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http://physics.stackexchange.com/questions/48048/how-can-crystal-structures-be-determined-using-x-ray-diffraction	# How can crystal structures be determined using X-ray diffraction? You have the intensity peaks and the diffraction angles. Let's say you suspect the material is cubic, how do I find if it's simple cubic or BCC or FCC? I've googled and all my textbooks just state that xrd is really nifty for determining crystal structure but no further information is presented. - Is this single crystal XRD, or powder XRD? The exact method is a bit different for the two. – Colin McFaul Feb 22 at 1:21 ## 2 Answers X-ray diffraction ends up being a beautiful application of Fourier theory. You are measuring the $k$-space (reciprocal lattice space) points with the x-ray diffraction. The Fourier transform of these points (provided you measured enough of them) gives you exactly the Bravais lattice. The Bravais lattice is the crystal stucture! You then have to look at the Bravais lattice and construct a Wigner-Seitz unit cell, then that construction will determine your crystal type. Alternatively, you could analyze the reciprocal lattice space "unit cell," find what it looks like, then Fourier transform that cell to get the Bravais lattice cell type. The Fourier transforms here is on an infinite periodic array, if that wasn't clear. The point arrangement actually determines the FCC or BCC type...I have included a diagram of the two different types (FCC and BCC) shown in the reciprocal lattice space below, taken from Introduction to the Physics of Electrons in Solids by Henri Alloul. You can see that they are not the same... - For BCC and FCC some of the lines are missing, so if you manage to index all the lines in your diffraction pattern you can tell immediately what type of cubic lattice you have. A quick Google found http://www.matter.org.uk/diffraction/x-ray/indexing_powder_pattern.htm and this is a pretty good introduction to the subject though anyone who has tried to do this in anger will knw that things are rarely as clear cut as the article suggests! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9138855934143066, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/35038/lorentz-invariance-of-maxwell-equations/35073	# Lorentz Invariance of Maxwell Equations I am curious to see a simple demonstration of how special relativity leads to Lorentz Invariance of the Maxwell Equations. Differential form will suffice. - ## 2 Answers Due to the constraint $\nabla \cdot B =0$, there exists a vector potential $A$ such that $B = \nabla \times A$ and $E_j = \partial_0 A_j - \partial_j A_0$ (up to a sign I forget). In other words, $E$ and $B$ assemble into a "field strength tensor" $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu = dA_{\mu \nu}$. This is the correct object to reason about when thinking relativistically. It's just a 2-form, so its transformation rules are simple. We can write Maxwell's equations as $d * F = 0$, here $*$ is the Hodge star. This is clearly invariant under isometries (Lorentz transformations). - 1 – Nick Kidman Aug 27 '12 at 18:56 I feel this answer, although accepted, doesn't relate to the role of the Lorentz transformations, does it? – Nick Kidman Aug 27 '12 at 21:30 Lorentz transformations are just isometries. What I meant above is that the equations can be written so that they only depend on the metric structure, so a diffeomorphism fixing the metric gives an automorphism of the space of solutions. These equations have this property for any metric, not just Minkowski's. Did I misunderstand your comment? – user404153 Aug 27 '12 at 21:40 2 "a diffeomorphism fixing the metric"? Anyway, you rewrote it geometrically, and so it's coordinate intedepend, i.e. diffeomorphism invariant. But this doesn't in any way highlight the role of Lorentz transformations for the Maxwell equations in special relativity. (Btw. the isometries are actually the whole Poincare group.) – Nick Kidman Aug 27 '12 at 22:23 Arbitrary diffeomorphisms don't induce automorphisms of the space of solutions. Take for example a diffeomorphism reversing the signature. I think this does not even fix the dimension. This is just a notational difficulty in the physics literature. If you make a change of coordinates and say that the metric transforms in the way you'd expect, this is an isometry generated by a diffeomorphism. Lorentz invariant (or Poincare invariant, if it matters) should just mean isometries => automorphisms of the space of solutions. Sometimes the easiest way to show it is to just rewrite the equations. – user404153 Aug 28 '12 at 9:44 show 1 more comment Another way to see it is when deriving the EM wave equation from Maxwell equations. The Lorentz Invariance means that the amplitude should be symmetric under translations of space and time and rotations. In the wave eq. you have only 2nd derivatives (time and space), hence symmetry under Lorentz is preserved. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9032312035560608, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/116679/minimax-theorems-v-s-fixed-point-theorems?answertab=active	# Minimax Theorems V.S. Fixed Point Theorems. Is there any relationship between the minimax theorems $$\mbox{Regularly hypothesis on } f:U\times V\to\mathbb{R} \implies \min_{u}\max_{v}f(u,v)=\max_{v}\min_{u}f(u,v)$$ and fixed point theorems $$\mbox{Regularly hypothesis on } F:X\to X \implies F(x)=x, \mbox{ for some } x\in X ?$$ More precisely, there is some kind teorem minimax implies some fixed point theorem? Or fixed point theorem implies some type minimax theorem? - ## 1 Answer One direction is given historically in game theory. Nash's result on the existence of Nash equilibria, essentially a fixed point theorem, implies the minimax theorem of von Neumann. Here is the coarse structure of the argument. We want to rewrite $$f:U\times V\to\mathbb{R} \implies \min_{u}\max_{v}f(u,v)=\max_{v}\min_{u}f(u,v)$$ as the solution to a fixed point problem. For this we have to generalize the idea of a fixed point to correspondences, or set-valued mappings. If $\phi:S\to 2^S$ maps points in $S$ to subsets of $S$, we say that $s$ is a fixed point of $\phi$ if $s\in\phi(s)$. One can often prove fixed point theorems for correspondences from fixed point theorems for functions by using selection or approximation theorems. So we define a correspondence $F:U\times V\to 2^{U\times V}$ by letting $$F(u,v)=\{u'\in U:u'\text{ minimizes } f(\cdot,v)\}\times\{v'\in V:v'\text{ maximizes } f(u,\cdot)\}.$$ Now the fixed points of $F$ are exactly the solutions to the minimax problem. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8694159984588623, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/130695/show-that-the-following-is-logically-valid	# show that the following is logically valid Let $x,y$ be variables, $A(x,y)$ a formula in which both $x$ and $y$ occur free. Show that $$\forall x \Big(\forall y\big(A(x,y)\big)\Big) \to \forall y \Big(\forall x\big(A(x,y)\big)\Big)$$ is logically valid - 7 In what sense: Any model for the left is also a model for the right? Or do you have some deduction rules you are supposed to use for this proof? (We have no way of guessing them...) – GEdgar Apr 12 '12 at 1:11 3 @GEdgar: You want to tell me that you're not a psychic?! I demand to talk to your supervisor! This is an outrage!! Professional mathematicians that cannot read minds?! Wait until the press hears about this! – Asaf Karagila Apr 12 '12 at 1:13 1 – jwodder Apr 12 '12 at 1:30 1 Mark13426, before the comments of others you may not have known that how you prove this depends on what formal proof system you use. Please don't feel discouraged that you didn't realize this. We all didn't know this at some point. That said, the very nature of a formal proof system requires that proofs in the object language stick rigorously to the rules and/or axioms of the system, or that slightly informal proofs can very easily get made into formal proofs (conversion of abbreviated wffs to actual wffs). Otherwise, the purported formal proof system is simply not formal. – Doug Spoonwood Apr 12 '12 at 2:00 ## 1 Answer In any structure $M$, both formulas hold if and only if $M \models A(x,y)$ for all $x,y \in \|M\|$. Thus $M \models (\forall x)(\forall y)[A(x,y)] \leftrightarrow (\forall y)(\forall x)[A(x,y)]$. Because that last formula is satisfied by every structure, by definition it is logically valid. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9486870765686035, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=396489	Physics Forums ## Coupled oscillators - mode and mode co-ordinates For this question I'm not going to introduce the particular problem I am working on, rather, I am merely wanting some explanation of a concept which I can't seem to find in any of my textbooks. I suspect the authors think it is just too obvious to bother explaining . I'm revising for a test and have the full worked solutions for this problem in front of me. I can follow the mathematics, but not the reasoning behind it. The question: Two masses M1 and M2 are connected by springs as in my expertly drawn diagram attached. Show that the amplitude of the displacement of the masses is described by expressions of the form: $$\psi _{1}\left( t \right)=A_{0}\cos \omega _{+}t$$ $$\psi _{2}\left( t \right)=A_{0}\cos \omega _{+}t$$ My worked solutions now say: Notice that when the system is in mode 1, the quantity (x2 - x1) is always zero, and (x1 + x2) varies harmonically. In mode 2 the reverse is true. Let us define a set of variables: $$q_{1}=\sqrt{\frac{m}{2}}\left( \psi _{2}+\psi _{1} \right)\; -->\; \dot{q}_{1}=\sqrt{\frac{m}{2}}\left( \dot{\psi }_{2}+\dot{\psi }_{1} \right)$$ $$q_{2}=\sqrt{\frac{m}{2}}\left( \psi _{2}-\psi _{1} \right)\; -->\; \dot{q}_{2}=\sqrt{\frac{m}{2}}\left( \dot{\psi }_{2}-\dot{\psi }_{1} \right)$$ My question: What exactly are q1 and q2, and why should these be equal to $$\sqrt{\frac{m}{2}}\left( \psi_{2}+\psi_{1} \right)$$ etc? Why $$\sqrt{\frac{m}{2}}$$? Is there a more specific name for this law that I could look up? I hope my question is easily understandable! Thank you for your help. (note: for the sets of equations relating q1 and q2 to m and x, there should be a "≡" sign rather than an "=" sign - for some reason my TEX formatting comes out with "8801;" rather than a "≡" sign. Odd.) Attached Thumbnails PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Formulae now fixed. I hope. Sorry if I confused anyone while I was editing things Thread Tools \| \| \| \| \|-----------------------------------------------------------------------\|------------------------\|---------\| \| Similar Threads for: Coupled oscillators - mode and mode co-ordinates \| \| \| \| Thread \| Forum \| Replies \| \| \| Classical Physics \| 3 \| \| \| General Physics \| 1 \| \| \| Electrical Engineering \| 0 \| \| \| Cosmology \| 2 \| \| \| General Physics \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9329304099082947, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/38615/proof-of-canonical-commutation-relation-ccr/38620	# Proof of Canonical Commutation Relation (CCR) I am not sure how $QP-PQ =i\hbar$ where $P$ represent momentum and $Q$ represent position. $Q$ and $P$ are matrices. The question would be, how can $Q$ and $P$ be formulated as a matrix? Also, what is the proof of this canonical commutation relation? - 1 This is a key axiom of matrix mechanics. It can be demonstrated in wave mechanics, but then you ask "why is p a differential operator?". The derivation is heuristic, it only works within the old quantum theory, and it is reproduced on Wikipedia's "matrix mechanics" page. – Ron Maimon Sep 29 '12 at 16:55 ## 3 Answers As Lubos has mentioned $QP-PQ=i\hbar$ is one of the basic requirements of quantum mechanics. Classically observables are functions of variables $q$, and $p$ and Poisson bracket relation read $\{q,p\}=1$ (note that $\{q,p\}$ is unitless quantity ) In QM observables are required to be hermitian operators (so that they can have real eigenvalues). In particular for position we have an operator $Q$, and for momentum we have an operator $P$. Poisson bracket is replaced by commutator and we require $[Q,P]=i\hbar$ In analogy with classical Poisson brackets we would have required $[Q,P]=1$ But this is not possible since i) We have already required that $Q,P$ be hermitian. So $[Q,P]^\dagger=(QP-PQ)^\dagger=(QP)^\dagger-(PQ)^\dagger=PQ-QP=-[Q,P]$. So if we require $[Q,P]$ to be a constant (i.e. constant multiple of Identity matrix) it should be purely imaginary. ii) $[Q,P]$ has units of $ML^2T^{-1}$. You can see this by noting that $Q$ is a position operator, so it has units of $L$, and that $P$ is a momentum operator, so it has units of $MLT^{-1}$. Two natural choices are $[Q,P]=i\hbar$ and $[Q,P]=-i\hbar$. They are both equivalent and choice of $[Q,P]=i\hbar$ is just a convention. No two finite dimensional matrices can satisfy $[Q,P]=i\hbar$. This can be seen by taking trace on both sides. However this relation can be satisfied by infinite dimensional matrices. More explicitly take vector space to be space of functions of $q$. Define $Q$ as $Qf=qf$, and $P$ as $Pf=-i\hbar\partial f/\partial q$. Then it can be seen that these operators satisfy required commutation relation. Moreover if we define our inner product as $(f,g)=\int f^* g\: dq$ then $Q$ and $P$ defined as above will also be hermitian. - Just to check: how do we get that $[Q,P]$ has units of $ML^2T^{-1}$? – War Sep 29 '12 at 10:58 @War $Q$ is position operator so it is required to have units of length $L$. Similarly $P$ is required to have units of momentum $MLT^{-1}$. So $QP-PQ$ should have dimensions of $ML^2T^{-1}$. – user10001 Sep 29 '12 at 11:06 oh, stupid me. thanks. – War Sep 29 '12 at 11:11 1 @War There is a correction to be made in last para of above answer. Definition of $P$ should be $Pf=-i\hbar\partial f/\partial q$. – user10001 Sep 29 '12 at 11:44 @dushya, I incorporated the content of your comments into the answer. I hope that this is okay. – Colin McFaul Sep 29 '12 at 16:01 show 1 more comment One may formulate more general rules that determine the commutators of operators in a quantum theory obtained from a Lagrangian or a Hamiltonian. However, there always have to be some axioms. In the simplest models of quantum mechanics, it's most legitimate to say that $xp-px=i\hbar$ is simply a key axiom of quantum mechanics, so it can't be derived from anything "deeper". The fact that they don't commute – that their product depends on the order – implies that $x,p$ cannot be ordinary numbers. Instead, they are operators: operators don't have to commute. An $\hat L$ operator is something that assigns to each vector $\|\psi\rangle$ of a space – in the case of quantum mechanics, Hilbert space – the result $\hat L \|\psi \rangle$. If one chooses a basis of the space with finitely or countably many elements, all the information about a linear operator may be expressed in terms of matrix elements $\langle i\|\hat L \|j\rangle = b(j,L(i))$ and the set of these inner products i.e. matrix elements is a matrix. This picture of physics is consistent but the consistency has many aspects so it's a broad question. In essence, you want to explain why everything in quantum mechanics works. Well, it does but it isn't quite a 1-line proof. Quantum mechanics is the right theory of everything so it would be unwise to expect 1-line proofs of its consistency or validity. If you have some more particular concern or hypothetical consistency, please feel to update your question. - In a matrix representation, $Q$ and $P$ are infinite-dimensional matrices. (Finite-dimensional matrices wouldn't do; the trace of the two sides of the commutation relation gives a contradition.) Many possible pairs of matrices qualify; the nicest ones are obtained when expressing position and momentum in a basis of eigenstates of the harmonic oscillator. See http://en.wikipedia.org/wiki/Matrix_mechanics#Harmonic_oscillator Indeed, this is Heisenberg's original representation for ''matrix mechanics''.) The more frequently used position representation (or momentum representation) takes $Q$ (resp. $P$) as a multiplication operator on wave functions depending on position (or momentum), and $P$ (resp. $Q$) as a first order differential operator chosen to match the commutation relation. To prove the relation when $Q$ and $P$ are given, simply apply both sides to an arbitrary state vector and check that one gets the same result. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 61, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9299567937850952, "perplexity_flag": "head"}
http://mathoverflow.net/questions/116808?sort=newest	## visualizing singularities of maps from sphere to R^2 ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a classification of singularities from $S^2 \to \mathbb{R}^2$ ? The critical points of the map $(x,y) \mapsto (f_1(x,y),f_2(x,y))$ where the matrix: $\left[\begin{array}{cc}\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y}\\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{array} \right]$ has less than full rank. Locally, can we draw a picture of what singularities look like? In the case of maps $S^2 \to \mathbb{R}^1$, we just get the critical points where $f(x,y) = f(x_0,y_0) + (x-x_0,y-y_0)^T (D^2 f )(x-x_0,y-y_0)$ - ## 1 Answer I think you want to look at Guillemin and Golubitsky's book Stable mappings and their singularities, which has a thorough description of what the singularity types of stable mappings are between surfaces. Basically, the only stable singularities for smooth maps between surfaces (i.e., $2$-manifolds) are folds and cusps, and these cannot usually perturbed away by small perturbations. If you don't restrict to stable mappings or some similar class, the kinds of singularities that can occur can be extremely complicated, and I doubt that there is any workable classification. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.928385853767395, "perplexity_flag": "middle"}
http://en.wikiversity.org/wiki/Talk:Gravity	# Talk:Gravity From Wikiversity ## Exercise The Apollo Command Module On July 20, Neil Armstrong and Buzz Aldrin became the first humans to land on the Moon. While Armstrong and Aldrin did important research on the surface of the moon, the third astronaut, Michael Collins orbited in the command module above. Determine the height (read: distance from the surface of the moon) at which Collins orbited in the command module by exploiting the following data. The moon has been approximated to weigh 7,3477×1022kg, and the radius to 1 737,10 km. The mass of the command module was 30,320 kg. Collins stayed in lunar orbit for 59 h 30 m. The command module did exactly 30 laps around the moon in this time. Assume that the orbit was circular and that the command module was very light compared to the moon. [Solution] The command module which stays in lunar orbit is constantly accelerating towards the moon, since a force acts on it. However, since the module has a large velocity component which is right to the gravitational force component (see schematic), it manages to overcome the acceleration towards the moon, so that the distance to the surface remains constant. It is then constantly falling without ever reaching the surface. Imagine throwing a rock. Eventually it will hit the ground since a gravitational force acts on it. But if you were to throw the rock fast enough the earth would "curve away" from the stone as it moves forward, and thus the rock would never hit the ground. The rock would then be in orbit. Let $m_1$ be the command module's mass, and $m_2$ the moon's mass. From Newton's Law of Universal Gravitation we get $F_G=G\frac{m_1m_2}{r^2}\,$ Since the command module is in an orbit around the moon, we can also express the gravitational force which acts on the command module as $F_G=m_1a_n$ By substituting this into Newton's Law of universal Gravitation we get $m_1a_n=G\frac{m_1m_2}{r^2}\,$ By dividing the equation with $m_1$ we get. $a_n=G\frac{m_2}{r^2}$ By rewriting the centripetal acceleration $a_n$ we get $\frac{v^2}{r}=G\frac{m_2}{r^2}\,$ By solving the equation on $v^2$ we get $v^2=G\frac{m_2}{r}$ We can express the velocity of the command module in an other way. From the relation of velocity and angular velocity we get $v=r\omega$ And from the relation of orbital period and angular velocity we get $T=\frac{2\pi}{\omega}$ (The orbital period $T$ is equal to the time it takes for the Command Module to to one lap around the moon.) By combining both expressions and solving for $v$ we get $v=\frac{2\pi r}{T}$ By substituting this expression into the expression we have from Newton's Law of Universal Gravitation we get $\mathbf{\left(\frac{2\pi r}{T}\right)}^2=G\frac{m_2}{r}$ By solving for $r$ we get $r=\sqrt[3]{\frac{Gm_2T^2}{(2\pi)^2}}$ Notice that the height at which the Command Module orbits the moon doesn't depend on the modules mass. As seen in the schematic, the value $r$ in the equation stands for the distance between the center of the moon and the center of the Command Module. We want to know the distance "$h$" to the surface, which naturally is $h=r-radius_{moon}$ Notice that we are making a small mistake by stating the above. We are actually calculating the distance from the center of the command module to the surface, not the distance from the side closest to the surface. This error is, however, so small that it makes no difference. This gives $h=\sqrt[3]{\frac{Gm_2T^2}{(2\pi)^2}}-radius_{moon}$ Inserting values gives $h=\sqrt[3]{\frac{(6,67428 \times 10^{-11} \ \mbox{m}^3 \ \mbox{kg}^{-1} \ \mbox{s}^{-2}) (7,3477 \times 10^{22}kg) \mathbf{\left(\frac{59 \cdot 60^2 \text{s}+30\cdot 60 \text{s}}{30}\right)}^2}{(2\pi)^2}} -1 737,10 \times 10^{3}m$ $\approx{113} \text{ km}$ --Ofey 20:23, 7 November 2008 (UTC)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9317294359207153, "perplexity_flag": "head"}
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I have decided to fix this lacuna once for ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 66, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9255909323692322, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/26665/given-an-alphabet-with-26-letters-how-many-reduced-alphabets-are-possible	# Given an alphabet with 26 letters, how many reduced alphabets are possible? I have an alphabet with 26 English letters. I can reduce it to 25 letters by representing two of the letters (e.g. O and X) with a new letter (e.g. $\otimes$). # 25 To get down to a 25 letter alphabet I have $26 \choose 2$ possibilities. So 325 possible reductions to alphabet of size 25. # 2 Getting down to an alphabet of size 2 is also pretty easy - I just need to count all possible ways of dividing the alphabet into 2 partitions. That would be $26 \choose 1$ + $26 \choose 2$ + ... + $26 \choose 13$. In the same way I can reduce my alphabet to any size from 1 to 25. That's 38,754,731 (computed in R like this `sum(choose(26,1:13))`). ## The Rest How can do the counting for size 3 or 24? What is the total number of such alphabet reductions? - You have double counted the ways to get to 2. $\binom{26}{2}$ and $\binom{26}{24}$ are counting the same thing. And what happens at 13? – Ross Millikan Mar 13 '11 at 15:57 Ross, you're exactly right, at 13 you are splitting into 2 partitions of the same size. I realized that too after sleeping it over. – Aleksandr Levchuk Mar 13 '11 at 18:48 So you have to divide the $\binom{26}{13}$ cases by 2. A simpler way to look at it is that you have to pick some subset to be the first one, everything else goes in the second, and neither the first or second can be empty. This is $2^{26}-2$ choices for the first set, but you have counted every split exactly twice. So the number of ways is $2^{25}-1$ – Ross Millikan Mar 13 '11 at 18:52 ## 1 Answer You're probably looking for the Bell numbers. - I need a slight variation of the Bell numbers. In bell numbers $B_3 = 5$ as $\{a, b, c\}$ partitions into (#1) $\{\{a\}, \{b\}, \{c\}\}$, (#2) $\{\{a\}, \{b, c\}\}$, (#3) $\{\{b\}, \{a, c\}\}$, (#4) $\{\{c\}, \{a, b\}\}$, and (#5) $\{\{a, b, c\}\}$. I need $B'_3 = 4$ because for my case partition #1 {{a}, {b}, {c}} is not meaningful - I need the size of the derived alphabet to be strictly less. – Aleksandr Levchuk Mar 13 '11 at 7:41 1 Then subtract 1. – Yuval Filmus Mar 13 '11 at 7:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9206717014312744, "perplexity_flag": "middle"}
http://mathpages.blogspot.com/2008/02/counting-points.html	Math Pages Blog God used beautiful mathematics in creating the world. Wednesday, February 13, 2008 Counting the points In the previous post, I wrote about an ancient paradox. The author of this paradox, Zeno of Elea, presented 7 paradoxes. In this post I will focus on another of his paradoxes. This paradox arises from the following situation. Suppose you need to go to a shop to buy some milk. You live very near to the shop so you decide to take a walk. Lets suppose the shop is 100 meters away. Photo by lime*monkey If you read my previous post, you probably already guessed that Zeno claimed that it is impossible for you to get to the shop... Where is the problem? Lets look on it in the following way - in order to get to the shop you need firstly to pass half of the distance to the shop. But even before this you have to pass a quarter of the distance. And before this an eighth of the distance... And so on. It turns out that in order to get to the shop you need to pass an infinite number of such mid points. Zeno reasoned that this is impossible, because you would need an infinite amount of time for this. Therefore you cannot move at all. No matter how small the distance you move we can divide it in the some way and get an infinite number of mid points again. Conclusion - movement is only an illusion. There is a simple solution to this paradox. Firstly, if we will sum all the time intervals the sum will be a finite number. This is the accepted solution to the paradox. But this is not a full solution. In order to sum the time intervals we need to know two things - distance and speed. We know the distance, it is given in the problem. But we don't know the speed. We can decide that the man in the problem has a speed of ten meters per minute. If we would do so, it is trivial to find how much time it would take him to get to the shop: $%5Cfrac%7B100%7D%7B10%7D=10%20minutes$ And for all the mid points we would get the following infinite series: 5+2.5+1.25+ .... =$%5Cfrac%7B5%7D%7B1-%5Cfrac%7B1%7D%7B2%7D%7D=10$ As you can see the sum of all the intervals is indeed a finite number. But can we assume that speed exists at all? The whole point of the paradox is to show that motion is only an illusion. By assuming that something called speed exists we assume that the paradox is wrong from the beginning. What is interesting in this paradox is that it doesn't describe our reality as we know it. There is a hidden assumption in the paradox that the distance can always be divided. But this is not true from a physical prospective. Our world is discrete - there is a basic unit of length called Planck length. Below this length distance is simply meaningless. So in our world we have only a finite number of points between us and the milk shop, so there is no problem to get there in a finite time. Posted by Anatoly Labels: math, paradoxes	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 2, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9640993475914001, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/50988-zeros.html	# Thread: 1. ## Zeros Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there? How do I figure this out? 2. There can be as many zeros as the number of the degree so there are 3 or less zeros the descartes rule of signs states that if you count the sign changes of the function, the number of sign changes (or minus two of that number) is the number of positive roots (or roots at 0) so x cubed - 2x has 1 sign change, so there must be 1 positive zero (if there are 3 sign changes, however, there could be 3 OR 1 positive zero) take f(-x) and count the sign changes to find out the number of negative zeros it becomes -x cubed + 2x and there is 1 sign change again, so there must be one negative zero because there must be 3 zeros total, and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero EDIT oh you said rational and irrational (not complex) woops i wasn't paying attention might not have answered your question IN THAT CASE: we have x cubed - 2x factor out an x x (x squared - 2) x=0 is a rational root x squared=2 absolute value of x = sq. root of 2 x = positive or negative sq. root of 2 so there is a root at 0, sq. root of 2, and negative sq. root of 2 so 2 irrational and 1 rational roots wait a minute where is the complex root? can anyone answer my question now? i've never had this happen before... 3. Hello, Originally Posted by juldancer Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there? How do I figure this out? $x^3-2x=x(x^2-2)$ use the difference of 2 squares : $x^2-2=(x-\sqrt{2})(x+\sqrt{2})$ Hence $x^3-2x=x(x-\sqrt{2})(x+\sqrt{2})$ ---------------------------------------- In order to know how many there are, you can use Descartes' rule of signs : http://www.purplemath.com/modules/drofsign.htm 4. Originally Posted by juldancer Given a cubic equation, such as x^3-2x, how many rational and irrational zeros are there? How do I figure this out? $x^3-2x=x(x-\sqrt{2})(z+\sqrt{2})$ So now you tell us how many rational and irrational roots there are. RonL 5. Can anyone answer my error? I factored correctly and I've used DesCartes rule of Signs perfectly billions of times before...maybe I've just forgotten how to do it correctly but I can't find where I made my error 6. Hi mikedwd, and we know there is 1 positive and 1 negative zero, we also know that there is one complex zero This is not possible. If the coefficients of a polynomial are real, then if there is a complex zero, there is another complex zero, namely its conjugate. The mistake is, in my opinion, in the fact that you count 0 as +0 and not -0. but I'm sorry, I don't know this rule enough to be able to spot more precisely the error... 7. Originally Posted by mikedwd Can anyone answer my error? I factored correctly and I've used DesCartes rule of Signs perfectly billions of times before...maybe I've just forgotten how to do it correctly but I can't find where I made my error There is one change of sign and one positive root. No problem. (a zero root added to a polynomial does not alter the number of times that the signs change, and so is undetectable by the rule of signs) RonL 8. hmm I'm sure I was told that a zero root is included with positive roots in the rule of signs, but I suppose that was wrong anyway yes I just realized (im a bit slow today apparently) that you cannot have 1 complex zero...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9567078351974487, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Discriminant_function	# Discriminant function analysis From Wikipedia, the free encyclopedia (Redirected from Discriminant function) Jump to: navigation, search Discriminant function analysis is a statistical analysis to predict a categorical dependent variable (called a grouping variable) by one or more continuous or binary independent variables (called predictor variables). The original dichotomous discriminant analysis was developed by Sir Ronald Fisher in 1936[1] It is different from an ANOVA or MANOVA, which is used to predict one (ANOVA) or multiple (MANOVA) continuous dependent variables by one or more independent categorical variables. Discriminant function analysis is useful in determining whether a set of variables is effective in predicting category membership[2] Discriminant analysis is used when groups are known a priori (unlike in cluster analysis). Each case must have a score on one or more quantitative predictor measures, and a score on a group measure.[3] In simple terms, discriminant function analysis is classification - the act of distributing things into groups, classes or categories of the same type. Moreover, it is a useful follow-up procedure to a MANOVA instead of doing a series of one-way ANOVAs, for ascertaining how the groups differ on the composite of dependent variables. In this case, a significant F test allows classification based on a linear combination of predictor variables. Terminology can get confusing here, as in MANOVA, the dependent variables are the predictor variables, and the independent variables are the grouping variables.[2] ## Assumptions The assumptions of discriminant analysis are the same as those for MANOVA. The analysis is quite sensitive to outliers and the size of the smallest group must be larger than the number of predictor variables.[3] • Multivariate normality: Independent variables are normal for each level of the grouping variable.[2][3] • Homogeneity of variance/covariance (homoscedasticity): Variances among group variables are the same across levels of predictors. Can be tested with Box's M statistic.[2][page needed] It has been suggested, however, that linear discriminant analysis be used when covariances are equal, and that quadratic discriminant analysis may be used when covariances are not equal.[3] • Multicollinearity: Predictive power can decrease with an increased correlation between predictor variables.[3] • Independence: Participants are assumed to be randomly sampled, and a participant’s score on one variable is assumed to be independent of scores on that variable for all other participants.[2][3] It has been suggested that discriminant analysis is relatively robust to slight violations of these assumptions,[4] and it has also been shown that discriminant analysis may still be reliable when using dichotomous variables (where multivariate normality is often violated).[5] ## Discriminant functions Discriminant analysis works by creating one or more linear combinations of predictors, creating a new latent variable for each function. These functions are called discriminant functions. The number of functions possible is either Ng-1 where Ng = number of groups, or p (the number of predictors), whichever is smaller. The first function created maximizes the differences between groups on that function. The second function maximizes differences on that function, but also must not be correlated with the previous function. This continues with subsequent functions with the requirement that the new function not be correlated with any of the previous functions. Given group $j$, with $\mathbb{R}_j$ sets of sample space, there is a discriminant rule such that if $x \in\mathbb{R}_j$, then $x\in j$. Discriminant analysis then, finds “good” regions of $\mathbb{R}_j$ to minimize classification error, therefore leading to a high percent correct classified in the classification table.Hardle, W., Simar, L. (2007). Applied Multivariate Statistical Analysis. Springer Berlin Heidelberg. pp. 289-303. Each function is given a discriminant score to determine how well it predicts group placement. • Structure Correlation Coefficients: The correlation between each predictor and the discriminant score of each function. This is a whole[clarification needed] correlation.Garson, G. D. (2008). Discriminant function analysis. http://www2.chass.ncsu.edu/garson/pa765/discrim.htm . • Standardized Coefficients: Each predictor’s unique contribution to each function, therefore this is a partial correlation. Indicates the relative importance of each predictor in predicting group assignment from each function. • Functions at Group Centroids: Mean discriminant scores for each grouping variable are given for each function. The farther apart the means are, the less error there will be in classification. ## Discrimination rules • Maximum likelihood: Assigns x to the group that maximizes population (group) density.[6] • Bayes Discriminant Rule: Assigns x to the group that maximizes $\pi_i f_i(x)$, where $f_i(x)$ represents the prior probability of that classification, and πi represents the population density.[6] • Fisher’s linear discriminant rule: Maximizes the ratio between SSbetween and SSwithin , and finds a linear combination of the predictors to predict group.[6] ## Eigenvalues An eigenvalue in discriminant analysis is the characteristic root of each function.[clarification needed] It is an indication of how well that function differentiates the groups, where the larger the eigenvalue, the better the function differentiates.[3] This however, should be interpreted with caution, as eigenvalues have no upper limit.[2][3] The eigenvalue can be viewed as a ratio of SSbetween and SSwithin as in ANOVA when the dependent variable is the discriminant function, and the groups are the levels of the IV[clarification needed].[2] This means that the largest eigenvalue is associated with the first function, the second largest with the second, etc.. ## Effect size Some suggest the use of eigenvalues as effect size measures, however, this is generally not supported.[2] Instead, the canonical correlation is the preferred measure of effect size. It is similar to the eigenvalue, but is the square root of the ratio of SSbetween and SStotal. It is the correlation between groups and the function.[2] Another popular measure of effect size is the percent of variance[clarification needed] for each function. This is calculated by: (λx/Σλi) X 100 where λx is the eigenvalue for the function and Σλi is the sum of all eigenvalues. This tells us how strong the prediction is for that particular function compared to the others.[2] Percent correctly classified can also be analyzed as an effect size. The kappa value[clarification needed] can describe this while correcting for chance agreement.[2] ## Variations • Multiple discriminant analysis (MDA): related to MANOVA. Has more than two groups, and uses multiple dummy variables.[7] • Sequential discriminant analysis: assesses the importance of a set of IVs over and above a set of controls. In this case, the controls are entered first, and then the IVs.[7] • Stepwise discriminant analysis: Selects the most correlated predictor first, removes that variance in the grouping variable then adds the next most correlated and continues until the change in canonical correlation is not significant. Of course, both forward and backward stepwise procedures may be performed.[7] ## Comparison to logistic regression Discriminant function analysis is very similar to logistic regression, and both can be used to answer the same research questions.[2] Logistic regression does not have as many assumptions and restrictions as discriminant analysis. However, when discriminant analysis’ assumptions are met, it is more powerful than logistic regression.[citation needed] Unlike logistic regression, discriminant analysis can be used with small sample sizes. It has been shown that when sample sizes are equal, and homogeneity of variance/covariance holds, discriminant analysis is more accurate.[3] With all this being considered, logistic regression is the common choice nowadays, since the assumptions of discriminant analysis are rarely met.[1][3] ## See also Wikiversity has learning materials about Discriminant function analysis ## References 1. ^ a b Cohen et al. Applied Multiple Regression/Correlation Analysis for the Behavioural Sciences 3rd ed. (2003). Taylor & Francis Group. 2. Green, S.B. Salkind, N. J. & Akey, T. M. (2008). Using SPSS for Windows and Macintosh: Analyzing and understanding data. New Jersey: Prentice Hall. 3. BÖKEOĞLU ÇOKLUK, Ö, & BÜYÜKÖZTÜRK, Ş. (2008). Discriminant function analysis: Concept and application. Eğitim araştırmaları dergisi, (33), 73-92. 4. Lachenbruch, P. A. (1975). Discriminant analysis. NY: Hafner 5. Klecka, William R. (1980). Discriminant analysis. Quantitative Applications in the Social Sciences Series, No. 19. Thousand Oaks, CA: Sage Publications. 6. ^ a b c Hardle, W., Simar, L. (2007). Applied Multivariate Statistical Analysis. Springer Berlin Heidelberg. pp. 289-303. 7. ^ a b c	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.859724760055542, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/38401/spectrum-of-sum-of-operators-on-banach-spaces	# Spectrum of sum of operators on Banach spaces Let $A$ and $B$ be two operators on a Banach space $X$. I am interested in the relationship between the spectra of $A$, $B$ and $A+B$. In particular, are there any set theoretic inclusions or everything can happen in general like: $\sigma(A)\subset\sigma(A+B)$, and conversely, $\sigma(B)\subset\sigma(A+B)$, and conversely? If we know the spectra $\sigma(A)$, $\sigma(B)$ of $A$ and $B$, can we determine the spectrum of $A+B$? I would appreciate any comment or reference. - 1 – Qiaochu Yuan May 11 '11 at 5:36 ## 1 Answer You can find a lot of information about the spectrum of sums of operators in the book: • Tosio Kato: "Perturbation Theory of Linear Operators" Chapter Four "Stability theorems", paragraph 3 "Perturbation of the spectrum" (that's about linear operators on infinite dimensional Banach spaces, other cases are treated as well in the book). As shown there, it is possible to prove some theorems about the spectrum of $A + B$, where A is a known operator (you know something about its spectrum) and B is a perturbation of A in the sense that it is small compared to A in a certain way, which can be made precise in different ways with different uses. These theorems are the best results that I know which show that what happens to the spectrum of the sum is not completey arbitrary in certain settings. - www.jstor.org/stable/1997004 – ktata.khaled May 11 '11 at 14:29 Thanks, more precisely I would ask these two questions: 1. Theorems 3.1 and 3.2 concern results for the essential spectra of A+B and AB for A closed and B bounded. Does this reference (or some other reference) contain results like Theorem 3.1 and 3.2 for the USUAL spectrum (\sigma(A+B), \sigma(AB)) for A closed and B bounded? 2. What exactly are the results of this paper for essential spectra of A+B and AB in the case when A and B are bounded? – ktata.khaled May 11 '11 at 14:33 the reference is www.jstor.org/stable/1997004 – ktata.khaled May 11 '11 at 14:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9358564615249634, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/7253-convergents-fixed-points.html	# Thread: 1. ## Convergents and Fixed Points Hey all. Could someone check my work here and get back to me before 9 central time. THanks, GREEN! ***Part A: Let f(x) = 1 / (x^2 + 3x -2) Find the fixed points for f. I got x = 0.85106383, x = -0.3191489, and x = -3.510638. Are these correct? *Part B: Try different seeds to find out which, if any are convergent. To me it seems that it converges at -0.343379569, so according to this none are convergent. Could someone please check my answers. GREEN 2. Originally Posted by Mr_Green [B]Hey all. Could someone check my work here and get back to me before 9 central time. THanks, GREEN! ***Part A: Let f(x) = 1 / (x^2 + 3x -2) Find the fixed points for f. I got x = 0.85106383, x = -0.3191489, and x = -3.510638 Are these correct? A fixed point has to exist by the Brouwer Fixed Point theorem (somebody correct me if I am wrong). If we solve, $x=\frac{1}{x^2+3x-2}$ We get, $x^3+3x^2-2x-1=0$ The solutions to this equation are roots. 3. ok i checked my work, and i believe these are my new and improved answers ... part a: x = 0.834 x = -0.343 x = -3.491 part b: remains the same, except now one of the fixed points are convergent. How's this look guys? Here is part c now: Zoom in on a convergent point until the graph of f looks like a straight line (this phenomenon is known as "local linerity.") Find the coordinates of two points on f - one to the left and one to the right of the convergent point. Then find the slope (rise over run) of the segment connecting these two points. I would also like to check my slope with someone, so if you find it, please let me know. Thanks again, GREEN	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8980677127838135, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/287866/fat-cantor-set	# “Fat” Cantor Set So the standard cantor set has an outer measure equal to 0, but how can you construct a "fat" cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well... Are there other constraints that need to be made in order to accomplish this? - 4 The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.) – Rahul Narain Jan 27 at 4:06 If the bits you remove at each stage have total length less than $1$, then what's left has positive measure. – Gerry Myerson Jan 27 at 4:06 You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness. – user53153 Jan 27 at 4:09 You can, however, have outer measure arbitrarily close to one. – Brian M. Scott Jan 27 at 4:10 – Martin Jan 27 at 4:13 show 1 more comment ## 1 Answer Say you delete the middle third. Then delete middle sixths from the two remaining intervals. Then delete middle twelfths from the four remaining intervals. And so on. The amount you delete is $\displaystyle\frac13+\frac16+\frac{1}{12}+\cdots= \frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9613978266716003, "perplexity_flag": "middle"}
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Also, ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 92, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9183998703956604, "perplexity_flag": "head"}
http://programmingpraxis.com/2011/07/29/approximating-pi/?like=1&source=post_flair&_wpnonce=dd1ec55e91	Programming Praxis A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer Approximating Pi July 29, 2011 In the two previous exercises we have seen three different methods for computing the prime-counting function π(n), all based on a formula of Legendre. We have also seen the difficulty of the calculation, as three eminent mathematicians got their sums wrong, and the difficulty continues even today with the use of computers, as Xavier Gourdon had to stop an attempt to compute π(1023) because of an error. Given the difficulty of calculation, in some cases it might make sense to calculate an approximate value of π, instead of an exact value. In today’s exercise we look at two methods for approximating the prime-counting function. The first method dates to the German mathematician Carl Friedrich Gauss, who gives an estimate π(x) ∼ Li(x), the logarithmic integral—this is the celebrated prime number theorem, which Gauss figured out when he was fifteen years old! The logarithmic integral can be calculated by expanding the infinite series $\mathrm{Li}(x) = \int_0^x \frac{d\ t}{\log_e t} = \gamma + \log \log x + \sum_{k=1}^\infty \frac{(\log x)^k}{k!\ k}$ to the desired degree of accuracy; somewhere around a hundred terms ought to be sufficient for arguments up to 1012. Beware the singularity at x=0. In 1859, Bernhard Riemann, whose hypothesis about the prime numbers is one of the greatest open questions in mathematics, described a different, and much more accurate, approximation to the prime counting function: $\mathrm{R}(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \mathrm{Li}(x^{1/n})$ where μ(n) is the Möbius function that takes the value 1 when n is 1, 0 when the factorization of n contains a duplicated factor, and (−1)k, where k is the number of factors in the factorization of n, otherwise. There is no better way to compute the Möbius function than to compute the factors of n. As with the logarithmic integral, about a hundred terms of the infinite series ought to be sufficient to get a good approximation for arguments up to 1012. Your task is to write functions to compute the logarithmic integral, the Möbius function, and Riemann’s R function, and to compare the results to the values you calculated for the prime-counting function in the previous exercise. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Like this: Pages: 1 2 Posted by programmingpraxis Filed in Exercises 2 Comments » 2 Responses to “Approximating Pi” 1. Graham said August 1, 2011 at 2:51 PM Here’s my Python solution; it reuses `primes`, a variation of the Python Cookbook’s version of the Sieve of Eratosthenes. ```from math import log from operator import mul GAMMA = 0.57721566490153286061 def li(x): return GAMMA + log(log(x)) + sum(pow(log(x), n) / (reduce(mul, xrange(1, n + 1)) * n) for n in xrange(1, 101)) def factors(n): return (p for p in primes(n + 1) if n % p == 0) def mu(n): m = lambda p: -1 if (n / p) % p else 0 return reduce(mul, (m(p) for p in factors(n)), 1) def r(x): return sum(mu(n) / n * li(pow(x, 1.0 / n)) for n in xrange(1, 101)) ``` 2. David said August 25, 2012 at 3:36 AM Clojure version, uses O’Neil sieve to get list of primes. Results for Riemann function matches table above. ```(load "lazy-sieve") (defn factors "list of factors of n" [n] (loop [n n, prime lazy-primes, l (vector-of :int)] (let [p (first prime)] (cond (= n 1) l (= (mod n p) 0) (recur (/ n p) prime (conj l p)) :otherwise (recur n (rest prime) l))))) (defn calc-mu "Calculate mu(n) for use in building a static array, n > 1" [n] (let [fac (factors n)] (if (not= (count fac) (count (distinct fac))) 0 (reduce * (map (fn [x] -1) fac))))) (def mu (into (vector-of :int 0 1) (map calc-mu (range 2 101)))) (defn powers [x] (iterate (partial * x) x)) (defn float-factorials [] (map first (iterate (fn [[n,i]] [(* n i) (inc i)]) [1.0 2]))) (defn logint "Compute gauss' logarithmic integral to N terms" [n, x] (let [gamma 0.5772156649015329, ln-x (Math/log x), sum (reduce + (map #(/ %1 (* %2 %3)) (take n (powers ln-x)) (take n (float-factorials)) (range 1 (inc n))))] (+ gamma (Math/log ln-x) sum))) (defn riemann "Compute riemann function, max 100 terms" [x] (reduce + (map #(* (/ (mu %1) %1) (logint 100 (Math/pow x (/ 1 %1)))) (range 1 101)))) ``` %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8936589360237122, "perplexity_flag": "middle"}
http://electronics.stackexchange.com/questions/22291/why-exactly-cant-a-single-resistor-be-used-for-many-parallel-leds	Why exactly can't a single resistor be used for many parallel LEDs? Why can't you use a single resistor for a number of LEDs in parallel instead of one each? - 4 Answers The main reason is because you can't safely connect diodes in parallel. So when we use one resistor, we have a current limit for the whole diode section. After that its up to each diode to control the current that goes through it. The problem is that real world diodes don't have same characteristics and therefore there's a danger that one diode will start conducting while others won't. So you basically want this and you in reality get this. As you can see, in the first example, all diodes are conducting equal amounts of current and in the second example one diode is conducting mot of the current while other diodes are barely conducting anything at all. The example itself is a bit exaggerated so that the differences will be a bit more obvious, but nicely demonstrate what happens in real world. The above is written with assumption that you will chose the resistor in such way that is sets the current so that the current is n times the current you want in each diode where n is the number of diodes and that the current is actually larger than the current which a single diode can safely conduct. What then happens is that the diode with lowest forward voltage will conduct most of the current and it will wear out the fastest. After it dies (if it dies as open circuit) the diode with next lowest forward voltage will conduct most of the current and will die even faster than first diode and so on until you run out of diodes. One case that I can think of where you can use a resistor powering several diodes would be if the maximum current going through the resistor is small enough that a single diode can work with full current. This way the diode won't die, but I myself haven't experimented with that so I can't comment on how good idea it is. - 1 In practice, diodes are apt to have voltage drops which are fairly close to each other, and a voltage sufficient to push excessive current through the diode with the lowest drop will almost certainly push some current through the ones with higher drops. Depending upon how much variation there is in the LEDs and how "close to the edge" one wants to run them, it may be practical to place a few in parallel with a resistor. – supercat Nov 15 '11 at 22:20 I understand now, thank you! – Dataflashsabot Nov 15 '11 at 22:22 7 To make things worse, you can have thermal run away issues where as the diode starts to get hot from current going through it, its characteristics change such that it will have more current flow through it, thus making it even hotter. – Kellenjb Nov 15 '11 at 22:23 See my recent detailed answer here Current will be divided unequally duet to spread in LED characteristics. Those that draw more than their share will get hotter and draw even more. Those that draw less than their share will get cooler and draw less. If you have say 10 LEDs and you connect them in parallel and drive them with a single LED at about the rated current for all 10 then: • With typical low cost LEDs the Vf/If matching will be poor enough that the lowest Vf LEDs may draw 2 or 3 or 4 times their rated current. • The over current LEDs will rapidly die. • Now there are 9 LEDS to share enough current for 10. The AVERAGE current is 110%. The lowest Vf LED will again be overloaded and fail but this time it will happen even quicker as there is more current available per LED. • The next ... :-) - chain reaction. Look at a typical cheap Asian* multi-LED torch. Note the LEDs which are brightest. Operate the torch for a while then re-observe. After not too long the brightest LEDs will be dimmer or dead. Observe the brightest LEDs ... • I say "cheap Asian" as most multi LED torches are of Chinese origin and most are low cost so that their price is hard to beat. They largely build to "what the market will bear" and the market will bear rubbish. LED lights from other countries with multiple LEDs OR lights made in China with proper design cost more and are less popular. There is a reason for the extra cost. LEDs in series (2 groups). Constant current drive. Costs more. - OK, let's do the calculation. A simplified model for a LED is a fixed voltage source in series with a small resistor. Let's pick this LED from Kingbright. The slope is 20mA/100mV, so the internal resistance is 5$\Omega$. The the intrinsic LED voltage is 1.9V. Let's assume that the LEDs need 20mA and that our power supply is 5V. Then the LED voltage is 1.9V + 5$\Omega$ $\cdot$ 20mA = 2V. Our single series resistor $R = \dfrac{5V - 2V}{2 \cdot 20mA} = 75 \Omega$. That's if both LEDs are equal. Now suppose that there's a slight discrepancy between the LEDs, and that the 1.9V for the second LED is actually 1.92V, just a 1% difference. Now it's not immediately clear what the voltage across the LEDs will be. Let's find out, and call that $V_L$. There's a single current $I_R$ through the 75$\Omega$ resistor: $I_R = \dfrac{5V - V_L}{75 \Omega}$ The current through the first LED: $I_1 = \dfrac{V_L - 1.9V}{5 \Omega}$ and, likewise for LED 2: $I_2 = \dfrac{V_L - 1.92V}{5 \Omega}$ Now $I_R = I_1 + I_2$, so $\dfrac{5V - V_L}{75 \Omega} = \dfrac{V_L - 1.9V}{5 \Omega} + \dfrac{V_L - 1.92V}{5 \Omega}$ From this we find that $V_L$ = 2.01 V. Then, filling in this value in the above equations for the LED currents we find $I_1 = 21.94 mA$ and $I_2 = 17.94 mA$ conclusion Just the smallest discrepancy in LED voltage (1%) already results in a 18% difference in LED current. IRL the difference may be larger and there may be a visible difference in brightness. The effect will be worse for lower internal resistances. - That would work if the LEDs all had identical characteristics. Unfortunately, that isn't the case. and they will have different currents flowing through them. Several LEDs in series can have a single current-limiting resistor, of course. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9553210735321045, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/41532/basic-relations-theorem/41537	# Basic relations theorem I need to prove that If relations $\rho$, $\sigma$ are both reflexive and symmetric, and their composition $\rho\sigma$ is symmetric, then $\rho\sigma = \rho\vee\sigma$. It's obvious that $\rho\sigma$ includes all relations from both $\rho$ and $\sigma$, but I'm having hard time proving that it doesn't include extra relations. Example of such case on set {1,2,3,4,5}: $\rho$ (mark on $i$-th row and $j$-th column denotes $i\rho j$): ```` 1 2 3 4 5 1 • • 2 • • 3 • • 4 • • 5 • ```` $\sigma$: ```` 1 2 3 4 5 1 • • 2 • • 3 • 4 • 5 • ```` $\rho\sigma$: ```` 1 2 3 4 5 1 • • • 2 • • • 3 • • • 4 • • • 5 • ```` $3\rho\sigma1$, but $3\bar{\rho}1$ and $3\bar{\sigma}1$. I've tried reduction to the absurd, assuming that $\rho\sigma$ is symmetric and that there are some $x$,$y$,$w$ for which $x \bar{\rho} y$ and $x\bar{\sigma} y$, but $x \rho w\sigma y$. From symmetry of $\rho\sigma$ it follows that $y\rho q\sigma x$ for some $q$. But I can't figure out what to do next. - ## 1 Answer How is the join operation defined? Is it really just set union? If so, consider the following example: $$\begin{align} \rho &= \{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,4),(4,3)\}, \\ \sigma &= \{(1,1),(2,2),(3,3),(4,4),(3,2),(2,3),(1,4),(4,1)\}. \end{align}$$ We have $$1\rho2\sigma3, 3\rho4\sigma1, 2\rho1\sigma4, 4\rho3\sigma2.$$ Therefore $\rho\sigma$ is the full relation. In particular, it's symmetric. But it's different from $\rho \cup \sigma$. - Yes, join operation is just that. So, the statement I've tried to prove was wrong. Thank you very much. – furikuretsu May 26 '11 at 20:01 Actually the join operation is not just set union: the join of $\rho$ and $\sigma$ is the smallest relation containing both $\rho$ and $\sigma$. And the statement you tried to prove is correct! – Tara B Feb 15 at 16:31 (I realise this is an old question and the OP hasn't been on the site for some time, so the above comment was mainly intended for anyone else who might stumble across this.) – Tara B Feb 15 at 16:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9642634987831116, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/40750/vector-calculus-question	# Vector Calculus Question My question is: How can I solve the following problem? $C$ is the curve resulting from the intersection of $x+y+z=0$ with $x^2+y^2+z^2=4$, parameterized counterclockwise as seen from the positive $x$-direction. Evaluate $\int { ydx + zdy + xdz }$ Thanks. - 1 Have you drawn the graph of both equations to determine $C$? That's the first step. If you already have graphed the equations, could you be more specific about where you're struggling? The more specific you can be about what you've done, where you're stuck, etc., the more helpful we can be. It's also a good idea, when posting homework on this site, to include a sketch of what you've already done: e.g., if you've graphed the equations, can you determine the region(s) over which you need to integrate? How have you approached similar problems in the past, etc. – amWhy May 23 '11 at 1:00 ## 1 Answer Hint: You have the vector field $F(x,y,z)=(y,z,x)$, and you want to evaluate $$\int_C F\cdot ds.$$ What is the Curl of your vector field? Notice that it must be a constant since $F$ is linear. Let $K=\nabla \times F$ denote this constant. Now, apply Stokes Theorem. What is the area of the enclosed curve? (Hint: it is a circle of radius $2$). If we call that area $A$, then the line integral is equal to $$K A.$$ Hope that helps, -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9473427534103394, "perplexity_flag": "head"}
http://mathoverflow.net/questions/21491/abc-conjecture-meets-catalan-conjecture/21617	## abc-conjecture meets Catalan conjecture? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm looking to efficiently zero-test "sparse integers", i.e. integers of the form $\sum C_i \cdot A_i^{X_i}$ (where $A_i, C_i, X_i$ are integers); equivalently test if a given (integer or rational) point is a zero of a sparse polynomial. For example, a randomised algorithm would be to compute the sum modulo a random prime since Fermat's Little Theorem reduces computations to a "manageable" level and with good probability there won't be a false positive. I'm wondering if this can be derandomised. So beside the obvious does anyone know anything relevant about the general problem, I have more of a first step question: Is (are) there (infinitely many) positive integer(s) $M$ such that there are integers $A,B,C,D,E,F,X,Y,Z$ with $\|A\|,\|B\|,\|C\| < M$, $D,E,F,X,Y,Z>M$ and $$0 < A\cdot D^X+B\cdot E^Y+C \cdot F^Z< \log M?$$ - 4 Umm...just choose D,X,E,Y,F,Z very large, with D and E and F pairwise coprime, and then gcd(D^X,E^Y,F^Z)=1 so by Euclid you can find A,B,C such that AD^X+BE^Y+CF^Z=1. Now if A,B,C happen to be too small just change them until they're not, e.g. A-->A+E^Y,B-->B-D^X. – Kevin Buzzard Apr 15 2010 at 18:53 1 ...alternatively, make X,Y,Z all very negative ;-) – Kevin Buzzard Apr 15 2010 at 19:42 ... so the answer is: Yes, because all integers greater than 1 satisfy this property. Could you could give some explanation of what this had to do with the ABC or Catalan conjectures? The connection is not obvious from here. – S. Carnahan♦ Apr 16 2010 at 0:47 I don't understand your definition of "sparse integer". – Qfwfq Apr 16 2010 at 9:23 Thanks and d'oh. My intention was that A,B,C were all about the same size as D,E,F,X,Y,Z (possibly smaller). I will have to rethink the question. (also X,Y,Z are meant to be positive) The relation to the ABC conjecture is possibly better illustrated with the (zero-testing) question: Given A,B,C,X,Y,Z, is A^X+B^Y=C^Z? (if Z is relatively large [compared with A,B,C], then C^Z > rad(A^X B^Y C^Z) = rad(ABC)) The relation to the Catalan conjecture is in the attempt to find "small" combinations of integer powers. – Paul Hunter Apr 16 2010 at 17:51 show 3 more comments ## 2 Answers Regarding Paul's "first step question", I believe the answer may well be that the set of such M is finite. This follows under some coprimality hypothesis from the n-term abc conjecture of Browkin and Brzezinski [Math. Comp. 62 (1994), 931--939]. This states that, if we have $a_1, a_2, \ldots, a_n$ integers, for $n \geq 3$, with $\gcd(a_1, \ldots, a_n)=1$, $a_1 + \cdots + a_n=0$, and no vanishing subsums, then $$\limsup \frac{\log \max \|a_i\|}{\log \mbox{Rad} (a_1 a_2 \cdots a_n)} = 2n-3.$$ There may be examples with small M (like 3 or 4).....I'm not sure. - Thanks, I read that paper earlier but obviously failed to register its usefulness! So for sparse integers of fixed number of terms we can (subject to coprimality constraints and the generalized abc conjecture) efficiently zero-test sparse integers as there are only finitely many cases where the exponents can be "large". I wonder if it's possible to remove the reliance on the abc conjecture (and fixed number of terms) since the general question is such a special instance of the conjecture... [but since that wasn't the question I asked, I deem this to be an answer] – Paul Hunter Apr 20 2010 at 12:23 ### 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 looks like you're trying to examine the Tijdeman-Zagier conjecture which is usually referred to as Beal's conjecture because Andrew Beal has offered 100000 USD for its solution. Conjecture (Tijdeman, Zagier) If $(a,b,c,x,y,z)$ are positive integers such that $$a^x + b^y = c^z$$ and $x,y,z$ are all $> 2$ then $a,b,$ and $c$ share a common prime factor. This is a natural generalization of Fermat's Last Theorem. The ABC conjecture implies that there are only finitely many counterexamples to the above conjecture. If you want to know more about the ABC conjecture these expositions of Mazur and Elkies are good places to start. - Lang also has a pretty good section in Algebra all about the ABC conjecture. – Harry Gindi Apr 18 2010 at 21:51 Jamie, see mathoverflow.net/questions/28764/…. Perhaps you could provide the explicit reference for the conjecture's proper attribution to Tijdeman & Zagier? – Halfdan Faber Jun 26 2010 at 16:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9230652451515198, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/137397-polar-coordinate.html	# Thread: 1. ## polar coordinate Hi please, can you help me to solve this integral using polar coordinate Thank you $<br /> \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy<br />$ 2. $x^2+y^2=r^2$ Now you have $e^{-r^2}$ which is the same as doing polar since you have $x=rcos and y=rsin$ and squaring both removes the trig 3. $<br /> \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy=\int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2}r d \theta dr = 2\pi \int_0^{\infty} e^{-r^2}r dr = \pi [e^{-r^2}]_0^{\infty} = \pi<br /> <br />$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9382538795471191, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/41611?sort=oldest	## $\omega$-triviality of knots? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) From the theory of finite type invariants of knots comes the concept of $n$-triviality. A knot is said to be $n$-trivial if there is some projection of the knot and $n$ pairwise-disjoint sets of crossings $S_1,S_2,\ldots,S_n$, such that changing the crossings in every nontrivial subset of `$S=\{S_1,\ldots, S_n\}$` turns the knot into the unknot. An easy example is the trefoil knot with a three-crossing projection. Let $S_1$ and $S_2$ each contain just one crossing from this projection. Than changing $S_1$, $S_2$ or $S_1\cup S_2$ yields the unknot. This notion of $n$-triviality is quite interesting because Goussarov first proved that a knot is $n$-trivial if and only if all finite type invariants vanish up to degree $n-1$. My question is about generalizing this notion to $\omega$-triviality. For this purpose, it is probably better to regard the crossing changes in $S_i$ as homotopies supported in neighborhoods of arcs connecting the knot to itself (finger moves.) Then one can define a knot to be $\omega$-trivial if there is a pairwise disjoint collection `$\{S_1,S_2,....\}$` where each $S_i$ is a set of finger moves on $K$, such that doing any nonempty subcollection `$\{S_i: i\in J\}$` for $\emptyset\neq J\subset \mathbb N$, of the finger moves gives you the unknot. Question: If a knot is $\omega$-trivial, must it be the unknot? A knot which is $\omega$-trivial would have vanishing finite type invariants of all degrees, so it shouldn't exist, but this question should be a lot easier than the question of whether finite type invariants detect knottedness. Does anybody have any ideas on this question? - Habiro (in talks, perhaps not in print) has asked this question for links, where C_n-moves take the place of crossing-changes. – Daniel Moskovich Oct 10 2010 at 19:55 ## 1 Answer I would guess that you would need to understand the transfinite lower central series of the mapping class group. A knot complement which could not be distinguished from a solid torus by any finite lower central series quotient of the mapping class group Torelli group of a Heegaard surface would be n-trivial for all n, which if I'm not mistaken follows from work of [a subset of] Ted Stanford, Habiro, and Garoufalidis-Goussarov-Polyak. I don't see offhand why such an example could not exist. Therefore, I think this question is open, and I can't see why it would be easy. - I assume you mean the lower central series of the Torelli group here. I agree that an example such as you describe would be n-trivial for all $n$. But I was hoping that the extra geometric content of $\omega$-triviality would make this an easier question to tackle than the obviously difficult question of whether a knot could be $n$-trivial for all $n$. – Jim Conant Oct 10 2010 at 13:34 corrected- thanks. I don't have any ideas on how the extra geometric content might be used. – Daniel Moskovich Oct 10 2010 at 15:40 Yeah, I agree it's probably a hard question. – Jim Conant Oct 10 2010 at 19:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9363038539886475, "perplexity_flag": "head"}
