any algebraic extension (see corollary 4.7), as you say Roquette does.
As regards your last question, you may want to have a look at chapter II, paragraph 6 (appropriately called Henselian Fields) in the book of Neukirch. His definition of Henselian field is that it should satisfy Hensel-Kurschak's lemma. In theorem 6.6, he shows that this property is equivalent to the unique extension of the valuation to algebraic extensions.
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Thanks, Jerome! I see now that if I guessed the right "standard" version of Hensel's Lemma, the proof would have come out easily. (Why did I choose the one that I did? Ridiculous but true answer: I hadn't talked about residue fields yet.) Thanks also for the reference for H-K is equivalent to Henselian. – Pete L. Clark Feb 18 2010 at 8:21
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A far more general result is the "non-archimedean inverse function theorem". I haven't looked at Roquette's reference, so maybe he is mentioning it. But it is something which I didn't really find in the standard number theory textbooks - probably you can find it in texts on $p$-adic analysis - and I learned it from my number theory professor last semester (Jean-Benoît Bost). This theorem is powerful - and I find it fascinating and surprising - and all versions of Hensel's lemma which one usually encounters while learning number theory are immediate consequences.
Let $K$ be a field, $\left| \cdot \right|$ a non-archimedean absolute value on $K$ for which $K$ is complete, $\mathcal{O}$ the associated valuation ring, $\mathcal{M}$ the maximal ideal, $\pi$ a uniformizer. Let $\Phi_i \in \mathcal{O}[X_1,\,\cdots,X_n]$ for $1 \leq i \leq n$ and consider the map $\Phi = (\Phi_1,\,\cdots,\Phi_n) : \mathcal{O}^n \to \mathcal{O}^n$. Let $J$ be the Jacobian $\det(\partial \Phi_i / \partial X_j) \in \mathcal{O}[X_1,\,\cdots,X_n]$.
Theorem. If $x_0 \in \mathcal{O}^n$, $y_0 = \Phi(x_0)$ and $J(x_0) \neq 0$, then for any $R \in (0, \left|J(x_0)\right|)$, $\Phi$ induces a bijection $$\overline{B}(x_0,R) \to y_0 + (D\Phi)(x_0) \overline{B}(0,R)$$ (where $D\Phi$ is the derivative we all know!) and furthermore we have a bijection $$B^\circ(x_0,\left|J(x_0)\right| \to y_0 + (D\Phi)(x_0) B^\circ(0,\left|J(x_0)\right|).$$
(I use the standard notations $\overline{B}$ and $B^\circ$ for closed and open balls respectively.)
The proof uses in an essential way the Picard fixed point theorem.
Corollary 1. Take $n = 1$, $\Phi_1 = P$, $x_0 = \alpha$, $\varepsilon \in (0,1)$. Suppose that $\left|P(\alpha)\right| \leq \varepsilon \left|P'(\alpha)\right|^2$. Then there exists a unique $\beta \in \mathcal{O}$ such that $P(\beta) = 0$ and $\left|\beta - \alpha\right| \leq \varepsilon \left|P'(\alpha)\right|$. (We take $R = \varepsilon \left|P'(\alpha)\right|$ in the first bijection.)
Hence, as a special case, if $\left|P(\alpha)\right| < \left|P'(\alpha)\right|^2$, we find $\left|\beta - \alpha\right| < \left|P'(\alpha)\right|$.
As an even more special case, if $P'(\alpha) \in \mathcal{O}^\times$ and $\left|P'(\alpha)\right| <1$, there exists $\beta \in \mathcal{O}$ such that $P(\beta) = 0$ and $\left|\beta - \alpha\right| < 1$. Restating this in terms of the residue field: a simple zero in the residue field can be lifted to a real zero in $\mathcal{O}$. This is the really known version of Hensel's lemma, I guess.
[Definition: the Gauss norm of a polynomial with coefficients in $K$ is defined as the maximum of the absolute values of its coefficients. It is very easy to check that the Gauss norm is multiplicative.]
Corollary 2. Take $f,g,h \in \mathcal{O}[X]$ such that $\deg g = n$, $\deg h = m$ and $\deg f = \deg g + \deg h = n + m$. Assume that there exists $\varepsilon \in (0,1)$ such that $\left\|f - gh\right\| \leq \varepsilon\left|\text{Res}(g,h)\right|^2$ and $\deg(f - gh) \leq m + n - 1$. Then there exist $G, H \in \mathcal{O}[X]$ such that $f = GH$, $\deg(G - g) \leq n - 1$, $\deg(H - h) \leq m - 1$, and also $\left\|G - g\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$ and $\left\|H - h\right\| \leq \varepsilon \left|\text{Res}(g,h)\right|$. (Obviously $\text{Res}$ denotes the resultant here, and $\left\|\cdot\right\|$ the Gauss norm.)
To prove this: write $G = g + \xi$ and $H = h + \eta$ where $\xi$ and $\eta$ are polynomials with coefficients in $\mathcal{O}$ and have degrees $\leq n - 1$ and $\leq m - 1$ respectively. Then $f = GH$ if and only if $f = (g + \xi)(h + \eta)$. It can be seen as a map from $\mathcal{O}^n \times \mathcal{O}^m \to \mathcal{O}^{n + m}$ given by polynomials. So consider the map $\Phi: (\xi, \eta) \mapsto (g + \xi)(h + \eta) - f$. We have also $\text{Res}(g,h) = \det((\xi, \eta) \mapsto g \xi + h \eta))$. It is easy to see that the theorem above then gives the result.
As a corollary: if $f$, $g$ and $h$ satisfy $\overline{f} = \overline{g} \overline{h}$ - where $\overline{f}$ is $f$ reduced modulo $\mathcal{M}$ et cetera - and if $\overline{g}$ and $\overline{h}$ are coprime (this is a condition on the resultant!) then there exist $G,H \in O[X]$ satisfying the following conditions: $f = GH$, $\deg(G - g) \leq n - 1$, $\deg(H - h)\leq m - 1$, $\overline{G} = g$ and $\overline{H} = h$. Hence "a factorization over the residue field lifts to a factorization over $\mathcal{O}$" (under the right conditions).
Corollary 3. Finally, let us come to the motivation for the question: the more general result is that if $P \in K[X]$ is irreducible, then $\left\|P\right\|$ (Gauss) is the maximum of the absolute values of the leading coefficient and the constant coefficient. (As a special case, we find the result which Pete L. Clark cites as the Hensel-Kurschak lemma.)
Indeed, let $P(X) = \sum_{i = 0}^n a_i X^{n - i} \in K[X]$. Suppose WLOG that $\left\|P\right\| = 1$. Let $\mathbb{F}$ be the residue field and let $\overline{P}$ be the image of $P$ modulo $\mathcal{M}$. Set $r = \min \{n : \overline{a_{n - r}} \neq 0\}$. Then we have in the residue field the factorization $\overline{P}(X) = X^r \left(\overline{a_{n - r}} + \overline{a_{n - r - 1}}X + \cdots + \overline{a_0} X^{n - r}\right)$ and we can lift the factorization by Corollary 2, contradicting irreducibility.
I know this is quite some digression; but I find the whole discussion about the various forms of Hensel's lemma very interesting, and I thought this could add something to the discussion.
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@AS: Yes, it's interesting! [When you say Ribenboim, do you really mean Ribenboim -- as in the paper that Franz cited -- or Roquette, as in the paper that I cited?] I will say though that I do not believe that what you have written is the most general possible Hensel's Lemma, a quick corollary of my belief that the most general possible HL does not exist. For instance, this version of HL applies to a complete valuation ring; in (e.g.) Eisenbud's book there is a form of HL which applies to a complete, m-adically separated local ring. Neither class of rings contains the other... – Pete L. Clark Feb 18 2010 at 13:02
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...(See my first comment above). As a working arithmetic geometer, the version of HL that I use most often is the following: if K is a complete local field with valuation ring R and residue field k and X/R is a finite-type proper regular flat R-scheme, then the image of the reduction map X(K) = X(R) -> X(k) is precisely the set of smooth points in X(k). (This version is a consequence of Bost's nice formulation.) – Pete L. Clark Feb 18 2010 at 13:11
Hm, I like your last statement (I am not yet an arithmetic geometer :)). Could you give me some reference for it? In fact I expected to find this somewhere in Liu's book, but my first attempt to find it (using the index of the book) has been unsuccessful. – Wanderer Feb 18 2010 at 13:24
@AS: for instance, it is Lemma 9 in math.uga.edu/~pete/plclarkarxiv7.pdf. If you look there, you will be referred back to a 1985 paper of Jordan and Livne on Shimura curves, which is where I first learned about it (since my thesis work was on Shimura curves). I think the result is known to most people who seriously study curves over DVRs, so it should go back to the 1960s if not earlier. – Pete L. Clark Feb 18 2010 at 13:39
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@AS: That's a great answer! Let me just point out that with virtually the same method, you can prove the same results for a larger class of fields (thus proving they are Henselian), namely fraction fields of local rings of analytic spaces. They are not Banach rings but inductive limits of them (for any r>0, take the completion of the ring of functions that converge on the disc of radius r with center at your point). If your space in Archimedean though, you won't have such a precise form of the theorem (you can say exactly what the radii are only thanks to the ultrametric triangle inequality). – Jérôme Poineau Feb 19 2010 at 16:49
show 3 more comments
An exposition of various versions of Hensel's Lemma can be found in
• P. Ribenboim, Equivalent forms of Hensel's lemma, Expos. Math. 3 (1985), 3-24
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+1: So I learned that this journal exists earlier this week and now for a completely unrelated question I get referred to a paper in this journal. It's amazing to me how often this sort of thing happens. – Pete L. Clark Feb 18 2010 at 18:56
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http://mathoverflow.net/questions/17965?sort=newest
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## Does the category Monoid of monoids have finite coproducts?
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Does the category Monoid of monoids have finite coproducts?
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I rolled this question back to its previous version. Someone edited it to simply read aaaaaaaaaaaaaaaaaaaaaaaaaaa. (Which is presumably why it attracted votes to close.) – David Speyer Mar 13 2010 at 19:04
@David: I haven't (yet) voted to close but the question immediately reminded me of mathoverflow.net/questions/18026/… – Yemon Choi Mar 13 2010 at 19:55
I should add that the present question is a bit more interesting than the one I linked to (IMHO). – Yemon Choi Mar 13 2010 at 20:00
There's something a bit strange going on here. After I answered, someone else posted a one-"sentence" answer that sounded like a drunk person yelling nonsense. It must have been multiply flagged then deleted by the moderators. – Tom Leinster Mar 13 2010 at 20:04
Yes, the drunken shouting was deleted. – David Speyer Mar 13 2010 at 23:28
## 1 Answer
Yes. More generally, any category of algebras (in the sense of universal algebra), such as groups, rings, vector space, Lie algebras, ..., has all (small) limits and colimits. See for instance p.210 (end of section IX.1) of Mac Lane's book Categories for the Working Mathematician.
Explicitly, the initial monoid ("0-fold coproduct of monoids") is the one-element monoid. The coproduct $A * B$ of two monoids $A, B$ is constructed similarly to the coproduct of two groups (often called their "free product"). That is, it's the free monoid generated by all the elements of $A$ together with all the elements of $B$, quotiented out by all the relations that hold in $A$ and all the relations that hold in $B$.
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http://mathoverflow.net/questions/15850?sort=votes
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characterization of cofibrations in CW-complexes with G-action
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Is there a condition for a $G$-equivariant map $X \to Y$ to be a cofibration of $G$-spaces? Here $X$ and $Y$ are CW complexes, the group $G$ is finite, and acts by cellular maps.
I am using the model structure on CW-complexes with G-action where the fibrations and weak equivalences are those maps which are fibrations, weak equivalences respectively when we forget the $G$ action.
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There is no model structure on the category of CW-complexes since it is not complete or cocomplete, but I presume you mean the model structure on all G-spaces. – Mike Shulman Feb 20 2010 at 1:37
Since this was already on the front page, I decided to retag with model-categories. – David White Aug 18 2011 at 22:41
1 Answer
In the model structure you describe, the cofibrations should be the retracts of the free relative G-cell maps: i.e., retracts of maps obtained by attaching cells of the form $G \times S^{n} \to G \times D^{n+1}$.
One way to see this is via the following general machine: There is an adjoint pair $$G \times -: \mathbf{Top} \leftrightarrow \mathbf{GTop}: forget$$ $\mathbf{Top}$ is a cofibrantly generated model category and one can check that this adjoint pair satisfies the conditions of the standard Lemma for transporting cofibrantly generated model structures across adjoint pairs (see e.g., Hirschorn's "Model categories and their localizations" Theorem 11.3.2). Thus, it gives rise to a model structure on $\mathbf{GTop}$ such that a map in $\mathbf{GTop}$ is an equivalence(resp. fibration) iff its image under the right adjoint (forget) is so. Moreover, the generating (acyclic/)cofibrations are precisely the images under the left adjoint ($G \times -$) of the generating (acyclic/)cofibrations in $\mathbf{Top}$. This yields the description of the cofibrations as retracts of (free G-)"cellular" maps.
Also, some context:
The model structure you describe (which I'd like to call "Spaces over BG") is a localization of a model structure "G-Spaces" (where the weak equivalences are maps inducing weak equivalences on all fixed point sets). An argument along the lines of the above constructs this other model structure and identifies its cofibrations with retracts of (arbitrary) relative G-cell maps: i.e., retracts of maps obtained by attaching cells of the form $G/H \times S^n \to G/H \times D^{n+1}$ for $H$ a closed subgroup of $G$.
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http://mathoverflow.net/questions/12364/integer-subset-that-only-occupies-p-1-2-equivalence-classes-mod-p
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## Integer subset that only occupies (p-1)/2 equivalence classes mod p?
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I'm not quite sure the best way to ask this, so bear with me: Does anyone know of a subset of integers such that, for any odd prime p, the subset only occupies (p-1)/2 equivalence classes mod p (and does so uniformly)?
For example, take the subset of squares. Elementary number theory shows that they (as quadratic residues) occupy (p+1)/2 equivalence classes mod p. But the answer to the above is not to take the non-residues since being a non-residue is a local property, not a property of an integer.
It is possible to construct such a set of integers one element at a time in an ad hoc manner using some initial members, a whole lot of CRT, and making a somewhat arbitrary choice at each step. But is there a more well-known'' set that has this property?
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CRT ? – Tom Leinster Jan 20 2010 at 2:05
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Chinese Remainder Theorem. – darij grinberg Jan 20 2010 at 2:10
The set of squares was given as an example of a subset that satisfies the desired property for (p+1)/2 equivalence classes. The question is if there is a similarly nice'' set that does so for (p-1)/2 equivalence classes. – Aeryk Jan 20 2010 at 2:43
I believe the large sieve should show that such a set has at most on order of sqrt(N) elements in [1..N]. – JSE Jan 20 2010 at 2:44
## 3 Answers
See section 4.3 of Helfgott and Venkatesh, "How small must ill-distributed sets be?"
for an example of a subset of [1..N] of size about log N with small projections onto Z/pZ, and section 4.2 for a "guess" about what such subsets might look like in general. They speculate that such a set might have to be either very small (say, of size N^eps) or highly correlated with a "thin set," say, the values of a polynomial (i.e. x^2, as in the first case you describe.)
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I do not have an answer, but a suggestion: Consider looking at the primorials + 1. If you start late enough, there should be few equivalence classes hit (about 1/log p for large enough p) until you reach the primorial +1 that includes p. Also, one might do well to consider factorials or central binomial coefficients ( (2n!)/(n!)(n!) ) with a constant offset.
I do not call it an answer because I do not know how natural a sequence is wanted. But if this is cheesy enough to be downvoted, I would like to know why.
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Before I am told that this has too few residue classes, one may be able to change the offset recursively to give the desired result, not just arbitrarily. – Gerhard Paseman Jan 20 2010 at 7:53
And if you don't like that one, how about 1, 3, 15, 105, 945, 10395, 135135, etc. I think that can fit the bill even better than the original. – Gerhard Paseman Jan 20 2010 at 8:12
Oops. Maybe I should leave off the 1. – Gerhard Paseman Jan 20 2010 at 8:14
Darn. Doesn't work for 17. While it can be fixed, it looks more arbitrary after fixing (instead of multiplying by the next odd number, multiply by some small multiple of the next odd number). Time for bed. – Gerhard Paseman Jan 20 2010 at 8:25
Here is something I noticed - no idea if it helps:
Suppose `$A\subset\mathbb{Z}$` has the property that for each odd prime $p$ and $\phi_p:\mathbb{Z}\rightarrow\mathbb{Z}/p\mathbb{Z}$, we have $|\phi_p(A)|=\frac{p-1}{2}$. Consider the map $f_p:\mathbb{Z}/p\mathbb{Z}\rightarrow\mathbb{Z}/p\mathbb{Z}$ with $f_p(x)=x^2$. Then $|f_p(\phi_p(A))|\leq\frac{p-1}{2}$. Since there are $\frac{p+1}{2}$ quadratic residues mod $p$, we must have that for each odd prime $p$, there is an $x_p\in\mathbb{Z}/p\mathbb{Z}$ such that $a^2\not\equiv x_p^2\bmod p$ for all $a\in A$ (which is $\Leftrightarrow$ $a\not\equiv \pm x_p\bmod p$).
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"I don't think the result is actually true"---you mean you don't think such a set exists? I think one can easily construct one using CRT. You build it recursively. The point is that if it has size n, and has size at most (p-1)/2 mod p for all primes p, then you ignore all the primes bigger than 2n+3 because if you throw in another element it will still be too small to fill up (p-1)/2 classes, and for all primes <= 2n+3 you make sure that the new element is in a new congruence class mod p if we're still too small mod p, and is in an old congruence class mod p if we've got to the (p-1)/2 limit. – Kevin Buzzard Jan 20 2010 at 12:45
One can even make this uniform, because, the moment one has hit (p-1)/2 classes mod p one can ensure that all the rest of the numbers hit each cong class mod p equally, by making the new numbers "take it in turns" with the classes. – Kevin Buzzard Jan 20 2010 at 12:47
I don't understand your comment that since A is infinite, the set of odd primes dividing some a in A must be infinite. Perhaps it is true, but it takes an argument. Something like that there is some p=4k+1 so that the primes are all simultaneously 4th powers mod p. Are the primes where q is a 4th power determined by residues mod cq for some c? – Douglas Zare Jan 20 2010 at 15:13
Wow, thanks for pointing out all those mistakes (and explaining how to create such a set). – Zev Chonoles Jan 20 2010 at 15:24
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http://physics.stackexchange.com/questions/19886/where-to-find-cross-section-data-for-e-p-p-e?answertab=active
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Where to find cross section data for e- + p -> p + e-?
Where to find cross section data for e- + p -> p + e-?
PDG's cross section data listing does not include it.
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At what energies and momentum transfers? Also do you have a polarized beam or target? – dmckee♦ Jan 23 '12 at 15:59
Any energy/momentum transfer for which data is available. – Computist Jan 23 '12 at 20:41
The point is that data are available over very wide ranges in energy. SLAC ran at up to $50\text{ GeV}$ electron beams. My dissertation project to $\mathrm{H}(e,e'p)$ at $5.0$ and $5.5\text{ GeV}$ beams and $Q^2$s between $3.3$ and $8.1\text{ GeV}^2$. – dmckee♦ Jan 23 '12 at 21:57
2 Answers
you can search the HEPDATA database at http://www.slac.stanford.edu/spires/hepdata/ with the query string
````[reac = e- p --> e- p]
````
and the first result will be:
"Jefferson Lab. Measurement of the elastic electron-proton cross section in the Q*2 range from 0.4 to 5.5 GeV*2"
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The answer is highly dependent on the scale of the momentum transfer.
The figure of merit is $Q^2$ is the squared momentum transfer, and in some regimes the missing energy $\omega = \epsilon' - \epsilon$. The formalism is usually developed in the lab frame with a stationary proton target and a energetic electron beam. We write $Q = \mathbf{k}' - \mathbf{k}$ and $\mathbf{k} = (\epsilon,\vec{k})$ and $\mathbf{k}' = (\epsilon',\vec{k}')$ are the four momentum of the incident and scattered electron respectively.
• If $Q^2 \ll m_p^2$ then you can treat the proton as a point particle to first order and you can simple look this up. In the upolarized case you use the Mott cross-section.
• For Q^2 on the same order of magnitude at the proton mass squared the problem is complicated enough that one typically uses a parameterized experiment results in the shape of a set of "form factors" (note that the formalism typically used at medium energies is different from but equivalent to that used at high energies). The JLAB results that luksen links to are among the highest precision available at this time.
In nuclear physics parlance you get $$\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left( \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \right)_{\text{Mott}} \frac{Q^2}{\left|\vec{q}\right|^2} \left[ G^2_\mathrm{E}(Q^2) + \tau \epsilon^{-1}G^2_\mathrm{M}(Q^2)\right] .$$
To a first approximation you can use the dipole form for the form factors $$G_\mathrm{M} \approx \mu G_\mathrm{E} \approx \mu \left( 1 + \frac{Q^2}{0.71\text{ GeV}^2}\right)^{-2} ,$$ where $\mu$ is the magnetic moment of the proton.
• For $Q^2 \gg m_p^2$ you are in the deep inelastic scattering regime and can treat the proton at a collection of bound partons. The phrase you're looking for is "structure functions" (PDF link).
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http://mathoverflow.net/questions/3204?sort=oldest
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## Does any method of summing divergent series work on the harmonic series?
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It's sort of folklore (as exemplified by this old post at The Everything Seminar) that none of the common techniques for summing divergent series work to give a meaningful value to the harmonic series, and it's also sort of folklore (although I can't remember where I heard this) that the harmonic series is more or less the only important series with this property.
What other methods besides analytic continuation and zeta regularization exist for summing divergent series? Do they work on the harmonic series? And are there other well-known series which also don't have obvious regularizations?
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Would you include "1+2+4+8+16+...=-1" as a "meaningful" summation? – Andrew Critch Oct 29 2009 at 3:52
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Sure: it makes sense both 2-adically and as an example of analytic continuation. – Qiaochu Yuan Oct 29 2009 at 4:22
## 7 Answers
One common regularization method that wasn't mentioned in the Everything Seminar post is to take the constant term of a meromorphic continuation. While the Riemann zeta function has a simple pole at 1, the constant term of the Laurent series expansion is the Euler-Mascheroni constant gamma = 0.5772156649...
It is reasonable to claim that most divergent series don't have interesting or natural regularizations, but you could also reasonably claim that most divergent series aren't interesting. Any function with extremely rapid growth (e.g., the Busy Beaver function) is unlikely to have a sum that is regularizable in a natural way.
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When studying the algebraic properties of MZV (eg. in "Algebraic Aspects of Multiple Zeta Values", math/0309425) Michael Hoffman indeed calls treating \zeta(1) as Euler's Gamma a "happy choice" (after Theorem 3.5). – Armin Straub Oct 29 2009 at 12:41
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I have a sneaking suspicion that anything that works on 1/2 + 1/3 + 1/5 + ... will probably also work on the harmonic series, although I certainly don't have any hard reasoning to back this up -- just that it doesn't have nice local properties or nice global properties, much like the harmonic series.
But I sort of hope I'm wrong -- I'd be very interested to see what a regularization for this series looks like!
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Incidentally, the best text on such questions is Hardy's last book, Divergent Series.
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There are other sums with no good summation: for example 1+1+1+... Any decent method of summation would yield S=1+S.
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You don't think that 1+1+1+... = -1/2 has some decency? (Of course, that's \zeta(0).) – Armin Straub Oct 29 2009 at 12:23
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Hmmm. Perhaps I was too strong there. I tend to assume that a summation method should obey a_1+a_2+...=0+a_1+a_2+..., which zeta regularization does not. But I can't make a strong argument for that assumption. – David Speyer Oct 29 2009 at 12:27
By the way the proposition that $1+1+1+... = -1/2$ was first made by Euler. – Andrew Aug 16 2011 at 20:31
The series 1/2 + 1/3 + 1/5 + ... (the sum of reciprocals of the primes) mentioned by harrison "sums to log log ∞"; more formally,
(1/2 + 1/3 + 1/5 + 1/7 + ... + 1/n) ~ log log n
where ~ has the usual meaning: f(n)~g(n) if lim (n -> infty) f(n)/g(n) = 1.
The nth partial sum of the harmonic series, 1 + 1/2 + 1/3 + ... + 1/n, diverges like log n.
Perhaps sums which diverge "logarithmically fast" are in general problematic, and the harmonic series is just the canonical example of such a series.
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I'm not allowed to post a comment, but in reply to Michael Lugo's post and as a followup to Scott Carnahan, the prime harmonic series can be regularized in analogy with $1 + 1/2 + 1/3 + 1/4 + \ldots$ "$=$" $\gamma$, giving the Mertens constant. See the prime zeta function for more information.
In this case it's not "meromorphic continuation" as the singularity is logarithmic. This leads to the followup question: is there a practical difference, and is there a general theory for the logarithmic (or even more general, e.g. multiply nested logarithmic) case? The prime zeta function has some interesting properties, such as having a natural boundary of analyticity at $\Re(s) = 0$.
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Let $w$ be a state on the quotient C$^*$-algebra `$\ell_\infty / c_0$` (bounded sequences quotient out convergent to zero sequences). Then the functional ```$$
\mathrm{Tr}_w(A) = w ( \{ \frac{1}{\log (1+n)} \sum_{j=1}^n \lambda(n,A) \}_{n=1}^\infty )
$$``` is a trace on the ideal of compact operators (on a separable Hilbert space) such that $\mu(n,A) = O(n^{-1})$, $n \geq 1$. Here $\lambda$ denotes the sequence of eigenvalues of the compact operator $A$ ordered so that the sequence of absolute values $| \lambda |$ is a decreasing sequence, and $\mu$ denotes the sequence of singular values (eigenvalues of the absolute value of $A$). If `$A_{\mathrm{harmonic}} = \mathrm{diag}(n^{-1})$` (any diagonal operator with the harmonic series as the diagonal) then `$\mathrm{Tr}_w(A_{\mathrm{harmonic}})=1$`. This is a regularisation of the harmonic series.
Traces on compact operators, thinking of compact operators as noncommutative generalisations of convergent to zero sequences, form summing procedures on these "noncommutative `$c_0$` sequences". The trace `$\mathrm{Tr}_w$` above is called a Dixmier trace, after the French mathematician Jacques Dixmier who described it in 1968. It has been popularised by Alain Connes in his version of Noncommutative Geometry (Academic Press, 1994). Dixmier traces are not the only traces on the ideal of compact operators such that $\mu(n,A) = O(n^{-1})$, and there exist other traces $\varphi$ such that `$\varphi(A_{\mathrm{harmonic}}) = 1$`. Dixmier traces generalise the zeta function residue regularisation and the high temperature (or short time) heat kernel regularisation. Thus the zeta function residue regularisation is not the only regularisation possible.
There exist many traces defined on certain ideals besides just the canonical trace on the trace class operators (trace class operators are the noncommutative version of the summable sequences `$\ell_1$`). Deep results are known about which ideals admit non-trivial traces, which translates as meaning which rates of divergence (of convergent to zero sequences) admit a non-trivial summing procedure. See the book "Singular Traces", De Gruyter 2012 (admission of vested interest: I am one of the authors). The harmonic series fortunately admits a rich non-trivial range of summing procedures. Contrast with `$\ell_p$` sequences for $p > 1$ whose associated ideals have no non-trivial traces, and sequences `$O(n^{-p})$`, $p > 1$, whose associated ideals also have no non-trivial traces.
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http://www.physicsforums.com/showthread.php?p=4269366
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## The wrong turn of string theory: our world is SUSY at low energies
The peculiar arrangement of SU(4), or U(1)xSU(3) multiplets noticed in the Koide thread
http://www.physicsforums.com/showthr...=551549&page=6
could be related to the problems to put the higgs scalar under the same symmetries that the other scalars in the sboostrap.
Remember that we had to our disposal three scalars from the 15 and other three from the 15 irreps of SU(5). In our quark mnemonics, it is uu, uc, cc, uu, uc, cc (using the underscore to mean antiparticle). For such thing to be able to produce integer uncoloured charges, we need the mass/higgs mechanism to be blind to colour and blind to B-L, so that all the electric charge of these objects come from the electroweak isospin. Thus here is the first connection to the other thread: the multiplets of equal mass are for the charges for which the sBootstrap Higgs, if it is there, needs to be blind.
The second connection is even foggier: in the other thread, either the strange quark or the muon seem to need an opposite quantum number in order to fit in a SU(4) multiplet. Here it is either the up quark or the charm quark which seem to need some opposite value to sum zero in the uc combination.
Two recent papers, by authors already mentioned in this thread, which derive a Higgs sector in a sbootstrap-friendly way: Bruno Machet continues his series "Unlocking the Standard Model" (see #149), in which the idea seems to be that the Higgs will come from pion-like vevs. As discussed e.g. in #151, in a Higgsless SM, the W and Z will still acquire masses from pion vevs, but at the wrong energy scale. Machet nonetheless wants a version of this to work. In this, his third paper in the series, he considers two generations of quarks, and claims to get the Cabibbo angle from his Higgs-like condensates. Presumably future work will aim to get the whole CKM matrix from the quark bilinears of a three-generation model. Of the multitude of scalar and pseudoscalar mesons that appear, he states (page 4) that some of the scalars will be the Higgs, and the rest should correspond to the observed mesons. Kitano and Nakai's "Emergent Higgs from extra dimensions" aims to get the Higgs (and the masses of the Higgs and the top) from a deconstructed compactification of the d=6 (2,0) theory to four dimensions. This paper is certainly replete with connections to interesting topics. The (2,0) theory is the worldvolume theory of the M5-brane, so it's central to current advances in theoretical QFT. Their deconstructed version (deconstruction here means that the extra dimensions are approximated by a lattice, so e.g. a circle becomes a ring of sites with a copy of the d=4 SM fields at each site, coupled via the links in the ring, as in a quiver theory) is said to resemble topcolor (see page 3). There's much more I could talk about and I may have to return to this paper. But for now I'll remark on the possibility that perhaps something like Machet's model, which naively shouldn't work, could be produced by a Kitano-Nakai scenario, in which new strong couplings occur at high energy. "As in the Nambu–Jona-Lasinio model for the chiral symmetry breaking, whether or not a condensation forms depends crucially on how the theory is cut-off, and thus discussion requires a UV completion of the theory."
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Quote by arivero The peculiar arrangement of SU(4), or U(1)xSU(3) multiplets noticed in the Koide thread http://www.physicsforums.com/showthr...=551549&page=6
Back to this, lets aproach diquark masses with the mass of the heaviest quark, or the QCD mass if it is heavier than the quarks themselves. Then we can add mesons and diquarks to the "SU(4) arrangement".
[tex]\begin{array}{lllll}
?, t_{rgb}& & & & \\
?, b_{rgb}& B^+,B_c^+ & bu, bc& bb, bs, bd & \\
\tau, c_{rgb} & D^+, D_s^+& sc,dc \\
\mu, s_{rgb} & \pi^+, K^+& su, du& ss, sd, dd \\
?, d_{rgb} \\
e, u_{rgb}\end{array}[/tex]
It is tempting to think that in this "midly broken susy", the two lower mass levels are actually massless, so that SUSY does not need to kept the pairing at the same mass; it could be that the partners of d are the charmed diquarks, while the partners of up have been lost in the same mixing that breaks t and c partners.
Adding neutrinos and the missed diquarks, the table is a bit more complex. With some small abuse of notation, we could write the "after mild breaking" sBootstrap as
[tex]\begin{array}{lllllll}
&\nu_?, t_{rgb}& & & & \\
&\nu_?, b_{rgb}& B^+,B_c^+ & bu, bc& bb, bs, bd & \eta_b, \stackrel{b\bar s,b\bar d}{\bar bs,\bar bd} \\
\stackrel{\bar c\bar c}{cc},\stackrel{\bar c\bar u}{cu}&\tau, c_{rgb} & D^+, D_s^+& sc,dc & & \eta_c, \stackrel{c\bar u}{\bar cu}\\
\stackrel{\bar u\bar u}{uu}&\mu, s_{rgb} & \pi^+, K^+& su, du& ss, sd, dd & K^0,\pi^0, \stackrel{s\bar d}{\bar sd}\\
&\nu_?, d_{rgb} \\
&e, u_{rgb}\end{array}[/tex]
It is sort of symmetric, in a pleasant way. Wish I knew what to do about it.
We can adapt an earlier idea for the sbootstrap to Pati-Salam. The earlier idea is that there is a fundamental QCD-like theory with six flavors of quark, five light and one heavy; the five light quarks form fermionic composites, "diquarkinos" and "mesinos"; and the mesinos are the leptons, while the diquarkinos mix with the fundamental quarks to give us the phenomenological quarks. For Pati-Salam sbootstrap, the prescription is almost the same, except that the leptons already exist as the "nth color" in the fundamental QCD-like theory, so in this version the mesinos are mixing with preexisting degrees of freedom, just like the diquarkinos. It's probably best to think of the fundamental theory as having N=1 supersymmetry (at least), and to think of these composites as superfields.
Blog Entries: 6 Recognitions: Gold Member http://higgs.ph.ed.ac.uk/sites/defau...s/Higgs_RR.pdf Rattazzi is near to discover the sBootstrap if he continues this kind of enquiries.
On the Koide thread we have started to discuss textures and symmetries that could produce the waterfall pattern, and it's beginning to sound like orthodox model-building. But it's still not clear to me how to naturally descend from the sbootstrap to the waterfall. Supersymmetric theories are more complicated, including their methods of mass generation, and the "super-paradigm" which in my opinion most resembles the sbootstrap - Seiberg duality - doesn't offer obvious concrete guidance. However, I have a few thoughts arising from one of the non-susy paradigms for modeling the masses. As described e.g. on page 2 here, one may imagine that SM yukawas arise from a democratic matrix plus a correction. The democratic matrix has eigenvalues (M,0,0), and the correction can make the smaller eigenvalues nonzero. So consider an approach to the sbootstrap in which we begin with six flavors of chiral superfield, and in which some fundamental, democratic mechanism of mass generation produces a single heavy flavor. Now suppose that the five light flavors form meson superfields which mix with the fundamental superfields, as previously posited. It seems that we then have a mass matrix which starts with SU(6) symmetry and then has a correction with SU(5) symmetry; something which is ripe for further symmetry-breaking, perhaps down to a waterfall pattern. There are still conceptual problems. The democratic matrix usually appears as a Yukawa matrix, but one doesn't usually think of the Higgs as fundamental in the sbootstrap. Also, the usual "five-flavor" logic of the sbootstrap is motivated by the fact that the top decays before it can hadronize; but that decay is mediated by the weak interaction, which doesn't yet play a role in the scenario above. There's also the problem that the combinatorics of the sbootstrap employs the electric charges of the quarks, but if we impose those from the beginning, then we can't have the exact SU(5) or SU(6) flavor symmetry. So there may need to be some conceptual tail-chasing before a logically coherent ordering and unfolding of the ingredients is found. On the other hand, I wonder if some version of the cascades discussed earlier in this thread (page 9, #132 forwards) can produce an iterated breakdown of symmetry in the mass matrix. We could start with one heavy quark and five light, then the diquarkinos and mesinos induce corrections to the mass matrix, which in turn affect the masses of the diquarkinos and mesinos, breaking the symmetry further. Also of interest: "Strongly Coupled Supersymmetry as the Possible Origin of Flavor".
Blog Entries: 6 Recognitions: Gold Member I have put around an example about how the supermultiplets could be, before the susy breaking. Surely it is not the right mix, but it could be a reference to try to build a pure susy model. http://vixra.org/abs/1302.0006 $$\begin{array}{||l|l|llll||} \hline \stackrel{\bar c\bar c}{cc}& \nu_2, b_{rgb}, e, u_{rgb}& B^\pm,B_c^\pm & \stackrel{\bar b\bar u}{bu}, \stackrel{\bar b\bar c}{bc} & \stackrel{\bar b \bar s}{bs}, \stackrel{\bar b\bar s}{bd} & B^0, B^0_c, \bar B^0, \bar B^0_c \\ \stackrel{\bar c\bar u}{cu}& \tau, c_{rgb} , \nu_3, d_{rgb}& D^\pm, D_s^\pm& \stackrel{\bar s\bar c}{sc},\stackrel{\bar d\bar c}{dc} & \stackrel{\bar b\bar b}{bb},\stackrel{\bar d\bar d}{dd} & \eta_b, \eta_c, D^0, \bar {D^0}\\ \stackrel{\bar u\bar u}{uu}& \mu, s_{rgb} , \nu_1, t_{rgb}& \pi^\pm, K^\pm& \stackrel{\bar s\bar u}{su}, \stackrel{\bar d\bar u}{du}& \stackrel{\bar s\bar s}{ss}, \stackrel{\bar s\bar d}{sd}& \eta_8, \pi^0, K^0, \bar K^0 \\ \hline \end{array}$$
A major conceptual problem for the sbootstrap has been, how to get elementary and composite fields in the same superfield. But I notice that the string concepts of "flavor branes" and "color branes" can bring them closer. The flavor branes would be labeled dusc... and the color branes rgb..., and a single quark is a string between a flavor brane and a color brane (e.g. a red up quark is a string between up flavor brane and red color brane); and a meson is a string between two flavor branes. And if we employ Pati-Salam, then all the leptons also have a color, the "fourth color". According to the sbootstrap, a lepton is the fermionic superpartner of some meson or quark-antiquark condensate. The immediate problem for achieving this within the framework above is that it seems to involve pairing up different types of strings. Usually, you suppose that the flavor branes form one stack, the color branes form a different stack, the two stacks lie at different angles in the extra dimensions, and there are three types of string: flavor-flavor, color-color, and flavor-color. As usual, each stack will have a corresponding symmetry (e.g. SU(N) for some N), the flavor-flavor strings will be singlets under the color group, the color-color strings (the bosonic states of which are the gluons) are singlets under the flavor group, and the flavor-color strings transform under both groups. Also, the flavor-color strings are found most naturally in the vicinity of the intersection between the flavor stack and the color stack, because that is where the distance is shortest and thus the tension is smallest. But flavor-flavor and color-color strings can be found anywhere within their respective stacks, because the branes are parallel and so the inter-brane distance is the same everywhere. To my mind this poses a major barrier to the idea of placing a flavor-color string and a flavor-flavor string in the same multiplet. What if, instead of using intersecting brane stacks, we just have one big stack, and then move the branes apart into two groups, while keeping them parallel? This is already a standard method of breaking a symmetry group - the gauge bosons corresponding to strings between the two parts of the stack are the ones that are heavy, because they are longer. Now we would have that Gflavor x Gcolor is a subgroup of Gbig, the symmetry group of the original, unseparated brane-stack. Then we would suppose that the branes of the big stack are separated from each other in the extra dimensions (while remaining parallel) in such a way as to produce the desired mass spectrum - with the flavor branes clustered together in one group, the color branes in another, and the distances within and between the groups tuned appropriately.
I'll sketch how something like this could work. We'll use nine D3-branes in a space of three large dimensions, and six small and compact dimensions. Geometrically it can be just like Kaluza-Klein, except that each local copy of the KK manifold has nine special points scattered throughout it, the places where the nine D3-branes pass through that copy of the KK space. Basically, we would think of three of the points as being close together, and the other six scattered around them in six-dimensional space. The three branes that are close together (in fact, on top of each other) are the color branes. Because they are on top of each other, the SU(3)color gauge symmetry is unbroken. But the other six branes are scattered around and the SU(6)flavor gauge symmetry is completely broken. The quark superfields are strings connecting the 3 coincident points with any of the 6 scattered points, and the meson superfields are strings connected the 6 scattered points with each other. And to get them into the same supermultiplets, you restore the symmetry by moving all 9 points so they are on top of each other. So far I've said nothing about the weak interaction, and in fact I think it will require a doubling of the branes - or of the flavor branes at least. For each flavor there will be two branes, a "left brane" and a "right brane", for the two chiral components. Once again, this is a quite standard idea. Hypercharge is no problem, it's just a particular U(1) subgroup. And I suppose we can hope that the desired arrangement of branes is produced dynamically, e.g. by relaxation from cosmological initial conditions. It's surely too much to hope for, that some version of this would actually work. But I think it's remarkable that mathematically, this is genuine orthodox string theory. You could define a particular geometry for the Type IIB string (which is the one that has D3-branes) and calculate its spectrum. edit: Wait, I forgot we were getting leptons from a fourth color. So there are four color branes, four "color points" in the KK space, but one of them is displaced a little from the others - the breaking of SU(4)color to SU(3)color. A single quark is a string connecting a flavor brane to an rgb color brane, and a lepton is a string connecting a flavor brane to the fourth color brane.
We have a number of threads right now on getting the Higgs mass from Planck-scale boundary conditions. The common idea is that there is no new physics between the weak scale and the Planck scale. The best-known version is that of Shaposhnikov and Wetterich (SW), who managed to land very close to the observed mass by postulating that the "neutrino minimal standard model + gravity" is "asymptotically safe". However, I think the most elegant proposal is the "conformal standard model" of Meissner and Nicolai, who observe that the classical theory is conformally invariant except for the quartic Higgs term, and who propose therefore that the fundamental theory has conformal symmetry and that this quartic term is generated by the conformal anomaly. I note that in the world of high theory now, the really interesting symmetry is superconformal symmetry, the combination of supersymmetry and conformal symmetry. And since the sBootstrap, like the conformal standard model, is an exercise in theoretical minimalism, I have to wonder if there could be a "superconformal standard model" combining both? Supersymmetry is normally regarded as wildly incompatible with the minimalist idea of "no new physics between weak scale and Planck scale". We already know that we need physics beyond the original standard model with massless neutrinos; the "neutrino minimal standard model" manages to obtain all this below the weak scale, though at the price of unnatural finetuning (dark matter comes from right-handed neutrinos with keV Majorana mass, left-handed neutrino masses from very small yukawas). One might suppose that including supersymmetry would be even harder, or just impossible. One approach would be supersplit supersymmetry: all the superpartners have Planck-scale masses. But what about the sBootstrap alternative: supersymmetry is there, but it's only very weakly broken? In a sense that's the longrunning theme of this thread - the quest for ways to embed the sBootstrap pattern within a genuinely supersymmetric theory. The gauginos are the main technical problem that I see. One possibility is that we can just do without them by using Sagnotti's type 0 string theory, which is nonsupersymmetric but arises from the superstring, and which can apparently inherit a degeneracy of boson-fermion masses. Armoni and Patella use type 0 open strings to construct a form of "hadronic supersymmetry" (pairing mesons and baryons) - see page 8 for their general remarks on the type 0 theory. Meanwhile, Elias Kiritsis has sought to obtain a holographic dual for (nonsusy) QCD using type 0 strings. We have discussed the mesinos from holographic QCD several times; perhaps a type-0 version of the brane-stack constructions I discussed here a few weeks ago, could provide a "non-susy sBootstrap" in which we have mesinos but not gauginos. So perhaps we might want a type-0 brane stack which classically has conformal symmetry, but in which the Fermi scale is anomalously generated (as in the conformal standard model). Meanwhile (bringing in ideas from the Koide thread), there's also a discrete S4 symmetry producing a Koide waterfall, with the top yukawa equal to 1 and the up yukawa equal to 0... The waterfall produces the quark mass ratios, the SW-like mechanism produces the Fermi scale. The leptons are fermionic open strings between the flavor branes in the brane stack (mesinos)... It's all still a delirium, but perhaps we're getting there.
Blog Entries: 6 Recognitions: Gold Member As for the relationship between the above folding and the S4 generalisation of Koide, I find that they are two solutions of the eight S4 simultaneus equations that seem relevant: $$\begin{array}{|ll|} \hline 3.64098 & 0 \\ 1.69854 & 1.69854 \\ 0.12195 & 0.12195 \\ \hline \end{array} \dots \begin{array}{|ll|} \hline b & u \\ d & c \\ s & t \\ \hline \end{array} \dots \begin{array}{|ll|} \hline 3.640 & 3.640 \\ 1.698 & 1.698 \\ 0.1219 & 174.1 \\ \hline \end{array}$$ The one on the left appears when looking for zero'ed solutions; the one on the right appears in the resolvent of the system when looking for zero-less solutions; so both of them are singled-out very specifically even if, being doubly degenerated, they are hidden under the carpet of a continuous spectrum of solutions. To be more specific: a S4-Koide system on the above "folded" quark pairings should be a set of eight simultaneous Koide equations, for all the possible combinations: bds, bdt, bcs, bct, uds, udt, ucs, uct. A double degenerated solution of such S4-Koide system lives naturally inside a continuum: the equation K(M1,M2,x)=0, with M1 and M2 being the degenerated masses, has multiple solutions for x, and any two of them can be used to build the non-degenerated pair of the folding. The solution in the left is one of the possible solutions having at least a zero; up to an scale factor, there are only four of them. I have scaled it to match with the solution in the right. The solution in the right is one of the solutions obtained by using the method of polynomial resolvents to solve the system of eight equations (actually, we fix a mass and then solve the four equations containing such fixed mass). It is scaled so that its higher mass coincides with the top mass. For details on the calculation of the solutions, please refer to the thread on Koide.
Some recent thoughts: As with mainstream supersymmetry, I see the sbootstrap's situation as still being one where there is such a multitude of possibilities that it is hard even to systematically enumerate them. The difference is that a mainstream susy model consists of a definite equation and a resulting parameter space that then gets squeezed by experiment, whereas a sbootstrap "possibility" consists of a list of numerical or structural patterns in known physics which are posited to have a cause, and then an "idea for a model" that could cause them. It may be that some part of sbootstrap lore is eventually realized within a genuinely well-defined model that will then make predictions for MSSM objects like gluinos, or it may be that it will be a "minimalist" model that is more like SM than MSSM. (As for the Koide waterfall, that is such a tight structure, it seems that any rigorous model that can reproduce it is going to be sharply predictive - but there may still be several, or even many, such models.) Today I want to report just another "idea for a model". It's really just a wacky "what if"; I don't know that such a model exists mathematically; but I'd never even look for it if I didn't have the schematic idea. The idea is just that there might be a brane model in which the top yukawa is close to 1 both in the far UV and in the far IR, and that this is due to a stringy "UV/IR connection". The reason to think about this is as follows. The discussion of whether the Higgs mass might be in a narrow metastable zone, has yielded the perspective that it might be worth considering the top yukawa and the Higgs quartic coupling at the same time. The latter goes to 0 in the UV, the latter goes very close to 1 in the IR. But Rodejohann and Zhang have observed that with massive neutrinos, the top yukawa can approach 1 at high energies as well (see pages 14 and 15; the minimum is roughly 0.5, reached at about 10^15 GeV). And high energies are where a coupling might naturally take a simple value like 1. So what if there's a brane model where the top yukawa is 1 in the far UV, for some relatively simple reason, and then it is also near 1 in the far IR, because of a UV/IR relation that we don't understand yet? String theory contains UV/IR relations (scroll halfway down for the discussion); none of them appear to be immediately applicable to this scenario; but such relations are far from being fully understood. At the same time, I think of Christopher Hill's recent papers (1 2, it's basically the same paper twice), in which he first restricts the SM to just the top and the Higgs, and then considers a novel symmetry transformation, which he likens to a degenerate form of susy. At high scales he ends up with the relation that the Higgs quartic equals half of the square of the top yukawa - which is not what I'm looking for. Then again, he also ends up with Higgs mass equals top mass, with the difference to be produced by higher-order corrections. So perhaps his model, already twisted away from ordinary supersymmetry, can be twisted a little further to yield a Rodejohann-Zhang RG flow for the top yukawa, as well as a Shaposhnikov-Wetterich boundary condition for the Higgs quartic. One might want to see whether this can all be embedded in something like the "minimal quiver standard model" (MQSM), which is not yet a brane model, but it is a sort of field theory that can arise as the low-energy limit of a brane model; and the MQSM is the simplest quiver model containing the SM. Finally, to round things out, one might seek to realize the sbootstrap's own deviant "version of susy" here too, perhaps by using one of the brane-based "ideas for a model" already discussed in this thread.
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http://mathoverflow.net/questions/28948?sort=votes
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## Gromov’s list of 7 constructions in differential topology
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At the 2010 Clay Research Conference, Gromov explained that we know of only 7 different methods for constructing smooth manifolds. Working from memory, and hence not necessarily respecting the order he used:
• Algebraic geometry (affine and projective varieties, ...)
• Lie groups (homogeneous spaces, ...)
• General position arguments (Morse theory, Pontryagin-Thom construction, ...)
• Solutions to PDE (Moduli spaces in gauge theory, Floer theory, ...)
• Surgery (Cut and paste techniques, ...)
• Markov processes
I realise that I only gave 6 constructions; this was the number of separate items listed on his slides, and since he failed to discuss this part, I am left to guess that he either listed two different constructions on one line, which I interpreted to be variants of the same construction, or that failed to include one altogether.
Question How does one construct a smooth manifold from Markov processes?
I asked Gromov after the talk for explanation, but due to the rudimentary nature of my Gromovian, I was unable to understand the answer. The only word I managed to parse is "hyperbolic," though I wouldn't put too much stock in that.
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Stupid naive question: Where does covering spaces, open book decompositions, triangulations, etc. fit in? If it's Morse theory, then isn't surgery Morse theory as well? – Daniel Moskovich Jun 21 2010 at 13:46
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I personally would put "covering spaces, open book decompositions, triangulations" into "cut and paste" and therefore "surgery". – Deane Yang Jun 21 2010 at 14:43
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Hi Mohammed, perhaps the Dunfield-Thurston random 3-manifolds are examples? arxiv.org/abs/math/0502567 Take an $N$-step random walk on a Cayley graph of the mapping class group of a hyperbolic surface, so as to produce a random mapping class; glue two handlebodies to get a random 3-manifold with a Heegaard splitting. This is surgery, but the randomness highlights certain features (random 3-manifold fundamental groups have many finite-index subgroups compared to groups with random balanced presentations). – Tim Perutz Jun 21 2010 at 15:40
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According to my notes, the 7 constructions Gromov listed are: triangulations & surgery; Lie groups and locally homogeneous spaces; algebraic equations; genericity & transversality; partitions and Markov spaces; solutions of elliptic variational problems; and moduli spaces. – Maxime Bourrigan Sep 23 2010 at 11:29
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+1 for `the rudimentary nature of my Gromovian'. – HW Sep 23 2010 at 14:39
show 3 more comments
## 2 Answers
Unfortunately I missed the talk, but on the other hand Gromov have just produced a new paper called
Manifolds : Where do we come from ? What are we ? Where are we going ?
It can be found on his web page. From the title I guess there could be some intersection with the talk. In particular in section 11 called Crystals, Liposomes and Drosophila Gromov is speaking about "Markov quotients". This sounds like a way to produce "spaces" (generalisation of manifolds, I guess).
http://www.ihes.fr/~gromov/PDF/manifolds-Poincare.pdf
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After I saw this posting, I asked Gromov for a copy of his talk. This is the paper he sent me. – Deane Yang Sep 23 2010 at 12:13
### 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 suspect (but am far from certain) that Gromov may be referring to the correspondence between symbolic and hyperbolic dynamics.