http://cs.stackexchange.com/questions/78/natural-candidates-for-the-hierarchy-inside-npi	# Natural candidates for the hierarchy inside NPI Let's assume that $\mathsf{P} \neq \mathsf{NP}$. $\mathsf{NPI}$ is the class of problems in $\mathsf{NP}$ which are neither in $\mathsf{P}$ nor in $\mathsf{NP}$-hard. You can find a list of problems conjectured to be $\mathsf{NPI}$ here. Ladner's theorem tells us that if $\mathsf{NP}\neq\mathsf{P}$ then there is an infinite hierarchy of $\mathsf{NPI}$ problems, i.e. there are $\mathsf{NPI}$ problems which are harder than other $\mathsf{NPI}$ problems. I am looking for candidates of such problems, i.e. I am interested in pairs of problems - $A,B \in \mathsf{NP}$, - $A$ and $B$ are conjectured to be $\mathsf{NPI}$, - $A$ is known to reduce to $B$, - but there are no known reductions from $B$ to $A$. Even better if there are arguments for supporting these, e.g. there are results that $B$ does not reduce to $A$ assuming some conjectures in complexity theory or cryptography. Are there any natural examples of such problems? Example: Graph Isomorphism problem and Integer Factorization problem are conjectured to be in $\mathsf{NPI}$ and there are argument supporting these conjectures. Are there any decision problems harder than these two but not known to be $\mathsf{NP}$-hard? - 1 So you are looking for problems $P \in \mathrm{NP}$ such that $P_1 \leq_p P \leq_p P_2$ with $P_1 \in \mathrm{NPI}$ and $P_2 \in \mathrm{NPC}$? – Raphael♦ Mar 7 '12 at 7:46 1 Yes but there is no known reduction from P to P1 (similarly no known reduction from P2 to P). – Mohammad Al-Turkistany Mar 7 '12 at 7:58 2 – Marcos Villagra Mar 7 '12 at 8:42 8 – Marcos Villagra Mar 7 '12 at 8:43 2 why is the list that Marcos links to not the answer to your question ? – Suresh Mar 7 '12 at 16:21 show 13 more comments ## 1 Answer I've found a nice problem called ModularFactorial. Take as input two $n$-digit integers $x$ and $y$, and output $x! \mod y$. This problem is at least as hard as Factoring and not know to be hard for FNP. The reference is the recent (and beautiful) book by Cristopher Moore and Stephan Mertens The Nature of Computation, page 79. - I believe the OP is looking for problems in NP. Can you reformulate this as a decision problem? – Zach Langley Mar 9 '12 at 12:42 FNP is the function version (i.e.,search problems) of NP. In fact, factoring is not in NP, it is FNP. For instance, the decision problem for factoring is trivial, the complexity is just O(1), but the search problem is the difficult part. Since the OP gave factoring as an example, I think this is also a valid answer. – Marcos Villagra Mar 9 '12 at 13:19 Factoring can be reformulated into a decision problem as follows: given an integer $n$ and an integer $k$, does $n$ contain a factor $d$ with $1 < d \le k$? Is there an analogous decision version of the ModularFactorial problem? – Zach Langley Mar 9 '12 at 14:46 @Marcos, Thanks. I'm interested in decision problems in NP. – Mohammad Al-Turkistany Mar 9 '12 at 15:31 @ZachLangley, yes of course I agree, but I was thinking in another decision version, namely, "does x has a factor?". The answer there is just, "yes" always. You can do the same with modularfactorial, give an integer k and decide if $x!\mod y$ is greater than $k$ or not. – Marcos Villagra Mar 9 '12 at 22:43 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9361609816551208, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/45208/would-a-gauss-rifle-based-on-generated-magnetic-fields-have-any-kickback?answertab=active	# Would a gauss rifle based on generated magnetic fields have any kickback? In the case of currently developing Gauss rifles, in which a slug is pulled down a line of electromagnets, facilitated by a micro-controller to achieve great speed in managing the switching of the magnets, does the weapon firing produce any recoil? If so, how would you go about calculating that recoil? - 4 The equal and opposite reaction has to go somewhere, but it may be interesting to try to devise a dissipation mechanism. – kojiro Nov 27 '12 at 13:15 1 Note that momentum is not recoil. Recoil is the effect that firing has on the wielder, especially their ability to maintain aim. So the force distribution over time would seem to be what is important, and a Gauss rifle would have a lot more control over this than a firearm. Not to mention the ability to reduce it simply by increasing the length of the barrel, which gives almost linear increase in velocity with the same force, while rifles have diminishing returns with barrel length. – Paul Hutton Nov 27 '12 at 21:10 @paulhutton has the basics of what I was hoping to ask. It seems to me that the reverse momentum would be transferred to the coils as each on would be active, so you would be generating a lot of small recoils rather one large one. – sarge_smith Nov 28 '12 at 8:05 ## 2 Answers If the Gauss rifle shoots a projectile with exit speed of $v_1$ and mass $m_1$, then its momentum will be: $p=m_1v_1$. Because of momentum conservation law, the rifle will have the same momentum in opposite direction. If the rifles mass is $m_2$, the rifle will start moving in the opposite direction with end speed of: $v_2 = \frac{m_1 v_1}{m_2}$. But, as the projectile is accelerated for longer time than in a gun, the force acting from rifle on its holder will be lower because $F=\frac{dp}{dt}$ - Forgive my ignorance, but I can't help but think there has to be a difference in recoil between a Gauss rifle and a conventional rifle firing the same mass at the same speed -- isn't some of the energy of the normal round's explosion lost (via the cycling mechanism, or because the explosion force is omnidirectional)? And therefore, the charge must be more powerful to provide the same exit velocity, which means slightly greater recoil? Also, doesn't the gas exiting the barrel behind the projectile also exert a force (kind of like thrust in a rocket engine)? Physics was never my strong suit. :) – Justin ᚅᚔᚈᚄᚒᚔ Nov 27 '12 at 15:10 2 @Justinᚅᚔᚈᚄᚒᚔ the projectile exerts an equal force back on the coils which provides the recoil. The gas exiting the barrel after the bullet counts as the momentum of the bullet - the recoil is the total of everything going forward, it doesn't matter if it's bullet/propellant. So yes for a given projectile mass/speed you will have a lower recoil in a Gauss rifle because you don't have the extra momentum of the forward gases to recoil from – Martin Beckett Nov 27 '12 at 16:19 2 As Paul notes in the comments on the question itself there is also the question of the distribution of the momentum transfer in time. For a fixed projectile velocity the Gauss rifle can (presumably) have a lower peak force. Only experiment will demonstrate how this trades off in terms of single shot accuracy and the difficulty of keeping on target in rock-n-roll mode. – dmckee♦ Nov 27 '12 at 23:52 Simple answer when you think about it: You are imparting a force to accelerate the slug, so you're going to get an equal and opposite reaction. In a normal rifle, the explosion accelerates the bullet rapidly and you get recoil. In a gauss rifle, the acceleration will be a bit lower, but for a slightly longer time (the entire length of the barrel), so for the same muzzle velocity you will be able to calculate the recoil in the exact same way. - 4 AFAIK unless you're using low power ammo in a rifle (eg .22Long), the bullet will still be accelerating by a significant amount until it exits the barrel. You will have different acceleration curves and the gauss gun's will presumably be flatter than that of the conventional firearms. The latter will have a sharp peak near the start reaching when the propellant is fully burned and then falling off as the bullets forward movement reduces the pressure of the combustion gas. – Dan Neely Nov 27 '12 at 16:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9497725963592529, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/11449/conversion-of-mass-to-energy-in-chemical-nuclear-reactions?answertab=oldest	# Conversion of mass to energy in chemical/nuclear reactions Is mass converted into energy in exothermic chemical / nuclear reactions? My (A Level) knowledge of chemistry suggests that this isn't the case. In a simple burning reaction, e.g. $C+O_2\to \text{CO}_2$, energy is released by the CO bonds forming; the atoms lose potential energy when they pull themselves towards each other, in the same way that a falling object converts GPE to KE. There are the same number of protons, electrons etc. in both the reactants and products. I would have assumed that this reasoning extends to nuclear fission/fusion as well, but one physics textbook repeatedly references very small amounts of mass being converted into energy in nuclear reactions. So I just wanted to know if I was wrong about either of these types of reactions, and if so, what mass is lost exactly? - ## 5 Answers This is actually a more complex question than you might think, because the distinction between mass and energy kind of disappears once you start talking about small particles. So what is mass exactly? There are two common definitions: 1. The quantity that determines an object's resistance to a change in motion, the $m$ in $\sum F = ma$ 2. The quantity that determines an object's response to a gravitational field, the $m$ in $F_g = mg$ (or equivalently, in $F_g = GMm/r^2$) The thing is, energy actually satisfies both of these definitions. An object that has more energy - of any form - will be harder to accelerate, and will also respond more strongly to a given gravitational field. So technically, when computing the value of $m$ to plug into $\sum F = ma$ or $F_g = mg$ or any other formula that involves mass, you do need to take into account the chemical potential energy, thermal energy, gravitational binding energy, and many other forms of energy. In this sense it turns out that "mass" is effectively just a word for the total energy of an object (well, divided by a constant factor: $m_\text{eff} = E/c^2$). If mass is just another word for energy, why do we even talk about it? Well, for one thing, people got used to using the word "mass" before anyone knew about all its subtleties ;-) But seriously: if you really look into all the different forms of energy that exist, you'll find that figuring out how much energy an object actually has can be very difficult. For instance, consider a chemical compound - $\mathrm{CO}_2$ for example. You can't just figure out the energy of a $\mathrm{CO}_2$ molecule by adding up the energies of one carbon atom and two oxygen atoms; you also have to take into account the energy required to make the chemical bond, any thermal energy stored in vibrational modes of the molecule or nuclear excitations of the atoms, and even slight adjustments to the molecular structure due to the surrounding environment. For most applications, though, you can safely ignore all those extra energy contributions because they're extremely small compared to the energies of the atoms. For example, the energy of the chemical bonds in carbon dioxide is one ten-billionth of the total energy of the molecule. Even if adding the energies of the atoms doesn't quite get you the exact energy of the molecule, it's often close enough. When we use the term "mass", it often signifies that we're working in a domain where those small energy corrections don't matter, so we can just add the masses of the parts to get the mass of the whole. Obviously, whether the "extra" energies matter or not depends on what sort of process you're dealing with, and specifically what energies are actually affected by the process. In chemical reactions, the only changes in energy that really take place are those due to breaking and forming of chemical bonds, which as I said are a miniscule contribution to the total energy of the particles involved. But on the other hand, consider a particle accelerator like the LHC, which collides protons with each other. In the process, the chromodynamic "bonds" between the quarks inside the protons are broken, and the quarks then recombine to form different particles. In a sense, this is like a chemical reaction in which the quarks play the role of the atoms, and the protons (and other particles) are the compounds, but in this case the energy involved in the bonds (by this I mean the kinetic energy of the gluons, not what is normally called the "binding energy") is fully half of the energy of the complete system (the protons) - in other words, about half of what we normally consider the "mass" of the proton actually comes from the interactions between the quarks, rather than the quarks themselves. So when the protons "react" with each other, you could definitely say that the mass (of the proton) was converted to energy, even though if you look closely, that "mass" wasn't really mass in the first place. Nuclear reactions are kind of in the middle between the two extremes of chemical reactions and elementary particle reactions. In an atomic nucleus, the binding energy contributes anywhere from 0.1% up to about 1% of the total energy of the nucleus. This is a lot less than with the color force in the proton, but it's still enough that it needs to be counted as a contribution to the mass of the nucleus. So that's why we say that mass is converted to energy in nuclear reactions: the "mass" that is being converted is really just binding energy, but there's enough of this energy that when you look at the nucleus as a particle, you need to factor in the binding energy to get the right mass. That's not the case with chemical reactions; we can just ignore the binding energy when calculating masses, so we say that chemical reactions do not convert mass to energy. - Thanks for this great answer. I had assumed that mass could somehow be converted, but you've made it clear that it's the other way round - mass is reduced because energy is released. – myk Jun 23 '11 at 18:56 Chemical reactions of any kind ---> No conversion. All mass (and energy) is conserved. Nuclear reactions ---> Yes, mass is converted to energy, in both fission and fusion reactions. What mass? Well, going by my college instruction, I would have said that a neutron is lost (converted to energy). But a quick surf of the internet shows me that what I learned in college is now considered an old wives' tale. The new answer, though obviously not what I learned, defies my efforts to understand, let alone condense into easy reader format for you. Maybe someone else will give it a go. - 1 really, if you don't understand and know it, why answer? – user2963 Jun 22 '11 at 22:57 1 @zephyr: not everyone knows everything. I don't think we need to discourage people from posting answers even if they aren't completely sure; the risk of getting downvoted should be enough to do that. – David Zaslavsky♦ Jun 23 '11 at 0:11 @Vintage: just in case you didn't know, if you realize that an answer you posted is wrong, you should be able to delete it. Of course you're not obligated to do so. – David Zaslavsky♦ Jun 23 '11 at 0:12 @David. Thank you for the tip. I'll stick with my answer, though, downrated as it is. I don't see anything in it that is untrue, in the classical mechanics sense. If someone were to tell me that mass is lost or gained in a chemical reaction, I'd respond, "How much mass?" If they tell me they are not sure, or can't even measure it, then I'd respond that my answer is perfectly fine in the practical world. Sort of reminds me of Pluto: It's not a planet anymore, but it was 10 years ago. Makes me wonder if it was really a planet when science said it was a planet. What changed on Pluto? – Vintage Jun 23 '11 at 20:18 2 @Vintage - it is misleading because it ignores the universality of the process. There is a mass loss in a nuclear reaction and a chemical reaction, it is only the magnitude that differs. The physics is exactly the same. This is not a philosophical position, it is a straightforward application of our currently known laws of physics. There are many implications of these laws which can be neglected in the practical world, that does not make them any less real or true, and you do a disservice to learners by dismissing the truth rather than seeing the unified whole. – user2963 Jun 28 '11 at 3:45 show 5 more comments The potential energy of the chemical bonds do correspond to an increase of mass proportionate to that energy. So, for example the energy in a CO-O bond is 110 kcal / mol or about 7.6 x 10^-19 Joules per bond. Dividing by c^2 gives us a mass of about 8.5 x 10^-33 grams. In a mole of CO2, that amounts to 5.1 x 10^-9 grams. Since a mole of CO2 has a mass of about 44g, you're going to have a hard time measuring the mass difference. - Theoretically, even chemical reactions do convert mass to energy or vice versa. But the amount of mass converted is so small that we are generally unable to detect it. - First you have to understand, there are two definitions for mass: rest mass and motional mass. Rest mass is the mass measured in the frame where the object is stationary. Motional mass is $m_1=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$. The old school calls the motional mass mass, while the new school call rest mass mass. Rest mass is not additive. The rest mass of a molecule is not the sum of its atoms, but the sum plus the bond energy divided by $c^2$. Therefore in a chemical reaction, though the number of atoms (or electrons, protons, neutrons) are conserved, the rest mass is not, since the bond energy is changed. And this can be called mass being converted into energy. In Motional mass, however, is just another name for energy, and therefore always conserved. In combustion, the molecular bond energy is converted to the additional kinetic energy of $\mathrm{CO_2}$ molecules, so the mass(or energy) is conserved. In nuclear reaction, the electromagnetic and strong potential energy between nucleons are converted into the kinetic energy of new nuclei and (sometimes) flying neutrons. Sometimes the new nuclei are in excited state, and will soon emit gamma radiation, which are basically high energy photons with zero rest mass but nonzero motional mass. A lot of old school books I read use double standards. When they refer to mass generically, they mean motion mass, which increases with speed; but when they refer to mass in nuclear reaction, they mean rest mass. I guess this is how you got confused. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9571700692176819, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/210900/is-t2-a-prime-element-of-mathbbf-2t2-s2?answertab=votes	# Is $t^{2}$ a prime element of $\mathbb{F}_{2}(t^{2},s^{2})$? I wish to find out if $t^{2}$ is a prime element of $\mathbb{F}_{2}(t^{2},s^{2})$ so I can justify the use of Eisenstein on the polynomial $x^{2}-t^{2}\in\mathbb{F}_{2}(t^{2},s^{2})[x]$ I believe that it does since $\mathbb{F}_{2}(t^{2},s^{2})/\langle t^{2}\rangle\cong\mathbb{F}_{2}(s^{2})$ is an integral domain (since it is a field) Is my argument correct and I may use Eisenstein ? (I already noted that $t^{4}$ does not divide $t^{2}$) - 1 $\mathbb{F}_2(t^2, s^2)$ is a field. It has no prime elements, and the ideal generated by $t^2$ is the whole thing, so the quotient is zero. – Qiaochu Yuan Oct 11 '12 at 5:24 @QiaochuYuan - I see, thank you. It seems that if I replace the brackets with $[ ]$ then it is a prime elemnt. Can I justify using Eisenstein now ? – Belgi Oct 11 '12 at 5:26 Yes, but you don't need to. You can give a direct proof. – Qiaochu Yuan Oct 11 '12 at 5:38 @QiaochuYuan If you can think of such an argument I would love to hear it, please take the time to write it. Its about the principle (I 'feel' Eisenstein and I can't argue the conditions). The question itself is not as important. if I can't seem to think of how to justify the use of Eisenstein (and you say it can be justified) then I either missing something important or not understanding some concept and this troubles me – Belgi Oct 11 '12 at 5:43 ## 2 Answers Two comments. 1. A general form of Eisenstein's criterion shows that $x^2 - t^2$ is irreducible in $\mathbb{F}_2[t^2, s^2][x]$. Since $\mathbb{F}_2[t^2, s^2]$ is a UFD, a general form of Gauss's lemma shows that $x^2 - t^2$ is irreducible in $\mathbb{F}_2(t^2, s^2)[x]$, and primeness is equivalent to irreducibility in a UFD. 2. Neither of the above general facts are necessary in this case. If you only want to show primeness in $\mathbb{F}_2[t^2, s^2][x]$, it suffices to observe that the quotient by $x^2 - t^2$ is $\mathbb{F}_2[s^2, x]$, which is an integral domain. If you want to show primeness in $\mathbb{F}_2(t^2, s^2)[x]$, you can observe directly that the only form a nontrivial factorization can take is $(x - a)(x + a)$ for some $a \in \mathbb{F}_2(t^2, s^2)$, and by degree considerations no such $a$ satisfies $a^2 = t^2$. This shows irreducibility and, again, irreducible is equivalent to prime in a UFD. - Can you please add explanation about the use of Eisenstein ? Where exactly do I need $t^2$ to be a prime element, if I say that $\mathbb{F}_{2}[t^{2},s^{2}][x]/\langle t^{2}\rangle\cong\mathbb{F}_{2}[s^{2}]$ then I have what I need ? I really appriciate your help! – Belgi Oct 11 '12 at 5:58 @Belgi: I don't really understand the question. – Qiaochu Yuan Oct 11 '12 at 6:00 I will try to be more clear: I wish to use Eisenstein in its general form to show that $x^{2}-t^{2}\in\mathbb{F}_{2}(t^{2},s^{2})[x]$ is irreducible . I believe we have shown $x^{2}-t^{2}\in\mathbb{F}_{2}[t^{2},s^{2}][x]$ using Eisenstein with the integral domain $R=\mathbb{F}_{2}[t^{2},s^{2}]$ and the prime ideal $\langle t\rangle$ of $R$. But I want $x^{2}-t^{2}$ to be irreducible in $\mathbb{F}_{2}(t^{2},s^{2})[x]$ not just in $\mathbb{F}_{2}[t^{2},s^{2}]$. – Belgi Oct 11 '12 at 6:08 – Qiaochu Yuan Oct 11 '12 at 6:13 Oh, So I have to use both Eisnstien and Gauss's lemma and not only Eisnstien , right ? thank you for the help (+1) – Belgi Oct 11 '12 at 6:15 show 1 more comment Hint $\$ prime $\rm t\in R[t]\iff$ domain $\rm\:R[t]/(t)\cong R$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9331730008125305, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/42500/mathbbrn-contains-no-subspace-homeomorphic-to-sn-borsuk-ulam	$\mathbb{R}^n$ contains no subspace homeomorphic to $S^n$ (Borsuk-Ulam) I want to prove that $\mathbb{R}^n$ contains no subspace homeomorphic to $S^n$, as per the wikipedia article I'm trying to set up a function $g:S^n \to \mathbb{R}^n$ that would be injective, which cannot occur by Borsuk-Ulam. If I have a subspace, $A \subset \mathbb{R}^n$, homeomorphic to $S^n$ then I can have a map $h:S^n \to A$. Can I compose this with the inclusion $i:\hookrightarrow \mathbb{R}^n$ to give an (injective) map $S^n \to \mathbb{R}^n$, and am done? (As an aside, I am sure there are other point-set topology ways to prove this problem, maybe using invariance of domain) Edit: Since the above works, to make this a little more interesting - can anyone provide any other proofs without using algebraic topology? (Although this is probably debatable, because some other methods to use it are probably based on algebraic topology) - Yes. ${}{}{}{}$ – Jonas Meyer Jun 1 '11 at 4:15 @Jonas - ok, thanks. I guess I'll keep the question here in case anyone else ever needs it – Juan S Jun 1 '11 at 4:17 @Qwirk: OK, but you shouldn't take my word for it! I'm honestly not sure what part you have a question about, but you outlined how the existence of $A$ would imply the existence of an injective continuous map $S^n\to\mathbb{R}^n$, which is impossible by Borsuk-Ulam. – Jonas Meyer Jun 1 '11 at 4:20 @Jonas - I guess there is no question! Sometimes I have trouble working out if what I did is correct (a lack of faith perhaps?). Also writing it out properly (not just my scribbles on a pad) probably made me write it out better – Juan S Jun 1 '11 at 4:27 @Jonas - well I edited it, since it was, as stands, not a very interesting question! I would be interested to see other ways to prove this then – Juan S Jun 1 '11 at 4:30 show 2 more comments 2 Answers Another way to see this is to use dimension theory (part of general topology, see e.g. Engelking's book "Theory of dimensions, finite and infinite", or others). One important theorem is that (for separable metrisable spaces, say, where the theory is nice and well-understood) $\dim(X) \le n$ is equivalent to "every continuous function $f$ from a closed subset $A$ of $X$ to $\mathbb{S}^n$ can be continuously extended over $X$". So Tietze's extension theorem with the $n$-sphere as the image. This theorem follows from Brouwer's fixed point theorem (eventually) and can be used to give a nice proof of the invariance of domain, that you already mentioned. In fact, Borsuk-Ulam, the sphere-extension theorem above, invariance of domain, the no retract theorem ($\mathbb{S}^n$ is not retract of $\mathbb{B}^{n+1}$) and Brouwer's fixed point theorem are all closely related, and having one we can relatively easily prove the others. If now a subset $A$ of $\mathbb{R}^n$ were homeomorphic to $\mathbb{S}^n$, then by compactness of the latter, $A$ would be closed in $\mathbb{R}^n$, which has dimension $n$, so we would have an extension of the homeomorphism to a continuous function, and this means we have a retraction (composing with the inverse of the homeomorphism) of $\mathbb{R}^n$ onto $A$, which is supposedly homeomorphic to $\mathbb{S}^n$, and this cannot be. - Thank you for this very nice answer – Juan S Jun 1 '11 at 12:04 When you compose $h$ with the inclusion, you're using the fact that the inclusion is continuous due to the definition of the subspace topology: The preimage of any open set $U$ of $\mathbb R^n$ under the inclusion is the intersection of $U$ with $A$, and that's by definition an open set of the subspace. You could also use this to show directly (i. e. without appealing to the fact that composition preserves continuity) that $h$ considered as a map to $\mathbb R^n$ is continuous, since the preimage of any open set $U$ of $\mathbb R^n$ under $h$ is the preimage of its intersection with $A$, which is open. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9588003158569336, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/209278/gaussian-elimination-and-matrix-question?answertab=votes	# Gaussian Elimination and Matrix question Given the system of equations: $$x_1 + 5x_3 = 1$$ $$2x_1 – x_2 + 7x_3 = 7$$ $$2x_2 + cx_3 = – 1$$ where $c$ is a constant. (the number behind the $x$ is supposed to be a number lower than the $x$, I just don't know how to use the system in here to write equations yet). (a) Write down the extended matrix. Find the solution using Gaussian elimination method. For what values of $c$ and equation unique solution? Interpret the meaning of other values for $c$ (e.g. forbidden values). (b) Assume that the system of equations above can be written in matrix form $Ax = B$. Let $c = 3$ Set up $A, x, B$. Calculate the determinant of $A$. (c) Compute the inverse of $A$ by means of the adjoint matrix (matrix of cofactors). Use the result for $A -1$ to find the solution to equation set. I'm not very good at math but I've tried this one. My books is really bad at explaining things and I don't really know anyone who is good at math. I would very much appreciate help with this problem. I've read up on homework questions in here and also sent my professor an email, he said it was fine as long as I afterwards were able to show that I understood it, not just copy pasted of someone else. (b) and (c) I'm really in the dark about these two, but I think I've understood somewhat how to do (a). The extended matrix: $$\begin{bmatrix} 1 & 0 & 5 & 1 \\ 2 & -1 & 7 & 7 \\ 0 & 2 & c & -1 \\ \end{bmatrix}$$ And in (a) I multiplied the first line and mixed it with the second to remove ($2x_1$), and in the third line (where my problem is) I got $(c-6)x_3 = 9$. So, I'd love some help with $a-c$, but if that's not allowed It would be very helpful if someone could explain what to do with the $c$ in the third line. What I understood is that if I manage to get $x_3 = "something"$ I could use that in the other lines to find $x_1$ and $x_2$ and that would mean that I've solved it using Gaussian elimination right? - I believe this question should have a label "homework" not to confuse the readers. "and in the third line (where my problem is) I got (c−6)x3=9" It is not a problem, you just have to express all the x_i variables as a function of c. To be more specific once you have x_3 as you do, you do the upwards sweeping portion of the Gaussian elimination i.e. eliminate the remaining off diagonal terms line by line. As a result all the x_i variables should be expressed as functions of c. About all the other parts you should consult your lecture notes / Wikipedia on how to compute determinants etc. they ar – peterm Oct 8 '12 at 15:49 ## 1 Answer Your system of equation give the following matrix as you did: $$\begin{bmatrix} 1 & 0 & 5 & \|1 \\ 2 & -1 & 7 & \|7 \\ 0 & 2 & c & \| -1 \\ \end{bmatrix}$$ Now by doing $-2R_1+R_2\rightarrow R_2$ you have $$\begin{bmatrix} 1 & 0 & 5 & \|1 \\ 0 & -1 & -3 &\| 5 \\ 0 & 2 & c & \|-1 \\ \end{bmatrix}$$ and by $2R_2+R_3\rightarrow R_3$ you have $$\begin{bmatrix} 1 & 0 & 5 & \|1 \\ 0 & -1 & -3 &\| 5 \\ 0 & 0 & c-6 & \|9 \\ \end{bmatrix}$$ Now you can arrange a new system of equations by using the final matrix exactly as you get the first extended matrix. I mean: $$x_1+5x_3=1\\-x_2-3x_3=5\\(c-6)x_3=9$$ Suppose that $c=6$ then what happens? A contradiction! In this case the system has no solution as form $(x_1,x_2,x_3)$. On the other hand you can read the determinant from the final matrix. It is $1(-1)(c-6)=6-c$. If you want your non homogenous system has a unique solution; the determinant should not be zero. It means that $c$ must be any number than $6$. - Opss! I made a mistake. Sorry. Fixed it. $-x_3$ should have been $-3x_3$ in the second equation.:) – Babak S. Oct 8 '12 at 19:16 @FrankLen: I used the elementary row operations to reduce a matrix to what is called triangular form (the last matrix). Always, when you want to find the determinant of a square matrix $A$; you can do these processes and you did it right for this system. When the matrix becomes upper triangular; you can multiply the diagonal elements and find it 6-c. – Babak S. Oct 8 '12 at 19:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9595231413841248, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=2923660	Physics Forums Thread Closed Page 1 of 2 1 2 > ## Has this equation got a solution? I worked hard to figure out whether this could be solvable or not x^0.5= -1 x belongs to any set in all of the mathematical world you prefer This can't be a homework I think. Note: it isn't x^2 = -1 Read up on imaginary numbers: http://en.wikipedia.org/wiki/Imaginary_numbers I think the notation is just fooling you, x^0.5 is the same as sqrt(x), so your problem is: sqrt(x)=-1 Does that help? ## Has this equation got a solution? You are right, no solutions exist. Quote by llamaprobe5 You are right, no solutions exist. Not true ... no REAL solutions exist. (There is, however, a complex solution). Quote by zgozvrm Not true ... no REAL solutions exist. (There is, however, a complex solution). Um, could you provide an example, I couldn't find one... Mentor Quote by llamaprobe5 Um, could you provide an example, I couldn't find one... Search Imaginary Number on wikipedia.org, for example... Quote by berkeman Search Imaginary Number on wikipedia.org, for example... I know what an imaginary number is...I mean, could you provide an example of a solution for this equation. How do you give an example of a solution? I guess "5" is an example of a solution, albeit not a very good one, because it is wrong! Quote by llamaprobe5 I know what an imaginary number is.. Apparently, you don't. Apparently you don't either if you're example solution is 5... I simply asked for an example solution...I know that you think that you are extremely intelligent, but you have failed to answer a rather simple question... Just to clarify, I am asking for a solution... This is what you are looking for: http://en.wikipedia.org/wiki/Exponen...tive_nth_roots The solution is x = 1, but it depends on how you make your definitions. If you define the square root to be a relation instead of a function, you can say that sqrt(1) is both 1 and -1. Quote by zgozvrm Not true ... no REAL solutions exist. (There is, however, a complex solution). Oh ok thank you for the direct answer, that makes sense... I guess that does work if you say +/- 1 is the square root. So there ARE real solutions... Actually, let me clarify: Let R be the space of real numbers. R x R (pronounced "R cross R") is the set of all ordered pairs (x,y) where x and y are both in R. When you graph a function f(x) = ???, you are looking at every x and finding the unique y value that lies above it. In fact, the ability to graph f(x) like this is what makes it a function! Another way to describe the graph of a function, however, is the set of ordered pairs (x,y) in R x R such that y = f(x). The top diagram in this image is the graph of a function, since there is exactly one point in the graph for each x value. Notice that I'm using "graph" in a very specific sense: it refers to the set of blackened points (which in this case form a wavy line). We can think of the color black as representing "TRUE" and white as representing "FALSE". So for any point (x,y), being in the graph (a.k.a. being black) is the same as it being TRUE that y = f(x). If (x,y) is not in the graph (in other words, it is white), then it is FALSE that y = f(x). So really, the graph of a function is just the set of points where it is TRUE that f(x) = y. With this understanding of a graph, we can generalize the idea of a function. The middle diagram shows the graph of what's known as a "relation". A point (x,y) is blackened if and only if x ~ y (read, "x related to y"). In this case, each x value has TWO possible y values where x ~ y. Just as a function can be defined by its graph, a relation can be defined by its graph. Just as you might ask "Does y = f(x)?", you can ask "Is x ~ y?" for any particular ordered pair (x,y). In fact, a function IS a special kind of relation. It's important to note that with most relations, order is importat! So just because x ~ y, doesn't mean y ~ x! The bottom image (thrown in for good measure), shows that any graph can be a relation. In this case, there are an infinite number of y values such that x ~ y. Now, we can define the square root to be a relation! We will say that x ~ y if x is "a square root of" y. And by this, we mean that x ~ y if and only if x^2 = y. This is in fact a function of x, whether x is positive or negative (or imaginary, for that matter!), because for every x, there is exactly one y satisfying x ~ y. It turns out that you are asking "Is there a y such that -1 is a square root of y?" Well, we just determined that for any x (including x = -1), we can find a unique such y! That is, we are looking for a y such that (-1)^2 = y. That y is just the number 1. Now, the reason I brought up all this stuff about relations is that while the "is square root of" relation is a function of x, it does NOT have a natural inverse function. To create an inverse function, people just restrict the domain of the "is square root of" function to only non-negative x. Attached Thumbnails Recognitions: Gold Member Science Advisor Staff Emeritus This is an odd situation. Actually, there is a real number solution to $x^{0.5}= -1$, but you have to extend to the complex number system to find it! In the real number system, the square root function is "single valued" (as are all function in the real numbers) and we define $\sqrt{x}= x^{0.5}$ to be the non-negative number, a, such that $a^2= x$. By that definition, a square root is never negative and we cannot have real x such that $x^{0.5}= -1$. However, in the complex number system, for a number of reasons, including the fact that the complex numbers cannot be made into an ordered field, we must allow "multivalued" functions. In the complex number system, $\sqrt{x}= x^{0.5}$ is any number, a, satisifying $a^2= x$. In the complex number system, since $1^2= 1$ and $(-1)^2= 1$, $\sqrt{-1}$ is both 1 and -1. Thus, $\sqrt{x}= x^{0.5}= -1$, in the complex number system, is satisfied by x= 1, which happens to be a real number. (Although x= 1 is a real number, x= 1 would NOT be a solution to that equation in the real number system. In the real number system $1^{0.5}= 1$, only.) Quote by zgozvrm Not true ... no REAL solutions exist. (There is, however, a complex solution). My bad ... I misread the question to mean that the OP was looking for [itex]x = \sqrt{-1}[/tex], or [itex]x^2 = -1[/tex] It was explicitly stated that he was NOT looking for the latter: "Note: it isn't x^2 = -1" That's what I get for trying to help out when I've got a head cold ... I can't think straight! My apologies... Thread Closed Page 1 of 2 1 2 > Thread Tools \| \| \| \| \|--------------------------------------------------------\|----------------------------------\|---------\| \| Similar Threads for: Has this equation got a solution? \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus \| 1 \| \| \| General Math \| 2 \| \| \| Calculus & Beyond Homework \| 6 \| \| \| Precalculus Mathematics Homework \| 6 \| \| \| General Math \| 3 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9367396831512451, "perplexity_flag": "middle"}
http://www.quantumdiaries.org/tag/lhc/	# Quantum Diaries Thoughts on work and life from particle physicists from around the world. ## Posts Tagged ‘LHC’ ### Tweeting the Higgs Wednesday, January 23rd, 2013 Back in July two seminars took place that discussed searches for the Higgs boson at the Tevatron and the LHC. After nearly 50 years of waiting an announcement of a $$5\sigma$$ signal, enough to claim discovery, was made and all of a sudden the twitter world went crazy. New Scientist presented an analysis of the tweets by Domenico et al. relating to the Higgs in their Short Sharp Scient article Twitter reveals how Higgs gossip reached fever pitch. I don’t want to repeat what is written in the article, so please take a few minutes to read it and watch the video featured in the article. The distribution of tweets around the July 2nd and July 4th announcements (note the log scale) Instead of focusing on the impressive number of tweets and how many people were interested in the news I think it’s more useful for me as a blogger to focus on how this gossip was shared with the world. The Higgs discovery was certainly not the only exciting physics news to come out of 2012, and the main reason for this is the jargon that was used. People were already familiar with acronyms such as CERN and LHC. The name “Higgs” was easy to remember (for some reason many struggled with “boson”, calling it “bosun”, or worse) and, much to physicists’ chagrin, “God particle” made quite a few appearances too. It seems that the public awareness was primed and ready to receive the message. There were many fellow bloggers who chose to write live blogs and live tweet the event (I like to think that I started bit of a trend there, with the OPERA faster than light neutrinos result, but that’s probably just wishful thinking!) Following the experiences of December 2011, when the webcast failed to broadcast properly for many users had twitter on standby, with tweets already composed, hungry for numbers. The hashtags were decided in advance and after a little jostling for the top spot it was clear which ones were going to be the most popular. Despite all the preparation we still saw huge numbers of #comicsans tweets. Ah well, we can’t win them all! The point is that while the world learned about the Higgs results I think it’s just as important that we (the physicists) learn about the world and how to communicate effectively. This time we got it right, and I’m glad to see that it got out of our control as well. Our tweets went out, some questions were asked and points clarified and the news spread. I’m not particularly fond of the phrase “God particle” , but I’m very happy that it made a huge impact, carrying the message further and reaching more people than the less sensational phrase “Higgs boson”. Everyone knows who God is, but who is Higgs? I think that this was a triumph in public communication, something we should be building on. Social media technologies are changing more quickly each year, so we need to keep up. A map of retweets on July 4th, showing the global spread. I’m glad to see more physicists using Twitter and youtube and other sites to spread the word because that’s where we can build audiences faster. (Incidentally if you want to see why we should be creating new audiences rather than addressing existing ones then see this video by Vihart.) It takes more work and it’s more experimental, but it’s worth the effort. Why did I make an advent calendar? Why tell physics jokes on Twitter? Just to see what works and what doesn’t. I’m not the first person to do these things, and I’m certainly not going to be the last. All I can hope to do is try new ideas out and give other people ideas. I don’t know the people I inspire and those I am inspired by, but that’s also part of the experiment. A lot of my ideas come from people who leave comments or send E-mails or tweets. Occasionally it gets heated and controversial, but if it’s not worth fighting for then it’s not worth saying in the first place. Many comments come from other bloggers too, and we can learn from each other. When I first started to blog someone sent me a few paragraphs of advice and I forgot most of it except one part “Ignore other people’s expectations. Some people will want you to always write about physics, some people will hate that. Write what matters to you.” When I combine that with what Vihart says (essentially “If your content is worth attention then people will pay attention to it.”) then rest is easy. Well, not easy, but less stressful. But moving back to the main point, the Higgs tweets went global and viral because they were well prepared and the names were simple. Other news included things like the search for the $$B_s$$ meson decaying to two muons and the limits that places on SUSY, but how does one make a hashtag for that? I would not want to put the hashtag #bs on my life’s work. It’s always more exciting to announce a discovery than an exclusion too. The measurement of $$\theta_{13}$$ was just as exciting in my opinion, but that also suffered the same problem. How is the general public supposed to interpret a Greek character and two numbers? I should probably point out that this is all to do with finding the right jargon for the public, and not about the public’s capacity to understand abstract concepts (a capacity which is frequently underestimated.) Understanding how $$\theta_{13}$$ fits in the PMNS mixing matrix is no more difficult than understanding the Higgs mechanism (in fact it’s easier!) It’s just that there’s no nice nomenclature to help spread the news, and that’s something that we need to fix as soon as possible. As a side note, $$\theta_{13}$$ is important because it tells us about how the neutrinos mix. Neutrino mixing is beyond the Standard Model physics, so we should be getting more excited about it! If $$\theta_{13}$$ is non-zero then that means that we can put another term into the matrix and this fourth term is what gives us matter-antimatter asymmetry in the lepton sector, helping to explain why we still have matter hanging around in the universe, why we have solid things instead of just heat and light. Put like that is sounds more interesting and newsworthy, but that can’t be squeezed into a tweet, let alone a hashtag. It’s a shame that result didn’t get more attention. It’s great fun and a fine challenge to be part of this whole process. We are co-creators, exploring the new media together. Nobody knows what will work in the near future, but we can look back what has already worked, and see how people passed on the news. Making news no longer stops once I hit “Publish”, it echoes around the world, through your tweets, and reblogs, and we can see its journey. If we’re lucky it gets passed on enough to go viral, and then it’s out of our control. It’s this kind of interactivity that it so rewarding and engaging. You can read the New Scientist article or the original paper on the arXiV. Thanks for reading! Posted in Latest Posts \| 5 Comments » ### Gluon Walls: A New Form of Matter? Tuesday, January 8th, 2013 Theoretical physicist Raju Venugopalan We sat down with Brookhaven theoretical physicist Raju Venugopalan for a conversation about “color glass condensate” and the structure of visible matter in the universe. Q. We’ve heard a lot recently about a “new form of matter” possibly seen at the Large Hadron Collider (LHC) in Europe — a state of saturated gluons called “color glass condensate.” Brookhaven Lab, and you in particular, have a long history with this idea. Can you tell me a bit about that history? A. The idea for the color glass condensate arose to help us understand heavy ion collisions at our own collider here at Brookhaven, the Relativistic Heavy Ion Collider (RHIC)—even before RHIC turned on in 2000, and long before the LHC was built. These machines are designed to look at the most fundamental constituents of matter and the forces through which they interact—the same kinds of studies that a century ago led to huge advances in our understanding of electrons and magnetism. Only now instead of studying the behavior of the electrons that surround atomic nuclei, we are probing the subatomic particles that make up the nuclei themselves, and studying how they interact via nature’s strongest force to “give shape” to the universe today. We do that by colliding nuclei at very high energies to recreate the conditions of the early universe so we can study these particles and their interactions under the most extreme conditions. But when you collide two nuclei and produce matter at RHIC, and also at the LHC, you have to think about the matter that makes up the nuclei you are colliding. What is the structure of nuclei before they collide? We all know the nuclei are made of protons and neutrons, and those are each made of quarks and gluons. There were hints in data from the HERA collider in Germany and other experiments that the number of gluons increases dramatically as you accelerate particles to high energy. Nuclear physics theorists predicted that the ions accelerated to near the speed of light at RHIC (and later at LHC) would reach an upper limit of gluon concentration—a state of gluon saturation we call color glass condensate.* The collision of these super-dense gluon force fields is what produces the matter at RHIC, so learning more about this state would help us understand how the matter is created in the collisions. The theory we developed to describe the color glass condensate also allowed us to make calculations and predictions we could test with experiments. (more…) Tags: Brookhaven National Laboratory, detector, heavy ion physics, LHC, physics, RHIC Posted in Latest Posts \| 4 Comments » ### Higgs update, HCP 2012 Thursday, November 22nd, 2012 Last week, Seth and I met up to discuss the latest results from the Hadron Collider Physics (HCP) Symposium and what they mean for the Higgs searches. We have moved past discovery and now we are starting to perform precision measurements. Is this the Standard Model Higgs boson, or some other Higgs boson? Should we look forward to a whole new set of discoveries around the corner, or is the Higgs boson the final word for new physics that the LHC has to offer? We’ll find out more in the coming months! Tags: ATLAS, CERN, CMS, HCP2012, Higgs, Higgs boson, LHC Posted in Latest Posts \| 5 Comments » ### Mixing it up Wednesday, November 14th, 2012 One of the other results presented at the Hadron Collider Physics Symposium this week was the result of a search for $$D^{0}–\bar{D}^{0}$$ mixing at LHCb. Cartoon: If a $$D^0$$ is produced, at some time t later, it is possible that the system has "oscillated" into a $$\bar{D}^0$$. This is because the mass eigenstates are not the same as the flavor eigenstates. Neutral meson mixing is predicted for any neutral meson system, and has been verified for the $$K^0–\bar{ K}^0$$, $$B^0–\bar{B}^0$$ and $$B_s^0–\bar{B_s}^0$$ systems. However, for the $$D^0–\bar{D}^0$$ system, no one measurement has provided a result with greater than $$5\sigma$$ significance that mixing actually occurs, until now. The actual measurement is of $$R(t)=R$$, which is effectively the Taylor expansion of the time dependent ratio of $$D^0 \rightarrow K^+ \pi^-$$ (“Wrong Sign” decay) to $$D^0\rightarrow K^- \pi^+$$ (“Right Sign” decay). Charge conjugates of these decays are also included. There is a “Wrong Sign” and a “Right Sign” because the Right Sign decays are much more probable, according to the standard model. The mixing of the $$D^0–\bar{D}^0$$ system is described by the parameters $$x = \Delta m /\Gamma$$ and $$y = \Delta \Gamma / 2\Gamma$$, where $$\Delta m$$ is the mass difference between the $$D^0$$ and $$\bar{D}^0$$, $$\Delta \Gamma$$ is the difference of widths of the mass peaks, and $$\Gamma$$ is the average width. What appears in the description of $$R$$, however, is $$x’$$ and $$y’$$, which give the relations between the $$x$$ and $$y$$ with added information about the strong phase difference between the Right Sign and Wrong Sign decays. The important part about $$x’$$ and $$y’$$ are that they appear in the time dependent terms of the Taylor expansion of $$R$$. If there were no mixing at all, then we would expect the ratio to remain constant, and the higher order time dependence to vanish. If mixing does occur, however, then a clear, non-flat trend should be seen, and hence a measurement of $$x’$$ and $$y’$$. That is why the time dependent analysis is so important. Fit of ratio of WS and RS decays as a function of decay time of the D meson. Flat line would be no mixing, sloped line indicates mixing. From http://arxiv.org/pdf/1211.1230.pdf Result of the mixing parameter fit of the neutral D meson system. 1,3 and 5 standard deviation contours are shown, and the + represents no mixing. From http://arxiv.org/pdf/1211.1230.pdf The result is a 9.1 $$\sigma$$ evidence for mixing, which is also in agreement with previous results from BaBar, Belle and CDF. On top of confirming that the neutral D meson system does mix, this result is of particular importance because, coupled with the result of CP violation in the charm system, it begs the question whether or not there is much more interesting physics beyond the standard model involving charm just waiting to be seen. Stay tuned! Tags: charm, D0-D0bar, HCP2012, LHC, LHCb, mesons, mixing, oscillation Posted in Latest Posts \| 1 Comment » ### The coolest and hottest fluid Friday, October 19th, 2012 In September, the Large Hadron Collider (LHC) operators at CERN attempted a new trick: putting in collisions protons in one beam and lead ions in the other. Usually, the LHC operates with two beams of identical particles (protons or ions) circulating in opposite directions in the accelerator. Here is what is expected from this new setup. These ions are atoms stripped of all their electrons, leaving only the nucleus. Lead ions contain 82 protons plus 126 neutrons, all held together by the nuclear force. Protons are also composite objects made of three quarks and bound together by “gluons”, the particles carrying the nuclear force. So when two such heavy ions collide at nearly the speed of light, I dare anyone to describe where each quark and each gluon will end up. Already, trying to predict where fifteen billiard balls go after breaking the pack is tough enough. But when each projectile is made of hundreds of particles, it becomes impossible. At first glance, it would seem all we could get out of this is just a mess. But this turns out to be the coolest and hottest mess one will ever see. From the most energetic collisions comes a new form of matter called the quark-gluon plasma. There are three very well known state of matter: solid, liquid and gaseous. Lesser known is the fourth state of matter called plasma. This is what one finds inside a neon tube when the electric current applied is strong enough to strip the gas of its electrons. Positively charged ions and negatively charged electrons float around freely, having enough energy not to recombine. The quark-gluon plasma is just one step above this. Imagine there is enough energy around that not only the atoms but the nucleons (the name given to protons and neutrons, the particles found inside the nucleus) break apart and coexist in some sort of an extremely energetic fluid. This is as hot as it got instants after the Big Bang. What is so cool about it though, is that this plasma exhibits collective behavior, meaning quarks and gluons do not float freely but have collective properties. The most spectacular of them is that this fluid has no viscosity and behaves as a perfect fluid. If you try to confine it in a container, it just flows up the container’s wall and spread all over the place. The ALICE experiment is dedicated to the study of the quark-gluon plasma. Each year, the LHC operates for a few weeks with lead ions instead of protons. ALICE collects data both during proton-proton collisions and heavy ions collisions. Even when only protons collide, the projectiles are not solid balls like on a billiard table but composite objects. By comparing what can is obtained from heavy ion collisions with proton collisions, the ALICE physicists must first disentangle what comes from having protons in a bound state inside the nucleus as opposed to “free protons”. So far, it appears that the quark-gluon plasma only formed during heavy-ion collisions since they provide the necessary energy density over a substantial volume (namely, the size of a nucleus). Some of the effects observed, such as the number of particles coming out of the collisions at different angles or momenta, depend in part on the final state created. When the plasma is formed, it reabsorbs many of the particles created, such that fewer particles emerged from the collision. By colliding protons and heavy ions, scientists hope to discern what comes from the initial state of the projectile (bound or free protons) and what is caused by the final state (like the suppression of particles emitted when a quark-gluon plasma forms). Already, with only one day of data taken in this new mode, the ALICE collaboration just released two papers. The first one presents the measurements of the charged hadrons density produced in proton-ion collisions and compares the result with the same measurement after proper normalization performed in proton-proton and ion-ion collisions. The second compares the transverse momentum distributions of charged hadrons measured in proton-ions and proton-proton collisions. The ultimate goal is to study the so-called “structure function”, which describes how quarks and gluons are distributed inside protons, when they are free or embedded inside the nucleus. More will be studied during the two-month running period with protons colliding on heavy ions planned for the beginning of 2013. A “snapshot” of the debris coming out of a proton-lead ion collision captured by the ALICE detector showing a large number of various particles created from the energy released by the collision. Pauline Gagnon Tags: ALICE, CERN, LHC, proton-ion collision, quark-gluon plasma Posted in Latest Posts \| 2 Comments » ### Torride et cool à la fois Friday, October 19th, 2012 En septembre, les opérateurs du Grand Collisionneur de Hadrons (LHC) au CERN on réussi un truc nouveau : mettre en collision un faisceau de protons avec un faisceau d’ions de plomb. Habituellement, le LHC fonctionne avec deux faisceaux de particules identiques (protons ou ions) circulant en sens inverse dans l’accélérateur. Pourquoi cette nouvelle configuration ? Ces ions sont des atomes auxquels on a arraché tous les électrons, ne laissant que le noyau atomique. Les ions de plomb contiennent 82 protons plus 126 neutrons, le tout maintenu ensemble par la force nucléaire. Les protons sont eux aussi des particules composites puisqu’ils sont faits de trois quarks « collés » ensemble grâce aux « gluons », les particules associées à la force nucléaire. Alors quand de tels noyaux entrent en collision à presque la vitesse de la lumière, qui pourrait prédire où chaque quark et chaque gluon aboutira? Même avec seulement quinze balles de billard, il est pratiquement impossible de deviner où elles iront après la casse. Si, de surcroit, chaque projectile est fait de centaines de particules, cela devient totalement imprévisible. A première vue, il semblerait que tout ce qui peut sortir de collisions ions-ions est un fouillis incroyable. Mais en fait, ces collisions super énergétiques produisent le fouillis le plus torride et le plus cool qui soit : un plasma de quarks et gluons. Tout le monde connaît les trois états de la matière: solide, liquide et gazeux mais le quatrième état, le plasma, est lui bien moins connu. C’est ce qu’on retrouve dans un tube au néon quand la différence de potentiel appliquée est assez forte pour arracher tous les électrons du gaz. Les ions chargés positivement ainsi que les électrons flottent librement, ayant suffisamment d’énergie pour ne pas se recombiner. Le plasma de quarks et gluons est juste l’étape suivante. Imaginez qu’on fournisse suffisamment d’énergie pour pouvoir dissocier non seulement les atomes mais aussi les noyaux et mêmes les nucléons (le nom générique donné aux neutrons et protons à l’intérieur des noyaux atomiques). On obtient alors une soupe extrêmement énergétique de quarks et de gluons. Il n’y a pas plus chaud et ce serait l’état dans lequel se trouvait toute la matière immédiatement après le Big Bang. Fait étonnant : le plasma de quarks et gluons se comporte comme un fluide ayant des propriétés collectives et non comme un ensemble de particules indépendantes. C’est en fait un fluide parfait ayant une viscosité nulle. Si on essayait de le confiner dans un contenant, le fluide remonterait les parois du contenant et se répandrait au maximum. Plus cool que ça et tu meurs… L’expérience ALICE se consacre justement à l’étude de ce plasma. Chaque année, le LHC opère pour quelques semaines avec des ions de plomb au lieu des protons. ALICE accumule des données durant les collisions protons-protons et celles d’ions lourds. Même lorsque ce sont seulement des protons qui entrent en collision, les projectiles ne sont pas des balles pleines comme au billard mais bien des objets composites. En comparant ce que l’on obtient à partir de collisions d’ions ou de protons, les physicien-ne-s d’ALICE doivent d’abord distinguer ce qui vient du fait que les projectiles sont des protons liés dans le noyau ou bien à l’état libre. Jusqu’à maintenant, il semble que le plasma de quarks et gluons ne se forme que dans les collisions d’ions puisqu’ils sont les seuls à fournir la densité d’énergie requise sur un volume assez substantiel (le volume d’un noyau atomique). Certains des effets observés, comme le nombre de particules à émerger du plasma de quarks et gluons à différents angles ou vitesses dépend en partie de la nature de l’état final créé. Quand un plasma se forme, il réabsorbe une partie des particules émises, de telles sortent qu’on en voit beaucoup moins sortir de ces collisions. Les collisions de protons sur des ions lourds permettront peut-être de démêler ce qui est attribuable à l’état initial (protons libres ou liés dans le noyau) et l’état final (comme lorsque le plasma réabsorbe une partie des particules émises). Déjà, avec une seule journée d’opération à ce régime, la collaboration ALICE vient de publier deux articles scientifiques. Le premier article donne la mesure de la densité de hadrons chargés produits dans des collisions proton-ions comparée aux mêmes mesures effectuées avec des collisions protons-protons ou ions-ions, après avoir normalisé le tout. Le second article porte sur la comparaison des distributions de quantités de mouvement des hadrons chargés pour des collisions protons-protons et ions-ions. Le but ultime est d’étudier les fonctions de structure des projectiles utilisés, c’est-à-dire décrire comment les quarks et les gluons sont distribués à l’intérieur des protons quand ils sont libres ou liés dans le noyau des ions de plomb. Bien d’autres études suivront au début de 2013 durant la période de deux mois consacrée aux collisions protons-ions. « Cliché » des débris d’une collision de proton-ion de plomb capturé par le détecteur ALICE montrant un grand nombres de particules diverses crées à partir de l’énergie dégagée. Pauline Gagnon Pour être averti-e lors de la parution de nouveaux blogs, suivez-moi sur Twitter: @GagnonPauline ou par e-mail en ajoutant votre nom à cette liste de distribution Tags: LHC, proton-ion collisions, quark-gluon plasma Posted in Latest Posts \| No Comments » ### “Snowmass” (Not Snowmass) Saturday, October 13th, 2012 Every so often, perhaps once or twice a decade, particle physics in the United States comes to some kind of a crossroads that requires us to think about the long-term direction of the field. Perhaps there is new experimental data that is pointing in new directions, or technology developments that make some new facility possible, or we’re seeing the end of the previous long-term plan and it’s time to develop the next one. And when this happens, the cry goes up in the community — “We need a Snowmass!” Snowmass refers to Snowmass Village in Colorado, just down the road from Aspen, the home of the Aspen Center for Physics, a noted haunt for theorists. During the winter, Snowmass a ski resort. During the summer, it’s a mostly empty ski resort, where it’s not all that expensive to rent some condos and meeting rooms for a few weeks. Over the past few decades there have been occasional “summer studies” held at Snowmass, typically organized by the Division of Particles and Fields of the American Physical Society (and sponsored by a host of organizations and agencies). It’s a time for the particle-physics community to come together for a few weeks and spend some quality time focusing on long-range planning. The last big Snowmass workshop was in 2001. At the time, the Fermilab Tevatron was just getting started on a new data run after a five-year shutdown for upgrades, and the LHC was under construction. The top quark had been discovered, but was not yet well characterized. We were just beginning to understand neutrino masses and mixing. The modern era of observational cosmology was just beginning. A thousand physicists came to Snowmass over the course of three weeks to plot the future of the field. (And I was a lot younger.) Flash forward eleven years: the Tevatron has been shut down (leaving the US without a major high-energy particle collider), the LHC is running like gangbusters, we’re trying to figure out what dark energy is, and just in the past year two big shoes have dropped — we have measured the last neutrino mixing angle, and, quite famously, observed what could well be the Higgs boson. So indeed, it is time for another Snowmass workshop. This week I came to Fermilab for a Community Planning Meeting for next year’s Snowmass workshop. Snowmass 2013 is going to be a bit different than previous workshops in that it will not actually be at Snowmass! Budgetary concerns and new federal government travel regulations have made the old style of workshop infeasible. Instead, there will be a shorter meeting this summer hosted by our colleagues at the University of Minnesota (hats off to thee for having us), so this time we won’t have as much time during the workshop to chew over the issues, and more work will have to be done ahead of time. (But I suspect that we’re still going to call this workshop “Snowmass”, just as the ICHEP conference was “the Rochester conference” for such a long time, even if it’s now the “Community Summer Study”.) This Snowmass is being organized along the three “frontiers” that we’re using to classify the current research efforts in the field — energy, intensity and cosmic. As someone who works at the LHC, I’m most familiar with what’s going on at the energy frontier, and certainly there are important questions that have only come into focus this year. Did we observe the Higgs boson at the LHC? What more do we have to know about it to believe that it’s the Higgs? What are the implications of not having observed any other new particles yet for particle physics and for future experiments? The Snowmass study will help us understand how we answer these questions, and specifically what experiments and facilities are needed to do so. There are lots of interesting ideas that are out there right now. Can the LHC tell us what we need to know, possibly with an energy or luminosity upgrade? Is this the time to build a “Higgs factory” that would allow us to study measure Higgs properties precisely? If so, what’s the right machine for that? Or do we perhaps need an accelerator with even greater energy reach, something that will help us create new particles that would be out of reach of the LHC? What kind of instrumentation and computing technologies are needed to make sense of the particle interactions at these new facilities? The intensity and cosmic frontiers have equally big and interesting questions. I would posit that the scientific questions of particle physics have not been so compelling for a long time, and that it is a pivotal time to think about what new experiments are needed. However, we also have the bracing reality that we are looking at these questions in a budget environment that is perhaps as constrained as it has ever been. Presentations from our champions and advocates at the Department of Energy and the National Science Foundation, the agencies that fund this research (and that sponsor the US LHC blog) were encouraging about the scientific opportunities but also noted the boundary conditions that arise from the federal budget as a whole, national research priorities, and our pre-existing facilities plan. It will continue to be a challenge to make the case for our work (compelling as it may be to us, and to someone who might be interested in looking at the Quantum Diaries site) and to envision a set of facilities that can be built and used given the funding available. The first (non-native) settlers of Snowmass, Colorado, were miners, who were searching for buried treasure under adverse conditions. They were constrained by the technology of the time, and the facilities that were available for their work. I shouldn’t suggest that what we are doing is exactly like mining (it’s much safer, for one thing), but hopefully when we go to Snowmass (or really “Snowmass”) we will be figuring out how to develop the technology and facilities that are needed to extract an even greater treasure. Tags: Higgs, LHC, Snowmass Posted in Latest Posts \| 4 Comments » ### Read-Set-Go: The LHC 2012 Schedule Thursday, September 20th, 2012 From Now Until Mid-December, Expect One Thing from the LHC: More Collisions. Figure 1: Integrated luminosity for LHC Experiments versus time. 8 TeV proton-proton collisions began in April 2012. Credit: CERN Hi All, Quick post today. That plot above represents the amount of 8 TeV data collected by the LHC experiments. As of this month, the ATLAS and CMS detector experiments have each collected 15 fb-1 of data. A single fb-1 (pronounced: inverse femto-barn) is equivalent to 70 trillion proton-proton collisions. In other words, ATLAS and CMS have each observed 1,050,000,000,000,000 proton-proton collisions. That is 1.05 thousand-trillion, or 1.05×1015. To understand how gargantuan a number this is, consider that it took the LHC’s predecessor, the Tevatron, 24 years to deliver 12 fb-1 of proton-antiproton collisions. The LHC has collected this much data in five months. Furthermore, proton-proton collisions will officially continue until at least December 16th, at which time CERN will shut off the collider for the holiday season. Near the beginning of the calendar year, we can expect the LHC to collide lead ions for a while before the long, two-year shut down. During this time, the LHC magnets will be upgraded in order to allow protons to run at 13 or 14 TeV, and the detector experiments will get some much-needed tender loving care maintenance and upgrades. To estimate how much more data we might get before the New Year, let’s assume that the LHC will deliver 0.150 fb-1 per day from now until December 16th. I consider this to be a conservative estimation, but I refer you to the LHC’s Performance and Statistics page. I also assume that the experiments operate at 100% efficiency (not so conservative but good enough). Running 7 days a week puts us at a little over 1 fb-1 per week. According to the LHC schedule, there about about 10 more weeks of running (12 weeks until Dec. 16 minus 2 weeks for “machine development”). By this estimation, both ATLAS and CMS will have at least 25 fb-1 of data each before shut down! 25 fb-1 translates to 1.75 thousand-trillion proton-proton collisions, more than four times as much 8 TeV data used to discover the Higgs boson in July. Fellow QDer Ken Bloom has a terrific breakdown of what all this extra data means for studying physics. Up-to-the-minute updates about the LHC’s performance are available via the LHC Programme Coordinate Page, @LHCstatus, and @LHCmode. There are no on-going collisions at the moment because the LHC is currently under a technical stop/beam recommissioning/machine development/scrubbing, but things will be back to normal next week. Happy Colliding - richard (@bravelittlemuon) 10 fb-1 were recorded each by CDF and DZero, but to be fair, it also took Fermilab about 100 million protons to make 20 or so antiprotons. ** The Higgs boson discovery used 5 fb-1 of 7 TeV data and 5.5 fb-1 of 8 TeV data Tags: @bravelittlemuon, data, December, fb, fb-1, inverse femtobarn, LHC, schedule Posted in Latest Posts \| No Comments » ### Beyond the Higgs: Training PanDA to Tackle Astrophysics, Biology Tuesday, September 18th, 2012 The art of data mining is about searching for the extraordinary within a vast ocean of regularity. This can be a painful process in any field, but especially in particle physics, where the amount of data can be enormous, and ‘extraordinary’ means a new understanding about the fundamental underpinnings of our universe. Now, a tool first conceived in 2005 to manage data from the world’s largest particle accelerator may soon push the boundaries of other disciplines. When repurposed, it could bring the immense power of data mining to a variety of fields, effectively cracking open the possibility for more discoveries to be pulled up from ever-increasing mountains of scientific data. Advanced data management tools offer scientists a way to cut through the noise by analyzing information across a vast network. The result is a searchable pool that software can sift through and use for a specific purpose. One such hunt was for the Higgs boson, the last remaining elementary particle of the Standard Model that, in theory, endows other particles with mass. With the help of a system called PanDA, or Production and Distributed Analysis, researchers at CERN’s Large Hadron Collider (LHC) in Geneva, Switzerland discovered such a particle by slamming protons together at relativistic speeds hundreds of millions of times per second. The data produced from those trillions of collisions—roughly 13 million gigabytes worth of raw information—was processed by the PanDA system across a worldwide network and made available to thousands of scientists around the globe. From there, they were able to pinpoint an unknown boson containing a mass between 125–127 GeV, a characteristic consistent with the long-sought Higgs. An ATLAS event with two muons and two electrons - a candidate for a Higgs-like decay. The two muons are picked out as long blue tracks, the two electrons as short blue tracks matching green clusters of energy in the calorimeters. ATLAS Experiment © 2012 CERN. The sheer amount of data arises from the fact that each particle collision carries unique signatures that compete for attention with the millions of other collisions happening nanoseconds later. These must be recorded, processed, and analyzed as distinct events in a steady stream of information. (more…) Tags: ATLAS, Brookhaven National Laboratory, computing, Higgs, LHC Posted in Latest Posts \| No Comments » ### Understanding the Higgs search Wednesday, August 15th, 2012 It’s been over a month since CERN hosted a seminar on the updated searches for the Higgs boson. Since then ATLAS and CMS and submitted papers showing what they found, and recently I got news that the ATLAS paper was accepted by Physics Letters B, a prestigious journal of good repute. For those keeping score, that means it took over five weeks to go from the announcement to publication, and believe it not, that’s actually quite fast. Crowds watch the seminar from Melbourne, Australia (CERN) However, all this was last month’s news. Within a week of finding this new particle physicists started on the precision spin measurement, to see if it really is the Higgs boson or not. Let’s take a more detailed look at the papers. You can see both papers as they were submitted on the arXiv here: ATLAS / CMS. ### The Higgs backstory In order to fully appreciate the impact of these papers we need to know a little history, and a little bit about the Higgs boson itself. We also need to know some of the fundamentals of scientific thinking and methodology. The “Higgs” mechanism was postulated almost 50 years ago by several different theorists: Brout, Englert, Guralnik, Hagen, Higgs, and Kibble. For some reason Peter Higgs seems to have his name attached to this boson, maybe because his name sounds “friendliest” when you put it next to the word “boson”. The “Brout boson” sounds harsh, and saying “Guralnik boson” a dozen times in a presentation is just awkward. Personally I prefer the “Kibble boson”, because as anyone who owns a dog will know, kibble gets everywhere when you spill it. You can tidy it up all you like and you’ll still be finding bits of kibble months later. You may not find bits often, but they’re everywhere, much like the Higgs field itself. Anyway, this is all an aside, let’s get back to physics. It helps to know some of history behind quantum mechanics. The field of quantum mechanics started around the beginning of the 20th century, but it wasn’t until 1927 that the various ideas started to get resolved into a consistent picture of the universe. Some of the greatest physicists from around the world met at the 1927 Solvay Conference to discuss the different ideas and it turned out that the two main approaches to quantum mechanics, although they looked different, were actually the same. It was just a matter of making everything fit into a consistent mathematical framework. At that time the understanding of nature was that fields had to be invariant with respect to gauge transformation and Lorentz transformations. The Solvay Conference 1927, where some of the greatest physicists of the 20th century met and formulated the foundations of modern quantum mechanics. (Wikipedia) A gauge transformation is the result of the kind of mathematics we need to represent particle fields, and these fields must not introduce new physics when they get transformed. To take an analogy, imagine you have the blueprints for a building and you want to make some measurements of various distances and angles. If someone makes a copy of the blueprints, but changes the direction of North (so that the building faces another direction) then this must not change any of the distances or angles. In that sense the distances and angles in blueprint are rotation-invariant. They are rotation-invariant because we need to use Euclidean space to represent the building, and a consequence of using Euclidean space is that any distances and angles described in the space must be invariant with respect to rotation. In quantum mechanics we use complex numbers to represent the field, and a gauge transformation is just a rotation of a complex number. The Lorentz transformation is a bit simpler to understand, because it’s special relativity, which says that if you have a series of events, observers moving at different speeds and in different directions will agree on the causality of those events. The rest of special relativity is just a matter of details, and those details are a lot of fun to look at. By the time all of quantum mechanics was coming together there were excellent theories that took these symmetries into account. Things seemed to be falling into place, and running the arguments backwards lead to some very powerful predictions. Instead of observing a force and then requiring it to be gauge and Lorentz invariant, physicists found they could start with a gauge and Lorentz invariant model and use that to predict what forces can exist. Using plain old Euclidean space and making it Lorentz invariant gives us Minkowski space, which is the perfect for making sure that our theories work well with special relativity. (To get general relativity we start with a space which is not Euclidean.) Then we can write the most general description of a field we can think of in this space as long as it is gauge invariant and that’s a valid physical field. The only problem was that there were some interactions that seemed to involve a massive photon-like boson. Looking at the interactions gave us a good idea of the mass of this particle, the $$W$$ boson. In the next few decades new particles were discovered and the Standard Model was proposed to describe all these phenomena. There are three forces in the Standard Model, the electromagnetic force, the weak force, and the strong force, and each one has its own field. ### Inserting the Higgs field The Higgs field is important because it unifies two of the three fundamental fields in particle physics, electromagnetism and the weak fields. It does this by mixing all the fields up (and in doing so, it mixes the bosons up.) Flip Tanedo has tried to explain the process from a theorist’s point of view to me privately on more than one occasion, but I must admit I just ended up a little confused by some of the finer points. The system starts with three fields which are pretty much all the same as each other, the $$W_1$$, $$W_2$$, and the $$W_3$$. These fields don’t produce any particles themselves because they don’t obey the relevant physical laws (it’s a bit more subtle in reality, but that’s a blog post in itself.) If they did produce their own fields then they would generate massless particles known as Goldstone bosons, and we haven’t seen these, so we know there is something else going on. Instead of making massless bosons they mix amongst themselves to create new fields, giving us massive bosons, and the Goldstone bosons get converted into extra degrees of freedom. Along comes the Higgs field and suddenly these fields separate and mix, giving us four new fields. The Higgs field, about to break the symmetry and give mass (Flip Tanedo) The $$W_1$$ and $$W_2$$ mix to give us the $$W^+$$ and $$W^-$$ bosons, and then the $$W_3$$ field meets the $$B$$ field to give us the $$Z$$ boson and the photon. What makes this interesting is that the photon behaves well on its own. It has no mass and this means that its field is automatically gauge invariant. Nature could have decided to create just the electromagnetic field and everything would work out fine. Instead we have the photon and three massive bosons, and the fields of these massive bosons cannot be gauge invariant by themselves, they need something else to make it all balance out. By now you’ve probably guessed what this mystery object is, it’s the Higgs field and with it, the Higgs boson! This field fixes it all up so that the fields mix, we get massive bosons and all the relevant laws (gauge invariance and Lorentz invariance) are obeyed. Before we go any further it’s worth pointing a few things out. The mass of the $$W$$ boson is so large in comparison to other particles that it slows down the interactions of a lot of particles, and this is one of the reasons that the sun burns so “slowly”. If the $$W$$ boson was massless then it could be produced in huge numbers and the rate of fusion in the sun would be much faster. The reason we have had a sun for billions of years, allowing the evolution of life on Earth (and maybe elsewhere) is because the Higgs field gives such a large mass to the $$W$$ boson. Just let that thought sink in for a few seconds and you’ll see the cosmic significance of the Higgs field. Before we get ahead ourselves we should note that the Higgs field leads to unification of the electromagnetic and weak forces, but it says nothing about the strong force. Somehow the Higgs field has missed out one of the three fundamental forces of the Standard Model. We may one day unite the three fields, but don’t expect it to happen any time soon. ### “Observation” vs “discovery”, “Higgs” vs “Higgs-like” There’s one more thing that needs to be discussed before looking at the papers and that’s a rigorous discussion of what we mean by “discovery” and if we can claim discover of the Standard Model Higgs boson yet. “Discovery” has come to mean a five sigma observation of a new resonance, or in other words that probability that the Standard Model background in the absence of a new particle would bunch up like this is less than one part in several million. If we see five sigma we can claim a discovery, but we still need to be a little careful. Suppose we had a million mass points, what is the probability that there is one five sigma fluctuation in there? It’s about $$20\%$$, so looking at just the local probability is not enough, we need to look at the probability that takes all the data points into account. Otherwise we can increase the chance of seeing a fluctuation just by changing the way we look at the data. Both ATLAS and CMS have been conscious of this effect, known as the “Look Elsewhere Effect”, so every time they provide results they also provide the global significance, and that is what we should be looking at when we talk about the discovery. Regular readers might remember Flip’s comic about me getting worked up over the use of the word “discovery” a few weeks back. I got worked up because the word “discovery” had been misused. Whether an observation is $$4.9$$ or $$5.1$$ sigma doesn’t matter that much really, and I think everyone agrees about that. What bothered me was that some people decided to change what was meant by a discovery after seeing the data, and once you do that you stop being a scientist. We can set whatever standards we like, but we must stick to them. Burton, on the other hand, was annoyed by a choice of font. Luckily our results are font-invariant, and someone said “If you see five sigma you can present in whatever durn font you like.” Getting angry over the change of goalposts. Someone has to say these things. In addition to knowing what we mean by “discovery” we also need to take hypothesis testing into account. Anyone who claims that we have discovered the Higgs boson is as best misinformed, and at worst willingly untruthful. We have discovered a new particle, there’s no doubt about that, but now we need to eliminate things are not the Higgs until we’re confident that the only thing left is the Higgs boson. We have seen this new particle decay to two photons, and this tells us that it can only only have spin 0 or spin 2. That’s eliminated spin 1, spin 3, spin 4… etc for us, all with a single measurement. What we are doing now trying to exclude both the spin 0 and spin 2 possibilities. Only one of these will be excluded, and then will know for sure what the spin is. And then we know it’s the Standard Model Higgs boson, right? Not quite! Even if we know it’s a spin 0 particle we would still need to measure its branching fractions to confirm that it is what we have been looking for all along. Bear this in mind when thinking about the paper- all we have seen so far is a new particle. Just because we’re searching for the Higgs and we’ve found something new it does not mean that it’s a the Higgs boson. ### The papers Finally we get to the papers. From the titles we can see that both ATLAS and CMS have been suitably agnostic about the particle’s nature. Neither claim it’s the Higgs boson and neither even claim anything more than an “observation”. The abstracts tell us a few useful bits of information (note that the masses quoted agree to within one sigma, which is reassuring) but we have to tease out the most interesting parts by looking at the details. Before the main text begins each experiment dedicates their paper to the memories of those who have passed away before the papers were published. This is no short list of people, which is not surprising given that people have been working on these experiments for more than 20 years. Not only is this a moving start to the papers, it also underlines the impact of the work. Both papers were dedicated to the memories of colleagues who did not see the observation. (CMS) Both papers waste no time getting into the heart of the matter, which is nature of the Standard Model and how it’s been tested for several decades. The only undiscovered particle predicted by the Standard Model is the Higgs boson, we’ve seen everything else we expected to see. Apart from a handful of gauge couplings, just about every prediction of the Standard Model has been vindicated. In spite of that, the search for the Higgs boson has taken an unusually long time. Searches took place at LEP and Tevatron long before the LHC collided beams, and the good news is that the LEP limit excluded the region that is very difficult for the LHC to rule out (less than $$114GeV$$). CDF and D0 both saw an excess in the favored region, but the significance was quite low, and personally I’m skeptical since we’ve already seen that CDF’s dijet mass scale might have some problems associated with it. Even so we shouldn’t spend too long trying to interpret (or misinterpret) results, we should take them at face value, at least at first. Next the experiments tell us which final states they look for, and this is where things will get interesting later on. Before describing the detectors, each experiment pauses to remind us that the conditions of 2012 are more difficult than those of 2011. The average number of interactions per beam crossing increased by a factor of two, making all analyses more difficult to work with (but ultimately all our searches a little more sensitive.) At this point both papers summarize their detectors, but CMS goes out of their way to show off how the design of their detector was optimized for general Higgs searches. Having a detector which can reconstruct high momentum leptons, low momentum photons and taus, and also tag b-jets is not as easy task. Both experiments do well to be able to search for the Higgs bosons in the channels they look at. Even if we limit ourselves to where ATLAS looked the detectors would still have trigger on leptons and photons, and be able to reconstruct not only those particles, but also the missing transverse energy. That’s no easy task at a hadron collider with many interactions per beam crossing. The two experiments have different overall strategies to the Higgs searches. ATLAS focused their attention on just two final states in 2012: $$\gamma\gamma$$, and $$ZZ^$$, whereas CMS consider five final sates: $$\gamma\gamma$$, $$ZZ^$$, $$WW^$$, $$\tau\tau$$, and $$b\bar{b}$$. ATLAS focus mostly on the most sensitive modes, the so-called “golden channel”, $$ZZ^$$, and the fine mass resolution channel, $$\gamma\gamma$$. With a concerted effort, a paper that shows only these modes can be competitive with a paper that shows many more, and labor is limited on both experiments. CMS spread their effort across several channels, covering all the final states with expected sensitivities comparable to the Standard Model. ### $$H\to ZZ^$$ The golden channel analysis has been presented many times before because it is sensitive across a very wide mass range. In fact it spans the range $$110-600GeV$$, which is the entire width of the Higgs search program at ATLAS and CMS. (Constraints from other areas of physics tell us to look as high as $$1000GeV$$, but at high masses the Higgs boson would have a very large width, making it extremely hard to observe. Indirect results favor the low mass region, which is less than around $$150GeV$$.) Given the experience physicists have had with this channel it’s no surprise that the backgrounds are very well understood at this point. The dominant “irreducible” background comes from Standard Model production of $$Z/\gamma$$ bosons, where there is one real $$Z$$ boson, and one “off-shell”, or virtual boson. This is called irreducible because the source of background is the same final state as the signal, so we can’t remove further background without also removing some signal. This off-shell boson can be an off-shell $$Z$$ boson or an off-shell photon, it doesn’t really matter which since these are the same for the background. In the lower mass range there are also backgrounds from $$t\bar{t}$$, but fortunately these are well understood with good control regions in the data. Using all this knowledge, the selection criteria for $$8TeV$$ were revisited to increase sensitivity as much as possible. The invariant mass spectrum for ATLAS's H→ZZ* search (ATLAS) Since this mode has a real $$Z$$ boson, we can look for two high momentum leptons in the final state, which mames things especially easy. The backgrounds are small, and the events are easy to identify, so the trigger is especially simple. Events are stored to disk if there is at least one very high momentum lepton, or two medium momentum leptons which means that we don’t have to throw any events away. Some triggers fire so rapidly that we can only store some of the events from them, and we call this prescaling. When we keep $$1$$ in $$n$$ events then we have a prescale of $$n$$. For a Higgs search we want to have a high efficiency as possible so we usually require a prescale of $$1$$. Things are not quite so nice for the $$\gamma\gamma$$ mode, as we’ll see later. The invariant mass spectrum for CMS's H→ZZ* search (CMS) After applying a plethora of selections on the leptons and reconstructing the $$Z$$ and Higgs boson candidates the efficiency for the final states vary from $$15\%-37\%$$, which is actually quite high. No detector can cover the whole of the solid angle, and efficiencies vary with the detector geometry. The efficiency needs to be very high because the fraction of Higgs bosons that would decay to these final states is so small. At a mass of $$125GeV$$ the branching fraction to the $$ZZ^$$ state is about $$2\%$$, and then branching fraction of $$Z$$ to two leptons is about $$6\%$$. Putting that all together means that only $$1$$ in $$10,000$$ Higgs bosons would decay to this final state. At a mass of $$125GeV$$ the LHC would produce about $$15,000$$ Higgs bosons per $$fb^{-1}$$. So for $$10fb^{-1}$$ we could expect to have about $$11$$ Higgs bosons decaying to this final state, and we could expect to see about $$3$$ of those events reconstructed. This is a clean mode, but it’s an extremely challenging one. The selection criteria are applied, the background is estimated, and the results are shown. As you can see there is a small but clear excess over background in the region around $$125GeV$$ and this is evidence supporting the Higgs boson hypothesis! CMS see slightly fewer events than expected, but still see a clear excess (CMS) ### $$H\to\gamma\gamma$$ Out of the $$H\to ZZ^$$ and $$H\to\gamma\gamma$$ modes the $$\gamma\gamma$$ final state is the more difficult one to reconstruct. The triggers are inherently “noisy” because they must fire on something that looks like a high energy photon, and there are many sources of background for this. As well as the Standard Model real photons (where the rate of photon production is not small) there are jets faking photons, and electrons faking photons. This makes the mode dominated by backgrounds. In principle the mode should be easy: just reconstruct Higgs candidates from pairs of photons and wait. The peak will reveal itself in time. However ATLAS and CMS are in the middle of a neck and neck race to find the Higgs boson, so both collaborations exploit any advantage they can, and suddenly these analyses become some of the most difficult to understand. A typical H→γγ candidate event with a striking signature (CMS) To get a handle on the background ATLAS and CMS each choose to split the mode into several categories, depending on the properties of the photons or the final state, and each one with its own sensitivity. This allows the backgrounds to be controlled with different strategies in each category, leading to increased overall sensitivity. Each category has its own mass resolution and signal-to-background ratio, each is mutually independent of the others, and each has its own dedicated studies. For ATLAS the categories are defined by the presence of two jets, whether or not the photon converts (produces an $$e^-e^+$$ pair) in the detector, the pseudorapidity of the photons, and a kinematic quantity called $$p_{T_T}$$, with similar categories for CMS. When modelling the background both experiments wisely chose to use the data. The background for the $$gamma\gamma$$ final state is notoriously hard to predict accurately, because there are so many contributions from different backgrounds, from real and fake photon candidates, and many kinematic or detector effects to take into account. The choice of background model even varies on a category by category basis, and choices of model vary from simple polynomial fits to the data, to exponential and skewed Gaussian backgrounds. What makes these background models particularly troublesome is that the background has to be estimated using the signal region, so small deviations that are caused by signal events could be interpreted by the fitting algorithm as a weird background shape. The fitting mechanism must be robust enough to fit the background shapes without being fooled into thinking that a real excess of events is just a slightly different shape. ATLAS's H→γγ search, where events are shown weighted (top) and unweighted (bottom) (ATLAS) To try to squeeze even more sensitivity out of the data CMS use a boosted decision tree to aid signal separation. A boosted decision tree is a sophisticated statistical analysis method that uses signal and background samples to decide what looks like signal, and then uses several variables to return just one output variable. A selection can be made on the output variable that removes much of the background while keeping a lot of the signal. Using boosted decision trees (or any multivariate analysis technique) requires many cross checks to make sure the method is not biased or “overtrained”. CMS's H→γγ search, where events are shown weighted (main plot) and unweighted (inset) (CMS) After analyzing all the data the spectra show a small bump. The results can seem a little disappointing at first, after all the peak is barely discernable, and so much work has gone into the analyses. Both experiments show the spectra after weighting the events to take the uncertainties into account and this makes the plots a little more convincing. Even so, what matters is the statistical significance of these results, and this cannot be judged by eye. The final results show a clear preference for a boson with a mass of $$125GeV$$, consistent with the Higgs boson. CMS see a hint at around $$135GeV$$, but this is probably just a fluctuation, given that ATLAS do not see something similar. ATLAS local significance for H→γγ (ATLAS) (If you’ve been reading the blog for a while you may remember a leaked document from ATLAS that hinted at a peak around $$115GeV$$ in this invariant mass spectrum. That document used biased and non peer-reviewed techniques, but the fact remains that even without these biases there appear to be a small excess in the ATLAS data around $$115GeV$$. The significance of this bump has decreased as we have gathered more data, so it was probably just a fluctuation. However, you can still see a slight bump at $$115GeV$$ in the significance plot. Looking further up the spectrum, both ATLAS and CMS see very faint hints of something at $$140GeV$$ which appears in both the $$ZZ^$$ and $$\gamma\gamma$$ final states. This region has already been excluded for a Standard Model Higgs, but there may be something else lurking out there. The evidence is feeble at the moment, but that’s what we’d expect for a particle with a low production cross section.) ### $$H\to WW^$$ One of the most interesting modes for a wide range of the mass spectrum is the $$WW()$$ final state. In fact, this is the first mode to be sensitive to the Standard Model Higgs boson searches, and exclusions were seen at ATLAS, CMS, and the Tevatron experiments at around $$160GeV$$ (the mass of two on-shell $$W$$ bosons) before any other mass region. The problem with this mode is that it has two neutrinos in the final state. It would be nice to have an inclusive sample of $$W$$ bosons, including the hadronic final states, but the problems here are the lack of a good choice of trigger, and the irreducible and very large background. That mean that we must select events with two leptons and two neutrinos in them. As the favored region excludes more and more of the high mass region this mode gets more challenging, because at first we lose the mass constraint on the second $$W$$ boson (as it must decay off-shell), and secondly because we must be sensitive in the low missing transverse energy region, which starts to approach our resolution for this variable. While we approach our resolution from above, the limit on the resolution increases from below, because the number of interactions per beam crossing increases, increasing the overall noise in the detector. To make progress in this mode takes a lot of hard work for fairly little gain. Both papers mention explicitly how difficult the search is in a high pileup scenario, with CMS stating “The analysis of the $$7TeV$$ data is described in [referenced paper] and remains unchanged, while the $$8TeV$$ analysis was modified to cope with more difficult conditions induced by the higher pileup of the 2012 data taking.” and ATLAS saying “The analysis of the $$8TeV$$ data presented here is focused on the mass range $$110<m_H<200GeV$$ It follows the procedure used for the $$7TeV$$ data described in [referenced paper], except that more stringent criteria are applied to reduce the $$W$$+jets background and some selections have been modified to mitigate the impact of the high instantaneous luminosity at the LHC in 2012.” It’s not all bad news though, because the final branching fraction to this state is much higher than that of the $$ZZ^$$ final state. The branching fraction for the Standard Model Higgs boson to $$WW^$$ is about $$10$$ times higher than that for $$ZZ^$$, and the branching fraction of the $$W$$ boson to leptons is also about $$3$$ times higher than the $$Z$$ boson to leptons, which gives another order of magnitude advantage. Unfortunately all these events must be smeared out across a large spectrum. There is one more trick we have up our sleeves though, and it comes from the spin of the parent. Since the Standard Model Higgs boson has zero spin the $$W$$ bosons tend to align their spins in opposite directions to make it all balance out. This then favors one decay direction over another for the leptons. The $$W^+$$ boson decays with a neutrino in the final state, and because of special relativity the neutrino must align its spin against its direction of motion. The $$W-$$ boson decays with an anti-neutrino, which takes its spin with its direction of motion. This forces the two leptons to travel in the same direction with respect to the decay axis of the Higgs boson. The high momenta of the leptons smears things out a bit, but generally we should expect to see one high momentum lepton, and a second lower momentum lepton n roughly the same region of the detector. The transverse mass for ATLAS's H→WW* search (ATLAS) ATLAS did not actually present results for the $$WW^$$ final state on July 4th, but they did show it in the subsequent paper. CMS showed the $$WW^$$ final state on July 4th, although it did somewhat reduce their overall significance. Both ATLAS and CMS spend some of the papers discussing the background estimates for the $$WW^$$ mode, but ATLAS seem to go to more significant lengths to describe the cross checks they used in data. In fact this may help to explain why ATLAS did not quite have the result ready for July 4th, whereas CMS did. There’s a trade off between getting the results out quickly and spending some extra time to understand the background. This might have paid off for ATLAS, since they seem to be more sensitive in this mode than CMS. The invariant mass for CMS's H→WW search (CMS) After looking at the data we can see that both ATLAS and CMS are right at the limits of their sensitivity in this mode. They are not limited by statistics, they are limited by uncertainties, and the mass point $$125GeV$$ sits uncomfortably close some very large uncertainties. The fact that this mode is sensitive at all is a tribute to the hard work of dozens of physicists who went the extra mile to make it work. CMS's observed and expected limits for H→WW, showing the dramatic degradation in sensitivity as the mass decreases (CMS) ### $$H\to b\bar{b}$$ At a mass of $$125GeV$$ by far the largest branching fraction of the Standard Model Higgs boson is to $$b\bar{b}$$. CDF and D0 have both seen a broad excess in this channel (although personally I have some doubts about the energy scale of jets at CDF, given the dijet anomaly they see that D0 does not see) hinting at a Higgs boson of $$120-135GeV$$. The problem with this mode is that the background is many orders of magnitude larger than the signal, so some special tricks must be used to remove the background. What is done at all four experiments is to search for a Higgs boson that is produced in associated with a $$W$$ or $$Z$$ boson, and this greatly reduces the background. ATLAS did not present an updated search in the $$b\bar{b}$$ channel, and taking a look at the CMS limits we can probably see why, the contribution is not as significant as in other modes. The way CMS proceed with the analysis is to use several boosted decision trees (one for each mass point) and to select candidates based on the output of the boosted decision tree. The result is less than $$1$$ sigma of significance, about half of what is expected, but if this new boson is the Higgs boson then this significance will increase as we gather more data. A powerful H→bb search requires a boosted decision tree, making the output somewhat harder to interpret (CMS) It’s interesting to note that the $$b\bar{b}$$ final state is sensitive to both a spin 0 and a spin 2 boson (as I explained in a previous post) and it may have different signal strength parameters for different spin states. The signal strength parameter tells us how many events we see compared to how many events we do see, and it is denoted with the symbol $$\mu$$. A there is no signal then $$\mu=0$$, if the signal is exactly as large as we expect then $$\mu=1$$, and any other value indicates new physics. It’s possible to have a negative value for $$\mu$$ and this would indicate quantum mechanical interference of two or more states that cancel out. Such an interference term is visible in the invariant mass of two leptons, as the virtual photon and virtual $$Z$$ boson wavefunctions interfere with each other. ### $$H\to\tau\tau$$ Finally, the $$\tau\tau$$ mode is perhaps the most enlightening and the most exciting right now. CMS showed updated results, but ATLAS didn’t. CMS’s results were expected to approach the Standard Model sensitivity, but for some reason their results didn’t reach that far, and that is crucially important. CMS split their final states by the decay mode of the $$\tau$$, where the final states include $$e\mu 4\nu$$, $$\mu\mu 4\nu$$, $$\tau_h\mu 3\mu$$, and $$\tau_h e3\nu$$, where $$\tau_h$$ is a hadronically decaying $$\tau$$ candidate. This mode has at least three neutrinos in the final state, so like the $$WW^$$ mode the events get smeared across a mass spectrum. There are irreducible backgrounds from $$Z$$ bosons decaying to $$\tau\tau$$ and from Drell-Yan $$\tau\tau$$ production, so the analysis must search for an excess of events over these backgrounds. In addition to the irreducible backgrounds there are penalties in efficiency associated with the reconstruction of $$\tau$$ leptons, which make this a challenging mode to work this. There are dedicated algorithms for reconstructing hadronically decaying $$\tau$$ jets, and these have to balance out the signal efficiency for real $$tau$$ leptons and background rejection. CMS's H→τtau; search, showing no signal (CMS) After looking at the data CMS expect to see an excess of $$1.4$$ sigma, but they actually see $$0$$ sigma, indicating that there may be no Standard Model Higgs boson after all. Before we jump to conclusions it’s important to note a few things. First of all statistical fluctuations happen, and they can go down just as easily as they can go up, so this could just be a fluke. It’s a $$1.5$$ sigma difference, so the probability of this being due a fluctuation if the Standard Model Higgs boson is about $$8\%$$. On its own that could be quite low, but we have $$8$$ channels to study, so the chance of this happening in any one of the channels is roughly $$50\%$$, so it’s looking more likely that this is just a fluctuation. ATLAS also have a $$\tau\tau$$ analysis, so we should expect to see some results from them in the coming weeks or months. If they also don’t see a signal then it’s time to start worrying. CMS's limit of H→ττ actually shows a deficit at 125GeV. A warning sign for possible trouble for the Higgs search! (CMS) ### Combining results Both experiments combine their results and this is perhaps the most complicated part of the whole process. There are searches with correlated and uncorrelated uncertainties, there are two datasets at different energies to consider, and there are different signal-to-background ratios to work with. ATLAS and CMS combine their 2011 and 2012 searches, so they both show all five main modes (although only CMS show the $$b\bar{b}$$ and $$\tau\tau$$ modes in 2012.) When combining the results we can check to see if the signal strength is “on target” or not, and there is some minor disagreement between the modes. For the $$ZZ^$$ and $$WW^$$ modes, the signal strengths are about right, but for the $$\gamma\gamma$$ mode it’s a little high for both experiments, so there is tension between these modes. Since these are the most sensitive modes, and we have more data on the way then this tension should either resolve itself, or get worse before the end of data taking. The $$b\bar{b}$$ and $$\tau\tau$$ modes are lower than expected for both experiments (although for ATLAS the error bars are so large it doesn’t really matter), suggesting that this new particle may a non-Standard Model Higgs boson, or it could be something else altogether. Evidence of tension between the γγ and fermionic final states (CMS) While the signal strengths seem to disagree a little, the masses all seem to agree, both within experiments and between them. The mass of $$125GeV$$ is consistent with other predictions (eg the Electroweak Fit) and it sheds light on what to look for beyond the Standard Model. Many theories favor a lower mass Higgs as part of a multiplet of other Higgs bosons, so we may see some other bosons. In particular, the search for the charged Higgs boson at ATLAS has started to exclude regions on the $$\tan\beta$$ vs $$m_{H^+}$$ plane, and the search might cover the whole plane in the low mass region by the end of 2012 data taking. Although a mass of $$125GeV$$ is consistent with the Electroweak Fit, it is a bit higher than the most favored region (around $$90GeV$$) so there’s certainly space for new physics, given the observed exclusions. The masses seem to agree, although the poor resolution of the WW* mode is evident when compared to the ZZ* and γγ modes (ATLAS) To summarize the results, ATLAS sees a $$5.9$$ sigma local excess, which is $$5.1$$ sigma global excess, and technically this is a discovery. CMS sees a $$5.0$$ sigma local excess, which is $$4.6$$ sigma global excess, falling a little short of a discovery. The differences in results are probably due to good luck on the part of ATLAS and bad luck on the part of CMS, but we’ll need to wait for more data to see if this is the case. The results should “even out” if the differences are just due to fluctuations up for ATLAS and down for CMS. ATLAS proudly show their disovery (ATLAS) ### Looking ahead If you’ve read this far then you’ve probably picked up on the main message, we haven’t discovered the Standard Model Higgs boson yet! We still have a long road ahead of us and already we have moved on to the next stage. We need to measure the spin of this new boson and if we exclude the spin 0 case then we know it is not a Higgs boson. If exclude the spin 2 case then we still need to go a little further to show it’s the Standard Model Higgs boson. The spin analysis is rather complicated, because we need to measure the angles between the decay products and look for correlations. We need to take the detector effects into account, then subtract the background spectra. What is left after that are the signal spectra, and we’re going to be statistically limited in what we see. It’s a tough analysis, there’s no doubt about it. We need to see the five main modes to confirm that this is what we have been looking for for so long. If we get the boson modes ($$ZZ^$$, $$WW^$$, $$\gamma\gamma$$) spot on relative to each other, then we may have a fermiophobic Higgs boson, which is an interesting scenario. (A “normal” fermiophobic Higgs boson has already been excluded, so any fermiophobic Higgs boson we may see must be very unusual.) There are also many beyond the Standard Model scenarios that must be studied. As more regions of parameter space are excluded, theorists tweak their models, and give us updated hints on where to search. ATLAS and CMS have groups dedicated to searching for beyond the Standard Model physics, including additional Higgs bosons, supersymmetry and general exotica. It will be interesting to see how their analyses change in light of the favored mass region in the Higgs search. A favored Higgs mass has implications for physics beyond the Standard Model. Combined with the limits on new particles (shown in plot) many scenarios can be excluded (ATLAS) 2012 has been a wonderful year for physics, and it looks like it’s only going to get better. There are still a few unanswered questions and tensions to resolve, and that’s what we must expect from the scientific process. We need to wait a little longer to get to the end of the story, but the anticipation is all part of the adventure. We’ll know is really happening by the end of Moriond 2013, in March. Only then can we say with certainty “We have proven/disproven the existence of the Standard Model Higgs boson”! I like to say “We do not do these things because they are easy. We do them because they are difficult”, but I think Winston Churchill said it better: This is not the end. It is not even the beginning of the end, but it is perhaps the end of the beginning.” W. Churchill ### References etc Plots and photos taken from: “Webcast of seminar with ATLAS and CMS latest results from ICHEP”, ATLAS Experiment, CERN, ATLAS-PHO-COLLAB-2012-014 Wikipedia “Observation of a new particle in the search for the Standard Model Higgs boson with the ATLAS detector at the LHC”, ATLAS Collaboration, arXiv:1207.7214v1 [hep-ex] “Observation of a new boson at a mass of 125 GeV with the CMS experiment at the LHC”, CMS Collaboration, arXiv:1207.7235v1 [hep-ex] Flip Tanedo It’s been a while since I last posted. Apologies. I hope this post makes up for it! Tags: ATLAS, CERN, CMS, Higgs, Higgs boson, LHC Posted in Latest Posts \| 34 Comments »	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 191, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9144985675811768, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/38282/find-flow-rate-of-the-aqueous-current-needed-to-process-benzene-hydrochloric-aci	# Find flow rate of the aqueous current needed to process benzene-hydrochloric acid solution [closed] I am trying to understand if my reasoning is right, as my exam is getting closer and I would really appreciate any feedback, I'd be very pleased to know if I have made any mistakes in solving this exercise. ## The problem A solution made of benzene-hydrochloric acid at 5.5% in weight needs to be purified by lowering the amount of acid to 3% in weight. In order to do that the above-mentioned solution is sent to a machinery, whereas it is dipped down in an aqueous solution of sodium hydroxyde. Part of the hydrochloric acid is solubilized in the aqueous current and reacts to the sodium hydroxyde by wearing it out according to the following formula: $$\begin{aligned} HCl+NaOH \rightarrow NaCl + H_2O\end{aligned}$$ Two liquid currents come out of the machinery: the first one contains the remaining hydrochloric acid and all of benzene, the second one contains water and sodium chloride. If we know that the sodium hydroxyde concentration in water is at 4%, calculate the flow rate of the aqueous current needed to process 137 lb/h of benzene-hydrochloric acid solution. # My solution to the problem First thing from the benzene-hydrochloridric acid solution I managed to calcuate the incoming and outcoming flow rate of the hydrochloridric acid, respectively 7.54 lb/h and 4.52 lb/h. The difference between these two amounts is actually the amount of hydrochloric acid reacting to the sodium hydroxyde. Due to stoichiometry the amount of hydrochroric acid involved in the reaction is equal to the amount of sodium hydroxyde. Therefore given the data of my problem, the newly found sodium hydroxyde amount is the 4% of the aqueous solution of the sodium hydroxyde going in the machinery. So by proportion I can calculate the water amount of the solution and then the total flow rate of the solution: $$\begin{aligned} 3.02 : 4\% = x : 96\% \Rightarrow x = 75.5\%\end{aligned}$$. - ## closed as off topic by David Zaslavsky♦Sep 25 '12 at 11:17 Questions on Physics Stack Exchange are expected to relate to physics within the scope defined in the FAQ. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about closed questions here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9408517479896545, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/91373/list	## Return to Question 4 edited tags 3 added 1 characters in body I have the following problem: Let $\mathcal{G} = (G_{i}){i} G_{i})_{i}$ be a collection of graphs. I would like to find a "system of representative edges" $f : \mathcal{G} \rightarrow \bigcup{i} bigcup_{i} E(G_{i})$ such that $f(G_{i}) \in E(G_{i})$ and for each $i, j$ either $f(G_{i}) = f(G_{j})$ or $f(G_{i}) \cap f(G_{j}) = \emptyset$. In other words, out of each graph we choose one representative edge. Edges $e_{1}$ and $e_{2}$ chosen for any two graphs need to either be non-overlapping or identiacal. This clearly looks very similar to special case of a system of distinct representatives for hypergraphs. However, as noted above, in my case the edges need not be distinct. An even more specialized case of this problem, where each $G_{i}$ is a biclique naturally emerges when considering CNF formulae (in this case the two partitions contain positive and negative literals respectively). Questions: Has such or similar problem been considered in the literature? What theorems apart from Kőnig's theorem and Hall's marriage theorem could serve as existence criteria or analysis tools for the presented problem. I will be most grateful for your help. 2 deleted 9 characters in body; edited tags; added 1 characters in body I have the following problem: Let $\mathcal{G} = { G_{i} }(G_{i}){i=1..n} i}$ be a collection of graphs. I would like to find a "system of representative edges" $f : \mathcal{G} \rightarrow \bigcup{i} E(G_{i})$ such that $f(G_{i}) \in E(G_{i})$ and for each $i, j$ either $f(G_{i}) = f(G_{j})$ or $f(G_{i}) \cap f(G_{j}) = \emptyset$. In other words, out of each graph we choose one representative edge. Edges $e_{1}$ and $e_{2}$ chosen for any two graphs need to either be non-overlapping or identiacal. This clearly looks very similar to special case of a system of distinct representatives for hypergraphs. However, as noted above, in my case the edges need not be distinct. An even more specialized case of this problem, where each $G_{i}$ is a biclique naturally emerges when considering CNF formulae (in this case the two partitions contain positive and negative literals respectively). Questions: Has such or similar problem been considered in the literature? What theorems apart from Kőnig's theorem and Hall's marriage theorem could serve as existence criteria or analysis tools for the presented problem. I will be most gratful grateful for your help. 1 # A non-distinct system of representative edges. I have the following problem: Let $\mathcal{G} = { G_{i} }{i=1..n}$ be a collection of graphs. I would like to find a "system of representative edges" $f : \mathcal{G} \rightarrow \bigcup{i} E(G_{i})$ such that $f(G_{i}) \in E(G_{i})$ and for each $i, j$ either $f(G_{i}) = f(G_{j})$ or $f(G_{i}) \cap f(G_{j}) = \emptyset$. In other words, out of each graph we choose one representative edge. Edges $e_{1}$ and $e_{2}$ chosen for any two graphs need to either be non-overlapping or identiacal. This clearly looks very similar to special case of a system of distinct representatives for hypergraphs. However, as noted above, in my case the edges need not be distinct. An even more specialized case of this problem, where each $G_{i}$ is a biclique naturally emerges when considering CNF formulae (in this case the two partitions contain positive and negative literals respectively). Questions: Has such or similar problem been considered in the literature? What theorems apart from Kőnig's theorem and Hall's marriage theorem could serve as existence criteria or analysis tools for the presented problem. I will be most gratful for your help.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9444983601570129, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/108456/list	## Return to Question 2 typo corrected By standard homological algebra we know that $Ext(A,B)$ of $R$-modules classifies certain equivalence classes of short exact sequences $0\rightarrow A\rightarrow B\rightarrow C \rightarrow B A \rightarrow 0$ of $R$-modules, where $R$ is a commutative ring. I now would like to understand this fact in geometry. 1. Let $X$ be a variety (or a scheme if you want), how should I understand $Ext(O_{Y},O_{Z})$ for subvarieties $Y,Z\subset X$? Of course it classifies extensions of $O_{Z}$ by $O_{Y}$, but are there any geometric or intuitive way to understand $Ext(O_{Y},O_{Z})$? 2. More generally are there any geometric way to understand $Ext(\mathcal{E},\mathcal{F})$ for coherent $O_X$-modules $\mathcal{E},\mathcal{F}$? I would appreciate any idea about "seeing" these extensions. 1 # How to understand $Ext(\mathcal{O}_{Y},\mathcal{O}_{Z})$ for subvarieties $Y,Z\subset X$? By standard homological algebra we know that $Ext(A,B)$ of $R$-modules classifies certain equivalence classes of short exact sequences $0\rightarrow A\rightarrow C \rightarrow B \rightarrow 0$ of $R$-modules, where $R$ is a commutative ring. I now would like to understand this fact in geometry. 1. Let $X$ be a variety (or a scheme if you want), how should I understand $Ext(O_{Y},O_{Z})$ for subvarieties $Y,Z\subset X$? Of course it classifies extensions of $O_{Z}$ by $O_{Y}$, but are there any geometric or intuitive way to understand $Ext(O_{Y},O_{Z})$? 2. More generally are there any geometric way to understand $Ext(\mathcal{E},\mathcal{F})$ for coherent $O_X$-modules $\mathcal{E},\mathcal{F}$? I would appreciate any idea about "seeing" these extensions.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8995922803878784, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/201104-power-series-center-convergence.html	# Thread: 1. ## power series and center of convergence I have a function g(x)=(2a)/(x^2-a^2) where a is some number greater than zero. I need to find a power series with a center of convergence at zero and then find the radius and the interval of convergence. Any help would be appreciated. Thanks 2. ## Re: power series and center of convergence Write it as $\frac{-2a}{a^2-x^2}=\frac{-2}a\frac 1{1-\left(\frac xa\right)^2}$ in order to use the expansion of $\frac 1{1+t^2}$ in power series.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9183530211448669, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/11233/outcomes-of-earth-slowing-down-spinning-on-its-own-axis/11692	# Outcomes of earth slowing down spinning on its own axis What are the possible outcomes if earth slows down spinning on its own axis? To be specific: Can the decrease in the internal centrifugal (or centripetal) force due to slowing down earth's spin: a. change earth's structure given the fact that major part of earth is not solid and the earth's oblong shape may be in dynamic equilibrium of gravity vs this internal centrifugal force. b. alter the sense of the "vertical" as indicated by the plumb line especially when we are off equator, given the fact that the direction of centrifugal force acting on a physical body has a non-zero component on the horizontal direction except on equator and poles? If this is true, will the altered horizontal component be sufficient enough to affect high rise buildings? - ## 4 Answers I would say: a - yes, it will change the shape of the Earth and b - no, it will not change the vertical, because of a. Now, this needs a clarification. a - Yes, because the Earth is globally in hydrostatic equilibrium, and the change in rotational speed is so slow that it cannot drive the Earth out of this equilibrium. Faster excitations, like plate tectonics or glaciations, do drive some out of equilibrium behavior, but the rotation slow down is just too slow. Just to get an idea of how slow this is: The earth flattening is now $f = 1/298$. Assuming it is proportional to the rotational speed of the Earth, and assuming a slowdown of 2 ms/d/century as suggested before, we get $$\frac{df}{dt} = - f \times 2.3\times 10^{-8}/\text{cy} = - 7.8\times10^{-11}/\text{cy}$$ That means that the equatorial radius is getting shorter by 0.16 mm per century, while the polar radius is increasing at twice this rate. b - It actually depends on what you are comparing the vertical to. If you compare the new vertical with what the old vertical was at the same, fixed geocentric latitude, then yes, it will change. If you compare at the same position on the crust, it will also change, but for another reason: plate tectonics will move your reference point far from where it was originally. If you compare at a fixed geographic latitude, it will not change, simply because the geographic latitude is defined by the direction of it's vertical. In any case, the vertical will stay perpendicular to the geoid because of the definition of the vertical (gradient of potential) and of the geoid (equipotential). It will also stay roughly perpendicular to the Earth's crust because of point a. Comparing with tall buildings will not be possible because the buildings we build today will not be here in a billion years. - +1 - I was reading the other answers and planned to correct a few misconceptions, but here, you gave the correct and complete answer! – ysap Jun 30 '11 at 14:17 This is not a problem of "misconceptions", but of severe editing of the question. To see, which part of an answer referred to which status of the question is difficult meanwhile. – Georg Jun 30 '11 at 17:14 For point b: Yes, but very slightly. (I can't properly answer a, since I'm not a geophysicist). Let's do some back of the envelope calculations. The centripetal acceleration at the equator is given by $\omega^2 r$, where $\omega$ is the angular velocity of the earth, and $r$ is the radius. $\omega$ is pretty small, somewhere around $7 \times 10^{-5} \frac{rad}{s}$, $r$ is about $6.4 \times 10^6 m$, which makes the centripetal acceleration somewhere around $4 \frac{cm}{s^2}$. Given that: • the gravitional acceleration is $980 \frac{cm}{s^2}$ • the slowing down is very gradual (the change in Earth's rotational period (a day in other words) is 2 milliseconds per century according to Wikipedia) you're not going to notice anything over your lifetime, and I doubt whether it could be measured experimentally over the course of a hundred years. At the very least, high rise buildings won't notice it as much as the forces they are continually subjected to; think thermal expansion/contraction, wind (the Empire State Building sways up to 3 cm in bad storms), earthquakes and even passing traffic!. If the earth's rotation would suddenly scream to a halt, things are different. For starters, the rotational energy is $\frac{m \omega^2 r^2}{5}$, about $2.9 \times 10^{29} J$ (if I didn't mess up too much :) which is 'equivalent' to $2 \times 10^{11}$ Japan 2011 earthquakes, or 'enough' to raise the sealevel temperature by several thousand degrees. All that energy has to go somewhere, but fortunately it also has to come from somewhere. - Whoops, should have read a bit more. Thank you for the correction, I'll edit it in. – yatima2975 Jun 29 '11 at 13:22 EDIT: Answer below is talking about a radical sudden slow-down of angular velocity, not a gradual one. Question a is actually pretty interesting. We can see a very extreme example of this in the case of a neutron star, some of which rotate with surface speeds of 1/3 the speed of light due to conservation of angular momentum, making it an oblate shape. The basic structure of a neutron star is analogous to earth, it has a hard "crust" (which is significantly stronger than earth's crust) and a fluid like core. As it slows down, the crust holds firm against the inner fluid, until suddenly it ruptures and the neutron star undergoes a "stellar quake," which create small gamma ray bursts. I imagine if for some reason or other the earth is significantly slowed down, a similar event would occur. At first there would be rather little seismic activity, but sometime the pressure would be so great that somewhat catastrophic fissures would occur in the earth's crust as it ruptures and re-shapes itself to be more spherical. - I don't think anything catastrophic should be expected. The change in centrifugal force is extremely slow, and the Earth's crust has plenty of time to adjust to it's new equilibrium (it is not a perfect solid, it can still flow). It will certainly be much smoother that the current post-glacial rebound. – Edgar Bonet Jun 29 '11 at 8:16 Oh, for a second I was thinking the question was asking about the effects of a radical slow-down, not a gradual one. – Benjamin Horowitz Jun 29 '11 at 20:32 Yarkovsky Effect quoting wikipedia - "The Yarkovsky effect is a force acting on a rotating body in space caused by the anisotropic emission of thermal photons, which carry momentum." http://en.wikipedia.org/wiki/Yarkovsky_effect Essentially, the Earth absorbs momentum from Sun photons and remits this momentum as the Earth rotates, for retrograde planetary motion this causes the planet to move Sunwards, and oppositely for prograde motion (Earth has a prograde motion). While this effect is minimal over a short period of time, it will sum to measurable effects over a great period (to wit millions of years). It can be easily imagined that for sufficient time frames, any rotating-orbiting body will move closer or further from the Sun until it reaches an orbit where the momentum released via the Yarkovsky effect will balance that gained from Sun. So in answer to your question (from the Yarkovsky effect) any change in rotation will cause an eventual change in orbit, and depending on the planets retrograde or prograde rotation will cause the planet to either heat up or cool down depending on the new orbit. Noting once more - this may take 'billions of years' - The topic here is outcome from slowing down of rotation, not another (and irrelevant) mechanism of slowing down. -1 – Georg Jun 19 '11 at 16:02 the question is pertaining to earth's spinning on its own axis, not the planetary rotation round the sun. – pongapundit Jun 21 '11 at 17:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9346561431884766, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-statistics/203609-how-does-autocorrelation-statisticians-relate-same-but-engineers.html	# Thread: 1. ## How does autocorrelation for statisticians relate to the same but for engineers? Hi, I was wondering if you could help me. I'm a little stuck with marrying up the ideas of autocorrelation from a statisticians point of view and an engineers point of view. I am neither of these but need to understand how they relate to one another. The statisticians point of view: Chris Chatfield defines in 'The Analysis of Time Series' p23 the autocovariance coefficients at lag k for a discrete time series $\left\{x_t\right\}_{t=1}^N$, $c_k=\frac{1}{N}\sum_{t=1}^{N}(x_t-\overline{x})(x_{[t+k]mod N}-\overline{x})$ I'm taking my time series to have period N here. Then the autocorrelation at lag k is defined as, $r_k=\frac{c_k}{c_0}$ The autocorrelation is normalised with respect to the autocorrelation at lag 0 and centred about the mean. I understand that the method looks to identify any linear relationships present in the data at differing lag times. Another definition in the same book is the autocorrelation function, $\frac{Cov[X(t),X(t+\tau)]}{Var[X(t)]}$ Is this the form of the autocorrelation at lag k, but for lag $\tau$with an infinite time series? Can I use both forms for unrealised random variables? The engineers define the autocorrelation in 'Digital Signal Processing' by Sanjit Mitra p105, at lag k to be, $r_k(x)=\frac{1}{N}\sum_{t=1}^Nx_tx_{[t-k]modN}$ This form isn't centred about the mean or normalised and is defined with the lag backwards in time. I believe the engineers are trying to check for patterns in the time series just the same, but only wish to check whether the result is associated with white noise or not. Does anyone know if this is really what is being done and I'm not missing something? Thanks :-) 2. ## Re: How does autocorrelation for statisticians relate to the same but for engineers? Hey alanaj5. It's hard to say what the case is (with the mean being assumed to be zero). As a guess, you might want to check the book if it's referring to a distribution with mean 0 and variance 1 (which fits a residual model for normal distributions especially in the context of errors). Can you give a bit more context and information that surrounds the buildup to this definition?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9375889301300049, "perplexity_flag": "middle"}
http://cogsci.stackexchange.com/questions/252/what-is-an-effective-metric-of-complexity-for-an-artificial-neural-network/253	# What is an effective metric of complexity for an Artificial Neural Network? After asking the question What is the most complex neural network... I realized I don't really have a good metric of "complexity" in a general sense. The simplest measure would likely be count of neurons or number of synapses, but that fails to take into account the structure of the network. A couple measures of complexity are discussed in the paper Complexity of Predictive Neural Networks but they are very specific to a single task. One is the amount of work needed to learn a certain thing, and the other is how many neurons are needed to approximate a certain function. Rough, animal based measures are often employees for the sake of grabbing headlines; such as the incorrect claims that The Blue Brain Project had emulated a neural network "as complex as" a cat's brain. C. elegans is a common and seemingly attainable level of complexity for an artificial neural network. Animal based measures are relateable to the layman but seem questionable, especially when comparing a neural network to that of an animal who's neural network has not been completely mapped (as C. elegans has). What is a meaningful measure by which artificial neural networks can be measured? How are such networks currently compared? Can any such metric appropriately measure complexity of such a system? - 1 – Artem Kaznatcheev Feb 8 '12 at 23:51 ## 1 Answer The standard complexity metric in theoretical computer science and machine learning, in particular in statistical learning theory, is the Vapnik–Chervonenkis (VC) dimension. It is of interest because it gives us a very good tool to measure the learning ability of a neural network (or any other statistical learner, in general). A good introduction to the use of VC dimension for studying neural nets is: Eduardo D. Sontag [1998] "VC dimension of neural networks" [pdf]. There, the author shows (for instance) that a network with one hidden layer, $n$ inputs, and $\tanh$ neurons has VC dimension of $n + 1$. He also explain some basic technique for how to upper-bound the VC dimension, and for how to use it for dynamic neural nets. - 1 – Artem Kaznatcheev Feb 8 '12 at 23:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9500749707221985, "perplexity_flag": "middle"}