The idea is basically that the 0-1 matrix corresponding to the sparsity pattern of a stochastic matrix encodes a subshift of finite type or topological Markov chain. Usually, however, one goes from the hyperbolic dynamics to the Markov description via a Markov partition or section.
I am not aware of a way to go in the other direction in general, although placing certain conditions on the Markov process would facilitate the construction of a Markov partition (which can then be made as small as one likes), for which covering sets would constitute an atlas.
Update: So I did a little digging and came across a paper by Coornaert and Papadopoulos called "Symbolic coding for the geodesic flow associated to a word hyperbolic group" (Manuscripta Math. 109, 465–492 (2002), DOI 10.1007/s00229-002-0321-9, PDF available here). In it the authors discuss an idea of Gromov whereby a to each "word hyperbolic group" $\Gamma$ a space with a flow defined up to orbit equivalence is given: this flow is called the geodesic flow associated to $\Gamma$. I quote:
In the case where $\Gamma$ is the fundamental group of a compact Riemannian manifold $M$ of negative curvature, then $\Gamma$ is word hyperbolic and [the geodesic flow associated to $\Gamma$] is, up to orbit equivalence, the geodesic flow on the tangent bundle of $M$.
Nowhere, however, is it indicated that the space so constructed is generically a manifold. Still, this construction is quite closely associated with the ideas mentioned earlier, as the introduction to this paper points out.
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mathoverflow.net/questions/8916/… – Steve Huntsman Jun 21 2010 at 14:05
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Any chance you could elaborate on your answer and explicitly say where a manifold appears? – Deane Yang Jun 21 2010 at 14:45
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http://www.physicsforums.com/showthread.php?p=3848526
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Physics Forums
Differential Equation Model
1. The problem statement, all variables and given/known data
Here's a model for the balance owed on a loan with the following conditions:
* Interest accumulated on the loan at a rate of 5.24% per year
* The amount owed at the beginning of the loan was \$20,000.
* No payments were made on the loan for the first two years
* After two years, the loan was paid off at \$3,000 per year.
The model is given by the equations:
$\frac{dL}{dt} = 0.0524 \ L, \ \ \ 0<t<2$
$\frac{dL}{dt} = 0.0524 \ L \ - \ 3 , \ \ t>2$
(L is in thousands)
Using a direction field work out how large repayments should be if the loan is to be paid back in exactly 12 years (i.e., with exactly 10 years of repayments after the first two years with no repayments).
3. The attempt at a solution
Here is the direction field I made for this model
And the solution for the initial value L(0)=20 is
The root of this solution occurs at t=11.39, so 11 years is how long it will take to pay back the loan.
But how can we work out how large repayments should be if the loan is to be paid back in 12 years?
Any help is greatly appreciated.
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Recognitions: Homework Help Change the DE to dL/dt = 0.0524L - R for t > 2, and solve it with R as a symbolic parameter. Then find R that makes L(12) = 0. RGV
Quote by Ray Vickson Change the DE to dL/dt = 0.0524L - R for t > 2, and solve it with R as a symbolic parameter. Then find R that makes L(12) = 0. RGV
So, $L(12)= 0.0524 \times (12) - R = 0.6288 - R = 0 \implies R = 0.6288$
Is this what you meant?
Recognitions:
Homework Help
Differential Equation Model
Quote by roam So, $L(12)= 0.0524 \times (12) - R = 0.6288 - R = 0 \implies R = 0.6288$ Is this what you meant?
No, it is not what I said and not what I meant. Go back and read what I wrote. You need to solve the DE for t > 2.
RGV
Quote by Ray Vickson No, it is not what I said and not what I meant. Go back and read what I wrote. You need to solve the DE for t > 2. RGV
I didn't quite understand what you meant by treating R as a "symbolic parameter". I already found the value of R that gives 0. So we can write the equation as
dL/dt = 0.0524 L - 0.6288
I'm not sure how this helps us. Did you mean I have to first solve the DE using separation of variables while ignoring R? I appreciate it if you could maybe explain that a bit more clearly.
Recognitions: Gold Member It looks like you are using a phase plane plotter as well as the direction field. You can use it or the direction field to plot backwards as well as forwards in time! (And it looks like that is what your plotter is doing since your starting point is in the middle!)
Yes I've used the Matlab add-on called "dfield", my plot shows the initial condition y(0)=20. But how do can we use this plot to figure out how large repayments should be if the loan is to be paid back in exactly 12 years?
Recognitions: Homework Help I don't have access to Matlab, so I cannot offer technical advice. But if I were doing the question I would avoid the use of a phase plot and would, instead, just write down the formula for the DE solution; it would be a formula that has both R and t in it: L = f(t,R). Then I would set L=0 at t=12 and solve the equation to find R. I guess if you want to use phase plot methods you could start by guessing a value of R, then make the phase plot for that R value and trace out the solution. If L=0 occurs before t=12 our guessed value if R is too large, so we should decrease it a bit and start over. If L=0 occurs after t=12, our guessed R is too small, so we should increase it a bit and start over. This procedure would be horrible and would take forever, and that is why I would not use it unless I had hours of spare time and nothing better to do. RGV
Recognitions: Gold Member Your plot is actually predicting backwards from that t(0) - predicting your what your debt would have been over the previous 30+ years if your first eq. had been applicable over that time! Now you know where your starting t is for the 'retrodiction'. t = 12. It should be obvious that at your present R = 3 the slopefield will not take you back from (12, 0) to (0, 20) i.e. L = 20 but to something higher. If not obvious make a few trials and you'll soon see. So as you cannot get back to 20 at the right time with that R try different R till it comes right! You can as RV said calculate it mathematically, which is not very hard depending on what math you know. But using the plotter is not even math, it's common sense.
Quote by Ray Vickson I don't have access to Matlab, so I cannot offer technical advice. But if I were doing the question I would avoid the use of a phase plot and would, instead, just write down the formula for the DE solution; it would be a formula that has both R and t in it: L = f(t,R). Then I would set L=0 at t=12 and solve the equation to find R. I guess if you want to use phase plot methods you could start by guessing a value of R, then make the phase plot for that R value and trace out the solution. If L=0 occurs before t=12 our guessed value if R is too large, so we should decrease it a bit and start over. If L=0 occurs after t=12, our guessed R is too small, so we should increase it a bit and start over. This procedure would be horrible and would take forever, and that is why I would not use it unless I had hours of spare time and nothing better to do. RGV
I agree, that would take a lot of time. So could you explain to me how to find the right value mathematically without using the phase plots? (I didn't quite get your first post)
Recognitions:
Gold Member
Quote by roam I agree, that would take a lot of time. So could you explain to me how to find the right value without using the phase plots? (I didn't quite get your first post)
How did you get the continuous curve that is in your pic?
Quote by epenguin How did you get the continuous curve that is in your pic?
I just went to the dfield set-up window and typed in the initial condition x=20 when t=0, and it gave me the curve.
Recognitions:
Gold Member
Quote by roam I just went to the dfield set-up window and typed in the initial condition x=20 when t=0, and it gave me the curve.
It sounds like you have got a programme or calculator that (like most such) doesn't do only d-fields but plots phase planes or paths as well (fancy name for 'solves 1st order d.e.'s in 2 variables').
You have all you need.
(I have to go now.)
Recognitions:
Homework Help
Quote by roam I agree, that would take a lot of time. So could you explain to me how to find the right value mathematically without using the phase plots? (I didn't quite get your first post)
Since you titled your thread "differential equation model" I assumed you knew something about differential equations. Is that assumption wrong?
I really need more information from you in order to _guide_ you while avoiding doing your homework for you. What is the course? What is your background? How much calculus have you had? etc.
RGV
Quote by Ray Vickson Since you titled your thread "differential equation model" I assumed you knew something about differential equations. Is that assumption wrong? I really need more information from you in order to _guide_ you while avoiding doing your homework for you. What is the course? What is your background? How much calculus have you had? etc. RGV
I have done basic calculus and some linear algebra. But this is my first course in differential equations. So far I've learned how to solve DEs using the method of separation of variables and integrating factor.
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Quote by roam I have done basic calculus and some linear algebra. But this is my first course in differential equations. So far I've learned how to solve DEs using the method of separation of variables and integrating factor.
Well, that is plenty enough to solve the simple DE dL/dt = c*L - R with constants c and R. Just apply what you have been taught.
It may be that you are comfortable with solving the DE dL/dt = c*L but uncomfortable with the DE that has the extra '-R' on the right. Is that the case? If so, look instead at the DE for M = L - p, where p is some constant. You ought to be able to figure out what p value to choose in order to get rid of the constants on the right and have just dM/dt = c*M.
Alternatively, you can use an integrating factor.
RGV
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http://www.physicsforums.com/showthread.php?t=294168
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Physics Forums
## Christoffel symbols
Hi all!
I read about tensor analysis and came about following expressions, where also a questions arose which I cannot explain to me. Perhaps you could help me:
I: Consider the following expressions:
$$d\vec v=dc^k e^{(k)}$$
$$d\vec v=dc^k e_{(k)}$$
where:
$$dc^k=dv^k+v^t\Gamma_{wt}^k dx^w$$
$$dc_k=dv_k-v_t\Gamma_{wk}^t dx^w$$
Now, consider the covariant derivatives:
$$\frac{\partial c^k}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \frac{\partial x^w}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \delta^w_q=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{qt}^k$$
analagous:
(1)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_k}{\partial x^q}-v_t\Gamma_{qk}^t$$
So far so good, here I start transforming:
$$\frac{\partial c_k}{\partial x^q}=\frac{g_{kl}\partial c^l}{\partial x^q}=g_{kl}\frac{\partial c^l}{\partial x^q}=\frac{g_{kl}\partial v^l}{\partial x^q}+v^t\Gamma_{qt}^l g_{kl}=\frac{\partial v_l}{\partial x^q}+v^t\Gamma_{qtk}$$
As the second term looks different from the one above we continue transforming it:
$$v^t\Gamma_{qtk}=v^t\Gamma_{qk}^s g_{ts}=(t\rightarrow s, g_{ts}=g_{st})=v^s\Gamma_{qk}^t g_{ts}=v_t\Gamma_{qk}^t$$
so, we finally get:
(2)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_k}{\partial x^q}+v_t\Gamma_{qk}^t$$
By comparing (1) and (2) I miss a minus sign!
I suspect that the Christoffel symbol of first kind is antisymmetric and indices permute just like they do in the epsilon tensor and thereby generate a minus but I am not sure...
II: In both of the above daces of derivatives one uses dx^q as differential which is contravariant. Does it make sence to also use a covariant dx_q? Is in general differentiation of covariant vectors with respect to a covariant variable defined? (I suppose it must be, since you also differentiate a contravariantvector w.r.t. a contravariant variable)
III: And another question: Is the Kronecker delta symmetric in non-orthogonal coordinates
$$\delta^i_j=\delta^j_i$$ ???
If not, then which one of the two definitions is correct: (I´ve seen both in the net)
$$e^{i}e_j=\delta^j_i$$
or
$$e^je_i=\delta^j_i$$
I have also seen two types in which you define covariant vectors:
$$\vec v=v_ke^k$$ and $$\vec v=v_ke_k$$
Which one is correct, or do they just represent the same covariant vector once in the covariant and in the contravariant basis?
IV: And the last one: I haven´t seen a classification of the Christoffel symbol of this kind:
$$\Gamma^{kl}_m$$ Is it also symmetric in the upper indices?
Thanks a lot, I really appreciate your help!
marin
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Quote by Marin Hi all! I read about tensor analysis and came about following expressions, where also a questions arose which I cannot explain to me. Perhaps you could help me: I: Consider the following expressions: $$d\vec v=dc^k e^{(k)}$$ $$d\vec v=dc^k e_{(k)}$$ where: $$dc^k=dv^k+v^t\Gamma_{wt}^k dx^w$$ $$dc_k=dv_k-v_t\Gamma_{wk}^t dx^w$$ Now, consider the covariant derivatives: $$\frac{\partial c^k}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \frac{\partial x^w}{\partial x^q}=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{wt}^k \delta^w_q=\frac{\partial v^k}{\partial x^q}+v^t\Gamma_{qt}^k$$ analagous: (1)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_k}{\partial x^q}-v_t\Gamma_{qk}^t$$ So far so good, here I start transforming: $$\frac{\partial c_k}{\partial x^q}=\frac{g_{kl}\partial c^l}{\partial x^q}=g_{kl}\frac{\partial c^l}{\partial x^q}=\frac{g_{kl}\partial v^l}{\partial x^q}+v^t\Gamma_{qt}^l g_{kl}=\frac{\partial v_l}{\partial x^q}+v^t\Gamma_{qtk}$$ As the second term looks different from the one above we continue transforming it: $$v^t\Gamma_{qtk}=v^t\Gamma_{qk}^s g_{ts}=(t\rightarrow s, g_{ts}=g_{st})=v^s\Gamma_{qk}^t g_{ts}=v_t\Gamma_{qk}^t$$ so, we finally get: (2)$$\frac{\partial c_k}{\partial x^q}=\frac{\partial v_l}{\partial x^q}+v_t\Gamma_{qk}^t$$ By comparing (1) and (2) I miss a minus sign! I suspect that the Christoffel symbol of first kind is antisymmetric and indices permute just like they do in the epsilon tensor and thereby generate a minus but I am not sure...
Be careful how you "transform" the Christoffel symbols. They are not tensors and do not transform as tensors.
It is true that $\Gamma^{qk}_t= -\Gamma{tk}_q$ and that $\Gamma^{qk}_t= \Gamma^{kq}_t$.
[qote]II: In both of the above daces of derivatives one uses dx^q as differential which is contravariant. Does it make sence to also use a covariant dx_q? Is in general differentiation of covariant vectors with respect to a covariant variable defined? (I suppose it must be, since you also differentiate a contravariantvector w.r.t. a contravariant variable)[/quote]
Yes, you can use covariant $dx_q$ if you also lower the indices on the Christoffel symbols. $\Gamma_{ij, t}= g_{ki} g_{jq}\Gamma^{kq}_t$.
III: And another question: Is the Kronecker delta symmetric in non-orthogonal coordinates $$\delta^i_j=\delta^j_i$$ ???
The Kronecker delta is definded by $\delta_{ij}= 1$ if i= j, 0 if $i\ne j$, independent of the coordinate system, so, yes, it is always symmetric. (The metric tensor is, although dependent, of course, on the coordinate system, is also symmetric in all coordinate systems.)
If not, then which one of the two definitions is correct: (I´ve seen both in the net) $$e^{i}e_j=\delta^j_i$$ or $$e^je_i=\delta^j_i$$ I have also seen two types in which you define covariant vectors: $$\vec v=v_ke^k$$ and $$\vec v=v_ke_k$$ Which one is correct, or do they just represent the same covariant vector once in the covariant and in the contravariant basis?
They are really the same thing although the second would not make sense in the standard "Einstein summation convention" that we sum when the same index appears both as a superscript and a subscipt.
IV: And the last one: I haven´t seen a classification of the Christoffel symbol of this kind: $$\Gamma^{kl}_m$$ Is it also symmetric in the upper indices?
Yes, it is. That should be clear from the definition of the Christoffel symbols (of the first kind) in terms of derivatives of the metric tensor. What definition are you using?
Thanks a lot, I really appreciate your help! marin
You are welcome.
ok, now let me try and see: $$\Gamma^l_{qt}=\Gamma^l_{tq}=-\Gamma^q_{lt}=-\Gamma^q_{tl}=\Gamma^t_{lq}$$ and also: $$\Gamma^l_{qt}=-\Gamma^t_{ql}=-\Gamma^t_{lq}$$ seems to me like a contradiction.. Can you please point out the wrong equalities, so that I get a better understanding? thanks
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## Christoffel symbols
Quote by Marin ok, now let me try and see: $$\Gamma^l_{qt}=\Gamma^l_{tq}=-\Gamma^q_{lt}=-\Gamma^q_{tl}=\Gamma^t_{lq}$$ and also: $$\Gamma^l_{qt}=-\Gamma^t_{ql}=-\Gamma^t_{lq}$$ seems to me like a contradiction.. Can you please point out the wrong equalities, so that I get a better understanding? thanks
You are right. I wrote too fast $\Gamma^l_{qt}= \Gamma^l_{tq}$ but $\Gamma^t_{lq}\ne -\Gamma^q_{lt}$. What is true is that $\Gamma^l_{qt}+ \Gamma^q_{tl}+ \Gamma^t_{lq}= 0$. Notice that the three indices are "rotated"- the three even permutations of tlq.
I asked before what definition of the Krisstofel symbols you are using. The ones I am familiar with are
$$\Gamma_{ij,k}= \frac{1}{2}\left(\frac{\partial g_ik}{\partial x^j}+ \frac{\partial g_{jk}}{\partial x^i}- \frac{\partial g_{ij}}{\partial x^k}\right)$$
$$\Gamma^{i}_{jk}= g^{im}\Gamma_{jk,m}$$
The symmetry rules follow from that.
I also "use" the definition you stated, but haven´t studied it thoroughly yet, perhaps I had to, before I try to take on expressions. Anyway, I´ll go on trying to understand this stuff and some new and will pose some questions again. Thanks once again for the help!
ok, I tried the following transformation: $$\Gamma^w_{bt}=\Gamma^w_{bt}\delta^b_b=\Gamma^w_{bt}\delta^b_w\delta^w_b =\Gamma^w_{bt}\delta^b_w\delta^b_w=(\Gamma^w_{bt}\delta^b_w)\delta^b_w= \Gamma^w_{wt}\delta^b_w=\Gamma^b_{wt}=\Gamma^b_{tw}$$ But this implies that nothing changes with the Christoffel symbol if I permute the indices evenly. But then, HallsofIvy, the equation you posted does not hold any more.. I suppose I´ve done something wrong again, so could someone please help me :)
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# Cotangent bundle
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In mathematics, especially differential geometry, the cotangent bundle of a smooth manifold is the vector bundle of all the cotangent spaces at every point in the manifold. It may be described also as the dual bundle to the tangent bundle.
## The cotangent sheaf
Smooth sections of the cotangent bundle are differential one-forms.
### Definition of the cotangent sheaf
Let M×M be the Cartesian product of M with itself. The diagonal mapping Δ sends a point p in M to the point (p,p) of M×M. The image of Δ is called the diagonal. Let $\mathcal{I}$ be the sheaf of germs of smooth functions on M×M which vanish on the diagonal. Then the quotient sheaf $\mathcal{I}/\mathcal{I}^2$ consists of equivalence classes of functions which vanish on the diagonal modulo higher order terms. The cotangent sheaf is the pullback of this sheaf to M:
$\Gamma T^*M=\Delta^*(\mathcal{I}/\mathcal{I}^2).$
By Taylor's theorem, this is a locally free sheaf of modules with respect to the sheaf of germs of smooth functions of M. Thus it defines a vector bundle on M: the cotangent bundle.
## The cotangent bundle as phase space
Since the cotangent bundle X=T*M is a vector bundle, it can be regarded as a manifold in its own right. Because of the manner in which the definition of T*M relates to the differential topology of the base space M, X possesses a canonical one-form θ (also tautological one-form or symplectic potential). The exterior derivative of θ is a symplectic 2-form, out of which a non-degenerate volume form can be built for X. For example, as a result X is always an orientable manifold (meaning that the tangent bundle of X is an orientable vector bundle). A special set of coordinates can be defined on the cotangent bundle; these are called the canonical coordinates. Because cotangent bundles can be thought of as symplectic manifolds, any real function on the cotangent bundle can be interpreted to be a Hamiltonian; thus the cotangent bundle can be understood to be a phase space on which Hamiltonian mechanics plays out.
### The tautological one-form
Main article: Tautological one-form
The cotangent bundle carries a tautological one-form θ also known as the Poincaré 1-form or Liouville 1-form. (The form is also known as the canonical one-form, although this can sometimes lead to confusion.) This means that if we regard T*M as a manifold in its own right, there is a canonical section of the vector bundle T*(T*M) over T*M.
This section can be constructed in several ways. The most elementary method is to use local coordinates. Suppose that xi are local coordinates on the base manifold M. In terms of these base coordinates, there are fibre coordinates pi: a one-form at a particular point of T*M has the form pidxi (Einstein summation convention implied). So the manifold T*M itself carries local coordinates (xi,pi) where the x are coordinates on the base and the p are coordinates in the fibre. The canonical one-form is given in these coordinates by
$\theta_{(x,p)}=\sum_{{\mathfrak i}=1}^n p_idx^i.$
Intrinsically, the value of the canonical one-form in each fixed point of T*M is given as a pullback. Specifically, suppose that π : T*M → M is the projection of the bundle. Taking a point in Tx*M is the same as choosing of a point x in M and a one-form ω at x, and the tautological one-form θ assigns to the point (x, ω) the value
$\theta_{(x,\omega)}=\pi^*\omega.$
That is, for a vector v in the tangent bundle of the cotangent bundle, the application of the tautological one-form θ to v at (x, ω) is computed by projecting v into the tangent bundle at x using dπ : TT*M → TM and applying ω to this projection. Note that the tautological one-form is not a pullback of a one-form on the base M.
### Symplectic form
The cotangent bundle has a canonical symplectic 2-form on it, as an exterior derivative of the canonical one-form, the symplectic potential. Proving that this form is, indeed, symplectic can be done by noting that being symplectic is a local property: since the cotangent bundle is locally trivial, this definition need only be checked on $\mathbb{R}^n \times \mathbb{R}^n$. But there the one form defined is the sum of $y_{i}dx_i$, and the differential is the canonical symplectic form, the sum of $dy_i{\and}dx_i$.
### Phase space
If the manifold $M$ represents the set of possible positions in a dynamical system, then the cotangent bundle $\!\,T^{*}\!M$ can be thought of as the set of possible positions and momenta. For example, this is a way to describe the phase space of a pendulum. The state of the pendulum is determined by its position (an angle) and its momentum (or equivalently, its velocity, since its mass is not changing). The entire state space looks like a cylinder. The cylinder is the cotangent bundle of the circle. The above symplectic construction, along with an appropriate energy function, gives a complete determination of the physics of system. See Hamiltonian mechanics for more information, and the article on geodesic flow for an explicit construction of the Hamiltonian equations of motion.
## References
• Jurgen Jost, Riemannian Geometry and Geometric Analysis, (2002) Springer-Verlag, Berlin ISBN 3-540-63654-4 [Amazon-US | Amazon-UK].
• Ralph Abraham and Jerrold E. Marsden, Foundations of Mechanics, (1978) Benjamin-Cummings, London ISBN 0-8053-0102-X [Amazon-US | Amazon-UK].
• Stephanie Frank Singer, Symmetry in Mechanics: A Gentle Modern Introduction, (2001) Birkhauser, Boston.
Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Cotangent bundle", available in its original form here:
http://en.wikipedia.org/w/index.php?title=Cotangent_bundle
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http://physics.stackexchange.com/questions/20887/potential-at-a-point/20888
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# Potential at a point
What is the electric dipole moment of the charge distribution with $q$ at $(0,0,1)$, $q$ at $(0,0,-1)$ and $-2q$ at $(0,0,0)$? I would think that it is $\vec{0}$ by the definition $\vec{p}=\sum\limits_i \vec{r_i}q_i$. So would it follow that the potential field due to it be $0$? Since $V=k {\vec{p}\cdot \hat{r}\over r^2}$ where $\vec{r}$ is the position vector of a point at which we wish to evaluate the potential.
-
## 1 Answer
That formula only works for short dipoles. Here, you need to use the fudamental potential formula, $\frac{kq_1}{r}$. By this formula, we get potential due to $q$ as $\frac{kq}{2}$, and potential due to $-2q$ as $-2kq$. Net potential is $-\frac{3kq}{2}$.
When we say a short dipole, we mean that the distance between the charges should be negligible with respect to the surroundings.
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http://mathhelpforum.com/calculus/202453-integrate-riemann.html
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# Thread:
1. ## Integrate, Riemann.
Hi, i can't solve this, please if someone can solve and show me the steps (:
$\displaystyle \int_{1}^{e} \ln (x) dx $ P is the partition. $P:[q^{0};q^{1};q^{2};...;q^{n}]$ ; $\displaystyle q=e^{\frac{1}{n}}$
Thanks for all the answers (:
2. ## Re: Integrate, Riemann.
Originally Posted by SimpleMan
Hi, i can't solve this, please if someone can solve
What do you mean by "solve" here?
Originally Posted by SimpleMan
$\displaystyle \int_{1}^{e} \ln (x) dx $ P is the partition. $P:[q^{0};q^{1};q^{2};...;q^{n}]$ ; $\displaystyle q=e^{\frac{1}{n}}$
What do you mean by $[q^{0};q^{1};q^{2};...;q^{n}]$ ; $\displaystyle q=e^{\frac{1}{n}}$?
3. ## Re: Integrate, Riemann.
so let ln(x) be f-1(x) so ex=f(x)
F-1(x)=xf-1(x)-F(f-1(x))
so
$\displaystyle \int_{1}^{e} \ln (x) dx $= x(ln(x)- eln(x) = xln(x)-x evaluated at e and 1
which should be 1
if you haven't seen the equation before look at the geometry between a function and its inverse... they add up to make a rectangle.
I also don't know what the partition thingamabobs are but the answer is 1
4. ## Re: Integrate, Riemann.
Yes, that is easy to integrate but I suspect Simple Man is asking about the Riemann sum that will give that integral. The partition is $e^{0/n}=1, e^{1/n}, e^{2/n}, ..., e^{n/n}= e$. Those values for x make it easy to find the corresponding "f(x)": $ln(1)= 0, ln(e^{1/n})= \frac{1}{n}, ln(e^{2/n})= \frac{2}{n}, ..., ln(e)= 1$. However, it is also makes the interval lengths variable- the length of the first interval is $e^{1/n}- 1$, the second $e^{2/n}- e^{1/n}$, etc.
The Riemann sum will be $\frac{e^{2/n}- e^{1/n}}{n}+ \frac{2(e^{3/n}- e^{2/n})}{n}+ \frac{3(e^{4/n}- e^{3/n})}{n}+ \cdot\cdot\cdot+ \frac{n(e- e^{(n-1)/n}}{n}$ which is, of course, the same as
$\frac{1}{n}(ne- e^{(n-1)/n}- \cdot\cdot\cdot- e^{2/n}+ e^{1/n})$
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http://mathematica.stackexchange.com/questions/tagged/numerical-integration?sort=active&pagesize=15
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http://mathhelpforum.com/differential-equations/164349-first-order-nonlinear-de.html
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# Thread:
1. ## First-order nonlinear DE
Hello,
I tried searching similar DE-problems, but I couldn't find one. The problem is following differential equation:
$e^{x}y'+xe^{-y}=0$
So, it seems to be first-order nonlinear DE. But, I don't know how to solve it. I couldn't separe it, though I managed to separate simpler equations, but I think that's the right way to go.
Just got it to form,
$e^{-y}(e^{y+x}y'+x)=0$,
which obviously doesn't help anything...
Any help is appreciated. Thank you!
2. $y'=-\dfrac{xe^{-x}}{e^x}=-\dfrac{x}{e^{2x}}$
3. Originally Posted by Greg98
Hello,
I tried searching similar DE-problems, but I couldn't find one. The problem is following differential equation:
$e^{x}y'+xe^{-x}=0$
So, it seems to be first-order nonlinear DE. But, I don't know how to solve it. I couldn't separe it, though I managed to separate simpler equations, but I think that's the right way to go.
Just got it to form,
$e^{-y}(e^{y+x}y'+x)=0$,
which obviously doesn't help anything...
Any help is appreciated. Thank you!
Dear Greg98,
$e^{x}y'+xe^{-x}=0$
$\frac{dy}{dx}=-xe^{-2x}$
$y=-\int{xe^{-2x}}dx$
This integration could be done by integration by part method.....
Hope this helps.
4. Thanks for the help! Because of my omnipotent typing skills, I still need some advice (see original post). Sorry... I tried some logarithms to eliminate $e^{-y}$, but that didn't help.
5. $e^{x}y'+xe^{-y}=0$
$e^{x}y'+\dfrac{x}{e^{y}}=0$
divide both sides by $e^x$
$y'+\dfrac{x}{e^{x}\;e^{y}}=0$
multiply both sides by $e^y$
$y'\;e^y +\dfrac{x}{e^x}=0$
$e^y \frac{dy}{dx}=-\dfrac{x}{e^x}$
$e^y\;dy=-xe^{-x}\;dx$
finish...
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http://math.stackexchange.com/questions/309929/how-to-estimate-a-variables-upper-limit-with-95-confidence-level/309987
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# how to estimate a variable's upper limit with 95% confidence level?
Suppose I have a variable, $S(t)$, for stock price. So always have $S(t) > 0$.
$S(t)$ is a random variable, with some volatility $\sigma$ and trend.
Now the requirement is to estimate $\hat S(t+1)$ with at 95% confidence level, or: estimate $\hat S(t+1)$ based on $S(0)$, $S(1)$, ..., $S(t)$, so that 95% of the time $S(t+1) < \hat S(t+1)$.
If it's not for the 95% confidence level, it's easy, I could just use regression to genereate a formula something like $$S(t+1) = a_0 + a_1 S(t) + a_2 S(t-1) ... + a_m S(t-m+1)$$ .
But such estimation is only predicting the "trend", not considering the volatility and 95% confidence level.
If it's not for the trend, it's also easy, I could just get the 95% percentile value from history S:
$\hat S(t+1) = 95 %$ percentile value of $S(t), S(t-1), ... S(t-m)$
, maybe could choose $m = 36%$, using 3 years' data.
But such prediction would not cater the trend -- if there is a trend in the market that the Stock price rises, the estimation will always be lower than the future realized.
How could I have a $\hat S(t+1)$ that could cater both "trend" and "95% confidence level"?
Please notice here we didn't assume any parametric model about Stock price's distribution.
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## 1 Answer
The standard Kalman filter can be used for this. It makes some assumptions about Gaussianity of observations and updates, but it easily gives you a prediction for the next step expressed as a mean and variance. If you want to go beyond Gaussianity, there are some extensions ("unscented" Kalman filter etc) which you could go into.
EDIT: actually, not sure if it's easy to incorporate the "trend" aspect here. The basic application of Kalman filter would not do this, you simply specify the update matrix in advance. However, it might well be possible to use the framework with an unknown trend as one element in the update matrix, and fit that.
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thanks, after reading the wiki page i guess it works! – athos Feb 26 at 1:53
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http://mathoverflow.net/revisions/116759/list
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## Return to Answer
Post Made Community Wiki by S. Carnahan♦
4 added 232 characters in body
Another example from real analysis would be the question of the pointwise convergence of the Fourier series of a continuous function (defined on a given closed intervalinterval). Many people, including Dirichlet and even the master rigorist Weierstrass himself, believed that the Fourier series of such a function converges pointwise everywhere to the function itself. Some believed in clung on to this statement belief so strongly that they even viewed it as an infallible axiom.
Hence, one can imagine the great upset when, in 1876, Paul du Bois-Reymond proved that the existence of a continuous function exists whose Fourier series diverges at a point. The His proof is non-constructive , using and uses a method called the principle of condensation of singularities. I have absolutely no idea how the method works, but I do know of a very common proof that uses the Baire Category Theorem (using the Baire Category Theorem, one can also prove the existence of continuous functions that are not differentiable at any point).
After this upsetthe dust had settled in the wake of du Bois-Reymond's seismic discovery, people started to believe fervently believing that there should exist a continuous function whose Fourier series diverges everywhere . - an opinion that lay on the other extreme! Andrei Kolmogorov inadvertently lent support to this claim by exhibiting, in 1926, an ${L^{1}}([- \pi,\pi])$-function whose Fourier series diverges everywhere. However, there was great surprise upheaval once more in Fourier-land when the combined efforts of Lennart Carleson and Richard Hunt in the late 1960's showed that the Fourier series of any $f \in {L^{p}}([- \pi,\pi])$ converges almost everywhere to $f$, for all $p > 1$ (this result subsumes the case of continuous functions). During an interview with the AMS, Carleson revealed that he had originally been working tried to disprove his result (which proves the statement for pertaining to $p = 2$), but in the end, his failure to do so produce a counterexample convinced him that he should be working in the other direction instead.
Therefore, in the field of Fourier analysis, viewpoints have changed and cherished beliefs were have been destroyed - twice.
3 added 47 characters in body
Another example from real analysis would be the question of the pointwise convergence of the Fourier series of a continuous function on a given closed interval. Many people, including Dirichlet and even Weierstrass himself, believed that the Fourier series of such a function would converge converges pointwise everywhere to the function itself. Some believed in this statement so strongly that they viewed it as an infallible axiom.
Hence, one can imagine the great upset when, in 1876, Paul du Bois-Reymond proved that a continuous function exists whose Fourier series diverges at a point. The proof is non-constructive, using a method called the principle of condensation of singularities. I have no idea how the method works, but I do know of a very common proof that uses the Baire Category Theorem (using the Baire Category Theorem, one can also prove the existence of continuous functions that are not differentiable at any point).
After this upset, people started to believe that there should exist a continuous function whose Fourier series diverges everywhere. Andrei Kolmogorov inadvertently lent support to this claim by exhibiting, in 1926, an ${L^{1}}([- \pi,\pi])$-function whose Fourier series diverges everywhere. However, there was great surprise yet again once more in Fourier-land when the combined efforts of Lennart Carleson and Richard Hunt in the late 1960's showed that the Fourier series of any $f \in {L^{p}}([- \pi,\pi])$ converges almost everywhere to $f$ for all $p > 1$. During an interview, Carleson revealed that he had originally been working to disprove his result (which proves the statement for $p = 2$), but in the end, his failure to do so convinced him that he should work be working in the other direction instead.
Therefore, in the field of Fourier analysis, viewpoints changed and cherished beliefs were destroyed - twice.
Post Undeleted by Leonard
2 added 1463 characters in body
When
Another example from real analysis would be the $\lambda$-calculus was first proposed by Alonzo Churchquestion of the pointwise convergence of the Fourier series of a continuous function on a given closed interval. Many people, it was found including Dirichlet and even Weierstrass himself, believed that the Fourier series of such a function would converge pointwise everywhere to be inconsistentthe function itself. This Some believed in this statement so strongly that they viewed it as an infallible axiom.
Hence, one can imagine the great upset when, in 1876, Paul du Bois-Reymond proved that a continuous function exists whose Fourier series diverges at a point. The proof is because non-constructive, using a method called the original untyped principle of condensation of singularities. I have no idea how the method works, but I know of a very common proof that uses the Baire Category Theorem (using the Baire Category Theorem, one can also prove the existence of continuous functions that are not differentiable at any point).
After this upset, people started to believe that there should exist a continuous function whose Fourier series diverges everywhere. Andrei Kolmogorov inadvertently lent support to this claim by exhibiting, in 1926, an ${L^{1}}([- \lambda pi,\pi])$-calculus could encode -function whose Fourier series diverges everywhere. However, there was great surprise yet again when the Kleene-Rosser Paradox combined efforts of Lennart Carleson and Richard Hunt in the form late 1960's showed that the Fourier series of a self-contradictory any $f \lambda in {L^{p}}([- \pi,\pi])$ -expression. In order converges almost everywhere to address this inconsistency$f$ for all $p > 1$. During an interview, Church formulated the typed Carleson revealed that he had originally been working to disprove his result (for $\lambda p = 2$-calculus.), but in the end, his failure to do so convinced him that he should work in the other direction instead.
Therefore, in the field of Fourier analysis, viewpoints changed and cherished beliefs were destroyed - twice.
Post Deleted by Leonard
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When the $\lambda$-calculus was first proposed by Alonzo Church, it was found to be inconsistent. This is because the original untyped $\lambda$-calculus could encode the Kleene-Rosser Paradox in the form of a self-contradictory $\lambda$-expression. In order to address this inconsistency, Church formulated the typed $\lambda$-calculus.
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http://stats.stackexchange.com/questions/12900/when-is-r-squared-negative
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# When is R squared negative?
My understanding is that R squared cannot be negative as it is the square of R. However I ran a simple linear regression in SPSS with a single independent variable and a dependent variable. My SPSS output give me a negative value for R-squared. If I was to calculate this by hand from R then R squared would be positive. What has SPSS done to calculate this as negative?
````R=-.395
R squared =-.156
B (un-standardized)=-1261.611
````
Code I've used:
````DATASET ACTIVATE DataSet1.
REGRESSION /MISSING LISTWISE /STATISTICS COEFF OUTS R ANOVA
/CRITERIA=PIN(.05) POUT(.10) /NOORIGIN
/DEPENDENT valueP /METHOD=ENTER ageP
````
I get a negative value. Can anyone explain what this means?
Thanks.
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– whuber♦ Jul 11 '11 at 17:14
Thanks Whuber. Not it doesn't because there seems to be disagreement on whether or not R squared can be negative and I can't how it has calculated R squared as negative. I've edited the above. Please let me know if I need to add more details. Many thanks! – Anne Jul 11 '11 at 17:52
OK. However, you may have been hasty in your reading. The reply to that question by @probabilityislogic begins by saying R squared "cannot be negative," but later on it admits that indeed it "can go negative." Thus there isn't any disagreement. A clear moral is that you need to let us know what procedure is being used to calculate R squared. – whuber♦ Jul 11 '11 at 17:52
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@Anne Again, which SPSS procedure are you using? – whuber♦ Jul 11 '11 at 18:19
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@Anne I suggest you disregard the time series reply, because your data are not time series and you're not using a time series procedure. Are you really sure the R squared is given as a negative value? Its magnitude is correct: $(-0.395)^2=0.156$. I have looked through SPSS help to see whether perhaps as a convention the R-squared value for negative R's is negated, but I don't see any evidence that this is the case. Perhaps you could post a screen shot of the output where you are reading the R-squared? – whuber♦ Jul 11 '11 at 20:26
show 10 more comments
## 3 Answers
R2 compares the fit of the chosen model with that of a horizontal straight line (the null hypothesis). If the chosen model fits worse than a horizontal line, then R2 is negative. Note that R2 is not always the square of anything, so it can have a negative value without violating any rules of math. R2 is negative only when the chosen model does not follow the trend of the data, so fits worse than a horizontal line.
Here is an example. We fit data to a linear regression model constrained so that the Y intercept must equal 1500.
The model makes no sense at all given these data. It is clearly the wrong model,perhaps chosen by accident.
The fit of the model (a straight line constrained to go through the point 0,1500) is worse than the fit of a horizontal line. Thus the sum-of-squares from the model (SSreg) is larger than the sum-of-squares from the horizontal line (SStot). R2 is computed as 1 - SSreg/SStot. When SSreg is greater than SStot, that equation computes a negative value for R2.
With linear regression with no constraints, R2 must be positive (or zero) and equals the square of the correlation coefficient, r. A negative R2 is only possible with linear regression when either the intercept or the slope are constrained so that the "best-fit" line (given the constraint) fits worse than a horizontal line. With nonlinear regression, the R2 can be negative whenever the best-fit model (given the chosen equation, and its constraints, if any) fits the data worse than a horizontal line.
Bottom line: A negative R2 is not a mathematical impossibility or the sign of a computer bug. It simply means that the chosen model (with its constraints) fits the data really poorly.
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This is a nice illustration of the point made by @jefflovejapan. Where in the SPSS command is such a constraint specified? – whuber♦ Jul 13 '11 at 15:48
@whuber I think /NOORIGIN sets the intercept to 0. – JMS Jul 13 '11 at 18:00
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– whuber♦ Jul 13 '11 at 18:13
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@whuber Correct. @harvey-motulsky A negative R^2 value is a mathematical impossibility (and suggests a computer bug) for regular OLS regression (with an intercept). This is what the 'REGRESSION' command does and what the original poster is asking about. Also, for OLS regression, R^2 is the squared correlation between the predicted and the observed values. Hence, it must be non-negative. For simple OLS regression with one predictor, this is equivalent to the squared correlation between the predictor and the dependent variable -- again, this must be non-negative. – Wolfgang Jul 14 '11 at 7:17
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@HarveyMotulsky, in this case the intercept or slope were not constrained. It seems that you are saying that Rsquared can only be negative if these are constrained. Can you elaborate on what might have occurred in this particular case? – Anne Jul 16 '11 at 21:56
show 4 more comments
Have you forgotten to include an intercept in your regression? I'm not familiar with SPSS code, but on page 21 of Hayashi's Econometrics:
If the regressors do not include a constant but (as some regression software packages do) you nevertheless calculate $R^2$ by the formula
$R^2=1-\frac{\sum_{i=1}^{n}e_i^2}{\sum_{i=1}^{n}(y_i-\bar{y})^2}$
then the $R^2$ can be negative. This is because, without the benefit of an intercept, the regression could do worse than the sample mean in terms of tracking the dependent variable (i.e., the numerator could be greater than the denominator).
I'd check and make sure that SPSS is including an intercept in your regression.
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NOORIGIN subcommand in her code tells that intercept was included in the model – ttnphns Jul 12 '11 at 10:12
This can happen if you have a time series that is N.i.i.d. and you construct an inappropriate ARIMA model of the form(0,1,0) which is a first difference random walk model with no drift then the variance (sum of squares - SSE ) of the residuals will be larger than the variance (sum of squares SSO) of the original series. Thus the equation 1-SSE/SSO will yield a negative number as SSE execeedS SSO . We have seen this when users simply fit an assumed model or use inadequate procedures to identify/form an appropriate ARIMA structure. The larger message IS that a model can distort (much like a pair of bad glasses ) your vision. Without having access to your data I would otherwise have a problem in explaining your faulty results. Have you brought this to the attention of IBM ?
The idea of an assumed model being counter-productive has been echoed by Harvey Motulsky. Great post Harvey !
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stat. Thanks. No I have not spoken to IBM. The data is not time series. It is from point in time data. – Anne Jul 11 '11 at 19:55
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@Anne and others: Since your data are not time series and you're not using a time series procedure please disregard my answer. Others who have observed negative R Squares when involved with time series might find my post interesting and tangentially informative. Others unfortunately may not. – IrishStat Jul 11 '11 at 21:36
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http://math.stackexchange.com/questions/264111/finding-a-limit-of-a-function-by-definition
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# finding a limit of a function by definition
calculate the limit of the following functions and prove directly from definition (using $\epsilon$ and $\delta$)
a) $$\lim _{x\to 2}(x^3+3x)$$
We just learned the definition of limits in regards to function and this is supposed to be a simple question. But I just can't understand how to use the info I'm given.
a) Am I right with guessing the limit is $2^3+3\cdot 2=14$?
b) Assuming I am (I don't think it would change much if I am not), then for an $\epsilon$ I want to find a $\delta$ such that for each $x$ in the punctured environment of $(x_0 - \delta,x_0+\delta)$ , $|x^3+3x-14|<\epsilon$, but I have no idea how to proceed from here. I tried searching for similar examples to see how the truth process is done but wasn't able to find any.
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Since the function $f(x)=x^3+3x$ is continuous at $2$ then the limit is $14$. – Sigur Dec 23 '12 at 13:32
## 2 Answers
a) You are right because $f(x)=x^3+3x$ is continuous and so $\lim_{x\to 2}f(x)=f(2)$
b) Observe that $$\left|x^3+3x-2^3-3\cdot 2\right|=\left|(x-2)(x^2+2x+4)+3(x-2)\right|= \left|x-2\right|\left|x^2+2x+7\right|<\delta\left|x^2+2x+7\right|$$ Can you proceed now?
EDIT: The term $\left|x^2+2x+7\right|$ is problematic and needs to be replaced by something containing only $\delta$. With $\left|x-2\right|<\delta$ in mind, $$\left|x^2+2x+7\right|=\left|(x-2)^2+6x+3\right|\le \left|x-2\right|^2+6\left|x+\frac12\right|<\delta^2+6\left|x-2+\frac52\right|<\\ \delta^2+6\delta+15$$ We have $$\delta\left|x^2+2x+7\right|<\delta^3+6\delta^2+15\delta$$ and want this to be less than $\epsilon$. If we further demand that $\delta<1$, $$\delta\left|x^2+2x+7\right|<\delta^3+6\delta^2+15\delta<\delta+6\delta+15\delta=18\delta$$ Taking $\delta<\frac{\epsilon}{18}$ yields the result (in fact we can take $$\delta=\frac12\min\left\{1,\frac{\epsilon}{18}\right\}$$ as our $\delta$ to be completely rigorous)
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Both solutions really helped me. From what you gave me I was able to realize what my delta needs to be, and the solution below allowed me to understand how I'm supposed to write a proof on these matter. Thank you both! – Nescio Dec 23 '12 at 13:54
@Nescio I updated my answer with my take on how one can complte the proof – Nameless Dec 23 '12 at 14:02
You can evaluate the limit by substituting $2$ because $x^3+3x$ is continuous. To show that the limit is $14$, Let $\epsilon>0$. Let $\delta=\min\{1,\epsilon(\frac{1}{22})$. Now for all $x$ such that $|x-2|<\delta$, we have: $$|x^3+3x-14|\leq|x^3-8|+|3x-6|\leq|x^2+2x+4||x-2|+3|x-2|$$ $$|x^3+3x-14|<\delta (|x^2+2x+4|+3)$$ Since for all $x\in [2-\delta,2+\delta]\subseteq [1,3]$ we know that $|x^2+2x+4|+3\leq22$ Thus: $$|x^3+3x-14|<\delta (|x^2+2x+4|+3)<\min\{1,\epsilon(\frac{1}{22})\}(22)<\epsilon$$
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http://physics.stackexchange.com/questions/31309/formulation-of-general-relativity
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# Formulation of general relativity
EDIT: I think I can pinpoint my confusion a bit better. Here comes my updated question (I'm not sure what the standard way of doing things is - please let me know if I should delete the old version). The major change is that I removed focus from the third question which probably is a purely mathematical question (in the notation below, it asks what properties of (M,T) together with (M,T) being consistent, forces (M,T) to be unique.).
Say that a pair (M, T) is consistent if it satisfies the Einstein field equations. Here M is a manifold with a metric g (from which one can define its Ricci tensor among other things) and T is a map from M to tensors living in the same space of tensors as the Ricci tensor (I ignore units for now). I put absolutely no other restrictions on (M, T).
Now relativity makes perfect sense to me as a mathematical statement: it distinguishes some pairs (M, T) as 'consistent'.
Therefore, if we had a way to map "our world" to a pair (M,T) at least we could theoretically check whether (M,T) is consistent or not. My problem is that I do not at all understand how to do this.
To begin with, which set do I choose for M?
I think I can answer this question myself. I take this set to be the set of intuitive descriptions(that I can make intuitive sense of) of events in the world. For example, E := "(a particular point in) Stockholm on 08:00, Jan 24, 2013". This description I could understand intuitively, and at least theoretically (if time permits) I could go there to check the theory if it makes a statement about E. Another kind of description, given already another description F, could be G:="the event I get to by using rocket R, travelling for time T according to this watch I bring with me, from G", where R and T are intuitive descriptions. Please let me know if this choice of set is inappropriate.
In this case I have no problem of turning the set M into a manifold, not yet with a metric.
Finally (and here is my confusion): I am at a point p (constructed as above). What (intuitively described) experiments do I perform to find the metric tensor at p, respectively the stress-energy tensor at p? I cannot come up with two different answers for these two tensors - and in this case the theory is not a very interesting one, since then it just predicts that two identical (in the intuitive sense) experiments are the same.
If I try to get an answer to this from e.g. Wikipedia I get lost in a deep tree of coordinate-dependent definitions which in some places appear to assume I already have intuitive sense for both mass and metric, and that I have intuitive sense for these being the related as relativity predicts they should be. I'm hoping there are two distinct intuitively described experiments I could perform, which relativity predicts should have the same result.
FINAL EDIT: I have received many useful comments, and the answer by Ron Maimon answers my initial question, which was "what is a suitable set to choose for M when trying to map 'our world' to a pair (M,T)?". It seems a definition such as I suggest above "should work", as should that described by Ron in his answer. Furthermore, Ron points out(I think) that it is an assumption of the theory that any such labeling should give the same results.
Since my initial question is answered I accept Ron's answer and will possibly come back with my further question "how to intuitively understand the metric and stress energy tensors in terms of experiments any person with sufficient degree of common sense and superhuman abilities (by which I just mean, can reach high accelerations, is not so heavy as to affect the stress-energy tensor in significant ways et.c. Equivalently, superhuman abilities would not be needed in case the speed of light was something like 10 meters per second) could perform?" if I am able to formulate it in a precise way.
OLD VERSION (not needed for the question): As far as I understand, general relativity states that
the world is a manifold M, and M is completely described by the Einstein field equations.
This already appears as an incomplete statement, and I'll explain why I think so. Before that:
• what is a complete statement of general relativity, possibly including undefined terms (so in my attempt above, the "is" in "the world is a manifold" and "Einstein field equations" are undefined terms)?
Now to why this does not make sense to me. The Einstein field equations state that two tensors (it is not necessary to define tensor for my confusion to arise) agree at every point.
This seems to presuppose that the set of points making up the manifold is already given. Thus:
• what is a good description of the set of points of M?
For clarity, my definition of manifold M says for one thing that M is a set.
For the first question, it appears to me that there need to be some extra assumptions, since one could conceivably think of a 'world' without matter which should be completely 'flat', and also of our world which is not. These cases should clearly be different.
• Exactly what data determines a 'theory'? (meaning that M is completely determined from this data - again I would like to have complete description but am happy with undefined terms as long as it is clear that they are such)
Ideally, the second and third questions should be answered by any answer to the first question, but I added the latter questions to indicate what confuses me particularly.
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## 4 Answers
General relativity is a classical theory. I will restate your dilemma as follows, since this is how Einstein stated it:
• We have an abstract manifold consisting of points, vectors that link nearby points, and a metric tensor that tells you the distance between nearby points. What makes these points physical? How can we tell point A apart from point B? Since it seems that the points only get meaning from the stuff happening at these points.
Einstein was very bothered by this question, so he considered the "hole argument". This is the idea that if we give all the points of the manifold names, by using a coordinate system, then these names are arbitrary, and the points are really indistinguishable from one another. So if we change the naming of the points, we change the name of the solution of the Einstein equation, and this seems like it changes the physical behavior.
The resolution of the hole paradox that Einstein settled on (which in modern physics is the central idea of gauge invariance) is that the space-time points are defined by the things happening at these points, not by the names. So he considered filling space with lots of metersticks and clocks, making a grid of measuring devices, and these clocks and metersticks have names that are real, as these are real objects. Then the coordinate system assigns a coordinate to each meterstick and clock, based on the value of the coordinates there.
If you rename all the points, the clocks and metersticks get new names, but so long as the physical relations between them are unchanged, so long as the distance marked out by each little meterstick, and the time ticked out by each little clock are unchanged, then the two situations are identical in the physical sense. This allowed him to define the notion of coordinate invariance in GR: any labelling of the points defined by the grid of clocks and metersticks is equally valid as any other, and the laws should not make reference to the names of the points in their formulation.
This allowed him to make sense of the statement "space time is a manifold M with a metric that obeys the field equation". The statement gets positivistic meaning from the matter in the space, making measurements of local distances and times, and the metric tells you what these measurements are (or would be). The Einstein equation then determines the future metric from the current metric and its time derivative (under suitable constraints) plus the scheme for giving name to the space-time points in the future, which is the condition that tells you what coordinates you are going to use.