http://motls.blogspot.com.au/2012/09/raphael-bousso-is-right-about-firewalls.html?m=1	# The Reference Frame Our stringy Universe from a conservative viewpoint ## Saturday, September 29, 2012 ### Raphael Bousso is right about firewalls Five days ago, I reviewed the discussions on black hole firewalls, started by AMPS in July 2012. Joe Polchinski wrote a guest blog for Cosmic Variance yesterday. During the days, I had enough time to sort all these issues and I am confident that Raphael Bousso is right and Polchinski and others are wrong. Here is Raphael's 33-minute talk from Strings 2012 and here is his July 22nd paper: Observer Complementarity Upholds the Equivalence Principle (The two men who disagree about firewalls were playing accomplices of one another back in 2000 when they were helping to ignite the anthropic coup d'état in string theory by their Bousso-Polchinski landscape (then: discretuum) paper.) In fact, I would say that the paper is very clear, crisp, and even shows the author's understanding of some basic features of quantum mechanics – features that others unfortunately keep on misunderstanding. That's a very different verdict from my verdict about the nonsensical MWI-inspired Bousso-Susskind hypermultiverse, isn't it? Bousso defends the complementarity principle. What this principle really means has often been misinterpreted. For example, some people said that the black hole interior contains all the degrees of freedom that one may measure outside the black hole. This is clearly nonsense. The interior contains at most a scrambled version of a part of the exterior degrees of freedom. Raphael nicely avoids many of the confusions by introducing a refined version of the complementarity principle, the so-called observer complementarity. It's a typically Boussoian concept – and one could argue that he has greatly contributed to this sort of thinking. In the firewall discussion, this Boussoian thinking is extremely natural and arguably right. If I add some "foundations of quantum mechanics" flavor to the principle, it says: Quantum mechanics is a set of rules that allows an observer to predict, explain, and/or verify observations (and especially their mutual relationships) that he has access to. An observer has access to a causal diamond – the intersection of the future light cone of the initial moment of his world line and the past light cone of the final moment of his world line (the latter, the final moment before which one must be able to collect the data, is more important in this discussion). No observer can detect inconsistencies within the causal diamonds. However, inconsistencies between "stories" as told by different observers with different causal diamonds are allowed (and mildly encouraged) in general (as long as there is no observer who could incorporate all the data needed to see an inconsistency). While the complementarity grew out of technical features of quantum gravity, you may see that this observer complementarity version of it sounds just like some Bohr's (or Heisenberg's) pronouncements. It shouldn't be surprising because it was Bohr who introduced the term "complementarity" into physics and the general idea was really the same as it is here. Bohr has said that physics is about the right things we can say about the real world, not about objective reality, and it has to be internally consistent. However, in the context of general relativity, the internal consistency doesn't imply that there has to be a "global viewpoint" or "objective reality" that is valid for everyone. This is analogous to the statement in ordinary quantum mechanics of the 1920s that a complete physical theory doesn't have to describe the position and momentum (or particle-like and wave-like properties) of a particle at the same moment. ..... Polchinski who is the most senior figure behind the recent crazy firewall proposal is not only the father of D-branes and other discoveries but also the author of perhaps the most standardized graduate textbook of string theory. Bousso shows that AMPS are inconsistently combining the perspectives of different observers in order to deduce their desired contradictions but this is illegal because no observer has access to things outside his causal diamond, and therefore no observer can operationally demonstrate any contradiction. In the Penrose diagrams below, you see that no observer may observe the matter right before it is destroyed by the singularity as well as the late Hawking radiation. You must either sacrifice your life and jump to the black hole or you must stay out: you can't do both things simultaneously. I recommend you to read the whole Bousso's paper which is just 7-9 pages long, depending on how you count it. However, a sufficient screenshot that explains all his resolutions is Figure 1: Bousso's caption for Figure 1: The causal diamond of an infalling (outside) observer is shown in red (blue); the overlap region is shown purple. Observer complementarity is the statement that the description of each causal diamond must be self-consistent; but the (operationally meaningless) combination of results from different diamonds can lead to contradictions. • (a) Unitarity of the Hawking process implies that the original pure state $$\Psi$$ is present in the final Hawking radiation. The equivalence principle implies that it is present inside the black hole. If we consider both diamonds simultaneously, then these arguments would lead to quantum xeroxing. However, no observer sees both copies, so no contradiction arises at the level of any one causal diamond. • (b) Unitarity implies that the late Hawking particle B is maximally entangled with the early radiation A (see text for details). At the earlier time when the mode B is near the horizon, the equivalence principle implies that it is maximally entangled with a mode C inside the black hole. Since B can be maximally entangled only with one other system, this constitutes a paradox. However, no observer can verify both entanglements, so no contradiction arises in any single causal diamond. Therefore, it is not necessary to posit a violation of the equivalence principle for the infalling observer. LM: Let me repeat those observations again. There are two possible paradoxes we may face but both of them are resolved by a careful application of observer complementarity: the xeroxing paradox and the firewall paradox. The xeroxing paradox is the observation that the matter that has collapsed into a black hole carries information and the information may get imprinted to two places – somewhere inside the black hole and in the Hawking radiation. These two places may even belong to the same spatial slice through the spacetime. Such a doubling of information is prohibited by the linearity of quantum mechanics. Despite the existence of the "unifying" spatial slice, there is no contradiction because there is no observer who could have access to both copies of the same quantum information, no causal diamond that would include both versions of the same qubit. That's why no particular observer can ever discover a contradiction and this is enough for the consistency of the theory according to the quantum mechanical, "subjective" standards. The firewall paradox on the right picture was proposed by AMPS. A late Hawking particle B may be shown to be maximally entangled both with some early Hawking radiation's degrees of freedom A as well as with some degrees of freedom inside the black hole C. In quantum mechanics, a system can't be maximally entangled with two other systems. But this is not a problem because no single observer able to access one causal diamond is able to verify both maximal entanglements. In fact, the "old" version of the complementarity could be a legitimate – although less accurate – explanation of the resolution here, too: the degrees of freedom in C aren't really independent from those in A. You may (approximately or accurately?) say that C is a scrambled subset of A so when you say that B is maximally entangled both with A and C, it is not entangled with two things because the relevant degrees of freedom in A and C are really the same! It's a similar situation as if you considered a "zig-zag" spatial slice of the spacetime that happens to contain 2 or 3 copies of the same object at about the same time. All "xeroxing-like paradoxes" would be artifacts of this slice that pretends the independence of things that aren't independent. So there is no paradox, one doesn't have to sacrifice the equivalence principle, the infalling observer may still enjoy a life that continues even after he crosses the event horizon of a young or old black hole, and everything makes sense. It really looks to me as though Polchinski et al. have really denied the very essence of the complementarity – whatever precise formulation of it you choose. Maybe they were also misled by some of the usual "realist" misinterpretations of the wave functions and density matrices – for example by the flawed opinion that it must be "objectively and globally decided" whether some system is described by a pure or mixed state, by qubits that are entangled with someone else or not. Such questions can't be answered "objectively and globally": they should be answered relatively to the logic and facts of a particular observer and whether a system is in a pure state or a mixed state is really just a question about the subjective knowledge of this observer, not an "objective question" that must admit a "universally and globally valid answer". In the case of general relativity, especially in spacetimes that are causally nontrivial, this subtlety is very important because individual observers can't transcend their causal diamonds so mutually incompatible perspectives that cannot be "globalized" are not only allowed but, in fact, common. So applause to Raphael Bousso, ladies and gentlemen. Update: Polchinski responds to Bousso Joe Polchinski thinks that Bousso's picture has a bug. A comment on CV: Bousso (and others) want to say that an infalling observer sees the mode entangled with a mode behind the horizon, and the asymptotic observer sees it entangled with the early radiation. This is an appealing idea, and was what I initially expected. The problem is that the infalling observer can measure the mode and send a signal to infinity, giving a contradiction. Bousso now realizes this, and is trying to find an improved version. The precise entanglement statement in our paper is an inequality known as strong subadditivity of entropy, discussed with references in the wikipedia article on Von Neumann entropy. I would need a picture and details. Where is the measurement taking place? What is the new contradiction? If he measures C inside the black hole, then he clearly can't send it to infinity for causal reasons. If he sends it right before he falls in, in the B phase (old hole's radiation), the information comes out redshifted and hardly readable as a part of the information in the Hawking radiation; it's the same information that we already count as B. If the infalling observer makes the measurement in the stage A, i.e. even earlier than that, then it's irrelevant because that's really the initial state in which the existence of the black hole doesn't play any role yet. If the measurement is done so that it's available to observers near singularity as well as those after the black hole evaporates, it's just a fact that both of these observers share. It makes no sense to say that this piece of information is later entangled with anything else: once a qubit or another piece of quantum information is measured, it's no longer entangled with anything else! When I measure a spin to be up, the state of the whole system is $$\ket{\rm up}\otimes \ket{\psi_{\rm rest}}$$ and similarly for the density matrices. No entanglement here. So whatever method I choose to read Polchinski's reply to Bousso, it makes no sense to me. #### 33 comments: 1. gradstudent I'm sorry, but I feel that Busso didn't invent anything at all actually. This principle of complementarity, if I understand it well, always been there since special relativity and generalized (in some sense) with general relativity. One can think about the uniformly accelerated observer in special relativity, or about the radiation of an infalling electrical charge in general relativity. In both cases, there are no inconsistencies because all the naive paradoxes one can encounter by qualitative reasonning disappear in a simple way : the creation of an horizon, where all the "bad stuff" turns out to be confined behind it. So, well, is that legit to say that Susskind, Polchinski & al never got it whereas every decent graduate graduate students did ? 2. Dear grad student, first of all, his name is Bousso. Second, it's complete nonsense that the complementarity principle was there in (classical) relativity. Complementarity of any form is a feature of quantum mechanics. The old Bohr's complementarity was coined by Niels Bohr in the context of QM. But here we're talking about a more specific complementarity, one related to black holes, and Bousso of course didn't invent this, either. However, he refined the way to formulate it – he invented "observer complementarity" which is actually somewhat closer to the complementarity of Bohr. Pretty much all your comment implies that you are completely unfamiliar with quantum mechanics – which automatically implies that you are unfamiliar with all issues concerning complementarity etc. – so your breathtaking arrogance isn't justifiable by anything at all. Classical physics assumed objective and unique values of all fields or other degrees of freedom whether or not they were behind the horizon. Only in quantum mechanics, such an assumption became subtle and challenged because of the constraints that quantum mechanics itself imposes, especially the "conservation of information" in the Hawking radiation that follows from unitarity of quantum mechanics. Before you actually learn at least some basic things in this quantum mechanical discussion, and if possible, also things like the Hawking radiation, information preservation, and so on, could you please exploit your opportunity to shut up? This is surely not a debate that can be resolved by comments about classical physics, which is what you seem to think. 3. gradstudent Well, maybe I don't understand what is the complementary principle then, but I really feel that its already here in relativity in the sense that the set of accessible events of an observer depends of the observer (and there time order too as everybody knows, still causality holds). Anyway, fair enough, i'm not familiar with black holes information loss, so maybe I misunderstood what is this complementary principles : any references ? :} 4. Hi, could you please just read the articles and trace to the original references - or textbooks - yourslef? I can't plan learning of all the basic physics for you. at least not if you don't pay me \$100,000 a year for doing this service for you. 5. Dilaton Dear Lumo, can you put the link to this article into your answer to my SE question too? I'd like to see it included in yout answer there but I'd never in my life dare to touch (edit) anything you have written ... ;-). Anyway, thanks for this update I've not yet read it but I'll enjoy it very soon? (At the moment I'm still trying to see what I can get out for myself of the exotic branes article (just for the heck and the fun of it), but for this I'll have to review certain things I have forgotten (darn !) etc ...) 6. Hi Dilaton, I added the link to the SE answer. 7. Vlad I'm layman when it comes to these things but that firewall zone is ridiculous to me.Black hole is a black hole regardless of its' size and age. 8. Robert Graham Thank you for this decisively clear summary of Bousso's great contribution to the firewall-debate. Could not the contradictons incurred by disregarding observor-complementarity be used to establish it as a fundamental principle in QM? 9. Dilaton Thanks Lumo :-) Maybe after having read this article I'll finally be able to give Raphael Bousso's Munich talk a try ... Cheers 10. lemon Polchinski has an example in the post on CV where he sends a half of \|+->+\|-+> into the BH. But this does not come out as a thermal radiation, no? The thermal radiation is a manifestation of how the IN vacuum is seen in the OUT region (considering asymptotic stationary regions). But an initial field with just one particle (from Polchinski's example) gives rise to a non-thermal corrections in the OUT field. This is enough to show that the information comes out unharmed after the BH evaporates. 11. Dear Lemon, if you know something about the initial state, e.g. if you know the pure state of two qubits you mentioned, no one in this discussion has any doubts that the final state after the BH evaporates is pure and encodes the initial state. More generally, even before the BH evaporates, it has deviations from thermality. It is not maximally mixed - thermal. As the BH starts to evaporate, the entanglement between the interior and exterior starts to grow but this entanglement is really unphysical because entanglement is correlation and no one can measure both the interior and the whole Hawking radiation. You could think that you may circumvent this a little bit because you may catch "most of the Hawking radiation" and assume it has "almost all the information" but what you omit is actually essential for the paradox to be absent. The real debate is whether any of these QM principles imply anything dramatic for the infalling observer crossing the event horizon. Polchinski et al. think that the answer is Yes, Bousso and others think it's No, and so do I. AMPS say that the interior isn't really allowed for a black hole because the later Hawking radiation B could be maximally entangled with two different systems, which is impossible in QM, namely the interior C and the early Hawking radiation A. But that's a wrong argument because to find a paradox, there must be an observer, and no observe can see A and C simultaneously. Effectively we can't even show that A and C are different degrees of freedom. 12. Borun D. Chowdhury Observer complementarity is a nice idea by itself. I must thank Harlow for explaining it to me in some detail. If two observers cannot communicate, they do not have to agree on the results of their experiment. However, I find it hard to believe that the experience of such observers might be so different outside the stretched horizon. I believe this is an ongoing discussion right now and I have nothing to add on that. However, I would like to point out something somewhat tautological about the statement of observer complementarity. If Alice's experience can be so different while falling in that she might see B maximally entangled with C (instead of A), I don't see the point in asking if she can communicate this to Bob. Communication depends on a predetermined algorithm. However, if the physics is so different and Alice and Bob do not know what it is when they were together, then there is no way to come up with such an algorithm. 13. Lemon Ok, thanks for the explanation. I was a bit confused what is this discussion all about. At one point it's about what an infalling observer sees but at the same time people talk about the information loss (Polchinski mentions it too). But nobody really disputes these days that the information is preserved. Am I right? Or did this problem crawl back again by the (suspicious) firewal proposal? 14. anony Bousso is on the right track. There is an additional serious flaw in the Polchinski argument. If the black hole emits radiation before the observer jumps in, you have to allow for the possibility for it to interact with the observer in some way prior to him jumping in. If the radiation that is emitted is maximally entangled with the black hole, the interaction with the observer prior to the observer jumping in allows for mixing of the information for both internal and external degrees of freedom with the observer. The only concern for the observer is whether there is consistency between the measured information from the outgoing radiation with the measured information he finds in the black hole. The only pure system at that point is the combined system of outgoing, observer, and black hole. IOW, the inconsistency that Polchinski found is because of an arbitrary assertion of separability of the blackhole information and the observer, even after the blackhole has emitted and before the observer jumps in. Basically he is trying to deny the initial interaction terms of the interaction hamiltonian, effectively imposing some sort of arbitrary cutoff in a region where those terms are significant. 15. Peter F. I think it is a fair assessment (of 'the relevant situation' in much of Science) to conclude that: A carefully conservative attitude [such as Lumo's - and to some extent also one such as mine (that I partly developed because of what I've learned during my many visits to TRF)] can help to keep imaginative and strongly driven (endogenously motivated) intellectual people from foraging too deeply into, and sometimes getting permanently lost in, 'theoretical Fairyland'. 16. Shannon I find this observer-complementarity absolutely awesome. I think you are holding something here ! It's like we are part of the blackhole picture as well... 17. Ruz Nice that you believe in Bousso's argument, now we just have to convince him to believe in it too ;) (He retracted his objections to the Polchinski et al. paper, mostly...) 18. Dear Ruz, I find this kind of "arguing" bizarre. I haven't seen any blunder in that argument so those comments about "retractions" – which I have seen as well – are completely vacuous for me so far. 19. Luke Lea So even the best of physicists can lose it. 20. cynholt It seems to me that describing a black hole as though it were behind some sort of information-based firewall is just a lame excuse for not figuring out how black holes work and how they fit into the evolution of the universe. If we didn't know how the Sun was formed and how it gets its energy, or why the sky is blue and why grass is green, or when Earth formed and when life formed on Earth, we'd simply say that firewalls are preventing us from knowing any of these and other sorts of puzzling things. Once we learn most of the hows, whys, and whens of black holes, all of this talk of firewalls will disappear. Then again, we'll move on to describing something else, which has got us collectively scratching our heads over, as being behind a firewall. Perhaps that something else will be the world beyond our world in the multi-universe. Death is the only thing that'll stop us from asking questions, after questions, in hopes of finding answers to at least some of them. 21. Eugene S Dear Cynthia, I probably know even less of physics than you but I beg to differ. The uncertainty principle in quantum mechanics is a well-known example where in trying to measure two non-commuting variables simultaneously, we run up against a fundamental limitation of nature: we cannot determine both their values accurately. This is how nature everywhere works on the smallest, the quantum, level. Black holes, on the other hand, are how nature works at its most extreme, where spacetime curvature becomes infinite. From what little I understand of the present topic, I would say that the scientists investigating this "firewall" business want to know if equally fundamental constraints on what can be known are placed by nature in the case of black holes. I do not view this as a "lame excuse" but as a noble endeavor in pursuit of an exciting goal. 22. Dilaton Ha ha, this interpretation of people hiding behind the firewall to avoid having to properly figure out how (parts of) nature works made me chuckle :-D (even though I do not exactly agree and, as Eugene, rather think they are trying to find out serious things) But I always thought people apply the anthropic principle as a lame excuse to avoid properly figuring out how things work :-P 23. Shannon Cynthia, I guess the main thing is that there are no inconsistencies in this theory on BH. That in itself is a step forward understanding how they work. Death might give us the answer to all this, or it might just freeze our knowledge... 24. Gene Lubos, Thanks for yet another crystal-clear explanation. I am still amazed that people as smart as Polchinski can be confused about the very essence of quantum mechanics. How do you explain this phenomenon? Is the disease contagious? 25. Dear Gene, there are many people confused by QM but so far, I didn't dare to think that Joe could be among them, so I still prefer to think this is just a more technical thing dependent on subtleties of quantum gravity and Joe would give the same answers to non-gravitational QM problems as Bohr - or two of us. ;-) 26. Newman What was the Dvali's question in the end about? Black hole is large N system? Does someone take his (Dvali's) recent papers seriously? 27. Hi, well, I personally don't... 28. Trimok This inside/outside observer complementary, seems to be closely related to all the Ads/CFT or Quantum. Gravity/QFT correspondences. Let me try some speculations. Maybe, very roughly, one may say that the (free falling) inside observer see Quantum gravity or String theory , while the (accelerated) outside observer see QFT and CFT (with one dimension less, at the (exterior) surface of the horizon). So, maybe, is some sense, our universe is just a boundary surface where we use standard QFT/CFT. We don't see the inside of the domain whose boudary (and some kind of horizon) is our universe. This inside domain could be described by String theory, and this description should be complementary to the standard CFT/QFT description at the boundary. But we should not use the two theories at the same time, for instance, to examine some paradox. 29. Dilaton Hm, I've some place read the argument that our inflating universe could be described as stuff falling towards a black hole horizon with the black hole lying outside the cosmic horizon. Do you mean something like this? 30. Trimok No, my speculation, was to consider our entire 4-D universe as an horizon/boundary of some higher 5-dimensional space. So CFT/QFT apply to our 4-D universe, and then String theory/Quantum gravity apply to the 5-dimensional space. They are just different expressions of the same physics, which are complementary, but you can't use the 2 theories together to express a paradox. 31. Clayton Hi Lubos, Have you seen that Bousso has now recanted his position? The latest version of the paper you link to above has an extremely different tone than the one taken here. Very interested to know what you think! 32. Thanks for telling me. I am going to study it soon. 33. Why are people so loath to give up omniscience? ## Who is Lumo? Luboš Motl Pilsen, Czech Republic View my complete profile ← by date	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.952223002910614, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Stochastic_process	All Science Fair Projects Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. Stochastic process In the mathematics of probability, a stochastic process can be thought of as a random function. In practical applications, the domain over which the function is defined is a time interval (a stochastic process of this kind is called a time series in applications) or a region of space (a stochastic process being called a random field). Familiar examples of time series include stock market and exchange rate fluctuations, signals such as speech, audio and video; medical data such as a patient's EKG, EEG, blood pressure or temperature; and random movement such as Brownian motion or random walks. Examples of random fields include static images, random topographies (landscapes), or composition variations of an inhomogeneous material. Contents Definition A stochastic process is an indexed collection of random variables, each of which is defined on the same probability space W and takes values on the same codomain D (often the reals $\R$). An important case is the discrete set $f_i: W \to D$, where i runs over some discrete index set I - for example if the probability distributions of the fi satisfy the Markov property the process is a Markov chain. fi is often called (stochastic) transition function or stochastic kernel. In a continuous stochastic process the index set is continuous (usually space or time), resulting in an infinite number of random variables. Each point in the sample space Ω; corresponds to a particular value for each of the random variables and the resulting function (mapping a point in the index set to the value of the random variable attached to it) is known as a realisation of the stochastic process. A particular stochastic process is determined by specifying the joint probability distributions of the various random variables f(x). Stochastic processes may be defined in higher dimensions by attaching a multivariate random variable to each point in the index set, which is equivalent to using a multidimensional index set. Indeed a multivariate random variable can itself be viewed as a stochastic process with index set $I = \{ 1 \ldots n \}$ Examples The paradigm continuous stochastic process is that of Brownian motion. In its original form the problem was concerned with a particle floating on a liquid surface, receiving "kicks" from the molecules of the liquid. The particle is then viewed as being subject to a random force which, since the molecules are very small and very close together, is treated as being continuous and, since the particle is constrained to the surface of the liquid by surface tension, is at each point in time a vector parallel to the surface. Thus the random force is described by a two component stochastic process; two real-valued random variables are associated to each point in the index set, time, (note that since the liquid is viewed as being homogeneous the force is independent of the spatial coordinates) with the domain of the two random variables being $\R$, giving the x and y components of the force. A treatment of Brownian motion generally also includes the effect of viscosity, resulting in an equation of motion known as the Langevin equation. As another example, take the domain to be $\N$, the natural numbers, and our range to be $\R$, the real numbers. Then, a function $f: \N \ \to \ \R$ is a sequence of real numbers, and a stochastic process with domain $\N$ and range $\R$ is a random sequence. The following questions arise: 1. How is a random sequence specified? 2. How do we find the answers to typical questions about sequences, such as 1. what is the probability distribution of the value of f(i)? 2. what is the probability that f is bounded? 3. what is the probability that f is monotonic? 4. what is the probability that f(i) has a limit as $i \to \infty$? 5. if we construct a series from f(i), what is the probability that the series converges? What is the probability distribution of the sum? Another important class of examples is when the domain is not a discrete space such as the natural numbers, but a continuous space such as the unit interval [0,1], the positive real numbers $[0, \infty]$ or the entire real line, $\R$. In this case, we have a different set of questions that we might want to answer: 1. How is a random function specified? 2. How do we find the answers to typical questions about functions, such as 1. what is the probability distribution of the value of f(x) ? 2. what is the probability that f is bounded/integrable/continuous/differentiable...? 3. what is the probability that f(x) has a limit as $x \to \infty$? 4. what is the probability distribution of the integral $\int_a^b f(x)\,dx$? There is an effective way to answer all of these questions, but it is rather technical (see Constructing Stochastic Processes below). Interesting special cases • Homogeneous processes : processes where the domain has some symmetry and the finite-dimensional probability distributions also have that symmetry. Special cases include stationary processes, also called time-homogeneous. • Processes with independent increments: processes where the domain is at least partially ordered and, if $x_1 < \ldots < x_n$, all the variables f(xk + 1) - f(xk) are independent. Markov chains are a special case. • See also continuous-time Markov chain. • Markov processes are those in which the future is conditionally independent of the past given the present. • Point processes : random arrangements of points in a space S. They can be modelled as stochastic processes where the domain is a sufficiently large family of subsets of S, ordered by inclusion; the range is the set of natural numbers; and, if A is a subset of B, $f(A) \le f(B)$ with probability 1. • Gaussian processes: processes where all linear combinations of coordinates are normally distributed random variables. • Poisson processes • Gauss-Markov processes: processes that are both Gaussian and Markov • Martingales -- processes with constraints on the expectation • Galton-Watson processes • Elevator paradox • Branching processes • Bernoulli processes • Many stochastic processes are Lévy processes. Constructing stochastic processes In the ordinary axiomatization of probability theory by means of measure theory, the problem is to construct a sigma-algebra of measurable subsets of the space of all functions, and then put a finite measure on it. For this purpose one traditionally uses a method called Kolmogorov extension. There is at least one alternative axiomatization of probability theory by means of expectations on C-star algebras of random variables. In this case the method goes by the name of Gelfand-Naimark-Segal construction. This is analogous to the two approaches to measure and integration, where one has the choice to construct measures of sets first and define integrals later, or construct integrals first and define set measures as integrals of characteristic functions. The Kolmogorov extension The Kolmogorov extension proceeds along the following lines: assuming that a probability measure on the space of all functions $f: X \to Y$ exists, then it can be used to specify the probability distribution of finite-dimensional random variables [f(x1),...,f(xn)]. Now, from this n-dimensional probability distribution we can deduce an (n-1)-dimensional marginal probability distribution for [f(x1),...,f(xn - 1)]. There is an obvious compatibility condition, namely, that this marginal probability distribution be the same as the one derived from the full-blown stochastic process. When this condition is expressed in terms of probability densities, the result is called the Chapman-Kolmogorov equation. The Kolmogorov extension theorem guarantees the existence of a stochastic process with a given family of finite-dimensional probability distributions satisfying the Chapman-Kolmogorov compatibility condition. Separability, or what the Kolmogorov extension does not provide Recall that, in the Kolmogorov axiomatization, measurable sets are the sets which have a probability or, in other words, the sets corresponding to yes/no questions that have a probabilistic answer. The Kolmogorov extension starts by declaring to be measurable all sets of functions where finitely many coordinates [f(x1),...,f(xn)] are restricted to lie in measurable subsets of Yn. In other words, if a yes/no question about f can be answered by looking at the values of at most finitely many coordinates, then it has a probabilistic answer. In measure theory, if we have a countably infinite collection of measurable sets, then the union and intersection of all of them is a measurable set. For our purposes, this means that yes/no questions that depend on countably many coordinates have a probabilistic answer. The good news is that the Kolmogorov extension makes it possible to construct stochastic processes with fairly arbitrary finite-dimensional distributions. Also, every question that one could ask about a sequence has a probabilistic answer when asked of a random sequence. The bad news is that certain questions about functions on a continuous domain don't have a probabilistic answer. One might hope that the questions that depend on uncountably many values of a function be of little interest, but the really bad news is that virtually all concepts of calculus are of this sort. For example: all require knowledge of uncountably many values of the function. One solution to this problem is to require that the stochastic process be separable. In other words, that there be some countable set of coordinates {f(xi)} whose values determine the whole random function f. 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 17, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.914128839969635, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/45734/kinetic-energy-and-rotational-motion	# Kinetic Energy And Rotational Motion The problem is, "A metal can containing condensed mushroom soup has mass 220 g, height 11.0 cm and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at 30.0° to the horizontal and is then released to roll straight down. It reaches the bottom of the incline after 1.50 s. (a) Assuming mechanical energy conservation, calculate the moment of inertia of the can. (b) Which pieces of data, if any, are unnecessary for calculating the solution? My attempt at solving (a): I figured that I could use the equation $\Sigma W=\Delta K=1/2I\omega_f^2-1/2I\omega_i^2$ Since the force of gravity that is acting along the incline is applied constantly over a distance, $W_g=mg\cos(60^{\circ})(3.00~m)$; and since the can rolls down the incline in 1.50 s, $v=3.00/1.5 \implies 2~m/s$, which means that $\omega_f=2/0.0319 \implies 62.695925~rad/s$ With this, and knowing that $\omega_i=0$, $mg\cos(60^{\circ})(3.00)=1/2I(62.695925)^2 \implies I=\frac{(6.00)mg\cos(60^{\circ})}{(62.695925)^2}$ When I calculated this, I got $I=0.00165~kg\cdot m^2$; however, the true answer is $I=0.000187~kg\cdot m^2$ I've re-worked my solution several times, what am I doing incorrectly? As for (b), the answer is that the height of the can is an irrelevant piece of information, why is that? - ## 1 Answer 1) The first thing I notice is that you have stated that the velocity at the end of the ramp is $2\textrm{ m/s}$. Remember that the can is accelerating as it rolls down the ramp, so the equation $v=\textrm{d}s/\textrm{d}t$ is not applicable here for finding the instantaneous velocity at the bottom. The can does indeed average $2\textrm{ m/s}$ during its trip, but this is not the final velocity of the can. Use this new corrected value to calculate angular frequency. 2) I find this problem simpler to solve using energy analysis. Take the can's initial potential energy: $$E_\textrm{pot} = m g h = 3.234\textrm{ J}\quad.$$ We also know that the final kinetic energy of the can must equal this due to the conservation of energy, but the final energy of the can must be broken into translational kinetic energy (due to the can's movement) and rotational kinetic energy (due to the rotation). (This is why your solution above was giving incorrect answers as it didn't take translational kinetic energy into account.) Thus, we also know that: $$E_\textrm{pot}(t=0) = E_\textrm{kin,trans}(t=1.5\textrm{ s}) + E_\textrm{kin,rot}(t=1.5\textrm{ s})\quad,$$ which, for our case, is $$3.234\textrm{ J} = \frac{1}{2} mv^2 + \frac{1}{2}Iω^2\quad.$$ Plugging in known values to this equation with the correct value for angular velocity $\vec \omega = \vec r \times \vec v$ gives the accepted answer: $$I = 0.000187\textrm{ kgm}^2\quad.$$ As for part B, the height of the can is irrelevant because as long as we know the mass and radius of the can, we can solve the problem. The ‘extra mass’ resulting from lengthening the can would be centered about the can's original center of mass, and as such the moment of inertia would not be affected for this problem. - So, the only problem I had with my approach is the miscalculation of the translational velocity? Would I still have to use a force approach to find the acceleration of the can? – Mack Dec 3 '12 at 13:35 You shouldn't have to use a force approach to calculate acceleration, no. You can use the fact that the average velocity of the can was 2 m/s and that it accelerated at a constant rate over 1.5 seconds. Thus, because the can started from 0 m/s, its final velocity would be 4 m/s such that it averages 2 m/s for the trip. – Mik Cox Dec 3 '12 at 23:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9621010422706604, "perplexity_flag": "head"}
http://mathhelpforum.com/math-software/61731-separating-points-algorithm-print.html	# separating points algorithm Printable View • November 26th 2008, 01:41 AM Janik separating points algorithm Hi, I'm a belgian student in applied economics: commercial engineer. We have to write a paper including an algorithm for an optimization problem. Part of the exercise is to give and explain examples of software that already does what our algorithm does. Even if it is just a step in the program. The separating points problem: Consider a set of points in the plane, each point being determined by an x-coordinate and a y-coordinate. We assume, for reasons of convenience, that no two points have a common x-coordinate or a common y-coordinate. The problem that we face is to draw horizontal and/or vertical lines such that each pair of points is separated by some line. Thus, no cell in the subdivision of the plane induced by the lines, contains more than one point. The challenge is to use as few lines as possible to achieve this requirement. Does anyone have an idea of which software incorporates this process? Or does anyone know if this can be done with some functions in Matlab, because we are not familiar with it. Thx • November 27th 2008, 06:45 AM shawsend 1 Attachment(s) Hi. My approach would be to first just draw a minimal excess of line: sort the x and y coordinates, take half the difference between each successive point, then draw lines at those points. That will segregate the points into blocks but will of course be an excess of lines. I don't have Matlab, but the Mathematica code below does that. Perhaps you can convert it to Matlab if you wish. Note I can remove the two red lines. Ok, now go through the line arrays and for each line, decide if that line can be removed. I'm not sure how to do that but you may wish to see if you can come up with some type of algorithm to do that. Code: ```nmax = 15; size = 10; pointlist = Table[{RandomReal[ {-size, size}], RandomReal[ {-size, size}]}, {nmax}]; plist = Point[pointlist]; xvals = Sort[(#1[[1]] & ) /@ pointlist]; yvals = Sort[(#1[[2]] & ) /@ pointlist]; xnew = Table[(xvals[[i]] + xvals[[i + 1]])/ 2, {i, 1, nmax - 1}]; ynew = Table[(yvals[[i]] + yvals[[i + 1]])/ 2, {i, 1, nmax - 1}]; xlines = (Line[{{#1, -size}, {#1, size}}] & ) /@ xnew; ylines = (Line[{{-size, #1}, {size, #1}}] & ) /@ ynew; Show[Graphics[{xlines, ylines, plist}]]``` • November 27th 2008, 12:14 PM shawsend 1 Attachment(s) I have some further thoughts about this. First I'll define the square ceiling function: $s(n)=\lceil\lceil n \rceil\rceil$ to be the square closest to n but not less than n. So $s(24)=25$. Then I initially suspect the smallest number of lines meeting the above requirement for n points (not including the outer framing lines) is $(\sqrt{s(n)}-1)^2$. Note the figure below for n=8. It may or likely is not this simple but it's a start. :) All times are GMT -8. The time now is 02:53 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9465099573135376, "perplexity_flag": "head"}
http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/103031	## Not especially famous, long-open problems which anyone can understand ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Question: I'm asking for a big list of not especially famous, long open problems that anyone can understand. Community wiki, so one problem per answer, please. Motivation: I plan to use this list in my teaching, to motivate general education undergraduates, and early year majors, suggesting to them an idea of what research mathematicians do. Meaning of "not too famous" Examples of problems that are too famous might be the Goldbach conjecture, the $3x+1$-problem, the twin-prime conjecture, or the chromatic number of the unit-distance graph on ${\Bbb R}^2$. Roughly, if there exists a whole monograph already dedicated to the problem (or narrow circle of problems), no need to mention it again here. I'm looking for problems that, with high probability, a mathematician working outside the particular area has never encountered. Meaning of: anyone can understand The statement (in some appropriate, but reasonably terse formulation) shouldn't involve concepts beyond (American) K-12 mathematics. For example, if it weren't already too famous, I would say that the conjecture that "finite projective planes have prime power order" does have barely acceptable articulations. Meaning of: long open The problem should occur in the literature or have a solid history as folklore. So I do not mean to call here for the invention of new problems or to collect everybody's laundry list of private-research-impeding unproved elementary technical lemmas. There should already exist at least of small community of mathematicians who will care if one of these problems gets solved. I hope I have reduced subjectivity to a minimum, but I can't eliminate all fuzziness -- so if in doubt please don't hesitate to post! To get started, here's a problem that I only learned of recently and that I've actually enjoyed describing to general education students. http://en.wikipedia.org/wiki/Union-closed_sets_conjecture Edit: I'm primarily interested in conjectures - yes-no questions, rather than classification problems, quests for algorithms, etc. - 1 You might get more success if you sampled certain open problem lists and indicated which ones fit your list and which ones did not. I could mention various combinatorial problems such as integer complexity, determinant spectrum, covering design optimization, but I can't tell from your description if they would be suitable for you. Gerhard "They Are Suitable For Me" Paseman, 2012.06.21 – Gerhard Paseman Jun 21 at 19:11 2 Here is some collection of some other "collect open problems" quests. on MO: mathoverflow.net/questions/96202/… PS Nice question ! PSPS may be add tag "open-problems" – Alexander Chervov Jun 21 at 20:53 1 Nice question!! – S. Sra Jun 22 at 3:25 11 To save the search for explanation of cryptic acronyms for those of us outside US, K-12 means high school. @Mahmud: You are using a wrong meaning of the word “problem”. The TSP is not an unproved mathematical statement, it is a computational task. – Emil Jeřábek Jun 22 at 12:05 7 More precisely, K-12 means anything up to high school (K = Kindergarten, 12 = 12th grade, and K-12 covers this range). – Henry Cohn Jun 22 at 13:05 show 1 more comment ## 73 Answers Does the series $\sum_{n=1}^{\infty} \frac{1}{n^3 \sin^2 n}$ converge? (Taken from http://math.stackexchange.com/questions/20555/are-there-any-series-whose-convergence-is-unknown where there are more such examples) - 1 and, in my answer at math.SE which you link here, I refer to the mathoverflow question mathoverflow.net/questions/24579. – George Lowther Jun 24 at 0:35 show 1 more comment ### 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 basically copied from my answer on this question, which I've now updated some. Let's let $\\|n\\|$ denote the smallest number of 1's needed to write n using an arbitrary combination of addition and multiplication. For instance, \|\|11\|\|=8, because $11=(1+1)(1+1+1+1+1)+1$, and there's no shorter way. This is sequence A005245. Then we can ask: For n>0, is $\\|2^n\\|=2n$? Since it is known that for m>0, $\\|3^m\\|=3m$, we can ask more generally: For n, m not both zero, is $\\|2^n 3^m\\|=2n+3m$? Attempting to throw in powers of 5 will not work; \|\|5\|\|=5, but $\\|5^6\\|=29<30$. (Possibly it could hold that $\\|a^n\\|=n\\|a\\|$ for some yet higher choices of a, but I don't see any reason why those should be any easier.) Jānis Iraids has checked by computer that this is true for $2^n 3^m\le 10^{12}$ (in particular, for $2^n$ with n≤39), and Joshua Zelinsky and I have shown that so long as $n\le 21$, it is true for all m. (Fixed powers of 2 and arbitrary powers of 3 are much easier than arbitrary powers of 2!) In fact, using an algorithmic version of the method in the linked preprint, I have computed that so long as $n\le 41$, it is true for all $m$, though I'm afraid it will be some time before I get to writing that up... That seems to be the best known. - The Cerny conjecture says that if X is a collection of mappings on an n element set such that some iterated composition (repetitions allowed) of elements of X is a constant map then there is a composition of at most $(n-1)^2$ mappings from X which is a constant mapping. This comes from automata theory. See http://en.m.wikipedia.org/wiki/Synchronizing_word. - 3D Version Of Blaschke-Lebesgue(1914) Theorem The planar, compact.convex set of constant width, say 1, of minimal area is the Reuleaux triangle: Blaschk-Lebesgue(1914). The 3D set of constant width and mimimal volume is unknown. - Is there an upper bound of quotients in the continued fraction representation of $\sqrt[3]{2}=[ 1; 3, 1, 5, 1, 1, \dots]$? - • Is Hilbert's tenth problem for Diophantine equations in rational numbers decidable? • Is Hilbert's tenth problem for Diophantine equations of power $3$ decidable? • Is there a universal Diophantine equation of power $3$? • Is there a universal Diophantine equation containing less than $9$ variables? If so, what is the minimal number of variables? What minimal power can be achieved for that number of variables? • Is there a universal Diophantine equation that can be written using less than $100$ arithmetic operations (additions or multiplications)? If so, what is the minimal number of operations? - Let ${^n a}$ denote tetration: ${^0 a}=1, {^{n+1} a}=a^{({^n a})}$. • It is unknown if ${^5 e}$ is an integer. • It is unknown if there is a non-integer rational $q$ and a positive integer $n$ such that ${^n q}$ is an integer. • It is unknown if the positive root of the equation ${^4 x}=2$ is rational (ditto for all equations of the form ${^n x}=2$ with integer $n>3$) • It is unknown if the positive root of the equation ${^3 x}=2$ is algebraic. - I get the feeling that you will enjoy reading about the Simonyi and Chvatal conjectures described here by some guy called Gil Kalai. Anyone know who that is? ;) - Some pages: Open Problem Garden The Open Problems Project - 2 [Removed the link to my open-problem page, which is more than a decade old; most of those problems are now solved.] – JeffE Jun 22 at 12:06 Easy-to-Explain but Hard-to-Solve Problems About Convex Polytopes slides by Jes´us De Loera contains 7 open problems (Hirsch conjecture is also there so it is out-of-date). - In an oriented graph, is there always a vertex from which there are at least as many vertices that one can access by moving along exactly two edges, than there are vertices that one can access by moving along one edge? This is known as Seymour's second neighborhood conjecture, and might be on the verge to being too famous (but it seems few of my colleagues know it). - Is there a dense subset of a plane having only rational distances between its points? - 4 This problem is given in Klee, Victor; Wagon, Stan (1991), ["Problem 10 Does the plane contain a dense rational set?"](books.google.com/…), Old and New Unsolved Problems in Plane Geometry and Number Theory, Dolciani mathematical expositions, 11, Cambridge University Press, pp. 132–135, ISBN 978-0-88385-315-3, and attributed to Ulam and Erdős. – Vladimir Reshetnikov Jul 25 at 13:39 2 It was also discussed here: mathoverflow.net/questions/19127/… – Vladimir Reshetnikov Jul 28 at 20:30 Does every nonseparating planar continuum have the fixed point property? - 12 For those of us whose general topology is rusty: what is the K-12 level formulation of this question? – Yemon Choi Jun 21 at 22:50 2 Uniform continuity is a K-12 concept now? – Douglas Zare Jun 22 at 14:25 1 Thanks Yemon. I agree with Doug that uniform continuity is not a stand alone K-12 concept. I mentioned it in order to avoid trying to decode its meaning in K-12 terms in this context, as follows: – Paul Fabel Jun 22 at 17:56 2 f is a set of 4-tuples (x1,y1,x2,y2) so that each point (x1,y,1) in K appears precisely once as the 1st two coordinates of some point in f, and we also require that the last two coordinates of each point of f is a point of K. For each positive radius R we can find a smaller positive radius r(R) so that if (x1,y1,x2,y2) is in f, then if (w1,z1) is both in K and also in the disk of radius r(R) centered at (x1,y1), and if (w1,z1,w2,z2) is in f, then (w2,z2) is in the disk of radius R centered at (x2,y2). – Paul Fabel Jun 22 at 17:56 show 3 more comments The Happy Ending Problem • Says that any set of five points in the plane in general position has a subset of four points that form the vertices of a convex quadrilateral. More generally, Erdös and Szekeres proved that for any positive integer $N$, there is a minimal integer $f(N)$ such that any set of $f(N)$ points in the plane in general position has a subset of $N$ points that form the vertices of a convex polygon, and it is known that $f(N)$ is at least $1+2^{N-2}$. An open question is: does $f(N)=1+2^{N-2}$ hold?. Taken from this MO link. - 2 Could you please not use BIG FONT for your headings? – Yemon Choi Jun 24 at 9:04 1 @Yemon: Why whats wrong? – Chandrasekhar Jun 24 at 9:08 Can one prove the infinitude of the primes without employing any functions of super-polynomial growth? (Of course I confess I have in mind Paris and Wilkie's more precise and sophisticated question concerning primes in the theory of bounded induction, but I think a high school student could think about looking for a positive answer without that background.) - 1 Where are functions of exponential growth used in the classical proof, or the Euler product + divergence of harmonic series? – Douglas Zare Jun 24 at 14:06 1 In the classical proof, the exponential growth is in the product of the known primes. See mathoverflow.net/questions/59262/… for a more precise discussion. In the Euler product proof, I imagine the growth occurs in the Chinese-remainder-theorem-based coding tricks necessary to express this proof in the language of first-order arithmetic, but I haven't thought this through carefully so maybe I'm totally off base. – Henry Cohn Jun 24 at 16:49 3 See also mathoverflow.net/questions/76058 . David, contrary to what you write, it is possible to define in bounded arithmetic a function computing rational approximations of logarithm. This does not imply that exponentiation is total, since there may be numbers greater that all values of logarithm. As for divergence of harmonic series, the problem is even to express this statement in bounded arithmetic: you cannot in general define $\sum_{n\le x}f(n)$ by a bounded formula, unless $x$ is logarithmic. – Emil Jeřábek Jun 25 at 11:35 1 Even if you restrict attention to small $x$, there is the question how do you formulate “divergence”. Bounded arithmetic certainly cannot prove that for every $y$, there exists $x$ such that $\sum_{n\le x}n^{-1}$ is defined and larger than $y$. However, I think that with appropriate formulations, it can prove that for every $y$, either there exists such an $x$, or all sums $\sum_{n\le x}n^{-1}$ that are defined have value less than, say, $y/2$ (which means the sum does not converge to $y$). Anyway, this is largely irrelevant. The real show-stopper in the proof using the Euler product is... – Emil Jeřábek Jun 25 at 11:52 show 6 more comments Here is a nice question due to John Conway. In a magical 4x4 square, show that the XOR composition of the four numbers, written in base 2, in every row and in every column is zero. This applies to a square in which the numbers 0 to 15 are used (rather than 1 to 16). For instance, a typical row might be 0 15 14 1, which in binary is 0000 1111 1110 0001, and in each of the four positions there happen to be two entries 0 and two entries 1, so the binary sum is zero. Of course there are only finitely many possible magic 4x4 squares, and you can give proof by "complete inspection" (aka brute force). In fact, that has been done, so the result is true. But neither he nor I know a conceptual proof. Should be easy to understand about a classical problem -- and yet seems not obvious. Try it! (Incidentally, the binary sum along the diagonals need not always be zero; that's not part of the question.) - What is the least $V$ such that any convex body of unit volume can be fit into a tetrahedron of volume $V$? It is known that $V \ge 9/2$ and conjectured that $V = 9/2$. - Is there such $n\in\mathbb{N}$ that ${^n\pi}\in\mathbb{N}$? (see tetration) - What is the largest possible volume of the convex hull of a space curve having unit length? - 1 It was discussed here: mathoverflow.net/questions/83026/… – Vladimir Reshetnikov Jul 28 at 21:10 The Polya--Szego conjecture for polygonal drums: among the polygonal drums with $n$ sides and given area, the regular one has the slowest vibration (and therefore the lowest tone). As far as I know, this remains open for $n\geq 5$. - The Kurepa conjecture : For every odd prime $p$, one has $$0!+1!+\cdots+(p-1)!\not\equiv0\pmod p$$ A proof was claimed and published in 2004 but the claim was withdrawn in 2011. See also my comment on the accepted answer to MO24265. - Do there exist five positive integers such that the product of any two of them increased by 1 is a perfect square? The same question for seven distinct nonzero rationals. Diophantine m-tuples pages - The complexity of matrix multiplication (i.e. the asymptotic number of steps required to multiply two n-by-n matrices). This is an important problem in CS theory, but is non-famous enough in other fields that a mathematician (Andrew Stothers) made a significant advance in it in 2010 (beating a 20-year-old bound of Coppersmith and Winograd), and wrote up the result on page 71 of his PhD thesis without bothering to state it as a theorem or otherwise call attention to it. Word of it only got around a year or so later, when a computer scientist (Virginia Vassilevska Williams) independently made a further improvement. The obvious multiplication algorithm takes $O(n^3)$ steps, and a well-known Karatsuba-like rearrangement gets the exponent $\omega$ down to about 2.8. There is a simple proof that the smallest possible $\omega$ is $\ge 2$. Coppersmith and Winograd got an exponent of 2.376 and the more recent results have it at 2.373. Apparently nobody has even shown that the minimum is not equal to 2: there are some who believe there's an algorithm faster than $O(n^{2+\epsilon})$ for any $\epsilon>0$ but not an $O(n^2)$ algorithm, but this is not known. - 1 I don't think this counts as "not especially famous"... – Harry Altman Jul 24 at 4:24 From Rick Kenyon's open problem list: What are the minimal number of squares needed to tile an $a \times b$ rectangle? Kenyon showed the correct order is $\log a$ assuming $a/b$ is bounded with $b \leq a$. However, there is plenty of room for improvement in the constant factor, and an exact formula seems far, far away. - Enumeration of meanders. (See also meander). Problem is to find some formula for the number of meanders or at least some good asymptotic. As far as I understand the attention to it has been attracted by V.I. Arnold. The problem is so "everyone can understand" that there is an article by him in the math. journal for shool-children "Quant" (sorry it is in Russian. I remember it from my school years): djvu file from the site. There are plenty papers in arXiv on the problem. E.g. http://arxiv.org/abs/cond-mat/0003008 Exact Meander Asymptotics: a Numerical Check Philippe Di Francesco, Emmanuel Guitter (SPHT-Saclay), Jesper Lykke Jacobsen (LPTMS-Orsay) As far as I understand from the nice book (or) by S. Lando and A. Zvonkin the problem is still open. - Is there a rectangle that can be cut into $3$ congruent connected non-rectangular parts? - Is there a positive integer which is both triangular and factorial except these obvious examples: $1, 6, 120$? (Tomaszewski conjecture, http://oeis.org/A000217) - Bonnesesn—Fenchel conjecture: Which body of constant width has the least volume? Is it Meissner's tetrahedron? - Alexander's Conjecture, and by extension a lot of open problems about combinatorial subdivision, are as easy to understand as they are maddening. To quote Melikhov: Alexander's 80-year old problem of whether any two triangulations of a [3-dimensional] polyhedron have a common iterated-stellar subdivision. They are known to be related by a sequence of stellar subdivisions and inverse operations (Alexander), and to have a common subdivision (Whitehead). However the notion of an arbitrary subdivision is an affine, and not a purely combinatorial notion. It would be great if one could show at least that for some family of subdivisions definable in purely combinatorial terms (e.g. replacing a simplex by a simplicially collapsible or consructible ball), common subdivisions exist... Stellar subdivision (and arbitrary subdivisions) can be explained to a K-12 student with a picture. For a stellar subdivision, choose a face F, take its midpoint, and connect it to all vertices of tetrahedra of which F is a face. For arbitrary subdivision, invent some silly triangulation of a simplex, and just plug it inside. refining heighbouring simplexes as needed. - The continuum hypothesis. Of course it's extremely famous, but everyone thinks it's resolved. I was astonished to find out that some serious set theorists apparently consider it (I mean in the present, decades past Cohen's proof) to be an important open problem that people should be working on solving (for some meaning of "solve"). P. Koellner ( http://logic.harvard.edu/EFI_CH.pdf ) describes some current approaches. - 1 You seem to allude that it has not been resolved yet. But in that case you should explain what you mean by that. On the other hand, the continuum hypothesis is so well-known that it does not fit to the question. – Martin Brandenburg Jul 19 at 12:17 3 By the way, I think your statement that "everyone thinks it's resolved" is a little misleading. Maybe it would be better to say it this way: "most people think there's nothing left to say about it, because it's been proven independent of ZFC." – Vectornaut Jul 22 at 18:26 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 72, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9363099336624146, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/121131/list	## Return to Answer 3 added 19 characters in body $\pi_n(S^n)=[S^n,S^n]=$ cobordism \pi_n(S^n)=[S^n,S^n]=\lbrace$cobordism classes of framed submanifolds of dimension zero (0-submanifolds$\rbrace$by the Pontrjagin-Thom construction)construction. These are collections of points (with sign), and sign) which add up to give the degree of the mapmaps, so this set is isomorphic to precisely$\mathbb{Z}\$. 2 deleted 2 characters in body $\pi_n(S^n)=[S^n,S^n]=$ equivalence cobordism classes of framed submanifolds of dimension zero (by the Pontrjagin-Thom construction). These are points (with sign), and add up to give the degree of the map, so is isomorphic to $\mathbb{Z}$. Post Made Community Wiki by S. Carnahan♦ 1 $\pi_n(S^n)=[S^n,S^n]=$ equivalence classes of framed submanifolds of dimension zero (by the Pontrjagin-Thom construction). These are points (with sign), and add up to give the degree of the map, so is isomorphic to $\mathbb{Z}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8781113624572754, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/55118/how-can-we-know-the-time-frames-for-events-in-the-early-universe	# How can we know the time frames for events in the early universe? I just finished watching Into the Universe with Stephen Hawking (2010). Specifically the third episode titled 'The Story of Everything.' In the episode Hawking is explaining the mainstream theories of the events just after the Big Bang. What confuses me is that he just said at the beginning that the laws of physics were different and time (among other things) did not exist as we know it during this early stage in the life of the universe, but then goes on to say certain events took certain amounts of time. I don't know if the numbers match, but the same issue exists on this Wikipedia article. How can we even guess the time frames of these events if time itself is largely dependent on the current state of the universe? - The short--short version is that particle physicists have measured the physics that applies under certain temperature and density conditions and can extrapolate to a certain degree and then you can just roll the physics backward. How far back you can go depends on exactly what theories you trust, but we can certainly go back to the weak freeze out (temperatures around the Z mass) and can say something about the physics of quark--gluon plasmas. – dmckee♦ Feb 25 at 20:19 This question isn't really a skeptic claim, but more of a request to clarify the physics behind the theory and how it was developed, as such I think that this question should be migrated to Physics.SE – Ilya Melamed Feb 25 at 21:14 @Ilya Migrate it if you want but things like this are what make people skeptical of what might be good science. – fredsbend Feb 25 at 21:26 1 @fredsbend, your question is a perfectly legitimate and good question, however, the feeling is that you want a better explanation of the physics behind the theory, you don't really want to see a list of links to published articles and quotes from their abstracts saying "we calculated that this and that phenomena happened at this and that time", but rather you want the physics behind those papers explained to you, that is what Physics.SE is for. – Ilya Melamed Feb 25 at 21:40 ## 1 Answer When you see the history of the universe plotted against time, the time used is the comoving time i.e. the time measured by a clock that is at rest with respect to the universe around it. This is the time co-ordinate used in the FLRW metric, which is a solution to the equations of GR that, as far as we can tell, gives a good description of the universe back to very early times. Earlier than around the Planck time after the Big Bang we expect the notion of time to become imprecise because it isn't possible to measure times shorter than the Planck time. Without a working theory of quantum gravity it isn't possible to comment further. However for all times later than the Planck time we expect time to be a good co-ordinate and be well behaved. This allows physicists to calculate at what time the various stages in the evolution of the universe happened. Response to comment Let me attempt to phrase my answer more broadly. The first point is that in relativity (both Special and General) you need to be careful talking about time. For example you've probably heard that time runs more slowly when you move at speeds near the speed of light. However there is a well defined standard time that cosmologists use for describing the history of the universe. We call this comoving time. So when you hear statements like "the universe is 13.7 billion years old" we mean it's 13.7 billion years old in comoving time. You don't need to know how comoving time is defined, just that it gives us a good timescale for describing the history of the universe back to the Planck time. Which brings us to ... You've probably also heard of Heisenberg's uncertainty principle. Again I'll gloss over the details, but one side effect of the uncertainty principle is that it's impossible to measure times less than about 5 $\times$ 10$^{-44}$ seconds. I don't know of a simple way to explain this to someone who isn't familiar with quantum mechanics, so I'm afraid you'll have to take this on faith. And this brings us back to Hawking's programme. As long as the times we are interested in are greater than 5 $\times$ 10$^{-44}$ seconds we can define the time using comoving time so we can assign reliable times to cosmological events like the electroweak transition. But for times close to 5 $\times$ 10$^{-44}$ seconds the whole notion of a "time when something happens" becomes meaningless because it's fundamentally impossible to measure times that short. I'd guess this is what Hawking means when he says time ceases to exist. - I am having a hard time understanding your answer. Maybe you could dumb it down a bit. No need to be exact just help me understand. – fredsbend Mar 7 at 8:14 @fredsbend: I've tried to simplify my answer. Have a look and see what you think. – John Rennie Mar 7 at 18:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9589778780937195, "perplexity_flag": "head"}
http://mathhelpforum.com/math/209443-where-should-i-post-my-findings.html	Thread: 1. Where should I post my findings? I've been messing around with formulas, and although my findings aren't really new knowledge I feel like I should share them anyway. Would the correct place to share them be on a blog? Should I post them here on the forum somewhere? One of the formulas I found is a bit tough to structure using normal latex packages, I had to use amsmath to get it right. Is this going to be a problem when trying to post it online? Should I just post the pdf instead? 2. Re: Where should I post my findings? Originally Posted by jdfm I've been messing around with formulas, and although my findings aren't really new knowledge I feel like I should share them anyway. Would the correct place to share them be on a blog? Should I post them here on the forum somewhere? One of the formulas I found is a bit tough to structure using normal latex packages, I had to use amsmath to get it right. Is this going to be a problem when trying to post it online? Should I just post the pdf instead? amsmath is a normal $\LaTeX$ package. . 3. Re: Where should I post my findings? So, that means I can just post it anywhere and it will work? I was under the impression that it might not be available everywhere. 4. Re: Where should I post my findings? Originally Posted by jdfm I've been messing around with formulas, and although my findings aren't really new knowledge I feel like I should share them anyway. Would the correct place to share them be on a blog? Should I post them here on the forum somewhere? There are a few places depending on just what level you wish to post for. 1. Lounge - This is the least formal, but would probably find a good sized audience 2. Math - Similar to above and you would be more likely to get constructive advice 3. Peer Math Review - High end constructive advice but likely to get the smallest audience. -Dan 5. Re: Where should I post my findings? So I tried using the \LaTeX code that I had, and it seems to be too long, so I guess I'll just post it in PDF form. I'll do this later when I get home.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9626933336257935, "perplexity_flag": "middle"}
http://www.askmehelpdesk.com/mathematics/how-find-surface-area-cone-terms-pi-683733.html	Browse All Topics Find questions to answer Find today's questions Find unanswered questions Search Search Forums Show Threads Show Posts Advanced Search Home ▸ Science ▸ Mathematics » how to find surface area of a cone in terms of pi how to find surface area of a cone in terms of pi Asked Jul 17, 2012, 04:11 AM — 2 Answers I am struggling to find out how to find the surface area of a cone giving all answers in terms of pi ...radius is 8 and slant height is 12... Please help?! Thread Summary 2 Answers Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #2 Jul 17, 2012, 05:38 AM Do you know the formula for the surface area of a cone? We'd like to understand what you find wrong with Unknown008's answer: What's inaccurate about this answer? Say it in 25 words or less here and/or reply in the thread with more detail. Please focus on the content not the person! Link to a credible and well-known source. You can provide a URL or simply describe the source. ebaines Posts: 10,033, Reputation: 5529 Expert #3 Jul 17, 2012, 05:43 AM Given the radius of the base =R and the slant length S: the area of the slanted side of the cone is $\pi R S$ and the area of the base is $\pi R^2$. The total surface area is the sum of these. We'd like to understand what you find wrong with ebaines's answer: What's inaccurate about this answer? Say it in 25 words or less here and/or reply in the thread with more detail. Please focus on the content not the person! Link to a credible and well-known source. You can provide a URL or simply describe the source. Thread Tools Search this Thread Search this Thread: Advanced Search Check out some similar questions! find the lateral area and surface area of a right cone with radius of 9 in. and height of 12 in. give the answer in terms of pi well im doing a project of an icosahedron, and i have no idea what im doing. My icosahedron has 12 verticies 29 edges and 20 faces. i just need to know what the volume and the surface area is. can u explain the whole equation step by step. i looked it up and it said something about 5 divide it by... Find the surface area of a cyclinder? 9 is height. 4 is radius. Pi is 3.14. I need some help on the way to solve for the surface area of a cylinder with a cone on the top. I think that I solve for the surface area of the cone, then solve for the surface of the cylinder. I have the height from the center of the bottom circle to the apex of the cone, and I have the radius... View more Mathematics questions Search	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 2, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9046017527580261, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/99483/positive-fourier-coefficients	## Positive Fourier coefficients ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi all, Is there any general way to construct a smooth 2pi periodic function which vanishes on an interval and has positive Fourier coefficients? And if that was too specific I can make this more general by asking if there is any general characterization for functions on the torus with positive Fourier coefficients? Thanks. - ## 2 Answers Sure. Let $f$ be a smooth real-valued function supported on $A \subseteq [0,2\pi]$. Then, denoting convolution by $$, we have: 1. ff s supported on the Minkowski sum $A+A$ 2. ff is smooth 3. $\widehat{ff}(n) = \|\hat{f}(n)\|^2 \geq 0$. From 1, if the support of $f$ is a small interval, the support of $ff$ will be a slightly bigger interval. Clearly, the support of $f$ can be chosen appropriately so that $ff$ will vanish on any specified interval. - 3 Minor point --- don't you want $f * g$ where $g(t) = \bar{f}(-t)$? The Fourier coefficients of $f * f$ are $(\hat{f}(n))^2$, not $\|\hat{f}(n)\|^2$. – Nik Weaver Jun 13 at 19:52 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. To make your more general question even more general, on any locally compact abelian group $G$, Bochner's theorem says a continuous function $f$ is the inverse Fourier transform of a positive measure on the dual group if and only if $f$ is positive definite, i.e. $\sum_{n,m=1}^N c_n \overline{c_m} f(x_n - x_m) \ge 0$ for all $c_1 \ldots c_N \in \mathbb C$ and $x_1 \ldots x_N \in G$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8875110745429993, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/67878/list	## Return to Answer 2 added 287 characters in body No, there does not exist such a $V$. Let $W = V \cup (V \oplus r)$, and suppose $W = F \Delta Q$. Note that for $s \in {\mathbb Q}$, $W \cap (W \oplus s)$ is nonempty if and only if $s$ or $s-r$ or $s+r$ is an integer. But if $F$ contained an interval of positive length, $W \cap (W \oplus s)$ would be nonempty for all sufficiently small $\|s\|$. Thus $F$ is nowhere dense, and $W$ is of first category. But this is impossible, because $[0,1] = \bigcup_{r \in {\mathbb Q}} (W \oplus r)$. For your second question, the properties of Vitali sets (and finite unions of their translates) that this uses are: 1) there is a dense set of $r$ such that $V \cap (V \oplus r) = \emptyset$. 2) there is a countable set $S$ such that $\bigcup_{r \in S} (V \oplus r) = [0,1)$. 1 No, there does not exist such a $V$. Let $W = V \cup (V \oplus r)$, and suppose $W = F \Delta Q$. Note that for $s \in {\mathbb Q}$, $W \cap (W \oplus s)$ is nonempty if and only if $s$ or $s-r$ or $s+r$ is an integer. But if $F$ contained an interval of positive length, $W \cap (W \oplus s)$ would be nonempty for all sufficiently small $\|s\|$. Thus $F$ is nowhere dense, and $W$ is of first category. But this is impossible, because $[0,1] = \bigcup_{r \in {\mathbb Q}} (W \oplus r)$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9136302471160889, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3347406	Physics Forums Page 5 of 6 « First < 2 3 4 5 6 > Has this idea been explored? Dark matter as matter in parallel universes... Quote by DavidMcC But that is nothing in comparison with the added complexity of making the "one universe" cosmology fit with the rest of astronomical and cosmological data, such as lop-sided particle-physics laws, dark energy that implies that the cosmological constant isn't even constant, etc, etc, not to mention the non-discovery of dark-matter particles. I strongly suspect that this isn't the case, and if you get a multiple universe theory to the point where you can fit the data, then it will be as complex if not more complex than what we have now. If you have a complex theory, then adding universes to the theory makes things more complex and not less complex. I'd be interested in hearing why you would think otherwise. How can a multiple universe theory of dark matter POSSIBLY be LESS complex? You'll need to include all of the elements of the a one-universe cosmology, and in addition you have to figure out how two spatially and temporally separated bubbles interact in what seems like a pretty uniform manner. Seems a bit mad to me as a matter of fact, and again, against the spirit of a multiverse if not every possible practice of one. Quote by Chalnoth What are you going on about? There is no evidence as of yet that the cosmological constant isn't constant, and dark matter is expected to be extremely hard to detect, so it's hardly a surprise we haven't yet. This, for example: http://www.scribd.com/doc/4787671/Th...nd-Dark-Energy "... This alleviates the classical problem of the curious energy scale of order a millielectronvolt associated with a constant lambda." Recognitions: Science Advisor Quote by Cosmo Novice I was under the impression the cosmoloigcal contant is theoretically constant. It is not temporally constant (ie: it changes over time) but is spatially constant. The cosmological constant is constant in both time and space. Perhaps you were thinking of the misnamed Hubble constant? Recognitions: Science Advisor Quote by DavidMcC This, for example: http://www.scribd.com/doc/4787671/Th...nd-Dark-Energy "... This alleviates the classical problem of the curious energy scale of order a millielectronvolt associated with a constant lambda." There are indeed many speculative alternatives to the cosmological constant that vary in time. But there is as yet no evidence of time-variation of dark energy. And the "fine tuning" argument here is a non-argument because the anthropic selection effect guarantees that the cosmological constant be small anyway. Quote by twofish-quant I strongly suspect that this isn't the case, and if you get a multiple universe theory to the point where you can fit the data, then it will be as complex if not more complex than what we have now. If you have a complex theory, then adding universes to the theory makes things more complex and not less complex. I'd be interested in hearing why you would think otherwise. "Smolin-esque" LQG-based BH cosmology only requires a few reasonable additions to at least provide a framework for making sense of what is otherwise just bizarre. (The main one is that "space-loops" are only linked within a space that is generated from the collapse of a single body, and that may already have been in Smolin's own version.) I've listed the rest several times before on this and other sites. The pro-matter, ant-anti-matter bias of the universe is one, as it suggests that what banged was somehow contaminated, as if having been condensed previously from part of a previous, much bigger universe. The apparent "fine-tuning" of the laws to the possibility of abiogenesis is another - this is the only way we might find ourselves in a universe in which the fundamental constants were just so, otherwise it would have to have been extraordinary lucky. Etc. Quote by Chalnoth There are indeed many speculative alternatives to the cosmological constant that vary in time. But there is as yet no evidence of time-variation of dark energy. And the "fine tuning" argument here is a non-argument because the anthropic selection effect guarantees that the cosmological constant be small anyway. You probably just ignore the evidences, in fact there are plenty experimental data supporting non-constant agenda. Specifically, the varying "alpha" has been reported for 15+ years, the recent report (see below) for the spacial alpha anisotropy explains the inconsistencies of previous reports. Refs: arxiv.org/abs/1008.3907: Evidence For Spatial Variation Of The fiFine Structure Constant arxiv.org/abs/1008.3957: Manifestations Of A Spatial Variation Of Fundamental Constants On Atomic Clocks, Oklo, The popular overview: http://www.sciencedaily.com/releases...0909004112.htm BTW, the dark matter flow correlates with alpha gradient (see http://en.wikipedia.org/wiki/Dark_flow) Stefan I was under the impression that 'Dark Flow' is at the level of, "might be something, might be an irregularity on the image."? It seems everyone uses this one to justify some claim, from colliding universes and more. In terms of established science, you seem to be going on with a bit of nonsense there stefanbanev, or at least grossly overreaching. @DavidMcC: Or, while I don't believe this, the 'eternal inflationists' could be right and we're part of an infinite set of universes, no more or less unique than any other part of an infinite grouping. When there is NOTHING to point one way or another, what is the point in all of this? Recognitions: Science Advisor Quote by stefanbanev You probably just ignore the evidences, in fact there are plenty experimental data supporting non-constant agenda. Specifically, the varying "alpha" has been reported for 15+ years, the recent report (see below) for the spacial alpha anisotropy explains the inconsistencies of previous reports. Refs: arxiv.org/abs/1008.3907: Evidence For Spatial Variation Of The fiFine Structure Constant arxiv.org/abs/1008.3957: Manifestations Of A Spatial Variation Of Fundamental Constants On Atomic Clocks, Oklo, The popular overview: http://www.sciencedaily.com/releases...0909004112.htm BTW, the dark matter flow correlates with alpha gradient (see http://en.wikipedia.org/wiki/Dark_flow) Stefan Um, that's a completely separate issue from the cosmological constant. $\alpha$ and $\Lambda$ are completely different parameters. But it's largely shown to be bunk. The basic idea behind the varying alpha is that if the fine structure constant were to vary, then atoms would not just have redshifted or blueshifted spectra, but the entire pattern of spectral lines changes, especially for heavier atoms. So the experimental team looked for these changes in the patterns of the more massive elements, such as Carbon and Oxygen, in distant quasars. The difficulty here is that the signatures of these elements are really, really faint, so they can only barely see them against the background. And the spectral signatures of these elements are also quite complex, with lots and lots of spectral lines, so that it's not at all clear which line belongs to which atom. So, in the end, it turns out that they're just fitting the background noise. This is supported by the fact that there is no consistency between the measurements of $\alpha$ between different quasars, and different experimental teams trying to replicate their results have come up with completely different results. Mentor Quote by Cosmo Novice I was under the impression the cosmoloigcal contant is theoretically constant. It is not temporally constant (ie: it changes over time) but is spatially constant. Quote by Chalnoth The cosmological constant is constant in both time and space. Perhaps you were thinking of the misnamed Hubble constant? Cosmo Novice, perhaps you were thinking of http://www.physicsforums.com/showthr...35#post3330035. Or did you really mean the cosmological constant? Chalnoth> But it's largely shown to be bunk. Please be more specific; is it shown by whom (reference please)? Chalnoth> So, in the end, it turns out that they're Chalnoth> just fitting the background noise. No offence, but may you buck it by something more tangible then just your opinion? Regards, Stefan Recognitions: Science Advisor Quote by stefanbanev Chalnoth> But it's largely shown to be bunk. Please be more specific; is it shown by whom (reference please)? Chalnoth> So, in the end, it turns out that they're Chalnoth> just fitting the background noise. No offence, but may you buck it by something more tangible then just your opinion? Regards, Stefan It's backed up by their very own work: http://arxiv.org/abs/1008.3907 We previously reported observations of quasar spectra from the Keck telescope suggesting a smaller value of the fine structure constant, alpha, at high redshift. A new sample of 153 measurements from the ESO Very Large Telescope (VLT), probing a different direction in the universe, also depends on redshift, but in the opposite sense, that is, alpha appears on average to be larger in the past. Inconsistent results are a hallmark of badly-done science. Quote by George Jones Cosmo Novice, perhaps you were thinking of http://www.physicsforums.com/showthr...35#post3330035. Or did you really mean the cosmological constant? Thankyou Chalnoth and George, I was a little confused and thinking of the hubble constant! Quote by Chalnoth Inconsistent results are a hallmark of badly-done science. Thanks for reference, it's not 100% definitive but clearly supports your position. Stefan Quote by stefanbanev Thanks for reference, it's not 100% definitive but clearly supports your position. Stefan It's not 100%, but it's pretty fat nail in that coffin. A lot of the "Dark" stuff other than matter is used to justify any number of pet theories; tread with care. Quote by Misericorde It's not 100%, but it's pretty fat nail in that coffin. A lot of the "Dark" stuff other than matter is used to justify any number of pet theories; tread with care. It's very true. I'm bias for any experimental evidence for "multiverse" support because it's its weakest spot. I still think that the traditional scientific method should work for such "metaphysics" frontier even it may be flexed quite a bit. The proposed "statistical" methods are indirect and prone to interpretations; therefore, those direct observation for alpha appeals a lot... Recognitions: Science Advisor Quote by stefanbanev It's very true. I'm bias for any experimental evidence for "multiverse" support because it's its weakest spot. I still think that the traditional scientific method should work for such "metaphysics" frontier even it may be flexed quite a bit. The proposed "statistical" methods are indirect and prone to interpretations; therefore, those direct observation for alpha appeals a lot... I was pretty excited about it the first time I heard about it too. I've just become a bit jaded after learning more about it. Page 5 of 6 « First < 2 3 4 5 6 > Thread Tools \| \| \| \| \|--------------------------------------------------------------------------------------------------\|-----------------\|---------\| \| Similar Threads for: Has this idea been explored? Dark matter as matter in parallel universes... \| \| \| \| Thread \| Forum \| Replies \| \| \| Astrophysics \| 2 \| \| \| Cosmology \| 2 \| \| \| Astrophysics \| 4 \| \| \| Cosmology \| 21 \| \| \| General Physics \| 4 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9397168755531311, "perplexity_flag": "middle"}
http://gilkalai.wordpress.com/2012/11/17/a-few-mathematical-snapshots-from-india-icm2010/	Gil Kalai’s blog ## A Few Mathematical Snapshots from India (ICM2010) Posted on November 17, 2012 by Can you find Assaf in this picture? (Picture: Guy Kindler.) In my post about ICM 2010 and India I hardly mentioned any mathematics. So here are a couple of mathematical snapshots from India. Not so much from the lectures themselves but from accidental meetings with people. (Tim Gowers extensively live-blogged from ICM10.) First, the two big problems in analysis according to Assaf Naor as told at the Bangalore airport waiting for a flight to Hyderabad. Next, a lecture on “proofs from the book” by Günter Ziegler. Then, some interesting things I was told on the bus to my hotel from the Hyderabad airport by François Loeser, and finally what goes even beyond q-analogs (answer: eliptic analogs) as explained by Eric Rains. (I completed this post more than two years after it was drafted and made major compromises on the the quality of my understanding of the things I tell about. Also, I cannot be responsible today for the 2-year old attempts at humor.) ### The two big problems in analysis according to Assaf, and one bonus problem The day before the ICM both Assaf Naor and I visited Microsoft Research at Bangalore hosted by Ravi Kannan (who had to fly early to make it to the grand rehearsal of the ICM opening ceremonies,) and Nisheeth Vishnoi. In the evening while we both waited for the flight to Hyderabad, Assaf told me what he regards as the two big problems in analysis. 1) The two-dimensional Carleson problem Carleson’s famous theorem asserts that the Fourier expansion of a function in $L_2$ converges to the function almost everywhere. High dimensional analogs of the question itself or of some basic ingredients in the argument are very important and very hard. See here. 2) Quantitative versions of Hardy-Littlewood maximal principle The story here starts with the Hardy-Littlewood maximal principle. Well, while I had heard about the maximal principle I really did not know what it was. The formulation is very simple. Start with a real function f in $L_1$. So the expected value of \|f(x)\| is bounded. Then define g(x) to be the supremum over radius r of the expectation of f(x) over a ball of radius r around x. As it turns out g is not necessarily in $L_1$ but it comes close. Let me say what “comes close” means here. A functions f in $L_1$ has the property that the measure of the points x where \|f(x)\|>t is at most C/t. (Where C is the $L_1$ norm of f.) This is Markov’s inequality which is extremely useful in probability and it is extremely important to find better inequalities or much better inequalities for various specific cases. (The importance of going beyond Markov’s inequality is something I was interested in a while, so this aspect of the story appealed to me.) Markov’s inequality is a consequence of being in $L_1$, but it is a somewhat weaker condition. Hardy and Littlewood proved that the maximum function g satisfies Markov’s inequality! The question is to find quantitative bounds for this theorem by Hardy and Littlewood depending on the dimension. Here is a related paper by Naor and Tao and a related post on “what’s new.” 3) Bonus question: Pisier’s dichotomy conjecture. A few days later I had a little chat with Assaf and he told me yet another very interesting problem about polytopes and about projections. This is Pisier’s dichotomy conjecture. (Gilles Pisier himself was also in Hyderabad). Let me simply quote Assaf on this one: Pisier’s dichotomy problem is: If K is a centrally symmetric polytope in $R^n$, with $e^{o(n)}$ faces, then for every ε>0 there is a linear subspace E of $R^n$, of dimension k that tends to infty with n (depending on the o(n) in the bound on the number of faces), and a parallelepiped $P=T(B_\infty^k)\subset E$, such that the section $E\cap K$ contains P, and is contained in (1+ε)P. Replacing ε by a universal constant (independent of dimension), and weakening the requirement that P is an image of a cube by just asking that it has polynomially (in k) many faces, is already interesting. Update: Half a year later I met Leonard Schulman in Caltech and he told me about a project he is involved with and a question of isoperimetric type regarding Boolean functions. Having the conversation with Assaf fresh in my memory I noticed that the question (without any modification) is a about maximal principle for the Boolean cube, which turned out to be a useful “nudge”. See this paper by Aram Harrow, Alexandra Kolla and Leonard Schulman which is also featured in this post on GLL. ### Proofs from the book lecture by Günter. On the way from Hyderabad airport to our hotels I set next to an Indian number theorist, and Günter Ziegler and Assaf were sitting one row ahead. Günter showed us the slides for his wide-audience “proofs from the book” talk which contain quite a few beautiful proofs. The slides of this lecture are now available online here. (Update: link fixed) ### Bercovich, Ngo and Model theory according to François I met François Loeser a couple of times in New Haven and in Jerusalem, and after Assaf and Günter left the bus for their hotel, François and I discovered that we were heading to the same hotel. François is collaborating with Ehud Hrushovski in several projects that combine model theory and algebraic geometry. And François told me about a paper that they are going to arXive in just in a few days about Berkovich spaces. Now, when I talked to François I really felt that I gained some understanding about this project. I remember that there was one aspect in which the new theory is a large extension of what Berkovich was doing while from another angle it is a special case. The theory is also related to real algebraic geometry (which is quite close to discrete geometry). But the truth of the matter is that I probably cannot tell you more about this paper and about the whole model theory – algebraic geometry endeavor than what you can get by looking at the paper. If there are good survey articles or popular accounts about this direction I will be happy to learn about them. “So does this model theory connection have anything to do with… ehh… say… the fundamental lemma?” I tried my luck? “Yes”, said François, not so much the work with Ehud but rather another work that he had with some partners. It turns out that it was important to know how to transfer results from finite fields to p-adic fields and model theory can help. Although by now more precise proofs of the required results can be proved using “pure” algebraic geometry. A couple of days later I met by chance (well, to meet another mathematician in an ICM is not a sheer coincidence) another famous figure in the connection between Model theory and algebra: Anand Pillay. Sixteen years earlier in Zurich ICM 1994 we were walking with several people and Lenore Blum said how wonderful Pillay’s talk was. I heard about his mathematics even before so I was quite curious to meet him then, and I even had a (bogus) mental image of how he looks but it waited 16 years. Update: From what I hear Hrushovski and Loeser’s approach to Berkovich spaces is a great hit, and model-theoretic thinking in algebraic geometry breaks new grounds. Unrelatedly, the recent proposed proof of the ABC conjecture also has a dose of logic-theoretic aspects. ### Elliptic analogs according to Eric Rains q-Analogs are not the end of the road – there are also elliptic analogs! We are familiar with q-analogs. The q analog of n! is $1(1+q)(1+q+q^2) ... (1+q+...q^{n-1})$. The q analogs of binomial coefficients are Gaussian coefficients (just replace the factorials by their q-analogs). Among the names associated with q-analogs, Euler, Ramanujan, Macdonald, Andrews (who was at Hyderabad) Zeilberger, Cherednick and many more. A famous younger guy is Eric Rains, and since I did not met Eric Rains before, I invented a mental image of how he looks which turns out to be completely wrong (but it was based on a wrong spelling “Reins”). One evening, while shoving my way together with Synthia Dwork, Ursula Hamenstädt, and Salil Vadhan, to the bus from the convention center to our hotel, Eric, whom we just met, told us about even more general objects called “elliptic analogs”. The idea is very simple: you replace the parameter q with a parameter representing elliptic curves. This goes back to a 1997 paper by Frenkel and Turaev. Of course, as simple as it sounds it made no sense to me at all. But after listening a to Eric and looking a little at some papers I got a very vague picture about what it is about. So here are some references: 1, 2, 3, 4, 5. Updates: When I was in Caltech half a year later I met Eric and he told me in greater detail about elliptic analogs, and I even took careful notes for the purpose of explaining the matter a little more professionally, but somehow I cannot find them now, but will welcome comments from people who can explain this further (and also the other stuff mentioned in this post). Cynthia Dwork gave a beautiful talk at the ICM about the mathematics and computer science of privacy. (Mathematical notions related to questions about privacy are connected to discrepancy questions, This connection played a role in the recent determination by Nikolov and Talwar of the hereditary discrepancy analog for Erdos’ discrepancy problem. See also this post in Gowers’s blog.) Being at an ICM is a very special experience. There are many friends you want to catch up with and many people you know and would like to talk to, and things are too hectic for this. Here you see some well-known distinguished people who you always wanted to see, and here, lo and behold, a few people treat you this way. I like these random and short meetings that leaves you unsatisfied, but I know some mathematicians that cannot stand it. The next ICM is in Seoul 2014. About these ads ### Like this: Like Loading... This entry was posted in Conferences, Open problems and tagged Assaf Naor, Eric Rains, François Loeser, Günter Ziegler, ICM2010. Bookmark the permalink. ### One Response to A Few Mathematical Snapshots from India (ICM2010) 1. Gil Kalai says: The link to Günter Ziegler’s slides on proofs from the book was fixed. • ### Blogroll %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 12, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9466468095779419, "perplexity_flag": "middle"}
http://planetmath.org/chebyshevfunctions	# Chebyshev functions There are two different functions which are collectively known as the Chebyshev functions: $\displaystyle\vartheta(x)=\sum_{{p\leq x}}\log p.$ where the notation used indicates the summation over all positive primes $p$ less than or equal to $x$, and $\displaystyle\psi(x)=\sum_{{p\leq x}}k\log p,$ where the same summation notation is used and $k$ denotes the unique integer such that $p^{k}\leq x$ but $p^{{k+1}}>x$. Heuristically, the first of these two functions measures the number of primes less than $x$ and the second does the same, but weighting each prime in accordance with their logarithmic relationship to $x$. Many innocuous results in number theory owe their proof to a relatively simple analysis of the asymptotics of one or both of these functions. For example, the fact that for any $n$, we have $\displaystyle\prod_{{p\leq n}}p<4^{n}$ is equivalent to the statement that $\vartheta(x)<x\log 4$. A somewhat less innocuous result is that the prime number theorem (i.e., that $\pi(x)\sim\frac{x}{\log x}$) is equivalent to the statement that $\vartheta(x)\sim x$, which in turn, is equivalent to the statement that $\psi(x)\sim x$. # References • 1 Ireland, Kenneth and Rosen, Michael. A Classical Introduction to Modern Number Theory. Springer, 1998. • 2 Nathanson, Melvyn B. Elementary Methods in Number Theory. Springer, 2000. Type of Math Object: Definition Major Section: Reference Groups audience: 11A41 Primes ## Recent Activity May 17 new image: sinx_approx.png by jeremyboden new image: approximation_to_sinx by jeremyboden new image: approximation_to_sinx by jeremyboden new question: Solving the word problem for isomorphic groups by mairiwalker new image: LineDiagrams.jpg by m759 new image: ProjPoints.jpg by m759 new image: AbstrExample3.jpg by m759 new image: four-diamond_figure.jpg by m759 May 16 new problem: Curve fitting using the Exchange Algorithm. by jeremyboden new question: Undirected graphs and their Chromatic Number by Serchinnho ## Info Owner: Mathprof Added: 2003-08-11 - 16:36 Author(s): Mathprof broken by yark ✓ ## Versions (v11) by Mathprof 2013-03-22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 15, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.880554735660553, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/287864/why-the-zeta-function	# Why the zeta function? Why is the zeta function, $\zeta(s)$ used to obtain information about the primes, namely giving explict formula for different prime counting functions, when there are many other functions that encode information about primes? For example, the relation: $$\frac{-\zeta'(s)}{\zeta(s)}=\sum_{n=1}^\infty \frac{\Lambda(n)}{n^s}$$ Could be seen as a special case of a polylogarithm identity: $$\frac{d}{ds}\text{Li}_s(x)=\sum_{n=1}^\infty\frac{\Lambda(n)}{n^s}\text{Li}_s(x^n)$$ Where $x=1$. - Are you going to finish the question? "Where $x=1$," --- what? – Gerry Myerson Jan 27 at 4:08 I was going to list other examples, but I don't think it really matters, so I deleted the coma – Ethan Jan 27 at 4:28 You might want to replace it with a period. It looks like you are in the middle of editing the question (and also it's typographically offensive). – Potato Jan 27 at 4:39 ## 3 Answers This is a slightly misguided question. The zeta is used to find information about the primes because it gives information about the primes. There is no real reason a priori why one should look at it, and people have looked at every other function from which they could extract information about primes. Zeta is famous because people have been particularly successful at extracting information from it. With time, a whole mass of related results have been obtained, and nowadays zeta functions appear everywhere in various forms, but this is not because of a desire to shun other information-encoding functions, but simply for the reason that there is an immense toolkit of methods, techniques, heuristics and so one that were developed in order to extract information from functions of that type. - Why is it easyier to extract information from the zeta function then say some other function? Also what do you mean by a priori? – Ethan Jan 27 at 5:18 I have no idea, i am no expert. But your asking me that somehow makes me think that you have the impression that the actual experts have a weird intention of making their own life more complicated! In view of the evidence available, it is clear that people find it easier to use zeta functions than... other sort of functions. Maybe the next century this will change and prizes will be set on conjectures about some other kind of function. – Mariano Suárez-Alvarez♦ Jan 27 at 5:21 People use whatever they can to do what they want to do. And given choices, they will use what will give best results with least effort. They have been using zeta functions for almost 2 hundred years... so what can we conclude? – Mariano Suárez-Alvarez♦ Jan 27 at 5:23 2 Mathematicians are extremely pragmatic people. – Mariano Suárez-Alvarez♦ Jan 27 at 5:31 The Euler product encodes unique factorization of natural numbers. A generalization of the zeta function exist for number fields and in that situation the Euler product there encodes the unique factorization of ideals. Euler's debut was the solution of the Basel problem - he in fact evaluated the zeta function at even values. Euler proved the divergence of $\sum\frac{1}{p}$ using the Euler product in 1737, which was the first genuinely new proof that there are infinitely many primes in decades. He never (explicitly?) analytically continued the function. Dirichlet proved that there are infinitely many primes in arithmetic progressions in 1837 as well as the class number formula using a version of the zeta function that has a periodic phase in the numerator, both of which require analytic continuation. Riemann of course studied the function in much more depth, derived its functional equation and published On the Number of Primes Less Than a Given Magnitude in 1859 which insight into prime numbers and led to the proof of the prime number theorem. A lot of difficult results in number theory have been strengthened on the assumption of the Riemann hypothesis, and there have been many many deep results proved using techniques related to zeta functions. So there is clearly something interesting going on there. Good read: https://terrytao.wordpress.com/2009/09/24/the-prime-number-theorem-in-arithmetic-progressions-and-dueling-conspiracies/ and http://www.dpmms.cam.ac.uk/~wtg10/zetafunction.ps tries to motivate it from "you could have discovered zeta" point of view. http://www.math.harvard.edu/~elkies/M229.09/index.html • February 8: psi.pdf: Complex analysis enters the picture via the contour integral formula for \psi(x) and similar sums - Ok so to ask you a question, I might ask: "Why don't we use category theory to study primes?". Your answer would probably be something along the lines of "There aren't many known connections between these two objects that links them in a useful way". This answers your question essentially, yes there are other mathematical objects connected with primes that tell us nice things about them but the Riemann zeta function has just turned out to be the best at telling us useful things so far. Why would we want to study objects which don't appear to be connected with them (unless we had reason to believe that there should be a link)? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9443104863166809, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/51125-solved-sat-geometry.html	# Thread: 1. ## [SOLVED] SAT geometry In the figure below, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the darkened region is a) $8+3pi$ b) $10+3pi$ c) $14+3pi$ d) $1+6pi$ e) $12+6pi$ Can you please show the steps? Thanks in advance hey i forgot a major info. The graph stated that SR is 6, sorry i didn't include it on the graph!! Attached Thumbnails 2. Originally Posted by fabxx In the figure below, arc SBT is one quarter of a circle with center R and radius 6. If the length plus the width of rectangle ABCR is 8, then the perimeter of the darkened region is a) $8+3pi$ b) $10+3pi$ c) $14+3pi$ d) $1+6pi$ e) $12+6pi$ Can you please show the steps? Thanks in advance i will start you off. Let $RC$ be $w$, that is, the width of the rectangle. Let $AR$ be $l$, that is, the length of the rectangle. Let $P$ be the perimeter of the darkened region. Note that the diagonal $AC = 6$ we are given $l + w = 8$ ............(1) by Pythagoras' theorem, $l^2 + w^2 = 6^2$ .............(2) you have two simultaneous equations with two unknowns, you can solve for $l$ and $w$. the rest should be straight forward $P = AS + AC + CT + \text{ arc}SBT$ now, $AS + l = 6 \implies \boxed{AS = 6 - l}$ $\boxed{AC = 6}$ $CT + w = 6 \implies \boxed{CT = 6 - w}$ $\boxed{\text{arc}SBT = \frac {90}{360} \cdot 2 \pi r}$, and you know $r$. so you can find all the components and hence the perimeter 3. Hey i forgot to include a major info in the graph. The radius is 6 is meant that SR is 6. I already corrected it in my previous thread. (: 4. Originally Posted by fabxx Hey i forgot to include a major info in the graph. The radius is 6 is meant that SR is 6. I already corrected it in my previous thread. (: as you can see, we already figured that out. you told us that R was the center and the circle had a radius of 6 (RT = 6 also ). nothing changes in what i did 5. Fabxx, This is SATs. Pythagoras is too long for SAT. You are supposed to get this answer in not more than 30 seconds, without a calculator. The worst thing a candidate make in a SAT exam is take it with a Texas Instruments. Really, this is not a joke. To solve this question, follow me carefully: It is given that AR + RC =8 ...(1) SR=RB=CA=6 ...(2) Perimeter = SA + arc ST + TC + CA (From above, we already know CA=6) SA = 6-AR TC= 6 - RC Add the two equations above: SA + TC = 12 - (AR+RC) = 12 - 8 =4 Therefore, Perimeter = arc ST + 4 + 6 = $\frac{1}{4}2\pi6$ + 4 +6 = 10+3 $\pi$ As you can see, you save time by not solving simultaneous equations and not evaluating pythagoras expression. Hope that helps. 6. Originally Posted by shailen.sobhee Fabxx, Perimeter = arc ST + 4 + 6 = $\frac{1}{4}2\pi6$ + 4 +6 = 10+3 $\pi$ how did you get that arc ST is $\frac{1}{4}2\pi6$ ? Thanks in advance (: 7. Originally Posted by fabxx how did you get that arc ST is $\frac{1}{4}2\pi6$ ? Thanks in advance (: $2 \pi * 6$ gives the circumference. you have 1/4 of it (it's a quarter circle) 8. Thanks to shailen.sobhee and Jhevon! 9. Originally Posted by shailen.sobhee Fabxx, SR=RB=CA=6 ...(2) How did you get that CA=6? CA isn't the radius, and from the graph, you can easily see that CA≠SR Thanks in advance 10. Originally Posted by fabxx How did you get that CA=6? CA isn't the radius, and from the graph, you can easily see that CA≠SR Thanks in advance you are correct. but RB is the radius, and CA = RB since they are both diagonals of the rectangle (sobhee mentioned this)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 33, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9300132393836975, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Group_cohomology	# Group cohomology In abstract algebra, homological algebra, algebraic topology and algebraic number theory, as well as in applications to group theory proper, group cohomology is a way to study groups using a sequence of functors Hn. The study of fixed points of groups acting on modules and quotient modules is a motivation, but the cohomology can be defined using various constructions. There is a dual theory, group homology, and a generalization to non-abelian coefficients. These algebraic ideas are closely related to topological ideas. The group cohomology of a group G can be thought of as, and is motivated by, the singular cohomology of a suitable space having G as its fundamental group, namely the corresponding Eilenberg–MacLane space. Thus, the group cohomology of Z can be thought of as the singular cohomology of the circle S1, and similarly for Z/2Z and P∞(R). A great deal is known about the cohomology of groups, including interpretations of low dimensional cohomology, functorality, and how to change groups. The subject of group cohomology began in the 1920s, matured in the late 1940s, and continues as an area of active research today. ## Motivation A general paradigm in group theory is that a group G should be studied via its group representations. A slight generalization of those representations are the G-modules: a G-module is an abelian group M together with a group action of G on M, with every element of G acting as an automorphism of M. In the sequel we will write G multiplicatively and M additively. Given such a G-module M, it is natural to consider the submodule of G-invariant elements: $M^{G} = \lbrace x \in M \ \| \ \forall g \in G : \ gx=x \rbrace.$ Now, if N is a submodule of M (i.e. a subgroup of M mapped to itself by the action of G), it isn't in general true that the invariants in M/N are found as the quotient of the invariants in M by those in N: being invariant 'modulo N ' is broader. The first group cohomology H1(G,N) precisely measures the difference. The group cohomology functors H* in general measure the extent to which taking invariants doesn't respect exact sequences. This is expressed by a long exact sequence. ## Formal constructions In this article, G is a finite group. The collection of all G-modules is a category (the morphisms are group homomorphisms f with the property f(gx) = g(f(x)) for all g in G and x in M). This category of G-modules is an abelian category with enough injectives (since it is isomorphic to the category of all modules over the group ring Z[G]). Sending each module M to the group of invariants MG yields a functor from this category to the category Ab of abelian groups. This functor is left exact but not necessarily right exact. We may therefore form its right derived functors; their values are abelian groups and they are denoted by Hn(G, M), "the n-th cohomology group of G with coefficients in M". H0(G, M) is identified with MG. ### Long exact sequence of cohomology In practice, one often computes the cohomology groups using the following fact: if $0 \to L \to M \to N \to 0$ is a short exact sequence of G-modules, then a long exact sequence $0\to L^G\to M^G\to N^G\overset{\delta^0}{\to} H^1(G,L) \to H^1(G,M) \to H^1(G,N)\overset{\delta^1}{\to} H^2(G,L)\to \cdots$ is induced. The maps δn are called the "connecting homomorphisms" and can be obtained from the snake lemma.[1] ### Cochain complexes Rather than using the machinery of derived functors, the cohomology groups can also be defined more concretely, as follows.[2] For n ≥ 0, let Cn(G, M) be the group of all functions from Gn to M. This is an abelian group; its elements are called the (inhomogeneous) n-cochains. The coboundary homomorphisms $d^n : C^n (G,M) \rightarrow C^{n+1}(G,M)$ are defined as $\left(d^n\varphi\right)(g_1,\dots,g_{n+1}) = g_1\cdot \varphi(g_2,\dots,g_{n+1})$ ${} + \sum_{i=1}^n (-1)^{i} \varphi(g_1,\dots,g_{i-1},g_i g_{i+1},g_{i+2},\dots,g_{n+1})$ ${} + (-1)^{n+1} \varphi(g_1,\dots,g_n)$ The crucial thing to check here is $d^{n+1} \circ d^n = 0$ thus we have a cochain complex and we can compute cohomology. For n ≥ 0, define the group of n-cocycles as: $Z^n(G,M) = \operatorname{ker}(d^n)$ and the group of n-coboundaries as $\begin{cases} B^0(G,M) = {0} \\ B^n(G,M)= \operatorname{im}(d^{n-1}), \ n \geq 1 \end{cases}$ and $H^n(G,M) = Z^n(G,M)/B^n(G,M).\$ ### The functors Extn and formal definition of group cohomology Yet another approach is to treat G-modules as modules over the group ring Z[G], which allows one to define group cohomology via Ext functors: $H^{n}(G,M) = \operatorname{Ext}^{n}_{\mathbf{Z}[G]}(\mathbf{Z},M),$ where M is a Z[G]-module. Here Z is treated as the trivial G-module: every element of G acts as the identity. These Ext groups can also be computed via a projective resolution of Z, the advantage being that such a resolution only depends on G and not on M. We recall the definition of Ext more explicitly for this context. Let F be a projective Z[G]-resolution (e.g. a free Z[G]-resolution) of the trivial Z[G]-module Z: $\dots \to F_n\to F_{n-1} \to\dots \to F_0\to \mathbf Z.$ e.g., one may always take the resolution of group rings, Fn = Z[Gn+1], with morphisms $f_n : \mathbf Z[G^{n+1}] \to \mathbf Z[G^n], \quad (g_0, g_1, \dots, g_n) \mapsto \sum_{i=0}^{n} (-1)^i(g_0, \dots, \widehat{g_i}, \dots, g_n).$ Recall that for Z[G]-modules N and M, HomG(N, M) is an abelian group consisting of Z[G]-homomorphisms from N to M. Since HomG(–, M) is a contravariant functor and reverses the arrows, applying HomG(–, M) to F termwise produces a cochain complex HomG(F, M): $\cdots \leftarrow \operatorname{Hom}_G(F_n,M)\leftarrow \operatorname{Hom}_G(F_{n-1},M)\leftarrow \dots \leftarrow \operatorname{Hom}_G(F_0,M)\leftarrow \operatorname{Hom}_G(\mathbf Z,M).$ The cohomology groups H(G, M) of G with coefficients in the module M are defined as the cohomology of the above cochain complex: $H^n(G,M)=H^n({\rm Hom}_{G}(F,M))$ for n ≥ 0. This construction initially leads to a coboundary operator that acts on the "homogeneous" cochains. These are the elements of HomG(F, M) i.e functions φn: Gn → M that obey $g\phi_n(g_1,g_2,\ldots, g_n)= \phi_n(gg_1,gg_2,\ldots, gg_n).$ The coboundary operator δ: Cn → Cn+1 is now naturally defined by, for example, $\delta \phi_2(g_1, g_2,g_3)= \phi_2(g_2,g_3)-\phi_2(g_1,g_3)+ \phi_2(g_1,g_2).$ The relation to the coboundary operator d that was defined in the previous section, and which acts on the "inhomogeneous" cochains φ, is given by reparameterizing so that $\begin{align} \varphi_2(g_1,g_2) &= \phi_3(1, g_1,g_1g_2) \\ \varphi_3(g_1,g_2,g_3) &= \phi_4(1, g_1,g_1g_2, g_1g_2g_3), \end{align}$ and so on. Thus $\begin{align} d \varphi_2(g_1,g_2,g_3) &= \delta \phi_3(1,g_1, g_1g_2,g_1g_2g_3)\\ & = \phi_3(g_1, g_1g_2,g_1g_2g_3) - \phi_3(1, g_1g_2, g_1g_2g_3) +\phi_3(1,g_1, g_1g_2g_3) - \phi_3(1,g_1,g_1g_2) \\ & = g_1\phi_3(1, g_2,g_2g_3) - \phi_3(1, g_1g_2, g_1g_2g_3) +\phi_3(1,g_1, g_1g_2g_3) - \phi_3(1,g_1,g_1g_2) \\ & = g_1\varphi_2(g_2,g_3) -\varphi_2(g_1g_2,g_3)+\varphi_2(g_1,g_2g_3) -\varphi_2(g_1,g_2), \end{align}$ as in the preceding section. ### Group homology Dually to the construction of group cohomology there is the following definition of group homology: given a G-module M, set DM to be the submodule generated by elements of the form g·m − m, g ∈ G, m ∈ M. Assigning to M its so-called coinvariants, the quotient $M_G:=M/DM, \,$ is a right exact functor. Its left derived functors are by definition the group homology $H_n\left(G,M\right)$. Note that the superscript/subscript convention for cohomology/homology agrees with the convention for group invariants/coinvariants, while which is denoted "co-" switches: • superscripts correspond to cohomology H and invariants XG while • subscripts correspond to homology H∗ and coinvariants XG := X/G. The covariant functor which assigns MG to M is isomorphic to the functor which sends M to Z ⊗Z[G] M, where Z is endowed with the trivial G-action. Hence one also gets an expression for group homology in terms of the Tor functors, $H_n(G,M) = \operatorname{Tor}_n^{\mathbf{Z}[G]}(\mathbf{Z},M)$ Recall that the tensor product N ⊗Z[G] M is defined whenever N is a right Z[G]-module and M is a left Z[G]-module. If N is a left Z[G]-module, we turn it into a right Z[G]-module by setting a g = g−1 a for every g ∈ G and every a ∈ N. This convention allows to define the tensor product N ⊗Z[G] M in the case where both M and N are left Z[G]-modules. Specifically, the homology groups Hn(G, M) can be computed as follows. Start with a projective resolution F of the trivial Z[G]-module Z, as in the previous section. Apply the covariant functor ⋅ ⊗Z[G] M to F termwise to get a chain complex F ⊗Z[G] M: $\dots \to F_n\otimes_{\mathbf{Z}[G]}M\to F_{n-1}\otimes_{\mathbf{Z}[G]}M \to\dots \to F_0\otimes_{\mathbf{Z}[G]}M\to \mathbf Z\otimes_{\mathbf{Z}[G]}M.$ Then Hn(G, M) are the homology groups of this chain complex, $H_n(G,M)=H_n(F\otimes_{\mathbf{Z}[G]}M)$ for n ≥ 0. Group homology and cohomology can be treated uniformly for some groups, especially finite groups, in terms of complete resolutions and the Tate cohomology groups. ## Functorial maps in terms of cochains ### Connecting homomorphisms For a short exact sequence 0 → L → M → N → 0, the connecting homomorphisms δn : Hn(G, N) → Hn+1(G, L) can be described in terms of inhomogeneous cochains as follows.[3] If c is an element of Hn(G, N) represented by an n-cocycle φ : Gn → N, then δn(c) is represented by dn(ψ), where ψ is an n-cochain Gn → M "lifting" φ (i.e. such that φ is the composition of ψ with the surjective map M → N). ## Non-abelian group cohomology Using the G-invariants and the 1-cochains, one can construct the zeroth and first group cohomology for a group G with coefficients in a non-abelian group. Specifically, a G-group is a (not necessarily abelian) group A together with an action by G. The zeroth cohomology of G with coefficients in A is defined to be the subgroup $H^{0}(G,A)=A^{G},\,$ of elements of A fixed by G. The first cohomology of G with coefficients in A is defined as 1-cocycles modulo an equivalence relation instead of by 1-coboundaries. The condition for a map φ to be a 1-cocycle is that φ(gh) = φ(g)[gφ(h)] and $\ \varphi\sim \varphi'$ if there is an a in A such that $\ a\varphi'(g)=\varphi(g)\cdot(ga)$. In general, H1(G, A) is not a group when A is non-abelian. It instead has the structure of a pointed set – exactly the same situation arises in the 0th homotopy group, $\ \pi_0(X;x)$ which for a general topological space is not a group but a pointed set. Note that a group is in particular a pointed set, with the identity element as distinguished point. Using explicit calculations, one still obtains a truncated long exact sequence in cohomology. Specifically, let $1\to A\to B\to C\to 1\,$ be a short exact sequence of G-groups, then there is an exact sequence of pointed sets $1\to A^G\to B^G\to C^G\to H^1(G,A) \to H^1(G,B) \to H^1(G,C).\,$ ## Connections with topological cohomology theories Group cohomology can be related to topological cohomology theories: to the topological group G there is an associated classifying space BG. (If G has no topology about which we care, then we assign the discrete topology to G. In this case, BG is an Eilenberg-MacLane space K(G,1), whose fundamental group is G and whose higher homotopy groups vanish). The n-th cohomology of BG, with coefficients in M (in the topological sense), is the same as the group cohomology of G with coefficients in M. This will involve a local coefficient system unless M is a trivial G-module. The connection holds because the total space EG is contractible, so its chain complex forms a projective resolution of M. These connections are explained in (Adem-Milgram 2004), Chapter II. When M is a ring with trivial G-action, we inherit good properties which are familiar from the topological context: in particular, there is a cup product under which $H^(G;M)=\bigoplus_n H^n(G;M)\,$ is a graded module, and a Künneth formula applies. If, furthermore, M = k is a field, then H(G; k) is a graded k-algebra. In this case, the Künneth formula yields $H^(G_1\times G_2;k)\cong H^(G_1;k)\otimes H^(G_2;k).\,$ For example, let G be the group with two elements, under the discrete topology. The real projective space P∞(R) is a classifying space for G. Let k = F2, the field of two elements. Then $H^(G;k)\cong k[x],\,$ a polynomial k-algebra on a single generator, since this is the cellular cohomology ring of P∞(R). Hence, as a second example, if G is an elementary abelian 2-group of rank r, and k = F2, then the Künneth formula gives $H^*(G;k)\cong k[x_1, \ldots, x_r]$, a polynomial k-algebra generated by r classes in H1(G; k). ## Properties In the following, let M be a G-module. ### Functoriality Group cohomology depends contravariantly on the group G, in the following sense: if f : H → G is a group homomorphism, then we have a naturally induced morphism Hn(G,M) → Hn(H,M) (where in the latter, M is treated as an H-module via f). Given a morphism of G-modules M→N, one gets a morphism of cohomology groups in the Hn(G,M) → Hn(G,N). ### H1 The first cohomology group is the quotient of the so-called crossed homomorphisms, i.e. maps (of sets) f : G → M satisfying f(ab) = f(a) + af(b) for all a, b in G, modulo the so-called principal crossed homomorphisms, i.e. maps f : G → M given by f(a) = am−m for some fixed m ∈ M. This follows from the definition of cochains above. If the action of G on M is trivial, then the above boils down to H1(G,M) = Hom(G, M), the group of group homomorphisms G → M. ### H2 If M is a trivial G-module (i.e. the action of G on M is trivial), the second cohomology group H2(G,M) is in one-to-one correspondence with the set of central extensions of G by M (up to a natural equivalence relation). More generally, if the action of G on M is nontrivial, H2(G,M) classifies the isomorphism classes of all extensions of G by M in which the induced action of G on M by inner automorphisms agrees with the given action. ### Change of group The Hochschild–Serre spectral sequence relates the cohomology of a normal subgroup N of G and the quotient G/N to the cohomology of the group G (for (pro-)finite groups G). ### Cohomology of finite groups is torsion The cohomology groups of finite groups are all torsion. Indeed, by Maschke's theorem the category of representations of a finite group is semi-simple over any field of characteristic zero (or more generally, any field whose characteristic does not divide the order of the group), hence, viewing group cohomology as a derived functor in this abelian category, one obtains that it is zero. The other argument is that over a field of characteristic zero, the group algebra of a finite group is a direct sum of matrix algebras (possibly over division algebras which are extensions of the original field), while a matrix algebra is Morita equivalent to its base field and hence has trivial cohomology. ## History and relation to other fields The low dimensional cohomology of a group was classically studied in other guises, long before the notion of group cohomology was formulated in 1943–45. The first theorem of the subject can be identified as Hilbert's Theorem 90 in 1897; this was recast into Noether's equations in Galois theory (an appearance of cocycles for H1). The idea of factor sets for the extension problem for groups (connected with H2) arose in the work of Hölder (1893), in Issai Schur's 1904 study of projective representations, in Schreier's 1926 treatment, and in Richard Brauer's 1928 study of simple algebras and the Brauer group. A fuller discussion of this history may be found in (Weibel 1999, pp. 806–811). In 1941, while studying H2(G, Z) (which plays a special role in groups), Hopf discovered what is now called Hopf's integral homology formula (Hopf 1942), which is identical to Schur's formula for the Schur multiplier of a finite, finitely presented group: $H_2(G,\mathbf{Z}) \cong (R \cap [F, F])/[F, R]$, where G ≅ F/R and F is a free group. Hopf's result led to the independent discovery of group cohomology by several groups in 1943-45: Eilenberg and Mac Lane in the USA (Rotman 1995, p. 358); Hopf and Eckmann in Switzerland; and Freudenthal in the Netherlands (Weibel 1999, p. 807). The situation was chaotic because communication between these countries was difficult during World War II. From a topological point of view, the homology and cohomology of G was first defined as the homology and cohomology of a model for the topological classifying space BG as discussed in #Connections with topological cohomology theories above. In practice, this meant using topology to produce the chain complexes used in formal algebraic definitions. From a module-theoretic point of view this was integrated into the Cartan–Eilenberg theory of homological algebra in the early 1950s. The application in algebraic number theory to class field theory provided theorems valid for general Galois extensions (not just abelian extensions). The cohomological part of class field theory was axiomatized as the theory of class formations. In turn, this led to the notion of Galois cohomology and étale cohomology (which builds on it) (Weibel 1999, p. 822). Some refinements in the theory post-1960 have been made, such as continuous cocycles and Tate's redefinition, but the basic outlines remain the same. This is a large field, and now basic in the theories of algebraic groups. The analogous theory for Lie algebras, called Lie algebra cohomology, was first developed in the late 1940s, by Chevalley–Eilenberg, and Koszul (Weibel 1999, p. 810). It is formally similar, using the corresponding definition of invariant for the action of a Lie algebra. It is much applied in representation theory, and is closely connected with the BRST quantization of theoretical physics. Group cohomology theory also has a direct application in condensed matter physics. Just like group theory being the mathematical foundation of spontaneous symmetry breaking phases, group cohomology theory is the mathematical foundation of a class of quantum states of matter -- short-range entangled states with symmetry. Short-range entangled states with symmetry are also known as Symmetry protected topological states. ## Notes 1. Page 62 of Milne 2008 or section VII.3 of Serre 1979 ## References • Adem, Alejandro; Milgram, R. James (2004), Cohomology of Finite Groups, Grundlehren der Mathematischen Wissenschaften 309 (2nd ed.), Springer-Verlag, ISBN 3-540-20283-8, MR 2035696, Zbl 1061.20044 • Brown, Kenneth S. (1972), Cohomology of Groups, Graduate Texts in Mathematics 87, Springer Verlag, ISBN 0-387-90688-6, MR 0672956 • Hopf, Heinz (1942), "Fundamentalgruppe und zweite Bettische Gruppe", Comment. Math. Helv. 14 (1): 257–309, doi:10.1007/BF02565622, JFM 68.0503.01, MR 6510, Zbl 0027.09503 • Chapter II of Milne, James (5/2/2008), Class Field Theory, v4.00, retrieved 8/9/2008 • Rotman, Joseph (1995), An Introduction to the Theory of Groups, Graduate Texts in Mathematics 148 (4th ed.), Springer-Verlag, ISBN 978-0-387-94285-8, MR 1307623 • Chapter VII of Serre, Jean-Pierre (1979), Local fields, Graduate Texts in Mathematics 67, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90424-5, MR 554237, Zbl 0423.12016 • Serre, Jean-Pierre (1994), Cohomologie galoisienne, Lecture Notes in Mathematics 5 (Fifth ed.), Berlin, New York: Springer-Verlag, ISBN 978-3-540-58002-7, MR 1324577 • Shatz, Stephen S. (1972), Profinite groups, arithmetic, and geometry, Princeton, NJ: Princeton University Press, ISBN 978-0-691-08017-8, MR 0347778 • Chapter 6 of Weibel, Charles A. (1994), An introduction to homological algebra, Cambridge Studies in Advanced Mathematics 38, Cambridge University Press, ISBN 978-0-521-55987-4, OCLC 36131259, MR1269324 • Weibel, Charles A. (1999), "History of homological algebra", History of Topology, Cambridge University Press, pp. 797–836, ISBN 0-444-82375-1, MR 1721123	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 36, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8875711560249329, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/schrodinger-equation+hamiltonian	# Tagged Questions 1answer 128 views ### How does a state in quantum mechanics evolve? I have a question about the time evolution of a state in quantum mechanics. The time-dependent Schrodinger equation is given as $$i\hbar\frac{d}{dt}\|\psi(t)\rangle = H\|\psi(t)\rangle$$ I am ... 2answers 402 views ### Two expressions for expectation value of energy I was looking up expectation value of energy for a free particle on the following webpage: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/expect.html It says that $E=\frac{p^2}{2m}$ and ... 1answer 303 views ### Solving time dependent Schrodinger equation in matrix form If we have a Hilbert space of $\mathbb{C}^3$ so that a wave function is a 3-component column vector $$\psi_t=(\psi_1(t),\psi_2(t),\psi_3(t))$$ With Hamiltonian $H$ given by H=\hbar\omega ... 1answer 83 views ### Where can I find hamiltonians + lagrangians? Where would you say I can start learning about Hamiltonians, Lagrangians ... Jacobians? and the like? I was trying to read Ibach and Luth - Solid State Physics, and suddenly (suddenly a Hamiltonian ... 1answer 265 views ### The Hermiticity of the Laplacian (and other operators) Is the Laplacian operator, $\nabla^{2}$, a Hermitian operator? Alternatively: is the matrix representation of the Laplacian Hermitian? i.e. \langle \nabla^{2} x \| y \rangle = \langle x \| ... 3answers 463 views ### What is the relationship between Schrödinger equation and Boltzmann equation? The Schrödinger equation in its variants for many particle systems gives the full time evolution of the system. Likewise, the Boltzmann equation is often the starting point in classical gas dynamics. ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9018546938896179, "perplexity_flag": "middle"}