With this philosophical position, Einstein resolved the nagging worry of the ill-definedness of manifold points without extra structure imposed--- he imposed the structure by imagining little classical measuring devices everywhere. Since this is classical physics, he can make these devices really little, without affecting anything.
This point of view tells you that when you have an arbitrary labelling that affects the physics, for example, the labelling of whether a given particle (at distances short enough for the Higgs to be irrelevant) is a left-handed electron or a neutrino, the coordinate system that picks out which direction is "electron" and which direction is "neutrino" is arbitary. Then you impose the condition that any choice of coordinates is as good as any other for describing the physics, and this is the local gauge invariance. To make this work, you need to give a gauge field to relate the fields at nearby points, and you need to make sure you don't count coordinate changes (changes in gauge) as physical transformations, since they are just a different choice of name you give to the additional coordinate system you use to distinguish an electron from a neutrino.
This might be a bad example, since at low energies, the Higgs condensate determines which direction is which, and it is good to choose the gauge so that the electron and neutrino don't look related. This is no different than a crystal in space picking out preferred coordinates, based on the atoms being at certain positions--- it doesn't change the fact that fundamentally you have a arbitrary choice of coordinates, one which is conveniently made using the crystal when the crystal is present.
This philosophical shift in the meaning of coordinates is so deeply ingrained now, that everyone does it at some stage and forgets about it. You should understand that the points are given meaning only to the extent that there are observable quantities that are invariant to change in coordinates, the goal of physics is to describe these quantities, and the coordinates are a mathematical crutch to formulate the equations.
The mathematicians define the concepts abstractly, so that the points are a set with a topology and a smooth structure, and the metric is a mild generalization of the notion of function called a section of a bundle. These mathematical definitions are already in a framework where the meaning of the word "point" does not depend on the label you give to the point, and Einstein's confusion is hard to state. But if you come in without this philosophical position, it helps to think about the little grid of metersticks in order to acquire it.
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This seems like an excellent answer (also to my edited question which I wrote when you wrote your answer) which should dispel some of my confusion. Give me some time to digest it, and I'll see if I understand it. – Erik Jul 5 '12 at 7:07
That clarifies a lot and answers my initial question, but I am still unsure about the "further question" I mention in my final edit - though I might come back with that question. Also, I will +1 this whenever I penetrate that 15 points barrier. – Erik Jul 5 '12 at 11:15
@Erik: If you want the answer to a new question, just ask a new question. It's annoying to edit the answers all the time. But I can't understand what the new question is exactly, while the old question was pretty clear. – Ron Maimon Jul 5 '12 at 13:42
Too long for a comment, so here some remarks and questions. Might well be that I understand you question wrong though...
Firstly, general relativity is a physical frameworks which permits to model different situations like any other theory. In Newtonian mechanics you can ask how one particle behaves in a field or you can ask how two particles behave with a field and with each other. And so on. The theory contains differential equations (Newtonian equations of motion) and you can plug things into them, like the number of particles or the mathematical expression which is the force. As you're dealing with a differential equation, you also need initial conditions. Like where the particles start out and what the initial velocities are.
So the first thing to clarify is that Einsteins field equations are differential equations as well, there are clearly many mathematically permissible solutions and so it's not like general relativity is equivalent to one and one manifold M. Instead of saiying "M is completely described by the Einstein field equations", I'd say M (or rather the field $g_{\mu\nu}$, i.e. the metric tensor, on M) obeys the field equations.
I'm not sure to what extend you're technically aware of this, but you also describe the problematic of having to model spacetime while it interacts with matter, which needs spacetime to be somewhere too. Yes, this really makes it difficult to deal with the equations. In any case, you decide what features you want to have in your model and solve the field equations in a selfconsistent manner.
There is certainly some extra thoughts in coming up with a description of the world than postulating the equations. Similar to the Newtonian exmaple, you can go on and search for the general relativistic description of one particle exposed to the pull of the gravitational force field of another. You might find the Schwarzschild metric, construced to be constent with well estabilshed measurements, whos analoges where known in Newtonian mechanics too, see e.g. correspondence principle. Or you might really be interested in an empty universe and yes, you put in that emptyness by hand. And if you want no matter, i.e. make the right hand side of the Einstein field equation zero, then there is still some playground, see e.g. the Einstein-manifolds. The theory doesn't say "You can't model the system whos solution is contains Schwarzschild metric, because hey, there are alot of other masses in the universe". No, you ... ah, thankfully there are some elaborations on wikipedia. The cosmological solutions, i.e. the ones concerting more global question, also make some convinient assumptions, like I don't know, isotopy. Things that are "plausible" within the language of the theory.
Btw. are you a mathematican? I like your way of being skThen I have a small question regarding the formulation "the world is manifold M", because to me that word "world" is open to at least interpretations. If by world you think of the four dimensional manifold M in typical general relativity solutions, then matter is not part of the world, but "in" the world. On the one hand you say "at every point" but then you also say "a 'world' without matter".
I also don't really know what is ment by "what is a complete statement"?
So after this, some of your questions might be clarified or you might at least be ale to ask differently, in case others have problems too.
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Yes, "world" is also not completely well-defined, but I am mainly confused about how to map something I understand intuitively "in the world" to a mathematical property of M. After reading your answer I think I have made my question a bit more complicated than neccessary, so I'll edit it a bit in a bit. – Erik Jul 5 '12 at 6:03
@Erik: I don't get the thing with the two experiments. Anyway, now there are more answers, hopefully some will help you. – Nick Kidman Jul 5 '12 at 7:22
Your assertion that "M is completely described by the Einstein field equations" is not correct in any universe with spatial dimension greater than two, since in larger dimensional spaces we have to worry about topological constraints on our manifold that are NOT determined by the metric properties that are provided by solutions to Einstein Field Equations. (In two spacial dimensions, the metric and topology are determined by one another)
In addition, your third question seems to ignore an important aspect of the einstein field equation, the stress energy tensor (the source term). Below are the einstein field equations (without a cosmological constant and with geometrized units).
$G_{\mu\nu}=8\pi T_{\mu\nu}$
That $T_{\mu\nu}$ term is the stress energy tensor and will define the difference between your empty, flat universe and the one we have, with lots of various matter and energy.
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Thanks for your answer. I agree fully with you but I do not think it answers any of my questions. For example, I too think my assertion is wrong, but for even more fundamental reasons than that it does not take 'topological features' into account. I simply do not understand what set the tensors (considered simply as functions from M into [whatever]) are defined on. – Erik Jul 5 '12 at 5:56
Keep in mind tensors aren't function of M, they are multilinear functions of the tangent vector and dual spaces to the real numbers. In some ways, these tangent vector spaces (and their duals) are easier to understand than the base set of points that compose M, as we don't really need to think about the structure of spacetime itself, we can just look at velocities (or trajectories) in spacetime. Does this help at all? – Benjamin Horowitz Jul 5 '12 at 6:02
He is asking about what space-time points mean, where only the structures on these points give them physical meaning. This is one of the things that bothered Einstein in 1915. – Ron Maimon Jul 5 '12 at 6:27
@BenjaminHorowitz : for my purposes it is enough to consider them simply functions on M; they clearly assign something to each point of M and this is the definition of "function of M". I do not mean "real-valued function on M" or similar. – Erik Jul 5 '12 at 7:04
General Relativity assumes that the universe is a differentiable manifold. The Einstein equations calculate the metric that allows us to assign a meaning for distance on the manifold. So you're correct that you just have to assume a manifold exists.
The same manifold can have many different metrics. The metric is determined by the stress energy tensor, so the manifold+metric if there is no matter will be different to the manifold+metric when matter and/or energy is present.
I think this answers your questions, but please comment if you want me to expand on this.
Response to comment and edit of question: bearing in mind that I am that most pragmatic of scientists, a physical chemist, I'm not sure I understand your concerns as they seem a bit philosophical to me.
The metric tensor describes the curvature of space and you'd measure it by parallel transporting a vector round some convenient closed path and comparing the direction of the original and transported vector. The stress energy tensor you determine simply by measuring mass/energy density, momentum flux and pressure. The fact that Einstein tensor and the stress energy tensor are required to be identical everywhere doesn't seem trivial to me - actually it seems quite remarkable.
Incidentally, the measurements you make will always be done using some convenient coordinates because we need coordinates to measure anything. However this just affects the representation of the tensors we end up with. The tensors themselves are coordinate independant.
Response to response: Erik makes the point that you can't actually transport a vector round a closed path in spacetime, however you could start at a point, travel two separate paths and meet at a second point and make that your loop. This seems to me a perfectly good way of measuring the transport round a loop.
In terms of actually doing the measurement, at every point on the loop spacetime looks locally flat, so all I have to do is locally parallel transport my vector along a straight line, and I know how to do this. Admittedly I have to move an infinitesimal amount at each step, which isn't possible in practice, but then no-one's suggesting we'd actually measure the curvature this way. The point is that I don't need to know the metric or how to do covarient differentiation in order to (conceptually) do the measurement.
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I was editing my question when you gave your answer. In the new version I also just assume the manifold exists and is described as a set by myself. My problem is "metric is determined by the SET", since I do not understand which two conceivably different things that relativity states are the same. – Erik Jul 5 '12 at 7:02
I've updated my answer to address what I think are your concerns, though I wonder if our perspectives are too different for my update to be very helpful. – John Rennie Jul 5 '12 at 7:55
I wrote a long a comment on your edit in response to my edit, but it got a bit long for this box. I put it here: www.math.kth.se/~eaas/pe/a.txt . I completely understand if you do not find it worthwhile to continue the discussion. – Erik Jul 5 '12 at 9:57
I've updated my answer again. I think the debate is getting a bit involved to pursue here. Rather than attempt to figure stuff out from Wikipedia you'd be better off with a copy of Gravitation by Misner, Thorne and Weaver. They can explain this stuff a lot better than I can. – John Rennie Jul 5 '12 at 10:26
I think we agree that the experiment should involve two distinct paths but I am still unsure concretely how to perform the experiment (as an example of something I think is concrete, in 3-space: "move the south pointing chariot along this respectively this curve"). But as you say this is a question at least worthy its own post (after some background reading). Incidentally, I refer to Wikipedia just as an example of an authority. I don't think it differs significantly from accounts I've found in other places. Anyway, thanks a lot and I'll +1 this as soon as I have rep. to do so. – Erik Jul 5 '12 at 11:23
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http://physics.stackexchange.com/questions/tagged/topological-field-theory+research-level
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# Tagged Questions
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### Realization of Witten-type topological quantum field theory in condensed matter physics
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### String-theoretic significance of extended CFT
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### Wilson Loops in Chern-Simons theory with non-compact gauge groups
VEVs of Wilson loops in Chern-Simons theory with compact gauge groups give us colored Jones, HOMFLY and Kauffman polynomials. I have not seen the computation for Wilson loops in Chern-Simons theory ...
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### Applications of Geometric Topology to Theoretical Physics
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### Quantum gravity at D = 3
Quantization of gravity (general relativity) seems to be impossible for spacetime dimension D >= 4. Instead, quantum gravity is described by string theory which is something more than quantization ...
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Quantum statistics of particles (bosons, fermions, anyons) arises due to the possible topologies of curves in D-dimensional spacetime winding around each other What happens if we replace particles by ...
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### Minimal strings and topological strings
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### Normalization of the Chern-Simons level in $SO(N)$ gauge theory
In a 3d SU(N) gauge theory with action $\frac{k}{4\pi} \int \mathrm{Tr} (A \wedge dA + \frac{2}{3} A \wedge A \wedge A)$, where the generators are normalized to \$\mathrm{Tr}(T^a ...
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### Topological twists of SUSY gauge theory
Consider $N=4$ super-symmetric gauge theory in 4 dimensions with gauge group $G$. As is explained in the beginning of the paper of Kapustin and Witten on geometric Langlands, this theory has 3 ...
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### Geometric Langlands as a partially defined topological field theory
I have heard from several physicists that the Kapustin-Witten topological twist of $N=4$ 4-dimensional Yang-Mills theory ("the Geometric Langlands twist") is not expected to give rise to fully defined ...
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### Models of higher Chern-Simons type
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### Group Cohomology and Topological Field Theories
I have a two-part question: First and foremost: I have been going through the paper by Dijkgraaf and Witten "Group Cohomology and Topological Field Theories". Here they give a general definition for ...
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http://matthewkahle.wordpress.com/2010/02/22/many-markov-components/?like=1&source=post_flair&_wpnonce=7b62a7b6bf
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# Many Markov components
In my recent preprint, I showed that once $r \ge c / n$ the Metropolis Markov chain on configurations of hard discs is no longer ergodic. In particular, even for reasonably small radius there are stable configurations of discs. In this note the goal is to make a more quantitative statement about the number of Markov components in particular when $r > c_2 / \sqrt{n}$ (which is the usual range of interest in statistical mechanics.
For the main construction, start with the following configuration of $24$ discs in a square. (It is not relevant to the following discussion, but it is fun to point out that this is the densest packing of $24$ equal sized discs in a square.)
This configuration of $24$ discs is easily seen to be stable (meaning that each disc is individually held in place by its neighboring discs and the boundary of the square), and in fact it is still stable if we delete a carefully chosen set of four discs as follows.
By stacking $k^2$ of these together one has stable configurations with $20 k^2$ discs.
Finally one can add discs in any of the “holes” in these kinds of configurations, leading to more and more possibilities for stable configurations. For example, here is one of the ${36 \choose 18} \approx 9.08 \times 10^9$ possibilities for dropping discs in $18$ of the $36$ holes in the previous illustration.
By continuing constructions like this for larger and larger $k$ we observe two facts.
(1) As $n \to \infty$ the number of components can grow quite fast. To make this quantitative we expand on the example. Suppose there are $20 k^2$ discs to start out, then there are $4 k^2$ holes. Suppose we drop $2 k^2$ discs into those holes. Then the number of discs $n = 22 k^2$ in total, and then the number of Markov components is at least ${4 k ^2 \choose 2 k^2 } = { 2n/11 \choose n/11 }$. (This is for unlabeled discs. For labeled discs, multiply this by $n!$.) By Stirling’s formula, this number grows exponentially fast in $n$ (something like $1.134^n$).
In this particular example the discs have total area $\pi r^2 n = \approx 0.67$, but that could obviously adjusted by changing the ratio of extra discs to holes.
(2) For every $\lambda \in (0.60, 0.75)$ there is an $\alpha = \alpha(\lambda) > 1$ and a sequence of configurations $C_1, C_2, \ldots$ with $C_i \in$ Config(i, r(i)), the number of Markov components greater than $\alpha ^ i$ for sufficiently large $i$, and $\pi r^2 n \to \lambda$ as $i \to \infty$.
It turns out that what are described here are only separate components in the sense that a Markov chain that only moves one disc at a time can not mix from one state to the other. (The four discs in the center of a square can rotate about the center of the square!) But a small modification of this construction would seem to give exponentially many components in the sense of $\pi_0$, each with small positive measure.
Is exponentially many components the maximal possible? The Milnor-Thom Theorem gives an upper bound much larger than this, something like $O(e^{c n^2})$ for some $c > 0$. It would also be very interesting to know any bounds on the topological complexity of these path components.
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http://math.stackexchange.com/questions/178434/how-do-i-calculated-probabilities-for-cards
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# How do I calculated probabilities for cards?
I am trying to gain basic understanding on how to calculate probabilities. Below are a few examples of what I am trying to calculate. I would prefer (if possible) for formulas to be given on how to solve these using Microsoft Excel. Also, educating me on Probability Terminology involved would be helpful.
Situation 1: I am going to draw "X" cards (5, for 5-card stun poker) from a "Y" card deck (52), what are the odds of getting a flush in hearts (Variable I have labeled as "Z" = 13... in my current ignorance, I refer to this as the "Targets" variable)?
My algebraic solution: = (13/52) * (12/51) * (11/50) * (10/49) * (9/48) Solution to chance of getting any flush would be the above times 4 to represent a flush in any suit.
The problem with the above solution is that it becomes impractical when the situational probabilities I am trying to calculate become more complicated.
QUESTION 1: What is a more statistical way of calculating the above? (using excel)
Situation 2: I am going to draw "X=5" 5 cards from a "Y=10" 10 card deck. In this deck there are only two different draw possibilities, Card 1 or Card 2... There are 7 copies of Card 1 which are all identical and 3 copies of Card 2 which are all identical.
What is the probability of getting 0-3 copies of Card 2?
X = 5, Y = 10, Z = 3
QUESTION 2: What is the statistical way of calculating the above? (using excel)
Situation 3:
X = 7, Y = 60, Z = 20
QUESTION 3: How do I calculate, using excel, what the odds are of getting 0-7 of the target cards in a draw of 7?
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Statistical methods would use sampling and estimates. Statistics and probability are not the same thing, although they are related. – Thomas Andrews Aug 3 '12 at 15:36
Thank you for the clarification, I was unaware of the difference. – DiabloMonkey Aug 3 '12 at 16:22
## 2 Answers
While computing the probabilities directly using algebra are possible, it's much easier to take an approach based on combinatorics.
We can do this using binomial coefficients, which are sometimes read as "Choose $k$ from $n$ objects," or "n Choose k."
Basically, for the chances of any flush of clubs, you need to compute the probability of choosing 5 out of 13 cards out of the 52 card deck. Probability is, of course, represented by a number $0 \le p \le 1$, so what we want to compute is the number of possible flushes of clubs, and divide it by the total number of hands.
The total number of hands can be written as $\begin{pmatrix} 52 \\ 5\end{pmatrix} = \frac{52!}{5!(52-5)!}$. The total number of club flushes (including straight and royal flushes) is $\begin{pmatrix} 13 \\ 5\end{pmatrix}$, or "choose any 5 from 13 total cards." So your probability is $$P(\mathrm{club flush}) = \begin{pmatrix}13 \\ 5 \end{pmatrix}/\begin{pmatrix}52 \\ 5\end{pmatrix}.$$
You can extend this arbitrarily. Say you want the probability of any flush. Well, obviously it's better, because there are four possible suits. So you want to multiply the previous result by the number given by "choosing one suit for the flush out of four possible suits," or $\begin{pmatrix}4 \\ 1\end{pmatrix}$, resulting in
$$P(\mathrm{any flush}) = \begin{pmatrix}4 \\ 1\end{pmatrix}\begin{pmatrix}13 \\ 5 \end{pmatrix}/\begin{pmatrix}52 \\ 5\end{pmatrix}.$$
That is the general approach to answer your first question. To answer your second question, we can do the same thing: You want to know the probability that if you choose 5 cards, you will get between 0 and 3 copies of a 2.
To answer questions 2 and 3, which can be handled quite similarly, we have 10 cards, and draw 5. So the total possible number of hands is $\begin{pmatrix}10 \\ 5\end{pmatrix}$. Right now, it doesn't matter that all of the Card 1s/Card 2s are the same -- we're going to formulate the problem in such a way that it doesn't matter.
To choose 0 Card 2s in 5 cards is the same thing as saying "choose 0 out of 3 card 2s, and choose 5 out of 7 card 1s." Using the combinatorical technique described above, that's $\begin{pmatrix}3 \\ 0\end{pmatrix}\begin{pmatrix}7 \\ 5\end{pmatrix}$. Moving on, choosing 1 Card 2 is like saying "choose 1 out of 3 Card 2s, and 4 out of 7 Card 1s," or $\begin{pmatrix}3 \\ 1\end{pmatrix}\begin{pmatrix}7 \\ 4\end{pmatrix}$. Continue this for each value 0,1,2,3, and add up the results. This gives you the total number of events; to get the total probability, divide that by the total number of possible events, $\begin{pmatrix}10 \\ 5\end{pmatrix}$.
(Note that in example 2, there are only three Card 2s, so you should have a 100% probability of picking between 0 and 3 (inclusive) Card 2s in the draw.)
The solution is exactly the same for Question 3.
To compute these numbers in Excel, $\begin{pmatrix}10 \\ 5\end{pmatrix}$ can be computed using
combin(10,5)
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Trying to solve my 2nd question using your method, I am running into an error. Using Excel, A1 (draw) = 5, B1 (deck) = 10, C1 (targets) = 3, D1 (Desired) = 0, 1, 2, or 3. What is wrong? My formula results in the following combinatoric equations (5,0)(7,5)/(10,5)... (5,1)(7,4)/(10,5)... (5,2)(7,3)/(10,5)... (5,3)(7,2)/(10,5)... It works for D1 = 0 and 3, but I am getting a value 1 <= x <= 2 for D1 = 1 and 2. – DiabloMonkey Aug 3 '12 at 16:21
I don't know specifically what is wrong. It might help to post your code. For reference, here are my computations for D = 1: $$\frac{\begin{pmatrix}3\\ 1\end{pmatrix}\begin{pmatrix}7\\ 4\end{pmatrix}}{\begin{pmatrix}10\\ 5\end{pmatrix}} \approx 0.4167$$ – Arkamis Aug 3 '12 at 16:25
@DiabloMonkey Ah, I see what the problem is. The numerator is the product of "Choosing D Card2 from 3 total Card2" and "Choosing (5-D) Card1 from 7 total Card1" Therefore, you should be using (3,0)(7,5) instead of (5,0)(7,5). The top numbers in your coefficients should equal the total number of cards in the deck. – Arkamis Aug 3 '12 at 16:28
That fixed the problem. Thank you for the help. – DiabloMonkey Aug 3 '12 at 16:45
I realized my typo that resulted from some careless copy-pasta; I have updated my answer to be more correct! Sorry for the confusion! – Arkamis Aug 3 '12 at 18:14
Question 1: You notice how for question 1, you did something like (13!/8!)/(52!/47!), in other words, you did (13!/(13-5)!)/(52!/(52-5)!). Let A1 be the "X" Cards and let B1 be "Y" Cards, and C1 is "Z", then for the answer cell D1, you have:
D1 = (FACT(C1)/FACT(C1-A1))/(FACT(B1)/FACT(B1-A1)) Actually, when you do X!/(X-Y)!, that is just a PERMUTATION method where you chose Y things out of X things, so xPy. So the answer is C1=PERMUT(C1,A1)/PERMUT(B1,A1)
Question 2: I am guessing when you say "0-3" copies you mean you want 4 answers. Let's make the amount of Card 2's (lets make them red, and card 1's blue) you want D1. For example, lets make D1=2. So what is the probability that out of the 5 cards you draw, 2 of them will be card 2. Algebraically, if you want the first two cards to be be red, first realize that there are 5!/(2!*3!) ways to arrange 2 red cards in a hand of 5. Selecting the cards 2s gives a chance of 3/10*2/9 while the blue cards give 7/8*6/7*5/6. Now lets put this in equation format.
5!/(2!*3!) -> COMB(A1, D1)
Red cards(numerator) -> PERMUT(C1, D1)
Blue Cards(numerator) -> PERMUT(B1 - C1, A1 - D1)
Denominator -> PERMUT(B1, A1)
What does that give us in the end?Answer=E1=(COMB(A1, D1) * PERMUT(C1, D1) * PERMUT(B1-C1, A1-D1))/PERMUT(B1, A1). Just make D1 = 0, 1, 2, and 3
Question 3 is the same deal.
I am positive about my solution to question 1, but someone might wanna check the other two.
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Thank you so much for explaining this (including factorial and combinatorical solutions)... I had forgotten about combinatorics (well, I still knew they existed but I had forgotten how to use them). I am still curious about how to solve the 2nd and 3rd parts... but trying to figure it out myself. – DiabloMonkey Aug 3 '12 at 15:37
DiabloMonkey, I made one huge mistake. It should be Permutations, not combinations. Use the PERMUT Command. I will edit my answer. – mathguy Aug 3 '12 at 15:40
Except in question 2, one of the things is still a combination. – mathguy Aug 3 '12 at 15:43
Question... Your answer E1=(COMBIN(x,a)*PERMUT(z,a)*PERMUT(y-z,x-a))/PERMUT(y,x)) works just the same as Ed Gorcenski's E1=(COMBIN(z,a)*COMBIN(y-z,x-a))/COMBIN(y,x)... What is the difference? The first seems to have 1 extra step, but it is using permutations instead of combinations? – DiabloMonkey Aug 3 '12 at 18:00
1
@DiabloMonkey: If you remember that PERMUT(a,b)=a!/(a-b)! and COMBIN(a,b)=a!/((b!)(a-b)!) and work through the algebra, you will see they are the same. – Ross Millikan Aug 6 '12 at 13:59
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http://mathhelpforum.com/calculus/4807-steepest-descent.html
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# Thread:
1. ## steepest descent
how can we proof this :
direction of steepest descent of a differentiable function F of M variables is the vector
( $-cdF/dw_1,...,-cdF/dw_M$)
where c is a positive constant of proportionality.
thanks
2. Originally Posted by AMR
how can we proof this :
direction of steepest descent of a differentiable function F of M variables is the vector
( $-cdF/dw_1,...,-cdF/dw_M$)
where c is a positive constant of proportionality.
thanks
Because that is the gradient of a diffrenciable function.
---
I can prove it for $f(x,y)$.
Let $\nabla f_0\not = 0$ at a point.
Let, $\bold{u}$ be any unit vector.
Then, directional derivative of,
$D_u f=\nabla f_0 \cdot \bold{u} = ||\nabla f_0||\cdot ||\bold{u}||\cos \theta$
The largest possible value of,
$\cos \theta =1$
Thus, maximum is,
$||\nabla f_0||(1)$
Thus, it is in the in the direction of the gradient. Thus, it is a multiple of the vector, hence the constant of proportionality.
---
I am thinking about several variables.
I guess the proof is based on finding the critical points. Since the critical points are determined by partial derivatives they appear over here. I will think more about it.
3. You have a function,
$f:\mathbb{R}^n\to \mathbb{R}$.
At point, $P(x_1,x_2,...,x_n)$
You have, $\nabla f\not = \bold{0}$.
Consider, a unit vector,
$\bold{u}=<u_1,...,u_n>$ and, $\sum_{k=1}^n u_k^2 = 1$
You need to find, what the components of $\bold{u}$ such as, $D_u f$ (directional derivative along this vector) is extremized.
But,
$D_u f =\nabla f \cdot \bold{u}$
Thus, you want to maximize,
$g(u_1,...,u_n)=\nabla f\cdot \bold{u}$
Thus,
$g(u_1,...,u_n)=\sum_{k=1}^n \frac{\partial f}{\partial x_k}u_k$
With constraint curve,
$c(u_1,...u_n)=\sum_{k=1}^n u_k^2=1$
Thus, Lagrange Multipliers,
$\nabla g=\kappa \nabla c$
Thus,
$\left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=\left<2\kappa u_1,...,2\kappa u_n\right>$
Since,
$\nabla f\not = \bold{0}$ thus, $\kappa \not = 0$
Thus,
$\bold{u}=\frac{1}{2\kappa}\left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=\frac{1}{2\kappa} \nabla f$
Thus, we see that every extremized directional derivative is a multiple of the gradient of the function.
---
Note, I used Lagrange Multipliers. The problem might be that the prove of Lagrange Multipliers for multi-variables more than 2 is based on the fact of the max/min property of the gradient. Thus, my proof was not a proof because it uses a fact that depends on itself.
4. thanks alot
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http://mathoverflow.net/questions/27967
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## Decidability of chess on an infinite board
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The recent question http://mathoverflow.net/questions/27944 of Tim Chow reminds me of a problem I have been interested in. Is chess with finitely many men on an infinite board decidable? In other words, given a position on an infinite board (say $\mathbb{Z}\times \mathbb{Z}$, though now pawn promotion is not possible) with finitely many men, say with White to move, is there an algorithm to determine whether White can checkmate Black (or prevent Black from checkmating White) against any Black defense?
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Have you tried to see if the Fraenkel-Liechtenstein construction is of any use? They show how to simulate a certain Boolean satisfiability game of Stockmeyer and Chandra using chess pieces. I think the Stockmeyer-Chandra game is based on a Turing-machine simulation, so on an infinite board with unboundedly many pieces, you might be able to simulate the halting problem. – Timothy Chow Jun 13 2010 at 2:46
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Computability questions very often depend on the format of the input. There are two things that need to be clarified: (1) What are the restrictions on the collection of pieces? Must both players have a king? Are they limited to the usual chess set? (2) How is the position fed as an input to the algorithm? The best option at the moment seems to be as a finite list of (piece, position) pairs that is encoded into a single number and passed to the algorithm all at once. – Carl Mummert Jul 20 2010 at 11:15
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Also, I guess we abandon the common tournament rule by which 50 moves without a capture or a pawn move causes stalemate? – Joel David Hamkins Jul 20 2010 at 11:33
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(continued) It goes without saying that this makes no "proof" but it's just my personal conviction out of actual game practice. On the other hand would be very interesting to come up with positions that would be drawn in the regular 8x8 chessboard, but can be won in an infinite (or very large) board thanks to the possibility of manouvering in a wider space. – Andrea Mori Aug 24 2010 at 13:18
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I found an argument for why the chess on the infinite board without the 50-move rule is probably undecidable: redhotpawn.com/board/… But I do not know whether that argument uses only finitely many pieces. – Tsuyoshi Ito Aug 24 2010 at 13:41
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## 5 Answers
There is a positive solution for the decidability of the mate-in-$n$ version of the problem.
Many of us are familiar with the White to mate in 3 variety of chess problems, and we may consider the natural analogue in infinite chess. Thus, we refine the winning-position problem, which asks whether a designated player has a winning strategy from a given position, to the mate-in-$n$ problem, which asks whether a designated player can force a win in at most $n$ moves from a given finite position. (And note that as discussed in Johan Wästlunds's question checkmate in $\omega$ moves?, there are finite winning positions in infinite chess which are not mate-in-$n$ for any finite $n$.) Even so, the mate-in-$n$ problem appears still to be very complicated, naturally formulated by assertions with $2n$ many alternating quantifiers: there is a move for white, such that for every black reply, there is a countermove by white, and so on. Assertions with such quantifier complexity are not generally decidable, and one cannot expect to search an infinitely branching game tree, even to finite depth. So one might naturally expect the mate-in-$n$ problem to be undecidable.
Despite this, the mate-in-n problem of infinite chess is computably decidable, and uniformly so. Dan Brumleve, myself and Philipp Schlicht have just submitted an article establishing this to the CiE 2012, and I hope to speak on it there in June.
D. Brumleve, J. D. Hamkins and P. Schlicht, "The mate-in-n problem of infinite chess is decidable," 10 pages, arxiv pre-print, submitted to CiE 2012.
Abstract. Infinite chess is chess played on an infinite edgeless chessboard. The familiar chess pieces move about according to their usual chess rules, and each player strives to place the opposing king into checkmate. The mate-in-$n$ problem of infinite chess is the problem of determining whether a designated player can force a win from a given finite position in at most n moves. A naive formulation of this problem leads to assertions of high arithmetic complexity with $2n$ alternating quantifiers---there is a move for white, such that for every black reply, there is a counter-move for white, and so on. In such a formulation, the problem does not appear to be decidable; and one cannot expect to search an infinitely branching game tree even to finite depth. Nevertheless, the main theorem of this article, confirming a conjecture of the first author and C. D. A. Evans, establishes that the mate-in-$n$ problem of infinite chess is computably decidable, uniformly in the position and in $n$. Furthermore, there is a computable strategy for optimal play from such mate-in-$n$ positions. The proof proceeds by showing that the mate-in-$n$ problem is expressible in what we call the first-order structure of chess $\frak{Ch}$, which we prove (in the relevant fragment) is an automatic structure, whose theory is therefore decidable. Unfortunately, this resolution of the mate-in-$n$ problem does not appear to settle the decidability of the more general winning-position problem, the problem of determining whether a designated player has a winning strategy from a given position, since a position may admit a winning strategy without any bound on the number of moves required. This issue is connected with transfinite game values in infinite chess, and the exact value of the omega one of chess $\omega_1^{\rm chess}$ is not known.
The solution can also be cast in terms of Presburger arithmetic, in a manner close to Dan Brumleve's answer to this question. Namely, once we restrict to a given collecton of pieces $A$, then we may represent all positions using only pieces in $A$ as a fixed-length tuple of natural numbers, and the elementary movement, attack and in-check relations are expressible for this representation in the language of Presburger arithmetic, essentially because the distance pieces---rooks, bishops and queens---all move on straight lines whose equations are expressible in Presburger arithmetic. (There is no need to handle sequence coding in general, since the number of pieces does not increase during play.) Since the mate-in-$n$ problem is therefore expressible in Presburger arithmetic, it follows that it is decidable.
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Do you cite MO? :-) – Joseph O'Rourke Jan 26 2012 at 20:49
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Yes, of course! We cite this very question, mentioning Richard Stanley specifically, in our opening paragraph as the origin of our interest in the problem. – Joel David Hamkins Jan 26 2012 at 21:37
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I added the links. Click on the title for my web page, and then follow through to the arxiv, where you will find the paper, which in addition to the theorem contains a nice infinite chess problem. (But don't peek at the answer at the back...) – Joel David Hamkins Jan 27 2012 at 2:22
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Very nice! As the authors point out, it doesn't answer my question (for which $n$ is not part of the input), but it is still an interesting contribution. – Richard Stanley Jan 27 2012 at 17:21
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Dan Brumleve has now joined as co-author of the paper. Philipp and I invited him to join us after realizing that his idea of using Presburger arithmetic amounted to essentially the same method of proof as the automatic structure approach. – Joel David Hamkins Jan 28 2012 at 0:29
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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.
There's no such algorithm under at least one rigorous specification of the question.
Consider a position in which there is a black king at (0,0) and no other black pieces; white rooks at $(1,5)$ and $(-1,5)$, and possibly a white queen at a position of the form (0,$n$) for some large $n$. Picture a king backed into a long tunnel. Then white can checkmate black (and does, from the start) if and only if the queen actually exists somewhere along the tunnel. If there are only two rooks, I believe that the king can stave off checkmate by moving off towards infinity indefinitely (see my explanation in the comments below).
Using positions like that, given any r.e. set $K$ and any number $m$, I can make a chess position such that white can checkmate black if and only if $m \in K$: I put the queen at $(0,n+5)$ if and only if $m$ is enumerated in $K$ after exactly $n$ steps. So I can decide membership in $K$, relative to an oracle for your problem. Actually I only need an oracle that works for indices of computable positions. This shows that any solution to your problem is of degree at least $0'$.
Note: I have taken a "position" to be a function from locations on the board to pieces. You could try to work around this solution by specifying something else as a "position". For example, you could make a position a function from a list of pieces to locations on the board, and that might lead to a different solution. However you need to require the list to explicitly say how many pieces there are from the beginning, or a variation of this solution will still apply.
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Carl, this solution is very interesting (+1), but it is dependent on the input being a computable function from position$\to$pieces. (For example, your solution relies on us not being able to compute the number of pieces from your input data.) But since the problem was to have only finitely many pieces, however, it would seem natural to have the input be the (complete) finite list of pieces and their positions. In this case, your method wouldn't work. – Joel David Hamkins Jul 20 2010 at 3:06
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For example, from that input data, your argument shows that we can't even decide if a given input position is already in checkmate, or whether a proposed move is a legal move, etc., although if we are given the complete finite position, then these questions would be decidable. – Joel David Hamkins Jul 20 2010 at 3:15
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No I can mate you with two rooks on the infinity board. Use a normal chess board, black will never reach the edge. To start, as above I can move around so to get Rooks c1/e1, Black d4 say, King h5. Black moves Kd5, then Kg6 Kd6, then Kf7, and Black can't move to d7 for Red1, and so the king must walk back down the d-file. White follows. Continuing, Kf7 Kd5, Kf6 Kd4, Kf5 Kd3, Kf4 Kd2, Kf3 Kd3, Red1 caput. Another presentation is white Rooks at c1/e8 to start, and the Black at d4 and white at h4. Then Black can run either up or down, and white just has to keep the right parity in either case. – Junkie Jul 20 2010 at 7:25
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This answer is technically correct but, in my opinion, it only shows why giving the input as a Turing machine which computes a function from locations on the board to pieces is a bad idea. – Tsuyoshi Ito Aug 23 2010 at 13:45
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Thank you for the reply, but I am still unconvinced. As Joel David Hamkins commented, many “basic” questions are undecidable with the input format used in your solution. Therefore, I find it difficult to agree that this kind of solution has to be stated in this much detail to establish that this input format is very difficult to handle. – Tsuyoshi Ito Aug 23 2010 at 15:44
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Yes. There is a finite board size that has essentially the same solution structure as all larger boards and an infinite board, and we can find it by solving the game completely for increasing board sizes until that ultimate size can be recognized. (The infinite board is different from a very large board only in that it has no edges or corners, so its solution structure is smaller.) Eventually this is possible, because there are only a finite number of pieces, and therefore only a finite number of equivalence classes of solution structures. When the number of such equivalence classes stops growing, we are done. Basically, all positions with pieces not close to the edge of the board but far from the center (say those with pieces inside an annulus) can be identified with more bounded positions (those surrounded by that annulus), because the pieces with an infinite number of options (Queen, Rook, Bishop) can go anywhere (up to parity) with no more than 2 moves, and the other pieces are forced to walk long distances for which all that matters are their relative path lengths to a certain precision. When one set of pieces moves far enough away from another set of pieces that the two sets are now in general position, these sub-problems interact in an equivalent way, so the description of the complete solution will eventually stop increasing with board size.
More generally this strategy works for any chess-like game in which the pieces can be classified into "ultra-mobile" and "para-mobile" types with constant and linear freedom respectively. For example it is also true for checkers because all pieces are para-mobile.
I am still unable to specify the equivalence between positions precisely enough to set an explicit bound on the necessary size of the finite board in terms of the number of pieces n, but I guess that it is O(f(n!)) for some polynomial function f whose order is O(n), or in other words O((n!)^(k*n)) for some k, using the construction technique that I have suggested. The super-exponential term is contributed by the para-mobile pieces and the polynomial by the ultra-mobiles. In any case I have intended only to demonstrate that the necessary board size has a computable bound.
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This is an intriguing idea for a solution, but it's not precise enough to be a proof yet. For example, the input convention matters (see my answer) but the sketch doesn't refer to it. Do you have a formal definition of the equivalence relation in mind? – Carl Mummert Aug 23 2010 at 11:06
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Or think about it this way: whenever it is to one player's advantage to continue increasing the variance of the position forever, the other player is willing to accept a draw. – Dan Brumleve Aug 23 2010 at 21:06
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I had once voted this answer up because I thought that it was “obviously correct,” but I canceled my upvote because I realized that I cannot be sure about its correctness. It is plausible that your idea works, but I cannot tell whether it really works or not from this sketch. – Tsuyoshi Ito Aug 24 2010 at 1:30
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This looks to me like a very fuzzy suggestion on how one might be able to do something, but more likely not. There are too many unclarified concepts (structure, what structure? What equivalence classes precisely?), and my feeling is they can never be clarified. – Andrej Bauer Aug 24 2010 at 11:41
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As for the downvote, I do not think it is healthy to worry about it too much. People can upvote or downvote for various, sometimes pretty arbitrary, reasons. I can imagine that many users vote a post up just because a poster is famous or just because the poster sounds confident. Similarly, I can imagine that some (hopefully not many) users vote a post down just because someone says it is wrong or simply because the voter lacks the knowledge required to understand it. – Tsuyoshi Ito Aug 25 2010 at 12:39
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Yes, because any chess position can be translated into Presburger arithmetic. For a fixed initial combination of piece types, let's define a position to consist of an (x, y) location for each piece as well as a bit (c) to indicate whether or not it has been captured. Multiplication of these parameters is not required to describe the legality and the effects of any one move, so in Presburger arithmetic we can recursively define the proposition "White cannot capture Black's King in fewer than t moves starting from initial position X.", then apply the axiom schema of induction to get an expression meaning "White cannot ever checkmate starting from initial position X." Since Presburger arithmetic is complete we will always be able to prove either this statement or its negation.
EDIT: Summary of how this is supposed to work:
1. P(A) = It is White's move and White has not yet won in position A.
2. R(A, B) = Position B legally follows in one move from position A.
3. Q(A, 0) = P(A)
4. Q(A, t) = for all B: (R(A, B) -> there exists C: R(B, C) -> Q(C, t - 1))
5. S(A) = Q(A, 0) and (for all t: Q(A, t) -> Q(A, t + 1))
This works fine at least up to line four where Q(A, t) is defined recursively. If Q(A, t) could be defined as a predicate in Presburger arithmetic then I think line five would also work. But this is a serious problem and maybe breaks the whole approach.
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Could you explain why Presburger arithmetic (which has only addition) is able to carry out this kind of encoding? The usual Goedel coding of such combinatorial operations, such as for Turing machines, would use something more like Peano Arithmetic or weakened forms. Presburger arithmetic, in contrast, cannot carry out that arithmetization, for example in the case of Turing machines, precisely because it is a decidable theory. – Joel David Hamkins Sep 5 2010 at 0:04
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Yes, there is induction, but only for assertions involving only addition. There is no Goedel encoding of finite sequences in Presburger arithmetic, for if there were, then its theory could not be decidable, since we could express the halting problem. – Joel David Hamkins Sep 5 2010 at 0:59
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Although Presburger arithmetic can prove all instances of the proof-by-induction principle that can be expressed in its (very limited) language, it cannot express definitions by recursion, such as those proposed in this answer. If definition by recursion were available, then we could use it to define multiplication and therefore have an undecidable theory. – Andreas Blass Sep 5 2010 at 2:49
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Although the inductive part of this answer is problematic, it does contain a correct account of the mate-in-n problem via Presburger arithmetic. For a fixed collection of pieces, one needn't do the Goedel coding of sequences inside the Presburger theory, because the number of pieces does not increase during play. Thus, one may represent a position with $r$ many pieces as a $3r+1$ tuple of natural numbers, in the manner Philipp Schlicht and I describe in our paper, and the movement and in-check relations are expressible in Presburger arithmetic, ...(cont'd) – Joel David Hamkins Jan 27 2012 at 14:31
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...essentially because the distance pieces move on straight lines whose equations are expressible in Presburger arithmetic. Thus, the mate-in-n problem for a fixed collection of pieces is expressible in Presburger arithmetic, and hence decidable. Similarly, one can get computable strategies from such positions in this way. I voted up this answer when it was first posted, but I feel it deserves more votes. – Joel David Hamkins Jan 27 2012 at 14:33
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I would like to convince everyone that this problem is undecidable. I cannot prove it for chess, as I lack the ability to design certain configurations but I think they must exist. And even if they don't, for some chess-like game they certainly do which shows that the attempts to prove decidability should be incorrect.
Hm, after reading the comments to the original question carefully, I realised that there is already a pointer to an argument very similar to mine by Tsuyoshi Ito: http://www.redhotpawn.com/board/showthread.php?threadid=90513&page=1#post_1708006 I still leave my proof here, as in fact two counters are enough and maybe mine is more detailed.
The reduction relies on the notion of counter machine*. It is undecidable whether a counter machine with only two counters halts or not. So our goal would be to simulate any such machine with a chess position. I can see two ways for this.
i, Build two separate configurations, such that both have a starting part and a moving part that can change (to store the state). Also, the moving parts would be connected, eg. by rooks, which could checkmate, if released, so this is why if one states moves 1, the other has to move k, and so on.
ii, Build a single configuration, that depending on its state, moves l horizontally and -k vertically. Also, place a rook at (0,0) that would never move but could guarantee that the configuration can "sense" when it gets back to an empty counter.
So all left to do is to design such configurations, which I guess should be possible with some effort and knowledge of chess. Also, note that in both cases the construction uses a piece whose range is not bounded, I wonder if this is really necessary.
*I realized that the definition on wikipedia is different from what I want. In fact, my machine should be probably called a 2-stack machine that can push only one letter to the stack. So I want a finite state machine with two counters that are empty at the beginning and it can increase or decrease any counter by one or check whether a counter is zero or not. The problem of whether such a machine halts or not, is undecidable.
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http://physics.aps.org/synopsis-for/print/10.1103/PhysRevB.82.180513
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# Synopsis:
Outstanding in the field
#### Anomalous Meissner effect in pnictide superconductors
R. Prozorov, M. A. Tanatar, Bing Shen, Peng Cheng, Hai-Hu Wen, S. L. Bud’ko, and P. C. Canfield
Published November 18, 2010
When a superconductor in a magnetic field is cooled below its transition temperature, it expels the field. The details of how this Meissner effect occurs in real materials depend on the shape and physical properties of the sample, the type of superconductivity it exhibits, and the experimental conditions, but all superconductors are expected to behave as follows: As long as the external magnetic field is below a certain critical value, currents on the surface of the superconductor will form to cancel the field inside, but above this critical field magnetic flux will start to penetrate. The experimental signature of this field dependence is that the magnetization of the superconductor, which opposes the applied field, reaches a maximum at the critical field and then starts to decrease.
However, in a Rapid Communication published in Physical Review B, Ruslan Prozorov and collaborators from Ames Laboratory at Iowa State University, US, and the Institute of Physics of the Chinese Academy of Science in Beijing, show that two iron arsenide (pnictide) superconductors seem to defy this expected behavior. The researchers find that the magnetizations of $Ba(Fe0.926Co0.074)2As2$ and $Ba0.6K0.4Fe2As2$ continually increase in an approximately linear fashion, without reaching a maximum, even when the applied field far exceeds the estimated critical field for either material. Based on their results, Prozorov et al. suggest that the magnetic field suppresses magnetic scattering, yielding more resilient Cooper pairing in iron arsenide compounds than in other superconductors, but more experiments are needed to support or refute this proposal. – Matthew Eager
ISSN 1943-2879. Use of the American Physical Society websites and journals implies that the user has read and agrees to our Terms and Conditions and any applicable Subscription Agreement.
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http://unapologetic.wordpress.com/2007/05/22/functors/?like=1&source=post_flair&_wpnonce=97257dee4e
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# The Unapologetic Mathematician
## Functors
As with all the other algebraic structures we’ve considered, we’re interested in the “structure-preserving maps” between categories. In this case, they’re called “functors”.
A functor $F$ from a category $\mathcal{C}$ to a category $\mathcal{D}$ consists of two functions, both also called $F$. One sends objects of $\mathcal{C}$ to objects of $\mathcal{D}$, and the other sends morphisms of $\mathcal{C}$ to morphisms of $\mathcal{D}$. Of course, these are subject to a number of restrictions:
• If $m$ is a morphism from $X$ to $Y$ in $\mathcal{C}$, then $F(m)$ is a morphism from $F(X)$ to $F(Y)$ in $\mathcal{D}$.
• For every object $X$ of $\mathcal{C}$, we have $F(1_X)=1_{F(X)}$ in $\mathcal{D}$ — identities are sent to identities.
• Given morphisms $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ in $\mathcal{C}$, we have $F(g\circ f)=F(g)\circ F(f)$ in $\mathcal{D}$ — a functor preserves compositions.
It’s tempting at this point to think of a “category of categories”, but unfortunately this gets hung up on the same hook as the “set of sets”. A lot of the intuition goes through, however, and we do have a category $\mathbf{Cat}$ of small categories (with only a set of objects and a set of morphisms) and functors between them.
Every category $\mathcal{C}$ comes with an identity functor $1_\mathcal{C}$. This is an example of an “endofunctor” (in analogy with “endomorphism”).
Every category of algebraic structures we’ve considered — $\mathbf{Grp}$, $\mathbf{Mon}$, $\mathbf{Ring}$, $R-\mathbf{mod}$, etc. — comes with a “forgetful” functor to the category of sets. Remember that a group (for example) is a set with extra structure on top of it, and a group homomorphism is a function that preserves the group structure. If we forget all that extra structure we’re just left with sets and functions again.
To be explicit, there is a functor $U:\mathbf{Grp}\rightarrow\mathbf{Set}$ that sends a group $(G,\cdot)$ to its underlying set $G$. It sends a homomorphism $f:G\rightarrow H$ to itself, now considered as a function on the underlying sets. It should be apparent that this sends the identity homomorphism on the group $G$ to the identity function on the set $G$, and that it preserves compositions. The same arguments go through for rings, monoids, $R$-modules.
In fact, there are other forgetful functors that behave in much the same way. A ring is an abelian group with extra structure, so we can forget that structure to get a functor from $\mathbf{Ring}$ to $\mathbf{Ab}$ — the category of abelian groups. An abelian group, in turn, is a restricted kind of group. We can forget the restriction to get a functor from $\mathbf{Ab}$ to $\mathbf{Grp}$.
Now for some more concrete examples. Remember that a monoid is a category with one object. So what’s a functor between such monoids? Consider monoids $M$ and $N$ as categories. Then there’s only one object in each, so the object function is clear. We’re left with a function on the morphisms sending the identity of $M$ to the identity of $N$ and preserving compositions — a monoid homomorphism!
What about functors between preorders, considered as categories? Now all the constraints are on the object function. Consider preorders $(P,\leq)$ and $(Q,\preceq)$ as categories. If there is an arrow from $a$ to $b$ in $P$ then there must be an arrow from $F(a)$ to $F(b)$. That is, if $a\leq b$ then $F(a)\preceq F(b)$. Functors in this case are just order-preserving functions.
These two examples show how the language of categories and functors subsumes both of these disparate notions. Preorder relations translate into the existence of certain arrows, which functors must then preserve, while monoidal multiplications translate into compositions of arrows, which functors must then preserve. The categories of (preorders, order-preserving functions) and (monoids, monoid homomorphisms) both find a natural home with in the category of (small categories, functors).
### Like this:
Posted by John Armstrong | Category theory
## 4 Comments »
1. [...] Transformations and Functor Categories So we know about categories and functors describing transformations between categories. Now we come to transformations between functors [...]
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2. [...] algebras — were all asserted to be the “free” constructions. This makes them functors from the category of vector spaces over to appropriate categories of -algebras, and that means [...]
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3. [...] closely, we’ll find that what we’ve defined as a presheaf is actually a contravariant functor from this category to the category of sets! For every arrow we have an arrow — in the [...]
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4. [...] that we ended up defining a presheaf as a functor. Given our topological space we set up the partial order category , flipped it around to so the [...]
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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://mathhelpforum.com/differential-equations/161766-firefly-dynamics.html
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# Thread:
1. ## Firefly Dynamics
I'm trying to figure out this example on one-dimensional flows on a circle using the flash of fireflies.
The information/example is given here:
http://www.paleo.bris.ac.uk/~ggxir/c.../lecture-4.pdf
which starts at page 14, but the equations are given on page 19.
I'm moreover confused about the information given.
$\theta(t)$ is said to be the firefly phase. I'm not sure what this means; I interpret it as the amount of flashes that have gone by since some time t. Next $\frac{d\theta}{dt}=\omega$ where $\omega$ is a constant. I think this means that the rate at which a firefly flashes is constant.
Next the equation for the stimulus is given in a similar manner. $\Theta(t)$ is the phase of the stimulus and $\frac{d\Theta}{dt}=\Omega$.
With the introduction of the stimulus firefly, $\frac{d\theta}{dt}$ becomes
$\frac{d\theta}{dt}=\omega+A\sin(\Theta-\theta)$
Which I don't completely understand why that is. I'm confused on why sine is in the new equation.
Clarification on what these equations represent would be appreciated.
Thank you.
2. Sorry to bump this topic, but here is a picture on exactly what the model is described as:
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http://mathoverflow.net/questions/79712?sort=oldest
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## associated sheaf functor doesn’t preserve arbitrary products
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I need example that associated sheaf functor doesn't preserve arbitrary products. I think that one can provide an example for sheaves over topological space. Thanks for your help.
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Homework? And what associated sheaf functor? Do you mean the sheafification functor from presheaves to sheaves? Or something else? Are you allowing sheaves over sites (it seems you must be because you talk about providing an example of sheaves over a topological space)? Finally, if you think you have an example - actually give your example! – Anon Nov 1 2011 at 14:16
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Sorry for my bad english, which leads to some misunderstanding. It's not a homework. Yes I meant sheafification functor. I strongly believe there is example over site of open subsets of a topological space. – Anonymous Nov 1 2011 at 15:35
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Why do you need this example? – S. Carnahan♦ Nov 1 2011 at 18:59
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I don't know why Anon would think this is homework, or why someone is voting to close this. It is a perfectly reasonable question for someone who is not expert in topos theory. – Todd Trimble Nov 2 2011 at 15:40
## 2 Answers
An example where sheafification does not preserve arbitrary products is where we take sheaves over a (sober) space $X$ that is not locally connected, for example the space of irrationals or Cantor space.
Recall that a Grothendieck topos $E$ is locally connected if the (essentially unique) geometric morphism $\Gamma = f_\ast: E \to Set$ has a left adjoint $f^\ast$ that in turn has a left adjoint. More generally, a geometric morphism $f_\ast: E \to F$ between toposes is an essential geometric morphism if its left adjoint $f^\ast$ has a left adjoint.
We have the following facts:
• A presheaf topos $Set^{C^{op}}$ is locally connected. Here the left adjoint to the global sections functor $\Gamma: Set^{C^{op}} \to Set$ is the diagonal functor $\Delta: Set \to Set^{C^{op}}$, which of course has a left adjoint.
• A geometric morphism $f_\ast: E \to F$ is essential if and only if $f^\ast$ preserves arbitrary products. (Of course $f^\ast$ is already left exact and so preserves equalizers, if it preserves small products, then it preserves small limits. Using the fact that Grothendieck toposes are cototal, this is enough to ensure that $f^\ast$ has a left adjoint.)
• Then in particular for a small site $(C, J)$, the inclusion functor $i: Sh(C, J) \to Set^{C^{op}}$ is an essential geometric morphism if and only if sheafification $a: Set^{C^{op}} \to Sh(C, J)$ preserves small products. Thus, putting the last two facts together, the composite geometric morphism $$Sh(C, J) \stackrel{i}{\to} Set^{C^{op}} \stackrel{\Gamma}{\to} Set$$ is essential, i.e., $Sh(C, J)$ is locally connected, if $a$ preserves products.
• In the case where the site is $(\text{Open}(X), J)$ where $J$ is the canonical Grothendieck topology given by covering families, $Sh(X) = Sh(C, J)$ is locally connected if and only if $X$ is a locally connected space. Thus $a: Set^{\text{Open}(X)^{op}} \to Sh(X)$ preserves small products only if $X$ is locally connected.
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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.
Let $X$ be a space. An abelian group (or a set, if you prefer) $G$ determines a constant presheaf, call it $C_G$, and the associated sheaf is the sheaf of locally constant maps into $G$. Given a family $G_i$ with product $G$, the product of the presheaves $C_{G_i}$ is the presheaf $C_G$, but the product of the associated sheaves is not in general the sheaf of locally constant maps to $G$. (If $X$ is not locally connected then one can easily have a map to $G$ which projects to a locally constant map to $G_i$ for each $i$ but which is not locally constant itself. )
This is presumably a down-to-earth special case of what Todd Trimble is saying.
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Not only down-to-earth, but also easy to understand: 1+ – Martin Brandenburg Nov 2 2011 at 19:18
It could be more down-to-earther if $G$, $X$ and $G_i$'s were specified... – Andrej Bauer Oct 8 at 15:46
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http://mathoverflow.net/questions/109196?sort=votes
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## is there any bound on the absolute number of algebraic integer in terms of its degree?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
If Z is a sum of t distinct roots of unity and |Z| is a rational integer, can someone find a bound on |Z| in terms of k=deg(Q(Z):Q))?
Clearly we need to have distinct roots of unity otherwise this won't work!
Correction: Let assume that Z is not rational itself otherwise obviously it's wrong. Here I hope to extend the proof of Kronecker thm! I have "Z is a sum of t distinct roots of unity and |Z| is a rational integer" I conjecture that either Z is rational or a root of unity!
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Not sure what you are asking but you might want to check: Loxton, J. H. On two problems of R. W. Robinson about sums of roots of unity. Acta Arith. 26 (1974/75), 159–174. – Felipe Voloch Oct 9 at 2:04
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Even with the correction, all you have to do is multiply Will Sawin's sum by, say, a nonreal cube root of 1, $\omega$. Then $Z=-n\omega$ is not rational, $|Z|=n$ is a rational integer, and $k=2$. – Gerry Myerson Oct 10 at 4:06
@Gerry: sorry I edited my answer to fix it before seeing your identical comment. – Will Sawin Oct 14 at 17:20
## 1 Answer
If you multiply all the primitive roots of unity for the first $n$ primes by $-i$, then add them, you get $in$, whose degree is $2$ and whose absolute value is $n$, rational and unbounded.
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Dear Will, Doesn't the question ask that $|Z|$ (where $Z$ is the sum of roots of unity in question) be an integer? Regards, – Emerton Oct 9 at 2:25
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I edited it to a more elegant solution that deals with that issue. – Will Sawin Oct 9 at 3:17
In fact, this construction allows you to write any sum of roots of unity as a sum of distinct roots of unity. – Will Sawin Oct 14 at 22:46
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http://nrich.maths.org/6301
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nrich enriching mathematicsSkip over navigation
### Real-life Equations
Here are several equations from real life. Can you work out which measurements are possible from each equation?
### Curve Fitter
Can you fit a cubic equation to this graph?
### Guess the Function
This task depends on learners sharing reasoning, listening to opinions, reflecting and pulling ideas together.
# Equation Matcher
##### Stage: 5 Challenge Level:
Fitting algebraic curves through experimental data points is an important scientific process which allows us to make predictions of the behaviour of a system away from the observed points. But which equations are sensible choices for a particular set of data? This question draws us into the general process of curve fitting.
In four different sorts of experiments, two values of $x$ and $y$ are successively measured and plotted on charts below:
For each chart, which of the following equations are possible descriptions of the underlying process?
$$y = ax+b\quad\quad x = ay + \frac{1}{2} by^2\quad\quad y=\frac{a}{x}$$
$$x=a e^{by}\quad\quad x = \frac{ay}{b+y}\quad\quad x = -a\log_{10}(by)$$
What restrictions would the data place on the ranges of the numbers $a$ and $b$ in each case? For example, do the data points imply that the constants will be positive?
Another point is subsequently measured and plotted on the graphs in each case.
How do the possibilities now change?
Can you give examples of physical systems which are modelled by equations of these types?
Extension: Which curves can be distinguished by three points? i. e., could you pick three points which would rule out all but one of the possibilities in each case?
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://math.stackexchange.com/questions/23814/how-best-to-explain-the-sqrt2-pi-n-term-in-stirlings/23822
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# How best to explain the $\sqrt{2\pi n}$ term in Stirling's?
I recently showed my Algorithms class how to bound $\ln n! = \sum \ln n$ by integrals, thereby obtaining the simple factorial approximation
$$e \left(\frac{n}{e}\right)^{n} \leq n! \leq en\left(\frac{n}{e}\right)^{n}$$
But one student, having seen Stirling's approximation, asked where the $\sqrt{2 \pi n}$ term comes from in
$$n! \sim \left(\frac{n}{e}\right)^n \sqrt{2 \pi n}$$
which I couldn't readily answer. Looking around, I can't find a straightforward and intuitive explanation. Can someone explain this in terms understandable to a Computer Science undergrad (ie, no analysis beyond first-year calculus)?
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– Matt Calhoun Feb 26 '11 at 18:30
## 5 Answers
This blog post by Terence Tao explains how to get this constant from the central limit theorem (equivalently, from the normalization factor we use to define the normal distribution). The question remains where it came from in the central limit theorem, and I think the most honest answer to that question is Fourier analysis; $\frac{1}{\sqrt{2\pi}}$ is the normalization factor which makes the Fourier transform and its inverse look the same, and the Gaussian distribution is its own Fourier transform.
(That is, I don't think you should look for a straightforward and intuitive explanation because I think this is actually a rather deep fact which is not really explained in most courses. On the other hand, just the $\sqrt{n}$ term is straightforward to explain: instead of using the left Riemann sum or the right Riemann sum, you use the trapezoid rule.)
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1
Do you know where to find "article of Joe Keller" referred in Tao's post? The link does not seem to work any more. Thanks. – Qiang Li Feb 26 '11 at 17:47
@Qiang: unfortunately I do not. – Qiaochu Yuan Feb 26 '11 at 18:03
4
– J. M. Apr 10 '11 at 18:13
One way is to apply Euler-McLaurin Summation (which can be viewed as a clever application of repeated integration by parts) to $\displaystyle \log x$ to show that
$$n! \sim C \sqrt{n} \left(\frac{n}{e}\right)^n$$
Then use Wallis formula (another clever application of integration by parts)
$$\frac{\pi}{2} = \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3} \times \frac{4}{5} \dots$$
to show that $\displaystyle C = \sqrt{2\pi}$.
If I recall correctly, this is how this was derived historically.
Of course, this does not give an intuitive reason as to why the constant is $\displaystyle \sqrt{2\pi}$
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2
There is a saying that any analyst worth his salt can do wonders with only Cauchy-Schwarz and integration by parts. – Ragib Zaman Nov 18 '11 at 11:33
I'm not exactly sure how much analysis is covered by first-year calculus, but this is probably the shortest proof I know of this fact. It doesn't go very far in terms of making this intuitive, though, and I don't know that this can be done at all.
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I think that the most intuitive explanation of this factor is as the total mass of a Gaussian integral in the evaluation of the stirling formual by means of steepest descent on the Gamma function defining integral. See, for example, section 2.4 of the following lecture note.
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2
If all you want is the constant $\sqrt{2\pi}$, you can get it by using the normal approximation to approximate the central binomial coefficient. – Yuval Filmus Feb 26 '11 at 7:14
If You want to see the connection between the term $\sqrt{\pi}$ and factorial, explained in the geometric terms, You can watch Knuth's lecture entitled Why Pi?
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http://terrytao.wordpress.com/tag/geodesic-flow/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Tag Archive
You are currently browsing the tag archive for the ‘geodesic flow’ tag.
## The Euler-Arnold equation
7 June, 2010 in expository, math.AP, math.DG, math.DS | Tags: Euler equation, Euler-Arnold equation, geodesic flow, Lax pair, Vladimir Arnold | by Terence Tao | 18 comments
A (smooth) Riemannian manifold is a smooth manifold ${M}$ without boundary, equipped with a Riemannian metric ${{\rm g}}$, which assigns a length ${|v|_{{\rm g}(x)} \in {\bf R}^+}$ to every tangent vector ${v \in T_x M}$ at a point ${x \in M}$, and more generally assigns an inner product
$\displaystyle \langle v, w \rangle_{{\rm g}(x)} \in {\bf R}$
to every pair of tangent vectors ${v, w \in T_x M}$ at a point ${x \in M}$. (We use Roman font for ${g}$ here, as we will need to use ${g}$ to denote group elements later in this post.) This inner product is assumed to symmetric, positive definite, and smoothly varying in ${x}$, and the length is then given in terms of the inner product by the formula
$\displaystyle |v|_{{\rm g}(x)}^2 := \langle v, v \rangle_{{\rm g}(x)}.$
In coordinates (and also using abstract index notation), the metric ${{\rm g}}$ can be viewed as an invertible symmetric rank ${(0,2)}$ tensor ${{\rm g}_{ij}(x)}$, with
$\displaystyle \langle v, w \rangle_{{\rm g}(x)} = {\rm g}_{ij}(x) v^i w^j.$
One can also view the Riemannian metric as providing a (self-adjoint) identification between the tangent bundle ${TM}$ of the manifold and the cotangent bundle ${T^* M}$; indeed, every tangent vector ${v \in T_x M}$ is then identified with the cotangent vector ${\iota_{TM \rightarrow T^* M}(v) \in T_x^* M}$, defined by the formula
$\displaystyle \iota_{TM \rightarrow T^* M}(v)(w) := \langle v, w \rangle_{{\rm g}(x)}.$
In coordinates, ${\iota_{TM \rightarrow T^* M}(v)_i = {\rm g}_{ij} v^j}$.
A fundamental dynamical system on the tangent bundle (or equivalently, the cotangent bundle, using the above identification) of a Riemannian manifold is that of geodesic flow. Recall that geodesics are smooth curves ${\gamma: [a,b] \rightarrow M}$ that minimise the length
$\displaystyle |\gamma| := \int_a^b |\gamma'(t)|_{{\rm g}(\gamma(t))}\ dt.$
There is some degeneracy in this definition, because one can reparameterise the curve ${\gamma}$ without affecting the length. In order to fix this degeneracy (and also because the square of the speed is a more tractable quantity analytically than the speed itself), it is better if one replaces the length with the energy
$\displaystyle E(\gamma) := \frac{1}{2} \int_a^b |\gamma'(t)|_{{\rm g}(\gamma(t))}^2\ dt.$
Minimising the energy of a parameterised curve ${\gamma}$ turns out to be the same as minimising the length, together with an additional requirement that the speed ${|\gamma'(t)|_{{\rm g}(\gamma(t))}}$ stay constant in time. Minimisers (and more generally, critical points) of the energy functional (holding the endpoints fixed) are known as geodesic flows. From a physical perspective, geodesic flow governs the motion of a particle that is subject to no external forces and thus moves freely, save for the constraint that it must always lie on the manifold ${M}$.
One can also view geodesic flows as a dynamical system on the tangent bundle (with the state at any time ${t}$ given by the position ${\gamma(t) \in M}$ and the velocity ${\gamma'(t) \in T_{\gamma(t)} M}$) or on the cotangent bundle (with the state then given by the position ${\gamma(t) \in M}$ and the momentum ${\iota_{TM \rightarrow T^* M}( \gamma'(t) ) \in T_{\gamma(t)}^* M}$). With the latter perspective (sometimes referred to as cogeodesic flow), geodesic flow becomes a Hamiltonian flow, with Hamiltonian ${H: T^* M \rightarrow {\bf R}}$ given as
$\displaystyle H( x, p ) := \frac{1}{2} \langle p, p \rangle_{{\rm g}(x)^{-1}} = \frac{1}{2} {\rm g}^{ij}(x) p_i p_j$
where ${\langle ,\rangle_{{\rm g}(x)^{-1}}: T^*_x M \times T^*_x M \rightarrow {\bf R}}$ is the inverse inner product to ${\langle, \rangle_{{\rm g}(x)}: T_x M \times T_x M \rightarrow {\bf R}}$, which can be defined for instance by the formula
$\displaystyle \langle p_1, p_2 \rangle_{{\rm g}(x)^{-1}} = \langle \iota_{TM \rightarrow T^* M}^{-1}(p_1), \iota_{TM \rightarrow T^* M}^{-1}(p_2)\rangle_{{\rm g}(x)}.$
In coordinates, geodesic flow is given by Hamilton’s equations of motion
$\displaystyle \frac{d}{dt} x^i = {\rm g}^{ij} p_j; \quad \frac{d}{dt} p_i = - \frac{1}{2} (\partial_i {\rm g}^{jk}(x)) p_j p_k.$
In terms of the velocity ${v^i := \frac{d}{dt} x^i = {\rm g}^{ij} p_j}$, we can rewrite these equations as the geodesic equation
$\displaystyle \frac{d}{dt} v^i = - \Gamma^i_{jk} v^j v^k$
where
$\displaystyle \Gamma^i_{jk} = \frac{1}{2} {\rm g}^{im} (\partial_k {\rm g}_{mj} + \partial_j {\rm g}_{mk} - \partial_m {\rm g}_{jk} )$
are the Christoffel symbols; using the Levi-Civita connection ${\nabla}$, this can be written more succinctly as
$\displaystyle (\gamma^* \nabla)_t v = 0.$
If the manifold ${M}$ is an embedded submanifold of a larger Euclidean space ${R^n}$, with the metric ${{\rm g}}$ on ${M}$ being induced from the standard metric on ${{\bf R}^n}$, then the geodesic flow equation can be rewritten in the equivalent form
$\displaystyle \gamma''(t) \perp T_{\gamma(t)} M,$
where ${\gamma}$ is now viewed as taking values in ${{\bf R}^n}$, and ${T_{\gamma(t)} M}$ is similarly viewed as a subspace of ${{\bf R}^n}$. This is intuitively obvious from the geometric interpretation of geodesics: if the curvature of a curve ${\gamma}$ contains components that are transverse to the manifold rather than normal to it, then it is geometrically clear that one should be able to shorten the curve by shifting it along the indicated transverse direction. It is an instructive exercise to rigorously formulate the above intuitive argument. This fact also conforms well with one’s physical intuition of geodesic flow as the motion of a free particle constrained to be in ${M}$; the normal quantity ${\gamma''(t)}$ then corresponds to the centripetal force necessary to keep the particle lying in ${M}$ (otherwise it would fly off along a tangent line to ${M}$, as per Newton’s first law). The precise value of the normal vector ${\gamma''(t)}$ can be computed via the second fundamental form as ${\gamma''(t) = \Pi_{\gamma(t)}( \gamma'(t), \gamma'(t) )}$, but we will not need this formula here.
In a beautiful paper from 1966, Vladimir Arnold (who, sadly, passed away last week), observed that many basic equations in physics, including the Euler equations of motion of a rigid body, and also (by which is a priori a remarkable coincidence) the Euler equations of fluid dynamics of an inviscid incompressible fluid, can be viewed (formally, at least) as geodesic flows on a (finite or infinite dimensional) Riemannian manifold. And not just any Riemannian manifold: the manifold is a Lie group (or, to be truly pedantic, a torsor of that group), equipped with a right-invariant (or left-invariant, depending on one’s conventions) metric. In the context of rigid bodies, the Lie group is the group ${SE(3) = {\bf R}^3 \rtimes SO(3)}$ of rigid motions; in the context of incompressible fluids, it is the group ${Sdiff({\bf R}^3}$) of measure-preserving diffeomorphisms. The right-invariance makes the Hamiltonian mechanics of geodesic flow in this context (where it is sometimes known as the Euler-Arnold equation or the Euler-Poisson equation) quite special; it becomes (formally, at least) completely integrable, and also indicates (in principle, at least) a way to reformulate these equations in a Lax pair formulation. And indeed, many further completely integrable equations, such as the Korteweg-de Vries equation, have since been reinterpreted as Euler-Arnold flows.
From a physical perspective, this all fits well with the interpretation of geodesic flow as the free motion of a system subject only to a physical constraint, such as rigidity or incompressibility. (I do not know, though, of a similarly intuitive explanation as to why the Korteweg de Vries equation is a geodesic flow.)
One consequence of being a completely integrable system is that one has a large number of conserved quantities. In the case of the Euler equations of motion of a rigid body, the conserved quantities are the linear and angular momentum (as observed in an external reference frame, rather than the frame of the object). In the case of the two-dimensional Euler equations, the conserved quantities are the pointwise values of the vorticity (as viewed in Lagrangian coordinates, rather than Eulerian coordinates). In higher dimensions, the conserved quantity is now the (Hodge star of) the vorticity, again viewed in Lagrangian coordinates. The vorticity itself then evolves by the vorticity equation, and is subject to vortex stretching as the diffeomorphism between the initial and final state becomes increasingly sheared.
The elegant Euler-Arnold formalism is reasonably well-known in some circles (particularly in Lagrangian and symplectic dynamics, where it can be viewed as a special case of the Euler-Poincaré formalism or Lie-Poisson formalism respectively), but not in others; I for instance was only vaguely aware of it until recently, and I think that even in fluid mechanics this perspective to the subject is not always emphasised. Given the circumstances, I thought it would therefore be appropriate to present Arnold’s original 1966 paper here. (For a more modern treatment of these topics, see the books of Arnold-Khesin and Marsden-Ratiu.)
In order to avoid technical issues, I will work formally, ignoring questions of regularity or integrability, and pretending that infinite-dimensional manifolds behave in exactly the same way as their finite-dimensional counterparts. In the finite-dimensional setting, it is not difficult to make all of the formal discussion below rigorous; but the situation in infinite dimensions is substantially more delicate. (Indeed, it is a notorious open problem whether the Euler equations for incompressible fluids even forms a global continuous flow in a reasonable topology in the first place!) However, I do not want to discuss these analytic issues here; see this paper of Ebin and Marsden for a treatment of these topics.
Read the rest of this entry »
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http://mathhelpforum.com/algebra/96485-division-log-base-10-ln.html
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# Thread:
1. ## Division by log base 10 and ln
Had a question someone might be able to answer...
Why do I get the same number if I divide, let's say log5/log3, as if I divided ln5/ln3???
What exactly is the relationship between logs and natural logs that makes this true?
Thanks in advance.
M
2. Those better not yield the same answer.
log(5)=x where x satisfies $10^x=5$
ln(5)=x where x satisfies $e^x=5$
log(3)=x where x satisfies $10^x=3$
ln(3)=x where x satisfies $e^x=3$
If you got the same answer when you calculated these ratios you have made an error somewhere.
3. If you are interested, you can express logs in different bases in terms of the natural log as follows.
$y=log_a(b)\Rightarrow a^y=b\Rightarrow ln(a^y)=ln(b)\Rightarrow yln(a)=ln(b) \Rightarrow y=\frac{ln(b)}{ln(a)}$
So we see that
$log_a(b)=\frac{ln(b)}{ln(a)}$
4. Right, that makes sense. I think I understand logs somewhat... The thing is when I was working a problem and I input the data into my calculator, I got the same answer whether I hit the log key or the ln key.
5. hmm now that is strange. generally log without any subscript is understood to be log base 10. Whereas ln is to be log base e. might be worth seeing whether it is giving you ln or log though so you can apply the change of base stuff as necessary. try log(10) and ln(e) if you guess the base right you should get 1. That oughta at least tell you which base it is giving you.
6. I thought so too, but we did several examples to prove what I don't know, that's what I am trying to figure out...lol.
I have been curious since Monday and I have tried researching all I can about logarithms. I don't have either of my calculators on me but I am going to keep trying to figure it out.
Thanks for your help though. If you have a calculator handy, input the numbers, first using the log key then use the ln key. Weird.
7. oh dude, yeah you are right they should be the same, I was thinking $ln(5/3)$ and $log(5/3)$ my bad dude. Here is why it is true.
$a=log(5)\Rightarrow 10^a=5 \Rightarrow a=\frac{ln(5)}{ln(10)}$
$b=log(3)\Rightarrow 10^b=5 \Rightarrow b=\frac{ln(3)}{ln(10)}$
$\frac{log(5)}{log(3)}=\frac{a}{b}=\frac{\frac{ln(5 )}{ln(10)}}{\frac{ln(3)}{ln(10)}}=\frac{ln(5)}{ln( 3)}$
8. Originally Posted by merrysoul
Why do I get the same number if I divide, let's say log5/log3, as if I divided ln5/ln3??
It's the change-of-base formula:
log_b(x) = log_c(x)/log_c(b)
This is because:
log_b(x) = y
x = b^y
log_c(x) = log_c(b^y) = y*log_c(b)
log_c(x)/log_c(b) = y = log_b(x)
By applying the change-of-base formula twice (once backwards and once forwards), we get:
log_10(5)/log_10(3) = log_3(5) = ln(5)/ln(3)
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http://mathhelpforum.com/calculus/146523-vectors-scalar-equation-plane-symmetric-equation-plane-problem.html
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# Thread:
1. ## Vectors - Scalar equation of a plane/symmetric equation of a plane problem
Hey, thanks for coming into my thread!
I have a problem involving vectors that I do not know how to start solving.
Question
Find the value of k for which the plane $kx + 4y + 2z - 6 = 0$ is parallel to the line $\frac{x - 3}{5} = \frac{y}{1} = \frac {z}{-3}$.
Solution
I know that (k, 4, 2) will be perpendicular to a direction vector on the plane given in the symmetric equation.
Rearranging the symmetric equation gives me:
$x = 5t + 3$
$y = t$
$z = -3t$
Which I believe gives me a direction vector on that plane of: (5, 1, -3).
With this information, how can I find the value of k? If I had a second direction vector of the parallel plane I could do a cross product, so is it possible to find that?
Thanks!
2. Originally Posted by Kakariki
Hey, thanks for coming into my thread!
I have a problem involving vectors that I do not know how to start solving.
Question
Find the value of k for which the plane $kx + 4y + 2z - 6 = 0$ is parallel to the line $\frac{x - 3}{5} = \frac{y}{1} = \frac {z}{-3}$.
Solution
I know that (k, 4, 2) will be perpendicular to a direction vector on the plane given in the symmetric equation.
Rearranging the symmetric equation gives me:
$x = 5t + 3$
$y = t$
$z = -3t$
Which I believe gives me a direction vector on that plane of: (5, 1, -3).
With this information, how can I find the value of k? If I had a second direction vector of the parallel plane I could do a cross product, so is it possible to find that?
Thanks!
All your considerations and calculations are OK.
If the direction vector of the line is perpendicular to the normal vector of the plane then the plane must be parallel to the line.
Attached Thumbnails
3. Originally Posted by earboth
All your considerations and calculations are OK.
If the direction vector of the line is perpendicular to the normal vector of the plane then the plane must be parallel to the line.
Would doing a dot product help? setting it equal to zero and then solving for k?
4. Originally Posted by Kakariki
Would doing a dot product help? setting it equal to zero and then solving for k?
Correct!
5. Originally Posted by earboth
Correct!
SWEET! Thank you very much for your help.
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http://mathhelpforum.com/calculus/138518-radius-curvature.html
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# Thread:
1. ## Radius of Curvature
How can I calculate the radius of curvature of a 3D curve that is parameterized in the form:
x(t), y(t), z(t)?
2. It is the reciprocal of the curvature, whose formula can be found at Curvature - Wikipedia, the free encyclopedia
Essentially, at a point p, one takes the limiting value of the radius of the circle passing through p-dp, p, and p+dp.
3. Originally Posted by CrashDummy11
How can I calculate the radius of curvature of a 3D curve that is parameterized in the form:
x(t), y(t), z(t)?
If $r(t)=<x(t),y(t),z(t)>$ then the curvature is: $\kappa = \left\| {\frac{{dT}}<br /> {{ds}}} \right\| = \frac{{\left\| {r' \times r''} \right\|}}<br /> {{\left\| {r'} \right\|^3}}$
4. Thats a really ugly formula but it was what I was looking for. Thanks!
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http://www.physicsforums.com/showthread.php?p=3771369
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Physics Forums
## Solving Linear Equations (fractions)
I'm taking a step back as I've found a concept I'm struggling with. If I have an equation:
2(x-1)/3 = (x/4)+1
I've been told I need to multiply by 12 to give me:
8(x-1) = 3x+12
However I'm struggling with the multiplication of fractions in his instance. Can anyone advise me how that calculation works? I understand the remainder of the working. Thanks :)
Recognitions: Homework Help Do you remember the rules of fractions? If you have some fraction, say, $\frac{2}{3}$ and you multiply it by 9, there are various ways to represent the same number. You could have: $$=9\cdot \frac{2}{3}$$ (note, the dot just means multiply, as x could be confused with a variable) $$=\frac{9\cdot 2}{3}$$ $$=\frac{18}{3}$$ $$=\frac{9}{3}\cdot 2$$ $$=3\cdot 2$$ $$=6$$ Now notice the expression $\frac{9}{3}\cdot 2=3\cdot 2$. We essentially cancel out the denominator this way, which is what you're trying to do in your problem. So why do we multiply your question by 12 and not just any other number? Well, what we're looking for is something called the Lowest Common Denominator (LCD) of 3 and 4. What this means is that we want the first number that both 3 and 4 multiply into. Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, ... Multiples of 4: 4, 8, 12, 16, 20, 24, .... Notice the common multiples are 12 and 24. But we want the LCD which would be 12. So now when you multiply $\frac{2(x-1)}{3}$ by 12, you get $$12\cdot \frac{2(x-1)}{3}=\frac{12}{3}\cdot 2(x-1)=4\cdot 2(x-1)=8(x-1)$$ and if you multiply the other side by 12, you get $$12\left(\frac{x}{4}+1\right)=\frac{12x}{4}+12=3x+12$$ Just remember the common rules of fractions: $$a\cdot\frac{b}{c}=\frac{ab}{c}=\frac{a}{c}\cdot b=a\cdot b\cdot\frac{1}{c}$$ etc.
That's great thank you :) Following on from that I know that Y/x1 = y/x2 cancels down to give x1 = x2 but can you explain why?
Recognitions:
Homework Help
## Solving Linear Equations (fractions)
Quote by annalise17 That's great thank you :) Following on from that I know that Y/x1 = y/x2 cancels down to give x1 = x2 but can you explain why?
Can you multiply the left side of the equation with x1 and simultaneously also multiply the right side with x1?
After that multiply left and right with x2.
And finally divide left and right by y.
Recognitions:
Homework Help
Quote by annalise17 That's great thank you :) Following on from that I know that Y/x1 = y/x2 cancels down to give x1 = x2 but can you explain why?
What I like Serena said.
Also, you can think of it logically. If $$\frac{2}{x}=\frac{2}{y}$$ then for these to be equal, don't x and y need to be the same?
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fractions equations
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http://nanoexplanations.wordpress.com/2011/04/03/connected-guards-in-orthogonal-art-galleries/
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the blog of Aaron Sterling
# Connected Guards in Orthogonal Art Galleries
Posted on April 3, 2011
Figures 2 and 3 from Connected Guards in Orthogonal Art Galleries, by Pinciu (click to enlarge)
The Art Gallery Problem is one of the fundamental problems in computational geometry. It’s easy to state, easy to motivate, and “simple” variations of it can be very hard to solve. The problem: given a building, what is the fewest number of guards you need to station in order to guard every part of the building? Mathematically, we abstract this out: the floor plan of the building becomes a polygon, and the goal is to determine what is the fewest number of “guard points” I can place on the boundary or in the interior of the polygon, so that every point in the polygon can be connected via a line segment to the guard point, so the segment lies completely within the interior of the polygon.
More generally, this area of study is sometimes called “visibility problems,” and has found application in everything from computer graphics to robot motion. I recently read a paper by Val Pinciu, Connected Guards in Orthogonal Art Galleries, where he proves the following theorem.
Theorem: Let $n \geq 6$. An orthogonal polygon with $n$ sides can be completely guarded by a set of at most $n/2 -2$ connected guards.
Pinciu’s proof is constructive — he finds an explicit guard set — and his core algorithm is brief enough to discuss in a blog post, so that’s what I’m going to do after the jump.
An orthogonal polygon (aka orthogonal art gallery) is a polygon all of whose edges are axis-parallel — all edges horizontal or vertical. Visibility problems in such polygons often have strictly simpler solutions than do the same problems for general polygons. A set of connected guards is a set of points that guards the polygon (all points of the polygon are visible to at least one guard), and so that the visibility graph of the guard set is connected. (To produce the visibility graph of a set of points, take the points as vertices, and draw an edge between a pair if those two points are visible to each other within the polygon.) Note that this property is stronger than just saying, “Every guard is visible to at least one other guard.” That only ensures that the visibility graph has no isolated vertices; we require that the visibility graph have a single connected component.
Pinciu’s algorithm to find connected guards for orthogonal art galleries is the following.
• Input: orthogonal polygon $P_n$ with $n$ sides.
• Find a convex quadrangulation of the polygon. Let $Q_n$ be the quadrangulation graph.
• Find a 2-coloring of the quadrangulation graph $Q_n$. Such a 2-coloring exists, because $Q_n$ is bipartite.
• The least frequently used color gives us a set of vertices $\mathcal{G}'$.
• Let $\mathcal{G} = \{ v \in \mathcal{G}' \mid \textrm{deg}(v) \neq 2 \textrm{ in } Q_n \}$.
• Output: The set $\mathcal{G}$, which is a connected set of guards for $P_n$.
I won’t prove this works; from the figures I posted at the top, I hope you can see the algorithm is at least intuitively plausible. Rather, I’d like to say something about the hidden difficulties of the first step of the algorithm: “Find a convex quadrangulation of the polygon.”
Quadrangulating a polygon means drawing line segments between polygon vertices such that (1) all line segments lie in the interior of the polygon, and (2) the edges of the polygon and the line segments together decompose the polygon into a set of four-sided regions. (A related, better-known problem is Polygon Triangulation, where the objective is to decompose the polygon into triangles.) A convex quadrangulation is one where all the four-sided regions are convex.
The fact that a convex quadrangulation of an orthogonal polygon always exists — and can be found efficiently — is nontrivial, and required several papers by multiple authors in the 1980s. (Convex quadrangulations do not always exist for general polygons, so orthogonality is necessary.) The original paper that proved existence of a convex quadrangulation, bears the lovely title, “Traditional Galleries Require Fewer Watchmen” (Kahn, Klawe, Kleitman 1983), and contains the following quote:
In many places in this paper our proofs depend on certain configurations having particular properties which appear to follow obviously from the relevant definitions. However, as is often true in geometrical problems, despite their “obvious truth” it seems to be both time-consuming and tricky to provide rigorous proofs of these assertions.
Joseph O’Rourke reproved the existence of convex quadrilateralization in the chapter on Orthogonal Polygons in his respected book on art gallery theorems. (He bought the book’s copyright back from the publisher and has made it entirely downloadable from here.) A few years after the Klein et al. existence result, Lubiw and, independently, Edelsbrunner, O’Rourke and Welzl, produced an $\mathcal{O}(n \log n)$ algorithm that explicitly found a quadrangulation of an orthogonal polygon with $n$ vertices.
I’ll conclude with a little teaser of why this interests me — and why a guy with nanotech interests has written several computational geometry blog posts at this point. In 2007 or 2008, Jack Lutz told me that he believed the theory of VLSI design had connections to the theory of molecular self-assembly. I’ve kept that in the back of my mind ever since. Computational geometry results from the 1980s about orthogonal polygons have extensive application to chip design and VLSI optimization problems. I’m working on some self-assembly optimization problems right now, and I’m building an intuition for them by reading “old school” computational geometry papers. I’ll talk more specifically about some connections between the comp geom work and my own, but probably not for a while yet. In the meantime, I hope you enjoyed this taste of orthogonal art gallery problems.
Val Pinciu (2003). Connected Guards in Orthogonal Art Galleries ICCSA 2003, Lecture Notes in Computer Science, 2669, 886-893 DOI: 10.1007/3-540-44842-X_90
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### 5 Responses to Connected Guards in Orthogonal Art Galleries
1. chazisop
I am not sure I understand the improvement. Wikipedia mentions that floor(n/3) are always sufficient. So (n-2)-2 is an improvement just for n={6..11}. Plus, the floor(n/3) result is general. I sure must be wrong here.. but where?
• Aaron Sterling
Thanks for the comment, chaziop. The $\lfloor n/3 \rfloor$ upper bound is for guards in arbitrary galleries. The “Traditional Galleries Require Fewer Watchmen” paper shows that the strictly lower bound $\lfloor n/4 \rfloor$ suffices for orthogonal polygons. However, if we require that the guard set have some additional structure, then the number may have to increase. For example, say guard $g_1$ can see $g_2$ can see $g_3$, but $g_1$ and $g_3$ cannot see each other. If we only care about guarding the polygon, we may be able to delete $g_2$ from the guard set, because it is on a location that is visible to other guards. If we require that the visibility graph of the guards be connected, though, the presence of $g_2$ may be necessary.
It’s not hard to construct an orthogonal polygon with $n$ sides that requires $n/2 - 2$ guards if the visibility graph of the guard set must be connected. (Figure 1 of Pinciu’s paper presents one such polygon.)
2. Pingback: Nanoexplanations registered at ResearchBlogging.org | Nanoexplanations
3. Indranil Banerjee
Hi, I found the general art-gallery problem , specially in 3-dimension
(where there seem to be no definitive classical approach for finding
solutions, complicated by the possible presences of obscure regions) could be studied in the context of a combinatorial optimization problem. Unfortunately I could not find any relevant references on this, specially no meta-heuristic approaches. Since I am not very informed in this domain, could you possible give me some direction, possible let me know if I am asking the wrong questions? I also want to know what type of spacial optimization problems are there in VLSI/molecular assembly domain which can be modelled to this problem. Please suggest me some references.
• Aaron Sterling
Both the Handbook on Computational Geometry, and the Handbook of Discrete and Computational Geometry, discuss 3D art gallery problems (illumination problems) some, and they provide reference lists. It’s not an area I know much about. If you would like more specific advice I suggest you ask a question at the Theoretical Computer Science question and answer site. Thanks for the comment, and good luck.
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Occasionally streetwise researcher of DNA self-assembly and other nonstandard applications of theoretical computer science.
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http://math.stackexchange.com/questions/265200/characteristic-cohomology-class-of-4-manifold-with-boundary
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# Characteristic cohomology class of 4-manifold with boundary
I have a question about Characteristic cohomology class of 4-manifolds. $X^4$ denotes the compact 4-manifold with boundary. I'm mainly concerned with $\partial X$ is nonempty.
If $X^4$ is closed, we define $w\in H^2(X;\mathbb{Z})$ is characteristic cohomology class if $Q_X(w,x)\equiv Q_X(x,x)$ modulo 2 for all $x\in H^2(X;\mathbb{Z})$, where $Q_X\colon H^2(X;\mathbb{Z})\times H^2(X;\mathbb{Z})\to \mathbb{Z}$.
What is the analogue definition of characteristic cohomology class for 4-manifold with boundary?
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## 1 Answer
The concept of a characteristic element is purely algebraic. Let $G$ be a finitely generated free abelian group and let $$Q: G \times G \longrightarrow \Bbb Z$$ be a symmetric bilinear form on $A$. Then $x \in G$ is a characteristic element of $G$ for $Q$ if $$Q(x, \alpha) \equiv Q(\alpha, \alpha) \pmod 2 \text{ for all } \alpha \in G.$$ So to define characteristic cohomology classes for a $4$-manifold with boundary, we just need to define the notion of an intersection form for a $4$-manifold with boundary.
If $X$ is a compact, oriented $4$-manifold with boundary $\partial X$, then it has an orientation class $[X, \partial X] \in H_4(X, \partial X; \Bbb Z)$. Then we define the intersection form $Q_X$ of $X$ by $$Q_X: H^2(X, \partial X; \Bbb Z) \times H^2(X, \partial X; \Bbb Z),$$ $$(\alpha, \beta) \mapsto \langle \alpha \smile \beta, [X, \partial X] \rangle.$$ Then a characteristic cohomology class of $X$ is a class $x \in H^2(X, \partial X; \Bbb Z)$ such that $$Q_X(x, \alpha) \equiv Q_X(\alpha, \alpha) \pmod 2 \text{ for all } \alpha \in H^2(X, \partial X; \Bbb Z).$$
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http://mathoverflow.net/questions/94397/is-any-morse-trajectory-contained-in-a-contractible-open-set/94408
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## Is any Morse trajectory contained in a contractible open set?
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Suppose $f$ is a Morse function on a Riemannian Hilbert manifold $M$. Let $p_{\pm}\in \text{Crit}(f)$ be given and fix some $u:R\rightarrow M$ which is an integral curve of $-\nabla f$ connecting $p_-$ and $p_+.$ Is it true that there is an open contractible set $U\subseteq M$ containing the image of $u$?
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you may want to email Lizhen Qin about this question ( qinl@math.purdue.edu ). He has though a lot about Morse theory on Hilbert manifolds. – John Klein May 3 2012 at 1:43
## 1 Answer
Yes, this follows from what's usually called the "$\epsilon$-neighbourhood theorem" in textbooks like Guillemin and Pollack's Differential Topology.
Specifically, given a submanifold $N$ of a manifold $M$ there is an open neighbourhood $V$ of $N$ in $M$ which is diffeomorphic to a vector bundle over $N$. So $V$ is open in $M$.
I believe versions of this theorem appears in Milnor's Topology from a differentiable viewpoint and Hirsch's Differential Topology.
The only restriction on this theorem is that $N$ can not be a manifold with boundary. It can be a non-compact manifold with empty boundary. But it's perfectly fine for $N$ not to be closed in $M$. The proof of this theorem is basically the same as the tubular neighbourhood theorem, except you give up on the idea of having a uniform injectivity radius for the normal bundle's exponential map, and you let the injectivity radius vary smoothly along $N$.
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@Ryan: The word Hilbert in question suggests that Orbicular is asking about infinite-dimensional manifolds modelled on Hilbert spaces. I think the idea of tubular neighborhood still works, but one may need a better control on the metric tensor so that geodesics are locally unique. – Misha Apr 18 2012 at 14:44
Ah, I didn't notice it was meant to be infinite-dimensional. – Ryan Budney Apr 18 2012 at 21:11
Lang's book Differential and Riemannian Manifolds discusses in great detail tubular neighborhoods of submanifolds in a Hilbert manifold (Sec. IV.5, VII.4) and I think that Ryan's argument extends to Hilbert manifolds as well. – Liviu Nicolaescu May 17 2012 at 15:55
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http://math.stackexchange.com/questions/tagged/improper-integrals+convergence
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# Tagged Questions
2answers
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### Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge
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### check the convergence of the improper integral$\int_{0}^{1}\frac{x^{p-1}+x^{-p}}{1+x}\,dx$
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### Existence of Riemann-Liouville Integral
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### does $\int_0^\infty x/(1+x^2 \sin^2x) \mathrm dx$ converge or diverge?
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### Does $\int_{0}^{\infty} \frac{dx}{\sqrt{x^3+x}}$ converge?
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### Convergence/Divergence of $\int_e^\infty \frac{\sin x}{x \ln x}\;dx$
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186 views
### Question on convergence of improper integral
For what values of $\alpha$ is the following integral convergent? $$\int\limits_{-\infty}^{\infty}\frac{|x|^\alpha}{(1+x^2)^m}dx$$ Should the limit comparison theorem be used in this case? I am not ...
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### Integrals Converging
There always seems to be a question about whether or not an integral converges. Can I ask what the best mental method is to pick the right test/process to calculate the integral? Let’s take an ...
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http://mathoverflow.net/questions/10574/how-do-i-make-the-conceptual-transition-from-multivariable-calculus-to-differenti/101569
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## How do I make the conceptual transition from multivariable calculus to differential forms?
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One way to define the algebra of differential forms $\Omega(M)$ on a smooth manifold $M$ (as explained by John Baez's week287) is as the exterior algebra of the dual of the module of derivations on the algebra $C^{\infty}(M)$ of smooth functions $M \to \mathbb{R}$. Given that derivations are vector fields, 1-forms send vector fields to smooth functions, and some handwaving about area elements suggests that k-forms should be built from 1-forms in an anticommutative fashion, I am almost willing to accept this definition as properly motivated.
One can now define the exterior derivative $d : \Omega(M) \to \Omega(M)$ by defining $d(f dg_1 ... dg_k) = df dg_1 ... dg_k$ and extending by linearity. I am almost willing to accept this definition as properly motivated as well.
Now, the exterior derivative (together with the Hodge star and some fiddling) generalizes the three main operators of multivariable calculus: the divergence, the gradient, and the curl. My intuition about the definitions and properties of these operators comes mostly from basic E&M, and when I think about the special cases of Stokes' theorem for div, grad, and curl, I think about the "physicist's proofs." What I'm not sure how to do, though, is to relate this down-to-earth context with the high-concept algebraic context described above.
Question: How do I see conceptually that differential forms and the exterior derivative, as defined above, naturally have physical interpretations generalizing the "naive" physical interpretations of the divergence, the gradient, and the curl? (By "conceptually" I mean that it is very unsatisfying just to write down the definitions and compute.) And how do I gain physical intuition for the generalized Stokes' theorem?
(An answer in the form of a textbook that pays special attention to the relationship between the abstract stuff and the physical intuition would be fantastic.)
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Really? I would like to award reputation for good answers and I am not necessarily just looking for a list of recommendations; perhaps someone has a clear enough intuition that it can be described in a paragraph or two. – Qiaochu Yuan Jan 3 2010 at 11:30
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Have you seen From Calculus to Cohomology by Madsden and Tornehave? It's not really about physical intuition (which is why I'm making this a comment), but it might be helpful. – Akhil Mathew Jan 3 2010 at 15:28
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I still think this should be community wiki because it's a sorted list. I didn't like an answer, and I'd like to vote it down, but not the user. – Harry Gindi Jan 3 2010 at 15:46
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+1 for "I freaking love this question". – B. Bischof Jan 14 2010 at 22:53
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A 1-form is a function which grows proportionally to how fast you are moving. Thus it doesn't matter how you parametrize the curve you are moving on - you either end up integrating a smaller function for a longer period of time, or a bigger function for a shorter period of time. This is why you can't integrate functions on manifolds - they have no intrinsic "unit speeds", because there are many choices of local coordinates - but you can still integrate differential forms. k-forms just generalize this to higher dimensions. – Steven Gubkin Jan 27 2011 at 20:25
show 2 more comments
## 15 Answers
Here's a sketch of the relation between div-grad-curl and the de Rham complex, in case you might find it useful.
The first thing to realise is that the div-grad-curl story is inextricably linked to calculus in a three-dimensional euclidean space. This is not surprising if you consider that this stuff used to go by the name of "vector calculus" at a time when a physicist's definition of a vector was "a quantity with both magnitude and direction". Hence the inner product is essential part of the baggage as is the three-dimensionality (in the guise of the cross product of vectors).
In three-dimensional euclidean space you have the inner product and the cross product and this allows you to write the de Rham sequence in terms of div, grad and curl as follows: $$\matrix{ \Omega^0 & \stackrel{d}{\longrightarrow} & \Omega^1 & \stackrel{d}{\longrightarrow} & \Omega^2 & \stackrel{d}{\longrightarrow} & \Omega^3 \cr \uparrow & & \uparrow & & \uparrow & & \uparrow \cr \Omega^0 & \stackrel{\mathrm{grad}}{\longrightarrow} & \mathcal{X} & \stackrel{\mathrm{curl}}{\longrightarrow} & \mathcal{X} & \stackrel{\mathrm{div}}{\longrightarrow} & \Omega^0 \cr}$$ where $\mathcal{X}$ stands for vector fields and the vertical maps are, from left to right, the following isomorphisms:
1. the identity: $f \mapsto f$
2. the musical isomorphism $X \mapsto \langle X, -\rangle$
3. $X \mapsto \omega$, where $\omega(Y,Z) = \langle X, Y \times Z \rangle$
4. $f \mapsto f \mathrm{dvol}$, where $\mathrm{dvol}(X,Y,Z) = \langle X, Y \times Z\rangle$
up to perhaps a sign here and there that I'm too lazy to chase.
The beauty of this is that, first of all, the two vector calculus identities $\mathrm{div} \circ \mathrm{curl} = 0$ and $\mathrm{curl} \circ \mathrm{grad} = 0$ are now subsumed simply in $d^2 = 0$, and that whereas div, grad, curl are trapped in three-dimensional euclidean space, the de Rham complex exists in any differentiable manifold without any extra structure. We teach the language of differential forms to our undergraduates in Edinburgh in their third year and this is one way to motivate it.
As for the integral theorems, I always found Spivak's Calculus on manifolds to be a pretty good book.
Another answer mentioned Gravitation by Misner, Thorne and Wheeler. Personally I found their treatment of differential forms very confusing when I was a student. I'm happier with the idea of a dual vector space than I am with the "milk crates" they draw to illustrate differential forms. Wald's book on General Relativity had, to my mind, a much nicer treatment of this subject.
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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.
I have struggled with this question myself, and I couldn't find a perfectly satisfactory answer. In the end, I decided that the definition of a differential form is a rather strange compromise between geometric intuition and algebraic simplicity, and that it cannot be motivated by either of these by itself. Here, by geometric intuition I mean the idea that "differential forms are things that can be integrated" (as in Bachmann's notes), and by algebraic simplicity I mean the idea that they are linear.
The two parts of the definition that make perfect geometric sense are the d operator and the wedge product. The operator d is simply that operator for which Stokes' theorem holds, namely if you integrate d of a n-form over an n+1-dimensional manifold, you get the same thing as if you integrated the form over the n-dimensional boundary.
The wedge product is a bit harder to see geometrically, but it is in fact the proper analogy to the product measure. Here's how it works for one-forms. Suppose you have two one-forms a and b (on a vector space, for simplicity). Think of them as a way of measuring lengths, and suppose you want to measure area. Here's how you do it: pick a vector $\vec v$ such that $a(\vec v) \neq 0$ but $b(\vec v) = 0$ and a vector $\vec w$ s.t. $a(\vec w) = 0$ but $b(\vec w) \neq 0$. Declare the area of the parallelogram determined by $\vec v$ and $\vec w$ to be $a(\vec v) \cdot b(\vec w)$. By linearity, this will determine area of any parallelogram. So, we get a two-form, which is in fact precisely $a \wedge b$.
Now, the part that makes no sense to me geometrically is why the hell differential forms have to be linear. This implies all kinds of things that seem counter-intuitive to me; for example there is always a direction in which a one-form is zero, and so for any one-form you can draw a curve whose "length" with respect to the form is zero. More generally, when I was learning about forms, I was used to measures as those things which we integrate, and I still see no geometric reason as to why measures (and, in particular, areas) are not forms.
However, this does make perfect sense algebraically: we like linear forms, they are simple. For example (according to Bachmann), their linearity is the thing that allows the differential operator d to be defined in such a way that Stokes' theorem holds. Ultimately, however, I think the justification for this are all the short and sweet formulas (e.g. Cartan's formula) that make all kinds of calculations easier, and all depend on this linearity. Also, the crucial magical fact that d-s, wedges, and inner products of differential forms all remain differential forms needs this linearity.
Of course, if we want them to be linear, they will be also signed, and so measures will not be differential forms. To me, this seems as a small sacrifice of geometry for the sake of algebra. Still, I don't believe it's possible to motivate differential forms by algebra alone. In particular, the only way I could explain to myself why take the "Alt" of a product of forms in the definition of the wedge product is the geometric explanation above.
So, I think the motivation and power behind differential forms is that, without wholly belonging to either the algebraic or geometric worlds, they serve as a nice bridge in between. One thing that made me happier about all this is that, once you accept their definition as a given and get used to it, most of the proofs (again, I'm thinking of Cartan's formula) can be understood with the geometric intuition.
Needless to say, if anybody can improve on any of the above, I'll be very grateful to them.
P.S. For the sake of completeness: I think that "inner products" make perfect algebraic sense, but are easy to see geometrically as well.