http://www.oilfieldwiki.com/wiki/Continuity_equation	# Continuity equation From Oilfield Wiki - the Oilfield Encyclopedia A continuity equation in physics is an equation that describes the transport of a conserved quantity. Since mass, energy, momentum, electric charge and other natural quantities are conserved under their respective appropriate conditions, a variety of physical phenomena may be described using continuity equations. Continuity equations are the (stronger) local form of conservation laws. All the examples of continuity equations below express the same idea, which is roughly that: the total amount (of the conserved quantity) inside any region can only change by the amount that passes in or out of the region through the boundary. A conserved quantity cannot increase or decrease, it can only move from place to place. Any continuity equation can be expressed in an "integral form" (in terms of a flux integral), which applies to any finite region, or in a "differential form" (in terms of the divergence operator) which applies at a point. Continuity equations underlie more specific transport equations such as the convection–diffusion equation, Boltzmann transport equation, and Navier-Stokes equations. ## General equation ### Preliminary description Illustration of how flux f passes through open surfaces S (vector S), flat or curved. Illustration of how flux f passes through closed surfaces S1 and S2. The surface area elements shown are dS1 and dS2, and the flux is integrated over the whole surface. Yellow dots are sources, red dots are sinks, the blue lines are the flux lines of q. As stated above, the idea behind the continuity equation is the flow of some property, such as mass, energy, electric charge, momentum, and even probability, through surfaces from one region of space to another. The surfaces, in general, may either be open or closed, real or imaginary, and have an arbitrary shape, but are fixed for the calculation (i.e. not time-varying, which is appropriate since this complicates the maths for no advantage). Let this property be represented by just one scalar variable, q, and let the volume density of this property (the amount of q per unit volume V) be φ, and the all surfaces be denoted by S. Mathematically, φ is a ratio of two infinitesimal quantities: $\varphi = \frac{{\rm d} q}{{\rm d} V},$ which has the dimension [quantity][L]-3 (where L is length). There are different ways to conceive the continuity equation: 1. either the flow of particles carrying the quantity q, described by a velocity field v, which is also equivalent to a flux f of q (a vector function describing the flow per unit area per unit time of q), or 2. in the cases where a velocity field is not useful or applicable, the flux f of the quantity q only (no association with velocity). In each of these cases, the transfer of q occurs as it passes through two surfaces, the first S1 and the second S2. Illustration of q, φ, and f, and the effective flux due to carriers of q. φ is the amount of q per unit volume (in the box), f represents the flux (blue flux lines) and q is carried by the particles (yellow). The flux f should represent some flow or transport, which has dimensions [quantity][T]−1[L]−2. In cases where particles/carriers of quantity q are moving with velocity v, such as particles of mass in a fluid or charge carriers in a conductor, f can be related to v by: $\mathbf{f} = \varphi \mathbf{v} .$ This relation is only true in situations where there are particles moving and carrying q - it can't always be applied. To illustrate this: if f is electric current density (electric current per unit area) and φ is the charge density (charge per unit volume), then the velocity of the charge carriers is v. However - if f is heat flux density (heat energy per unit time per unit area), then even if we let φ be the heat energy density (heat energy per unit volume) it does not imply the "velocity of heat" is v (this makes no sense, and is not practically applicable). In the latter case only f (with φ) may be used in the continuity equation. ### Elementary vector form Consider the case when the surfaces are flat and planar cross-sections. For the case where a velocity field can be applied, dimensional analysis leads to this form of the continuity equation: $\varphi_1 \bold{v}_1 \cdot \bold{S}_1 = \varphi_2 \bold{v}_2 \cdot \bold{S}_2$ where • the left hand side is the initial amount of q flowing per unit time through surface S1, the right hand side is the final amount through surface S2, • S1 and S2 are the vector areas for the surfaces S1 and S2 respectively. Notice the dot products $$\bold{v}_1 \cdot \bold{S}_1, \, \bold{v}_2 \cdot \bold{S}_2 \,\!$$ are volumetric flow rates of q. The dimension of each side of the equation is [quantity][L]-3•[L][T]-1•[L]2 = [quantity][T]-1. For the more general cases, independent of whether a velocity field can be used or not, the continuity equation becomes: $\bold{f}_1 \cdot \bold{S}_1 = \bold{f}_2 \cdot \bold{S}_2$ This has exactly the same dimensions as the previous version. The relation between f and v allows us to pass back to the velocity version from this flux equation, but not always the other way round (as explained above - velocity fields are not always applicable). These results can be generalized further to curved surfaces by reducing the vector surfaces into infinitely many differential surface elements (that is S → dS), then integrating over the surface: $\int\!\!\!\!\int_{S_1} \varphi_1\bold{v}_1 \cdot {\rm d}\bold{S}_1 = \int\!\!\!\!\int_{S_2} \varphi_2\bold{v}_2 \cdot {\rm d}\bold{S}_2$ more generally still: $$\int\!\!\!\!\int_{S_1} \bold{f}_1 \cdot {\rm d}\bold{S}_1 = \int\!\!\!\!\int_{S_2} \bold{f}_2 \cdot {\rm d}\bold{S}_2$$ in which • $$\int\!\!\!\!\int_S {\rm d}\bold{S} \equiv \int\!\!\!\!\int_S \bold{\hat{n}}{\rm d}S$$ denotes a surface integral over the surface S, • $$\mathbf{\hat{n}}$$ is the outward-pointing unit normal to the surface S N.B: the scalar area S and vector area S are related by $$\mathrm{d}\mathbf{S} = \mathbf{\hat{n}}\mathrm{d}S$$. Either notations may be used interchangeably. ### Differential form The differential form for a general continuity equation is (using the same q, φ and f as above): $$\frac{\partial \varphi}{\partial t} + \nabla \cdot \mathbf{f} = \sigma\,$$ where • ∇• is divergence, • t is time, • σ is the generation of q per unit volume per unit time. Terms that generate (σ > 0) or remove (σ < 0) q are referred to as a "sources" and "sinks" respectively. This general equation may be used to derive any continuity equation, ranging from as simple as the volume continuity equation to as complicated as the Navier–Stokes equations. This equation also generalizes the advection equation. Other equations in physics, such as Gauss's law of the electric field and Gauss's law for gravity, have a similar mathematical form to the continuity equation, but are not usually called by the term "continuity equation", because f in those cases does not represent the flow of a real physical quantity. In the case that q is a conserved quantity that cannot be created or destroyed (such as energy), this translates to σ = 0, and the continuity equation is: $\frac{\partial \varphi}{\partial t} + \nabla \cdot \mathbf{f} = 0\,$ ### Integral form In the integral form of the continuity equation, S is any imaginary closed surface that fully encloses a volume V, like any of the surfaces on the left. S can not be a surface with boundaries that do not enclose a volume, like those on the right. (Surfaces are blue, boundaries are red.) By the divergence theorem (see below), the continuity equation can be rewritten in an equivalent way, called the "integral form": {{Equation box 1 \|indent=: \|equation=$$\frac{{\rm d} q}{{\rm d} t} +$$ \| intsubscpt = $$\scriptstyle S$$ \| integrand = $$\bold{f} \cdot {\rm d}\bold{S} = \Sigma$$ }} \|cellpadding \|border \|border colour = #0073CF \|background colour=#F5FFFA}} where • S is a surface as described above - except this time it has to be a closed surface that encloses a volume V, • $$\scriptstyle S$$$${\rm d}\bold{S}$$ denotes a surface integral over a closed surface, • $$\int\!\!\!\int\!\!\!\int_V \, {\rm d}V$$ denotes a volume integral over V. • $$q = \int\!\!\!\!\int\!\!\!\!\int_V \varphi \, \mathrm{d}V$$ is the total amount of φ in the volume V; • $$\Sigma = \int\!\!\!\!\int\!\!\!\!\int_V \sigma \, \mathrm{d}V$$ is the total generation (negative in the case of removal) per unit time by the sources and sinks in the volume V, In a simple example, V could be a building, and q could be the number of people in the building. The surface S would consist of the walls, doors, roof, and foundation of the building. Then the continuity equation states that the number of people in the building increases when people enter the building (an inward flux through the surface), decreases when people exit the building (an outward flux through the surface), increases when someone in the building gives birth (a "source" where σ > 0), and decreases when someone in the building dies (a "sink" where σ < 0). ### Derivation and equivalence The differential form can be derived from first principles as follows. #### Derivation of the differential form Suppose first an amount of quantity q is contained in a region of volume V, bounded by a closed surface S, as described above. This is equal to the amount already in V, plus the generated amount s (total - not per unit time or volume): $q(t) = \int\!\!\!\!\int\!\!\!\!\int_V \varphi(\mathbf{r},t) \mathrm{d}V + \int^t\Sigma(t')\mathrm{d}t'.$ (t' is just a dummy variable of the generation integral). The rate of change of q leaving the region is simply the time derivative: $\frac{\partial q(t)}{\partial t} = -\frac{\partial }{\partial t} \int\!\!\!\!\int\!\!\!\!\int_V \varphi(\mathbf{r},t) \mathrm{d}V + \Sigma(t)$ where the minus sign has been inserted since the amount of q is decreasing in the region. (Partial derivatives are used since they enter the integrand, which is not only a function of time, but also space due to the density nature of φ - differentiation needs only to be with respect to t). The rate of change of q crossing the boundary and leaving the region is: $$\frac{\partial q(t)}{\partial t} =$$$$\scriptstyle S$$$$\mathbf{f}(\mathbf{r},t)\cdot\mathrm{d}\bold{S},$$ so equating these expressions: $$\scriptstyle S$$$$\mathbf{f}(\mathbf{r},t)\cdot\mathrm{d}\bold{S}=- \int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial \varphi(\mathbf{r},t)}{\partial t} \mathrm{d}V + \Sigma(t),$$ Using the divergence theorem on the left-hand side: $\int\!\!\!\!\int\!\!\!\!\int_V \nabla\cdot\mathbf{f}(\mathbf{r},t) \mathrm{d}V = - \int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial \varphi(\mathbf{r},t)}{\partial t} \mathrm{d}V + \int\!\!\!\!\int\!\!\!\!\int_V \sigma(\mathbf{r},t) \mathrm{d}V.$ This is only true if the integrands are equal, which directly leads to the differential continuity equation: $\begin{align} & \nabla\cdot\mathbf{f}(\mathbf{r},t) = - \frac{\partial \varphi(\mathbf{r},t)}{\partial t} + \sigma(\mathbf{r},t), \\ & \nabla\cdot\mathbf{f} + \frac{\partial \varphi}{\partial t} = \sigma \rightleftharpoons \nabla\cdot(\varphi \mathbf{v}) + \frac{\partial \varphi}{\partial t} = \sigma.\\ \end{align}$ Either form may be useful and quoted, both can appear in hydrodynamics and electromagnetism, but for quantum mechanics and energy conservation, only the first may be used. Therefore the first is more general. #### Equivalence between differential and integral form Starting from the differential form which is for unit volume, multiplying throughout by the infinitesimal volume element dV and integrating over the region gives the total amounts quantities in the volume of the region (per unit time): $\begin{align} & \int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial \varphi(\mathbf{r},t)}{\partial t} \mathrm{d}V + \int\!\!\!\!\int\!\!\!\!\int_V \nabla \cdot \mathbf{f}(\mathbf{r},t) \mathrm{d}V = \int\!\!\!\!\int\!\!\!\!\int_V \sigma(\mathbf{r},t) \mathrm{d}V \\ & \frac{\mathrm{d[[File:OiintLaTeX.png\|x45px\|alt=\oiint]]{\mathrm{d} t} \int\!\!\!\!\int\!\!\!\!\int_V \varphi(\mathbf{r},t)\mathrm{d}V + \int\!\!\!\!\int\!\!\!\!\int_V \nabla \cdot \mathbf{f}(\mathbf{r},t) \mathrm{d}V = \Sigma(t) \end{align} \,\!$ again using the fact that V is constant in shape for the calculation, so it is independent of time and the time derivatives can be freely moved out of that integral, ordinary derivatives replace partial derivatives since the integral becomes a function of time only (the integral is evaluated over the region - so the spatial variables become removed from the final expression and t remains the only variable). Using the divergence theorem on the left side $$\frac{\mathrm{d} q(t)}{\mathrm{d} t} +$$$$\scriptstyle S$$$$\mathbf{f}(\mathbf{r},t)\cdot{\rm d}\mathbf{S} = \Sigma(t)$$ which is the integral form. #### Equivalence between elementary and integral form Starting from $\int\!\!\!\!\int_{S_1} \bold{f}_1(\mathbf{r},t) \cdot {\rm d}\bold{S}_1 = \int\!\!\!\!\int_{S_2} \bold{f}_2(\mathbf{r},t) \cdot {\rm d}\bold{S}_2$ the surfaces are equal (since there is only one closed surface), so S1 = S2 = S and we can write: $\int\!\!\!\!\int_{S} \bold{f}_1(\mathbf{r},t) \cdot {\rm d}\bold{S} = \int\!\!\!\!\int_{S} \bold{f}_2(\mathbf{r},t) \cdot {\rm d}\bold{S}$ The left hand side is the flow rate of quantity q occurring inside the closed surface S. This must be equal to $\int\!\!\!\!\int_{S} \bold{f}_1(\mathbf{r},t) \cdot {\rm d}\bold{S} = \Sigma(t) - \frac{{\rm d}q(t)}{{\rm d}t}$ since some is produced by sources, hence the positive term Σ, but some is also leaking out by passing through the surface, implied by the negative term -dq/dt. Similarly the right hand side is the amount of flux passing through the surface and out of it, so $$\int\!\!\!\!\int_S \bold{f}_2(\mathbf{r},t) \cdot {\rm d}\bold{S} =$$$$\scriptstyle S$$$$\bold{f}(\mathbf{r},t) \cdot {\rm d}\bold{S}$$ Equating these: $$\Sigma(t) - \frac{{\rm d}q(t)}{{\rm d}t} =$$ \| intsubscpt = $$\scriptstyle S$$ \| integrand=$$\bold{f}(\mathbf{r},t) \cdot {\rm d}\bold{S}$$}} $$\frac{{\rm d}q}{{\rm d}t}+$$ \| intsubscpt = $$\scriptstyle S$$ \| integrand=$$\bold{f} \cdot {\rm d}\bold{S} =\Sigma$$}} which is the integral form again. ## Electromagnetism Main article: Charge conservation ### 3-currents In electromagnetic theory, the continuity equation can either be regarded as an empirical law expressing (local) charge conservation, or can be derived as a consequence of two of Maxwell's equations. It states that the divergence of the current density J (in amperes per square meter) is equal to the negative rate of change of the charge density ρ (in coulombs per cubic metre), $\nabla \cdot \mathbf{J} = - {\partial \rho \over \partial t}$ Maxwell's equations are a quick way to obtain the continuity of charge. Consistency with Maxwell's equations One of Maxwell's equations, Ampère's law (with Maxwell's correction), states that $\nabla \times \mathbf{H} = \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}.$ Taking the divergence of both sides results in $\nabla \cdot ( \nabla \times \mathbf{H} ) = \nabla \cdot \mathbf{J} + \frac{\partial (\nabla \cdot \mathbf{D})}{\partial t},$ but the divergence of a curl is zero, so that $\nabla \cdot \mathbf{J} + \frac{\partial (\nabla \cdot \mathbf{D})}{\partial t} = 0.$ Another one of Maxwell's equations, Gauss's law, states that $\nabla \cdot \mathbf{D} = \rho,\,$ substitution into the previous equation yields the continuity equation $\nabla \cdot \mathbf{J} + {\partial \rho \over \partial t} = 0.\,$ ### 4-currents Conservation of a current (not necessarily an electromagnetic current) is expressed compactly as the Lorentz invariant divergence of a four-current: $J^\mu = \left(c \rho, \mathbf{j} \right)$ where • c is the speed of light • ρ; the charge density • j the conventional 3-current density. • μ; labels the space-time dimension since $\partial_\mu J^\mu = \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j}$ then $\partial_\mu J^\mu = 0$ which implies that the current is conserved: $\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0.$ ### Interpretation Current is the movement of charge. The continuity equation says that if charge is moving out of a differential volume (i.e. divergence of current density is positive) then the amount of charge within that volume is going to decrease, so the rate of change of charge density is negative. Therefore the continuity equation amounts to a conservation of charge. ## Fluid dynamics In fluid dynamics, the continuity equation states that, in any steady state process, the rate at which mass enters a system is equal to the rate at which mass leaves the system.[1][2] The differential form of the continuity equation is:[1] ${\partial \rho \over \partial t} + \nabla \cdot (\rho \mathbf{u}) = 0$ where • ρ is fluid density, • t is time, • u is the flow velocity vector field. If ρ is a constant, as in the case of incompressible flow, the mass continuity equation simplifies to a volume continuity equation:[1] $\nabla \cdot \mathbf{u} = 0,$ which means that the divergence of velocity field is zero everywhere. Physically, this is equivalent to saying that the local volume dilation rate is zero. Further, the Navier-Stokes equations form a vector continuity equation describing the conservation of linear momentum. ## Energy Conservation of energy (which, in non-relativistic situations, can only be transferred, and not created or destroyed) leads to a continuity equation, an alternative mathematical statement of energy conservation to the thermodynamic laws. Letting • u = local energy density (energy per unit volume), • q = energy flux (transfer of energy per unit cross-sectional area per unit time) as a vector, the continuity equation is: $\nabla \cdot \mathbf{q} + \frac{ \partial u}{\partial t} = 0$ ## Quantum mechanics In quantum mechanics, the conservation of probability also yields a continuity equation. The terms in the equation require these definitions, and are slightly less obvious than the other forms of volume densities, currents, current densities etc., so they are outlined here: • The wavefunction Ψ for a single particle in the position-time space (rather than momentum space) - i.e. functions of position r and time t, Ψ = Ψ(r, t) = Ψ(x, y, z, t). • The probability density function ρ = ρ(r, t) is: $\rho = \Psi^{} \Psi \,\!$ • The probability that a measurement of the particle's position will yield a value within V at t, denoted by P = Pr ∈ V(t), is: $P=P_{\mathbf{r} \in V}(t) = \int_V \Psi^{} \Psi \mathrm{d} V = \int_V[[File:OiintLaTeX.png\|x45px\|alt=\oiint]]{\partial t} ,\\ \end{align}$ where U is the potential function. The partial derivative of ρ with respect to t is: $\frac{\partial \rho}{\partial t} = \frac{\partial \|\Psi \|^2}{\partial t } = \frac{\partial}{\partial t} \left ( \Psi^{} \Psi \right ) = \Psi^{} \frac{\partial \Psi}{\partial t} + \Psi \frac{\partial\Psi^{}}{\partial t} .$ Multiplying the Schrödinger equation by Ψ then solving for $$\scriptstyle \Psi^{} \partial \Psi/\partial t \,\!$$, and similarly multiplying the complex conjugated Schrödinger equation by Ψ then solving for $$\Psi \partial \Psi^ / \partial t \,\!$$; $\begin{align} & \Psi^\frac{\partial \Psi}{\partial t} = \frac{1}{i\hbar } \left [ -\frac{\hbar^2\Psi^}{2m}\nabla^2 \Psi + U\Psi^\Psi \right ], \\ & \Psi \frac{\partial \Psi^}{\partial t} = - \frac{1}{i\hbar } \left [ - \frac{\hbar^2\Psi}{2m}\nabla^2 \Psi^* + U\Psi\Psi^* \right ],\\ \end{align}$ substituting into the time derivative of ρ: $\begin{align} \frac{\partial \rho}{\partial t} & = \frac{1}{i\hbar } \left [ -\frac{\hbar^2\Psi^{[[File:OiintLaTeX.png\|x45px\|alt=\oiint]]{2m}\nabla^2 \Psi + U\Psi^{}\Psi \right ] - \frac{1}{i\hbar } \left [ - \frac{\hbar^2\Psi}{2m}\nabla^2 \Psi^{} + U\Psi\Psi^{} \right ] \\ & = \frac{\hbar}{2im} \left [ \Psi\nabla^2 \Psi^{} - \Psi^{}\nabla^2 \Psi \right ] \\ \end{align}$ The Laplacian operators (∇2) in the above result suggest that the right hand side is the divergence of j, and the reversed order of terms imply this is the negative of j, altogether: $\begin{align} \nabla \cdot \mathbf{j} & = \nabla \cdot \left [ \frac{\hbar}{2mi} \left ( \Psi^{} \left ( \nabla \Psi \right ) - \Psi \left ( \nabla \Psi^{} \right ) \right ) \right ] \\ & = \frac{\hbar}{2mi} \left [ \Psi^{} \left ( \nabla^2 \Psi \right ) - \Psi \left ( \nabla^2 \Psi^{} \right ) \right ] \\ & = - \frac{\hbar}{2mi} \left [ \Psi \left ( \nabla^2 \Psi^{} \right ) - \Psi^{} \left ( \nabla^2 \Psi \right ) \right ] \\ \end{align}$ so the continuity equation is: $\begin{align} & \frac{\partial \rho}{\partial t} = - \nabla \cdot \mathbf{j} \\ & \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{j} = 0 \\ \end{align}$ The integral form follows as for the general equation. </div> </div> Consistency with the wavefunction probability distribution The time derivative of P is $\frac{\mathrm{d}P}{\mathrm{d}t} = \frac{\partial}{\partial t} \int\!\!\!\!\int\!\!\!\!\int_V \|\Psi\|^2 \mathrm{d}V = \int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial \|\Psi\|^2}{\partial t} \mathrm{d}V = \int\!\!\!\!\int\!\!\!\!\int_V \left( \frac{\partial \Psi}{\partial t}\Psi^* + \Psi \frac{\partial \Psi^}{\partial t} \right ) \mathrm{d}V$ where the last equality follows from the product rule and the fact that the shape of V is fixed for the calculation and therefore independent of time - i.e. the time derivative can be moved through the integral. To simplify this further consider again the time dependent Schrödinger equation and its complex conjugate, in terms of the time derivatives of Ψ and Ψ respectively: $\begin{align} & \frac{\partial \Psi}{\partial t} = \frac{i \hbar}{2m} \nabla^2 \Psi - \frac{i}{\hbar} U \Psi \\ & \frac{\partial \Psi^}{\partial t} = - \left [ \frac{i \hbar}{2m} \nabla^2 \Psi^ - \frac{i}{\hbar} U \Psi^* \right ] \\ \end{align}\,\!$ Substituting into the preceding equation: $\int_V \frac{\partial \|\Psi\|^2}{\partial t} \mathrm{d}V = - \int\!\!\!\!\int\!\!\!\!\int_V \frac{\hbar}{2mi} \left(\Psi^* \nabla^2 \Psi - \Psi \nabla^2 \Psi^* \right) \mathrm{d}V$. From the product rule for the divergence operator $\begin{align} \nabla \cdot \left(\Psi^* \nabla \Psi - \Psi \nabla \Psi^* \right) & = \nabla \Psi^* \cdot \nabla \Psi + \Psi^* \nabla^2 \Psi - \nabla \Psi \cdot \nabla \Psi^* - \Psi \nabla^2 \Psi^* \\ & = \Psi^* \nabla^2 \Psi - \Psi \nabla^2 \Psi^* \\ \end{align}$ substituting: $\int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial \|\Psi\|^2}{\partial t} \mathrm{d}V = - \int\!\!\!\!\int\!\!\!\!\int_V \nabla \cdot \left [ \frac{\hbar}{2mi} \left(\Psi^* \nabla \Psi - \Psi \nabla \Psi^* \right ) \right ] \mathrm{d}V$ On the right side, the argument of the divergence operator is j, $\int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial \|\Psi\|^2}{\partial t} \mathrm{d}V = - \int\!\!\!\!\int\!\!\!\!\int_V \left( \nabla \cdot \mathbf{j} \right) \mathrm{d}V \,\!$ using the divergence theorem again gives the integral form: $$\int\!\!\!\!\int\!\!\!\!\int_V \frac{\partial[[File:OiintLaTeX.png\|x45px\|alt=\oiint]]\(\scriptstyle S$$$$(\mathbf{j}\cdot\mathbf{\hat{n)\mathrm{d}S$$}} To obtain the differential form: $\int\!\!\!\!\int\!\!\!\!\int_V \left( \frac{\partial \|\Psi\|^2}{\partial t} + \nabla \cdot \mathbf{j} \right) \mathrm{d}V = 0$ The differential form follows from the fact that the preceding equation holds for all V, and as the integrand is a continuous function of space, it must vanish everywhere: $\frac{\partial \|\Psi\|^2}{\partial t} + \nabla \cdot \mathbf{j} = 0.$ ## References 1. ↑ {{#invoke:citation/CS1\|citation \|CitationClass=book }} 2. Clancy, L.J.(1975), Aerodynamics, Section 3.3, Pitman Publishing Limited, London ## Further reading • Hydrodynamics, H. Lamb, Cambridge Univresity Press, (2006 digitalization of 1932 6th edition) ISBN 978-0-521-45868-9 • Introduction to Electrodynamics (3rd Edition), D.J. Griffiths, Pearson Education Inc, 1999, ISBN 81-7758-293-3 • Electromagnetism (2nd edition), I.S. Grant, W.R. Phillips, Manchester Physics Series, 2008 ISBN 0-471-92712-0 • Gravitation, J.A. Wheeler, C. Misner, K.S. Thorne, W.H. Freeman & Co, 1973, ISBN 0-7167-0344-0 ar:معادلة الاستمرارية ca:Equació de continuïtat cs:Rovnice kontinuity de:Kontinuitätsgleichung es:Ecuación de continuidad fr:Équation de conservation ko:연속 방정식 it:Equazione di continuità he:משוואת רציפות ka:უწყვეტობის განტოლება hu:Kontinuitási egyenlet ms:Persamaan keselanjaran nl:Continuïteitsvergelijking ja:連続の方程式 nn:Kontinuitetslikninga pl:Równanie ciągłości pt:Equação de continuidade ro:Ecuația de continuitate ru:Уравнение непрерывности sk:Rovnica kontinuity sl:Kontinuitetna enačba fi:Jatkuvuusyhtälö tr:Süreklilik denklemi uk:Рівняння неперервності zh:連續性方程式	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 42, "mathjax_display_tex": 35, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8877493143081665, "perplexity_flag": "middle"}
http://nrich.maths.org/5429/note	### Pebbles Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time? ### It Figures Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all? ### Bracelets Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads? # Multiples Grid ## Multiples Grid Here is a $100$ grid with some numbers shaded: What do all the numbers shaded blue have in common? What do you notice about all the numbers shaded pink? Can you work out why two of the numbers are shaded in a maroon colour? Now, here is part of a $100$ square shaded in a different way: Can you explain the shading this time? Here are some more parts of the $100$ square, each one shaded according to different rules. Can you work out what the rules are for each? Is there only one solution each time? This problem is featured in Maths Trails - Excel in Problem Solving, one of the books in the Maths Trails series written by members of the NRICH Team and published by Cambridge University Press. For more details about the other books in the series, please see our publications page . ### Why do this problem? This problem is an interesting way of reinforcing understanding of factors and multiples. ### Possible approach To start with, ask children to talk in pairs about why the numbers in the first 100 square are shaded blue, pink and maroon. Invite them to share their ideas and encourage correct use of vocabulary. Learners could continue to work in pairs, perhaps using this sheet of the four parts of differently-shaded 100 squares. As they work on the problem, trying to find out which factors have been chosen in order to produce the shading, encourage them to justify their solutions to their partners, and perhaps then to the whole class. How are they going about the task? It might be useful to discuss ways of working systematically so that no solutions are omitted. This spreadsheet ,which shades the squares according to the chosen factors, can be used to check their hypotheses. In a plenary session, you could use the second sheet of the spreadsheet to pre-prepare some shaded sections of the 100 grid without numbers. If you tell them which multiples have been shaded, can the class work out where the small part of the 100 grid is, i.e. which numbers it contains? ### Key questions What do the numbers shaded blue have in common? What do the pink numbers have in common? Can you rule out some factors straight away? How? How will you know you have found all the possible solutions? ### Possible extension Learners could explore the spreadsheet for themselves at a computer. Challenge them to make up their own questions to ask a friend. ### Possible support A multiplication square may be useful for those children who find instant recall of multiplication facts difficult. 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.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9387180209159851, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/2859-integers-print.html	# Integers Printable View • May 7th 2006, 09:50 AM mathlg Integers Find an integer n > 20 so that the equation 3x = b always have a unigue solution in Zn regardless of the value b Not to sure what Zn means Can the value of b be anything? • May 7th 2006, 10:07 AM ThePerfectHacker Quote: Originally Posted by mathlg Find an integer n > 20 so that the equation 3x = b always have a unigue solution in Zn regardless of the value b Not to sure what Zn means Can the value of b be anything? By, $\mathbb{Z}_n$ we mean the group of positive integers added modulo $n$. I believe the theorem goes that, $ax=b$ has a solution in $\mathbb{Z}_n$ when, $\gcd(b,n)=1$. For your problem you need to find all the integers relatively prime to $20$ and less. This is called the 'phi-function'. Thus, you need to find the smalles $n>20$ such as all the previous integers are relatively prime to it. Meaning a prime number, the smallest after 20 is 23. • May 7th 2006, 10:20 AM mathlg So you are saying that since n > 20 the next smallest prime number is 23. That would be the integer im looking for in the problem? I get confused when it says regardless of the value b. • May 7th 2006, 01:13 PM ThePerfectHacker Quote: Originally Posted by mathlg So you are saying that since n > 20 the next smallest prime number is 23. That would be the integer im looking for in the problem? I get confused when it says regardless of the value b. Yes All times are GMT -8. The time now is 01:56 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159101843833923, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/43336/blowups-of-cohen-macaulay-varieties	## Blowups of Cohen-Macaulay varieties ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose that $X$ is a Cohen-Macaulay normal scheme/variety and $\pi : Y \to X$ is a proper birational map with $Y$ normal. Question: Is $Y$ also Cohen-Macaulay? Are there common conditions which imply it is? If $Y$ is not normal I know of several ways to show that the answer to the first question is no. There are obvious spectral sequences but I don't see how to deduce what I want from them, perhaps I'm being dumb (or maybe there is an obvious example). - 1 Karl, just to clarify, when X is affine, you don't want the Rees algebra to be CM, but just the Proj to be CM, right? There are examples of the Rees algebra not CM. – Hailong Dao Oct 24 2010 at 6:26 Long, yes. I'm just interested in the Proj of the Rees algebra. – Karl Schwede Oct 24 2010 at 13:33 ## 2 Answers An example was given in Section 3 of this paper by Cutkosky: "A new characterization of rational surface singularities" (The scheme $Z$ in the last page, which is a blow up of some $m$-primary ideal of a regular local ring of dimension $3$, is normal but not Cohen-Macaulay). The algebraic side of this example has been studied quite a bit, so perhaps more explicit examples are known. I am not an expert here, but you can check out a paper by Huckaba-Huneke here, or papers by Vasconcelos (he has a book called "Arithmetic of Blow-up algebras" which discussed, among other things, Serre's condition $(S_n)$ on Rees algebras), and the references there. - Long thanks, that's quite interesting. I've looked at Vasconcelos' book a long time ago. This example seems to show that there is no reasonable condition on $X$ which would imply blow-ups are Cohen-Macaulay. – Karl Schwede Oct 24 2010 at 19:51 Actually, another reason this is interesting, is that I've certainly had people tell me several times that normal blow-ups of varieties with rational singularities still have rational singularities. This seems to debunk that assertion. – Karl Schwede Oct 25 2010 at 14:11 ### 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. Karl, my feeling is that this will not be true without further conditions. Here are some thoughts: 1) The obvious spectral sequences do not give anything clear, so the safe bet would be that it is not true. 2) I can't believe that being normal would make a difference at the end. Maybe in low dimensions, but after all the difference is simply the codimension of the singular set. So I would concentrate on $S_n$. 3) So, let's see how one could construct a counterexample. Perhaps cones will do... 4) Let $W$ be a projectively normal variety and $X$ the cone over it. This ensures that $X$ is normal. Under these conditions $X$ is $S_d$ if and only if $H^i(W,\mathcal O_W(n))=0$ for all `$0<i<d-1$` and $n\in \mathbb Z$. (This last statement is for instance Lemma 3.1 here). So, it seems that if we find a projective birational morphism $\phi:Z\to W$ such that $W$ satisfies the above condition for $d=\dim W$, but $Z$ only satisfies it for say $d=2<\dim W$ and it is also $R_1$, then $Y$, the cone over $Z$ maps to $X$ (is this right? I have not checked this, but it seems OK) and $Y$ is normal, but not CM. 5) So assuming that #4 is OK, we just need an example of $\phi:Z\to W$ as required there. It is easy to have $W$ satisfy the conditions: If we can keep $W$ a hypersurface of dimension at least $2$ with isolated singularities, then the condition is satisfied. So, let's try that and blow up a point on $W$ and hope for some cohomology group changing. The most obvious would be $\mathcal O_W$, but that will not change as long as we blow up a smooth point (or even a rational singularity). So either one plays around with the other sheaves os takes a non-rational singularity. So, how about taking $W$ to be a cone over a high degree plane curve. That gives us everything we wanted and a non-rtl singularity. Then if $Z$ is the blow-up, then $H^1(Z,\mathcal O_Z)\neq 0$. (This should also be checked. My thinking was that by choice $R^1\phi_\mathcal O_Z\neq 0$, but $H^1(W,\mathcal O_W)= H^2(W,\mathcal O_W)= 0$, so something has to give.) Anyway, I think this has a good chance. The only point it could break is that map between the cones, but it seems all right to me at the moment. It will probably not be a nice blow up but it seems to me that there should be a morphism. 6) So what condition should we ask for? I guess the first guess is something like $R^i\pi_\mathcal O_Y=0$ for $i>0$. I think this might give you what you want: Grothendieck duality gives that then $$R\pi_\omega_Y^\cdot\simeq_{qis}\omega_X^\cdot$$ Now if $X$ is CM, then the right hand side has only one non-zero cohomology. Now one could write $\omega_Y^\cdot$ as $$\omega_Y[n]\to \omega_Y^\cdot \to \omega_Y^+ \to^{+1}$$ So, if we also knew that $R^i\pi_\omega_Y=0$ for $i>0$, then it would follow that $R\pi_\omega_Y^+=0$ and I think that should imply that actually $\omega_Y^+=0$ which would imply that $Y$ is CM. So, this seems like a condition: If $R^i\pi_\mathcal O_Y=0$ and $R^i\pi_*\omega_Y=0$ for $i>0$, then what you want might follow. Then again, this might be more then what you would want to assume. Any thoughts? - S\'andor, I agree that it seems unlikely to be true. For 4), I'm slightly confused, because $Y$, the cone over $Z$ is also affine, so I don't think the map $f:Y\to X$ will be proper+birational (is there even a map in general). For the existence of $f$, I suppose it depends on the choice of a particular line bundle on $Y$ to take the sectionf. Start with the pull back of whatever ample on $X$ twisted by something relatively ample? My guess is that the map $Y \to X$ will not be defined everywhere, maybe resolving the indeterminacies will help... then do what you did? I'll think about this. – Karl Schwede Oct 24 2010 at 14:01 1 Karl, yes, regarding the map in #4, these are exactly my worries, but here is what I was thinking: It seems that the morphism is OK outside of the vertices of the cones, so if you resolve the indeterminacies, then you get a morphism extending this on a scheme that you get by blowing up an ideal supported at the vertex of $Y$ and the morphism maps the entire exceptional set to the vertex of $X$. But then, it seems that this morphism should actually factor through $Y$. In fact, this might be a place where $Y$ being normal makes a difference. – Sándor Kovács Oct 24 2010 at 14:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 62, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9669684171676636, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/99753/why-is-the-riemann-integral-only-defined-on-compact-sets	Why is the Riemann integral only defined on compact sets? Every text I look at says a function must be bounded and be defined on a compact set before one can even think about the Riemann integral. Boundedness makes sense, otherwise the Darboux sums could be undefined. However, I don't see where it becomes important that the integral be taken over a compact set. - 2 Because you need the function to obtain suprema and infima on any partition of the set. To ensure this, you need the subsets of the partition to be compact, so that your original set must be a finite union of compact sets. But this of course means that the set is compact to begin with. – user12014 Jan 17 '12 at 2:53 That doesn't really matter though. It will only affect the first and last subinterval in darboux integral if you chose to define it on, say, an open integral. The tagged partitions in the riemann integral avoids this altogether. – Stuart Jan 17 '12 at 3:02 You can modify the Darboux integral so as to apply to unbounded functions. If a function is unbounded, then for any partition $\mathcal{P}$ of $[a,b]$, either $U(f,\mathcal{P}) = \infty$ or $L(f,\mathcal{P}) = -\infty$ (or both). However it is still meaningful to talk about functions having upper and lower integral both equal to $+\infty$ (or $-\infty$). – Pete L. Clark Jan 17 '12 at 3:52 1 Answer If you use the definition without tagged partitions, the reason the interval needs to be compact is that you need the function to obtain suprema and infima on every subinterval on a partition. For example $f(x) = 1/x$ is continuous on $(0,1)$ (so it should be integrable), but it never attains a supremum in the first subinterval of any partition. Even if you use tagged partitions, this problem persists. Again, consider $f(x) = 1/x$ on $(0,1)$. Let $P_n$ be a sequence of partitions such that $P_{n+1}$ is a refinement of $P_n$ for all $n$, and let $t_n$ be the tagged point in the first subinterval of $P_n$. Then $t_n \to 0$ as $n \to \infty$. Hence $f(t_n) \to \infty$ so that the limit of $f(t_n)\Delta_1$ will be infinite for partitions whose mesh size tends to $0$ slower than $f(t_n)$ tends to $\infty$. Hence the Riemann sums will not converge to any finite limit which means, by definition, that $f$ is not integrable. One way to interpret this discussion is that the theorem "If $f$ is continuous on $I$, then $f$ is Riemann integrable on $I$" will no longer be true if we allow non-compact $I$. In fact, this is one of the "deficiencies" that made the Riemann integral unsuitable (along with the more pressing problems regarding convergence for sequences of functions). For Lebesgue integrals, you can use an open set without problems. - 1 But $f(x) = \frac{1}{x}$ is not Lebesgue integrable on $(0,1)$ either. – Pete L. Clark Jan 17 '12 at 3:49 @PeteL.Clark I thought the Lebesgue integral is defined for all measurable functions although the integral might be infinite. This is not the case for the Riemann integral. Or am I mistaken? Its been a while since I picked up Rudin... – user12014 Jan 17 '12 at 3:54 1 I suppose there is some difference of terminology here, but I think most people would say that "Lebesgue integrable" implies that the integral is finite. Anyway, note that the same thing holds for the Riemann integral: in several ways one can extract a meaning of the Riemann integral of $f(x) = \frac{1}{x}$ on $(0,1$) -- e.g. as an improper integral; or by extending $f$ to be defined at $0$ -- the problem is that since the function is unbounded, the Riemann integral cannot be finite. But one can make sense of the statement $\int_0^1 \frac{1}{x} = \infty$ just as well... – Pete L. Clark Jan 17 '12 at 4:03 3 By the way, it is not true (in any sense!) that every measurable function is Lebesgue integrable. The general definition of a Lebesgue integral involves subtracting the negative part from the positive part. When both are infinite, the Lebesgue integral is not defined (even though the improper Riemann integral might be!). – Pete L. Clark Jan 17 '12 at 4:08 1 @TutPoe: So what about $f(x) = \sin x$ on $\mathbb{R}$? – Nate Eldredge Jul 15 '12 at 0:24 show 3 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.900619387626648, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/203888/listing-all-derangements-for-a-given-n?answertab=oldest	# Listing all derangements for a given n How to find all the derangements of [n]. Specifically,if n is 4? What is the process to get all the derangments in general? - Sorry, what do you mean by 'derangement'? Give an example. – Berci Sep 28 '12 at 9:40 3 – user1210233 Sep 28 '12 at 9:43 ## 2 Answers A braindead method would be to list all the permutations of $[n]$ and then remove the ones with fixed points. You can list all the permutations of $[n]$ by listing them in lexicographic order, for example. Another method is as follows. The elements of $S_n$ with no fixed points are precisely those permutations which contain no $1$-cycles when written as a product of disjoint cycles. For $n=4$ these permutations must have cycle type $(2,2)$ or $(4)$, i.e. they look like $(\,\cdot\ \cdot\,)(\,\cdot\ \cdot\,)$ or $(\,{\cdot}\ {\cdot}\ {\cdot}\ {\cdot}\,)$. They're easy enough to find; then apply each of these to your set. - A general method is described in Gray code for derangements by Jean-Luc Baril and Vincent Vajnovszki, Discrete Applied Mathematics $140$ ($2004$) $207$–$221$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9065836071968079, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/11021/can-a-planet-cover-the-whole-sky-as-seen-from-its-satellite/11102	# Can a planet cover the whole sky as seen from its satellite Sorry if this is a trivial question, but I do not know enough physics to answer this on my own. Can there exist a satellite that revolves around a planet so large such that the planet covers the entire sky as seen from at least one point on the satellite? Added 1 I am sorry for being unclear earlier. But assume human eyes can't resolve less than $0.02^\circ$. Added 2 Assuming both the planet and satellite (ignoring atmosphere to make things simpler) are spheres and since spheres are locally euclidean, for large enough radii a planet could cover the whole sky as it appears to the eye. Consider this picture: - 1 Interesting problem in general relativity. – Carl Brannen Jun 14 '11 at 22:37 ## 3 Answers If the planet and satellite are perfect spheres, or any convex shape then the answer is NO. If the planet is close to the satellite and you stand on the point nearest to it then you will only be able to see a small band around the horizon, but the whole horizon cannot be blocked. If the satellite is not convex and you stand in a hole or crater on the satellite so that your view of the sky is reduced then the planet can fill the rest of the visible sky, so the answer is YES. If the satellite itself is a sphere then the planet cannot obscure the whole horizon unless the satellite somehow passes through a tunnel or at least a gorge in the planet, not very likely Update: Since the question has been updated there are a couple of extra points to be made If the observer on the satellite is limited by the resolution of his vision then a sufficiently big planet could obscure the whole sky with the only band of sky left near the horizon being too small for him to see. Also, regarding Carl's comment on the effect of GR. If the planet is sufficiently massive, but not the satellite, then light will bend towards the centre of the planet. In this case it is possible for the planet to cover the whole sky in the most extreme circumstances. As mentioned in the comments on this answer, an atmosphere on the satellite would bend the light towards the satellite making it harder for a plenet to cover the whole sky. If the atmosphere is on the planet it could only help if the satellite was inside the atmosphere, in which case the drag would slow it down so that it would not remain in orbit for very long. - 1 @Philip: What if the satellite had a thick atmosphere, causing the light from planet to be refracted? – qftme Jun 12 '11 at 18:39 @qftme Can the satellite keep its thick atmosphere when under the gravity of the near very large planet? – M.Sameer Jun 12 '11 at 19:42 @M.Sameer: Good point. I don't know, guess we'll have to wait for a planetary physicist to pitch in. I was merely pointing out that the refractive properties of an atmosphere would be an exception to both the "no" answers so far. The whole situation was pretty hypothetical from the start.. – qftme Jun 12 '11 at 22:50 @M.Sameer. In a hypothetical way, we could make the density of the planet low enough that tidal forces don't rip the atmosphere off the satellite. In any realistic case, atmospheric drag, and tidal effects would cause the satellite to impact the planet. – Omega Centauri Jun 12 '11 at 23:39 @Omega , In this case the thick atmosphere @qftme proposed might do the trick and make the planet cover the whole sky view. – M.Sameer Jun 12 '11 at 23:48 show 5 more comments No, because, if you assume spherical shapes for both satellite and main body, this would only be (geometrically) possible if they touch at the point where you stand. It would be possible if you stand in a hole on the satellite, the walls of which obscure part of the sky. As for the case of main body and satellite orbiting each other and nearly touching, I think this is theoretically possible (if neither has an atmosphere), but not a stable configuration. I'm no astronomer, though. - I think if the satellite gets that close you have to consider tidal drag (as well as atmospheric drag it there is one present), which would mean any super close orbit would decay on a timescale that is small compared to astronomical timescales. The planet must be rotating slower than the orbital speed at its equatorial radius, otherwise it would be unstable, and mass/angular momentum would be shed -presumably forming a ring which might condense into a moon. Also if there are any other bodies nearby, so that it isn't governed by two body dynamics, the orbit will vary with time, and unless the near grazing condition was the innermost approach of an oscillation, it probably wouldn't avoid an impact for long. Now if you want to select a special time, just before the satellite impacts the planet, then yes it can happen. In fact a satellite in a wrong way orbit will lose momentum due to tidal drag and spiral in until it impacts. I think that is your best scenario for reaching your scenario. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9430124759674072, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/10666/isomorphism-types-or-structure-theory-for-nonstandard-analysis	## Isomorphism types or structure theory for nonstandard analysis ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) My question is about nonstandard analysis, and the diverse possibilities for the choice of the nonstandard model R. Although one hears talk of the nonstandard reals R, there are of course many non-isomorphic possibilities for R. My question is, what kind of structure theorems are there for the isomorphism types of these models? Background. In nonstandard analysis, one considers the real numbers R, together with whatever structure on the reals is deemed relevant, and constructs a nonstandard version R, which will have infinitesimal and infinite elements useful for many purposes. In addition, there will be a nonstandard version of whatever structure was placed on the original model. The amazing thing is that there is a Transfer Principle, which states that any first order property about the original structure true in the reals, is also true of the nonstandard reals R* with its structure. In ordinary model-theoretic language, the Transfer Principle is just the assertion that the structure (R,...) is an elementary substructure of the nonstandard reals (R,...). Let us be generous here, and consider as the standard reals the structure with the reals as the underlying set, and having all possible functions and predicates on R, of every finite arity. (I guess it is also common to consider higher type analogues, where one iterates the power set ω many times, or even ORD many times, but let us leave that alone for now.) The collection I am interested in is the collection of all possible nontrivial elementary extensions of this structure. Any such extension R will have the useful infinitesimal and infinite elements that motivate nonstandard analysis. It is an exercise in elementary mathematical logic to find such models R* as ultrapowers or as a consequence of the Compactness theorem in model theory. Since there will be extensions of any desired cardinality above the continuum, there are many non-isomorphic versions of R. Even when we consider R of size continuum, the models arising via ultrapowers will presumably exhibit some saturation properties, whereas it seems we could also construct non-saturated examples. So my question is: what kind of structure theorems are there for the class of all nonstandard models R? How many isomorphism types are there for models of size continuum? How much or little of the isomorphism type of a structure is determined by the isomorphism type of the ordered field structure of R, or even by the order structure of R? - I haven't read the book, but it appears that "Super-Real Fields" by Dales & Woodin is really about that very question. – François G. Dorais♦ Jan 6 2010 at 19:14 Uh, oh. I hope this doesn't put me in the doghouse with Woodin! :-) (He was my advisor 15-20 years ago.) Actually, now I remember Dales giving a number of seminar talks in Berkeley on superreal fields, when he was visiting Woodin at that time. – Joel David Hamkins Jan 6 2010 at 22:18 ## 5 Answers Under a not unreasonable assumption about cardinal arithmetic, namely `$2^{<c}=c$` (which follows from the continuum hypothesis, or Martin's Axiom, or the cardinal characteristic equation t=c), the number of non-isomorphic possibilities for R of cardinality c is exactly 2^c. To see this, the first step is to deduce, from `$2^{<c} = c$`, that there is a family X of 2^c functions from R to R such that any two of them agree at strictly fewer than c places. (Proof: Consider the complete binary tree of height (the initial ordinal of cardinality) c. By assumption, it has only c nodes, so label the nodes by real numbers in a one-to-one fashion. Then each of the 2^c paths through the tree determines a function f:c \to R, and any two of these functions agree only at those ordinals $\alpha\in c$ below the level where the associated paths branch apart. Compose with your favorite bijection R\to c and you get the claimed maps g:R \to R.) Now consider any non-standard model R of R (where, as in the question, R is viewed as a structure with all possible functions and predicates) of cardinality c, and consider any element z in R. If we apply to z all the functions g for g in X, we get what appear to be 2^c elements of R. But R was assumed to have cardinality only c, so lots of these elements must coincide. That is, we have some (in fact many) g and g' in X such that g(z) = g'(z). We arranged X so that, in R, g and g' agree only on a set A of size `$<c$`, and now we have (by elementarity) that z is in A. It follows that the 1-type realized by z, i.e., the set of all subsets B of R such that z is in B, is completely determined by the following information: A and the collection of subsets B of A such that z is in B. The number of possibilities for A is `$c^{<c} = 2^{<c} = c$` by our cardinal arithmetic assumption, and for each A there are only c possibilities for B and therefore only 2^c possibilities for the type of z. The same goes for the n-types realized by n-tuples of elements of R; there are only 2^c n-types for any finite n. (Proof for n-types: Either repeat the preceding argument for n-tuples, or use that the structures have pairing functions so you can reduce n-types to 1-types.) Finally, since any R of size c is isomorphic to one with universe c, its isomorphism type is determined if we know, for each finite tuple (of which there are c), the type that it realizes (of which there are 2^c), so the number of non-isomorphic models is at most (2^c)^c = 2^c. To get from "at most" to "exactly" it suffices to observe that (1) every non-principal ultrafilter U on the set N of natural numbers produces a R of the desired sort as an ultrapower, (2) that two such ultrapowers are isomorphic if and only if the ultrafilters producing them are isomorphic (via a permutation of N), and (3) that there are 2^c non-isomorphic ultrafilters on N. If we drop the assumption that `$2^{<c}=c$`, then I don't have a complete answer, but here's some partial information. Let \kappa be the first cardinal with 2^\kappa > c; so we're now considering the situation where \kappa < c. For each element z of any R as above, let m(z) be the smallest cardinal of any set A of reals with z in A. The argument above generalizes to show that m(z) is never \kappa and that if m(z) is always < \kappa then we get the same number 2^c of possibilities for R as above. The difficulty is that m(z) might now be strictly larger than \kappa. In this case, the 1-type realized by z would amount to an ultrafilter U on m(z) > \kappa such that its image, under any map m(z) \to \kappa, concentrates on a set of size < \kappa. Furthermore, U could not be regular (i.e., (\omega,m(z))-regular in the sense defined by Keisler long ago). It is (I believe) known that either of these properties of U implies the existence of inner models with large cardinals (but I don't remember how large). If all this is right, then it would not be possible to prove the consistency, relative to only ZFC, of the existence of more than 2^c non-isomorphic R's. Finally, Joel asked about a structure theory for such R's. Quite generally, without constraining the cardinality of R to be only c, one can describe such models as direct limits of ultrapowers of R with respect to ultrafilters on R. The embeddings involved in such a direct system are the elementary embeddings given by Rudin-Keisler order relations between the ultrafilters. (For the large cardinal folks here: This is just like what happens in the "ultrapowers" with respect to extenders, except that here we don't have any well-foundedness.) And this last paragraph has nothing particularly to do with R; the analog holds for elementary extensions of any structure of the form (S, all predicates and functions on S) for any set S. - Andreas, Welcome to MO! I am very glad to see you here, and thank you very much for your thorough answer. – Joel David Hamkins Jun 17 2010 at 12:08 Andreas, I noticed that you may have multiple MO identities (see mathoverflow.net/users/6428). The moderators can merge these. I'll flag this post to bring the issue to their attention. – Joel David Hamkins Jun 18 2010 at 12:59 Thanks Joel! ... – Scott Morrison♦ Jun 18 2010 at 15:56 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I think that the nonstandard models of R will be fairly wild by most reasonable metrics, since the theory is unstable (the universe is linearly ordered). For instance, I don't think that arbitrary models will be determined up to isomorphism by well-founded trees of countable submodels (as they are in classifiable'' theories). EDIT: I'm not sure how many nonisomorphic models there are of cardinality c (the size of the continuum), but there are 2^{2^c} distinct nonisomorphic nonstandard models of theory of R* of size 2^c. A crude counting argument shows that this is the maximum number of nonisomorphic models of size 2^c that any theory with a language of cardinality 2^c could possibly have, which can be considered as evidence that the class of models of the theory of R* is wild.'' (This result follows from the proof of Theorem VIII.3.2 of Shelah's Classification Theory, one of his many-models'' arguments about unclassifiable theories. In fact, an argument from the second chapter of my thesis applied to this theory shows that you can even build a collection of 2^{2^c} models of size 2^c which are pairwise bi-embeddable but pairwise nonisomorphic.) It's a good question whether or not you can have two models of this theory which are order-isomorphic but nonisomorphic -- there must be somebody studying o-minimal structures with an answer to this. - Thanks for the answer! I'd love to hear about how the size of the language affects the situation with stability... – Joel David Hamkins Jan 6 2010 at 1:16 Isn't your crude bound off by a power set? After all, the language here has size 2^c, not c. By my crude calculations, then, a model is determined up to isomorphism by a list of 2^c many subsets of a set of size c. But this gives 2^{2^c} not 2^{2^omega}. ( c = continuum = 2^omega ) Does this problem affect the rest of your answer? – Joel David Hamkins Jan 6 2010 at 14:06 @Joel: You're right, I was thinking that the size of the language was only c, not 2^c, which does affect my answer (which I just edited). A lot of things seem to break down when you start thinking about structures that are smaller than their language (e.g. Lowenheim-Skolem won't work to produce such models). – John Goodrick Jan 6 2010 at 19:23 The following was useful in a recent paper on asymptotic cones with Kramer, Shelah and Tent. How many ultraproducts $\prod_{\mathcal{U}} \mathbb{N}$ exist up to isomorphism, where $\mathcal{U}$ is a non-principal ultrafilter over $\mathbb{N}$? If $CH$ holds, then obviously just one ... if $CH$ fails, then $2^{2^{\aleph_{0}}}$. In the case when $CH$ fails, the ultraproducts are already nonisomorphic as linearly ordered sets. The proof uses the techniques of Chapter VI of Shelah's book "Classification Theory and the Number of Non-isomorphic Models". - Let me offer one counterpoint to John's excellent answer. Under the Continuum Hypothesis, the ultrapower version of R* will be saturated in any countable language. That is, it will realize all finitely realizable countable types with countably many parameters. Thus, by the usual back-and-forth construction, if we take the reduct to any countable part of the language, such as the ordered field structure and more, then under CH there will be only one saturated model of size continuum. I'm not sure if John's construction can produce saturated models, but if so, then under CH this observation will answer his question at the end about whether one can have non-isomorphic R's that are isomorphic orders or even as ordered fields. - See Robinson's book Non-Standard Analysis (North-Holland 1966)... Section 3.1 has some remarks about the order type of the non-standard natural numbers. - In the case of countable nonstandard models of arithmetic, the order type is clearly omega+ZQ, since there will be a Z chain around each nonstandard integer, and the Z blocks will be densely ordered, hence Q many of them. Does he classify the order-type arising for the N* of R? I guess it has order type omega+Zt for some dense order type t, which looks closely related to R. – Joel David Hamkins Jan 6 2010 at 16:34 This is his Theorem 3.1.6 ... the order type is omega + (omega + omega) theta , where theta is a dense order type without first or last element. – Gerald Edgar Jan 7 2010 at 16:45	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 3, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9347581267356873, "perplexity_flag": "middle"}
http://cms.math.ca/10.4153/CJM-1999-052-4	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \|----------\|----\|-----------\|----\| \| \| \| \| \| \| \| \| Site map \| \| \| CMS store \| \| location: Publications → journals → CJM Abstract view # Reflection Subquotients of Unitary Reflection Groups Read article [PDF: 198KB] http://dx.doi.org/10.4153/CJM-1999-052-4 Canad. J. Math. 51(1999), 1175-1193 Published:1999-12-01 Printed: Dec 1999 • G. I. Lehrer • T. A. Springer Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: HTML LaTeX MathJax PDF PostScript ## Abstract Let $G$ be a finite group generated by (pseudo-) reflections in a complex vector space and let $g$ be any linear transformation which normalises $G$. In an earlier paper, the authors showed how to associate with any maximal eigenspace of an element of the coset $gG$, a subquotient of $G$ which acts as a reflection group on the eigenspace. In this work, we address the questions of irreducibility and the coexponents of this subquotient, as well as centralisers in $G$ of certain elements of the coset. A criterion is also given in terms of the invariant degrees of $G$ for an integer to be regular for $G$. A key tool is the investigation of extensions of invariant vector fields on the eigenspace, which leads to some results and questions concerning the geometry of intersections of invariant hypersurfaces. MSC Classifications: 51F15 - Reflection groups, reflection geometries [See also 20H10, 20H15; for Coxeter groups, see 20F55] 20H15 - Other geometric groups, including crystallographic groups [See also 51-XX, especially 51F15, and 82D25] 20G40 - Linear algebraic groups over finite fields 20F55 - Reflection and Coxeter groups [See also 22E40, 51F15] 14C17 - Intersection theory, characteristic classes, intersection multiplicities [See also 13H15]	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7582488656044006, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/175380-martingale-process.html	# Thread: 1. ## Martingale Process Hi everybody! $<br /> c_t^\gamma = \betarfE_t(c_{t+1}^{\gamma})<br />$ 2. If $\beta=0$ then this is a martingale Can't you be more precise ? It could be an interesting discussion but we weren't here 3. Originally Posted by Moo If $\beta=0$ then this is a martingale Can't you be more precise ? It could be an interesting discussion but we weren't here The condition for this process to be a mg... I thought $\gamma\ne 0$ because in this case the function is not integrable. My colleague doesn't agree. 4. But what are all these parameters, what are the conditions, what is c_t ??? 5. Originally Posted by Moo But what are all these parameters, what are the conditions, what is c_t ??? $c_t$ is a stochastic process $\beta$ is a discount factor $\in (0,1]$ $r_f$ is the risk free that is given (deterministic)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9359155297279358, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/02/18/how-to-use-the-ftoc/?like=1&source=post_flair&_wpnonce=f636fc8316	# The Unapologetic Mathematician ## How to Use the FToC I’ll get back to deconstructing comics another time. For now, I want to push on with some actual mathematics. After much blood, toil, tears, and sweat, we’ve proven the Fundamental Theorem of Calculus. So what do we do with it? The answer’s in this diagram: This is sort of schematic rather than something we can interpret literally. On the left we have real-valued functions — we’re being vague about their domains — and collections of “signed” points. We also have a way of pairing a function with a collection of points: evaluate the function at each point, and then add up all the values or their negatives, depending on the sign of the point. On the right we also have real-valued functions, but now we consider intervals of the real line. We have another way of pairing a function with an interval: integration! At the top of the diagram, we can take a function and differentiate it to get back another function. At the bottom, we can take an interval and get a collection of signed points by moving to the boundary. The interval $\left[a,b\right]$ has the boundary points $\{a^-,b^+\}$, where we consider $a$ to be “negatively signed”. Now, what does the FToC tell us? If we start with a function $F$ in the upper left and an interval $\left[a,b\right]$ in the lower right, we have two ways of trying to pair them off. First, we could take the derivative of $F$ and then integrate it from $a$ to $b$ to get $\int_a^b F'(x)dx$. On the other hand, we could take the boundary of the interval and add up the function values along the boundary to get $F(b)-F(a)$. The FToC tells us that these two give us the same answer! To write this in a diagram seems a little much, but keep the diagram in mind. We’ll come back to it later. For now, though, we can use it to understand how to use the FToC to handle integration. Say we have a function $f$ and an interval $\left[a,b\right]$, and we need to find $\int_a^bf(x)dx$. We’ve got these big, messy Riemann sums (or Darboux sums), and there’s a lot of work to compute the integral by hand. But notice that the integral is living on the right side of the diagram. If we could move it over to the left, we’d just have to evaluate a function twice and add up the results. Moving the interval to the left of the diagram is easy: we can just read off the boundary. Moving the function is harder. What we need is to find an antiderivative $F(x)$ so that $F'(x)=f(x)$. Then we move to the left of the diagram by switch attention from $f$ to $F$. Then we can evaluate $F(b)-F(a)$ and get exactly the same value as the integral we set out to calculate. So if we want to find integrals, we’d better get good at finding antiderivatives! This has an immediate consequence. Our basic rules of antiderivatives carry over to give some basic rules for integration. In particular, we know that integrals play nicely with sums and scalar multiples: $\displaystyle\int\limits_a^bf(x)+g(x)dx=\int\limits_a^bf(x)dx+\int\limits_a^bg(x)dx$ $\displaystyle\int\limits_a^bkf(x)dx=k\int\limits_a^bf(x)dx$ ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 16 Comments » 1. The FToC diagram is very cool! Could you please post a higher-definition version? Much appreciated. Comment by \| February 18, 2008 \| Reply 2. I’ll see what I can do about forcing TeXshop to make something bigger. But trust me: you haven’t seen anything yet in terms of cool diagrams. Comment by \| February 18, 2008 \| Reply 3. I couldn’t make my own TeX installationdo the enlargement for me – however, I could use some devious ghostscript trickery to boost the size without losing the prettiness. The result can be found at http://mpc723.mati.uni-jena.de/~mik/mathochist.png http://mpc723.mati.uni-jena.de/~mik/mathochist-big.png http://mpc723.mati.uni-jena.de/~mik/mathochist-huge.png Enjoy. Comment by \| February 18, 2008 \| Reply 4. By the way – the way I solved it was by 1) compiling with LaTeX instead of my usual PDFLaTeX. 2) dvips -x fact -y fact -O -5cm,-5cm mathochist.dvi 3) convert -trim mathochist.ps mathochist-size.png where the -5cm will need to be twiddled to coerce the diagram back onto the page, and where the factor in -x and -y should stay the same. The results in my pictures are for factors of 2074 and 2986. Comment by \| February 18, 2008 \| Reply 5. If that’s nothing in the way of cool diagrams, I can’t wait to see what’s coming up… Tangentially, does anyone here know why we us $\partial$ for both taking boundaries and partial derivatives? I’ve been mystified by this since I first saw the $\partial$-as-boundary used. Comment by \| February 20, 2008 \| Reply 6. I’m not sure why, actually. I tend to think of it as a “nonstandard d”. We use it for boundaries because the boundary operator is dual to exterior differentiation. It’s about homology, where differentiation is about cohomology. “What are homology and cohomology doing here?” All in good time… Comment by \| February 20, 2008 \| Reply 7. Thanks. At least that’s a start. Is the fact that the boundary operator is dual to exterior differentiation (part of) what makes the generalized Stokes theorem so pretty? Comment by \| February 20, 2008 \| Reply 8. Geoff: the duality is the Stokes theorem. In fact (a) The array of different versions of the Stokes theorem all look like the diagram above. (b) All these diagrams fit together in a very nice way. (c) We can make the whole picture rigorous, instead of just schematic like it is now. Comment by \| February 20, 2008 \| Reply 9. John @ 6.: YAAAAAY! My good old friends are returning! Let me guess – you’ll be leading up to de Rham? Comment by \| February 20, 2008 \| Reply 10. Eventually, yes Comment by \| February 20, 2008 \| Reply 11. [...] way to use the FToC Let’s look back at that diagram that encapsulated the FToC. I want to point out something that’s going to seem really silly [...] Pingback by \| February 21, 2008 \| Reply 12. [...] Since we’ve established the connection between integration and antidifferentiation, we’ll be concerned mostly with [...] Pingback by \| February 25, 2008 \| Reply 13. [...] by Parts Now we can use the FToC as a mirror to work out other methods of finding antiderivatives. The linear properties of differentiation were [...] Pingback by \| February 25, 2008 \| Reply 14. [...] of Variables Just like we did for integration by parts we’re going to use the FToC as a mirror, but this time we’ll reflect the chain [...] Pingback by \| March 6, 2008 \| Reply 15. [...] we can start evaluating these integrals. The innermost integral goes immediately, using the fundamental theorem of calculus. An antiderivative of with respect to is , and so we [...] Pingback by \| January 4, 2010 \| Reply 16. [...] Change of Variables in Multiple Integrals I In the one-variable Riemann and Riemann-Stieltjes integrals, we had a “change of variables” formula. This let us replace our variable of integration by a function of a new variable, and we got the same answer. This was useful because the form of the resulting integrand might have been simpler to work with in terms of using the fundamental theorem of calculus. [...] Pingback by \| January 5, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 22, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.912996232509613, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/71507/analytic-continuation-of-fourier-transform?answertab=active	# analytic continuation of Fourier transform let be the Fourier transform $$h(u)= \int_{-\infty}^{\infty} g(x) \exp(iux) \, \text{d}x$$ then let us suppose that for $\|x\| \to \infty$ the function $g$ goes as $$g(x) = \exp(-ax) \text{ for some positive $a$}$$ does it means that we can analytically continue the Fourier transform $h$ to the region of the complex plane where $-a < \text{Im}(z) < a$? Apart from being $h(u)$ defined for every real $u$? if $g(x)$ tends to $0$ sufficiently fast does it mean I can define the fourier transform $h(z)$ for every complex number $z$? in fact if we put $u = iz$ for real $z$ then $$h(iz)= \int_{-\infty}^{\infty}g(x) \exp(-zx) \, \text{d}x$$ is it defined for every real $z$ in this case if $g$ is a Gaussian for example? - I have tried to improve your post a bit, check if it is still what you are trying to say. Note the difference between the value $g(x)$ and the function $g$. – Jonas Teuwen Oct 10 '11 at 19:00 ## 1 Answer Yes, it is possible to continue $h$ analytically to $-a < \text{Im}(z) < a\$ if $\|g(x)\| \le C\exp(-a\|x\|)\$ for some $a>0$. The function is defined by putting $z$ instead of $x$ in the Fourier transform. If $g$ satisfy this estimate for any $a>0$ then $h$ is an entire function. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8938323259353638, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/113278/finding-the-random-r-in-a-paillier-encrypted-message-with-knowledge-of-the-pri	# Finding the random $r$ in a Paillier encrypted message with knowledge of the private key. In the Paillier cryptosystem, suppose that I know a Ciphertext encrypted with some unknown random $r$ i.e. $$C = (g^m r^n) \bmod n^2$$ I know $g, n$, the prime factorization of $n$, i.e., $pq$. I also know the private key $-$ hence I know the message $m$. Would there be a way for me to recover $r$ easily? - – mikeazo Sep 12 '12 at 14:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8393101096153259, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/tagged/moments?sort=votes&pagesize=15	# Tagged Questions For questions about the moments of a random variable, that is, the expectation of $X^k$ where $X$ is a random variable and $k$ an integer (or a real number with $X$ non-negative). 2answers 786 views ### What's so 'moment' about 'moments' of a probability distribution? I KNOW what moments are and how to calculate them and how to use the moment generating function for getting higher order moments. Yes, I know the math. Now that I need to get my statistics knowledge ... 3answers 958 views ### Moments of a distribution - any use for partial or higher moments? It is usual to use second, third and fourth moments of a distribution to describe certain properties. Do partial moments or moments higher than the fourth describe any useful properties of a ... 1answer 325 views ### Intuition for higher moments in circular statistics In circular statistics, the expectation value of a random variable $Z$ with values on the circle $S$ is defined as $$m_1(Z)=\int_S z P^Z(\theta)\textrm{d}\theta$$ (see wikipedia). 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http://www.euro-math-soc.eu/node/2891	# The European Mathematical Society Hosted by: ## Mathematics of Approximation August 28, 2012 - 10:17 — Anonymous Author(s): Johan de Villiers Publisher: Atlantis Press/Springer Verlag Year: 2012 ISBN: 978-94-91216-49-7 (P), 978-94-91216-50-3 (E) Price (tentative): 59.95 € (net) Short description: This is the first volume of a new series Mathematical Textbooks for Science and Engineering. It builds up standard approximation theory from scratch (requiring only some advanced calculus and linear algebra) up to a reasonably advanced level. No complex analysis or advanced functional analysis is needed. As a textbook it includes all the proofs and has many exercises following each chapter. URL for publisher, author, or book: <a href="http://www.springer.com/mathematics/book/978-94-91216-49-7">www.springer.com/mathematics/book/978-94-91216-49-7</a> MSC main category: 65 Numerical analysis MSC category: 65-01 Other MSC categories: 42A10, 42A16, 42C10, 65D05, 65D07, 65D10, 65D15, 65D32, 65T40 Review: The idea of the new series Mathematical Textbooks for Science and Engineering (MSTE) is to publish self contained textbooks on applied or applicable mathematics at undergraduate and graduate level of mathematics science and engineering. It should be easy reading, even for self study. This book is volume 1 and it is devoted to the standard basics of approximation in the broad sense. The subjects are not a surprise and line up as one would expect: Polynomial interpolation, best approximation (in $L_2$ and $L_\infty$-sense by polynomials), elementary orthogonal polynomials (Chebyshev, Legendre), numerical quadrature, Fourier series, and spline approximation. A more detailed table of contents can be found on the publisher's website. Note that it involves only approximation by polynomials, trigonometric polynomials or piecewise polynomials (no rationals or other function systems) and only real approximation (complex analysis is avoided completely). All the material is standard and fully proved in the book, and no historical notes are given about the origin of these results. Hence no references to research literature are needed and thus there is no list of references. As with most textbooks, it is an endeavour resulting from many years of experience while teaching the course to many students and it is not different in this case. Such a ripening process is needed to make all the things explicit that a teacher considers obvious but may actually be a source of wild and unexpected speculations by students. So it is a remarkable achievement to introduce the right amount of rigour to avoid any misunderstanding from the reader's side and yet not to overload the text and keep it very readable also for self studying. In that respect de Villiers has made some specific choices that allowed to deliver a thoroughly thought over product that serves these characteristics very well. Some of these choices have their consequences. Although the series aims at an audience from science and engineering', this book is in my opinion more oriented towards science than towards engineering. It is practical in the sense that a lot of attention goes to error estimates and that it handles approximation techniques that are important and actually used in practical applications, but it is not so practical that proper attention is paid to the pure numerical and algorithmic aspects. Many of the subjects discussed will also appear in a textbook on numerical analysis, but there much more attention will be given to rounding errors and efficient implementation. This particular choice also shows by a total absence of graphs. A simple graph of an orthogonal polynomial, the location of the Chebyshev points, a spline, or whatever could relax the reader a bit from the rigorous and sometimes complicated formulas to explain a relatively simple idea. Another exponent of this choice of approach is a lack of numerical examples. There are some examples included, but mostly quite simple and of an academic nature. So there are also no graphs or tables that illustrate the error or the convergence behaviour of some of these methods. Of course a particular phenomenon might require a numerical explanation that has been deliberately avoided in this book. Another example is the incidental mentioning of FFT in the chapter on Fourier series: [...] indeed (9.3.73)-(9.3.76) form the basis of the widely used Fast Fourier Transform (FFT), a detailed presentation of which is beyond the scope of this book.'