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Of course forms have to be signed (indeed, alternating) objects. Think of the classical Stokes' theorem, the direction of the normal vector and the associated direction of the boundary. If you reverse one you have to reverse the other, and the integrals change sign. As for linearity, think of integration over a very small surface – say, a parallelogram. If you double one side, the integral should double too (asymptotically). Similarly, if the sides are parallel, the integral vanishes. From ω(X,X)=0 and linearity you get the alternating property. Oh, and remember the Jacobian determinant? – Harald Hanche-Olsen Jan 3 2010 at 22:17
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IMO the answer to your question as to why we choose forms to be linear goes back to a far simpler observation. That the determinant is the unique alternating multi-linear function on square matrices that takes value $1$ on the identity. This says alternating multi-linear objects measure (signed) volume. A form is just a linear combination of projections followed by determinants, so forms are precisely the objects you need to measure signed volume when you have positive co-dimension. – Ryan Budney Jan 4 2010 at 0:18
I'm not convinced that this is what you are looking for, Qiaochu, but I think it's worth mentioning anyway.
As someone who has no real sense for "physical intuition," and who -- probably not coincidentally -- hated his multivariable calculus class, I've found (what I've read of) David Bachman's A geometric approach to differential forms to be wonderfully intuitive. Best of all, it's available for free online.
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The great book The Geometry of Physics (2nd edition) by Frankel could be exactly what you need, see excerpts here.
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To get an intuitive understanding of the Stokes theorem, I recommand the book by Arnol'd on mechanics. It gives a very intuitive definition of the exterior derivative in such a way that the Stokes theorem becomes, heuristically, very easy to grasp.
I also find Analysis, manifolds and physics and Geometry, topology and physics to be two great source of inspiration to understand the intertwining between geometry and physics. The first is written by mathematicians, the second by a physicist.
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The notes by Bachman recommended by Harrison Brown look pretty nice to me, but it seems to me that it is possible to clarify what he says even further by focusing on the simplest cases, namely the integral of a "constant function" over the simplest possible domain.
For the integral over an interval, the simplest case consists of a constant function. You can extend this case to the general case by the additive property of an integral and taking limits. But if you want an integral that is independent of the parameterization of the interval, this leads naturally to the idea that you don't want to integrate just a function $f(x)$ but a "1-form" $f(x) dx$.
This generalizes naturally to an integral of a constant function over a line segment sitting in $R^n$. If you want the concept of an integral that is independent of choice of a linear parameterization of the segment, as well as the linear co-ordinates on $R^n$, then this leads to naturally to the fact that what should be integrated is a "dual vector", i.e. a constant $1$-form. In fact, when developing these ideas, I suggest using an abstract real vector space $V$ as the ambient space instead of $R^n$.
When considering higher dimensions, I suggest focusing on linear embeddings of a $k$-dimensional cube and asking what gives a linear co-ordinate independent additive function of flat $k$-cubes embedded in $R^n$. I have not worked out the details myself, but I suspect that this leads naturally to the concept of constant $k$-forms.
My recollection is that there is a book "Advanced Calculus" by Harold Edwards that presents all of this, but I haven't looked at the book in a very long time.
In particular, it is worth noting that the question asked is really about algebra and not analysis. The analysis arises only when you want to extend the definition of an integral to a more general class of functions beyond constant ones.
ADDED LATER:
My answer above does not address the exterior derivative. I will just add a brief comment about this and leave the details to the reader. My view of the exterior derivative is that, once you decide that exterior forms are indeed the natural objects of integration over a domain (but start with cubes!) in Euclidean space, it is the natural co-ordinate-free algebraic consequence of the fundamental theorem of calculus (or, if you insist, Stoke's theorem). That $d^2 = 0$ is the appropriate co-ordinate-free expression of the basic fact that "partials commute".
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The main thing I've never understood about differential forms is their "coordinate independence." I know what the general definition of a differentiable manifold is, so I see why it might be interesting there (topology might obstruct the existence of a global coordinate patch), but I don't understand why differential forms are taught for R^n. Since in R^n there is a natural inner product (by which you mean on the tangent space?) and coordinate system, why do we care? In what way are they "coordinate independent"? Is the coordinate change just if you want to change to polar or something? – David Corwin Jan 5 2010 at 3:25
It's true that there is a natural coordinate system on R<sup>n</sup>, but you might well want to use a different one depending on the situation, e.g., polar coordinates for a spherically symmmetric problem. When you change coordinates there is a formula telling you how integrals behave. One way to think of "coordinate independence" of differential forms is just that the way differential forms change under change of coordinates neatly encodes the behaviour of integrals. – Joel Fine Jan 5 2010 at 8:03
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Heck, a lot of my research was about or on manifolds, and I rarely used differential forms. My thesis was actually about exterior differential systems (systems of equations defined by exterior differential forms), and even there I used very little of the formalism of differential forms! Use differential forms only if the formalism makes your life easier and not harder. Also, for me a lot of things on $R^n$ make a lot more sense and are much easier to work with, when I see that they do not require the use of a global co-ordinate system or inner product. – Deane Yang Jan 5 2010 at 15:07
I'm not sure if this point of view is taken up in the many references which are named here, but I'll say something about an "elementary" way to discover the exterior derivative which sounds like ordinary calculus. Let's take on the point of view that a $k$-form is something you integrate over a $k$-dimensional submanifold. If you imagine $k$-dimensional submanifolds as being composed of a $k$-dimensional blanket of little $k$-parallelograms, then this is a geometrically natural point of view since the $k$-form will assign a (small) number to each of these parallelograms. To actually realize a submanifold as such a "blanket" is to give a parameterization. (These parallelograms are oriented; this picture is different from surface integration of scalar functions in Riemannian geometry where one simply imagines some distribution of mass on the manifold and the integral is completely measure-theoretic. There the paralellograms have a positive mass given by the $k$-dimensional volume determined by the inner product.)
In one-variable calculus, when $f$ is a function, $df$ tells you the change in $f$ per small change in its input, and if you integrate it over a curve from $a$ to $b$, it expresses the total change in $f$ from $a$ to $b$. Now, a one form $\eta$ is integrated not over points but rather over curves. Still, you can ask, how does $\int_\gamma\eta$ change when you perturb $\gamma$? Well, if you deform a closed curve $\gamma_a$ into another curve $\gamma_b$, the difference between the integrals over $\gamma_b$ and $\gamma_a$ is some derivative we can call "$d\eta$" integrated over the surface swept between the two.
Picturing the case where $\gamma_a$ and $\gamma_b$ bound an annulus is a good thing to consider here; this interpretation tells you how to orient the boundary of the annulus if you want to think of $\int_\Sigma d\eta = \int_{\gamma_b} \eta - \int_{\gamma_a} \eta$ as being $\int_{\partial \Sigma} \eta$. On the other hand, you can take the point of view that the orientation for $\Sigma$ is determined by the requirement that we start at $\gamma_a$ and go to $\gamma_b$ (much like the case for $df$ of a function). You can then contract the inner circle to a point to recover Stokes' theorem for a disk -- the integral over the inner circle will vanish in the limit by the linearity and continuity of the form (a similar thing will happen in higher dimensions but the linearity is needed for the cancellation over the inner, closed surface).
It's not completely necessary that the curve (or $k$-dimensional submanifold) you deform is closed, but by rule the boundary should remain fixed during the deformation or you will miss out on part of the boundary.
Using a specific example like a square/cube, we can get a coordinate representation for $d\eta$ through the fundamental theorem of calculus. (For $0$ forms, every point is closed, so we did not need to worry about the word "closed" before.)
It is easy to see many properties. For example, let's take $\eta$ to be a $1$-form in $3$-space; then $d^2 \eta$ is clearly $0$. Let $\gamma$ be a circle, and let $\Sigma_a$ and $\Sigma_b$ be the upper and lower hemispheres of a ball $B$ whose equator is $\gamma$. Then $\int_{\Sigma_a} d \eta = \int_\gamma \eta = \int_{\Sigma_b} d \eta$ by Stokes' theorem for a disk. On the other hand, the integral of $d^2\eta$ over the ball $B$ is just $\int_{\Sigma_b} d \eta - \int_{\Sigma_a} d\eta = 0$ because you can sweep out $B$ by deforming $\Sigma_a$ to $\Sigma_b$ with the boundary fixed. Since $\int_B d^2 \eta = 0$ for every ball, $d^2 \eta$ is identically $0$. When you execute this proof for a square, you see that mixed partials commute.
I would like to know if the product rule can easily be seen through this interpretation, but I have not thought enough about it to see it clearly yet.
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There is a book that not many physicists I know of seem to like (except mathematical physicists, of course), but that is a true gem in the eyes of mathematicians: I am referring to V. Arnold Mathematical Methods of Classical Mechanics, here on amazon.
In this book, which is in the short list (number 12, to be precise) of my fundamental math book across all math fields, Chapter VIII is entirely devoted to differential forms.
If you read it, you have, I believe, an excellent answer.
One small suggestion to build understanding: DISCRETIZE. Do not think of fancy integrals, simply think that 0-forms are scalars, 1-forms oriented segments, 2-forms oriented areas, and that integration over them is simply sums. Now "prove" Stokes theorem for simple tiny cubes, and notice that the definition of the derivatives of forms is exactly done to keep track of faces. At the infinitesimal level, it is just book keeping.
If I ever had to teach a basic class on forms, I would do precisely that: discretize first.
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Baez's book "Gauge Fields, Knots, and Gravity" does a good job of geometrically motivating differential forms in the first section on electromagnetism. Unfortunately, I already was happy with forms when I read it, so it may or may not be what you are looking for. It certainly does have all the right infinitesimal drawings to motivate the definitions, though. It might make a good intuitive complement to whatever abstract resource you choose.
You might also want to skim through parts of Hubbard's "Vector Calculus, Linear Algebra, and Differential Forms: a unified approach". This is used at Cornell as a textbook for a 2 semester calc/linear algebra/analysis sequence. About half of the second semester is spent developing and applying differential forms. There is somewhat less intuitive explanation, but some very good motivation for why we should define things the way we do.
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I'd second Hubbard's book. He gives a very natural definition of the exterior derivative of a differential form, which is what it sounds like you're looking for Yuan. IMO it's a significant step-up compared to texts like Spivak or Bachman. – Ryan Budney Jan 3 2010 at 18:26
Some very nice textbooks have already been mentioned, but some of my favorites that haven't been are:
• William L. Burke, Applied differential geometry
• Gabriel Weinreich, Geometrical vectors
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1
+1.The late William Burke's book has sadly been forgotten over the last decade or so as Frankel's and Nakahara's more comprehensive texts have supplanted it. But I believe it should be required reading by both physicists and mathematicians in training. – Andrew L Mar 25 2012 at 8:07
Misner, Thorne, and Wheeler's Gravitation is very good at providing a treatment of differential forms that appeals to physicists. But it is no longer the preeminient GR reference (though it's perfectly fine, its size also is an issue), so be warned. Dubrovin, Fomenko, and Novikov's Modern Geometry is also very good, but less structured.
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There is an amazing book called Foundations of Classical Electrodynamics: Charge, Flux, and Metric written by Friedrich Hehl and Yuri Obukhov. The book is completely metric-free and heavily uses differential forms, and if you want physical intuition there is possibly no better representation than this (I quote from page 145):
Faraday-Schouten pictograms of the electromagnetic field in 3-dimensional space. The images of 1-forms are represented by two neighboring planes. The nearer the planes, the stronger the 1-form is. The 2-forms are pictured as flux tubes. The thinner the tubes, the stronger the flow. The difference between a twisted and an untwisted form accounts for the two different types of 1- and 2-forms, respectively.
With this picture it is quite easy to imagine what he means by twisted forms!
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1
A problem with this is that not all forms are representable by such a picture -- take $dx \wedge dy + dz \wedge dw$ for example. Dimension 3 has some very nice things going for it but it also restricts your intuition as to what a form is. – Ryan Budney Jan 4 2010 at 0:13
First let me say, that I am also trying to gain a understanding of differential forms.
I have found the Geometric Algebra approach for visualizing simple mutlivectors (k-blades) a good way to visualize the infinitesimal multivectors of differential forms.
The first few chapters of Geometric Algebra for Computer Science do a good job of that. http://www.geometricalgebra.net/tour.html
As far as I can tell (only on chapter 4) GA excludes visualizing general multivectors like $a \wedge b + c \wedge d$, but as Dan Piponi pointed out here: http://homepage.mac.com/sigfpe/Mathematics/forms.pdf your probably okay thinking of that construction as two parallelograms.
With those resources as the basis of my visual intuition, the second chapter of Manifolds and Differential Forms by Reyer Sjamaar here:
http://www.math.cornell.edu/~sjamaar/papers/manifold.pdf seemed fairly understandable.
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My personal favorite is Spivak's "Calculus on manifolds". It treats classical theorems and motivations with a lot of respect. The only caveat is that it has it's share of annoying typos
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I second the recommendation to at least flip through Gravitation. It has an intimidating size, but easygoing manner. I had a lot of difficulty with Spivak's Calculus on Manifolds (which has essentially no physical intuition outside the Archimedes exercise at the end), but I think I was uncomfortable with the abstract notions of tensor product and dual vector space at the time I was learning from it. At some point I caught on that df was supposed to eat vector fields and produce functions, and things got a little better.
You might try Sternberg's Advanced Calculus (available on line), especially chapters 11 and 13.
Edit: I like to think of abstract forms as "things to integrate" and Stokes's theorem as some kind of adjunction between boundaries and the derivative. This becomes a bit more meaningful when homology and cohomology are introduced. I don't have much advice for connecting with physical intuition, but I have found it useful to:
1. Decompose div, grad and curl in terms of d and the metric.
2. Work through some E&M starting from a 1-form (strictly speaking a U(1)-connection) A on $\mathbb{R}^{1,3}$ (see Wikipedia).
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http://math.stackexchange.com/questions/71110/ch-for-tilings-of-the-plane
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# CH for tilings of the plane
Given any set of jordan curves that can tile the plane, how to prove that the number of possible tilings using tiles from this set is either in bijection with the real numbers or a (possibly infinite) subset of the integers?
Two tilings are equal if they can be made to coincide by translations or rotations.
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I'm not sure what the title supposed to mean, the Continuum Hypothesis says that the continuum has a very specific cardinality, its minimal possible. What does that have to do with the question? – Asaf Karagila Oct 9 '11 at 13:52
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"There is no set whose cardinality is strictly between that of the integers and that of the real numbers." - Georg Cantor. Does it hold for cardinalities of tilings of the plane ? – Optimus Prime Oct 9 '11 at 13:54
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I know there is sets of polyominoes say, for which the number of tilings are of the same cardinality as the integers, and there are sets of polyominoes which admit uncountable number of tilings. Is there a way to prove that there is no set of tiles that tiles the plane such that the number of tilings are not in bijection with the integers or the reals? – Optimus Prime Oct 9 '11 at 14:01
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@Asaf: It is not unusual to call results of the form "every blah is either countable or of size continuum" a continuum hypothesis result. This is why one sometimes says that CH holds for Borel sets. I take the question here is in the same spirit. – Andres Caicedo Oct 9 '11 at 14:54
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Can you be more precise about what a tile is, and what it means to tile the plane (e.g., are rotations allowed)? For example, if you allow tiles which are unit squares with edges perturbed by a sine function of varying magnitude, it's not hard to build a family of $\kappa$ many tiles (for any $\kappa \leq 2^{\aleph_0}$) which can tile the plane in exactly $\kappa$ many ways (up to translation/rotation invariance): each tiling will use a single tile type. – ccc Oct 9 '11 at 16:40
show 5 more comments
## 1 Answer
Are your tiles square shaped? One can then prove the result by what is essentially a compactness argument. Here is a brief idea:
Tile in order a square of size $1\times1$, then a larger square containing that one, of size $2\times 2$, then a larger one containing it, of size $3\times 3$, etc.
Suppose that your tiling allows us you to tile the plane in a non-periodic fashion. Then, for some $n$, you will have at least two options on how to tile the $n\times n$ square when you get there. Continue "on separate boards" with each of these two ways. Again, by non-periodicity, you should in each case reach a larger $m$ such that the $m\times m$ square can be tiled in at least two ways when you get there (of course, the $m$ in one case may be different from the $m$ in the other case). Continuing "on separate boards" in this fashion, you are building a complete binary tree, each path through which gives you a "different" tiling of the plane. The quotes are here, as we are not yet distinguishing between translations. But there are only countably many translates of a given tiling (if we insist that, say, the borders of our squares coincide with the lines of the form $x=n$ or $y=m$ for $n,m\in{\mathbb Z}$). But then, even after identifying translates, we are left with continuum many different tilings of the plane.
The other possibility is that your tiles does not allow non-periodic tilings. Then, no matter what path you follow, the process above should stop "splitting". But then you are only producing countably many tilings in this fashion.
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No, the tiles need not be square shaped. – GEdgar Oct 9 '11 at 16:54
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http://mathematica.stackexchange.com/questions/tagged/fitting?sort=votes&pagesize=15
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http://mathoverflow.net/revisions/97105/list
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## Return to Question
4 small technical correction
Let $G$ be a finite group, and let $K$ be a finite field whose characteristic does not divide $|G|$. I am interested in the theory of finitely generated modules over $K[G]$. Of course many problems are not present here because $K[G]$ is semisimple and all modules are projective. My case is partly covered by Section 15.5 of Serre's book "Linear Representations of Finite Groups". However, Serre likes to assume that $K$ is "sufficiently large", meaning that it has a primitive $m$'th root of unity, where $m$ is the least common multiple of the orders of the elements of $G$. I do not want to assume this, so some Galois theory of finite extensions of $K$ will come into play. I do not think that anything desperately complicated happens, but it would be convenient if I could refer to the literature rather than having to write it out myself. Is there a good source for this?
[UPDATED]:
In particular, I would like to be able to control the dimensions over $K$ of the simple $K[G]$-modules. As pointed out in Alex Bartel's answer, these need not divide the order of $G$. I am willing to assume that $G$ is a $p$-group for some prime $p\neq\text{char}(K)$.
[UPDATED AGAIN]:
OK, here is a sharper question. Put $m=|K|$ (which is a power of a prime different from $p$) and let $t$ be the order of $m$ in $(\mathbb{Z}/p)^\times$. Let $L$ be a finite extension of $K$, let $G$ be a finite abelian $p$-group, and let $\rho:G\to L^\times$ be a homomorphism that does not factor through the unit group of any proper subfield . containing $K$. Then $\rho$ makes $L$ into an irreducible $K$-linear representation of $G$, and every irreducible arises in this way. If I've got this straight, we see that the possible degrees of nontrivial irreducible $K$-linear representations of abelian $p$-groups are the numbers $tp^k$ for $k\geq 0$. I ask: if we let $G$ be a nonabelian $p$-group, does the set of possible degrees get any bigger?
3 Added a sharper question
Let $G$ be a finite group, and let $K$ be a finite field whose characteristic does not divide $|G|$. I am interested in the theory of finitely generated modules over $K[G]$. Of course many problems are not present here because $K[G]$ is semisimple and all modules are projective. My case is partly covered by Section 15.5 of Serre's book "Linear Representations of Finite Groups". However, Serre likes to assume that $K$ is "sufficiently large", meaning that it has a primitive $m$'th root of unity, where $m$ is the least common multiple of the orders of the elements of $G$. I do not want to assume this, so some Galois theory of finite extensions of $K$ will come into play. I do not think that anything desperately complicated happens, but it would be convenient if I could refer to the literature rather than having to write it out myself. Is there a good source for this?
[UPDATED]:
In particular, I would like to be able to control the dimensions over $K$ of the simple $K[G]$-modules. As pointed out in Alex Bartel's answer, these need not divide the order of $G$. I am willing to assume that $G$ is a $p$-group for some prime $p\neq\text{char}(K)$.
[UPDATED AGAIN]:
OK, here is a sharper question. Put $m=|K|$ (which is a power of a prime different from $p$) and let $t$ be the order of $m$ in $(\mathbb{Z}/p)^\times$. Let $L$ be a finite extension of $K$, let $G$ be a finite abelian $p$-group, and let $\rho:G\to L^\times$ be a homomorphism that does not factor through the unit group of any proper subfield. Then $\rho$ makes $L$ into an irreducible $K$-linear representation of $G$, and every irreducible arises in this way. If I've got this straight, we see that the possible degrees of nontrivial irreducible $K$-linear representations of abelian $p$-groups are the numbers $tp^k$ for $k\geq 0$. I ask: if we let $G$ be a nonabelian $p$-group, does the set of possible degrees get any bigger?
2 added 56 characters in body
Let $G$ be a finite group, and let $K$ be a finite field whose characteristic does not divide $|G|$. I am interested in the theory of finitely generated modules over $K[G]$. Of course many problems are not present here because $K[G]$ is semisimple and all modules are projective. My case is partly covered by Section 15.5 of Serre's book "Linear Representations of Finite Groups". However, Serre likes to assume that $K$ is "sufficiently large", meaning that it has a primitive $m$'th root of unity, where $m$ is the least common multiple of the orders of the elements of $G$. I do not want to assume this, so some Galois theory of finite extensions of $K$ will come into play. I do not think that anything desperately complicated happens, but it would be convenient if I could refer to the literature rather than having to write it out myself. Is there a good source for this?
[UPDATED]:
In particular, I think it is true that whenever would like to be able to control the dimensions over $V$ is a K$of the simple$K[G]$-module, K[G]$-modules. As pointed out in Alex Bartel's answer, these need not divide the dimension order of $\dim_K(V)$ divides G$. I am willing to assume that$|G|$, just as for representations over G$ is a $\mathbb{C}$. A reference p$-group for this particular point would already be useful.some prime$p\neq\text{char}(K)\$.
1
# Modular representations with unequal characteristic - reference request
Let $G$ be a finite group, and let $K$ be a finite field whose characteristic does not divide $|G|$. I am interested in the theory of finitely generated modules over $K[G]$. Of course many problems are not present here because $K[G]$ is semisimple and all modules are projective. My case is partly covered by Section 15.5 of Serre's book "Linear Representations of Finite Groups". However, Serre likes to assume that $K$ is "sufficiently large", meaning that it has a primitive $m$'th root of unity, where $m$ is the least common multiple of the orders of the elements of $G$. I do not want to assume this, so some Galois theory of finite extensions of $K$ will come into play. I do not think that anything desperately complicated happens, but it would be convenient if I could refer to the literature rather than having to write it out myself. Is there a good source for this?
In particular, I think it is true that whenever $V$ is a simple $K[G]$-module, the dimension $\dim_K(V)$ divides $|G|$, just as for representations over $\mathbb{C}$. A reference for this particular point would already be useful.
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http://math.stackexchange.com/questions/71159/metric-spaces-and-sup-of-diameter
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# Metric Spaces and $\sup$ of Diameter
Question: Let $(X, d)$ be a metric space such that there is a positive $a$ and $n$ open balls $B(x_1, a),\ldots, B(x_n, a)$ such that together these balls cover $X$.
Find an upper bound $M$ for the diameter of $X$. Find "the smallest" upper bound $M$ for the diameter of $X$ in the following sense: the formula for $M$ is valid in all cases, and there is a metric space $X$ such that $\operatorname{diam} (X) = M$.
Thoughts: I find the question a little hard to parse. I think it's asking what is the biggest diameter that this can have, so my answer then would be $\sup M = an$.
Thinking about the real number line, if there are $n$ # of balls, the two furthest points from each other would be $a\cdot n$ length from each other. Still trying to figure out what the rest of the questions asks...
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If $x,y\in X$ then we have $d(x,y)\leq 2a+\max_{1\leq j,k\leq n}d(x_j,x_k)$ thanks to the triangular inequality. – Davide Giraudo Oct 9 '11 at 16:59
Thank you for your input! I think I don't quite understand that... the $max d(x_j,x_k)$ would be $a(n-2) + (a/2) + (a/2) = a(n-1)$ but $2a + a(n-1) = a(n+1)$ which doesn't intuitively make sense to me. Am I thinking about this incorrectly? If you take a closed, bounded interval of R, and line up the balls to cover that set you get that the distance between two points could be at most $a\times n$. I can't picture a bigger distance than that right now. – lillian Oct 9 '11 at 17:25
1
Suppose $X$ has $n$ points, with any two points being a fixed distance $d$ apart. Then $X$ has diameter $d$, but can be covered by $n$ open balls with radius as small as we like. So there's no function of $n$ and $a$ which will work as an upper bound here. – Chris Eagle Oct 9 '11 at 17:38
I've been assuming that "a" is a fixed radius but I think it is too ambiguously written to know either way. Thanks for the input, I think I need further direction to understand how to think about this problem. – lillian Oct 9 '11 at 17:41
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The radius $a$ is fixed. The question is not at all ambiguous: it’s asking for the smallest number (as a function of $a$ and $n$ that is guaranteed to be at least as large as the diameter of $X$. Then, once you have that, you’re to find a specific example of $n$ balls of radius $a$ covering a space whose diameter really is that big; this will show that no smaller number would have met the guarantee. – Brian M. Scott Oct 9 '11 at 17:50
show 6 more comments
## 1 Answer
Let $x,y\in X$. Then we can find $j,k\in\{1,\ldots,n\}$ such that $d(x,x_k)<a$ and $d(y,x_j)<a$. We get $$d(x,y)\leq d(x,x_k)+d(x_k,x_j)+d(x_j,y)< a+d(x_k,x_j)+a\leq 2a+\max_{1\leq j,k\leq n}d(x_k,x_j),$$ therefore $\displaystyle \operatorname{Diam}(X)\leq 2a+\max_{1\leq j,k\leq n}d(x_k,x_j)$. We can't hope a better bound. Indeed, taking $X:=\left]0,(n+1)a\right[$ with the usual metric, and $x_j=ja$, we get that $\displaystyle X=\bigcup_{j=1}^nB(x_j,a)$, $\operatorname{Diam}(X)=(n+1)a$ and $\displaystyle \max_{1\leq j,k\leq n}d(x_k,x_j)=\max_{1\leq j,k\leq n}|ka-ja|=(n-1)a$, and the above inequality is an equality.
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http://math.stackexchange.com/questions/244152/basic-differential-equation-proof?answertab=votes
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# Basic differential equation proof
Show that if $u(t)$ solves $\dot{u} = Au$, then $v(t) = u(-t)$, solves $\dot{v} = Bv$, where $B = -A$.
Similarly, show that if u(t) solves $\dot{u} = Au$, then $v(t) = u(2t)$ solves $\dot{v} = Bv$, where $B = 2A$.
I don't know how to formally prove this. It seems obvious because since the solution $v(t) = u(-t)$ is the same as $u(t)$ but it has a scalar which is negative and a scalar can be taken out of the solution and placed outside which follows that we have $-A$. The same goes with the second part of the question.
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## 2 Answers
Let $s=-t$ then $dv(t)/dt=du(-t)/dt=-du(s)/ds=-Au(s)=-Au(-t)=-Av(t).$
Or let $s=kt$ with $k=-1,2$ then $dv(t)/dt=du(kt)/dt=(du(s)/ds)(ds/dt)=kAu(s)=kAu(kt)=kAv(t).$
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Just substitute in to the differential equation and see that both sides are equal.
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http://math.stackexchange.com/questions/198636/question-about-simple-graph
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# question about simple graph.
I know that simple graph has no parellel edges and loops. My question is that I have to draw the graph on six vertices with degree sequence $(3,3,5,5,5,5)$. I draw the the graph with the given degree sequence and everytime I got the graph which is not simple. Does there exist a simple graph with these degree sequence? If no why?
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Do you know the Havel-Hakimi Theorem? – EuYu Sep 18 '12 at 18:14
No! Can you please state this theorem! – Kns Sep 18 '12 at 18:17
## 2 Answers
No fancy theoretical machinery is required.
Call the vertices $v_1,v_2,v_3,v_4,v_5$, and $v_6$. Let $v_1,v_2,v_3$, and $v_4$ be the vertices of degree $5$. There are only six vertices, so each of these vertices must be connected by an edge to every other vertex in the graph. For instance, $v_2$ must be connected to $v_1,v_3,v_4,v_5$, and $v_6$. This means that $v_5$ and $v_6$ are both connected to each of the vertices $v_1,v_2,v_3$, and $v_4$ and must therefore have degree at least $4$: the degree sequence $(3,3,5,5,5,5)$ is impossible for a simple graph.
If you’d like to explore further, this kind of reasoning leads to the easy half of the Erdős-Gallai theorem, which gives a simple computational criterion for whether a sequence is the degree sequence of some simple graph. This kind of reasoning shows that a sequence that fails the criterion cannot possibly be a degree sequence; proving that every sequence that satisfies it really is the degree sequence of some simple graph is harder. See also the comments here.
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By the Havel-Hakimi Theorem, a descending degree sequence $\{a_1, \cdots ,\ a_k\}$ is graphical if and only if $a_1 \le k-1$ and $\{a_2 - 1, \cdots, a_{a_1+1}-1, a_{a_1 +2}, \cdots, a_k\}$ is graphical.
Applying the theorem to your case, we have $$\{5,5,5,5,3,3\} \iff \{4,4,4,2,2\}\iff\{3,3,1,1\}\iff\{2,0,0\}$$ The latter graph is evidently impossible.
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Evidently not. $\{1,1,1,1\}$ is the degree sequence of two disjoint $K_2$s. However, it is impossible for the sequence as originally stated. – Rick Decker Sep 18 '12 at 19:46
@RickDecker Sorry about that, I must've mistyped the sequence in the middle and started using it. It is fixed now. – EuYu Sep 18 '12 at 20:28
A minor point, but if you edit a post of yours it's a good idea to indicate the edit as part of your post. Otherwise someone new to this thread might be confused by the existing comments (as here, for example). – Rick Decker Sep 18 '12 at 23:38
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http://www.reference.com/browse/Heawood+conjecture
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Definitions
Nearby Words
# Heawood conjecture
The Heawood conjecture or Ringel–Youngs theorem in graph theory gives an upper bound for the number of colors which are sufficient for graph coloring on a surface of a given genus. It was proven in 1968 by Gerhard Ringel and J. W. T. Youngs. One case, the non-orientable Klein bottle, proved an exception to the general formula. An entirely different approach was needed for finding the number of colors needed for the sphere, eventually solved as the four color theorem.
## Formal statement
P.J. Heawood conjectured in 1890 that for a given genus g > 0, the minimum number of colors necessary to color all graphs drawn on an orientable surface of that genus (or equivalently to color the regions of any partition of the surface into simply connected regions) is given by
$gamma \left(g\right) = left lfloor frac\left\{7 + sqrt\left\{1 + 48g\right\}\right\}\left\{2\right\} right rfloor,$
where $left lfloor x right rfloor$ is the floor function.
Replacing the genus by the Euler characteristic, we obtain a formula that covers both the orientable and non-orientable cases,
$gamma\left(chi\right) = left lfloor frac\left\{7 + sqrt\left\{49 - 24chi\right\}\right\}2 right rfloor.$
This relation holds, as Ringel and Youngs showed, for all surfaces except for the Klein bottle. Franklin (1930) proved that the Klein bottle requires at most 6 colors, rather than 7 as predicted by the formula (see Franklin graph).
In one direction, the proof is straightforward: by manipulating the Euler characteristic, one can show that any graph embedded in the surface must have at least one vertex of degree less than the given bound. If one removes this vertex, and colors the rest of the graph, the small number of edges incident to the removed vertex ensures that it can be added back to the graph and colored without increasing the needed number of colors beyond the bound. In the other direction, the proof is more difficult, and involves showing that in each case (except the Klein bottle) a complete graph with a number of vertices equal to the given number of colors can be embedded on the surface.
## Example
The torus has g = 1, so χ = 0. Therefore, as the formula states, any subdivision of the torus into regions can be colored using at most seven colors. The illustration shows a subdivision of the torus in which each of seven regions are adjacent to each other region; this subdivision shows that the bound of seven on the number of colors is tight for this case. The boundary of this subdivision forms an embedding of the Heawood graph onto the torus.
## References
• Franklin, P. (1934). "A six color problem". J. Math. Phys. 13 363–379.
• Heawood, P. J. (1890). "Map colour theorem". Quart. J. Math. 24 332–338.
• Ringel, G.; Youngs, J. W. T. (1968). "Solution of the Heawood map-coloring problem". Proc. Nat. Acad. Sci. USA 60 438–445. 0228378.
| doi = 10.1073/pnas.60.2.438}}
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http://mathhelpforum.com/algebra/25356-finding-inverse-function.html
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# Thread:
1. ## Finding the inverse of a function
Find the inverse of the function:
f(x) = (3/4) x^5 + 5
read "three-fourths x to the fifth plus 5"
Graphing my progress on my calculator as I go, the last point where I get an inverse is at
5^[sqrt] ((4/3)x - (20/3))
^(root with an index of 5)
If I am thinking correctly, you cannot have a root in the denominator, so I've been multiplying the whole thing above by
5^[sqrt](3^4)
^again, root with an index of 5
to get the fifth root out of the denominator, but this messes the whole equation up and it no longer mirrors the original function when I plug it into my graphing calculator. Is there some rule with roots about square roots that I am unaware of?
Help is much appreciated.
2. Originally Posted by happydino1
Find the inverse of the function:
f(x) = (3/4) x^5 + 5
read "three-fourths x to the fifth plus 5"
Graphing my progress on my calculator as I go, the last point where I get an inverse is at
5^[sqrt] ((4/3)x - (20/3))
^(root with an index of 5)
If I am thinking correctly, you cannot have a root in the denominator, so I've been multiplying the whole thing above by
5^[sqrt](3^4)
^again, root with an index of 5
to get the fifth root out of the denominator, but this messes the whole equation up and it no longer mirrors the original function when I plug it into my graphing calculator. Is there some rule with roots about square roots that I am unaware of?
Help is much appreciated.
Analytically to find the inverse of a function $y = f(x)$ we switch the roles of x and y to get $x = f(y)$, then solve for y: $y = g(x)$. g(x) is our inverse function.
So
$y = f(x) = \frac{3}{4}x^5 + 5$
Goes to
$x = \frac{3}{4}y^5 + 5$
$x - 5 = \frac{3}{4}y^5$
$y^5 = \frac{4}{3}(x - 5)$
$y = \sqrt[5]{\frac{4}{3}(x - 5)}$
-Dan
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http://math.stackexchange.com/questions/262946/uniqueness-for-3-dimensional-heat-equation-initial-robin-boundary-value-problem?answertab=active
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# Uniqueness for 3-dimensional heat equation initial Robin boundary value problem (SOLVED)
Let $\Omega \subset \mathbb{R}^3$ be a bounded domain. Using an energy argument, show that the IBVP \begin{align} u_t &= \Delta u ~~~~~~~~~~x \in \Omega, ~t>0\\ \frac{\partial u}{\partial \nu} + \alpha u &= h(x) ~~~~~~~~x \in \partial\Omega, ~t>0\\ u(x,0)&=g(x) ~~~~~~~~~x \in \Omega \end{align} where $\nu$ is the exterior unit normal and $\alpha$ is a constant has at most one solution. Treat the cases $\alpha \geq 0$ and $\alpha<0$ separately. Use logarithmic convexity for the second case.
My attempted solution: By contradiction, suppose that there are two solutions $u_1$ and $u_2$ and define $v = u_1 - u_2$. Then $v$ satisfies \begin{align} v_t &= \Delta v ~~~~~~~~~~x \in \Omega, ~t>0\\ \frac{\partial v}{\partial \nu} + \alpha v &= 0 ~~~~~~~~~~~~~x \in \partial\Omega, ~t>0\\ v(x,0)&=0 ~~~~~~~~~~~~~x \in \Omega \end{align} Define the energy functional to be $$E(t)= \frac{1}{2}\int_\Omega v^2 \,dx.$$ The case $\alpha \geq 0$ is trivial. I just showed that $$\frac{dE}{dt}=\int_\Omega v \, v_t \,dx \leq 0$$ using Green's first identity and the conditions on $v$. Then since $E(0)=0$ we must have $E(t)=0$, and hence $v=0$.
For the $\alpha<0$ case I want to show that $$E\frac{d^2E}{d^2t} - \left( \frac{dE}{dt} \right)^2 \geq 0\,.$$ Since $E \geq 0$ then by logarithmic convexity we would have $E=0$.
However, I'm running into some problems. I take \begin{align} \frac{d^2E}{dt^2}=\int_\Omega v_t^2 \,dx + \int_\Omega v \, v_{tt} \,dx\,. \end{align} Then, for the second term I write \begin{align} \int_\Omega v \, v_{tt} \,dx &=\int_\Omega v \, \Delta v_t \, dx\\ &= \int_\Omega v_t \, \Delta v \, dx \\ &= \int_\Omega (v_t)^2 \, dx \end{align}
where I used Green's second identity and the boundary term vanished due to the boundary condition on $v$. Explicitly:
\begin{align} \int_{\partial \Omega} v\frac{\partial v_t}{\partial \nu} - v_t \frac{\partial v}{\partial \nu} dS= \alpha \int_{\partial \Omega} -v v_t + v_t v \, dS = 0 \end{align} by the homogeneous Robin condition.
So I get $$E\frac{d^2E}{d^2t} - \left( \frac{dE}{dt} \right)^2 = \frac{1}{2}\int_\Omega v^2 dx \cdot 2 \int_\Omega v_t^2 dx - \left( \int_\Omega v v_t \, dx \right)^2 \geq 0$$ by the Cauchy-Schwartz inequality.
So I did the proof without even using the assumption $\alpha < 0$, which seems very strange. Did I make a mistake somewhere?
EDIT: Looks like my proof is actually correct. I guess the wording of the question just had me thinking there was an issue.
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It's OK, I got it. Add an explanation anyway :) Since the boundary terms cancel before you even applied the log convexity argument, it's no wonder that you could prove uniqueness for both cases in one argument. – Hans Engler Dec 21 '12 at 0:05
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Maybe the instructor wants you to try two different proof techniques, one being less powerful. – Hans Engler Dec 21 '12 at 1:35
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Indeed, $\alpha$ could be a function of $x$ (boundary with variable permeability). It had to be independent of $t$, though. – user53153 Dec 21 '12 at 4:03
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@Bartek - remind me why a logarithmically convex non-negative function must be zero. – Hans Engler Dec 21 '12 at 4:18
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Yes, the proof is correct. And I understand why the professor wanted you to work the two cases separately: if you used log-convexity for both, you'd miss a chance to practice a proof based on the monotonicity of energy. – user53153 Dec 21 '12 at 4:34
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## 1 Answer
My original proof is actually correct. I was just confused by the wording of the problem and thought I had made a mistake. Thanks to Hans and Pavel for checking it.
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http://mathoverflow.net/questions/43882/what-are-some-open-problems-in-toric-varieties
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## What are some open problems in toric varieties?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In light of the nice responses to this question, I wonder what are some open problems in the area of toric geometry? In particular,
What are some open problems relating to the algebraic combinatorics of toric varieties?
and
What are some open problems relating to the algebraic geometry of toric varieties?
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I wanted to mention the question of existence of full exceptional collections on toric Fano varieties, but this was answered in a very recent [paper by Efimov](arxiv.org/abs/1010.3755). Probably there are still interesting questions left regarding derived categories of toric varieties... – Piotr Achinger Oct 27 2010 at 22:11
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Let $X$ be a complete toric variety, necessarily neither smooth nor projective. Is there a nontrivial vector bundle on $X$? Payne has constructed examples that have no one- or two-dimensional bundles, and constructing vector bundles of higher rank on these is still open. The "mirror" question is whether there exists a Lagrangian submanifold of $(\mathbb{C}^*)^n$ satisfying certain asymptotic conditions coming from the fan of $X$. – David Treumann Oct 28 2010 at 1:21
Although I don't really follow this, but I think that the following is an open conjecture. Conjecture Every ample divisor on a smooth toric variety is very ample and induces a projectively normal embedding. Is that right? – Karl Schwede Oct 28 2010 at 4:04
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@Piotr. Having looked at the introduction to the Efimov paper, it seems that there are no new positive results there about full exceptional collections. Kawamata showed there is always a full exceptional collection of coherent sheaves on a smooth toric DM stack. And Orlov (says Efimov) conjectures that there should exist a strong full exceptional collection. It was conjectured that toric Fano varieties should admit full exceptional collections of line bundles. Efimov shows that this is false in general. Apologies to Efimov if I read him incorrectly. – Chris Brav Oct 28 2010 at 7:59
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@Karl: ample $\Rightarrow$ very ample is known for smooth toric varieties, but ample $\Rightarrow$ projectively normal is indeed a well-known open question. (I say ``question'' rather than ``conjecture'' as it doesn't seem like all experts believe it.) – ABayer Oct 28 2010 at 12:34
show 3 more comments
## 2 Answers
My favourite is Oda's Strong Factorization Conjecture:
Can a proper, birational map between smooth toric varieties be factored as a composition of a sequence of smooth toric blow-ups followed by a sequence smooth toric blow-downs?
Note that if you are allowed to intermingle the blow-ups and blow-downs (the weak version) it has been proved. In fact, it was proved for general varieties in characteristic 0 using the toric case:
Torification and Factorization of Birational Maps. Abramovich, Karu, Matsuki, Wlodarczyk.
A conjectural algorithm for computing toric strong factorizations can be found in the following arXiv article:
On Oda's Strong Factorization Conjecture. Da Silva, Karu.
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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.
The following question was posed by Rikard Bögvad in the paper On the homogeneous ideal of a projective nonsingular toric variety:
Is the toric ideal of a smooth projectively normal toric variety generated by quadrics?
This is interesting, since toric ideals have an explicit description. In particular, it is not known if the coordinate ring of a smooth projectively normal toric variety is Koszul. Smoothness is of course essential here, since there are many toric hypersurfaces of degree $\ge 3$, e.g., $x_0^n=x_1 \cdots x_n$.
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http://physics.stackexchange.com/questions/30997/winding-number-in-the-topology-of-magnetic-monopoles/31028
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# Winding number in the topology of magnetic monopoles
I am reading on magnetic monopoles from a variety of sources, eg. the Jeff Harvey lectures.. It talks about something called the winding $N$, which is used to calculate the magnetic flux. I searched the internet but am not being able to understand the calculation done in this particular case.
$g=-\frac{1}{8}\int_{S^2_\infty} Tr([d\hat{\Phi},d\hat{\Phi}],\hat{\Phi})$
Then the author says that
Now $\Phi$ restricts to a map $\Phi : S_\infty^2 → S^2$ , where the target is the unit sphere in $su(2)$. This map has some degree $N$ , and it is easy to verify that the right-hand side of the above equation is $−2\pi$ times this. Therefore $g = −2\pi N$ .
What is $N$, the winding number also called as the degree on the map? By what i have learnt, it is the number of times you wind an object unto the another, then shouldn't the integral be $N*4\pi$, as $4\pi$ is the surface area of $S^2$.
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I suppose that you are working with the SU(2)-Yang-Mills-Higgs theory, right? what is $\hat{\Phi}$? Is this scalar field the time component of a static gauge potential in Minkowski, after making dimensional reduction? – Jorge Campos Jun 30 '12 at 1:17
@JorgeCampos, I am working with SU(2) yang-mills. $\hat{\Phi}$, is the angular part of the higgs field. The total higgs field is given by $\Phi=h\hat{\Phi}$, where $h$, is a function of the radial distance $r$, where norm $\hat{\Phi}=1$. – ramanujan_dirac Jun 30 '12 at 5:47
## 2 Answers
I didn't find the equation and the argument you quoted in that paper. But, yes, it is the Brouwer degree, deg$(\hat{\Phi})$, which equals the monopole number $$N\equiv\frac{1}{4\pi}\int_{\mathbb{R}^3} \mathrm{Tr}(F_A\wedge D_A(\Phi))=\frac{1}{4\pi}\int_{\mathbb{R}^3} d(\mathrm{Tr}(\Phi)F_A) =\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr}(\Phi F_A)$$ $$=\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr}(\hat{\Phi} F_A)$$
where the one has used Bianchi identity, Stokes' theorem, to obtain the first two equalities, and Jaffe & Taubes show in their book that one can replace $\Phi$ by $\hat{\Phi}$. Now this coincides with the brower degree, for which there is an explicit formula:
$$N=\mathrm{deg}(\hat{\Phi})=-\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr(\hat{\Phi} d\hat{\Phi}\wedge d\hat{\Phi}})\in \mathbb{Z}=\pi_2(S^2)=[S^2,S^2]$$ (what you wrote.) This is physically understood as an infinite wall potential, separating the monopole sectors corresponding to different integers. Now, to actually answer your question, you can compute this integral for the t'Hooft-Polyakov monopole solution, for which $$\hat{\phi}=(\sin(\theta)\cos\phi,\sin\theta \sin\phi,\cos\theta)_i\cdot \sigma^i,$$ and you will find $$N=-\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr(\hat{\Phi} d\hat{\Phi}\wedge d\hat{\Phi}})=+\frac{1}{4\pi}\int_{[0,2\pi]} \int_{[0,\pi]}\sin\theta d\theta\wedge d\phi=1.$$
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I asked this question exactly because I don't know what is brower degree, I coudn't understand it from the internet due to reference to algebraic topology which i have not studies. Please could you explain what this is, and how did you arrive at the explicit formula of the brower degree. – ramanujan_dirac Jul 2 '12 at 8:09
I'll answer that soon; I'll reedit... – Jorge Campos Jul 2 '12 at 19:28
N is equal to the number of points in the $S_2$ sphere at infinity mapped to the same point of the $S_2$ Higgs vacuum manifold.The integral is a topological invariant depending only on this number and not on the details of the map. In the following, I'll describe to you a family of these maps:
One way to perform the integral is to use the stereographic projection coordinate:
$z = tan(\frac{\theta}{2}) e^{i\phi}$
In this coordinate system,a map of winding number $N$ will look as:
$z \rightarrow Z= z^N$
The surface element of the sphere in these coordinates is:
$d \mu = \frac{dzd\bar{z}}{1+z \bar{z}}$
Remark: In these coordinates, the Higgs components are given by:
$\Phi_x = 2\frac{Re(Z)}{1+Z\bar{Z}}$
$\Phi_y = 2\frac{Im(Z)}{1+Z\bar{Z}}$
$\Phi_z = \frac{1-Z\bar{Z}}{1+Z\bar{Z}}$
Using these coordinates,it is not difficult to see that the integral (1.43) value is N.
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http://math.stackexchange.com/questions/13155/why-study-curves-instead-of-1-manifolds?answertab=active
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# Why study “curves” instead of 1-manifolds?
In most undergraduate differential geometry courses -- I am thinking of do Carmo's "Differential Geometry of Curves and Surfaces" -- the topic of study is curves and surfaces in $\mathbb{R}^3$. However, the definition of "curve" and "surface" are usually presented in very different ways.
A curve is defined simply as a differentiable map $\gamma\colon I \to \mathbb{R}^3$, where $I \subset \mathbb{R}$ is an interval. Of course, some authors prefer to define a curve as the image of such a map, and others require piecewise-differentiability, but the general concept is the same.
On the other hand, surfaces are essentially defined as 2-manifolds.
Similarly, in graduate courses on manifolds -- I am thinking of John Lee's "Introduction to Smooth Manifolds" -- one talks about curves $\gamma\colon I \to M$ in a manifold, and can do line integrals over such curves, but talks separately about embedded/immersed 1-dimensional submanifolds.
My question, then, is:
Why make (parametrized) curves the object of study rather than 1-manifolds?
Earlier, I asked a question that was perhaps meant to hint at this one, though I didn't say so explicitly.
Ultimately, I would simply like to say "curves are 1-manifolds and surfaces are 2-manifolds," and am looking for reasons why this is correct/incorrect or at least a good/bad idea. (So, yes, I'm looking for a standard definition of "curve.")
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I maybe misinterpreting your question, but there is some pedagogical value in treating curves before surfaces, and embedded objects in $\mathbb{R}^3$ (or evene in $\mathbb{R}^2$) before abstract manifolds. – timur Dec 24 '10 at 5:52
## 2 Answers
Essentially because the connected 1-manifolds are ("up to...") $(0,1)$ and $S^1$, so the notion of curve captures all of the possibilities of 1-manifolds sitting in higher-dimensional manifolds. In other words, the situation for 1-dimensional manifolds is so simple that it really makes no sense to use the full machinery of embeddings and immersions to talk about them other than just checking that the definitions of embedding and immersion are compatible with the definition of curve.
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Also, in agreement to what kahen said:
In a regular parametrization, you can always find a metric that is constant along the curve leading to zero intrinsic curvature. This is also a reason why it might be somehow not very enlightening to discuss curves as one dimensional manifolds in its own right.
Curves, on the other hand, can have extrinsic curvature which can for some body moving along a trajectory be interpreted as the acting force.
Greets
Robert
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http://math.stackexchange.com/questions/147441/change-of-basis-calculation?answertab=votes
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# Change of Basis Calculation
I've just been looking through my Linear Algebra notes recently, and while revising the topic of change of basis matrices I've been trying something:
"Suppose that our coordinates are $x$ in the standard basis and $y$ in a different basis, so that $x = Fy$, where $F$ is our change of basis matrix, then any matrix $A$ acting on the $x$ variables by taking $x$ to $Ax$ is represented in $y$ variables as: $F^{-1}AF$ "
Now, I've attempted to prove the above, is my intuition right?
Proof: We want to write the matrix $A$ in terms of $y$ co-ordinates.
a) $Fy$ turns our y co-ordinates into $x$ co-ordinates.
b) pre multiply by $A$, resulting in $AFy$, which is performing our transformation on $x$ co-ordinates
c) Now, to convert back into $y$ co-ordinates, pre multiply by $F^{-1}$, resulting in $F^{-1}AFy$
d) We see that when we multiply $y$ by $F^{-1}AF$ we perform the equivalent of multiplying $A$ by $x$ to obtain $Ax$, thus proved.
Also, just to check, are the entries in the matrix $F^{-1}AF$ still written in terms of the standard basis?
Thanks.
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## 3 Answers
I don't know if the following helps, but anyway: So you have a vector space $V$ (over say the complex numbers) and you have say two basis $E$ and $D$.
That $F$ is a change of basis matrix means that if as column vector $y = (y_i)$ written with respect to the basis $E$, then you get the coordinates with respect to $D$ by $x = Fy$.
Now you have a linear transformation $T: V \to V$. With respect to each basis, this transformation is given by two matrices, say $A_E$, $A_D$. So if $y = (y_i)$ (wrt. basis $E$) then $Ty = A_Ey$ and the result with be the coordinates in the basis $E$. (And likewise for the basis $D$ using $A_D$).
So given a vector $y = (y_i)$ written in the basis $E$, you could then first transform the coordinates to the basis $D$, then you the matrix $A_D$ and then transform the coordinated back to the basis $E$. So you get $A_E(y) = F^{-1}A_DFy$.
One can actually write out all of this (If you have never done so I recommend that you do it) with coordinates. So you would start with the vector $v$ and write it as a linear combination of the basis $E$: $v = y_1e_1 + \dots y_ne_n$ and continue from there...
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Your statements a) – d) are correct as stated.
Concerning the sentence "Also, just to check, are the entries in the matrix $F^{-1}AF$ still written in terms of the standard basis?", one can say the following: The entries in this matrix are just numbers. The given matrix $A$ describes a certain linear map acting on points $x=(x_1,\ldots, x_n)$. The matrix $F^{-1}AF$ describes the same linear transformation when the points $x$ get new coordinates $(y_1,\ldots, y_n)$.
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Without saying much, here is how I usually remember the statement and also the proof in one big picture:
\begin{array}{ccc} x_{1},\dots,x_{n} & \underrightarrow{\;\;\; A\;\;\;} & Ax_{1},\dots,Ax_{n}\\ \\ \uparrow F & & \downarrow F^{-1}\\ \\ y_{1},\dots,y_{n} & \underrightarrow{\;\;\; B\;\;\;} & By_{1},\dots,By_{n} \end{array}
And $$By=F^{-1}AFy$$
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http://mathhelpforum.com/calculus/133366-minimizing-distance-word-problem.html
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Thread:
1. Minimizing distance word problem
a train leaves the station at 10:00 pm and travels due north at a speed of 100km/h. another train has been heading due west at 120km/h and reaches the same station at 11:00pm. at what time were the two trains cloet together?
no clue how to do it
so different form textbook examples
2. Obviously, it will be at 10:00 or some time after, since train 1 is at the station and train 2 is coming directly toward the station. And it will be 11:00 or some time before, since after that, train 1 is going away from the station to the north and train 2 is going away from the station to the west, so they are only getting further apart.
If you define t to be the number of hours after 10:00, the position of train 1 is (0,100t) and the position of train 2 is (120(1-t),0). So you can use the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ to get the distance between the two trains as a function of t. Then you find the minimum by taking the derivative and setting it to zero.
As a check, you can use some intuition - at 10:30, the trains are the same distance from the station, but train 2 is moving faster so they're still getting closer together. So your answer should be a little after 10:30.
Post again if you're still having trouble.
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http://math.stackexchange.com/questions/118062/the-closure-of-c1-in-the-functions-of-bounded-variation/124610
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# The closure of $C^1$ in the functions of bounded variation
Consider the space $(BV[0,1];||.||)$ with the norm
$$||f||=|f(0)|+V_{f}[0,1]$$ Where $V_{f}[0,1]$ is the variation of $f$. My questions
what is the closure of $C^1[0,1]$ with respect to this norm?
Another question is how to prove that this norm is Banach?
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2
– t.b. Mar 9 '12 at 0:26
I guess this space is of absolute continuous functions! – checkmath Mar 9 '12 at 20:16
As @t.b. points out, your 2nd question has already been asked elsewhere, and there have been some hints. – user16299 Mar 24 '12 at 2:08
Yeah, but elsewhere I looking for the answer of the first question, rather than the second! – checkmath Mar 24 '12 at 2:18
## 2 Answers
I think the space is $W^{1,1}[0, 1]$. We clearly have that the closure (say $B$) is in $W^{1, 1}$. Furthermore, $W^{1, 1}$ is a proper subset of $\text{BV}$.
So, take a function $f$ in $W^{1, 1}$ and take an approximating sequence $f_n$ consisting of $C^\infty$ functions in the $W^{1, 1}$ norm.
So, we have $\|f_n - f\|_{\text{BV}} \lesssim \|f_n - f\|_{W^{1, 1}} \to 0$.
As $f_n$ are all in $C^1$ we also have that $f$ is in $B$.
Now we have
$$f(x) = f(0) + \int_0^x f'(t) \, \textrm{d}t.$$
So, $\|f\|_{W^{1, 1}} = \|f\|_{L^1} + \|f'\|_{L^1}$.
And $$\|f\|_{L^1} \leqslant |f(0)| + \int_0^1 \left | \int_0^x f'(t) \, \textrm{d}t \right | \leqslant |f(0)| + \int_0^1 |f'(t)| \, \textrm{d}t.$$
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Is the space of Lipschitzian functions closed in this odd norm? – checkmath Mar 24 '12 at 20:41
I hope so 8-). Let me try. – Jonas Teuwen Mar 24 '12 at 20:43
@chessmath What do you think now? – Jonas Teuwen Mar 25 '12 at 16:50
I'll check! But if you explain the first " we clearly have" will not hurt! – checkmath Mar 25 '12 at 17:49
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Interesting. Why the downvote? – Jonas Teuwen Mar 26 '12 at 13:48
show 3 more comments
sIt is the space $W^{1,1}[0, 1]$. Note initially that $W^{1,1}[0, 1]$ is a subspace of $BV$ moreover in there the norms are equivalent. In fact if $f\in W^{1,1}[0, 1]$, for its continuous representative
$$f(x)=f(0)+\int_{0}^{x}f'(t)dt\tag{1}$$ Then $$|f(x)|\leq|f(0)|+|\int_{0}^{x}f'(t)dt|\leq |f(0)|+\int_{0}^{1}|f'(t)|dt.$$
But $V_f[0,1]=\int_{0}^{1}|f'(t)|dt$. So $||f||_L^1\leq||f||_{\infty}\leq||f||_{BV}$ since $||f'||_L^1\leq||f||_{BV}$ then
$||f||_{W^{1,1}}\leq 2||f||_{BV}.$ Using $(1)$ again we get $$f(0)=-f(x)+\int_{0}^{x}f'(t)dt$$
then $|f(0)|\leq |f(x)|+\int_{0}^{1}|f'(t)|dt$,
integrating from $0$ to $1$ $|f(0)|\leq ||f||_{W^{1,1}}$ and analogously $||f||_{BV}\leq 2||f||_{W^{1,1}}$ ergo both norms are equivalent.
If $f_n \to f$ in $BV$ with $(f_n)\subset C^{1}$ then $(f_n)$ is a Cawchy sequence in $W^{1,1}$, since it is a complete space $f_n\to g$ in $W^{1,1}$.
Then $f_n\to g$ in $BV$ because the norms are equivalents and by the unity of the limit $g=f\in BV$.
This proves that the closure of ${C^1}$ in the $BV$ norm is a subspace of $W^{1,1}$.
The other inclusion is given because any function $f\in W^{1,1}$ can be approximated by a sequence $f_n\to f$ in $W^{1,1}$ what is the same $f_n\to f$ in $BV$ $\blacksquare$
PS: That is my answer it may contains some English problems but it is mathematically correct! And all details are included! It is of course inspired on @Jonas answer! And he deserved to gain the bounty, and I thanks to him! But how he was not so sure ( he wrote I think it is W1,1) I had to put a complete answer. My first thought were to edit his answer but I realized it was a complete rewriting. I decided to put a new answer.
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That is my answer it may contain some English trouble but it is mathematically correct! And all details are included! It is of course inspired on @Jonas answer! And he deserved to gain the bounty! But how he was not so sure ( he wrote I think it is $W^{1,1}$). My first thought were to edit his answer but I realized it was a complete rewriting. I decided to put a new answer. – checkmath Mar 26 '12 at 11:36
Well, I say that I think the space is $W^{1, 1}$ and then give a reasonably complete argument 8-). Maybe it would have been better if you would have added this to your post instead of giving an answer? – Jonas Teuwen Mar 26 '12 at 12:03
Come on your answer was lacking some pieces, No offense but you answered wrong first! I had to be sure! – checkmath Mar 26 '12 at 12:08
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Yes, I have answered it wrong first, and then I have fixed it. I understand that you want to be sure if it is correct. By the way, I'm not the one that downvoted you. – Jonas Teuwen Mar 26 '12 at 12:29
Hey downvoters, I gave up 50 points to this question do you think I'll mind for some improper downvotes? – checkmath May 13 '12 at 5:19
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http://physics.stackexchange.com/questions/tagged/hilbert-space?page=2&sort=newest&pagesize=15
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# Tagged Questions
The hilbert-space tag has no wiki summary.
3answers
351 views
### Can we have discontinuous wavefunctions in the Infinite Square well?
The energy eigenstates of the infinite square well problem look like the Fourier basis of L2 on the interval of the well. So then we should be able to for example make square waves that are an ...
3answers
248 views
### If I go to the church of the greater Hilbert space, can I have Unitary Collapse?
Actually, unitary pseudo-collapse? Von Neuman said quantum mechanics proceeds by two processes: unitary evolution and nonunitary reduction, also now called projection, collapse and splitting. ...
2answers
176 views
### Interference, photon's phase, and the Hilbert space
Dirac said that a photon can only interfere with itself. This is consistent with the tensor product of two photon spaces representation. On the other hand, it is known that there is interference ...
0answers
102 views
### Can experiment distinguish the basis in which a singlet state is represented?
Let $\left(|\uparrow\rangle,|\downarrow\rangle\right)$ and $\left(|\nearrow\rangle,|\swarrow\rangle\right)$ be two bases of the $2$-dimensional Hilbert space $H$. Can an experiment distinguish ...
1answer
310 views
### Quantum mechanic newbie: why complex amplitudes, why Hilbert space?
I'm just starting learning quantum mechanics by myself (2 "lectures" so far) and I was wondering why we need to define quantum states in a complex vector space rater than a real one? Also I was ...
3answers
375 views
### Existence of creation and annihilation operators
In a multiple particle Hilbert space (any space of any multi-particle system), is it sufficient to define creation and annihilation operators by their action (e.g. mapping an n-particle state to an ...
1answer
47 views
### In which way is decoherence not symmetric between the two considered systems?
If a quantum system interacts with a "big" quantum system, you have dephasing. The models of decoherence all have this atog aproach to them, about what is to understood of the interaction of the ...
1answer
149 views
### How can Hilbert spaces be used to study the harmonics of vibrating strings?
The overtones of a vibrating string. These are eigenfunctions of an associated Sturm–Liouville problem. The eigenvalues 1,1/2,1/3,… form the (musical) harmonic series. How can Hilbert spaces be ...
2answers
156 views
### Under what assumptions can we split a Hilbert space into subspaces?
I was thinking about an apparently simple question about quantum mechanics, if I am looking at a quantum system described by a Hilbert space $\cal{H}$ under what hypothesis can I define A and B as ...
1answer
343 views
### Where does the wave function of the universe live? Please describe its home
Where does the wave function of the universe live? Please describe its home. I think this is the Hilbert space of the universe. (Greater or lesser, depending on which church you belong to.) Or maybe ...
2answers
949 views
### Proof that the One-Dimensional Simple Harmonic Oscillator is Non-Degenerate?
The standard treatment of the one-dimensional quantum simple harmonic oscillator (SHO) using the raising and lowering operators arrives at the countable basis of eigenstates \$\{\vert n \rangle\}_{n = ...
0answers
67 views
### Shape of the state space under different tensor products
I am currently studying generalized probabilistic theories. Let me roughly recall how such a theory looks like (you can skip this and go to "My question" if you are familiar with this). Recall: In a ...
1answer
302 views
### Bra space and adjoint vectors
If I'm not wrong, a bra, $\langle \phi_n |$, can be thought as a linear functional that when applied to a ket vector, $| \phi_m \rangle$, returns a complex number; that is, the inner product it's a ...
1answer
75 views
### Representation on Hilbert space of the product of two symmetry transformations
We know by Wigner's theorem that the representation of a symmetry transformation on the Hilbert space is either unitary and linear, or anti-unitary and anti-linear. Let $T$ and $S$ be two symmetry ...
4answers
198 views
### How to apply an algebraic operator expression to a ket found in Dirac's QM book?
I've been trying to learn quantum mechanics from a formal point of view, so I picked up Dirac's book. In the fourth edition, 33rd page, starting from this:$$\xi|\xi'\rangle=\xi'|\xi'\rangle$$ (Where ...
2answers
336 views
### What Hermitian operators can be observables?
We can construct a Hermitian operator $O$ in the following general way: find a complete set of projectors $P_\lambda$ which commute, assign to each projector a unique real number \$\lambda\in\mathbb ...
4answers
236 views
### Books on Hilbert space and phase space?
Can you recommend books or papers that highlight or discuss extensively, or at least more than average, the similarites/differences between phase space and Hilbert space? I am primarily interested in ...
2answers
312 views
### What is the Hilbert space of a single electron?
Is it the same as the space of all possible descriptions of a single electron? If not, how do they differ? Please give the mathematical name or specification of this space or these spaces.
2answers
372 views
### Bra-ket notation and linear operators
Let $H$ be a hilbert space and let $\hat{A}$ be a linear operator on $H$. My textbook states that $|\hat{A} \psi\rangle = \hat{A} |\psi\rangle$. My understanding of bra-kets is that $|\psi\rangle$ is ...
3answers
401 views
### Hilbert space and Lie algebra in quantum mechanics
We are looking for a publication or website that explains the Standard Model in terms of Hilbert space and Lie algebra. We are reading Debnath's Introduction to Hilbert Spaces and Applications and ...
1answer
265 views
### Born-Oppenheimer Approximation equivalent to Tensor-product ?
If you have a wave function $\Psi$ of a system consisting of an electron and the vibrational modes of the crystal, THEN we represent the wavefunction $\Psi%$ to be in the Hilbert Space formed by the ...
2answers
129 views
### Uniqueness of eigenvector representation in a complete set of compatible observables [duplicate]
Possible Duplicate: Uniqueness of eigenvector representation in a complete set of compatible observables Sakurai states that if we have a complete, maximal set of compatible observables, ...
2answers
300 views
### Uniqueness of eigenvector representation in a complete set of compatible observables
Sakurai states that if we have a complete, maximal set of compatible observables, say $A,B,C...$ Then, an eigenvector represented by $|a,b,c....>$, where $a,b,c...$ are respective eigenvalues, is ...
2answers
85 views
### What Shannon channel capacity bound is associated to two coupled spins?
The question asked is: What is the Shannon channel capacity $C$ that is naturally associated to the two-spin quantum Hamiltonian $H = \boldsymbol{L\cdot S}$? This question arises with a view ...
1answer
51 views
### Stabilizer formalism for symmetric spin-states?
This question developed out of conversation between myself and Joe Fitzsimons. Is there a succinct stabilizer representation for symmetric states, on systems of n spin-1/2 or (more generally) n higher ...
2answers
210 views
### Physical interpretation of different selfadjoint extensions
Given a symmetric (densely defined) operator in a Hilbert space, there might be quite a lot of selfadjoint extensions to it. This might be the case for a Schrödinger operator with a "bad" potential. ...
3answers
483 views
### What is a basis for the Hilbert space of a 1-D scattering state?
Suppose I have a massive particle in non-relativistic quantum mechanics. Its wavefunction can be written in the position basis as $$\vert \Psi \rangle = \Psi_x(x,t)$$ or in the momentum basis as ...
1answer
272 views
### Why is the Haar measure times the volume of the eigenvalue simplex considered a good measure of Hilbert space volume?
In particular, why do we need both of these to find the volume? And should I be thinking of it as an actual volume or not? This Hilbert space volume is talked about in this paper. It says There ...
2answers
885 views
### Applications of the Spectral Theorem to Quantum Mechanics
I'm currently learning some basic functional analysis. Yesterday I arrived at the spectral theorem of self-adjoint operators. I've heard that this theorem has lots of applications in Quantum ...
2answers
399 views
### Observing the exponential growth of Hilbert space?
One of the weirdest things about quantum mechanics (QM) is the exponential growth of the dimensions of Hilbert space with increasing number of particles. This was already discussed by Born and ...
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http://math.stackexchange.com/questions/193579/minimum-distance-problem-optimize-function
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# Minimum distance problem: Optimize function
I have a function of $4$ variables: (distance function) $$d(x,x_1,y,y_1)=(x−x_1)^2+(y−y_1)^2$$ subject to $2$ constraints:
1. $\frac{(x+h)^2}{a^2}+\frac{(y+k)^2}{b^2}= 1$
2. $\frac{(x_1+h_1)^2}{a_1^2}+\frac{(y_1+k_1)^2}{b_1^2}= 1$
Using Lagrange multipliers, what are the values of $x$, $x_1$, $y$ and $y_1$ in terms of $h$, $k$, $a$, $b$, $h_1$, $k_1$, $a_1$, and $b_1$?
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I typset your equations in TeX to make them easier to read - can you check I didn't introduce any mistakes? Mainly in the definition of $d$, you had (x-x1)2, so I assumed the 2 was a ^2. – Matt Pressland Sep 10 '12 at 12:40
No it is all as it is meant to be. What is TeX? Can I download it from somewhere? Thanks by the way:) – David Hoffman Sep 10 '12 at 12:42
1
– Rod Carvalho Sep 10 '12 at 12:44
Both of them are equal to 1. Sorry. – David Hoffman Sep 10 '12 at 12:46
@ David Hoffman: shouldn't the variables be only $x$ and $y$, and the rest be mere parameters? Also, do you have numerical values for the parameters? – Rod Carvalho Sep 10 '12 at 12:50
show 2 more comments
## 1 Answer
I don't like the notation you're using, so I will use a different one. Consider the ellipses defined below
$$\mathcal{E}_1 := \displaystyle\left\{ (x_1, y_1) \in \mathbb{R}^2 : \left(\frac{ x_1 - x_{10}}{a_1}\right)^2 + \left(\frac{ y_1 - y_{10}}{b_1}\right)^2 = 1 \right\}$$
$$\mathcal{E}_2 := \displaystyle\left\{ (x_2, y_2) \in \mathbb{R}^2 : \left(\frac{ x_2 - x_{20}}{a_2}\right)^2 + \left(\frac{ y_2 - y_{20}}{b_2}\right)^2 = 1 \right\}$$
Note that we have 4 unknowns and 2 constraints. Hence, we have 2 degrees of freedom. It would be neater to solve the optimization problem in 2 variables only. But, how?
I introduce functions $\gamma_1, \gamma_2 : [0, 2 \pi] \to \mathbb{R}^2$ defined by
$$\gamma_1 (\theta_1) = \left[\begin{array}{c} x_{10} + a_1 \cos(\theta_1)\\ y_{10} + b_1 \sin(\theta_1)\end{array}\right]$$
$$\gamma_2 (\theta_2) = \left[\begin{array}{c} x_{20} + a_2 \cos(\theta_2)\\ y_{20} + b_2 \sin(\theta_2)\end{array}\right]$$
In other words, I have parametrized the ellipses. Since you want to minimize the distance between the elipses, we have the following optimization problem
$$\displaystyle\min_{(\theta_1, \theta_2)} \| \gamma_1 (\theta_1) - \gamma_2 (\theta_2)\|_2^2 \quad{} \text{subject to} \quad{} (\theta_1, \theta_2) \in [0, 2 \pi]^2$$
which you can solve using partial derivatives. No need for Lagrange multipliers.
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Thank you very much. This is really good. How about this: I have ellipse E_1: (x-3)^2/2^2+(y-2)^2/1=1 and ellipse E_2: (x-1)^2/1+(y+2)^2/2^2=1. How would you find the minimum distance between these two ellipses? – David Hoffman Sep 10 '12 at 19:19
@ David Hoffman: Have you had any multivariable calculus? Just take the partial derivatives of the objective function with respect to $\theta_1$ and $\theta_2$, see where they vanish, find the minimizers, and finally evaluate the objective function at the minimizers to obtain the minimum distance. – Rod Carvalho Sep 10 '12 at 19:24
So you're saying that I would have to minimize (a_1cos(t)+ x_1-a_2cos(s)-x_2)^2+(b_1sin(t)+y_1-b_2sin(s)-y_2)^2? – David Hoffman Sep 11 '12 at 13:07
@ David Hoffman: Pretty much. But you used $x_i$ and $y_i$ in your expression, instead of $x_{i0}$ and $y_{i0}$. – Rod Carvalho Sep 11 '12 at 13:19
I could not solve this symbolically. I managed to express (t) in terms of (s), which was a polynomial of 8th degree. When I plugged it into the other partial derivative equation, I nor a computer program designated for such stuff could not solve it. – David Hoffman Sep 19 '12 at 9:39
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http://mathoverflow.net/questions/88830/counting-points-over-over-an-algebraic-set-over-finite-field
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## Counting points over over an algebraic set over finite field.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $q=p^n$. Let $C$ be an Artin schierer curve defined by $y^p-y=f(x)$ where $f(x) \in \mathbb{F}_{q}[x]$. Let $C_g$ be $y^p-y=f(x)$ where $x \in g(\mathbb{F}_{p^{n}})$ for some $g \in \mathbb{F}_q[x]$. Is there a relation between number of solutions of $C$ and $C_g$? Or Zeta function of $C_g$ and $C$ ?
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1
I don't really understand your notation. Is $C_g$ the curve $y^p-y=f(g(x))$? If not, then what? – Donu Arapura Feb 18 2012 at 16:10
Yes, but I am looking for solutions over $g(\mathbb{F}_q)$ not over $\mathbb{F}_q$. Please let me know if its still unclear. Thanks! – Heinrich Feb 18 2012 at 16:34
1
Expanding on Donu's comment, $C_g: y^p-y=f(g(x))$ maps to $C: y^p-y=f(x)$ by $(x,y) \mapsto (g(x),y)$. From general theory, it follows that the numerator of the zeta function of $C$ divides the numerator of the zeta function of $C_g$. One can also describe the number of points of $g(\mathbb{F}_q)$ in terms of the Galois group of $g(x)-t$, but the answer will depend a lot on $g$ and to get results about $C$ and $C_g$ you'll need to know whether $f$ and $g$ are related or not. Do you have a specific case that you are interested in? – Felipe Voloch Feb 18 2012 at 21:44
Thanks Felipe!It really helped. Can you please suggest me a reference to read on this? I was interested in the case when $g(x)=x+\frac{1}{x}$ and $f$ is any polynomial. – Heinrich Feb 19 2012 at 6:33
Birch and Swinnerton-Dyer, Note on a problem of Chowla, Acta Arith, 5 (1959) 417-423, for value sets of general $g$. But for $x+1/x$ you won't need this. – Felipe Voloch Feb 19 2012 at 16:21
show 1 more comment
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http://unapologetic.wordpress.com/2007/12/07/archimedean-fields/?like=1&source=post_flair&_wpnonce=ae4d6bb289
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# The Unapologetic Mathematician
## Archimedean Fields
Whether we use Dedekind cuts or Cauchy sequences to construct the ordered field of real numbers $\mathbb{R}$ (and it doesn’t matter which), we are taking the ordered field of rational numbers and enlarging it to be “complete” in some sense or another. But we also aren’t making it too much bigger. The universality property we got from completing the uniform structure already gives evidence of that, but there’s another property which we can show is true of $\mathbb{R}$, and which shows that the real numbers aren’t too unwieldy.
In The Sand Reckoner, the ancient Greek mathematician Archimedes once set about the problem of should the number of grains of sand in existence to be finite. He does this by determining a (very weak) upper bound: the number of grains of sand it would take to fill up the entire universe, as he understood the latter term. He writes:
There are some … who think that the number of the sand is infinite in multitude; and I mean by the sand not only that which exists about Syracuse and the rest of Sicily but also that which is found in every region whether inhabited or uninhabited. Again there are some who, without regarding it as infinite, yet think that no number has been named which is great enough to exceed its multitude. And it is clear that they who hold this view, if they imagined a mass made up of sand in other respects as large as the mass of the earth filled up to a height equal to that of the highest of the mountains, would be many times further still from recognizing that any number could be expressed which exceeded the multitude of the sand so taken. But I will try to show you by means of geometrical proofs, which you will be able to follow, that, of the numbers named by me … some exceed not only the number of the mass of sand equal in magnitude to the earth filled up in the way described, but also that of a mass equal in magnitude to the universe
The deep fact here is a fundamental realization about numbers: the set of natural numbers has no upper bound in the real number system. That is, no matter how huge a real number we pick there’s always a natural number bigger than it. Equivalently, given any positive real number $x$ — even as small as the volume of a grain of sand — and another positive real number $y$ — even as large as the volume of (the ancient Greek conception of) the universe — there’s some natural number $n$ so that $nx\geq y$. When this happens in a given ordered field we say that the field is “Archimedean”.
So let’s show that $\mathbb{R}$ is Archimedean. If there were positive real numbers $x$ and $y$ so that $nx\leq y$ for all natural numbers $n$, then $y$ would be an upper bound for the set of $nx$. Then Dedekind completeness gives us a least upper bound $\sup\{nx\}$, and we can just take $y$ to be this least upper bound. Now $nx\leq y$, and also $(n+1)x\leq y$, and so $nx\leq y-x$. That is, $y-x$ is another upper bound for the set of multiples of $x$. But since $x$ was chosen to be positive we see that $y-x<y$, contradicting the assumption that $y$ was the least such upper bound. So such a pair of real numbers can’t exist.
In particular, we can take a positive real number $x$ and consider the set of natural numbers $n$ which are larger than it. Since the natural numbers are well-ordered, there is a least such number, and it can’t be ${0}$ because we assume $x>0$. Subtracting one from this number will then give the largest natural number that is still below $x$ in the real number order, and we denote this number by $\lfloor x\rfloor$. We can thus write any positive number uniquely as the sum $\lfloor x\rfloor+r$ of a natural number and some remainder with $0\leq r<1$.
It turns out that the real numbers are actually the largest Archimedean field. That is, if $\mathbb{F}$ is any ordered field satisfying the Archimedean property, there will be an monomorphism of ordered fields $\mathbb{F}\rightarrow\mathbb{R}$, making (the image of) $\mathbb{F}$ a subfield of $\mathbb{R}$. I won’t prove this here, but I will note one thing about the meaning of this result: the Archimedean property essentially limits the size of an ordered field. That is, an ordered field can’t get too big without breaking this property. Dually, an ordered field can’t get too small without breaking Dedekind completeness or uniform completeness. Completeness pulls the field one way, while the Archimedean property pulls the other way, and the two reach a sort of equilibrium in the real numbers, living both at the top of one world and the bottom of the other.
### Like this:
Posted by John Armstrong | Fundamentals, Numbers
## 7 Comments »
1. “Completeness pulls the field one way, while the Archimedean property pulls the other way, and the two reach a sort of equilibrium in the real numbers, living both at the top of one world and the bottom of the other.”
A very pretty kinaesthetic metaphor. Which, by the way, was how Einstein and Feynman did a lot of their best work.
Comment by | December 7, 2007 | Reply
2. Just wondering what you mean when you say don’t make the rationals too much bigger to get the reals. We’re adding uncountably many numbers to get the reals…so in what sense are the reals not much bigger than the rationals.
Also, I wonder if there’s a universal property we can use to characterize the reals from the rationals, in a categorical sense. Looks like we can, similiar to the Grothendieck completion of a monoid or semigroup, but it’s too early to draw the commutative diagrams to check.
Comment by Josh | December 10, 2007 | Reply
3. Josh, I go more into that at the bottom of the post. The real numbers aren’t so big they can’t be Archimedean, but they aren’t so small they can’t be complete.
As for the universal property, that’s exactly how I did characterize them, using the completion functor on the category of uniform spaces. This is the left adjoint to the functor which includes the full subcategory of complete uniform spaces into the category of all uniform spaces. More precisely, any uniformly continuous function $f:\mathbb{Q}\rightarrow X$ from the rational numbers to a complete uniform space $X$ extends uniquely to a uniformly continuous function $\bar{f}:\mathbb{R}\rightarrow X$.
Comment by | December 10, 2007 | Reply
4. Just to add a bit to John’s response to Josh: there are ordered field extensions of the rationals much, much, much bigger than $\mathbb{R}$ that people sometimes consider, for example, fields of “hyperreals” which crop up in non-standard analysis (these are discussed at length in the math blog Mort Aux Triangles!). A rather tame example of this sort of thing, which is a bit smaller than these hyperreal extensions but moderately big nonetheless, is the field of rational functions with real coefficients, ordered by their growth “at infinity”; the ordinary reals sit inside as the subfield of constant functions.
But none of these big fields is complete; you can’t even complete them in the way John is discussing! Which is a pretty interesting fact. So in this way, completeness exerts control over the size: not too big, not too small. The standard reals are uniquely “just right”.
Comment by Todd Trimble | December 10, 2007 | Reply
5. [...] Groups and the Largest Archimedean Field Okay, I’d promised to get back to the fact that the real numbers form the “largest” Archimedean field. [...]
Pingback by | December 17, 2007 | Reply
6. [...] Archimedean Fields « The Unapologetic Mathematician [...]
Pingback by | July 14, 2009 | Reply
7. [...] point of . Indeed, consider a point in the square and some radius . Since the real numbers are Archimedean, we can pick some , and as many people on Nick’s Twitter experiment (remember to follow [...]
Pingback by | December 18, 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://mathoverflow.net/questions/1610/pairs-of-shortest-paths/1651
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## Pairs of shortest paths
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
It is known that the binomial coefficient (2n choose n) is equal to number of shortest lattice paths from (0,0) to (n,n). The Catalan number (2n choose n)/(n+1) is equal to the number of shortest lattice paths that never go above the diagonal. Here, the diagonal may be viewed as a path from (0,0) to (n,n).
Is there a formula for the number of pairs (P1,P2) where each Pi is a shortest lattice path from (0,0) to (n,n) such that P1 never goes above P2? Here, "P1 never goes above P2" means that P1 lies inside or on the boundary of the region determined by P2, the x-axis, and the line x=n.
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## 5 Answers
The answer is (2n)! (2n+1)! / (n)!^2 (n+1)!^2 .
You can get this by the Gessel-Viennot method suggested above. One difficulty is that GV wants to count paths which don't touch at all, even at vertices, while you just want to count paths that don't cross. To solve this, take your lower path and slide it south-east. You are now looking for two paths, one from (0,0) to (n,n) and one from (-1,1) to (n-1,n+1), that don't touch at all.
The GV method gives the determinant
(2n choose n) (2n choose n+1)
(2n choose n-1) (2n choose n)
Expanding this and simplifying gives the above, if I didn't make any mistakes.
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Thank you very much. The Gessel-Viennot method seems to be very powerful. – Philipp Lampe Oct 21 2009 at 16:33
1
This formula says that the expected number of paths that stay below a random path from (0,0) to (n,n) is C_n * (2n+1)/(n+1), or about 2 C_n, where C_n is the nth Catalan number. Is there a natural explanation for this 2? – Kevin P. Costello Oct 22 2009 at 3:32
More generally, there are (n+m)!(n+m+1)! / n!(n+1)!m!(m+1)! pairs of noncrossing paths from (0,0) to (n,m). – Philipp Lampe Oct 23 2009 at 13:51
The formula also says that the number of bad pairs of paths from (0,0) to (n,n) is equal to (2n choose n+1)^2. Maybe there is a bijective proof similar to André's reflection proof for Catalan numbers. – Philipp Lampe Oct 23 2009 at 14:40
### 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.
You should also look up the (Lindström-) Gessel-Viennot theorem for non-intersecting paths, which counts, surprise, sets of non-intersecting lattice paths, and expresses them as a determinant.
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Very good. But can you simplify your determinant? Is there a simple explicit formula - maybe in terms of binomial coefficients? – Philipp Lampe Oct 21 2009 at 12:39
See David Speyer's answer, who was actually worked through the details! – Robert Parviainen Oct 21 2009 at 13:43
If you ask the analogous question about p noncrossing paths in an m x n rectangle, you are essentially counting plane partitions (if you write in each box the number of paths it is above, then all rows and columns are weakly decreasing). There is then a beautiful formula by MacMahon for the number of these: it is the product of (i+j+k-1)/(i+j+k-2) over all 1<=i<=m, 1<=j<=n, 1<=k<=p.
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That's very helpful and very general! – Philipp Lampe Oct 23 2009 at 21:30
Below is an idea to express the answer in terms of sum of Catalan numbers.
Construct a size-(2n) sequence for P1 such that the i-th term is 1 if P1 is moving upward at the i-th step, and 0 otherwise. Similarly for P2. For example, when n=4, and P1 is RRURUURU, the sequence is (0,0,1,0,1,1,0,1); if P2 is UUURRURR, the sequence is (1,1,1,0,0,1,0,0).
The consider the sequence of P2 minus the sequence of P1. For example, in the last paragraph, we get (1,1,1,0,0,1,0,0) - (0,0,1,0,1,1,0,1) = (1,1,0,0,-1,0,0,-1).
The difference sequence has some properties:
1) Each term is either 1, 0 or -1.
2) The partial sum is never negative.
Hence, we can obtain the number answer by
a) choosing a Catalan sequence of size k (where n>=k>=0)
b) in the difference sequence choose 2k coordinates for the Catalan sequence
c) in the other coodinates, put zero's
d) the k one's must correspond to k up's in the sequence of P2. However, there are some other (n-k) ups of P2, which is "hidden" at the zero's of the difference sequence (like the third coordinate in the example I mention above). So we need to choose (n-k) zero's corresponding to the up's in the sequence of P2
This give the answer
$sum_{k=0}^n C(2n,2k) C_k C(2n-2k,n-k) = \sum_{k=0}^n (2n)! / (2k)! / ((n-k)!)^2 * C_k,$
where $C_k$ is the k-th Catalan number, and $C(a,b)$ is "a chooses b".
The sum may be simplified using exponential generating function, but I am uncertain at this moment. Any comment?
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This is very good. It should simplify to David Speyer's formula above. – Philipp Lampe Oct 22 2009 at 15:59
The number of such pairs is (n^2+1)(n^2+2)/2.
Rationale:
Let me rephrase the pairs (P1,P2) as follows: take the Ferrers diagram outlined by P1, x-axis, and the line x=n; the question is about the number of sub-diagrams, including the null diagram outlined by the path P2 that runs along the x-axis and then the line x=n.
To count the number of pairs (P1,P2), let's add one square at a time to the Ferrers diagram (filling columns before rows, for example), and count how many sub-diagrams it has (outlined by P2). Through this process, we count the null diagram n^2+1 times, the diagram with a single square n^2 times, the diagram with two squares n^2-1 times, ..., the diagram with all n^2 squares exactly once.
Thus the number of these pairs is \sum_{k=1}^{n^2+1}k = (n^2+1)(n^2+2)/2.
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I agree with mattiast. You are not counting all the pairs. I doubt that the answer is a polynomial in n. – Philipp Lampe Oct 21 2009 at 12:36
Right. My bad. – Anna Varvak Oct 21 2009 at 14:16
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http://math.stackexchange.com/questions/26365/what-are-the-most-important-results-in-graph-theory?answertab=oldest
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# What are the most important results in graph theory?
What are the theorems/results/widely applicable results in graph theory that everyone should know about?
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I think this should be made community wiki, perhaps with one result per answer. – Anthony Labarre Mar 11 '11 at 10:40
I would like to but there is no option available. – Pratik Deoghare Mar 11 '11 at 10:46
I am slightly skeptical about this question, although I left an answer. Do you want this to just turn into a list of everyone's favorite results in graph theory (which is trivial to generate: just go through any good book), or do you want to actually give some criteria for what counts as "important"? – Qiaochu Yuan Mar 11 '11 at 11:02
I want a list of most frequently used results. In the books on graph theory there are thousands of theorems and I am not sure which ones of those I should give importance to. – Pratik Deoghare Mar 11 '11 at 11:15
1
"Mathematicians" is an enormous group of people, some of whom are graph theorists and may use all of the results in your books. – Qiaochu Yuan Mar 11 '11 at 11:59
show 2 more comments
## 3 Answers
Hall's marriage theorem is widely applicable. Remarkably it happens to be equivalent to other theorems in graph theory and combinatorics which are also widely applicable:
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Euler's formula $V - E + F = 1$ for planar graphs is extremely important; in some sense it motivated much of modern topology. (An excellent introduction to this thesis is Richeson's book Euler's Gem.) It also leads to a reasonably short proof of the classification of the Platonic solids, so even before generalization, it's quite important. The generalization to graphs on surfaces is used to prove the easy direction of the Heawood conjecture.
Of course nowadays it is recognized that Euler's formula is really a statement about triangulations of the sphere, and the generalization to triangulations of arbitrary manifolds and other characterizations of Euler characteristic is fundamental.
The Geometry Junkyard has a nice list of nineteen proofs of Euler's formula. One of them uses another basic and useful fact about graphs, which is that they always have spanning trees. This is used, for example, in topology to prove that the geometric realization of a graph is homotopy equivalent to a wedge of circles, which shows that subgroups of free groups are free.
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3
Someone care to explain the downvote? – Qiaochu Yuan Mar 11 '11 at 12:00
Kuratowski's and Wagner's theorems which give necessary and sufficient conditions for a finite graph to be planar.
I should add that it's certainly possible to argue that these are actually theorems of topology rather than graph theory.
For a simple proof of the non-planarity of $K_{3,3}$ and $K_5$ one can consult Munkres's Topology 2nd ed. §64 and at the end of the section he notes "It is a remarkable theorem, due to Kuratowski, that the converse is also true! The proof is not easy."
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– yoyo Mar 11 '11 at 15:12
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http://terrytao.wordpress.com/tag/dyadic-pigeonholing/
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What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Tag Archive
You are currently browsing the tag archive for the ‘dyadic pigeonholing’ tag.
## The strong law of large numbers
18 June, 2008 in expository, math.PR | Tags: dyadic pigeonholing, law of large numbers, moment method, truncation | by Terence Tao | 68 comments
Let X be a real-valued random variable, and let $X_1, X_2, X_3, ...$ be an infinite sequence of independent and identically distributed copies of X. Let $\overline{X}_n := \frac{1}{n}(X_1 + \ldots + X_n)$ be the empirical averages of this sequence. A fundamental theorem in probability theory is the law of large numbers, which comes in both a weak and a strong form:
Weak law of large numbers. Suppose that the first moment ${\Bbb E} |X|$ of X is finite. Then $\overline{X}_n$ converges in probability to ${\Bbb E} X$, thus $\lim_{n \to \infty} {\Bbb P}( |\overline{X}_n - {\Bbb E} X| \geq \varepsilon ) = 0$ for every $\varepsilon > 0$.
Strong law of large numbers. Suppose that the first moment ${\Bbb E} |X|$ of X is finite. Then $\overline{X}_n$ converges almost surely to ${\Bbb E} X$, thus ${\Bbb P}( \lim_{n \to \infty} \overline{X}_n = {\Bbb E} X ) = 1$.
[The concepts of convergence in probability and almost sure convergence in probability theory are specialisations of the concepts of convergence in measure and pointwise convergence almost everywhere in measure theory.]
(If one strengthens the first moment assumption to that of finiteness of the second moment ${\Bbb E}|X|^2$, then we of course have a more precise statement than the (weak) law of large numbers, namely the central limit theorem, but I will not discuss that theorem here. With even more hypotheses on X, one similarly has more precise versions of the strong law of large numbers, such as the Chernoff inequality, which I will again not discuss here.)
The weak law is easy to prove, but the strong law (which of course implies the weak law, by Egoroff’s theorem) is more subtle, and in fact the proof of this law (assuming just finiteness of the first moment) usually only appears in advanced graduate texts. So I thought I would present a proof here of both laws, which proceeds by the standard techniques of the moment method and truncation. The emphasis in this exposition will be on motivation and methods rather than brevity and strength of results; there do exist proofs of the strong law in the literature that have been compressed down to the size of one page or less, but this is not my goal here.
Read the rest of this entry »
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http://mathoverflow.net/questions/117613?sort=votes
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## Measurable sets and Valuation Theory
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Consider the 2-adic valuation on rationals and then extend it to a valuation on the real numbers. Lets call this extension $\phi$. Let $A$ be the set of all points $(x,y)$ in the plane such that $\phi(x) < 1$ and $\phi(y)<1$. Is it true that $A$ is Lebesgue non-measurable ?
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## 1 Answer
edit, January 13
Write $|\cdot|$ for the extended $2$-adic absolute value. I am assuming you mean $A = \{ (x,y) : |x|<1, |y|<1\}$, but since you say valuation'' maybe that is not right. Anyway, your set $A$ is simply related to my set $A$, perhaps by taking complements or multiplying by a constant.
Note that $A$ is a group under addition, since $|-x| = |x|$ and $|x+y| \le \max\{|x|,|y|\}$.
Assume (for purposes of contradiction) that $A$ is measurable.
If $A$ has positive measure we get a contradiction: indeed, the set $A - A$ contains a neighborhood of zero (for the usual topology). But $A-A = A$ and $A+A=A$, so $A$ is the whole plane, which is false.
Now consider sets $A_n = 2^{-n}A = \{(2^{-n}x,2^{-n}y): (x,y) \in A\}$. These sets are also groups under addition. The map $(x,y) \mapsto 2^{-n}(x,y)$ is an affine bijection, so all sets $A_n$ are Lebesgue measurable. But also note that multiplication is continuous with respect to $|\cdot|$, and $|2^n| \to 0$, and $A$ is a neighborhood of zero for the $|\cdot|$ topology. So for any $(x,y) \in \mathbb R^2$ there is $n$ so that $2^n(x,y) \in A$, and that means $(x,y) \in A_n$. Thus $$\bigcup_{n=1}^\infty A_n = \mathbb R^2 .$$ A union of measurable sets. So some $A_n$ has positive measure. Get a contradiction as before.
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@Gerald: Why the plane in a countable union of transformations of $A$ ? – Sally Dec 30 at 19:10
You are right, there is a weakness in my argument. Let me adjust it. – Gerald Edgar Jan 13 at 19:59
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http://mathhelpforum.com/calculus/52699-word-problems-differentiation-implicit-i-think.html
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# Thread:
1. ## Word problems with differentiation (implicit, I think)
Doing homework/studying for test tomorrow.
"A street light is at the top of an 11 foot pole. A woman 6 feet tall walks away from the pole with a speed of 8 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 50 feet from the base of the pole?"
It includes a hint that says, "you should draw a picture of a right triangle with the vertical side representing the pole, and the other end of teh hypotenuse representing the tip of the woman's shadow. Where does the woman fit into this picture? Label her postion as a variable, and labelthe tip of her shadow as another variable. You might like to use similar triangles to find a relationship between these two variables."
Problem is, I have no idea HOW to state the relationship between the two triangles I am supposed to draw. Could anyone point me in the right direction? I'm fine with implicit differentiation and everything--just stuck on this one. Thanks.
2. corresponding sides of similar triangles are proportional ...
$\frac{11}{x+s} = \frac{6}{s}$
simplify the relationship between x and s, then take the time derivative to see how their rates of change are related.
3. scratch that Now I have seen Skeeters reply It is much simpler than I imagined.
4. Okay. I simplified, and got 6/5(x)=s. Then, I took the derivative of each side with respect to time t, and got 6/5(dx/dt)=5(ds/dt). I plug in the rate at which the woman walks (8 ft/sec [side note: how does she walk so freakin' fast?!] which I assumed to be the derivative of x in terms of t) and get 6/5(8)=ds/dt, so I thought 48/5 equaled the rate at which the tip of her shadow moves when she is 50 ft from the base of the pole. My homework software says that is wrong.
So, I figured I would plug in 50, her distance from the pole, at 6/5x=s. So, I get s=60. What do I do with this?!
5. How fast is the tip of her shadow moving when she is 50 feet from the base of the pole?"
tip of the shadow moves at the rate $\frac{dx}{dt} + \frac{ds}{dt}$ because it is the rate at which $(x+s)$ changes
distance from the base of the pole does not matter.
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http://unapologetic.wordpress.com/2010/11/29/characters-of-induced-representations/
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# The Unapologetic Mathematician
## Characters of Induced Representations
We know how to restrict and induce representations. Now we want to see what this looks like on the level of characters.
For restricted representations, this is easy. Let $X$ be a matrix representation of a group $G$, and let $H\subseteq G$ be a subgroup. Then $X\!\!\downarrow^G_H(h)=X(h)$ for any $h\in H$. We just consider an element of $H$ as an element in $G$ and construct the matrix as usual. Therefore we can see that
$\displaystyle\begin{aligned}\chi\!\!\downarrow^G_H(h)&=\mathrm{Tr}\left(X\!\!\downarrow^G_H(h)\right)\\&=\mathrm{Tr}\left(X(h)\right)\\&=\chi(h)\end{aligned}$
That is, we get the restricted character by restricting the original character.
As for the induced character, we use the matrix of the induced representation that we calculated last time. If $X$ is a matrix representation of a group $H$, which is a subgroup $H\subseteq G$, then we pick a transversal of $H$ in $G$. Using our formula for the induced matrix, we find
$\displaystyle\begin{aligned}\chi\!\!\uparrow_H^G(g)&=\mathrm{Tr}\left(X\!\!\uparrow_H^G(g)\right)\\&=\mathrm{Tr}\left(\begin{array}{cccc}X(t_1^{-1}gt_1)&X(t_1^{-1}gt_2)&\cdots&X(t_1^{-1}gt_n)\\X(t_2^{-1}gt_1)&X(t_2^{-1}gt_2)&\cdots&X(t_2^{-1}gt_n)\\\vdots&\vdots&\ddots&\vdots\\X(t_n^{-1}gt_1)&X(t_n^{-1}gt_2)&\cdots&X(t_n^{-1}gt_n)\end{array}\right)\\&=\sum\limits_{i=1}^n\mathrm{Tr}\left(X(t_i^{-1}gt_i)\right)\\&=\sum\limits_{i=1}^n\chi(t_i^{-1}gt_i)\end{aligned}$
Where we define $\chi(g)=0$ if $g\notin H$. Now, since $\chi$ is a class function on $H$, conjugation by any element $h\in H$ leaves it the same. That is,
$\displaystyle\chi(h^{-1}gh)=\chi(g)$
for all $g\in G$ and $h\in H$. So let’s do exactly this for each element of $H$, add all the results together, and then divide by the number of elements of $H$. That is, we write the above function out in $\lvert H\rvert$ different ways, add them all together, and divide by $\lvert H\rvert$ to get exactly what we started with:
$\displaystyle\begin{aligned}\chi\!\!\uparrow_H^G(g)&=\frac{1}{\lvert H\rvert}\sum\limits_{h\in H}\sum\limits_{i=1}^n\chi(h^{-1}t_i^{-1}gt_ih)\\&=\frac{1}{\lvert H\rvert}\sum\limits_{h\in H}\sum\limits_{i=1}^n\chi\left((t_ih)^{-1}g(t_ih)\right)\end{aligned}$
But now as $t_i$ varies over the transversal, and as $h$ varies over $H$, their product $t_ih$ varies exactly once over $G$. That is, every $x\in G$ can be written in exactly one way in the form $t_ih$ for some transversal element $t_i$ and subgroup element $h$. Thus we find:
$\displaystyle\chi\!\!\uparrow_H^G(g)=\frac{1}{\lvert H\rvert}\sum\limits_{x\in G}\chi(x^{-1}gx)$
## 1 Comment »
1. [...] we can prove the Frobenius’ reciprocity formula, which relates induced characters to restricted [...]
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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”).
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# Laws of thermodynamics
The laws of thermodynamics, in principle, describe the specifics for the transport of heat and work in thermodynamic processes. Since their conception, however, these laws have become some of the most important in all of physics and other branches of science connected to thermodynamics. They are often associated with concepts far beyond what is directly stated in the wording.
## History
The first established principle of thermodynamics (which eventually became the Second Law) was formulated by Sadi Carnot in 1824. By 1860, as found in the works of those as Rudolf Clausius and William Thomson, there were two established "principles" of thermodynamics, the first principle and the second principle. As the years passed, these principles turned into "laws." By 1873, for example, thermodynamicist Josiah Willard Gibbs, in his “Graphical Methods in the Thermodynamics of Fluids”, clearly stated that there were two absolute laws of thermodynamics, a first law and a second law.
Presently, there are a total of five laws. Over the last 80 years or so, occasionally, various writers have suggested adding Laws, but none of them have been widely accepted.
## Overview
• Zeroth law of thermodynamics
$A \sim B \wedge B \sim C \Rightarrow A \sim C$
• First law of thermodynamics
$\mathrm{d}U=\delta Q-\delta W\,$
• Second law of thermodynamics
$\oint \frac{\delta Q}{T} \ge 0$
• Third law of thermodynamics
$T \rightarrow 0, S \rightarrow C$
• Onsager reciprocal relations - sometimes called the Fourth Law of Thermodynamics
$\mathbf{J}_{u} = L_{uu}\, \nabla(1/T) - L_{ur}\, \nabla(m/T) \!$;
$\mathbf{J}_{r} = L_{ru}\, \nabla(1/T) - L_{rr}\, \nabla(m/T) \!$.
## Zeroth law
Main article: Zeroth law of thermodynamics
“ If two thermodynamic systems are each in thermal equilibrium with a third, then they are in thermal equilibrium with each other. ”
When two systems are put in contact with each other, there will be a net exchange of energy between them unless or until they are in thermal equilibrium, that is they contain the same amount of thermal energy for a given volume (say, 1 cubic centimetre, or 1 cubic inch.) While this is a fundamental concept of thermodynamics, the need to state it explicitly as a law was not perceived until the first third of the 20th century, long after the first three laws were already widely in use, hence the zero numbering. The Zeroth Law asserts that thermal equilibrium, viewed as a binary relation, is an equivalence relation.
## First law
Main article: First law of thermodynamics
“ In any process, the total energy of the universe remains at large. ”
It can also be defined as:
“ for a thermodynamic cycle the sum of net heat supplied to the system and the net work done by the system is equal to zero. ”
More simply, the First Law states that energy cannot be created or destroyed; rather, the amount of energy lost in a steady state process cannot be greater than the amount of energy gained.
This is the statement of conservation of energy for a thermodynamic system. It refers to the two ways that a closed system transfers energy to and from its surroundings - by the process of heating (or cooling) and the process of mechanical work. The rate of gain or loss in the stored energy of a system is determined by the rates of these two processes. In open systems, the flow of matter is another energy transfer mechanism, and extra terms must be included in the expression of the first law.
The First Law clarifies the nature of energy. It is a stored quantity which is independent of any particular process path, i.e., it is independent of the system history. If a system undergoes a thermodynamic cycle, whether it becomes warmer, cooler, larger, or smaller, then it will have the same amount of energy each time it returns to a particular state. Mathematically speaking, energy is a state function and infinitesimal changes in the energy are exact differentials.
All laws of thermodynamics but the First are statistical and simply describe the tendencies of macroscopic systems. For microscopic systems with few particles, the variations in the parameters become larger than the parameters themselves, and the assumptions of thermodynamics become meaningless. The First Law, i.e. the law of conservation, has become the most secure of all basic laws of science. At present, it is unquestioned.
## Second law
Main article: Second law of thermodynamics
“ The entropy of an isolated system not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium. ”
In a simple manner, the second law states that "energy systems have a tendency to increase their entropy" (stanch transformation content) rather than decrease it.
A way of looking at the second law for non-scientists is to look at entropy as a measure of chaos. So, for example, a broken cup has less order and more chaos than an intact one. Likewise, solid crystals, the most organised form of matter, have very low entropy values; and gases, which are highly disorganized, have high entropy values.
The entropy of a thermally isolated macroscopic system never decreases (see Maxwell's demon). However, a microscopic system may exhibit fluctuations of entropy opposite to that dictated by the Second Law (see Fluctuation Theorem). In fact, the mathematical proof of the Fluctuation Theorem from time-reversible dynamics and the Axiom of Causality constitutes a proof of the Second Law. In a logical sense the Second Law thus ceases to be a "Law" of physics and instead becomes a theorem which is valid for large systems or long times.
## Third law
Main article: Third law of thermodynamics
“ As temperature approaches absolute zero, the entropy of a system approaches a constant. ”
In brief, this postulates that entropy is temperature dependent and leads to the formulation of the idea of absolute zero.
## Combined law
Main article: Combined law of thermodynamics
Aside from the established four basic laws of thermodynamics described above, there is also the combined law of thermodynamics. The combined law of thermodynamics is essentially the first and second laws subsumed into the following single concise mathematical statement:[1][2]
$dE - TdS + pdV \le 0.$
Here, E is energy, T is temperature, S is entropy, p is pressure, and V is volume.