. In many computational issues, the linear algebra aspect takes an important role when it comes down to implementation (e.g. the computation of Gaussian quadrature formulas by solving an eigenvalue problem). This is another aspect that has not been included in this book. But even, if the contents has a strong theoretical component, the scrupulous attention paid to error estimates, estimates of Lebesgue constants, and the different ways in which a solution can be represented and computed have a definite `applicable' component as well. So, I am somewhat surprised that, besides Lagrange and Newton forms of interpolation polynomials, there is no mentioning of the barycentric representation which allows some nice theory and has definitely great practical importance. Another nice piece of theory could have been devoted to wavelets, which are absent as well (except for a brief note in an exercise about splines). Let me make it clear that I do not consider my enumeration of what is not in this book as a critique. After all, any course that is actually taught has to transfer knowledge in a finite number of teaching hours, and there is enough material left to teach a major course on approximation. My only intention is to inform potential buyers of what is and what is not to be expected. A completely different approach is for example taken in the matlab based package Chebfun developed by Nick Trefethen and his team. This is less broad, less theoretical, and obviously much more hands-on. Learn by experience, not by formulas. A book on Chebfun is announced as a SIAM publication for early 2013. That book will be in many aspects different from the present book under review: probably not the opposite but certainly a complement. Reviewer: A. Bultheel Affiliation: KU Leuven ## Post new comment • Web page addresses and e-mail addresses turn into links automatically. • Allowed HTML tags: <a> <em> <strong> <cite> <code> <ul> <ol> <li> <dl> <dt> <dd> • Lines and paragraphs break automatically.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9309242367744446, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/77681/isoperimetric-like-inequality-for-non-convex-sets	## Isoperimetric-like inequality for non-convex sets ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The classical isoperimetric inequality can be stated as follows: if $A$ and $B$ are sets in the plane with the same area, and if $B$ is a disk, then the perimeter of $A$ is larger than the perimeter of $B$. There are several ways to define the perimeter. Here is a unusual one: if $A \subset \mathbb{R}^2$ is a convex set, the Cauchy-Crofton formula says that the perimeter of $A$ equals the measure of the set of lines that hit $A$, or $$p(A) = \frac{1}{2} \int_{S^1} \lambda(P_{\theta} A) d\theta,$$ where $P_\theta$ is the orthogonal projection in the direction $\theta \in S^1$, and $\lambda$ the Lebesgue measure on any line. Now, this definition of $p(A)$ makes sense for non-necessarily convex sets, excepts that it is not the usual notion of perimeter, so let's call it rather "mean shadow". My question if whether the isoperimetric inequality holds for the mean shadow instead of perimeter: if $A,B$ are (open, say) subsets of the plane with equal area, and if $B$ is a disk, is the mean shadow of $A$ larger that the mean shadow of $B$ ? The inequality is true if $A$ is convex, and we can assume that $A$ is a disjoint union of convex sets (since taking the convex hull of a connected set does not change the mean shadow). - I don't quite understand the question. You can always compare an open set to its convex hull, it will have the same mean shadow, but a smaller area. So the isoperimetric inequality for open subsets should follow from the convex case? – Jean-Marc Schlenker Oct 10 2011 at 10:03 4 For connected sets, taking the convex hull does not change the mean shadow, but for non-connected sets it increases ... – Guillaume Aubrun Oct 10 2011 at 10:09 I'm not sure whether considering open sets changes the following argument substantially: Take two closed balls with volume one half and radii $r$. Put them with canters apart by a distance of $L$ in the plane. Then for $L\to\infty$ the mean shadow will be $4\pi r$. Further the mean shadow is monotone in $L$ and decreases for decreasing $L$ as the overlap in the projection becomes larger if the balls are nearer. Eventually, for $L=2r$ the balls will touch and you have a connected set, where the mean shadow conincides with the perimeter of the convexification. – André Schlichting Oct 11 2011 at 9:40 ## 2 Answers As you noticed, it is sufficient to consider the case $$F=\bigcup_{i=1}^n F_i$$ where $F_1$, $F_2,\dots, F_n$ are disjoint convex figures with nonempty interior. Let $s$ be mean shadow of $F$. Denote by $K$ the convex hull of all $F$. Note that $$\mathop{\rm length}(\partial K\cap F)\le s.$$ We will prove the following claim: one can bite from $F$ some arbitrary small area $a$ so that mean shadow decrease by amount almost $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s$ (say $\ge 2{\cdot}\pi{\cdot}\tfrac{a}s{\cdot}(1-\tfrac{a}{s^2})$ will do). Once it is proved, we can bite whole $F$ by very small pieces, when nothing remains you will add things up and get the inequality you need. The claim is easy to prove in case if $\partial F$ has a corner (i.e., the curvature of $\partial F$ has an atom at some point). Note that the total curvature of $\partial K$ is $2{\cdot}\pi$, therefore there is a point $p\in \partial K$ with curvature $\ge 2{\cdot}\pi{\cdot}\tfrac1s$. The point $p$ has to lie on $\partial F$ since $\partial K\backslash \partial F$ is a collection of line segments. Moreover, if there are no corners, we can assume that $p$ is not an end of segment of $\partial K\cap F$. This proof is a bit technical to formalize, but this is possible. (If I would have to write it down, I would better find an other one.) - I don't understand why the total length of $\partial K \cap F$ is at least as big as the mean shadow of $F$. It seems to me that $\partial K \cap F$ can be arbitrary small for a given mean shadow (consider the case when $F_1,F_2,F_3$ are very small balls around the vertices of a large equilateral triangle, and $F_4$ anything inside the triangle). Did I miss something ? – Guillaume Aubrun Oct 12 2011 at 12:00 @Guillaume, now it is fixed (if you would read further you would see that this is a misprint). – Anton Petrunin Oct 12 2011 at 17:19 @Anton, what if the curvature is concentrated at corners where $\partial K$ meets $F$? – Sergei Ivanov Oct 12 2011 at 21:34 OK, I got it now, that's indeed a nice argument although it may be hard to write down. Thanks ! This gives also a proof of the usual isoperimetric inequality which I didn't know. Is this proof standard, or written somewhere ? – Guillaume Aubrun Oct 13 2011 at 8:58 @Guillaume. I learned this idea from Kliener's proof of isoperimetric inequality for 3-dimensional Hadamard space. – Anton Petrunin Oct 13 2011 at 15:52 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I believe that the answer is positive. If $A$ is connected, then it has the same mean shadow as its convex hull $CH(A)$ so the isoperimetric inequality for $CH(A)$ shows that the mean shadow of $A$ is larger than the mean shadow of $B$. If $A$ is not connected I believe the same inequality holds. I'll sketch a proof when $A$ has finitely many connected components $A_1, \cdots, A_n$, the general case then follows by an approximation argument. Choose a point $x\in CH(A)$ and define a 1-parameter family of deformations of $A$ by making a parallel translate of each connected component $A_i$ so that its barycenter moves towards $x$, say at constant speed to reach it at time $t=1$. Stop this 1-parameter family of deformation as soon as a contact occurs between two connected components, then merge those two connected components and repeat. The point is that the area of $A$ does not change under this deformation, however the mean shadow is non-increasing -- actually the size of the shadow is non-increasing in every direction. At the end of this deformation one obtains a connected set to which the usual isoperimetric inequality can be applied, and the same inequality then also applies to $A$. - 2 That's a very interesting idea, but it's not clear to me that it works, and it does not seem true that the shadow in every direction is non-increasing when the sets move as you describe. For example take 2 (disjoint) congruent equilateral triangles, one pointing to the right, and one pointing to the left, such that their vertical shadows coincide. When you move them along the line passing though their barycenters, the vertical shadow is actually increasing ! – Guillaume Aubrun Oct 11 2011 at 14:21 1 You're right -- I think that the mean shadow probably decreases, but the argument I gave is not correct. I'll look into this later, unless some else can make it work... – Jean-Marc Schlenker Oct 11 2011 at 15:34 1 Note that your argument works (shadows in every directions decrease) in the case when $A_1,\dots,A_n$ have a center of symmetry, basically by the one-dimensional version of the Hadwiger-Kneser-Poulsen conjecture. – Guillaume Aubrun Oct 11 2011 at 17:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 70, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9387742280960083, "perplexity_flag": "head"}
http://mathhelpforum.com/business-math/69181-profit-function.html	# Thread: 1. ## Profit function! P(X) = -60X2 + 1500X - 4000 - 1 ≤ x ≤ 25 This is the profit function and its domain, 1) the value of X where there is a breakeven point 2) the value of X where profit be maximum thankyou so very much!!! 2. Originally Posted by karimwahab P(X) = -60X2 + 1500X - 4000 - 1 ≤ x ≤ 25 This is the profit function and its domain, 1) the value of X where there is a breakeven point 2) the value of X where profit be maximum thankyou so very much!!! The breakeven point is the point where there is no profit and no deficit. In other words, the poitns when $P(x) = 0$. You find x values of this by finding the roots of the equation. Using the quadratic formula! $P(x) = -60x^2+1500x-4000 = 0$ $x = \frac{-1500 \pm \sqrt{1500^2-4(-60)(-4000)}}{2(-60)}$ $x = \frac{-1500 \pm \sqrt{1290000}}{-120}$ $x = 3.035$ or $x = 21.964$ For the maximum, the first derivative of the function is 0. Solve for x, and then plug that x value into the original equation to find P(x) for that value of x. The 2nd derivative test will tell you whether you have a maximum or a minimum. (Although you will have a maximum, since your coefficients of x^2 is negative!)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8513041138648987, "perplexity_flag": "middle"}
http://www.conservapedia.com/Field_(mathematics)	# Field (mathematics) ### From Conservapedia A field is a commutative ring which contains a non-zero multiplicative identity and all non-zero elements have multiplicative inverses. Loosely, a field is a collection of entities with well-behaved and compatible addition and multiplication operations. A few examples serve to illustrate this point. ## Examples 1. The rational numbers $\mathbb Q$, with operations the usual addition and multiplication. 2. The real numbers $\mathbb R$, with operations the usual addition and multiplication. 3. The complex numbers $\mathbb C$, with operations the usual addition and multiplication. 4. The integers modulo p (denoted $\mathbb Z/p\mathbb Z$), where p is prime. Here the operations are addition and multiplication modulo p. Observe that if p is not prime, then $\mathbb Z/p\mathbb Z$ is not a field. For example, the element $2 \in \mathbb Z/6 \mathbb Z$ has no multiplicative inverse modulo 6! In this case, $\mathbb Z/p\mathbb Z$ has only the structure of a ring. 5. The field $\mathbb Q[\sqrt{3}]$ of real numbers of the form $a+b\sqrt{3}$, where both a and b are rational. 6. Finite fields: for each prime number p and positive integer n, there is a unique (up to isomorphism) finite field of cardinality is pn. This field is of characteristic p. 7. The set of meromorphic functions on a complex manifold, with pointwise addition and multiplication. For example, the set of meromorphic functions on $\mathbb C$ or the unit disk $\Delta \subset \mathbb C$. 8. The p-adic fields $\mathbb Q_p$ and Ωp, which play a prominent role in number theory. The characteristic of a field must be either 0 or a prime number p. A field of characteristic 0 is necessarily infinite. Fields play an important role in nearly every area of mathematics, and are one of the most basic objects studied by algebra. The study of the relationships between different fields, and in particular subfields of a given field, leads to the study of Galois theory, and makes possible the proof of Abel's theorem and was one of the motivations for the early study of fields and abstract algebra more generally. ## Axioms Technically, a field F consists of a set of elements and two binary operations, addition + and multiplication , which obey the following axioms: Addition 1. Closure: If a,b ∈ F then a + b ∈ F. 2. Associativity: For a,b,c ∈ F, (a + b) + c = a + (b + c). 3. Commutativity: For a,b ∈ F, a + b = b + a 4. Identity: There exists an element 0 ∈ F such that a + 0 = 0 + a = a for all a ∈ F. 5. Inverse: If a ∈ F then there exists an (-a) ∈ F such that a + (-a) = (-a) + a = 0. Multiplication 1. Closure: If a,b ∈ F then ab ∈ F. 2. Associativity: For a,b,c ∈ F. (ab)c = a(bc). 3. Commutativity: For a,b ∈ F, ab = ba 4. Identity: There exists an element 1 ∈ F such that a1 = 1a = a for all a ∈ F. 5. Inverse: If a ∈ F and a ≠ 0 then there exists an a-1 ∈ F such that aa-1 = a-1a = 1. Distributivity For a,b,c ∈ F, a(b + c) = ab + a*c. Sometimes the condition 0 ≠ 1 is also included.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8902168273925781, "perplexity_flag": "head"}
http://mathoverflow.net/questions/21214/particular-problem-solved-by-solving-a-more-general-problem/21231	## Particular problem solved by solving a more general problem. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is sometimes much easier to solve a general problem than a particular one. What are good examples in mathematics that are difficult to solve but solved successfully by solving more general ones? - 10 This should be community wiki, probably. – Bill Kronholm Apr 13 2010 at 15:34 ## 15 Answers For what it's worth, here's a trivial one: when explaining induction to students, I sometimes stress that it might be easier to prove a stronger result by induction than a weaker one---you're trying to get more out, but you're putting more in. As a concrete example I note that proving that the sum of the first 100 odd numbers is a square sounds like it might be tricky, proving that the sum of the first $n$ odd numbers is a square for all $n\geq1$ sounds like it might be accessible using induction but in fact it still too weak, and proving that the sum of the first $n$ odd numbers is $n^2$ is really rather easy to prove. In some sense the stronger the statements get, the easier they become. - The first problem on the first homework in my number theory class last semester had something similar, but it was with finding a bound on something. I don't remember the exact problem, but it ended up being easier to give an upper bound $2-\frac{1}{n}$ instead of just $2$. – Harry Gindi Apr 13 2010 at 17:01 10 @Harry: Probably it was $\sum_{j=1}^n 1/j^2$. – villemoes Apr 13 2010 at 17:04 Probably! Sounds right! – Harry Gindi Apr 13 2010 at 17:16 4 +1 for the sequence of three statements rather than just two. – gowers Sep 26 2010 at 8:41 ### 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. Another way of proving that something is nonzero is to prove that it is odd. One good example of that idea is the proof of Sperner's lemma More generally, one can prove that something is nonzero by proving that it is nonzero mod p. That is the idea used in Chevalley-Warning theorem (one proves that the number of solutions is 0 mod p, then proves there is a trivial solution), and in the proof of Cauchy's theorem. - Frequently in mathematics the best way to determine the value of a sequence at a particular index is to compute its value at every index, even though the latter seems on the surface like a harder problem. Here is one of my favorite examples of this phenomenon. Suppose you want to know how many vectors of a particular norm there are in some lattice $L$. On the surface, this seems like a hard problem - it involves figuring out how many times some quadratic form takes some value. One can solve this problem by solving the harder problem of determining the answer for every possible norm by writing down the theta function $$\Theta_L(\tau) = \sum_{v \in L} e^{\pi i \tau \left< v, v \right>}.$$ If $L$ satisfies certain technical properties, $\Theta_L$ is a modular form with respect to some congruence subgroup, and modular forms live in finite-dimensional vector spaces; moreover, a lot is known about how to write down modular forms. For example, the theta function of the $E_8$ lattice is a modular form of weight $4$ and level $1$. The space of such forms is one-dimensional - in fact, it's spanned by an Eisenstein series - and it then follows that $$\Theta_{E_8}(\tau) = 1 + 240 \sum_{n \ge 1} \sigma_3(n) q^n$$ where $q = e^{2\pi i \tau}$. Similar considerations lead to the well-known formulas for the number of ways to represent an integer as the sum of two or four squares. - 1 For more on the subject, consider Noam Elkies's beautiful lectures at the 2009 AWS: swc.math.arizona.edu/aws/09/index.html. – pmoduli Apr 13 2010 at 22:15 Some of the previous answers have said that sometimes the easiest way to prove a set it non-empty is to show that it's large or even infinite. A variation on that idea is to show that something exists by showing that the probability of selecting it at random from some larger set is positive. As one of my professors used to say, some things are so hard to find that the best way to look for them is at random. - 1 Indeed, entire textbooks have been written on the subject: amazon.com/… – Qiaochu Yuan Apr 13 2010 at 18:12 2 A nice example of looking at random is in the area of error correcting codes. There is lots of fine theory for generating codes involving all kinds of beautiful mathematics from group theory and algebraic geometry. But it turns out that if you want to get close to the Shannon limit, random codes (subject to some constraints) will do the job just fine: en.wikipedia.org/wiki/… – Dan Piponi Apr 13 2010 at 23:19 Cantor proved the existence of transcendental real numbers by proving that most numbers are transcendental. The set of algebraic real numbers is countable. - The only way to prove that there's at least one prime in every arithmetic progression is by proving that there are infinitely many primes in every arithmetic progression. This is intuitively a fairly tremendous jump in difficulty to get the initial rather modest result out. I imagine that most examples of this phenomenon take the form that the question as asked is "more difficult" only in the sense that it's been phrased in such a way as to mask what's "really going on." I think this is probably at the core of hundreds and thousands of problem-solving type puzzles -- the difficulty of the puzzle comes from masking the influence of the governing theorem, which is likely to be easier to see how to prove in its general form than it is to realize which parts of the puzzle are the important ones. In short, puzzles have red herrings, good theorems do not. - 2 When we come across a (rearch) problem, sometimes we may not see its full picture. And in each step for a good theorem, we may not need their full generality, but a special case is enough. – Sunni Apr 13 2010 at 16:09 8 A good example of such a puzzle is this: Let x = sqrt(2) and y = 2+sqrt(2). Let X be the set { floor(nx) \| n a positive integer }, and define Y similarly. Prove that X and Y are disjoint, and that their union is the set of positive integers. The not-so-obvious key is that x and y are (positive) irrational numbers satisfying 1/x + 1/y = 1; and in fact the statement holds (and is easier to prove) for all such pairs. – villemoes Apr 13 2010 at 17:01 4 You can delete the word "known": the two are equivalent, since if there's at least one prime in every arithmetic progression then given an arithmetic progression a mod n, there's a prime congruent to a+n mod n^2, a prime congruent to a+n^2 mod n^3, etc. – Qiaochu Yuan Apr 13 2010 at 21:04 Ah, good point. Interesting. – Cam McLeman Apr 13 2010 at 21:57 This might not be correct---perhaps someone can confirm---but I was once told (when I was a graduate student) that the way that Leopoldt's conjecture was proved for abelian number fields was as follows: first do the standard reduction to show that Leopoldt is true if certain special values of certain $p$-adic $L$-functions $L(1,\chi)$ are non-zero, and then prove that these numbers are non-zero by showing that they are transcendental! As I say, I don't know for sure if this is true, but my source was pretty reliable. The emphasis was on the observation that (at the time at least), apparently the only way of proving the numbers were non-zero was by showing they were transcendental. - Occasionally when trying to prove a certain type of object exists, it is easier to show that the set of those objects is very large. For instance, it's difficult to give an example of a transcendental number over the rationals. However, it is quite easy to show that the set of algebraic numbers is only countably infinite, so almost every real number is transcendental. - 2 A related example: it's not completely trivial to write down a language that can't be recognized by a Turing machine, but there are uncountably many languages and countably many Turing machines. – Qiaochu Yuan Apr 13 2010 at 18:13 4 Also related, it is difficult to give an example of a continuous real valued function on [0,1] that is not monotone on any interval, but it is not too difficult to prove using the Baire category theorem that almost all functions are like this. – TS Apr 14 2010 at 3:55 Matijasevic's theorem, that all recursively enumerable sets are diophantine, was needed to prove that the set of primes is diophantine. - 1 I don't know about "needed": my understanding is that people simply didn't believe that such things were possible before Matiyasevich's theorem was proven. The actual construction was done explicitly a few years later by Jones et al.: en.wikipedia.org/wiki/… – Qiaochu Yuan Apr 13 2010 at 21:00 It is easier to prove that almost all real numbers are normal than to prove that any particular real number is normal. Indeed, none of the most natural candidates, such as $\sqrt{2}$, $\pi$ or $e$, has yet been proved normal. - (John, the original said "almost real numbers". I made the obvious correction.) – François G. Dorais♦ Apr 14 2010 at 0:05 1 Thanks, François. That will teach me always to proofread, even a single sentence! – John Stillwell Apr 14 2010 at 0:11 Proving that almost all real numbers are normal does not solve the problem of whether $\sqrt{2}$ is normal, so this isn't really an example of what the question asks for. – gowers Sep 26 2010 at 8:38 2 On second thoughts, it does if you regard the problem as being "Prove that there exists a normal number." But then it's not obvious that proving that almost all real numbers are normal is easier than proving that 0.12345678910111213141516... is normal. – gowers Sep 26 2010 at 10:22 Actually 0.12345678910111213141516... is proven normal only in base 10 – Ostap Chervak Apr 9 2011 at 20:22 Proving the general solution for lambert series is much easier than dealing with specific examples c.f. Apostol - Modular Functions and Dirichlet Series in Number Theory ch. 1 Ex. 14. - I believe Fermat's Last Problem was solved by proving the Modularity Theorem (for the case of semistable elliptic curves), but I don't know the proof enough to say if the problem is just a direct corollary. The Modularity theorem is not in any sense easy anyway, but at least it's been proved successfully. :) - I wouldn't call the modularity theorem a "generalization" of FLT; it's just a large and diffficult theorem that happens to imply FLT, and this implication is itself a theorem: en.wikipedia.org/wiki/Ribet%27s_theorem . I would reserve the term "generalization" for when the specialization can be obtained by specializing some universal quantifiers. – Qiaochu Yuan Apr 14 2010 at 17:09 @Qiaochu: Would you say that ZMT is a generalization of the Nullstellensatz? – Harry Gindi Apr 14 2010 at 17:27 @Qiaochu: True. As I said, I didn't know the deduction from Modularity Theorem itself is a theorem. Btw, does MT imply information about solutions of a certain class of equations, or is FLT really one-of-a-kind type of thing? – elfking Apr 14 2010 at 18:16 The Carlson Lemmas in combinatorics. He told me at the time that he had struggled with the simple and natural way the problem was originally posed, but was able to push through a far more elaborate, stronger, and much less intuitive version. Timothy J. Carlson A dual form of Ramsey's theorem, (with S. Simpson) Advances in Mathematics 53 (1984), pp. 265-290. Some unifying principles in Ramsey theory, Discrete Mathematics Volume 68 , Issue 2-3, (February 1988), Pages: 117 - 169 - The generating function proof of Cayley's theorem counting labeled trees (e.g., the theorem that there are $n^{n-2}$ labeled trees on $n$ vertices) is a good example. In the lecture notes I linked to, the more general question is Theorem 1 and the particular question is Corollary 1. - There are perhaps many examples found in inequalities because a given inequality with a mass of variables and functions can easily hide "Simple Inequality" of which it is a special case. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.952900767326355, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/65021-p-extension.html	# Thread: 1. ## p-extension Let p be a prime and let $F$ be a field. Let $K/F, L/K$ be p-extensions. Then the Galois closure of $L$ over $F$ is a p-extension of $F$. ( $A$ is called a p-extension of $B$ if $A$ is a Galois extension of $B$ whose Galois group is a p-group.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.827629804611206, "perplexity_flag": "head"}
http://simple.wikipedia.org/wiki/Norm_(mathematics)	# Norm (mathematics) In mathematics, the norm of a vector is its length. For the real numbers the only norm is the absolute value. For spaces with more dimensions the norm can be any function $p$ with 1. Scales for real numbers $a$, that is $p(ax) = \|a\|p(x)$ 2. Function of sum is less than sum of functions, that is $p(x + y) \leq p(x) + p(y)$ or the triangle inequality 3. $p(x) = 0$ if and only if $x = 0$. ## Examples 1. The one-norm is the sum of absolute values: $\\|x\\|_1 = \|x_1\| + \|x_2\| + ... + \|x_N\|.$ This is like finding the distance from one place on a grid to another by summing together the distances in all directions the grid goes; see Manhattan Distance 2. Euclidean norm is the sum of the squares of the values: $\\|x\\|_2 = \sqrt{x_1^2 + x_2^2 + ... + x_N^2}$ 3. Maximum norm is the maximum absolute value: $\\|x\\|_{\infty} = \max(\|x_1\|,\|x_2\|,...,\|x_N\|)$ This short article about mathematics can be made longer. You can help Wikipedia by .	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8819395899772644, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/132599-chain-rule-leibniz-notation.html	# Thread: 1. ## Chain Rule Leibniz Notation Hi, I have been stuck on this question for a really long time any help would be appreciated. Using leibniz notation, apply the chain to determine (dy/dx) at the indicated value of x. $y= u(2-u^2), u= (1/x), x=2$ I think that I have to use the product rule and then thee chain rule but I'm lost. 2. Originally Posted by darksoulzero Hi, I have been stuck on this question for a really long time any help would be appreciated. Using leibniz notation, apply the chain to determine (dy/dx) at the indicated value of x. $y= <i>u</i>(2-<i>(u^2)</i>), <i>u</i>= (1/x), x=2$ I think that I have to use the product rule and then thee chain rule but I'm lost. First of all this is very hard to read. I assume that it's $y = u(2 - u^2)$ and $u = \frac{1}{x}$, where $x = 2$. You need to find $\frac{du}{dx}$ and $\frac{dy}{du}$. You are correct that to find $\frac{dy}{du}$ you should use the product rule. Another alternative is to expand the expression first. You should also note that $\frac{1}{x} = x^{-1}$. Once you have those derivatives, multiply them together to get $\frac{dy}{dx}$ and then evaluate it at $x = 2$. 3. Originally Posted by Prove It First of all this is very hard to read. I assume that it's $y = u(2 - u^2)$ and $u = \frac{1}{x}$, where $x = 2$. You need to find $\frac{du}{dx}$ and $\frac{dy}{du}$. You are correct that to find $\frac{dy}{du}$ you should use the product rule. Another alternative is to expand the expression first. You should also note that $\frac{1}{x} = x^{-1}$. Once you have those derivatives, multiply them together to get $\frac{dy}{dx}$ and then evaluate it at $x = 2$. ok I expanded it and got $\frac{dy}{dx} = 2u-u^3$ now how would I get the derivative in leibniz notation. 4. Originally Posted by darksoulzero ok I expanded it and got $\frac{dy}{dx} = 2u-u^3$ now how would I get the derivative in leibniz notation. No that doesn't look right. You have $y = 2u - u^3$. So now apply the power rule to each term to find $\frac{dy}{du}$. 5. Originally Posted by Prove It No that doesn't look right. You have $y = 2u - u^3$. So now apply the power rule to each term to find $\frac{dy}{du}$. ok I think I got it. $=2u-u^3$ $=2-3u^3$ $=2-3(-x^{-2})^2$ $=2-\frac{3}{-x^4}$ $=2-\frac{3}{-(2)^4}$ 6. No. $y = 2u - u^3$ $\frac{dy}{du} = 2 - 3u^2$ $= 2 - 3\left(\frac{1}{x}\right)^2$ $= 2 - \frac{3}{x^2}$. $u = \frac{1}{x}$ $= x^{-1}$. $\frac{du}{dx} = -x^{-2}$ $= -\frac{1}{x^2}$. $\frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$ $= -\frac{1}{x^2}\left(2 - \frac{3}{x^2}\right)$ $= \frac{3}{x^4} - \frac{2}{x^2}$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 42, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9583086967468262, "perplexity_flag": "head"}
http://mathoverflow.net/questions/84696/can-one-easily-pick-out-a-basis-of-a-simple-lie-algebra-after-picking-a-convex-or	## Can one easily pick out a basis of a simple Lie algebra after picking a convex order? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) One of the basic results of Lie theory is that if one picks a Cartan subalgebra of a simple Lie algebra, there there is a canonical decomposition of $\mathfrak{g}$ into the Cartan and a bunch of 1-dimensional subspaces (root spaces). Picking any basis of the Cartan (I'm not going to worry about which one), one is thus extremely close to having a canonically chosen basis of the Lie algebra, except that there are a bunch of scalar factors running around. I'd like to fix a particular choice of basis, while making a somewhat manageable number of choices. Choice 1: I'll pick a base of my root system, that is, a set of simple roots/positive roots/a Borel containing my chosen Cartan. I can now choose completely at random vectors $E_i$ in the simple root spaces (all of these are conjugate by inner automorphisms, so they are all "the same"). This fixes basis vectors $F_i$ in the negative simple root spaces that commute with the $E_i$'s as the standard generators of $\mathfrak{sl}_2$. Choice 2: Well, now I have to do something else non-canonical. What I'd like to choose is a convex order on my positive roots. This case, let be propose a basis for the other root spaces by $E_{\alpha+\beta}=[E_{\alpha},E_{\beta}]$ if $\alpha <\beta$ and similarly for the $F$'s. My question: is $E_{\alpha}$ uniquely defined? Has this construction appeared before in the literature? If this definition doesn't work, is there a variant on it which does? - ## 1 Answer What you are trying to do is done in complete detail in Leclerc's paper "Dual canonical bases, quantum shuffles and q-characters" based on the paper "Standard Lyndon bases of Lie algebras and enveloping algebras" by Lalonde and Ram. The root spaces are one dimensional, so you have to put conditions on $\alpha$ and $\beta$ to get a well defined $E_{\alpha+\beta}$. Roughly speaking, you start by observing that convex orderings on positive roots are equivalent to total orderings on simple roots. So fix one such ordering. In the (quantized) enveloping algebra you can start to think about words in the simple roots associated to (ordered) products of the corresponding $E_i$. For reasons explained in these papers, the monomial $E_{i_1}\cdots E_{i_n}$ corresponds to the linear combination of words obtained by (quantum) shuffling the symbols $i_1,i_2,\ldots,i_n$ together in all possible ways and summing the outcomes (Note that in the quantum context, this is a noncommutative product!). Write this shuffle product as $$i_1\cdotsi_n=\sum_{w\in S_n}q^{(\star)}\; i_{w(1)}\cdots i_{w(n)},$$ where $(\star)$ can be given explicitly. Let $\max(i_1\cdotsi_n)$ be the largest monomial (in the lexicographic order) in the sum $i_1\cdotsi_n$ (such a monomial is called \emph{good} or \emph{standard}). The set of all possible $\max(i_1\cdotsi_n)$ for $n\geq 1$ can be identified with the crystal $B(\infty)$. It is also worth pointing out that there is a nice triangularity property: $E_{i_1}E_{i_2}\cdots E_{i_n}$ corresponds to $$i_1i_2\cdots*i_n=i_1i_2\cdots i_n+\sum(words < i_1i_2\cdots i_n).$$ Anyway, given a word $w\in \mathcal{B}(\infty)$, call it "standard Lyndon" if it is larger than any subword ($w=w_1 w_2$ implies $w_1 < w$). The key fact is that the set of standard Lyndon words are in bijection with the set of positive roots. Leclerc explains how to calculate the set standard Lyndon words in Proposition 25 of his paper. For a positive root $\alpha$, let $w(\alpha)$ be the corresponding word. You construct the root vectors inductively by $E_\alpha=[E_{\alpha_1},E_{\alpha_2}]$, where $\bullet$ $w(\alpha)=w(\alpha_1)w(\alpha_2)$, $\alpha=\alpha_1+\alpha_2$, and $\alpha_1\neq 0$, $\alpha_2\neq0$ are positive roots, $\bullet$ $w(\alpha_1)$ is standard Lyndon, and $\bullet$ if $w(\alpha)=w_1w_2$, with $w_1$ standard Lyndon, then $w_1$ is a shorter word than $w(\alpha_1)$. This construction has also been carried out in the affine case by Beck-Chari-Pressley and Beck-Nakajima. In my paper with Melvin and Mondragon, we make these calculations explicitly in all types (well, type $E_8$ isn't exactly explicit, but almost). Note, however, we use the middle east reading convention when defining the lex. ordering. - 1 I should have thought of this; I've actually been reading that Leclerc paper a lot lately. It's funny how papers that go in the box for one project are harder than they should be to bring to mind for another project. – Ben Webster♦ Jan 2 2012 at 0:08	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9264355897903442, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/49927-solved-differential-equation-help-print.html	# [SOLVED] Differential Equation Help Printable View • September 20th 2008, 05:47 PM Maccaman [SOLVED] Differential Equation Help Hi all, Earlier I started a thread with a past exam question that I wanted to understand before I attempted a homework question. It is similar to the question I posted earlier, but Im still having trouble with it. Here it is: "find the equation of R(t) when $\frac{dE}{dt}= -\gamma_{b}E$ where $E(0) = E_0$ and $\gamma_b = \gamma_{b_1} + \gamma_{b_2} + \lambda_p$, $\frac{dS}{dt}= -\gamma_{s}S$ where $S(0) = S_0$ and $\gamma_s = \gamma_{s_1} + \gamma_{s_2} + \lambda_p$ and $\frac{dR}{dt} = \gamma_{b_1} E + \gamma_{s_1} MS$ where $M, \gamma_{b_1}, \gamma_{b_2}, \gamma_{b}, \gamma_{s_1}, \gamma_{s_2}, \gamma_{s}, \lambda_p$ are all constants. Now I know that $S(t) = S_0 e^{-\gamma_{s} t}$ $E(t) = E_0 e^{-\gamma_{b} t}$ and then by substituting these values into $\frac{dR}{dt}$ we obtain $\frac{dR}{dt} = \gamma_{b_1} E_0 e^{-\gamma_b t} + \gamma_{s_1} M S_0 e^{-\gamma_s t}$ this is where I get lost. How do I continue from here? What type of differential equation is this? Thanks • September 20th 2008, 05:49 PM Prove It Quote: Originally Posted by Maccaman Hi all, Earlier I started a thread with a past exam question that I wanted to understand before I attempted a homework question. It is similar to the question I posted earlier, but Im still having trouble with it. Here it is: "find the equation of R(t) when $\frac{dE}{dt}= -\gamma_{b}E$ where $E(0) = E_0$ and $\gamma_b = \gamma_{b_1} + \gamma_{b_2} + \lambda_p$, $\frac{dS}{dt}= -\gamma_{s}S$ where $S(0) = S_0$ and $\gamma_s = \gamma_{s_1} + \gamma_{s_2} + \lambda_p$ and $\frac{dR}{dt} = \gamma_{b_1} E + \gamma_{s_1} MS$ where $M, \gamma_{b_1}, \gamma_{b_2}, \gamma_{b}, \gamma_{s_1}, \gamma_{s_2}, \gamma_{s}, \lambda_p$ are all constants. Now I know that $S(t) = S_0 e^{-\gamma_{s} t}$ $E(t) = E_0 e^{-\gamma_{b} t}$ and then by substituting these values into $\frac{dR}{dt}$ we obtain $\frac{dR}{dt} = \gamma_{b_1} E_0 e^{-\gamma_b t} + \gamma_{s_1} M S_0 e^{-\gamma_s t}$ this is where I get lost. How do I continue from here? What type of differential equation is this? Thanks Just integrate with respect to t. • September 20th 2008, 05:50 PM topsquark Quote: Originally Posted by Maccaman Hi all, Earlier I started a thread with a past exam question that I wanted to understand before I attempted a homework question. It is similar to the question I posted earlier, but Im still having trouble with it. Here it is: "find the equation of R(t) when $\frac{dE}{dt}= -\gamma_{b}E$ where $E(0) = E_0$ and $\gamma_b = \gamma_{b_1} + \gamma_{b_2} + \lambda_p$, $\frac{dS}{dt}= -\gamma_{s}S$ where $S(0) = S_0$ and $\gamma_s = \gamma_{s_1} + \gamma_{s_2} + \lambda_p$ and $\frac{dR}{dt} = \gamma_{b_1} E + \gamma_{s_1} MS$ where $M, \gamma_{b_1}, \gamma_{b_2}, \gamma_{b}, \gamma_{s_1}, \gamma_{s_2}, \gamma_{s}, \lambda_p$ are all constants. Now I know that $S(t) = S_0 e^{-\gamma_{s} t}$ $E(t) = E_0 e^{-\gamma_{b} t}$ and then by substituting these values into $\frac{dR}{dt}$ we obtain $\frac{dR}{dt} = \gamma_{b_1} E_0 e^{-\gamma_b t} + \gamma_{s_1} M S_0 e^{-\gamma_s t}$ this is where I get lost. How do I continue from here? What type of differential equation is this? Thanks Well, as the right hand side doesn't depend on R all you need to do is integrate both sides with respect to t. -Dan • September 20th 2008, 05:53 PM Maccaman ahhhhhh, of course (Headbang) I feel so stupid now (Rofl) thanks • September 20th 2008, 05:56 PM topsquark Quote: Originally Posted by Maccaman ahhhhhh, of course (Headbang) I feel so stupid now (Rofl) thanks There is nothing more difficult than seeing the obvious after you have done something complicated. -Dan • September 20th 2008, 06:35 PM Maccaman Is this answer correct? (my integration rules are a bit dodgy) $R(t) = - \frac{\gamma_{b_1} E_0 e^{-\gamma_b t}}{\gamma_b} - \frac{\gamma_{s_1} M S_0 e^{-\gamma_s t}}{\gamma_s}$ • September 20th 2008, 06:48 PM topsquark Quote: Originally Posted by Maccaman Is this answer correct? (my integration rules are a bit dodgy) $R(t) = - \frac{\gamma_{b_1} E_0 e^{-\gamma_b t}}{\gamma_b} - \frac{\gamma_{s_1} M S_0 e^{-\gamma_s t}}{\gamma_s}$ Looks good to me. (Bow) -Dan All times are GMT -8. The time now is 01:37 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 38, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9469407200813293, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4206031	Physics Forums Thread Closed Page 10 of 10 « First < 7 8 9 10 Recognitions: Gold Member Science Advisor ## Julian Barbour on does time exist Quote by sshai45 Thanks for the response. I'm curious about that "bounce". Does it imply that the future of the universe is to recollapse ("Big Crunch") and bounce again? ... Thanks for the interesting discussion. In fact it does not imply recollapse. The Penn State people run many different cases on the computer, including Λ = 0 so they get a variety of behavior including that cyclic behavior you mentioned. But when they put in a realistic positive cosmological constant then they get just one bounce. This is similar to the classical DeSitter universe which has Lambda>0 and only one bounce. Personally I don't think of Λ > 0 as representing an "energy". I just think of Lambda as a constant which naturally occurs in the Einstein equation of GR (the symmetries of the theory permit two constants, G and Λ). And all the evidence so far is that Λ does not change over time. So if you think of it as an "energy" that energy density would not be changing. In the way it first appeared in the GR equation, Λ is not an energy density but simply a small inherent CURVATURE. That is to say, the reciprocal of an area. If you have a favorite force unit in mind you can always multiply reciprocal area by force and get a pressure and that is the same type of physcial quantity as an energy density. So you can convert Λ to an energy density by fiddling with it. Move it from left side (curvature) of equation to right side (matter) and make mysterious talk about "dark energy" but I think that is going out of style. More often now I hear cosmologists simply refer to the cosmo constant Λ. "Dark energy" is more for the media. All we know is there is this acceleration that appears exactly as if due to a constant curvature at the classical level. http://arxiv.org/abs/1002.3966 However one likes to think of it, including constant Λ > 0 in the picture with either classic DeSitter or Loop QC, you get a universe history with just one bounce. Hi Marcus, Thank you very much for your reply. This is pretty much my 4th version of a reply to you, just trying to keep it short. I have so much to say on this subject that my posts often end up being far too long unless I’m very careful. So excuse me if I don’t address each of your points specifically. But re QM and SpaceTime, here is my angle. I followed the links to the Do We Need a Physics of ‘Passage’? site, and looked at the Program of events. Very interesting , I`ll read through the abstracts. re your posts, There seems to be problems reconciling QM with ‘time’ or space time etc. I know very little about QM, but I accept its essence. I have read, thought, written and spoken a lot about the idea ‘time’, from the angle that time may be a complete misunderstanding. And I think I can show how ‘time’ is a completely unfounded idea, and thus how it does not exist. (In the same way that ‘phlogiston’ or the ‘Aether’ are just unfounded ideas). If I am correct in my view on time not existing, then we see there is no temporal direction and no temporal order to the universe. Seeing if this is the case, I assume that problems with QM such as Qm’s incompatibility with space time, the need for ‘retro’ causality to explain certain things – or energy being borrowed from ‘the future’ to make quantum leaps etc, can be seen in a new way by those who understand them. The problem with space-'time'. Einstein famously declared the distinctions between the past present and future to be illusions because he considered the validity of 'blocktime', in which 'before' and 'after' might be akin to 'over here' and 'over there' - so just as the distinction between 'here' and 'there' is arbitrary, so might be the distinction between 'times'. But I believe it can be shown that even blocktime is a highly misleading red herring. And that the distinctions between past and future simply do not exist because those places or concepts are simply invalid and nonexistent. Einstein’s work does show us that moving machines (atoms, life etc) run slow. And this is astounding and unexpected. But just ‘calling’ a machine a clock, then saying that a ‘clock’ is a thing that measures a thing called ‘time’, and then claiming this is a proof that ‘time’ exists, is imo absolutely not a proof that a thing called ‘time’ exists. However, if ‘time’ happens to exist, then it is agreed Einstein's work tells us something about it, and the concept of space-time may be fully or partially correct. But, if time does not exist, then Einstein’s SR only shows us is that ‘moving machines’ (etc) run slower than stationary ones.(And this is absolutely not the same as proving there is a thing called time, connecting past and future, who's passage can be dilated, such that thing 'sink into past' or 'surge into future' etc) So – it may be the case thatthere may be space, and things moving in space, and those moving things may be changing more slowly than other ‘stationary’ things. And that is all! If so, then we should not be insisting ‘space’ and ‘time’ should always be talked about together, or that what is true of one must a have a bearing on the other – unless we have shown both exist. ‘space’ is clearly a bit tricky mentally to grasp, but I have no deep problem with it. ‘Space’, it is the gap between railing in my local park etc. I have a big problem with ‘time’, because most people, seem to have the most paper thin logic for assuming the existence of a 4th dimension. Even someone of the stature of Stephen Hawking bases his belief that 'time' exists because... "we remember the past but not the future" (brief history of time p161) While in fact , surely we just look at some of the contents of our minds and 'call' them 'the past', and construct the 'idea' there is a 'future', and say we cant 'remember' 'it'. For a pattern to form and be fixed in your mind only requires that matter can exist, move , change and interact. Just because we can form useful and interesting patterns in our minds as we observe events (memories) - does not imo prove the the universe also creates a record of all events in a place called 'the past' - and so is not a proof that 'the past' and thus 'time' exist. QM and space-time. Therefore, what I am suggesting is this, ‘time’ does not exist. It is just a useful idea. If we have no reason or proof that ‘places’ such as the ‘past’ or ‘the future’ exist then we should not scientifically insist they do. If ‘the past’ and ‘the future’ were to exist, then it would make sense to consider that there is a thing called ‘time’ that has a flow and a direction (or any other description of time one prefers, ‘blocktime’ etc). But, If there is NOT a past and a future, then there is NOT a past and a future! And, if there is NOT a past and a future, then there is no such thing as ‘time’, and no such thing as a flow or direction of ‘time’. And, there is no such thing as ‘space-time’ to be compatible or incompatible with QM etc. This means that every bit of matter in the universe is just, ‘always’ somewhere, doing something ‘now’ (to use a thus redundant term). This means that causes do not have effects in a ‘temporal’ direction, but they very simply have effects in physical directions. A tennis racket hitting a ball has an effect in that direction, the opponents racket has an effect in the opposite physical direction. An explosion has effects in all directions. Likewise there are no ‘relativistic tilted planes of simultaneity’ to be accounted for. (if a space ship is heading at speed to ‘Andromeda’, then that’s what it’s doing – all its machines (‘clocks’) etc may be running slower than those on Andromeda, but nothing is ever experiencing a different ‘now’. So, if there is a discrepancy or conflict between QM and space-time, I suggest people go over what is and is not proven in Relativity, and double check, and be very, very clear as to whether there is space, motion, and a thing called ‘time’, to be accounted for. Or just space and motion to be accounted for. And these are very different positions. where you say... This does not mean that TIME AS A PROCESS OF CHANGE isn't real. - I agree, but also think things are even simpler that this, there may not be a thing called ‘time’ which is a process of change. There may just be change. Giving change a second name is confusing, and misleading. (I also think I have worked through every basic aspect of Relativity, and shown how does not describe a thing called 'time', but still makes full sense as a theory only describing the finer aspects of 'change'). What I am trying to suggest, (given the title of this discussion is ‘... does time exist’), is that one should be very careful and very clear as to the specific terms we use in any discussion aimed at clarifying some deep confusion, and precisely what we mean by them. I would agree that things ‘change’ – that is ‘things’ exist, and move, collide, interact, transform from matter to energy etc etc etc, and so the term ‘change’ is legitimate. But to say ‘time’ as a process of change is real, (for me) is to say ‘the process of change is (also) called time’ But using the word ‘time’ suggests a plethora of other mysterious, deep and intangible, ethereal and unseen ‘places or things’ also exist. E.g. a place or thing called the future, a place or thing called the past, a thing called time that flows between these mysterious places, and which may be slowed by motion, or possibly travelled through etc etc etc. (while change just means existing stuff may be moving and changing shape or place etc, simply, always just ‘here’ ‘now’) So, I am saying, consider that change is JUST change. Period. re So the QG program is aimed at constructing and verifying the most fundamental picture of the world. If you cannot incorporate a block space-time in QG, then that is as much to say a block spacetime does not exist, and time is not a 4th dimension. I can see the aim is to construct and verify the most fundamental (i.e. simplest sensible working model that addresses all observations without adding unnecessary 'frills') - and I think the picture I have reached does this in every way I have tested it. Change is not a thing that happens ‘over a thing called time’, and change is not a thing intrinsically linked to a thing called time. or a thing that 'proves the existence of a thing called time'. imo there is just change, and, if you can see the heart of this (and if i am correct) then we see there never 'is' or 'was 'a past', and never 'is' or 'was' a 'future'. These are both purely and only useful ideas. Deeper still we may see that if there is no past and no future, then there is no time, and specifically no 'direction' to a thing called time. This also means there is no 'temporal order'. As a simple example of this consider - With the idea of time fixed in our minds, and us habitually trying to make the idea of temporal progression fit what we see, it seems obvious that (say) the 'Wright Flyer' existed 'before' the jumbo jet. And thus it seems clear there is temporal order, temporal progression, and thus time. But with the idea that "perhaps everything is always just here 'now' and happening 'now' ", we can see a way to at least consider that the atoms that make the Write Flyer, are always doing something as are the atoms that make up jumbo jets. As are the laws of physics that allow write flyers and/or jumbo jets to work, always present. And thus the atoms that make the flyer are never doing something 'before' or 'after' the atoms that make a jumbo jet - and vice versa. Instead things, wherever they are, are just there, moving and changing, (not over a thing called time) but just where they are and in whatever physical direction they are heading in, with no ‘future’ ahead of them and no ’past’ behind them, i.e. timelessly so to speak. This is easier to see on a large scale, but perhaps various problems with QM may be seen in a different light if this directional ‘linear’ straight jacket of an idea of ‘time’ is seen to be completely dissolved. I say all of this because this physics form is specifically about the question ‘...does time exist’, and I think it can be shown ‘time’ does not exist, other than as a useful idea. (ok, I’ll stop here, that’s 1000 words, I never can keep this short :) M.Marsden. Recognitions: Gold Member Science Advisor Hi Matt, I looked over your post and it seems to me it could be clearer and more compact if your words were anchored to definite mathematical objects. Physics is a mathematical science which means people are looking for the simplest best fit model (in math language). When we talk with English words there is normally some underlying math the words can be reduced down to, or can be anchored to. It may be very simple but there is usually some nonverbal foundation. Talking English can be convenient and bridge people with different technical upbringing and help speed up acquiring intuition, but the verbal description is seldom the whole story. So you say HAVING TWO WORDS IS REDUNDANT AND CONFUSING we shouldn't have separate words "time" and "change". That makes a certain amount of sense on a purely verbal level. But the Tomita math model of time has a use for both words. This is because of a subtle difference in the way we TREAT intervals of time and the changes that correspond to them, mathematically. There is an algebra M consisting of all possible measurements or observations. It is an algebra because you can add two measurements X+Y and multiply them XY. And there is an extremely useful object alpha-sub-t called a ONE PARAMETER GROUP OF AUTOMORPHISMS. αt is the change corresponding to an interval of time of length t. For every real number t there is a change αt which stirs M around, it sends every element of M to a new element. X → αt(X) And ADDING TWO TIMES corresponds to doing first one change and then the other. If there are two real numbers s and t. then the change corresponding to s+t is what you get by changing by αs and then changing by αt. Doing one change and then the other change is thought of as group multiplication and so we write αs+t = αsαt The alphas would normally be large MATRICES of complex numbers, or something analogous. Their actual written form would vary depending on the problem. The matrix entries would depend on the time parameter t. So it is useful to have two words: time is the additive parameter, and you add time intervals together. Changes are matrices that stir the world around, and you multiply two matrices together to see what happens when you do one change and then the other. Changes correspond to passage of a certain amount of time. That is what a one parameter group of transformations is, or a one parameter group of automorphisms, or changes, is. Time is the additive real number parameter t, and αt is the change. From a math standpoint it would be inconvenient and confusing to have only one word. The words are NOT redundant, from a math standpoint. But you have written a purely verbal essay arguing that we should reform the way we speak and have only one word, because from your verbal perspective the two words are redundant. I hope I've clarified the difficulty somewhat. Quote by Mattmars But just ‘‘calling’’ a machine a clock, then saying that a ‘‘clock’’ is a thing that measures a thing called ‘‘time’’, and then claiming this is a proof that ‘‘time’’ exists, is imo absolutely not a proof that a thing called ‘‘time’’ exists. And "exists" is also a wooly word. Of course. I agree. But these are just nice words; remember 40 years of fruitless speculation about string theory! To connect words, or squiggles on paper (as Hardy called mathematics) with memorable physics, one needs to suggest something practical we can actually do with new ideas. Making something that can reveal part of the future, like a time machine, would be good! Even correctly predicting the fall of cards in a poker game would draw attention. So far in this thread no one, sadly including myself , has come anywhere near making such a useful suggestion. So far, it’s a futile story. Recognitions: Gold Member Science Advisor Hi Paulibus, glad to see you! Personally I think the topic has a certain beauty and excitement because of the prospect of doing general relativistic statistical mechanics (and thermodynamics) something not hitherto possible. You probably have seen the Einstein field equation dozens of times---relating curvature quantities on the left side to matter and energy quantities on the right side. Back in the 1990s Ted Jacobson DERIVED that equation from some thermodynamic law, some fact about entropy. That to me is a very mysterious thing. they seem like utterly separate departments of physics. The connection remains puzzling and incomplete to this day. this is one reason that I view the current interest in this Tomita time---the only universal time flow I know of (the cosmologist's Friedmann model time being a special case of it arising under simplifying assumptions)---as far from futile. I see it as pretty exciting. Another exciting thing has to do with what Matt just said: "Change is not a thing that happens ‘over a thing called time’, and change is not a thing intrinsically linked to a thing called time." When you read the Princeton Companion to Mathematics treatment of Tomita flow you see that the change matrix is a certain root matrix raised to the t power where time is measured in natural (Planck) units. So one can say time is the exponent of change. There is a matrix, or more generally a unitary operator S such that the automorphism corresponding to the passage of time t (in nature's units) is given by the matrix/operator St. This is why adding times corresponds to multiplying change (or doing one, followed by the other). That is how exponents behave. You always have Xs+t=Xs Xt. What this illustrates, to me, is that the world is more intrinsically unified at a math level than it is at a verbal level. Because Matt says "change is not intrinsically linked to time" but when I look at the world with Tomita's eyes I see immediately that TIME IS THE LOGARITHM OF CHANGE Time is the real number that you have to raise the Tomita base to, to get a given change process. (a stirring around or automorphism of the world M of measurements). Here is where the Princeton Companion describes how to find the Tomita base (like the number e, the base of the natural logarithms), it is what you raise to the power t to get the change corresponding to that passage of time. http://books.google.com/books?id=ZOf...20math&f=false Hi Marcus, Thank you very much for that reply. Thanks for clarifying the 'subtle difference in the way we TREAT intervals of time and the changes that correspond to them'. I see what you are saying + I will have to read up on this and some of the other details you mention to address them properly. (no point just blindly replying :) However, the essence of my point is that with the question '... does time exist' there may be some very simple basic 'truth' that is consistently missed -because- the mathematics works, and the science it leads to is so practical and useful. That is to say we may be confusing the usefulness of the mathematics with what it actually does and does not prove. For example, of course accountancy maths and scientific maths are somewhat different, but nonetheless, consider that no matter how perfectly one might balance the books of a multinational conglomerate, it would be foolish to think this high level of accuracy, or the usefulness of what you had done, related in anyway at all to how well you had proved that " 'Money' really is a thing that actually exists", (other than as a useful idea). Im trying to show that we may be making the same error with high level mathematics and the notion of 'time'. Thanks again for your reply, you've given me a couple of things to think about. Ill make sure I've understood your points then respond. mm Recognitions: Gold Member Science Advisor Matt you might be interested in some earlier parts of thread. Whether or not a concept emerges as meaningful useful real can depend on contextual things like e.g. SCALE. So we were talking earlier about how time could be emergent rather than fundamental. Like temperature of a gas, which is real enough but does not exist at the level of a single molecule---it is a property of the collective when you look largescale. Or the waterlevel in a pond, which is real enough at a large scale but at microscale the pond does not have a welldefined surface it is a wild fuzzy dance of molecules. So things can be emergent rather than fundamental (to use a verbal shorthand). I'll recall part of that earlier discussion. This is post #57 Quote by marcus ... Obviously the free energy in a situation depends on the scale you're able to manipulate. If you are molecule-size and live in a box of gas, then you can lasso molecules and can harness them (or play the Maxwell demon with them), and get energy. But whatever you do with the energy makes no difference to a large outsider. He looks in and sees no free energy---because he can't see or manipulate or benefit at your scale. He sees a uniform "temperature" throughout, which you do not. Whatever you accomplish with the free energy you see doesn't make a damn bit of difference to him---it still looks like gas in a box. So free energy depends on the scale at which the observer is interacting with it, and likewise the Boltzmann distribution, depending as it does on the free energy. So the idea of EQUILIBRIUM depends on scale... ... The reason it's relevant is that several of us in the thread seem to agree on looking at time as real but emergent either from local motions or thermodynamics. In particular e.g. Julian Barbour in his prize-winning FQXi essay showed clearly how time is emergent from local motions, at a certain level. One does not have to treat it as a quasi-spatial "extra" dimension. One wants to be able to generalize on both Barbour's time and thermodynamic or "thermal" time (which may, at root, be the same thing as Barbour's) to understand the emergence of time in a variety of contexts and at various scales. Paulibus said he liked some of post #57 but he didn't fully agree, and he made several other interesting points. I'll quote portions of his post #58. Quote by Paulibus .. As Niels Bohr pointed out, Physics is a matter of what we say about stuff, not what stuff “is”. ...say of hot and cold, or the maintenance of a status quo. When we try to extend such descriptions beyond scales familiar to us, a qualification as “emergent” can be useful for broadening context. So is the quantitative and logical extension provided to ordinary language by mathematics. But let’s not kid ourselves that the words and mathematical descriptions we use have absolute eternal meanings; they just conveniently communicate concepts between us. Like the mysterious word “time” that everybody knows. Although we cannot yet claim to accurately understand and describe time, one thing does stand out: using time as a parameter to characterise change works wherever physics rules. This, it seems, is all over the Universe. Therefore: time can’t just be some local quirky emergent thing; it must be related to something universal, like the observed red-shift and its cause, namely “expansion”. Or is this also just an "emergent" aspect of the “reality” that we try to describe? In the part I highlighted, Paulibus italicized the word works. That's key. In physics, as Niels Bohr indicated, we are less interested in what exists than in accurate consistent statements, predictions---the simplest best-fit model, testable stuff, measurements. As Paulibus just said: "exist" is a fuzzy word. You can waste a lot of time talking about whether this or that "exists". So we have this working distinction between more or less fundamental and emergent, and the notion that the reality or usefulness of concepts can depend on scale. Temperature can be very important at largescale and gradually lose meaning---become undefined or inapplicable---as you go to smaller and smaller scale. Concepts can be scale-dependent, observer-dependent, context-dependent, state-dependent---there is a lot of nuance in physics (as in other branches of language! ) Recognitions: Gold Member Science Advisor I'll repost a set of links useful for this discussion, mostly sources on thermal time (= Tomita flow time). ==from post #129== This is to page 517 of the Princeton Companion to Mathematics http://books.google.com/books?id=ZOf...20math&f=false It's a nice clear concise exposition of the Tomita flow defined by a state on a *-algebra. For notation see the previous post: #128. Here's the article by Alain Connes and Carlo Rovelli: http://arxiv.org/abs/gr-qc/9406019 Here is Chapter 1 of Approaches to Quantum Gravity (D. Oriti ed.) http://arxiv.org/abs/gr-qc/0604045 Page 4 has a clear account of the progressive weakening of the time idea in manifold-based physics, which I just quoted a couple of posts back. I see the inadequacy of time in manifold-based classical and quantum relativity as one of the primary motivations for the thermal time idea. The seminal 1993 paper, The Statistical State of the Universe http://siba.unipv.it/fisica/articoli....1567-1568.pdf This shows how thermal time recovers conventional time in several interesting contexts. Here's a recent paper where thermal time is used in approaches to general relativistic statistical mechanics and general covariant statistical QM. http://arxiv.org/abs/1209.0065 It can be interesting to compare the global time defined by the flow with a local observer's time. The ratio between the two can be physically meaningful. http://arxiv.org/abs/1005.2985 Jeff Morton blog on Tomita flow time (with John Baez comment): http://theoreticalatlas.wordpress.co...d-tomita-flow/ Wide audience essays--the FQXi "nature of time" contest winners: http://fqxi.org/community/essay/winners/2008.1 Barbour: http://arxiv.org/abs/0903.3489 Rovelli: http://arxiv.org/abs/0903.3832 Ellis: http://arxiv.org/abs/0812.0240 ==endquote== Interestingly, Tomita flow time is the only independent time-variable available to us if we want to study a general relativistic system. Observer-time is not well-defined unless we already have settled on a particular fixed geometry. If the underlying geometry is dynamic and undecided we can't specify an observer's world-line. Tomita time is independent of the observer. It depends only on what we think we know about the world---the correlations among measurements that embody physical theory and presumed initial conditions, along with our degree of confidence/uncertainty. That is, it depends on the state. In Bohr's words: "what we can SAY". As Wittgenstein put it: "The world is everything that is the case." Here's a Vimeo video of part of a talk on Tomita time by Matteo Smerlak: http://vimeo.com/33363491 It's from a 2-day workshop March 2011 at Nice, France. The link just missed being included in the above list. Recognitions: Gold Member Science Advisor But at my back I always hear Time's wingèd chariot hurrying near; And yonder all before us lie Deserts of vast eternity. Andrew Marvell, around 1650 I also want to recall this other passage, which is crucial to the discussion. This concisely summarized one of the troubles with time in a classical GR context. And indicates how the problem appears to get even more severe when one goes to a quantum version. But it is just at this point that the (M, ω) picture with its universally-defined Tomita flow becomes