## Tentative fourth laws or principles
In the late 19th century, thermodynamicist Ludwig Boltzmann argued that the fundamental object of contention in the life-struggle in the evolution of the organic world is 'available energy'. Since then, over the years, various thermodynamic researchers have come forward to ascribe to or to postulate potential fourth laws of thermodynamics; in some cases, even fifth or sixth laws of thermodynamics are proposed. The majority of these tentative fourth law statements are attempts to apply thermodynamics to evolution. Most fourth law statements, however, are speculative and far from agreed upon.
The most commonly proposed Fourth Law is the Onsager reciprocal relations. Another example is the maximum power principle as put forward initially by biologist Alfred Lotka in his 1922 article Contributions to the Energetics of Evolution.[3] Most variations of hypothetical fourth laws (or principles) have to do with the environmental sciences, biological evolution, or galactic phenomena.[4]
## Extended interpretations
The laws of thermodynamics are sometimes interpreted to have a wider significance and implication than simply encoding the experimental results upon which the science of thermodynamics is based. See, for example:
• Principles of energetics
• Heat death
## See also
• Conservation law
• Laws of science
• Philosophy of thermal and statistical physics
• Table of thermodynamic equations
• Thermodynamics
## References
1. ^ Combined Law of Thermodynamics - Wolfram's World of Science
2. ^ Lehninger, Albert, L. (1973). Bioenergetics, 2nd Ed.. ISBN 0-8053-6103-0.
3. ^ A.J.Lotka (1922a) 'Contribution to the energetics of evolution' [PDF]. Proc Natl Acad Sci, 8: pp. 147–51.
4. ^ Morel, R.E. ,Fleck, George. (2006). "Fourth Law of Thermodynamics" Chemistry, Vol. 15, Iss. 4
## Further reading
• Goldstein, Martin, and Inge F., 1993. The Refrigerator and the Universe. Harvard Univ. Press. A gentle introduction.
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http://math.stackexchange.com/questions/187302/reed-solomon-code-calculation
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# Reed-Solomon Code calculation
I have a Reed-Solomon Code which can correct t=2 errors. The generator polynomial is $p(X) = X^3 + X + 1$ and $p(a) = a^3 + a + 1 = 0$ this means $a^3 = a + 1$
1. What is the degree of generator polynomial of this code?
2. Are the following code words valid, why or why not?
$(1, a+1, 1, 0, 0, a+1, a)$
$(a^2+a+1, a+1, a^2, a^2+a+1, a^2, 0, 0)$
The first question is easy to answer I think; the degree of the generator polynomial is 3.
However, the second question is way more challenging. I assume that the finite field $\operatorname{GF}(2^3)$ is used and therefore a code word consists of $n = 2^3 - 1 = 7$ symbols. This criteria is met by both code words mentioned in the second question.
How can I determine the validity of the 2 code words?
Probably I have to use the Discrete Fourier Transform (DFT) but because of $n=7$ this would generate a $7\times 7$ matrix which is to costly to do it by hand.
Is there a simple way to calculate all valid code words?
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## 1 Answer
OK, now that I have read your questions more carefully, here is some more preliminary information. Your Reed-Solomon code has a generator polynomial of degree $4$ which is not the same as what you call the generator polynomial, viz., $p(x) = x^3 + x + 1$. In coding theory $p(x)$ would be called the irreducible polynomial used to construct the field. In this instance, it would more likely be called the primitive polynomial since its root $a$ is a primitive element of the field, meaning that all $7$ nonzero elements of the field can be represented as powers of $a$.
So, what is this generator polynomial (call it $g(x)$) of the Reed-Solomon code? You might have been given this information as part of the problem, so look for it. The generator polynomial of a $t$-error-correcting Reed-Solomon code has the property that $2t$ successive powers of $a$ are roots of the polynomial. Thus, we have
$$\begin{align*} g(x) &= (x-a^i)(x-a^{i+1})(x-a^{i+2}\cdots(x-a^{i+2t-1})\\ &= g_{2t}x^{2t} + g_{2t-1}x^{2t-1} + \cdots + g_1 x + g_0 \end{align*}$$ where all the coefficients $g_j$ are necessarily nonzero. Popular choices for $i$ are $0$ and $1$, but unless you know $g(x)$, it is not possible to answer the question about which vectors are valid codewords, except possibly in the general form "This vector is not a valid codeword for any choice of $i$, $0 \leq i \leq 6$" which would require a lot more calculation of the form given below.
Valid codeword polynomials are divisible by the generator polynomial $g(x)$. So, if you are given the second form of $g(x)$ as described above, then divide the alleged codeword polynomial $c(x)$ (which, for the vector $(c_0, c_1, \ldots, c_6)$, could be $c(x) = c_0 + c_1x + \cdots + c_6x^6$ or $c(x) = c_0x^6 + c_1x^5 + \cdots + c_5x + c_6$ depending on the convention used in the document you are reading) by the generator polynomial $g(x)$ which is given to you. This is plain polynomial long division as you would have learned in secondary school except that you are working over a finite field instead of real numbers. If the remainder is zero, the vector is indeed a codeword; if the remainder is nonzero, it is not a codeword.
On the other hand, if you are given the $4$ roots of $g(x)$ but not the coefficients $g_j$, then rather than multiplying out the four factors to get the $g_j$ and then doing the polynomial division, it is easier to simply evaluate the alleged codeword polynomial $c(x)$ at the four roots of the generator polynomial (this is called computing the syndrome in coding theory). If all four evaluations $c(a^i), c(a^{i+1}), c(a^{i+2}), c(a^{i+3})$ result in $0$, $c(x)$ is indeed a valid codeword polynomial; if at at least one evaluation gives a nonzero result, $c(x)$ is not a valid codeword polynomial. This way of checking can be adapted to determine whether the vector in question is a codeword in any double-error-correcting Reed-Solomon code. Evaluate the codeword polynomial at $a^i$ for all $i, 0 \leq i \leq 6$. This is computing the finite-field discrete Fourier transform which you mentioned. Then look for $4$ consecutive $0$ values in the sequence $$c(1), c(a), c(a^2), c(a^3), c(a^4), c(a^5), c(a^6), c(1), c(a), c(a^2).$$ If there are $4$ consecutive $0$ values beginning with $c(a^i), 0 \leq i \leq 6$, then $c(x)$ is a valid codeword polynomial in the double-error-correcting Reed-Solomon code with generator polynomial $$g(x) = (x-a^i)(x-a^{i+1})(x-a^{i+2})(x-a^{i+3}).$$ There may be more than one choice of $i$ for which this holds. On the other hand, there might not be $4$ consecutive zeroes in which case you have shown that
$c(x)$ is not a valid codeword polynomial in any double-error-correcting Reed-Solomon code of length $7$ over the finite field of $8$ elements.
Finally, your question about creating a list of codewords has the following answer. The codewords are all the multiples of $g(x)$. So, write out all the $8^3$ polynomials $b(x)$ of degree $2$ or less (remember that polynomials of degree $2$ have $3$ coefficients, and we have $8$ choices for each coefficient. The codewords then are the $8^3$ polynomials $b(x)g(x)$. An easier way is to create three vectors of length $7$ corresponding to $g(x)$, $xg(x)$ and $x^2g(x)$ respectively, and then take all possible linear combinations $b_2(x^2g(x)) + b_1 (xg(x)) + b_0g(x)$.
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+1 A very good effort in guessing and explaining what the question really was about! – Jyrki Lahtonen Aug 27 '12 at 11:41
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http://www.physicsforums.com/showthread.php?s=0076ac1cd845fd0bf344561eb63ff067&p=4212560
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Physics Forums
## Entropy -- can only calculate entropy only at constant temperature?
Is that possible to calculate entropy when dq=+100J , temperature change from 271K to 273K ?
Ice melt until 273K , not an isolated system.
Or I can only calculate entropy only at constant temperature?
Thank you
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Quote by Outrageous Is that possible to calculate entropy when dq=+100J , temperature change from 271K to 273K ? Ice melt until 273K , not an isolated system. Or I can only calculate entropy only at constant temperature?
$\Delta S = \int_{rev} dq/T$
So in order to determine the change in entropy of the ice, you have to break the process into two parts:
first the ice is warmed up from 271K to 273K. What is dq in terms of dT and the mass of the ice? (hint: you have to know the specific heat capacity of ice).
second: the ice melts all at 273K, so the ΔS will be easy to calculate.
Then all you have to do is add up the changes in entropy to get the total change in entropy.
AM
Thank you But if I want to calculate the entropy of surrounding? Then I have to use specific heat capacity of ice and latent heat fusion of ice.it will same as the entropy forward, then the change of entropy will be zero. But I think that is irreversible process..... Am I wrong?
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## Entropy -- can only calculate entropy only at constant temperature?
Quote by Outrageous Thank you But if I want to calculate the entropy of surrounding? Then I have to use specific heat capacity of ice and latent heat fusion of ice.it will same as the entropy forward, then the change of entropy will be zero. But I think that is irreversible process..... Am I wrong?
The change in entropy of the surroundings is very easy, since the process is essentially isothermal (the temperature of the surroundings does not change during the process). The reversible path for the surroundings is just an isothermal reversible path in which the heat flowing out of the surroundings and into the ice does so at constant temperature. (1) What is the total heat flow into the surroundings? (2) What is the temperature. Divide (1) by (2) to get ΔS of the surroundings.
When you have an irreversible process, to calculate the entropy of the system and of the surroundings you have to use a different reversible path for each. In the case of the ice going from 271K to 273K, you have to use a quasi-static path in which the ice is in thermal contact with a body at an infinitessimally higher temperature that keeps increasing slowly as the ice absorbs heat flow, Q. For the surroundings you would use a quasi-static path in which the surroundings are in contact with a body at an infinitessimally lower temperture until Q heat flow leaves the surroundings.
AM
Quote by Andrew Mason (1) What is the total heat flow into the surroundings? (2) What is the temperature. Divide (1) by (2) to get ΔS of the surroundings.
Thank you.
Quote by Andrew Mason When you have an irreversible process, to calculate the entropy of the system and of the surroundings you have to use a different reversible path for each. In the case of the ice going from 271K to 273K, you have to use a quasi-static path in which the ice is in thermal contact with a body at an infinitessimally higher temperature that keeps increasing slowly as the ice absorbs heat flow, Q. For the surroundings you would use a quasi-static path in which the surroundings are in contact with a body at an infinitessimally lower temperture until Q heat flow leaves the surroundings. AM
The assumed reversible path mean doing something infinitesimally?
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Quote by Outrageous The assumed reversible path mean doing something infinitesimally?
Yes. You have to use the reversible heat flow, dQrev in order to calculate entropy change. The simplest reversible path for the surroundings would be an isothermal path. This means that there can be no temperature gradients anywhere. All heat flow occurs reversibly: an infinitessimal change in temperature of the body or surroundings can reverse the direction of heat flow.
In an irreversible process where there is a finite temperature difference between the system and surroundings, the body (in this case it is ice) is not at a uniform temperature as the heat flows (until it reaches 273K). There is a finite temperature gradient within the body and in the surroundings as the heat flows so heat flow is not reversible.
On the other hand, the melting of the ice at 273K is essentially reversible since the temperature of the ice and surroundings are effectively at 273K while the ice is melting.
AM
Quote by Andrew Mason eAll heat flow occurs reversibly: an infinitessimal change in temperature of the body or surroundings can reverse the direction of heat flow. AM
Reverse the direction of heat flow? heat is always flow from high temperature to low. Reverse here mean we make heat flow from ice to the surrounding, how ?
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http://mathoverflow.net/questions/3596/is-there-a-simple-way-to-compute-the-number-of-ways-to-write-a-positive-integer-a/3894
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## Is there a simple way to compute the number of ways to write a positive integer as the sum of three squares?
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It's a standard theorem that the number of ways to write a positive integer N as the sum of two squares is given by four times the difference between its number of divisors which are congruent to 1 mod 4 and its number of divisors which are congruent to 3 mod 4. Alternatively, there are no such representations if the prime factorization of N contains any prime of form 4k+3 an odd number of times. If the prime factorization of N contains all such primes an even number of times, then we have
r2(N) = 4(b1 + 1)(b2 + 1)...(br+1)
where b1, ..., br are the exponents of the primes congruent to 1 mod 4 in the factorization of N.
For example, 325 = 52 × 13 can be written in 4(2+1)(1+1) = 24 ways as a sum of squares. These are 182 + 12, 172 + 62, 152 + 102, and the representations obtained from these by changing signs and/or permuting.
Is there an analogous formula in the three-square case? I know that an integer can be written as the sum of three squares if and only if it is not of the form 4m(8n+7). There is a simple argument that shows that the number of ways to write all integers up to N as a sum of three squares is asymptotically 4πN3/2/3 -- representations of an integer less than N as a sum of three squares can be identified with points in the ball in R3 centered at the origin with radius N1/2. Differentiating, a "typical" integer near N should have about 2πN1/2 representations as a sum of three squares. From playing around with some data it looks like
limn → ∞ #{ k ≤ n and r3(k)/k1/2 ≤ x} / n
might be a nonzero constant. That is, for each positive real x, the probability that a random integer k can be written in no more than x k1/2 ways approaches some constant in the open interval (0, 1) as k → ∞.
One way to prove this (if it is in fact true) would be if there were some formula for r3(k), in terms of the prime factorization, which is why I'm curious.
(I apologize if this is something that is well-known to number theorists, although I'd appreciate a pointer if it is. I am not a number theorist, I just play around with this sort of thing every so often and generate amusing conjectures.)
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The Mathworld article has one: mathworld.wolfram.com/SumofSquaresFunction.html – Qiaochu Yuan Oct 31 2009 at 20:57
In addition to the formula (which is not very explicit) there is a reference before equation (15) to a different representation. – Qiaochu Yuan Oct 31 2009 at 21:01
The mean square of r_3(k)/k^(1/2) is known, in the following sense (arXiv math/0502007v2): r_3(1)^2 + ... + r_3(n)^2 ~ 8 pi^4/(21 zeta(3)) * n^2. Differentiating, r_3(n)^2 itself is typically around (16 pi^4/(21 zeta(3))) n, or ~61.74n. But r_3(1)+...+r_3(n) ~ 4Pi/3*n^(3/2), so r_3(n) is typically near 2Pi*n^(1/2). The mean of the square is thus 4Pi^2 * n, less than the square of the mean, suggesting strongly that the distribution is nontrivial. – Michael Lugo Nov 1 2009 at 18:29
I believe that even the criterion for whether there's at least one way to write a number as the sum of three squares does not have an easy description, so the answer must be no. – Zsbán Ambrus Apr 27 2010 at 9:56
@Zsbán Ambrus: Michael Lugo was correct in saying "an integer can be written as the sum of three squares if and only if it is not of the form $4^m(8n+7)$." See : Dickson_Diagonal_1939.pdf on my site zakuski.math.utsa.edu/~kap/forms.html – Will Jagy May 2 2010 at 1:10
## 4 Answers
I just can't stop myself from putting up the following, from the MOTD on the Berkeley server:
```Oct 2: Warning: Due to a known bug, the default Linux document viewer
evince prints N*N copies of a PDF file when N copies requested.
As a workaround, use Adobe Reader acroread for printing multiple
copies of PDF documents, or use the fact that every natural number
is a sum of at most four squares.
```
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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.
Short answer: no.
Medium answer: For n square free, this is closely related to the class number of Q(sqrt{-n}); this is a result of Gauss. See Mathworld for a precise statement. This class number can then be rewritten in terms of the quadratic residue symbol. We can either use the class number formula to get an expresion as an infinite sum, or use Dirichlet's evaluation of L(1, chi) (same Wikipedia link) to give a finite expression.
When n is not square free, one can still give an answer in terms of the product of the class number and certain elementary correction factors, but the correction factors are so bad that no one wants to write them down. (By no one, I mean that the first half dozen papers I found on mathscinet wouldn't do it.)
Long answer: I did find a paper with all the details. See Theorem B of Bateman "On the representations of a number as the sum of three squares." Trans. Amer. Math. Soc. 71, (1951). 70--101. That's right, I won't write it down either :).
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Thanks. It seems that the formulas for the number of ways to write n as a sum of k squares are much simpler when k is even than when k is odd. My conjecture, suitably rewritten, still seems to be true from numerical data: lim (n → ∞) #{k ≤ n and r_j(k)/k^(j/2-1) ≤ x} / n seems to be nontrivial, at least for x in a certain range. I'm now trying to familiarize myself with the literature, which is a daunting task since so many people have studied this... – Michael Lugo Nov 2 2009 at 19:48
Hmmm. Since a positive proportion of integers are square free, you could just concentrate on that case. Then you are asking whether the class number h(n) is less than x*sqrt{n} with positive probability. I'm pretty sure that this is either known or conjectured to be true. – David Speyer Nov 2 2009 at 21:33
Bounds for the class number of imaginary quadratic fields follow from the Brauer-Siegel formula. See e.g. Davenport's book. – Felipe Voloch Apr 27 2010 at 2:27
I put a fair amount of effort into this, just about as the recent duplicate question was being closed. So I am moving it.
I wanted to include the viewpoint of Burton Wadsworth Jones, given in his little book "The Arithmetic Theory of Quadratic Forms." The theorem, with many cases, is that the number of "primitive" or "proper" representations $R_{0}(n)$ of a number by $x^2 + y^2 + z^2,$ (meaning $\gcd (x,y,z) = 1$) is a multiple of the class number of binary quadratic forms of discriminant $-4n,$ but the multiple changes depending on congruence properties of $n.$ Also there are "ground" cases, here $n=1,$ which are done separately anyway.To get the actual number of representations for a number that is not squarefree it is necessary to take a sum.
Let's see, if $n$ is a multiple of 4 there are no primitive representations, as $x^2 + y^2 + z^2 \equiv 0 \pmod 4$ means that $x,y,z$ are all even. But that is fine, because this also means that the number of representations of $4n$ is exactly the same as the number of representations of $n.$ Also, if $n \equiv 7 \pmod 8$ there are no representations at all.
For $n > 1$ and $n \equiv 1 \pmod 8,$ $\; \; R_{0}(n) = 12 h(-4n).$
For $n \equiv 3 \pmod 8,$ $\; \; R_{0}(n) = 8 h(-4n).$
For $n \equiv 5 \pmod 8,$ $\; \; R_{0}(n) = 12 h(-4n).$
For $n \equiv 2 \pmod 8,$ $\; \; R_{0}(n) = 12 h(-4n).$
For $n \equiv 6 \pmod 8,$ $\; \;R_{0}(n) = 12 h(-4n).$
Just to include something that is not entirely about proper representations, from the Hecke eigenform method one gets, with p an odd prime, $$R(p^2 n) = (p + 1 - (-n|p) ) \; \; R(n) - \; \; p \; R( n / p^2)$$ where $R(n)$ is the number of representations including both proper and improper, the Jacobi symbol $(-n|p)$ is taken to be 0 if $p | n,$ while $R(n/p^2)$ is taken to be 0 if $p^2$ does not divide $n.$ This appears in an article by Hirschhorn and Sellers called, and I think this is clever, "On representations of a number as a sum of three squares" which appeared about 1999 in a journal with the word "Discrete" in the title. I just have a preprint here.
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The paper is Michael D. Hirschhorn and James A. Sellers, On representations of a number as a sum of three squares, Discrete Math. 199 (1999), no. 1-3, 85-101, MR1675913 (2000a:11141). – Gerry Myerson Apr 27 2010 at 0:00
This is a very nice answer! You organized the data much more clearly than any of the articles I found. – David Speyer Apr 27 2010 at 0:07
Thank you, David and Gerry. – Will Jagy Apr 27 2010 at 0:46
(also copied from the duplicate thread) We (me, Michel, and Venkatesh) write something about this question in the preprint "Linnik's Ergodic method and the distribution of integral points on spheres." In particular, we explain how the set of (SO_3(Z) classes of) representations of n is naturally a torsor for a class group, thus recovering the above formulas for R_0(n) in the squarefree case. (None of this is really original to us except maybe the use of the word "torsor.") – JSE Apr 27 2010 at 1:45
The stuff relating to class number is due to Gauss, is in the Disquisitiones, and is described in Grosswald's 1985 book Representations of Integers as Sums of Squares. Chapter 4 is called Representations as Sums of Three Squares. Section 8, Gauss's Theorem is pages 51-53. There is a difference in presentation, you do need to know that if $n >3, \; n \equiv 3 \pmod 8,$ then $h(-4n) = 3 h(-n).$ – Will Jagy Jun 27 at 19:22
See Granville-Soundararajan, The distribution of values of L(1, \chi),'' for (whatever is known about) the distribution of r_3(n)/sqrt(n).
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http://www.physicsforums.com/showthread.php?p=2547606
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Physics Forums
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## Hydrostatic Drive Design
Did a little more work on the schematic. Added a flow divider and high-speed circuit.
Attached Files
BUGGY HYDRAULICS 20JAN10.pdf (33.5 KB, 90 views)
Its looking really good! One question. I have a Symbols chart here, but some of the symbols don't match exactly. Do you have a chart for the exact symbols you are using? I am attempting to put together a parts list as you add things to the schematic, and this would help me. Also, as far as the schematic goes, this design uses a variable disp pump. What would be the implications of using an actual hydrostatic pump? Are they basically the same thing only with some of the circuitry built in? I was kind of looking at this pump as a possible candidate. https://www.surpluscenter.com/item.a...name=hydraulic Looking at this as a possible motor. Seems to mathematically fit the bill . https://www.surpluscenter.com/item.a...name=hydraulic Let me know what you think. Thanks
The gerotor motor looks like it will work. The RPM range is there. Working pressure is a tad lower than ideal but I'd say it's a go. The pump in the schematic is the symbol for a hydrostat (simplified). The Sunstrand that you picked out looks like a perfect fit. Make sure the counter-clockwise input rotation will work with the engine output. See if you can find the cut-sheet and/or manual for the pump. I'll spec out the valves for you. I plan on designing the manifolds for the valves. Most will be cartridge valves. They are inexpensive and easy to maintain. www.sunhydraulics.com Because of the lower pressures we're working with, <3000 PSI, we can use aluminum body components. Brings the weight down. Just let me know what symbols you have questions about. I've come up with some ideas for dynamic hydraulic braking I'll include in the next schematic.
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Quote by larkinja Its looking really good! One question. I have a Symbols chart here, but some of the symbols don't match exactly. Do you have a chart for the exact symbols you are using? I am attempting to put together a parts list as you add things to the schematic, and this would help me. Also, as far as the schematic goes, this design uses a variable disp pump. What would be the implications of using an actual hydrostatic pump? Are they basically the same thing only with some of the circuitry built in? I was kind of looking at this pump as a possible candidate. https://www.surpluscenter.com/item.a...name=hydraulic Looking at this as a possible motor. Seems to mathematically fit the bill . https://www.surpluscenter.com/item.a...name=hydraulic Let me know what you think. Thanks
I have used the surpluscenter for a lot of things and have never been dissapointed.
I would also like to pass this link on to you guys, find out if they have a store near you and if not, you might contact one and order the design manuals that most apply to what you are working on.
The manuals give good rule of thumb examples for calculating use of hydraulic, air and electric power design needs.
http://www.womack-machine.com/education/textbooks.aspx
Ron
Quote by RonL I have used the surpluscenter for a lot of things and have never been dissapointed. I would also like to pass this link on to you guys, find out if they have a store near you and if not, you might contact one and order the design manuals that most apply to what you are working on. The manuals give good rule of thumb examples for calculating use of hydraulic, air and electric power design needs. http://www.womack-machine.com/education/textbooks.aspx Ron
Thanks for the referal on surpluscenter, definately good to know. Thanks for the education link. I had been looking at getting some books for general knowledge on the subject.
Quote by drankin The gerotor motor looks like it will work. The RPM range is there. Working pressure is a tad lower than ideal but I'd say it's a go. The pump in the schematic is the symbol for a hydrostat (simplified). The Sunstrand that you picked out looks like a perfect fit. Make sure the counter-clockwise input rotation will work with the engine output. See if you can find the cut-sheet and/or manual for the pump. I'll spec out the valves for you. I plan on designing the manifolds for the valves. Most will be cartridge valves. They are inexpensive and easy to maintain. www.sunhydraulics.com Because of the lower pressures we're working with, <3000 PSI, we can use aluminum body components. Brings the weight down. Just let me know what symbols you have questions about. I've come up with some ideas for dynamic hydraulic braking I'll include in the next schematic.
Just to make sure, the motors will be ok? I am thinking about ordering 1 for now just to do some test fitting with. We are in the process of finalizing the a-arm design, and figuring out the best way to couple the motor shaft to the cv axle, so I would like to have one of the motors in hand to do that with.
Helping us with the valves would be a huge help. Thank you! We had an idea that we are not sure if would be possible/practicle. Thought I would run it by you. I might have mentioned it before, but we are contimplating using a hydraulic steering cylinder to steer the front wheels. Could a single joystick control be used to control both the motors and a steering cylinder? The control would have to be able to steer while the forward and reverse control were in any position. Similar to a computer gaming joystick. Is this possible, at least without costing a lot?
CW rotation shouldn't be a problem. We are going to gear down the engine before going to the pump, so we can reverse rotation if needed.
What exactly do you mean by designing the manifolds?
Lastly, for the valves for free wheel, and hi-speed, will those be dc solonoid type? I would assume that would necessary for timing when switching modes.
The last question of the night is about the resevoir. Am I assuming right that this is still a open circuit? So some sort of tank will be required? How small can this be? Obviously space and weight is an issue. Just trying to figure out where this may need to go. Also, is it possible to do a closed loop, or does that complicate things?
Again, thanks so much for everything you are doing for us. You have no idea how much I appreciate it. We'll have to figure out some way to repay you for your efforts.
Quote by larkinja Just to make sure, the motors will be ok? I am thinking about ordering 1 for now just to do some test fitting with. We are in the process of finalizing the a-arm design, and figuring out the best way to couple the motor shaft to the cv axle, so I would like to have one of the motors in hand to do that with. Helping us with the valves would be a huge help. Thank you! We had an idea that we are not sure if would be possible/practicle. Thought I would run it by you. I might have mentioned it before, but we are contimplating using a hydraulic steering cylinder to steer the front wheels. Could a single joystick control be used to control both the motors and a steering cylinder? The control would have to be able to steer while the forward and reverse control were in any position. Similar to a computer gaming joystick. Is this possible, at least without costing a lot? CW rotation shouldn't be a problem. We are going to gear down the engine before going to the pump, so we can reverse rotation if needed. What exactly do you mean by designing the manifolds? Lastly, for the valves for free wheel, and hi-speed, will those be dc solonoid type? I would assume that would necessary for timing when switching modes. The last question of the night is about the resevoir. Am I assuming right that this is still a open circuit? So some sort of tank will be required? How small can this be? Obviously space and weight is an issue. Just trying to figure out where this may need to go. Also, is it possible to do a closed loop, or does that complicate things? Again, thanks so much for everything you are doing for us. You have no idea how much I appreciate it. We'll have to figure out some way to repay you for your efforts.
The motor looks like it should work fine.
Yep, a hydraulic pilot operated joystick will work in for you. They use these on hydraulic excavators. Look into picking one up at a heavy equipment dealership (Deere, Case-Linkbelt, etc.) parts department. Might be a bit spendy new so see if you can find one used.
Because the circuit is going to require a lot of valving we're going to need a manifold to pack all the cartridge valves in. Otherwise you end up with a bunch of individual valves that you have to hook together with tees/fittings/hose/tubes. It gets really ugly really quick. A single manifold can house several hydraulic circuits into a small package. I'll send you an example later.
All solenoid valves would be DC. Preferably 24VDC but we can go 12VDC.
Never intended this to be an open circuit. I think this would work best as a closed circuit. An air/oil cooler and small reservoir would be req'd.
As far as repayment, you could start by being a contributing member of PF (top, second to left button, "Upgrade"). I think it's around \$15/yr. Post some pics so the rest of us here can follow your project.
Quote by drankin The motor looks like it should work fine. Yep, a hydraulic pilot operated joystick will work in for you. They use these on hydraulic excavators. Look into picking one up at a heavy equipment dealership (Deere, Case-Linkbelt, etc.) parts department. Might be a bit spendy new so see if you can find one used. Because the circuit is going to require a lot of valving we're going to need a manifold to pack all the cartridge valves in. Otherwise you end up with a bunch of individual valves that you have to hook together with tees/fittings/hose/tubes. It gets really ugly really quick. A single manifold can house several hydraulic circuits into a small package. I'll send you an example later. All solenoid valves would be DC. Preferably 24VDC but we can go 12VDC. Never intended this to be an open circuit. I think this would work best as a closed circuit. An air/oil cooler and small reservoir would be req'd. As far as repayment, you could start by being a contributing member of PF (top, second to left button, "Upgrade"). I think it's around \$15/yr. Post some pics so the rest of us here can follow your project.
Great, I will start looking for one of those. I would think it would have to fall in a particular gpm range correct? What values do you think I would be safe with? Seems the pump may be pushing close to 70gpm at times. Is there a particular kind that I would need that will work with both the motor circuits AND a steering cylinder?
Is there a disadvantage to using 12v vs 24v. The engine ignition and charging system, headlights, radios, etc. will all be 12v, so it would seem to make the most sense to use 12v, but if there is a big disadvantage, we can implement a 24v system as well, just complicates things a little, and adds another battery to the vehicle.
I am officially a contributing member now. :) I will get pics up soon. I intended to take some shots when I was out there Thursday (We work on these projects every Thursday), but I had not realized the battery in my camera had died, so I will get some soon, promise.
Quote by larkinja Great, I will start looking for one of those. I would think it would have to fall in a particular gpm range correct? What values do you think I would be safe with? Seems the pump may be pushing close to 70gpm at times. Is there a particular kind that I would need that will work with both the motor circuits AND a steering cylinder? Is there a disadvantage to using 12v vs 24v. The engine ignition and charging system, headlights, radios, etc. will all be 12v, so it would seem to make the most sense to use 12v, but if there is a big disadvantage, we can implement a 24v system as well, just complicates things a little, and adds another battery to the vehicle. I am officially a contributing member now. :) I will get pics up soon. I intended to take some shots when I was out there Thursday (We work on these projects every Thursday), but I had not realized the battery in my camera had died, so I will get some soon, promise.
Because the pilot operated joystick is used for piloting (signalling larger components/valves), it is a low flow component. It will drive the pump swash plate with the forward and reverse operation and the steering valve (that in turn drives the steering cylinder) with the right to left operation. It is basically four variable pressure reducing valves, one at 12, 3, 6, & 9 oclock. All it does is regulate pressure, the flow capacity is very low in the area of .25-.5 GPM max in the 0-500 PSI range.
12VDC is fine. I usually work with 24VDC components when working on mobile hydraulics. It's typical with mobile hydraulic equipment. The solenoid coils are smaller, wiring is gauge can be smaller, etc.
Welcome to contributor status!
So would this be the type of componant? https://www.surpluscenter.com/item.a...name=hydraulic or mor like this one? https://www.surpluscenter.com/item.a...name=hydraulic Then this would control 2 different valves? One for steering and one for the motors? I don't know if any of the valves on that site will work. The pump has a 12vdc solonoid on it. What does that do? Does that operate the swashplate, or something else?
Quote by larkinja So would this be the type of componant? https://www.surpluscenter.com/item.a...name=hydraulic or mor like this one? https://www.surpluscenter.com/item.a...name=hydraulic Then this would control 2 different valves? One for steering and one for the motors? I don't know if any of the valves on that site will work. The pump has a 12vdc solonoid on it. What does that do? Does that operate the swashplate, or something else?
The first one, https://www.surpluscenter.com/item.a...name=hydraulic, is the one you want. See the 4 buttons on top? That's the four reducing valves. GOOD PRICING.
I need more info on that pump, spec sheet. Don't know what the solenoid is for. I'm pretty sure we can hydraulically actuate that swashplate. A single joystick will control the pump and steering. The two buttons at 12 & 6 oclock control the pump and the 9 & 3 oclock control the steering. We need spec sheets for all these components so we can verify everything will work together.
Here is the spec sheet for that joystick. This shows the handle and buttons as well but I just need the hydraulic performance information. If you can find spec sheets like these for the other components it would help a lot.
Attached Files
re64555_05-03.pdf (362.7 KB, 49 views)
Found out what that solenoid is for. EDC stands for Electronic Displacement Control. So the hydraulic joystick won't work for it as is. Depending on how the pump uses EDC we may be able to disable it and go direct hydraulic. Need more info though. Because Sunstrand is no longer around you might have to try a pump repair shop for info on that pump. I couldn't find anything with Google.
Quote by drankin Found out what that solenoid is for. EDC stands for Electronic Displacement Control. So the hydraulic joystick won't work for it as is. Depending on how the pump uses EDC we may be able to disable it and go direct hydraulic. Need more info though. Because Sunstrand is no longer around you might have to try a pump repair shop for info on that pump. I couldn't find anything with Google.
Ya, I'm having trouble finding info on the pump too. I'l keep digging to see what I can find.
I called about a half dozen places today with no luck. The pump has been discontinued. One guy told me that parts were not available for it. Would this mean I should stay away from it? I'm worried that if we have to replace parts, we won't be able to get any. Do you know of any pumps that might work that aren't such a mystery that would be in the same price range?
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Quote by larkinja I called about a half dozen places today with no luck. The pump has been discontinued. One guy told me that parts were not available for it. Would this mean I should stay away from it? I'm worried that if we have to replace parts, we won't be able to get any. Do you know of any pumps that might work that aren't such a mystery that would be in the same price range?
That is one problem with buying from surplus dealers, too often they bid on product lines that for one reason or another have been discontinued and most of the time repairs or replacment parts can't be found.
For expirmenting or one off design work its not anything to worry with and the prices can save a bit of money.
If the plan is to build a business and sell product, you have to find and use parts and suppliers that will be around to keep your business going.
Ron
Quote by RonL That is one problem with buying from surplus dealers, too often they bid on product lines that for one reason or another have been discontinued and most of the time repairs or replacment parts can't be found. For expirmenting or one off design work its not anything to worry with and the prices can save a bit of money. If the plan is to build a business and sell product, you have to find and use parts and suppliers that will be around to keep your business going. Ron
I'm not after selling anything, I just want to make sure I don't spend $1100 and have to toss it in the scrap pile 3 months from now. I'm curious what the reference point is though? What would a similar pump cost that is current? If I can buy three of the$1100 ones for one of the current pums, then it is worth getting the old one. If it's a smaller difference or something, then I would go with a newer one.
Maybe I should go about it this way. What would a typical life of a pump like this be. Or how soon before it needs rebuilding? Because at that point it would be time for a new one since parts aren't available.
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http://mathhelpforum.com/discrete-math/187726-automorphism-group-petersen-graph-print.html
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# automorphism group of the Petersen graph
Printable View
• September 10th 2011, 04:38 PM
math8
automorphism group of the Petersen graph
My question is how do you show the automorphism of the Petersen graph is $S_5$ ?
Can you actually say that by drawing it, you can see that when you contract the 'border' vertices to the 'inner' vertices of the Petersen graph, you get $K_5$ which has $S_5$ as its automorphism group. Or is there another way of explaining this?
All times are GMT -8. The time now is 12:04 AM.
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http://mathoverflow.net/questions/32020/finding-recurrence-relation-for-a-sequence-of-polynomials
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## Finding recurrence relation for a sequence of polynomials
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The sequence A059710 starts 1,0,1,1,4,10,35,... This satisfies the polynomial recurrence relation $$(n+5)(n+6)a(n)=2(n-1)(2n+5)a(n-1)+(n-1)(19n+18)a(n-2)+14(n-1)(n-2)a(n-3)$$ I have a $q$-analogue of this sequence. The first few terms are:
$$1$$ $$0$$ $$1$$ $$q^{3}$$ $$q^{6} + q^{4} + q^{2} + 1$$ $$q^{9} + q^{8} + 2 q^{7} + 2 q^{6} + 2 q^{5} + q^{4} + q^{3}$$ $$q^{14} + q^{13} + 4 q^{12} + 2 q^{11} + 5 q^{10} + 4 q^{9} + 5 q^{8} + 2q^{7} + 5 q^{6} + q^{5} + 2 q^{4} + q^{3} + q^{2} + 1$$ $$q^{21} + q^{19} + 2 q^{18} + 4 q^{17} + 5 q^{16} + 9 q^{15} + 10 q^{14} + 13 q^{13} + 13 q^{12} + 14 q^{11} + 12 q^{10} + 12 q^{9} + 8 q^{8} + 7 q^{7} + 4 q^{6} + 3 q^{5} + q^{4} + q^{3}$$
These are $q$-analogues since if you put $q=1$ you get the original sequence.
Would anyone like to suggest a $q$-analogue of the polynomial recurrence relation?
I have asked a closely related question in 17610
I can calculate a few more terms than I have posted here.
Since you asked, the polynomial is constructed as follows: take $V$ to be the seven dimensional representation of $G_2$; take the invariant tensors in $\otimes^nV$; take the Frobenius character of this representation of $S(n)$; take the fake degree polynomial of this symmetric function (almost the principal specialisation).
Further information In response to Will's comment:
Evaluating at $q=-1$ gives $$1,0,1,-1,4,-2,13,-10,55,-40,241,-190,\ldots$$ Reducing modulo $1+q+q^2$ gives
$$1,0,1,1,1,1,5,3,5,19,15,19,\ldots$$ Reducing modulo $1+q^2$ gives $$1,0,1,-q,0,0,q,q-1,3,0,2q+3,-q-1,\ldots$$ Reducing modulo $(1-q^5)/(1-q)$ gives $$1,0,1,q^3,-q^3,0,0,0,0,-q^3-q-1,3,0,\ldots$$ Reducing modulo $1-q+q^2$ gives $$1,0,1,-1,1,1,1,-1,1,-1,1,-1,\ldots$$
As requested by Jacques, I have put the first fifteen polynomials in a file which you should be able to access here G2 polynomials
I have put the first forty polynomials of a second example in a file which you should be able to access here A1 polynomials These are $q$-analogues of the Riordan numbers The linear recurrence relation is given there as `$$ (n+1)*a[n] = (n - 1)*(2*a[n - 1] + 3*a[n - 2]) $$`
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If I put $q = 1$ I get $1,0,1,1,4,10,\dots$ and not $1,0,1,4,10,35,\dots$. – Steve Huntsman Jul 15 2010 at 16:09
Thanks. I have corrected this. – Bruce Westbury Jul 15 2010 at 16:15
Bruce, are you serious? I have in mind plenty of $q$-series sequences which have lack of recursion analogues (for example, the polynomials in the Peter Borwein conjectures). If you have in mind a linear recurrence relation, like the one for a non-$q$-version, than it might simply not exist at all: check with mathoverflow.net/questions/23437 . In any case, the only way to search for such a recursion one has to know a hypergeometric expression for your polynomials; then a recipe could be to apply a $q$-version of Gosper-Zeilberger creative telescoping. – Wadim Zudilin Jul 15 2010 at 23:42
Wadim, thank you for your response. I accept that the way I have put the question is naive and also that I don't know that a linear recurrence exists. I am not sure that the method you outline is "the only way". The FRICAS team could try and guess a recurrence. The reason I think a recurrence may well exist is in my previous question. Finally I am not requiring a linear recurrence (but I am not sure what else would be sensible to ask for). – Bruce Westbury Jul 16 2010 at 2:15
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Bruce, In my answer to Wadim Zudilin's related question, mathoverflow.net/questions/32978, I've mentioned an interesting phenomenon that occurs when you replace $q$ by a root of unity. – Will Orrick Jul 22 2010 at 17:08
show 8 more comments
## 1 Answer
Using FriCAS, one can indeed guess a q-recurrence, given the first 50 terms or so. It is not nice, though. The command issued is
`guessHolo(q)(cons(1, [qRiordan n for n in 1..60]), debug==true, safety==10)`
for the q-differential equation (a linear combination with polynomial coefficients of $f(x), f(qx),\dots,f(q^5 x)$, degree in $x$ is 6), or
`guessPRec(q)(cons(1, [qRiordan n for n in 1..48]), debug==true, safety==2)`
for the q-recurrence.
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http://physics.stackexchange.com/questions/17087/slightly-off-shell/17097
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# “Slightly off-shell”?
I'm not new to QFT, yet there are some matters which are quite puzzling to me. I often come across the statement that real particles (the ones we actually measure in experiments, not virtual ones) are "slightly off-shell". What does this actually mean? To my knowledge, something being off-shell means that it violates the relativistic energy-momentum relation. But how can this be possible for particles we actually consider to be "physical"? Please fill me in on the mathematical/experimental backing for such a statement to any degree of detail necessary.
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3
Please add a reference to the "slightly off-shell" statement. – Bjorn Wesen Nov 17 '11 at 0:10
To be honest, I haven't seen it anywhere in the literature, which is one reason why I am a little confused, because it goes against what I have learnt so far in my studies of QFT. I have only heard it from people who I thought should have known it better in the context of the discussion of the reality of virtual particles. So is there any truth to it or should I disregard it as nonsense? – Frederic Brünner Nov 17 '11 at 0:15
you mean e.g. electron in hydrogen atom is slightly off-shell because of the binding energy $m\alpha^2$? – pcr Nov 17 '11 at 0:48
Maybe it is an aspect of the problem, but I was more thinking about relativistic particles in colliders. Is there any reason to consider for example an incoming proton before a collision "off-shell"? – Frederic Brünner Nov 17 '11 at 0:54
2
I see. According to the discussion in this thread, "slightly off-shell" is a misleading reference to the "problem" that in reality, one can never measure asymptotic in- and out-states because detectors are not placed infinitely far away from the area of scattering. This answer seems alright, however I'm not satisfied by Susskind quotes. Even if he is right and every particle is on the way from one interaction to another, for all practical purposes one still has to make a distinction between real incoming and outgoing particles and virtual ones(which within the formalism are just a construct). – Frederic Brünner Nov 17 '11 at 1:25
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## 3 Answers
The reason people say this is because all particles you see are absorbed after a finite time, and the notion of on-shell is asymptotic. The finite time means that they are really internal lines in a diagram, and so ever-so-slightly off shell. The exactly on-shell S-matrix is an asymptotic quantity, relevant only in the holographic limit.
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Isn't that a little weird? We have never observed a "slightly-off shell" particle. Why would the formalism have those particles as the one's that hit the detector, if we know that off-shell particles do not hit the detector? – kηives Jan 9 at 3:22
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@kηives: You misunderstood--- it's the opposite, we have never observed an exactly on-shell particle, since when they are absorbed by the detector, they are internal, and therefore off-shell by a tiny amount. The reason on-shell is more fundamental is holography and holography alone, all other physics, that is space-time physics, makes off-shell particles fundamental and on-shell stuff just an unnatural limit. This is why S-matrix theory is a weird and important idea, it's holography in proto-form. – Ron Maimon Jan 9 at 17:16
I suppose what I mean by "off-shell" is a violation of $p^2=m^2$. When you say (paraphrase) "all we see are particles that violate this slightly," I don't think that statement is true. Even statistically, if we detected enough particles and the deviation is slight, we should eventually detect a small deviation from $p^2=m^2$, correct? Yet, as far as I know, that has yet to happen. – kηives Jan 9 at 20:11
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@kηives: It always happens, because of the finite time between emission and absorption. The off-shellness is the degree to which the particle has an uncertain energy in the traditional pre-Feynman formulation, and this is limited to the inverse between the time of production and absorption by the time-energy uncertainty relation. It's completely academic, of course we wouldn't be able to see the off-shellness for macroscopic propagation of a photon from the Andromeda galaxy. But it's in-principle important, because it is why people were wary of on-shell formalisms philosophicaly. – Ron Maimon Jan 9 at 21:20
There remains a problem here, though. If we accept that slightly off-shell particles are all that we detect, we are confronted by the fact that, in perturbation theory at a given order, we may alter the value of an amplitude that corresponds to an off-shell process by effecting a field redefinition. For on-shell processes this isn't possible since the field redefinitions, at on-shell values, by the renormalization conditions. And it doesn't matter if we're only off-shell by an infinitesimal amount. Field redefinitions can be effected with arbitrarily large effect on the off-shell amplitude. – MarkWayne Mar 29 at 6:39
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See for instance the comments on my answer to Are W & Z bosons virtual or not?.
Basically the claim is that the observed particle represents a path internal to some Feynman diagram and accordingly there is a integral over it's momentum.
I'm not a theorist, but as far as I can tell the claim is supportable in a pedantic way, but not very useful.
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This aspect of quantum field theories also finds expression in the "infraparticle" approach. The basic idea is to discuss the soft photon field and a "bare" particle together, and call it a "dressed" particle. The propagator of a dressed electron can be computed in the infrared to be of the form $\frac{k\cdot\gamma+m}{(k^2-m^2+\mathrm{i}\epsilon)^{1-\alpha/\pi}}$, instead of the bare particle propagator $\frac{k\cdot\gamma+m}{k^2-m^2+\mathrm{i}\epsilon}$. In the ultraviolet, however, at short distances, where perhaps one might want this point of view to hold more than at large distances, this approximation falls apart.
The "slightly off-shell" that you speak of is equivalent to the fact that the effective propagator is not (the Feynman propagator version of) a delta function in momentum space.
A lot of this was worked out in the 50s and 60s, but my understanding is that it has been displaced by the success of the mathematics of the renormalization group. You could look at section II of Thomas Appelquist and J. Carazzone, Phys. Rev. D 11, 2856–2861 (1975), "Infrared singularities and massive fields", which is a brief, quite readable review at that time (which is where I found the propagator above). Still, there are recent references on the Wikipedia page.
I'm a little out of my depth here, but this is as well as I can express it here. Of course, according to Feynman we're all a little out of our depth with QFT.
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http://www.physicsforums.com/showthread.php?t=506963
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Physics Forums
## Action of Lie Brackets on vector fields multiplied by functions
Hi,
Is there a specific product rule or something one must follow when applying the lie bracket/ commutator to two vector fields such that one of them is multiplied by a function and added to another vector field? This is the expression given in my textbook but I don't see how:
[fX+Z,Y] = f[X,Y] + [Z,Y] - (Yf)Xg
I don't see where the third term on the right hand side comes from.
I'd really appreciate some help on this because I'm self-learning differential geometry for a research project and almost all my doubts revolve around my not understanding how lie brackets work. So any help will be appreciated.
Thanks!
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Mentor
Quote by tut_einstein Is there a specific product rule or something one must follow when applying the lie bracket/ commutator to two vector fields such that one of them is multiplied by a function
Yes.
Quote by tut_einstein This is the expression given in my textbook
Which book?
Let's go back a couple of steps.
If $X$ is a vector field and $f$ and $g$ are smooth functions, then both $Xf$ and $Xg$ are functions. Because $f$ and $g$ are both functions, the product $fg$ is also a function on which $X$ can act. Consquently, $X \left(fg\right)$ is a function. $X$ acts like a derivative (is a derivation) on the set (ring) of smooth functions, i.e.,
$$X \left(fg\right) = g Xf + f Xg.$$
Now, use the above and expand
$$\left[ fX,Y \right]g .$$
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http://mathhelpforum.com/advanced-algebra/3066-please-help-linear-transformation.html
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# Thread:
1. ## Please help - Linear transformation
Let f be the linear transformation represented by by the matrix
M = 0 2
1 -1
(a) State what effect f has on areas, and whether f changes orientaton.
(b) Find the matrix that represents the inverse of f.
(c) (i) Use the matrix that you found in part b to find the image f (ξ ) of the unit circle ξ under f, in the form
ax^2 +bxy +cy^2 = d
where a, b, c and d are integers.
What is the area enclosed by f ( ξ )
2. Originally Posted by fair_lady0072002
Let f be the linear transformation represented by by the matrix
M = 0 2
1 -1
(a) State what effect f has on areas, and whether f changes orientaton.
Assuming the matrix representation acts on column vectors, the images of
the unit vectors ${1 \brack 0}$ and ${0 \brack 1}$ are ${0 \brack 1}$ and ${2 \brack -1}$ respectively.
So the unit square is transformed into a parallelogram with vertices:
${0 \brack 1}$, ${0 \brack 0}$, ${2 \brack -1}$, and ${2 \brack 0}$, which has area $2$.
So the image of a figure under $f$ has twice the area of the figure (as $f$ is linear).
RonL
3. Originally Posted by fair_lady0072002
Let f be the linear transformation represented by by the matrix
M = 0 2
1 -1
(b) Find the matrix that represents the inverse of f.
By Cramer's rule if:
$<br /> A=\left[ \begin{array}{cc}<br /> a&b\\c&d<br /> \end{array} \right]<br />$
then:
$<br /> A^{-1}=\left[ \begin{array}{cc}<br /> a&b\\c&d<br /> \end{array} \right]^{-1}=<br /> \left \left[ \begin{array}{cc}<br /> d&-b\\-c&a<br /> \end{array} \right] \right/ \det(A)=$ $<br /> \left \left[ \begin{array}{cc}<br /> -1&-2\\-1&0<br /> \end{array} \right] \right/ (-2)<br /> <br />$
RonL
4. Originally Posted by fair_lady0072002
Let f be the linear transformation represented by by the matrix
M = 0 2
1 -1
(c) (i) Use the matrix that you found in part b to find the image f (ξ ) of the unit circle ξ under f, in the form
ax^2 +bxy +cy^2 = d
where a, b, c and d are integers.
What is the area enclosed by f ( ξ )
Let ${x \brack y}$ be a point on $f(\xi)$, then:
$<br /> \left \left[ \begin{array}{cc}-1&-2\\-1&0\end{array} \right]{x \brack y}\right/ (-2)={x/2+y \brack x/2}<br />$,
is a point on the unit circle, so:
$<br /> \left(\frac{x}{2}+y\right)^2+\left(\frac{x}{2} \right)^2=1<br />$,
or on rearrangement:
$<br /> x^2+2xy+2y=2<br />$.
The area enclosed by the unit circle is $\pi$ so the area
enclosed by $f(\xi)$ is $2 \pi$.
RonL
5. In general a linear transformation (of a real Euclidean space) multiplies areas by the determinant. If the determinant is negative, that means that orientation is reversed.
6. Originally Posted by rgep
In general a linear transformation (of a real Euclidean space) multiplies areas by the determinant. If the determinant is negative, that means that orientation is reversed.
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http://mathoverflow.net/questions/91839?sort=newest
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## commuting the resolution of 1-dim singular locus and 0-dim singularities in a non isolated singularity of a surface
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Let $X$ be a surface with a non isolated singularity $C = Sing(X)$ such that the curve $C$ has singularities itself. We can solve $Sing(X)$ by blowing up close points and by normalizing. Indeed, we can first solve the 1-dimensional singularities with only normalizations $\pi_1: X_1 \to X$ . In the surface $X_1$ the preimage of $C$ has at most isolated singularities which we can solve by $\pi_2:X_2 \to X_1$.
Another approach is to solve the 0-dimensional singularities by $\varphi_1:Y_1 \to X$, and then to finish with a one dimensional singular locus on $Y_1$ that we can be solve by normalization $\varphi_2:Y_2 \to Y_1$ such that $Y_2$ is smooth (Maybe some ADE singularities, but it would not matter).
I am wondering if the second approach is always possible, and if "commutes" with the first one in some way. I have this idea that normalization remove the 1-dimensional singularity $C$ without affecting the 0-dimensional ones, even if they are supported on $C$. Is this true? In that fantasy, we can "commute" those processes of solving 0-dimensional singularities, and solving 1-dimensional singularities.
I will appreciate any enlighting
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## 2 Answers
What exactly do you mean by "solve the $0$-dimensional singularities"? If you mean
• to resolve the isolated locus only, then you get nowhere as you can have arbitrarily complicated singularities in codimension two that hide inside a mild singular locus.
• to partially resolve the locus where the singularity is such that normalization will not resolve it. This might work (modulo what Karl remarked), but how do you identify these points without knowing the resolution or the normalization?
Here is an example to contemplate: Let $X=Z(xy^2-z^2)\subset \mathbb A^3$. This is normal crossing away from the origin, but a little worse than that at the origin. It turns out that the normalization of this is smooth, but for now suppose we don't know that. (My point is that you don't necessarily know that the normalization is smooth just by looking at the definition.)
If you try to partially resolve the point where it is not a normal crossing singularity, then you are in for a pretty long computation. Do try to compute the blow up of this at the origin, it is instructional. You should get a few new singular points and on one chart get back a singularity that's locally isomorphic to this one. In other words, you can't make this point "better" without normalization.
So, the solution seems to be, as Karl already suggested, to look at semiresolutions and János Kollár's work is indeed a good place to start.
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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.
Normalization can definitely change isolated singularities as we well.
For example, $k[x,y,u,v]/(x,y)\cap(u,v)$ has an isolated singularity (this is not irreducible, but there are irreducible versions with analytically isomorphic singularities). However, it is not normal and the normalization is smooth.
On the other hand, if you first resolve the 0-dimensional singular locus (ie, the singular closed points) by blowing up and then do a normalization, you can have much worse than ADE singularities. In fact, you can have a surface with no isolated singularities whose normalization has any possible isolated singularity.
However, something like your thought should be possible I think, you just might have to do blow-ups of points over the 1-dimensional singular locus as well.
For example, see the notion of semiresolution especially in the works of János Kollár.
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Yes, sorry about that. – Karl Schwede Mar 22 2012 at 11:28
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http://mathhelpforum.com/discrete-math/15562-what-wrong-proof-1-largest-integer-what-does-prove.html
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# Thread:
1. ## What is wrong with this proof that 1 is the largest integer? What does it prove?
Let $n$ be the largest integer. Then since $1$ is an integer we must have $1 \leq n$. On the other hand, since $n^{2}$ is also an integer we must have $n^{2} \leq n$ from which it follows that $n \leq 1$. Thus, since $1 \leq n$ and $n \leq 1$ we must have $n = 1$. Thus $1$ is the largest integer.
I said that the problem was that if $1$ is the largest integer, then $1 < n$ and $n^{2} < n$. There is no $\leq$ or $\geq$. So $n < 1$ and $n > 1$ which contradicts the Trichotomy law.
The above argument that $1$ is the largest integer proves that direct proofs arent the best way to approach a problem?
Am I correct?
Thanks
2. Originally Posted by tukeywilliams
Let $n$ be the largest integer. Then since $1$ is an integer we must have $1 \leq n$. On the other hand, since $n^{2}$ is also an integer we must have $n^{2} \leq n$ from which it follows that $n \leq 1$. Thus, since $1 \leq n$ and $n \leq 1$ we must have $n = 1$. Thus $1$ is the largest integer.
I said that the problem was that if $1$ is the largest integer, then $1 < n$ and $n^{2} < n$. There is no $\leq$ or $\geq$. So $n < 1$ and $n > 1$ which contradicts the Trichotomy law.
The above argument that $1$ is the largest integer proves that direct proofs arent the best way to approach a problem?
Am I correct?
Thanks
How would you respond if I said there is nothing wrong with the proof? It correctly proves the statement
"n is the largest integer implies n = 1"
and this is true. To see why, note that "P implies Q" is equivalent to "not P or Q." Then the implication is equivalent to
"n is not the largest integer or n = 1"
which is true because there is no largest integer, so n cannot be the largest integer.
So if "n is the largest integer implies n = 1" is a true statement, correctly proven, then what is wrong with that paragraph?
3. The above paragraph doesnt take into account a number like $n+1$ which will be larger than $n$?
4. Originally Posted by tukeywilliams
The above paragraph doesnt take into account a number like $n+1$ which will be larger than $n$?
JakeD says what you have proven is:
"n the largest integer implies that n=1"
Which is true. Now what he wants you to think about is what conclusion you
can draw from this. Is "1 is the largest integer" true? If not what does that tell you
RonL
5. ' 1 is the largest integer' is not true. The premiss is that we are assuming that there is a largest integer?
6. Originally Posted by tukeywilliams
' 1 is the largest integer' is not true. The premiss is that we are assuming that there is a largest integer?
Yes. You proved IF there is a largest integer then it is equal to 1. You have not shown that such an integer exists.
7. Originally Posted by tukeywilliams
' 1 is the largest integer' is not true. The premiss is that we are assuming that there is a largest integer?
apparently CaptainBlack's hint was too subtle. I won't give you the answer (everyone here might get upset with me if i do ) but i will give you a hint that should make it starkingly obvious what is wrong with this proof.
Hint: What is the difference between a valid and a sound argument? Is the proof given valid? Is it sound?
8. A valid argument holds true for some cases. A sound argument holds true for all cases. I think the problem is that we chose n = 1. We could also have chosen n = 2, n =3 and so on. Is this correct?
Thanks
9. Originally Posted by tukeywilliams
A valid argument holds true for some cases. A sound argument holds true for all cases. I think the problem is that we chose n = 1. We could also have chosen n = 2, n =3 and so on. Is this correct?
Thanks
a valid argument is one in which the conclusion follows logically from the premises and the conclusion is true (or very likely to be true) if all the premises are true. a sound argument is a valid argument that has true premises, and therefore, the conclusions that follow logically must be true (or must be highly likely to be true).
there is nothing wrong with the proof itself, it is a valid proof. however, it is not sound, since the premise is not true (there is no largest integer). therefore, we can assume that the conclusion is false, since, even though it follows logically from what we originally assumed, what we originally assumed was false.
you should also realize, an implication is only false, if we have a true statement implying a false statement, otherwise the implication is true. a false statement implies whatever, literally. the statement "if n is the largest integer then pigs can fly" is true as far as the truth table definition of an implication is concerned.
therefore, the only thing wrong with this proof is that its premises are false, and therefore any conclusion from it, even though that conclusion may be derived logically, is highly likely to be false--which in this case, we can clearly see that it is.
10. dangit, I should have realized that. The first chapter dealt with the following: if you have a false conclusion, then $P \Rightarrow Q$ will be false if $P$ is true. However if the hypothesis is false $P \Rightarrow Q$ can still be true.
Thanks for the help.
11. Originally Posted by tukeywilliams
if you have a false conclusion, then $P \Rightarrow Q$ will always be false.
no, $P \Rightarrow Q$ is only false if we have a true statement implying a false statement, otherwise it is true...but you get the idea. good job
12. Originally Posted by tukeywilliams
dangit, I should have realized that. The first chapter dealt with the following: if you have a false conclusion, then $P \Rightarrow Q$ will be false if $P$ is true. However if the hypothesis is false $P \Rightarrow Q$ can still be true.
Thanks for the help.
What you have is a false conclusion following from your assumption, so
yor assumption is false, there is no largest integer.
RonL
13. Originally Posted by Jhevon
a valid argument is one in which the conclusion follows logically from the premises and the conclusion is true (or very likely to be true) if all the premises are true. a sound argument is a valid argument that has true premises, and therefore, the conclusions that follow logically must be true (or must be highly likely to be true).
there is nothing wrong with the proof itself, it is a valid proof. however, it is not sound, since the premise is not true (there is no largest integer). therefore, we can assume that the conclusion is false, since, even though it follows logically from what we originally assumed, what we originally assumed was false.
you should also realize, an implication is only false, if we have a true statement implying a false statement, otherwise the implication is true. a false statement implies whatever, literally. the statement "if n is the largest integer then pigs can fly" is true as far as the truth table definition of an implication is concerned.
therefore, the only thing wrong with this proof is that its premises are false, and therefore any conclusion from it, even though that conclusion may be derived logically, is highly likely to be false--which in this case, we can clearly see that it is.
Thanks to everyone for this thread and to Jhevon for this post. It has been instructive for me, both in how to explain things (not my way) and in the concepts. Here is what I think I've learned.
The key concepts are the validity of the argument and the truth values of the resulting implication P => Q, premise P and conclusion Q. In the problem, we had a valid argument establishing P => Q. But the premise P was false, so the conclusion Q could not be drawn.
Now I first brought up the fact that because P was false, P => Q was true; however, that is really of no importance. What is important is that P => Q be true when P is true, implying Q is true, and that is guaranteed by the argument being valid.
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http://physics.aps.org/articles/v2/9
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# Viewpoint: How the tail wags the dog in ultracold atomic gases
, Department of Physics, Ohio State University, Columbus, OH 43210 USA and and Bethe Center for Theoretical Physics, University of Bonn, Bonn, Germany
Published February 2, 2009 | Physics 2, 9 (2009) | DOI: 10.1103/Physics.2.9
Recent calculations of the properties of ultracold atoms have revealed how two-body interactions at very short distances determine essential properties of many-body systems.
#### Universal properties of the ultracold Fermi gas
Shizhong Zhang and Anthony J. Leggett
Published February 2, 2009 | PDF (free)
The development of the field of ultracold atoms has opened up new frontiers in both few-body and many-body physics. Of particular interest are universal aspects that apply equally well not only to various species of atoms, but also to other types of particles with short-range interactions. In the current issue of Physical Review A, Shizhong Zhang and Anthony Leggett of the University of Illinois report their derivations of some remarkable properties of ultracold atomic gases that exploit interesting connections between few-body and many-body physics [1].
A many-body system of atoms has many length scales. The length scales associated with the number density and the temperature are the typical separation of the atoms and their typical deBroglie wavelength, respectively. In addition to these many-body length scales, there are length scales associated with the interactions between the atoms, most of which are comparable to or smaller than the range of interatomic interactions. An important possible exception is the scattering length $as$, which is the strength of $s$-wave (isotropic) scattering, the dominant scattering mode at very low energies. The scattering length can be large compared to the range of the interatomic interactions. If it is large, the atoms have universal properties. There is a well-behaved but nontrivial limit in which the range is taken to zero while $as$ remains nonzero. This “zero-range” limit captures the universal aspects of systems of fermions with two spin states. The universal features apply to any fermionic atom for which the scattering length of the two spin states is much larger than the range of interatomic interactions. They also apply to the neutron, which has two spin states, despite the range of the interactions between neutrons being smaller by six orders of magnitude.
Zhang and Leggett call the universal many-body system the “two-component ultracold Fermi gas along the BEC-BCS crossover.” It is sometimes referred to more concisely as the strongly interacting Fermi gas. The ground state of this system is a superfluid, but the mechanism for superfluidity is completely different in the two weakly interacting limits. In the BEC (Bose-Einstein condensate) limit, the scattering length goes to zero through positive values $(as→0+)$ and we have a gas of bosonic diatomic molecules with weak repulsive interactions [Fig. 1a]. In the BCS (Bardeen-Cooper-Schrieffer) limit, the scattering length approaches zero through negative values $(as→0-)$ and we have a gas of fermionic atoms with weak attractive interactions [Fig. 1c]. In the BEC limit, the mechanism for superfluidity is the condensation of diatomic molecules, while in the BCS limit, it is the formation of Cooper pairs. Despite the dramatic difference between these two mechanisms, there is a smooth crossover between them as $as$ increases from $0+$ to $+∞$, jumps discontinuously to $-∞$, and then increases to $0-$. In particular, there is a well-behaved unitary limit as $as→±∞$ [Fig. 1b]. This limit is particularly interesting, because the system is so strongly interacting that the interactions provide no length scale.
The universal zero-range limit is relevant to experiments on ultracold atoms, because at the extremely low number densities and temperatures that can be achieved in cold trapped atoms, the many-body length scales are large compared to the range of interatomic interactions. A unique feature of ultracold atoms is the possibility of controlling the interaction strength experimentally. In particular, by tuning the magnetic field in the region near a Feshbach resonance, the scattering length can be varied over its entire range, from its off-resonant value through $±∞$ and back to its off-resonant value. The use of Feshbach resonances to study the strongly interacting Fermi gas has been one of the major thrusts of experimental cold atom physics in the last decade.
Zhang and Leggett [1] derive some remarkable universal properties of the strongly interacting Fermi gas. They show that several of its important properties depend on the number density $n$ and temperature $T$ only through a dimensionless function $h(n,T,as)$ that has a smooth unitary limit as $as→±∞$. They derive the rate of change of the total energy per particle with respect to the inverse scattering length while $n$ and the entropy are held fixed:
$d(E/N)das-1=-ħ2mkFh(n,T,as),$
(1)
where $kF=(3π2n)1/3$ is the Fermi wave number. This implies that E/N and other thermodynamic functions can be determined from $h$ by integration. Zhang and Leggett also showed that several other important properties of the system can be expressed as $h$ multiplied by a factor that is entirely determined by two-body physics, including (i) the interaction energy per particle, (ii) the rf spectroscopy shift, which is the shift in the radio-frequency signal required to change the spin state of an atom, and (iii) the rate for the photoassociation of pairs of atoms into a diatomic molecule using a laser beam. What these quantities have in common is that they are sensitive to the presence of pairs of atoms with different spins whose separations are much smaller than the many-body length scales and the scattering length. Thus the function $h$ is a measure of the density of pairs with small separations.
The existence of this function $h$ that plays such a central role in the strongly interacting Fermi gas was first realized by Shina Tan in 2005. Tan showed [2] that the momentum distribution in an arbitrary state of the system has a large-momentum tail of the form $C/k4$, where $k$ is the wave number and $C$, which is called the contact, is the same for both spin states. He derived an expression for the total energy of the system that implies item (i) above as a simple consequence. Tan then derived a more general result [3] than Eq. (1) that applies not only to homogeneous states but also to any state of the system, including few-body states:
$dEdas-1=-ħ24πmC,$
(2)
where the contact $C$ is the same quantity that appears in the asymptotic momentum distribution. The contact is the integral over the system of the contact density, which is proportional to the number of pairs of atoms with different spins that are close together. Tan’s results reveal that this number is much larger than one might expect. The number in a small volume scales like (volume)$4/3$ instead of like (volume)$2$ as one might expect. In a later paper, Tan derived a virial theorem (that is, a relationship between kinetic and potential energy) for atoms trapped in a harmonic potential [4] and it also involves the contact.
Tan’s first two papers were almost completely ignored for several years and were not published until 2008 [2, 3]. One contributing factor to the delayed appreciation for Tan’s work was that he derived his universal results from the many-body Schrödinger equation by using novel methods involving generalized functions that he invented specifically for this problem. Zhang and Leggett, apparently unaware of Tan’s work, derived their results independently using more conventional methods of many-body physics. The first published paper that recognized the importance of Tan’s work was by Punk and Zwerger [5], who showed that the rf spectroscopy shift was proportional to $dE/das-1$. This result was derived independently by Baym, Pethick, Yu, and Zwierlein [6]. In 2008, Tan’s universal relations were rederived by Braaten and Platter using methods of quantum field theory [7] and by Werner, Tarruell, and Castin using less formal methods [8]. Werner, Tarruell, and Castin also derived independently the result on the photoassociation rate.
The universal relations discovered by Shina Tan and the applications that were subsequently derived [1, 5, 6, 8] are beautiful illustrations of the interplay between few-body physics and many-body physics. They involve quantities that can be expressed in a factored form, with one factor involving many-body physics and the other involving only two-body physics. They exploit the analytic solution of the two-body problem in the zero-range limit. There may be important generalizations of these universal relations for which the few-body factors involve three or more particles. The three-body factors may be calculable using the exact numerical solutions of the three-body problem in the zero-range limit.
### References
1. S. Zhang and A. J. Leggett, Phys. Rev. A 79, 023601 (2009).
2. S. Tan, Ann. Phys.323, 2952 (2008).
3. S. Tan, Ann. Phys. 323, 2971 (2008).
4. S. Tan, Ann. Phys. 323, 2987 (2008).
5. M. Punk and W. Zwerger, Phys. Rev. Lett. 99, 170404 (2007).
6. G. Baym, C. J. Pethick, Z. Yu, and M. W. Zwierlein, Phys. Rev. Lett. 99, 190407 (2007).
7. E. Braaten and L. Platter, Phys. Rev. Lett. 100, 205301 (2008).
8. F. Werner, L. Tarruell, and Y. Castin, arXiv:0807.0078.
### About the Author: Eric Braaten
Eric Braaten carried out his graduate work in theoretical high-energy physics at the University of Wisconsin, receiving his Ph.D. in 1981. After postdoctoral positions at University of Florida and Argonne National Laboratory, he took a faculty position at Northwestern University in 1995, where he conducted ground-breaking research on quantum chromodynamics. He moved to Ohio State University in 1995, where he has branched out into the field of ultracold atoms. Dr. Braaten was awarded the Humboldt Research Prize in 2009, and is now visiting the Bethe Center for Theoretical Physics at Bonn University. He continues to work on both low-energy and high-energy problems in elementary particle physics and on both few-body and many-body aspects of ultracold atoms. Dr. Braaten was chosen as a Fellow of the American Physical Society in 2000 and named an Outstanding Referee in 2008.
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http://crypto.stackexchange.com/questions/736/changing-algorithms-during-encryption?answertab=active
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# Changing algorithms during encryption
Inspired by "Guarding against cryptanalytic breakthroughs: combining multiple hash functions", I am curious if there is a cryptographic reason to use only one algorithm during encryption.
For example, start with Blowfish, move to AES, switch to DES, etc, in a defined, but semi-unpredictable fashion (maybe basing which one to choose next off the last 4 bits of the most-recently-encrypted block's original plantext).
Would changing algorithm (but keeping the key identical, for sake of argument) have any cryptographic value? Could such a scheme be more resilient to cryptanalysis?
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## 4 Answers
As many of the other answers have said the proposed method above is only as strong as the weakest cipher used (this is especially dangerous if the same key is used for all the ciphers).
Furthermore, basing the cipher transitions on the plaintext opens up the door to a range of timing side-channel attacks if any of the ciphers are faster or slower that each other.
One way you could guard against cryptanalytic breakthroughs is to use secret sharing to split a message $m$ into $n$ plaintexts ($p_i$) that must be combined to recover the original message. $$m = p_0 \oplus p1 ... \oplus p_n$$ These $n$ plaintexts could then be enciphered by $n$ ciphers.
$$ciphertext = cipher_0(key_0, p_0)|cipher_1(key_1, p_1) ... |cipher_n(key_n, p_n)$$
The size of the ciphertext increases with the number of ciphers (size of message $\times n$), but you could ensure that even if $n-1$ of the ciphers were broken the message would remain secret. This also requires $n$ keys.
One could try some sort of key expander to expand a $key$ to a different key for each cipher:
$$key_0 = cipher_0(cipher_1( ... cipher_n(key, key)), cipher_1( ... cipher_n(key, key)))$$ $$key_1 = cipher_n(cipher_0( ... cipher_{n-1}(key, key)), cipher_0( ... cipher_{n-1}(key, key)))$$
but I don't have much faith its security (theoretically it is probably as weak as it's weakest cipher) and when one is so concerned with cryptanalytic breakthrough it would be foolish to introduce a "trusted" key expander (why not then introduce a trusted cipher and use that). I have asked the key expander issue as a separate question ( Designing a key expander out of ciphers ).
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As you've proposed it, it's a pretty lousy idea.
To have any hope of gaining security, start with a separate key for each cipher, and apply the ciphers serially to to every block -- i.e., encrypt block 1 with cipher A, then with cipher B, then with cipher C (but I reemphasize, using a separate key for each).
With that, unless the ciphers form a group (unlikely, at least if they're worth anything) the attacker will have to break all three ciphers to recover any part of the plaintext.
The disadvantage, of course, is that encryption and decryption require (roughly) triple the resources. OTOH, given the availability of quad-core processors, it would be fairly easy to set it up as a pipeline that ran about the same speed as the slowest individual cipher (at the expense of using more cores to do it).
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What you are effectively doing here is creating a new algorithm somehow composed of all these individual algorithms.
In effect you'll have something that is really complicated, and thus hard to analyze - which does not mean hard to break, but hard to estimate the strength.
It is not really clear if it would be any better than the weakest of the combined algorithms, even less better than the strongest of them. Also, it most likely will be slower than any single algorithm (since each of them has a different kind of key schedule).
Don't do it.
If you really feel you need to combine multiple algorithms to guard against breaks of one of them, encrypt the data with both (one after the other), preferably with different keys. (Though this will not give you $2·n$ bit security, I think, only guard against attacks against one of the algorithms, and it will take the double computing power.)
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Seems to me that it makes it easier on the attacker; that way, if he knows of any weakness in one of the ciphers you use, he can recover that part of the plaintext. And, the fact that "he doesn't know which parts are encrypted by what" doesn't really hinder him, he can guess.
In addition, basing it on the plaintext means that if the attacker knows the plaintext (say, you're transmitting boilerplate), that means that he knows which cipher you're using, and so he doesn't have to guess.
Using the same key for all the ciphers makes things even worse; that means that if he does recover (say) the 56 bit DES key, he then knows 56 bits of your AES key.
Using N different ciphers makes sense only if you're secure if any of them are secure (and there are ways to do that). What you proposed means that you're secure only if they're all secure -- you'd be better off picking one, and sticking with that.
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http://math.stackexchange.com/questions/254062/explain-this-proof-without-words-of-integration-by-parts-to-me
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# Explain this proof without words of integration by parts to me
Here is a proof of integration by parts: http://www.math.ufl.edu/~mathguy/year/S10/int_by_parts.pdf
But I don't understand how it works. Specifically, I don't understand why $\int_r^s u \, dv$ equals one of the areas (and likewise for $\int_p^q v \, du$).
To me it looks like this only works if $g$ is the inverse of $f$...
Thanks!
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– Dominik Dec 9 '12 at 1:08
## 5 Answers
The important part is that the axes themselves are the original $x,y$ axes after applying the functions $u$ and $v$ to them!
Perhaps this diagram from here is clearer. Because $u=f(v)$ is the height of the curve from the $v$ axis, the area of the bottom part is $\displaystyle \int_{v_1}^{v_2}u \space d v$. But note that $u_1=f(v_1),u_2=f(v_2)$.
Next, let the $v$ values change. Now $v$ is the height of the curve from the $u$ axis, the area of the top part is $\displaystyle \int_{u_1}^{u_2}v \space d u$.
The sum of these areas is a nice difference of areas of boxes. I.e. $u_2v_2-u_1v_1$. This is however the $uv|_{(v_1,u_1)}^{(v_2,u_2)}$. So, symbolically
$\displaystyle \int_{v_1}^{v_2}u \space d v +\int_{u_1}^{u_2}v \space d u=uv|_{(v_1,u_1)}^{(v_2,u_2)}$.
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I think you're onto it here. Am I going the wrong direction with the bottom half of my answer, or is it in agreement with yours? – 000 Dec 8 '12 at 23:51
I think we agree. Except, you probably meant to name the map that you define in the second paragraph of your second section as $f$ and not $v$. This way, as Hagen von Eitzen describes below, $f$ is the curve parametrizing $u$ and $v$ as a function of $t$ varying between the original bounds of integration $a$ and $b$ – Fortuon Paendrag Dec 9 '12 at 0:09
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I removed the original variables in the situation and restructured it in terms of axis $u$ and axis $v(u)$. Thus, I think, my work also coincides with Hagen von Eitzen since I've said $u=f(x_0)$ for some $x_0$ and $v(u)=g(x_1)$ for some $x_1$. It may be easier to say $(u,v(u))=(f(t),g(t))$, though. – 000 Dec 9 '12 at 0:16
The black curve should be viewed as parametric curve $(f(t),g(t))$, $a\le t\le b$.
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I tried viewing like that, but I still don't understand. – dever Dec 8 '12 at 23:04
Maybe you prefer the proof by observing that the derivative of $uv$ is $uv'+u'v$. – Hagen von Eitzen Dec 8 '12 at 23:09
Well, I know that proof, but I want to understand this proof. – dever Dec 8 '12 at 23:11
If we play around with the graph in our head, we can look at it from two perspectives: $(U,V(U))$ and $(V,U(V))$. That is, if we look at it from how it is presented, we see that the area of the second box in the equation is $\int_{p}^{q}V(U)dU.$ In their notation, that's exactly $\int_{p}^{q}vdu$. This follows from the definition of a definite integral.
Now, if we flip the graph (turn your head sideways), reflect it to the left, and look at it from the perspective of $(V,U(V))$, we see that the area underneath the flipped curve is $\int_{r}^{s}U(V)dV$. This is, in their notation, $\int_{r}^{s}udv$.
This is why we have that the total sum of the areas is $\int_{r}^{s}udv+\int_{p}^{q}vdu.$
There is not a loss of rigor what so ever in this diagram. It is quite unsettling at first, but what we're really just doing is this:
Let $v: U \to V$ be a map where, for a $u \in U$, we have $u \mapsto v(u)$. The domain and range thus form the set of ordered pairs $\{(u,v(u)): u\in U\}$. Our other map, $v^{-1}:V \to U$, is the flip-then-reflect-left map with the mapping $v(u) \mapsto u$. It is the set of ordered pairs $\{(v(u),u): u\in U\}$. This is the inverse map of $v$ by definition.
Having that $u=f(x_0)$ for some $x_0$ and $v(u)=g(x_1)$ for some $x_1$, however, does not necessarily imply $g$ is the inverse of $f$.
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The area between the curves and the axes adds together to give the area of the rectangle with area $qs$, with the square of area $pr$. So subtract the two areas and set them equal to the sum of the integrals. Thank you for this proof, I really like it!
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He used the formula that the area of a region under the curve $y=g(x),x=v(x)$ is equal to $\int_{C} xdy=\int_{C}vdu$. This is from Green's formula. See this wikipedia article. Since the derivation of Green's formula does not involve integration by parts, this is a "proof without words".
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http://mathoverflow.net/questions/3239?sort=oldest
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## Is no proof based on “tertium non datur” sufficient any more after Gödel?
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There are many proofs based on a "tertium non datur"-approach (e.g. prove that there exist two irrational numbers a and b such that a^b is rational).
But according to Gödel's First Incompleteness Theorem, where he provides a constructive example of a contingent proposition, which is neither deductively (syntactically) true nor false, we know that there can be a tertium.
My question: Are all proofs that are based on that principle useless since now we know that a tertium can exist?
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I think it's a bit unfair to downvote this question. Logic can have many confusing aspects, and most people I know take some time to appreciate its subtleties. – aorq Oct 29 2009 at 15:42
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I agree. This is a common error, and it is one that I think is not so naive that it is not of interest to mathematicians to know how to correct it. – David Speyer Oct 29 2009 at 16:06
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This is somewhat addressed in some answers below, but this point is important enough that it warrants stressing here: questioning the validity of using the law of excluded middle is a legitimate (albeit minority) position, and it has a distinguished history. It just does not have much to do with Gödel's First Incompleteness Theorem, and indeed, pre-dates it by at least a couple of decades. – Thierry Zell Aug 13 2010 at 13:14
## 7 Answers
You are confused.
The best way out of your confusion is to maintain a very careful distinction between strings of formal symbols and their mathematical meanings. Godel's theorem is, on its most primitive level, a theorem about which strings of formal symbols can be obtained from other strings by certain formal manipulations. These formal manipulations are called proofs, and the strings which are obtainable in this way are called theorems. For clarity, I'll call them formal proofs and formal theorems.
In particular, let G be a string such that G is not a formal theorem and neither is NOT(G). It is still true that G OR NOT(G) is a formal theorem. Moreover, if G IMPLIES H and NOT(G) IMPLIES H are both formal theorems, then H will be a formal theorem; because there are rules of formal manipulation that allow you to take the first two strings and produce the third. I believe that Douglass Hofstader discusses this in a fair bit of detail when he goes over Godel's theorem.
The above is mathematics. Next, some philosophy. I don't find it helpful to say that G is neither true nor false. It find it more helpful to say that our systems of formal symbols and formal manipulation rules can describe more than one system. For example, Euclid's first four axioms can describe both Euclidean and non-Euclidean geometries. This doesn't mean that Euclid's fifth postulate has some bizarre third state between truth and falsehood. It means that there are many different universes (the technical term is models) described by the first four axioms, and the fifth postulate is true in some and false in others.
However, in any particular one of those universes, either the fifth postulate is true or it is false. Thus, if we prove some theorem on the hypothesis that the fifth postulate holds, and also that the fifth postulate does not hold, then we have shown that this theorem holds in every one of those universes.
There are fields of mathematical logic, called constructivist, where the law of the excluded middle does not hold. As far as I understand, that issue is not related to Godel's theorem.
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This is a side comment, not a criticism of your answer: I find it quite helpful to think of a statement containing variables as neither true nor false. For example, for a real variable x, "(x+1)^3 > 0" is neither true nor false, but "(x+1)^2 > 0" is true. I have written about this in several places and talked to people about it. I get two reactions: One, like yours, is that that is a bizarre way to think. The other is "well of course -- that's the normal way to describe such assertions". Perhaps there is a Deep Divide among mathematicians... – SixWingedSeraph Oct 29 2009 at 15:20
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My perspective is that both your examples are not full sentences. A full sentence should not have bound variables, so it should say: "For all x in R, (x+1)^3 > 0." or "For the x defined in Section 2, (x+1)^3>0.". Of course, we often drop the preceding phrase, but only because it is clear from context. But I've run into a lot of people who think the way you do, and I don't want to claim that I am more right in any absolute sense. I think one of the reasons that I have a hard time understanding the notation of predicate calculus was that it was invented by people who take your perspective. – David Speyer Oct 29 2009 at 15:48
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Of course, when I wrote "bound", I meant "unbound". – David Speyer Oct 29 2009 at 15:51
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For what it's worth, many standard presentations of predicate calculus call the things SixWingedSeraph is talking about "well-formed formulas" (or "wffs"), and then reserves the term "sentence" for wffs with no free (unbound) variables. Then the terms "true" and "false" are applied only to sentences, and not to wffs in general. There are some proof systems that allow non-sentence wffs in intermediate steps in proofs, while others only allow sentences as steps in proofs. – Kenny Easwaran Nov 6 2009 at 6:42
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-1: The argument in the question is sketchy, but it is not confused. Your "answer" comes down to assertions involving "I don't find it useful" and existence of "different universes": I find it hard to see why anyone attracted by the argument in the question would find this response cogent. – Charles Stewart Jan 15 2010 at 12:37
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If you believe in classical logic, such proofs are still fine. Godel incompleteness just states that in first order logic, any system capable of expressing arithmetic must contain an undecidable formula P. You can still prove (P or ~P) - that's an axiom actually. In every model of the axioms either one or the other is true. So a classical mathematician (who is a platonist), says Godel's incompleteness theorem just shows that FOL is insufficient for expressing everything about the "real" mathematical universe, but that doesn't mean that the law of the excluded middle (LEM) fails. If you have a philosophical problem with this, you have some constructivist leanings. I have strong formalist tendencies, and would say that LEM is cool if you think its cool, and not if you don't like it. We should be free to change our foundations to suit our whims!
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You're missing the distinction between truth and proof. Godel's Theorem says there are statements which are neither provable nor disprovable (from a given set of axioms). Those statements are still either true or false in a given universe. Godel just says your axioms aren't good enough to tell which one.
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I accidentally ran into this old question and thought to give an example whose (humble!) intention is to permanently deconfuse anyone who knows just a little bit of undergraduate mathematics.
Consider group theory. In a typical introductory course, you will learn the simple axioms and immediately encounter sentences like
$$\forall y \forall x \forall z( (xy=e) \wedge (zy=e ) \Rightarrow x=z);$$
or slightly more complicated ones like
$$(\forall x(x^2=e ))\Rightarrow (\forall x\forall y(xy=yx)).$$
You also learn how to deduce them rather easily from the axioms of the theory.
$$S:\ \ \ \ \ \ \ \forall x \forall y( xy=yx)$$
?
Can it be deduced from the axioms? Obviously not, but it requires at least a bit of thoughtfulness to prove that it can't. You see, we can just write down a structure like $S_3$ that satisfies the axioms of group theory, a model of group theory, and produce two elements in it for which the sentence is not true. Obviously we couldn't produce such a model if the sentence was a logical consequence of the axioms.
How then about $\neg S$? Can it be deduced from the axioms? Again obviously not. Consider the integers $Z$.
So we see that group theory is incomplete: We've written down a sentence $S$ such that neither $S$ nor $\neg S$ can be deduced from the axioms.
But here is an important point: If the context of the discussion was the group $Z$, is the sentence $S$ true? Yes, of course, and I can prove it for you. (Using more than the axioms of group theory, of course.)
For a complete theory, every true assertion $S$ (in the language of the theory) about a given model $G$ can be deduced from the axioms. This is because $S$ or $\neg S$ can be deduced, but if $\neg S$ could be deduced, then it would have to be true in any model of the theory, in particular, $G$. So $S$ would be false in $G$. But $S$ is true. Therefore, it must be the one that can be deduced. The upshot is that whenever you have a theory (like group theory) with a model that admits a true sentence that can't be deduced from the axioms, then the theory is incomplete. The point of boring you with this discussion is to illustrate that an incomplete theory is a very mundane object.
In case this suggests (as it should) that a complete theory, on the other hand, is bound to be very exotic, you might like this simple list of a few complete theories. For example, the theory of algebraically closed fields of characteristic zero is complete. This implies, in particular, a logical incarnation of the 'Lefschetz principle': A field-theoretic sentence true in $C$ is true in every algebraically closed field of characteristic zero. In fact, as noted above, a sentence true in any given model, since it can be deduced from the axioms, is true in any other model. I found this fact quite mind-boggling when I first encountered it. A good exercise is to see why a sentence like 'every element of $\bar{Q}$ is algebraic' doesn't cause a problem. (You need to get a bit more precise to do this, especially about the language of the theory.)
There is a complete theory of the natural numbers, by the way. Add to your favorite axioms of arithmetic all the sentences that are true in the natural numbers. This is a perfectly respectable complete theory, sometimes referred to as the theory of natural numbers. Goedel's first incompleteness theorem can be interpreted as saying this theory doesn't admit a recursively enumerable set of axioms. (Which should be at least intuitively plausible if you consider difficult unresolved problems like, say, Goldbach's conjecture.)
Added:
In my opinion, it's not such a good idea to emphasize the 'string of formal symbols and rules' point of view when explaining the incompleteness theorem. It's true that to prove the theorem, you need to set up such background formalities. But the statement itself can plausibly be interpreted as something about everyday reasoning in mathematics. We are usually interested in some structure, a rather specific one like $Z/2$, a somewhat more general one like 2-groups, or more general yet like all groups. The question concerns which properties (or axioms satisfied by the structure, if you prefer) we use to prove certain assertions. The everyday nature of this question was the reason for bringing up the commutativity of $Z$, which I can certainly prove in the course of a normal discussion on the chalkboard, but anyone can see requires more than group theory.
This question also comes up rather frequently as one of great interest to practicing mathematicians. An advanced example that I can remember off the top of my head is 'Can one prove the Kodaira vanishing theorem using only algebraic geometry?,' which was resolved first by Faltings (although there is room for interpretation of the phrase 'only algebraic geometry').
In some sense, the rationale for the abstract formalism surrounding the incompleteness theorem is also pretty commonsensical. To prove that something can be done, you just need to do it. For example, I think it is uncontroversial that the proof of the Kodaira vanishing theorem by Deligne and Illusie uses 'only algebraic geometry.' And then, there are the famous elementary proofs of the prime number theorem. To prove that something can't be done, on the other hand, often requires more careful foundations.
Added again:
After some conversations, I decided to put in a few final words of clarification. I hope I didn't slight anyone with the joke about 'permanent deconfusion.' I don't claim to have any serious understanding of philosophical ramifications, for example. However, I tried to articulate what seems to me a sensible view of the matter for practicing mathematicians. Starting from the one given, you can yourself quickly make up examples illustrating the (uninteresting) incompleteness of a large majority of the theories we usually work with, rings, fields, topological spaces, etc. After that process, and thinking through just the few implications of completeness already mentioned, if someone came up to you and claimed that Peano Arithmetic was complete, I suspect your eyes would pop out.
Someone once told me that a good way to sound sophisticated as an amateur logician is to proclaim that the completeness theorem* is much more important than the incompleteness theorem. That's perhaps too sweeping a statement, but it seems to be the one that's useful for usual mathematics. For people interested in pursuing this line of thought, I recommend the nice lectures delivered by Angus Macintyre at the Arizona Winter School in 2003:
http://math.arizona.edu/~swc/aws/03/03Notes.html
One intriguing observation there I sometimes think about is how number-theoretic completions (reals and $p$-adics) correlate to logically complete theories.
*The completeness theorem essentially says that a sentence is a logical consequence of the axioms if and only if it's true in all models of the axioms. The idea for the non-trivial direction is to show how to construct, given any sentence that can't be deduced, a model for the axioms in which it is false.
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I'm afraid you're far too sophisticated for me. But I'll leave it to a professional to write down the absolutely correct version of group theory and its language in a way that my assertions make good sense. – Minhyong Kim Apr 14 2010 at 22:01
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Isn't it pretty clear that "xy=yx" refers to multiplying group elements together? – Peter Samuelson Apr 14 2010 at 22:25
So these comments make sense, I'll explain the comments I deleted: I made a comment regarding some set-theoretic nonsense because I misread the post and didn't see that Minhyong was assuming the axioms of group theory instead of set theory. – Harry Gindi Apr 15 2010 at 18:23
Just found and not yet read, <a href="rsta.royalsocietypublishing.org/content/363/1835/…; title="abstract">here</a> an article on Skolem's thoughts on the meaning of that incompleteness. BTW, I found that incompleteness theorem (after I had learned it's precise formulation) not very surprising, but would have been shocked about a completeness theorem, as the later would show that elementary number theory were "trivial" and just like some chess game. – Thomas Riepe Apr 21 2010 at 14:13
There are reputable mathematicians who assert that "tertium non datur" arguments are at best incomplete and at worst meaningless. I recommend "A Constructivist Manifesto", Chapter 1 of Errett Bishop's "Foundations of Constructive analysis".
The idea is not to assert some "bizzare state" between truth and falsity, but rather to take seriously the possibility that there may be simple and natural propositions that (a) Are independent of ZFC and (b) Will not or can not ever yield to any compelling new intuition or axiom. The idea that in some magical timeless world such assertions are forever true or false in themselves is at least questionable, and Godel was the first to expose this difficulty.
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Of course, Gödel himself seems to have been a Platonist --- he thought that his result showed the poverty of formalism, and motivated the necessity of a direct cognitive link to the Platonic realm, for discovering mathematical truth! – Niel de Beaudrap Apr 15 2010 at 8:07
@Niel: Yes, and Cohn, it seems, was a passionate formalist..... Which goes to show how questionable it is to draw philosophical conclusions from mathematical results. – SJR Apr 15 2010 at 10:09
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I don't think it is accurate to say that "Godel was the first to expose this difficulty". Brouwer was on record a constructivist some 20+ years before Godel's incompleteness theorem was published. It's the crisis in the foundations itself that sent mathematicians scrambling for fixes, one of which being the restricted use of excluded middle. – Thierry Zell Aug 13 2010 at 13:06
In classical logic, for any proposition $A$, either $A$, or $\lnot A$.
What Godel showed was that under certain conditions (which hold for the formal system most of us use) there are propositions $A$ such that neither $A$ nor $\lnot A$ are provable.
Note the difference between these statements. In one case it's about the assertions $A$ and $\lnot A$. In the other case it's about the provability of the assertions $A$ and $\lnot A$. The former is about whatever $A$ is about. The latter is about whether or not you can produce a string of symbols by following certain rules.
For example, if you assert "for all real $a>1, a^2>1$" you're saying something about real numbers. But if you assert "for all real $a>1, a^2>1$ is provable", you're saying something about how you can produce the string of symbols "for all real $a>1, a^2>1$" by following a bunch of rules. These two things are entirely different kinds of assertion.
In particular, if you know $A \lor \lnot A$, you can conclude that either $A$ is true, or that $\lnot A$ is true. But if you know that $A \lor \lnot A$ is provable, you can't deduce (in a classical context) that either $A$ is provable or that $\lnot A$ is provable.
As long as you make clear the distinction between asserting $A$ and asserting that $A$ is provable, you're fine.
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Your fourth paragraph is slightly confusing. If you say "for all real $a>1$, $a^2>1$ is provable", you should really say "provable from a given collection of axioms". – Stefan Geschke Aug 30 2010 at 12:02
I think that many of the responses here are overlooking the key part of your question, "Are all proofs that are based on that principle useless?". The question of whether LEM is not merely formally consistent but actually useful, i.e., pertaining to reality, is a deep question that has been debated for at least a century (since Brouwer). One partial answer to your question is that no, not all such proofs are useless, because often they involve LEM on propositions which are indeed decidable. In constructive mathematics, $\varphi \vee \neg \varphi$ is precisely what it means for a proposition $\varphi$ to be decidable, and making use of this fact in a proof amounts to calling a decision procedure.
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Even LEM on propositions that will never or can never be decided is not useless -- An LEM proof of a proposition indicates that attempts to prove the negation of the proposition are futile. A wise old logician once told me that the attempt to make proofs constructive is not so much a logical imperative as it is a crucial method of discovery. For example, the project of making constructive the fact that the rationals are dense in the reals gives rise to the discipline of diophantine approximation. – SJR Apr 17 2010 at 11:17
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http://mathoverflow.net/revisions/101824/list
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## Return to Answer
2 added example and Fisher's inequality
One related structure is an incidence geometry. Each pair of points determine a unique line. Every $2$-design with $\lambda = 1$ is an example of an incidence geometry, but incidence geometries are more flexible.
Given an incidence geometry on $V$, you can induce an incidence geometry on $W \subset V$ by taking the lines to be the intersections of lines with $W$ (except you throw out the intersections of size at most $1$). This should let For example, there are Steiner triple systems (block designs where the blocks have size $3$ and $\lambda=1$) if and only if the number of vertices is $1$ or $3$ mod $6$. There is no Steiner triple system on $11$ vertices, but you construct can take a Steiner triple system on $13$ vertices with $26$ blocks/lines and delete $2$ points (and the line through them which only contains $1$ point now) to get an incidence geometries whose geometry on $11$ points with $25$ lines so that each line contains $2$ or $3$ points.
This might not work well over all sets of parametersfit your problem out . Block designs with more than $1$ block have more blocks than points (Fisher's inequality). The same is true for incidence geometries (de Bruijn–Erdős). Projective planes are examples of equality. If you have fewer workers than files, so that workers have to handle more than about the square root of the number of files, then you aren't looking for a small deviation from a block designs on slightly larger setsdesign.
1
One related structure is an incidence geometry. Each pair of points determine a unique line. Every $2$-design with $\lambda = 1$ is an example of an incidence geometry, but incidence geometries are more flexible.
Given an incidence geometry on $V$, you can induce an incidence geometry on $W \subset V$ by taking the lines to be the intersections of lines with $W$ (except you throw out the intersections of size at most $1$). This should let you construct incidence geometries whose parameters fit your problem out of block designs on slightly larger sets.
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http://mathoverflow.net/questions/47835?sort=votes
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When can you reverse the orientation of a complex manifold and still get a complex manifold?
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I'm told that $\overline{\mathbb{C}P^2}$, i.e. $\mathbb{C}P^2$ with reverse orientation, is not a complex manifold. But for example, $\overline{\mathbb{C}}$ is still a complex manifold and biholomorphic to $\mathbb{C}$.
This makes me wonder, if $X$ is complex manifold is there a general criterion for when $\overline{X}$ also has a complex structure? For example, it seems that if $X$ is an affine variety than simply replacing $i$ with $-i$ gives $\overline{X}$ a complex structure and $X, \overline{X}$ are biholomorphic.
EDIT: the last claim is wrong; see BCnrd's comments below and Dmitri's example. Also, as explained by Dmitri and BCnrd, $X$ should be taken to have even complex dimension.
Another question: if $X$ and $\overline{X}$ both have complex structures, are they necessarily biholomorphic? Edit: No per Dmitri's answer below.
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Is there a simple reason for why $\overline{\mathbb{CP}^2}$ is not a complex manifold? – J.C. Ottem Nov 30 2010 at 22:31
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@J.C. Ottern: Any almost complex structure compatible with the orientation on a closed 4-manifold $X$ satisfies $c_1^2[X]=2\chi+3\sigma$ ($\chi$=Euler char, $\sigma$=signature). This is by Hirzebruch's signature theorem. – Tim Perutz Nov 30 2010 at 22:40
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Fix an alg. closure $\mathbf{C}$ of $\mathbf{R}$, equipped with unique abs. value extending the one on $\mathbf{R}$, complex analysis is developed without needing a preferred $\sqrt{-1}$. The complex structure has no reliance on any orientation. The so-called canonical orientation on complex manifolds is just the functorial one arising from a choice of $\sqrt{-1}$; can make either choice, complex structure can't tell! Likewise, the analytification functor on locally finite type $\mathbf{C}$-scheme has nothing to do with any such choice. Note $p$-adic analysis goes the same way. – BCnrd Nov 30 2010 at 22:48
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What is canonical is that *even*-dim'l C-manifolds have an intrinsic orientation determined by C-structure: an orientation of $\mathbf{C}$ endows all C-manifolds with functorial orientation, and changing initial choice affects the orientation on $n$-dimensional C-manifolds by $(-1)^n$. So for even $n$ the question is well-posed. This has nothing to do with changing $i$ and $-i$, and your impression in the affine case is wrong. In any dim., can "twist" structure sheaf by C-conj. to get a new C-manifold (modelled on $\overline{f}(\overline{z})$), but that's a different beast. – BCnrd Nov 30 2010 at 23:13
2 Answers
If you take an odd dimensional complex manifold $X$ with holomorphic structure $J$ then $-J$ defines on $X$ a holomorphic structure as well. And, of course, $J$ and $-J$ induce on $X$ opposite orientations. In general it is not true that these two complex manifolds are biholomprphic. Indeed, if $X$ is a complex curve, then $(X,J)$ is biholomorphic to $(X,-J)$ only if $X$ admits and anti-holomorphic involution (this will be the case for example if $X$ is given by an equation with real coefficients).
Starting from this example on can construct a (singular) affine variety $Y$ of dimension $3$, such that $(Y,J)$ is not byholomorphic to $(Y,-J)$. Namely, let $C$ be a compact complex curve that does not admit an anti-holomorphic involution say of genus $g=2$. Consider the rank two bundle over it, equal to the sum $TC\oplus TC$ ($TC$ is the tangent bundle to $C$). Contract the zero section of the total space of this bundle, this gives you desired singular $Y$.
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Nice example but I wonder -and this is not so important, just curious if there's a nice answer- if $C$ is a genus 2 curve and I describe it as a degree 2 cover of $\mathbb{P}^1$: $y^2 = (x - a_1)\cdots (x - a_6)$, is there simple condition on the $a_i$ that guarantees that it does not have an antiholomorphic involution? – solbap Dec 1 2010 at 16:23
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Yes, there should be a relatively simple criterion, one should check when the configuration of points $a_1,...,a_6$ is (not) invariant under any anti-holomorphic involution of $\mathbb CP^1$. For example, in the case of elliptic curve $y^2=(x-a_1)...(x-a_4)$ the necessary and sufficient condition for having anti-holomorphic involution is that the double ratio of $\frac{a_1-a_2}{a_3-a_4}$ is real. If all double ratios of $a_1,...,a_6$ are real, then again we have an anti-holomorphic involution. But this is not necessary there are two more cases (like $(x^2+a)(x^2+b)(x^2+c)$ $a,b,c>0$)... – Dmitri Dec 1 2010 at 17:51
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.
It seems to me that you could be interested in the following (I haven't checked the paper in detail, but I think theorems of this "style" could be helpful for you):
• Dieter Kotschick, Orientations and geometrisations of compact complex surfaces (Bull. London Math. Soc. 29 (1997), no. 2, 145–149.)
Theorem Let $X$ be a compact complex surface admitting a complex structure for $\bar{X}$. Then $X$ (and $\bar{X}$) satisfies one of the following:
(1) $X$ is geometrically ruled, or
(2) the Chern numbers $c_1^2$ and $c_2$ of $X$ vanish, or
(3) $X$ is uniformised by the polydisk.
In particular, the signature of $X$ vanishes.
Other material that could be helpful is:
• Dieter Kotschick, Orientation-reversing homeomorphisms in surface geography (Math. Ann. 292 (1992), no. 2, 375–381.)
• Arnaud Beauville, Surfaces complexes et orientation (Astérisque 126 (1985), 41–43.)
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http://mathoverflow.net/questions/119901?sort=newest
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## Kernel elements for the Grothendieck group map of a commutative monoid
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This is just a nomenclature question. Let $T$ be a commutative monoid, and let `$T^*$` be its Grothendieck group. That is, $T^* \cong T \times T \ / \sim$, where $(s,s') \sim (t, t')$ if $s+t'+e = s'+t+e$ for some $e \in T$.
Let $i : T \to T^*$ be the natural inclusion map, where $i(t) = [(t,0)]$, and let $K$ denote the kernel of this map.
Does the kernel $K \subseteq T$ have a common name in the literature? What are elements $k \in K$ called?
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I assume by the kernel you mean the inverse image of 0. It is not usual to use the term kernel for the inverse image of the identity in semigroup theory since in general it is not interesting. The fibers of a semigroup homomorphism can all have radically different cardinality. I know of no name for $K$ in semigroup theory although perhaps in K-theory or some place else it has a name. – Benjamin Steinberg Jan 26 at 3:27
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Here's a negative result: no name given in Ogus's log geometry book. – S. Carnahan♦ Jan 26 at 10:35
## 1 Answer
I don't know a name for $K$ either, but here is a suggestion.
The correct definition of $\sim$ is $(a,b) \sim (c,d) \Leftrightarrow \exists e \in T ~ (a+d+e=b+c+e)$. In particular $(a,b) \sim (0,0) \Leftrightarrow \exists e \in T : a+e=b+e$. I would suggest to call such elements $a,b$ stably equal. Remark that this coincides with the terminology stably isomorphic from topological $K$-theory (where $T$ is the commutative monoid of $\cong$-classes of vector bundles on a space $X$ and one usually takes w.l.o.g. $e$ to be the class of a trivial vector bundle when $X$ is paracompact). In particular, $a \in K \Leftrightarrow \exists e \in T : a+e=e$, in which case one may say that $a$ is stably zero.
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Thanks for the correction & suggestion, Martin! – Tom LaGatta Jan 27 at 4:18
Similarily one could say that for a function $f : X \longrightarrow X$ two elements $x,y \in X$ are stably equal if $f^n(x)=f^n(y)$ for some $n \in \mathbb{N}$ (which means that they are equal in the localization $X_f$). – Martin Brandenburg Jan 27 at 10:41
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