available. So the problem contains the seeds of its own solution. This passage gives a concise motivation for the star-algebra state-dependent way of treating time evolution. ==quote page 4 http://arxiv.org/abs/gr-qc/0604045 == ... Therefore, properly speaking, GR does not admit a description as a system evolving in terms of an observable time variable. This does not mean that GR lacks predictivity. Simply put, what GR predicts are relations between (partial) observables, which in general cannot be represented as the evolution of dependent variables on a preferred independent time variable. This weakening of the notion of time in classical GR is rarely emphasized: After all, in classical GR we may disregard the full dynamical structure of the theory and consider only individual solutions of its equations of motion. A single solution of the GR equations of motion determines “a spacetime”, where a notion of proper time is associated to each timelike worldline. But in the quantum context a single solution of the dynamical equation is like a single “trajectory” of a quantum particle: in quantum theory there are no physical individual trajectories: there are only transition probabilities between observable eigenvalues. Therefore in quantum gravity it is likely to be impossible to describe the world in terms of a spacetime, in the same sense in which the motion of a quantum electron cannot be described in terms of a single trajectory. ==endquote== In the (M,ω) picture, M —essentially the set of all measurements— functions as a quantum-compatible replacement for spacetime, doing away with the need for it. Uncertainty, including geometric uncertainty, is built into every measurement in the set. And there's another very clear explanation of the problem here (to get the original paper just google "connes rovelli" ): ==page 2 of http://arxiv.org/abs/gr-qc/9406019 == In a general covariant theory there is no preferred time flow, and the dynamics of the theory cannot be formulated in terms of an evolution in a single external time parameter. One can still recover weaker notions of physical time: in GR, for instance, on any given solution of the Einstein equations one can distinguish timelike from spacelike directions and define proper time along timelike world lines. This notion of time is weaker in the sense that the full dynamics of the theory cannot be formulated as evolution in such a time.1 In particular, notice that this notion of time is state dependent. Furthermore, this weaker notion of time is lost as soon as one tries to include either thermodynamics or quantum mechanics into the physical picture, because, in the presence of thermal or quantum “superpositions” of geometries, the spacetime causal structure is lost. This embarrassing situation of not knowing “what is time” in the context of quantum gravity has generated the debated issue of time of quantum gravity. As emphasized in [4], the very same problem appears already at the level of the classical statistical mechanics of gravity, namely as soon as we take into account the thermal fluctuations of the gravitational field.2 Thus, a basic open problem is to understand how the physical time flow that characterizes the world in which we live may emerge from the fundamental “timeless” general covariant quantum field theory [9]. In this paper, we consider a radical solution to this problem. This is based on the idea that one can extend the notion of time flow to general covariant theories, but this flow depends on the thermal state of the system. More in detail, we will argue that the notion of time flow extends naturally to general covariant theories, provided that: i. We interpret the time flow as a 1- parameter group of automorphisms of the observable algebra (generalised Heisenberg picture); ii. We ascribe the temporal properties of the flow to thermodynamical causes, and therefore we tie the definition of time to thermodynamics; iii. We take seriously the idea that in a general covariant context the notion of time is not state- independent, as in non-relativistic physics, but rather depends on the state in which the system is. ==endquote== So they describe the problem, and they propose a solution. The problem is "In a general covariant theory there is no preferred time flow, and the dynamics of the theory cannot be formulated in terms of an evolution in a single external time parameter." But we HAVE to have a preferred time flow if we are going to do general relativistic statistical mechanics--stat mech and thermodynamics INCLUDING GEOMETRY. The temperature and entropy of the geometry as well, not merely of matter distributed on some pre-arranged fixed geometry. These and other types of analysis require a time flow. We want to comprehend the whole, including its dynamic geometry, not merely a part. The proposed solution was clearly a radical departure, namely to roll all you think you know about the world up into one ball of information, called the state function, and make that give you an inherent distinguished time flow. Make it do that. Force it to give you an intrinsic flow on the set of all observations/measurements. Tomita, a remarkable Japanese mathematician, showed how. For some reason this reminds me again of Andrew Marvell's words: "Let us roll all our strength and all our sweetness up into one ball" and basically just blast on through "the iron gates of life." It is a bold move. He was talking about something else, though. Hi Marcus, Thanks for those posts and I appreciate those links, I’ll look through them. (in particular the ‘emergent’ aspect etc). I want to air all my thoughts on many of the points you raise (because I genuinely think I have a meaningful answer to each of them), but that would create a massive post. So as I read up on the areas you’ve pointed to Ill post some thoughts on a select few. Im drawn to this post ‘Julian Barbour on does time exist’, because of it’s title, and because like Julian I have also written a book (eBook) on the nonexistence of time – but from a significantly different and far simpler angle. ( search for ‘A Brief History of Timelessness’ + youll find the site/videos/book etc) I appreciate your ‘angle’ on mathematics, but I really think the essence of the ‘problem’ can be expressed in simple plain English, and, what I see throughout all the discussions about ‘time’ I have read, (which is a lot), is the more sophisticated the conversation gets the more the basic issue gets irretrievably obscured. Effectively people end up considering questions that cannot be answered – because they may be based on very basic incorrectly made assumptions. (For example if we ask the question which is probably the more correct view of time, Newton’s fixed and universal time, or Einstein's dilatable space time? Many people would start considering that obviously Einstein's view of time has been theoretically and practically proven to be more correct. However, I would say both are ‘completely’ wrong in that there is no such thing as time, at all, period. Einstein's work is clearly correct and proven in many ways, (Mercury’s advancing perihelion, gps etc), but (imo) it is not about a thing called ‘time’, it is just and only about the way things move (‘now’ to use a redundant and slightly misleading word). Normally saying Einstein's view is better than Newton’s doesn’t matter. But if people assume this thus means it is a forgone conclusion that a thing called ‘time’ in some way exists, and Einstein's view of it is more refined, and now we just need to refine the view further- then they may never recheck the fact that the existence of the thing actually isn’t proven by either view. And I’ve got a stack of books all about time, virtually all of which are written as if ‘time’ exists, built on the idea that Einstein's work in some way proved times existence. For example, as I say Im drawn to this post because of the title, ‘Julian Barbour on does time exist’. I read Mr Barbours very well written and comprehensive book (‘The End of Time’) as careful as I could, in case his point was the same as mine, but it’s not. The End of Time p143, Mr Barbour says ‘The block universe picture is in fact close to my own’,(I think)... he then describes a kind of ‘phase space of infinite dimensions’ in which all possible configurations of all the matter in the universe constantly, and statically exist. And in which all possible ‘historys’ statically exist – such that as we ‘move’ through particular paths in this infinite phase space it appears to each of us that ‘time’ with a past and future exist. (Note that is just my own very rough understanding of what I think Julian is saying, the maths etc in the book is out of my league so I may have got that wrong). In my opinion, this is an example of just how complicated things get if we have incorrect, unchecked assumptions at the core of a ‘theory’, and, if we ‘follow’ the maths searching for an answer to a problem that does not exist. My approach... My approach very briefly (I’ve expressed the whole thing on the website), to me the idea of “’time’ being explained by all possibly histories constantly existing’, or the question ‘”is ‘time’ emergent (or not)?” become redundant and moot if we just consider the following question very carefully (as I mentioned in my 1st post) Do we actually have any legitimate reason to even suspect that (say) the term ‘the past’ in any way at all actually relates to some ‘thing’? It seems to me that the only reasons we even consider that a thing called ‘time’ might exist (or be emergent) is because we think our ‘memories’ indicate some kind of a ‘past’. But if we see that there is absolutely no reason at all to suspect that our ‘memories’ are anything at all other than ‘a collection of ions and electrons etc in a particular formation in our heads’ – then we can see that there is no reason at all to suspect ‘a (temporal) past’ exists in any way at all. And if we are wrong to think there is ‘a past’, then we have no reason to suspect there is ‘a future’. And thus wrong to suspect there is a thing called ‘time’, or that ‘it’ has a flow or a direction. i.e (imo) it just does not exist and we are wrong to think it may. Instead – the world is just as it appears to be – full of stuff moving in organised and/or chaotic ways. Whatever is true of QM is true of QM, but it needn’t have any component of a thing called time mixed in it. The universe may be expanding and heading for a heat death (or not) – but that may be ‘just’ what it is doing. Just because the universe is expanding in an ‘irreversible’ way, doesn’t mean there ‘is’ a past or a ‘future’ – or time, or an arrow of time. Likewise , if things ‘just’ move and change etc , then ‘time’ is not emergent... things jjust move and change. Calling this ‘time’ and asking if it is emergent is (to me) like asking if ‘movement is emergent from movement’. I appreciate what you say about the ‘Tomita flow time’ (tho I haven’t read up on it yet to be honest), but when I consider the .. “subtle difference in the way we TREAT intervals of time and the changes that correspond to them” My position is that there are no ‘intervals’ of ‘time’. We can sit in front of a motorised hand that is rotating around a numbered dial (a clock) – but just because we are breathing and that hand is rotating does not prove that there are such things as ‘intervals of time’ – or that a thing called ‘time’ is ‘passing’. Its a odd concept to see if we are deeply ingrained that ‘time’ makes sense in some way, but if theres no past and no future, then there’s no time and ‘passage’ or intervals of it. And useful as the maths is, we may say t1+t2 =t3, and if time exists this may be valid, but in itself I don’t think this shows that ‘time’ exists, -ok, i’ve hit 1000 words again, and im duplicating points, so ill stop. Please don’t think i`m ignoring the things you have pointed to, these are just my views at the moment – having written my book on the subject. Sorry I haven’t. Addressed the issues you pointed to directly, I will read them in detail, and you're right I should (will) read earlier points of the thread, (just busy due to xmas), I just wanted to send this post before I take the xmas break. Matt marsden (brief history of timelessness) Recognitions: Gold Member Science Advisor Interesting thing about Barbour is if you look at what he is actually saying, e.g. http://arxiv.org/abs/0903.3489 he shows that a time variable does not have to be put in "by hand" at the beginning of the analysis, but can be DERIVED from observation. In this case he derives the evolution parameter from observing a dynamical system---a bunch of planets, or other bodies---stars, satellites, whatever. This derivation us purely classical = non-quantum. The Tomita method is analogous in the sense that the evolution parameter is DERIVED (from correlations among observations) rather than put in by hand at start of analysis. However it is not limited to classical systems--can be applied to quantum ones as well. As I think I said earlier in thread, I don't think "time does not exist" is an accurate headline for what Barbour and Connes-Rovelli are saying. "Exist" is a kind of fuzzy word anyway---more philosophy than science. Time remains a crucial indispensable and all-important element in their analysis, what they show is that it can be derived from other stuff. In a sense that makes time all the more real because you cannot avoid it. It is implicit in what we observe. As I said earlier, the headline "time does not exist" would be primarily an attention-getter, not an accurate summary of what Barbour or the others are saying. Perhaps a better (but less strikingly worded) summary would be "time is inherent in natural processes and can be derived from observation". Since I'm currently very interested in the strategies used to derive time from observations, it actually does NOT get my attention to say "time does not exist". It may work for other people, though. I think a frank answer, by Barbour, to the question does t exist (in thread title) would be YES INDEED it exists and it is very interesting how one can mathematically derive it! Mentor Thread closed temporarily for Moderation... Thread Closed Page 10 of 10 « First < 7 8 9 10 Thread Tools \| \| \| \| \|--------------------------------------------------------\|-----------------------------\|---------\| \| Similar Threads for: Julian Barbour on does time exist \| \| \| \| Thread \| Forum \| Replies \| \| \| General Discussion \| 56 \| \| \| Science Textbook Discussion \| 0 \| \| \| General Physics \| 1 \| \| \| Science Textbook Discussion \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 4, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9406009912490845, "perplexity_flag": "middle"}
http://nrich.maths.org/2654&part=	### Geoboards This practical challenge invites you to investigate the different squares you can make on a square geoboard or pegboard. ### Polydron This activity investigates how you might make squares and pentominoes from Polydron. ### Tiles on a Patio How many ways can you find of tiling the square patio, using square tiles of different sizes? # Ten Hidden Squares ##### Stage: 2 Challenge Level: On the graph below there are $34$ marked points. These points all mark the vertices (corners) of ten hidden squares. Each of the $6$ red points is a vertex shared by two squares. The other $28$ points are each a vertex of just one square. All of the squares share at least one vertex with another square. All the squares are different sizes. There are no marked points on the sides of any square, only at the vertices. (There are two near misses!) Can you find the ten hidden squares? You might like to play the game Square It which uses some similar ideas. 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.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9021376967430115, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/145121/how-to-manage-the-absolute-value-on-a-differential-equation-txat-xbt	# How to manage the absolute value on a differential equation $\|T(x)'+A(T,x)+B(T,x)\| = f(T,x)$ Hi everyone I need to solve an equation of this type: $\|T(x)'+A(T,x)+B(T,x)\| = f(T,x)$ with boundaries conditions. The absolute value is my problem. Of course without it, the solution of these is well treated in the literature. After search in the questions I found this: Differential equation with absolute value So, can I do the same procedure? Or there is another way to solve this? Thanks. - – user17762 May 14 '12 at 18:35 – user17762 May 14 '12 at 18:37 1 ok I deleted the other one. – Nikko May 14 '12 at 18:45 ## 1 Answer Your equation does not determine $T'$ in terms of $T$ and $x$, so uniqueness of solutions may be a problem. You can say $T'$ is either $f(T,x) - A(T,x) - B(T,x)$ or $-f(T,x) - A(T,x) - B(T,x)$. Presumably there will be one region where it's the first and one region where it's the second, and (assuming $f,A,B$ are continuous) if you want $T'$ to exist everywhere it'll be impossible to switch between one and the other except when $T' = 0$. - Thank you Robert. But can you explain me why you only take two combinations and not more? I mean why not $f(T,x)+A(T,x)−B(T,x)$ for instance ? – Nikko May 14 '12 at 18:47 $\|t\|$ is either $t$ or $-t$ (assuming we're talking about real numbers). – Robert Israel May 14 '12 at 19:07 Does not matter if A(T,x) and B(T,x) are negatives right?. Thank you very much! – Nikko May 15 '12 at 21:03 Right. Why would it matter? – Robert Israel May 16 '12 at 0:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9175039529800415, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/101814/points-on-deligne-lusztig-varieties-interpreting-borels-in-relative-position-as	## Points on Deligne-Lusztig varieties: Interpreting Borels in relative position as flags with conditions ### Background I am studying the paper "On the Green polynomials of classical groups" by Lusztig, in which he computes the values of the Deligne-Lusztig representation, corresponding to a Coxeter element of minimal length in a classical group, on unipotent elements. I am interested in computing the values of representations not corresponding to a Coxeter element of minimal length. (Note that this is done in a vast generalisation in later work by Lusztig, and in the work of Shoji. But I am not in a place to be able to use their methods) ### Question Let $G$ is a classical group defined over a finite field with frobenius morphism $F$, $w$ an element of the Weyl group, and let $X(w)$ be the Deligne-Lusztig variety - all Borel subgroups $B$ of $G$ such that $B$ and $F(B)$ are in relative position $w$. My specific question is: how do I translate this definition into the language of flags? I.e. I would like an alternative definition for $X(w)$ as the variety of flags satisfying some conditions involving to $F$. In the original Deligne-Lusztig paper, my question is answered, for the case $w$ a Coxeter element of minimal length, in a short section. - A small question about the formulation: What do you mean by "a Coxeter element of minimal length"? In a finite Coxeter group such as a Weyl group, all Coxeter elements `$w$` have the same length (equal to the rank of the given group) and are conjugate. I'll have to look at the original papers to sort out better what is going on here, but the terminology confuses me at first. – Jim Humphreys Jul 9 at 22:56 I am aware of this fact, and it confuses me as well. The words "a Coxeter element of minimal length" are in the third paragraph of the article I cite. – Dror Speiser Jul 9 at 23:00 @Dror: To clarify further, Lusztig is working with a special set-up just for certain classical linear groups. One Weyl group might be a subgroup of another (fixed by a twisted Frobenius map); thus a Coxeter element (in Steinberg's twisted sense) would be defined relative to the larger Weyl group, etc. The paper has lots of ad hoc notation and may or may not be a useful place to start. It was published in Proc. LMS 33 (1976) at a very early stage of what turned out to be a vast project to find all character values of finite groups of Lie type. This paper focuses on a "Coxeter torus". – Jim Humphreys Jul 10 at 14:14 ## 1 Answer It might help if you specified what part of the translation your having trouble with (for example, it's unclear from what you wrote if you're comfortable with relative position). For $SL_n$, a Borel corresponds to a flag in affine $n$-space over an extension of your field (essentially, the one you need to diagonalize the elements of a torus in the Borel). Applying the Frobenius means doing it to the coordinates on $n$ space, and looking at the resulting flag. Relative position w just means that the dimensions of the intersections of the pieces of the flag are the same as those between the standard flag, and that gotten by per muting the basis by $w$. - Indeed, I was having trouble with what relative position means in terms of flags, but your last sentence solved that. Thank you very much. – Dror Speiser Jul 10 at 7:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9218564629554749, "perplexity_flag": "head"}
http://mathoverflow.net/questions/90875?sort=oldest	## Extreme points of unit ball in tensor product of spaces ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $B_1, B_2$ be unit balls in finite-dimensional normed spaces $X_1, X_2$ respectively. Let $e(B_1), e(B_2)$ be corresponding extreme points sets. Consider the unit ball $B$ in tensor product $X_1\otimes X_2$ with the largest (projective) cross-norm on it. Can we say that extreme points of $B$ in tensor product are exactly tensor products of extreme points for $B_1, B_2$, i.e. $e(B)=\{u\otimes w: u\in e(B_1), w\in e(B_2)\}$? This seems plausible, but things are not looking very straightforward. In particular, opposit pairs of extreme points produce the same point in tensor product, i.e. $(-u)\otimes (-w) = u\otimes w$. - ## 2 Answers See [11] Ruess, W.M. and Stegall, C.P., Extreme points in duals of operator spaces, Math. Ann., 261 (1982), 535–546. They prove what you want in a more general context: If $X$, $Y$ are Banach spaces s.t. either `$X^$` or `$Y^$` has the approximation property and either `$X^$` or `$Y^$` has the Radon-Nikodyn property, then the extreme points of the unit ball of the projective tensor product of `$X^$` and `$Y^$` are the tensor products of extreme points of the respective unit balls. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There is a hierarchy of notions related to the notion of extreme point, among them strong extreme point, point of continuity, denting point, as sketched in BOR-LUH LIN, PEI-KEE LIN AND S. L. TROYANSKI. CHARACTERIZATIONS OF DENTING POINTS. PROC. OF AMS. Volume 102, Number 3, March 1988, p.526-528 (http://www.ams.org/journals/proc/1988-102-03/S0002-9939-1988-0928972-1/S0002-9939-1988-0928972-1.pdf). The corresponding statement is true for denting points as shown in DIRK WERNER. DENTING POINTS IN TENSOR PRODUCTS OF BANACH SPACES. PROC. OF AMS. Volume 101, Number 1, September 1987, p. 122-126 (http://www.ams.org/journals/proc/1987-101-01/S0002-9939-1987-0897081-1/S0002-9939-1987-0897081-1.pdf). Now, it's enough to note that notions of extreme point and denting point are equivalent in my context. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8524847626686096, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/258304/how-to-determine-the-length-of-the-wider-base-of-a-trapezoid-from-the-height-s/258305	# How to determine the length of the wider base of a trapezoid from the height + shorter base length + angles? Given a trapezoid like the one shown below, how do I determine the length of the wider base? I'm looking for a formula based method rather than drawing the shape and measuring it. - ## 1 Answer Use the Law of Sines on the triangles (separately) on either end. For example, on the right side of the trapezoid, drop a perpendicular from the top right vertex. Call the length of the base of the resulting right triangle $x$. The other (non-hypotenuse) side is 3/4. The bottom (interior) angle is $70^\circ$ while the top (interior) angle is $20^\circ$. The Law of Sines says $${{3\over 4}\over \sin(70^\circ)}={x\over \sin(20^\circ)}.$$ Solve for $x$ to obtain $x={3\over 4}\cdot{\sin(20^\circ)\over \sin(70^\circ)}\approx 0.272978$. Play the same game on the other side and you will know the total length of the base of the trapezoid. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.842074453830719, "perplexity_flag": "head"}
http://www.sciforums.com/showthread.php?108001-Destructive-interference-where-does-the-energy-go	• Forum • New Posts • FAQ • Calendar • Ban List • Community • Forum Actions • Encyclopedia • What's New? 1. If this is your first visit, be sure to check out the FAQ by clicking the link above. You need to register and post an introductory thread before you can post to all subforums: click the register link above to proceed. To start viewing messages, select the forum that you want to visit from the selection below. # Thread: 1. ## Destructive interference - where does the energy go? Consider two counter-propagating waves on a string with equal frequencies but 180 degrees out of phase so that there is complete destructive interference. Each wave carries energy - one to the left and the other to the right, let's say. But when they interfere we get a stationary string and, apparently, no energy. Where did the energy go? I'd like to nail down the answer to this question. If you look on the internet, there are all kinds of different explanations. Let's see if we can do better at sciforums. 2. Just as a sky blue guess, I'll venture that it shows up as heat. Instead of string. lets use a flexible strip of superconducting materiel and measure any change in temperature. 3. The string doesn't move, so there's no mechanism for generating heat. In any case, let's consider an "ideal" string that doesn't have any friction forces on it if/when it moves. 4. Originally Posted by James R The string doesn't move, so there's no mechanism for generating heat. In any case, let's consider an "ideal" string that doesn't have any friction forces on it if/when it moves. I'll give you a hint: destructive interference always happens in conjunction with constructive interference. So, where does the energy go? 5. I don't want a hint, Tach. I want an answer. I've read other internet answers. 6. Haha, this is a fun question, I remember wondering about it a few years back. I was worrying about quantum interference but I guess it is a concern classically too. My first question would be how did you manage to create these waves in the first place? It doesn't take any energy to create the situation you describe, it is just a motionless string. You will not be able to create even a thought experiment in which you can achieve this configuration of waves from some sources I think. 7. Originally Posted by James R I don't want a hint, Tach. I want an answer. I've read other internet answers. You have your answer, you just need to think about it. It is right in front of your nose. 8. Originally Posted by AlexG Just as a sky blue guess, I'll venture that it shows up as heat. Instead of string. lets use a flexible strip of superconducting materiel and measure any change in temperature. No heat is being generated. 9. Originally Posted by James R The string doesn't move Are you sure about that? When you say two waves are propagating in opposite directions, I think of something like cos(x - ct) + cos(x + ct) = 2 cos(x) cos(ct), which does move. Am I missing something? 10. Originally Posted by przyk Are you sure about that? When you say two waves are propagating in opposite directions, I think of something like cos(x - ct) + cos(x + ct) = 2 cos(x) cos(ct), which does move. Am I missing something? Yeah I think he would prefer they travel in the same direction: $cos(kx+\omega t) + cos(kx+\omega t + \pi/2) = 0$ 11. przyk: Originally Posted by przyk Are you sure about that? When you say two waves are propagating in opposite directions, I think of something like cos(x - ct) + cos(x + ct) = 2 cos(x) cos(ct), which does move. Am I missing something? Hmm.... Maybe I need to alter the scenario to consider co-propagating waves that are out of phase by 180 degrees. So, to spell it out, what is your answer for the counterpropagating case you describe? And then, what about the co-propagating case? kurros: Originally Posted by kurros Haha, this is a fun question, I remember wondering about it a few years back. I was worrying about quantum interference but I guess it is a concern classically too. Yes. You can think about it with waves on a string, water waves, light, sound, matter waves, or whatever. My first question would be how did you manage to create these waves in the first place? It doesn't take any energy to create the situation you describe, it is just a motionless string. You will not be able to create even a thought experiment in which you can achieve this configuration of waves from some sources I think. Well, co-propagating case. At the left end of the string, I have a shaker that creates a sinusoidal oscillation on the string that moves to the right (say). A little way to the right of that, there's another shaker that applies a force to the string to create the other wave, which also propagates to the right. Turn on one shaker and leave the other one off and you have a wave moving right along the string. Turn both on and suddenly there's no wave to the right of the second shaker. But the shakers are still shaking away (I assume). So what's happening to the energy they are supplying? Or is this somehow impossible. Tach: Originally Posted by Tach You have your answer, you just need to think about it. It is right in front of your nose. Once again, I have to doubt that you actually have an answer. You seem quite incapable of giving anything but a vague response to physics questions. Just leave this thread alone if you don't know. 12. James I don't know if this is the same but if you get 2 people ( yes I know this is practically impossible, its a thought) in a vacume to push against eachother with the same force where does the energy go. Its not sound because they are in a vacume, its not light and I doubt its heat. If done properly it shouldn't deform. 13. Originally Posted by kurros Haha, this is a fun question, I remember wondering about it a few years back. I was worrying about quantum interference but I guess it is a concern classically too. My first question would be how did you manage to create these waves in the first place? It doesn't take any energy to create the situation you describe, it is just a motionless string. You will not be able to create even a thought experiment in which you can achieve this configuration of waves from some sources I think. I've no idea what you're trying to say. It's an easy experiment to do, it DOES require energy and the string most certainly DOES move. It's not just a thought experiment, any kid of 5 can do the real thing. The answer is that you will always wind up with standing waves - motionless nodes and sections between those nodes that vibrate back and forth. And the final bit is that YES, the energy in the string is converted into heat. 14. Originally Posted by James R Consider two counter-propagating waves on a string with equal frequencies but 180 degrees out of phase so that there is complete destructive interference. I don't quite get the picture. Do you mean this: Or this: 15. Originally Posted by Read-Only I've no idea what you're trying to say. It's an easy experiment to do, it DOES require energy and the string most certainly DOES move. It's not just a thought experiment, any kid of 5 can do the real thing. The answer is that you will always wind up with standing waves - motionless nodes and sections between those nodes that vibrate back and forth. And the final bit is that YES, the energy in the string is converted into heat. Once the wave is set up, it doesn't need energy to maintain, except to replace the damping of air resistance and internal friction. A standing wave is a (kind of inefficient)! energy storage medium. 16. Originally Posted by James R co-propagating case. At the left end of the string, I have a shaker that creates a sinusoidal oscillation on the string that moves to the right (say). A little way to the right of that, there's another shaker that applies a force to the string to create the other wave, which also propagates to the right. Turn on one shaker and leave the other one off and you have a wave moving right along the string. Turn both on and suddenly there's no wave to the right of the second shaker. But the shakers are still shaking away (I assume). So what's happening to the energy they are supplying? It might not work, depending on the way the shakers work. The second shaker has to be able to absorb the energy of the first, because the force it applies is always in the opposite direction to displacement, so the energy expended is negative. 17. Originally Posted by James R kurros: Well, co-propagating case. At the left end of the string, I have a shaker that creates a sinusoidal oscillation on the string that moves to the right (say). A little way to the right of that, there's another shaker that applies a force to the string to create the other wave, which also propagates to the right. Turn on one shaker and leave the other one off and you have a wave moving right along the string. Turn both on and suddenly there's no wave to the right of the second shaker. But the shakers are still shaking away (I assume). So what's happening to the energy they are supplying? Or is this somehow impossible. Ok well I have not gone off and solved the 1D wave equation with 2 dirac delta in x but sinusoidal in t source functions, but I expect that the solution will be that yes, the string does nothing outside the two shakers, but inside plenty happens so all the energy is contained there. You can only put the shakers at integer wavelength positions to get the waves to cancel on both sides. Admittedly the amplitude of the interior waves won't increase as the shakers move closer together, which seems bad. Hmm. I might have to simulate this to convince myself. Actually maybe the amplitude of the interior wave WILL increase, we have a standing wave in there with continual energy input, I guess the solution will blow up... yes I think this is what I am going to go with. Anyway if you place the two sources directly over each other they will cancel themselves out so you will be back to the homogeneous wave equation. I have mostly thought about 2d or 3d scenarios and disallowed myself infinite sources, which is what the 2d equivalent of your scenario would required. I'll get back to you. 18. Originally Posted by James R Where did the energy go? Into the string. See this website and look at the two waves travelling in opposite directions. One is green, the other is blue. When they're exactly out of phase the red superpositional wave is a straight line. At that instant the string is flat. Freeze that instant in your mind. If you imagine the string to be made up of a series of spheres connected by coil-springs under tension, then at the flat instant, the tension in the coil springs is reduced. The dimensionality of energy is pressure x volume, tension is negative pressure, and the energy in the string has reduced its tension. If the string was stretched over an archer's bow, the ends of the bow would be starting to move apart at this point. 19. Originally Posted by Pete Once the wave is set up, it doesn't need energy to maintain, except to replace the damping of air resistance and internal friction. A standing wave is a (kind of inefficient)! energy storage medium. Precisely. And it's created by the reflection at the far end of the string - which is not a proper terminator (which would allow the energy to pass out of the string) with even a single source of wave input. The second input, as stated in the OP, simply adds even more energy to the system. But nevertheless, the cancellation of the waves will result in creating standing waves - where the energy will reside (just as you said) until it's eventually converted to friction by the air resistance. Exactly what I said earlier, just with more detail. 20. Asguard: Originally Posted by Asguard James I don't know if this is the same but if you get 2 people ( yes I know this is practically impossible, its a thought) in a vacume to push against eachother with the same force where does the energy go. Its not sound because they are in a vacume, its not light and I doubt its heat. If done properly it shouldn't deform. This is different. With forces, no energy is ever expended unless the object the force acts on moves in some way. So, if two people push on each other with equal forces, no useful work is done and the people don't move. They will be straining their muscles, and with all the muscle twitching going on in their bodies there will be movement there. That movement ultimately produces heat. So, the net conversion in that situation is from chemical energy to mechanical to heat in the muscles. Pete: Originally Posted by Pete I don't quite get the picture. Do you mean this: Or this: I can't see the second image - it gives me a "no permission" message. Originally Posted by Pete It might not work, depending on the way the shakers work. The second shaker has to be able to absorb the energy of the first, because the force it applies is always in the opposite direction to displacement, so the energy expended is negative. That makes some kind of sense. So the "missing" energy is perhaps absorbed by the sources. Farsight: Originally Posted by Farsight Into the string. See this website and look at the two waves travelling in opposite directions. One is green, the other is blue. When they're exactly out of phase the red superpositional wave is a straight line. At that instant the string is flat. Freeze that instant in your mind. If you imagine the string to be made up of a series of spheres connected by coil-springs under tension, then at the flat instant, the tension in the coil springs is reduced. The dimensionality of energy is pressure x volume, tension is negative pressure, and the energy in the string has reduced its tension. If the string was stretched over an archer's bow, the ends of the bow would be starting to move apart at this point. I'm not sure exactly what you're saying here. In what form is the wave energy when the string is flat? ---- For those who are worried about infinite waves extending off in all directions, consider the following, perhaps simpler, case: Two identical pulses propagate along a string, but one is upside-down compared to the other, as here: (Thanks, Pete.) My question is now: In the centre image, where the pulses exactly overlap and the string is flat, where is the energy that was in the two pulses in the image immediately before that one? And where does the energy come from to reconstruct the two pulses after the time when the string is flat? In this situation, let us assume that the string is very long and the sources of the original pulses are very far away from where the pulses overlap. #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts • • BB code is On • Smilies are On • [IMG] code is On • [VIDEO] code is On • HTML code is Off Forum Rules All times are GMT -5. The time now is 04:20 AM. sciforums.com	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9521893262863159, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/131652/how-to-measure-how-erratic-is-a-function-between-a-and-b?answertab=oldest	# How to measure how erratic is a function between a and b I need to compare the outputs of some functions and to rate their "erratness". Given a function, the less erratic function between a and b would be a straight line and the more erratic function would probably be a triangular or sine wave, the bigger its amplitude or its frequency the bigger their erratness. I'm not sure if it's clear what I mean. I've thought that a way to calculate it could be to calculate the length of the line generated by the function between a and b. The smaller the length the less erratic will be the function. The questions are: 1. Do you think this is a good way to achieve what I need? 2. How can the length of the output of a function be calculated? Thanks in advance. - You could look up the arclength integral... – J. M. Apr 14 '12 at 9:21 No idea on what are you talking about :) (I'm not a mathematic) but I will take a look at wikipedia. Thanks! – SoMoS Apr 14 '12 at 9:29 Right, so type the words "arclength" and "integral" into your favorite search engine, and enjoy. Note that most such integrals do not have nice simple forms, and numerics would be needed. – J. M. Apr 14 '12 at 9:32 Do you mean that they cannot be calculated by an algorithm? I'm developing software. – SoMoS Apr 14 '12 at 9:38 1 Stock Trading ... – SoMoS Apr 14 '12 at 9:52 show 4 more comments ## 2 Answers There is something called the total variation of a function. If the function is differentiable, it's computed as $\int_a^b\|f'(x)\|\,dx$. There are standard methods for computing the derivative and the integral numerically. If the function is not differentiable, there's a different definition, which should also be suitable for numerical computation. See the discussion at http://en.wikipedia.org/wiki/Total_variation. - You rock man. That's what I was looking for and the wiki explanation is great. Thanks again. – SoMoS Apr 14 '12 at 20:25 If the end-use is an algorithm that must run quickly, your best bet is probably the sum squared difference (or SSD). If the line running between points $a$ and $b$ is: $$y = mx +n$$ Then you want to sum the squared difference between each point $(x_i,y_i)$ and the line: $$SSD\equiv\sum{(y_i - mx_i -n)^2}$$ The smaller the $SSD$, the less "erratic" your function. Also note that if you have many points and not many computer cycles available, you may want to choose a small random subset of the points instead of summing over all of them. - Looks as a good way. I specially like the quadratic feature to penalize more the more erratic points. Thanks, you rock too. – SoMoS Apr 14 '12 at 20:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9409230947494507, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/54481/why-is-linear-independence-of-harmonic-oscillator-solutions-important?answertab=active	# Why is linear independence of harmonic oscillator solutions important? The equation of motion for the harmonic oscillator (mass on spring model) $$\frac{d^2x}{dt^2} + \omega_0^2 x = 0$$ with $\omega_0^2 = D/m$, $D$ and $m$ being the force constant of the spring and the mass, has the solution $$x=ce^{\lambda t}$$ where $c$ and $\lambda$ are a constant and a parameter. Inserting $x$ leads to $$\lambda_1 = +i\omega_0$$ and $$\lambda_2 = -i\omega_0$$ and so the solutions are $$x_1(t) = c_1 e^{i \omega_0 t}$$ and $$x_2(t) = c_2 e^{-i \omega_0 t}.$$ In my book, I now read "...these solutions are linearly independent for $\omega_0 \neq 0$." What does this mean (i.e. how can I see this) and why is it important? - ## 4 Answers They would be linearly dependent if and only if there exist complex numbers $\alpha$ and $\beta$ such that $\alpha x_{1}(t) + \beta x_{2}(t) = 0 \forall t$ Clearly, if $\omega_{0}=0$ then this is the case for $\alpha = 1$ and $\beta = -c_{1}/c_{2}$. So then they are linearly dependent. However, if $\omega_{0}\neq0$, you can't find a combination of $\alpha$ and $\beta$ that fulfills this requirement for all $t$. The importance lies in the fact that (1) Any linear combination $\gamma x_{1}(t) + \delta x_{2}(t)$ of the two functions is also a solution. (Just plug the linear combination into the equation to see this.) (2) These are the only solutions. Namely, if you would find a solution $y(t)$ you could always write it as a combination of $x_{1}$ and $x_{2}$. So these are the only solution to care about, all the dynamics of the system is contained in them. - Linearly independent means that the equation $$\lambda_1x_1(t) + \lambda_2 x_2(t) =0$$ has for unique solution $\lambda_1=\lambda_2=0$ for $\omega_0 \ne 0$. Otherwise, your equation being linear you can see (and it maybe is the easiest view) the solution space as a vector space, of which your $x_1(t)$ and $x_2(t)$ are one basis. - It is like $\sin(\omega t)$ cannot be reduced to $\cos(\omega t),$ i.e., obtained by multiplying by a coefficient, because they are different functions - they are not proportional to each other. Existence of two independent solutions is also like existence of two independent integration constants for a second order differential equation. Using both of them guarantees that you can cover a general case and miss nothing. - It means, that you can't produce one solution out of a linear combination of the other. This is important because if you could gain a solution produced by superposition of other known solutions, it is actually not a new solution. It's information is already stored in the other known solution and therefor not relevant. So here the author wants to tell you: These are two different solutions. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 7, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9611107110977173, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/41379/how-do-i-interpret-the-results-of-a-regression-which-involves-interaction-terms	# How do I interpret the results of a regression which involves interaction terms? If I have a linear regression of the form: $$y \sim 1+\beta_1x_1 + \beta_2x_1x_2,$$ where $x_2$ is a Boolean variable depending on another variable $x_{2cont}$, a positive variable: $$x_2 = \begin{cases} 1 & \mbox{if } x_{2cont} > 5 \\ 0 & \mbox{if } 0 < x_{2cont} \le 5. \end{cases}$$ And I got regression results. How shall I interpret the regression results? esp. how shall I interpret the two coefficients I get from this regression, $\beta_1 \text{ and } \beta_2$? - 2 Perhaps you want to give this question a shot yourself and people can chime in with additional insight? I also recommend avoiding using the letter B as a variable name because it usually represents a coefficient, for consistency stick with x1, x2, x3, etc. Finally, you should be default include both terms in the interaction independently as well, which would give you 3 coefficients. – Michael Bishop Oct 28 '12 at 21:11 I fixed up your formatting, in part to match @MichaelBishop 's excellent suggestions. In addition, you shouldn't turn a continuous variable into a dichotomous one except under very unusual circumstances and you should include an intercept except under very unusual circumstances. – Peter Flom Oct 28 '12 at 21:20 Thanks guys for your suggestions. I've added the intercept terms. Yes, indeed I did it with the intercept term. But I couldn't figure the interpretation... could you please help me? Thank you! – Luna Oct 28 '12 at 21:49 ## 1 Answer $\beta_1$ describes how $y$ changes for a one-unit change in $x_1$ if $x_2 = 0$ The sum of $\beta_1$ and $\beta_2$ describes how $y$ changes for a one-unit change in $x_1$ if $x_2 = 1$ $\beta_2$ describes the difference in the change in $y$ per one-unit change in $x_1$ between $x_2 = 0$ and $x_2 = 1$ I think your notation is still not standard. Also, according to the principle of marginality you should include all main effects of the interactions you include, so here this means that a main effect for $x_2$ should be included (to estimate the part of the effect of $x_2$ that is independent of that of $x_1$). I think your model should look something like $E(Y\|X) = \beta_0 + \beta_1X_1 + \beta_2X_2 + \beta_3X_1X_2$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9462502598762512, "perplexity_flag": "middle"}
http://nrich.maths.org/6719	### Pebbles Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time? ### It Figures Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all? ### Bracelets Investigate the different shaped bracelets you could make from 18 different spherical beads. How do they compare if you use 24 beads? # Three Dice ##### Stage: 2 Challenge Level: Take a look at some ordinary dice. What do you notice about the way the numbers are arranged? Now look at these three dice in a row: The numbers on the tops of the dice read $6$, $1$ and $5$. What do the numbers on the top add up to? Can you use what you found out about the way the numbers are arranged to say what numbers are on the bottom of the dice? Were you correct? What is the sum of the numbers on the bottoms of the dice? Let's try that again. This time the numbers on the top read $1$, $4$ and $3$. Can you work out the total? Can you work out the numbers on the bottom and their total? Try out some arrangements yourself. Each time record the sum of the numbers on the top and the sum of the numbers on the bottom. Do you notice a relationship between the 'top sum' and the 'bottom sum'? Can you explain it? I experimented with arrangements where the top sum is a multiple of three, and find that in each case the bottom sum is also a multiple of three. Is it always true? I try to arrange the dice so that the top and bottom sums are both multiples of four, but can't seem to be able to do it. Can you? Can you explain what you find out? On the other hand, if I arrange four dice in a row it is easy to make the top and bottom sums both multiples of four. Can you arrange four dice so that the top and bottom sums are both multiples of three? Can you explain what you find out? 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.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9314746856689453, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/321543/proving-that-2-3-sqrt-2-is-reducible-in-mathbbz-sqrt-2	# Proving that $2 + 3\sqrt{-2}$ is reducible in $\mathbb{Z}[\sqrt{-2}]$ Prove that $2 + 3\sqrt{-2}$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ So far, I have let $2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2})$ I then took the norm and got $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2)$ I think I must then split 22 into $(2)(11)$ but I don't know how to proceed from there. Help is much appreciated! Note: I originally posed the question as proving it was irreducible. Apologies if I sent people down the wrong track in the answers below! Thank you again for the help. - 1 Are you certain it is irreducible? What if you divide by $\sqrt{-2}$? – trb456 Mar 5 at 16:03 Just checked with my lecturer - I was wrong, it is reducible. I'll edit the question title accordingly. Thanks anyway! – user49394 Mar 5 at 16:09 @user49394, I posted an answer that shows how you could prove irreducibility .. but it goes wrong because the thing can be factored: but the way it goes wrong produces a factorization! – user58512 Mar 5 at 19:10 ## 3 Answers Let $2 + 3\sqrt{-2} = (a + b\sqrt{-2})(c+ d\sqrt{-2})$ then $\mathbf{N}(2 + 3\sqrt{-2}) = 22 = (a^2 + 2 b^2)(c^2 + 2 d^2)$ That is a good way to start! Now we just need to show $a^2+2b^2 = 2$ and $c^2+2d^2 = 11$ is impossible, but the first part is possible so we need to show $c^2+2d^2 = 11$ is impossible: This is easy, let's just write out all numbers of the form $x^2+2y^2$: $$\begin{array}{\|c\|c\|c\|} \hline 0 & 2 & 8 & 18 \\ \hline 1 & 3 & 9 & 19 \\ \hline 4 & 6 & 12 & \\ \hline 9 & \color{red}{11} & & \\ \hline \end{array}$$ So we have a factorization from the $x=1,y=3$ box which is $3^2 + 2\cdot 1^2 = 11$. $$2(3+\sqrt{-2})(3-\sqrt{-2})$$ - I was trying to prove it was irreducible .. but then I noticed it factored.. I left this answer because it shows how to prove irreducible. – user58512 Mar 5 at 16:14 Find the elements in $\mathbb{Z}[\sqrt{-2}]$ of norm $2$ (they are $\pm i \sqrt{2}$) and $11$ (they are $\pm 3 \pm i \sqrt{2}$), and check whether one of the possibile products will give you $2 + 3\sqrt{-2}$. (Of course @trb456 has already give you a hint.) Then if you want you may use the fact that if the norm of an element is a prime integer, then the element is irreducible. This will show that the two factors that you have found are irreducible, so $2 + 3\sqrt{-2}$ is the product of two irreducibles. - Hint $\rm\,\ ad+b\sqrt{d}\, =\, \sqrt{d}\,(a\sqrt{d}+b)$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 24, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9538058042526245, "perplexity_flag": "head"}
http://mathoverflow.net/questions/51187?sort=newest	What is the generic poset used in forcing, really? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm not a set theorist, but I understand the 'pop' version of set-theoretic forcing: in analogy with algebra, we can take a model of a set theory, and an 'indeterminate' (which is some poset), and add it to the theory and then complete to a model with the desired properties. I understand the category theoretic version better, which is to take sheaves (valued in a given category $Set$) over said poset with a given Grothendieck topology (the double negation topology). The resulting topos is again a model of set theory, but now has the properties you want, absent from the original $Set$. But what is this poset, really? Is it the poset of subobjects of the set you want to append to your theory/model (say a set of a specified cardinality, or some tree with a property)? Is it related to a proof of the property you are interested in? To clarify, I'm not interested in the mechanical definition of an appropriate generic poset, but what it is morally. Bonus points for saying what it 'is' before and after forcing, if this even makes sense. - 1 Let me say something about the analogy that you mention (it is typical to compare forcing extensions to field extensions, and certainly it may help build some intuition to pursue the analogy). There is a serious discrepancy though: Given a typical field, there are several isomorphic fields that realize a concrete field extension. In set theory this does not happen, because (transitive) models are rigid. In particular, different generic objects for the same poset always give us non-isomorphic extensions. – Andres Caicedo Jan 6 2011 at 0:07 1 In regard to Stefan's answer, see related: mathoverflow.net/questions/48522/…. I think boolean-valued models form a very elegant and intuitive approach to forcing. – Jason Jan 6 2011 at 2:06 5 Answers The other answers are excellent, but let me augment them by offering an intuitive explanation of the kind you seem to seek. In most forcing arguments, the main idea is to construct a partial order out of conditions that each consist of a tiny part of the generic object that we would like to add; each condition should make a tiny promise about what the new generic object will be like. The generic filter in effect provides a way to bundle these promises together into a coherent whole having the desired properties. For example, with the Cohen forcing $\text{Add}(\omega,\theta)$, we want to add $\theta$ many new subsets of $\omega$, in order to violate CH, say. So we use conditions that specify finitely many bits in a $\omega\times\theta$ matrix of zeros and ones. Each condition makes a finite promise about how the entire matrix will be completed. The union of all conditions in the generic filter is a complete filling-in of the matrix. Genericity guarantees that each column of this matrix is a new real not present in the ground model and different from all other columns, since any finite condition can be extended so as to disagree on any particular column with any particular real or from any particular other column. With the collapse forcing $\text{Coll}(\omega,\kappa)$, we want to add a new surjective function $f:\omega\to\kappa$. So we use conditions consisting of the finite partial functions $p:\omega\to\kappa$, ordered by extension. Each such condition is a tiny piece of the generic function we want to add, describing finitely much of it. The union of the generic filter provides a total function $g:\omega\to \kappa$, and the genericity of the filter will guarantee that $g$ is surjective, since for any $\alpha<\kappa$, any condition $p$ can be extended to a stronger condition having $\alpha$ in the range. And similarly with many other forcing arguments. We design the partial order to consist of tiny pieces of the object that we are trying to add, each of which makes a small promise about the generic object. If $G$ is a generic filter for this partial order, then the union of $G$ is the joint collection of all these promises. In many forcing arguments, it is not enough just to build a partial order consisting of tiny pieces of the desired object, since one also wants to know that the forcing preserves other features. For example, we want to know that the forcing does not inadvertently collapse cardinals or that it can be iterated without collapsing cardinals. This adds a wrinkle to the idea above, since one wants to use tiny pieces of the generic object, but impose other requirements on the conditions that will ensure that the partial order has a nice chain-condition or is proper and so on. So the design of a forcing notion is often a trade-off between these requirements---one must find a balance between simple-mindedly added pieces of the desired generic object and ensuring that the partial order has sufficient nice properties that it doesn't destroy too much. In this sense, I would say that the difficult part of most forcing arguments is not the mastery of the forcing technology, the construction of the generic filter and of the model---although that aspect of forcing is indeed nontrivial---but rather it is the detailed design of the partial order to achieve the desired effect. - 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 add some view that hasn't really been mentioned above, namely that of Boolean valued models. Every poset has a completion, which is a complete Boolean algebra. To each poset we can define a forcing language that essentially looks like first order logic, only that we use so called names (which depend on the poset and stand for elements of the forcing extension) as constants. Now the elements of the complete Boolean algebra can be considered as truth values of statements in the forcing language. Generic filters of the poset and its completion are in 1-1 correspondence to each other. A filter in the complete Boolean algebra corresponds to a consistent theory in the forcing language. So, choosing a generic filter for the poset over a model of set theory corresponds to choosing a "generic" theory in the forcing language, the theory consisting of statements in the forcing language that are true in the corresponding generic extension. Actually, with this approach it is not even necessary to really go to a generic extension at all: If you want to show that CH is not provable in ZFC, all you need to do is construct a poset so that in the corresponding forcing language all ZFC axioms have truthvalue 1 in the corresponding complete Boolean algebra and CH does not. Since everything that follows from ZFC has truthvalue 1 as well, CH does not follow from ZFC. Now, of course, the computations and arguments you have to do to push through this program are just as difficult as in the case of the usual approach. - 1 Just to mention the connection between this and the topos-theoretic version: the topos of double-negation sheaves on the poset is equivalent to the topos of sheaves on its "completion" Boolean algebra regarded as a locale. – Mike Shulman Jan 5 2011 at 22:29 In topos-theoretic terms, I like to describe forcing as follows. You consider a geometric theory whose models are the objects you want to adjoin. For instance, if you want to adjoin an injection from a set A to a set B, then you consider the propositional geometric theory whose basic propositions are of the form $[f(a)=b]$" for some a∈A and b∈B, and whose axioms are • $\vdash \bigvee_b [f(a)=b]$ for each a (i.e. "f is defined everywhere") • $[f(a)=b_1]\wedge [f(a)=b_2] \vdash \bot$ whenever $b_1\neq b_2$ ("f is single-valued") • $[f(a_1)=b] \wedge [f(a_2)=b] \vdash\bot$ whenever $a_1\neq a_2$ ("f is injective") If A has a larger cardinality than B, then this theory has no models in Set. But we can still consider its classifying topos, which may have no points. Like the classifying topos of any theory, this classifying topos contains a "universal model" of the theory (if you like, by the Yoneda lemma—a representing object x of any functor F is determined by a universal element of F(x).). A convenient way to construct the classifying topos of a propositional geometric theory is to consider the poset of finite conjunctions of basic formulae (which is the free finitely-complete category generated by the theory) and equip it with a Grothendieck topology coming from the axioms. In the above example, a finite conjunction of basic propositions is a finite relation from A to B. We can discard those which are not single-valued and injective since the axioms tell us that those are covered by the empty family; thus we obtain a poset of "partial injections," looking like those that set-theorists usually talk about. Finally, if you're a classical set-theorist and want to also have the law of excluded middle, you can consider a Boolean subtopos of this classifying topos. One obvious Boolean topos to look at is the topos of double-negation sheaves on the above poset. Both this and the classifying topos are subtoposes of the presheaf topos on the poset, and if you're lucky (I don't know the exact conditions necessary), then the double-negation subtopos will be contained in the classifying topos, and therefore it will also contain a model of the theory in question (since it has a map to the classifying topos of the theory). - First, a point of clarification: you're not adding the poset but rather forcing with it, and the poset itself remains unchanged after the forcing. The generic object that you're adding is actually a filter on this set meeting every dense subset (or maximal antichain) of the partial order. The elements of the partial order, often called conditions in the context of forcing, form the possible components and the generic filter is responsible for collecting them in such a way to add an object of the desired form. In this way, the poset is highly correlated with the object you want to add because if you don't have the correct building blocks, there is no way to filter out the ones you want to use to construct an object of the desired form. The key property of the generic filter is that it consists of a compatible collection of conditions simultaneously meeting every dense subset. You can think of these dense subsets as the individual properties that we are going to want this filter to have. This is all better illustrated with examples: Forcing To Add $\kappa$ many Cohen Reals: Here your forcing poset consists of finite partial functions from $\omega \times \kappa$ into {0, 1} ordered by reverse inclusion (longer is stronger). The conditions are the finite pieces of the $\kappa$ many Reals (subsets of $\omega$) that you're adding. By virtue of consisting of a compatible set of conditions meeting all dense subsets, a generic filter $G$ will pick them out in such a way that $\bigcup G: \omega \times \kappa \rightarrow \{0, 1\}$ is a total function with all of its $\kappa$ columns representing a newly added distinct Real. Specifically, the filter part insures that these conditions can be put together to form a partial function while the generic part makes sure that we've met all of the conditions imposed by the dense subsets, which include making sure the function is total, and that every column defines a distinct Real that's different from every one in the ground model. Because of the simplicity of the forcing poset, the cardinals between the ground model and the forcing extension are the same so if $\kappa = \omega_2$, then you've really forced $\lnot$CH to hold in the extension. Forcing to Collapse a Cardinal $\lambda$ to have size $\kappa$: Here we force with the partial order consisting of partial functions from $\kappa$ into $\lambda$, each having size strictly less than $\kappa$. Again the partial order is ordered by reverse inclusion so the more the domain is filled in, the stronger the condition (lower in the partial order). A generic filter $G$ will again select these partial functions in such a way that $\bigcup G: \kappa \rightarrow \lambda$ is a total function. By virtue of meeting all dense subsets, $\bigcup G$ will have all elements from $\lambda$ in its range so it will be a newly added surjection. Consequently, if $\lambda > \kappa$ for $\kappa$ and $\lambda$ cardinals in the ground model (pre-forcing), then $\lambda$ will have been collapsed to be a $\kappa$-sized ordinal in the forcing extension. Because of the poset's closure, we didn't add any ${<} \kappa$-sized subsets of objects from the ground model or collapse any cardinals at or below $\kappa$, and we sometimes use these facts or even stronger properties to argue about truth in the ground model by virtue of truth in the forcing extension. Sacks forcing: Here we add a "minimal Real" by forcing with the partial order consisting of perfect trees of finite binary sequences ordered by inclusion. A generic filter $G$ will thus consist of a collection of trees with larger and larger stalks so that $\bigcap G$ is a new branch through the tree ${}^{\omega}2$ that can be associated with a new Real. Prikry forcing: If we have a nonprincipal normal $\kappa$-complete measure on a cardinal $\kappa$, then we can force with the collection of conditions of the form $\langle s, A\rangle$ where $s$ is a finite sequence of ordinals from $\kappa$ and $A$ is a subset of $\kappa$ from the normal measure. $\langle t, B\rangle \leq \langle s, A\rangle$ if $B \subseteq A$ and $t$ is an end extension of $s$ only adding ordinals from $A$ to the range of $t$. Here we won't take the union or the intersection of the generic filter $G$ but only the union of the finite sequences of ordinals in $\kappa$ from the first coordinates of the conditions in $G$. Again, the filter will make sure that this constructed object is a function, and by virtue of meeting all dense subsets, this union will form a countable unbounded sequence in $\kappa$. The importance of using a normal measure was making sure that no cardinals collapse, which is used for showing the relative consistency of the negation of the Singular Cardinal Hypothesis with a measurable cardinal. In this case, we winded up getting rid of unnecessary parts of the conditions, but they were used to guide the construction. Specifically, they were used to make sure our functions reached up high but did so without collapsing cardinals. However, despite the fact that all of these forcing posets (notions) guide what our construction will look like, we can have much less intuitive posets accomplishing the same things. Mainly, we can associate each of these partial orders with some strange ordering on the elements of a cardinal and still accomplish the same result. Specifically, all of these forcing extensions would be equivalent to forcing extensions adding a subset of some cardinal. There are obviously a number of other known forcing notions, but since this post is already long enough and maybe a little too technical for your question, let me just conclude with the main point. Mainly, we choose our posets looking ahead to the properties we want our generic object to have. The more complex the generic object that we want to add, the more complex the dense subsets we're going to need to have to make sure that the filter meeting them has all of the desired characteristics. This greater complexity often comes in terms of some combination of larger possible antichain sizes, more dense subsets, less closure, etc. While we can always define a descending sequence of conditions so that we meet any countable collection of dense subsets or maximal antichains, the genericity of the filter magically does this in a way to simultaneously meet all of them in a compatible way. - This is easier to understand in terms of locales. The double-negation locale associated to a forcing is the space in which the generic filter lives. The trouble is that non-trivial double-negation locales do not have points, so one has to be adjoined to form the generic extension. In other words, if $X$ is the double-negation locale we want to force with, we would like to have a pair of geometric morphisms $Set \to Sh(X) \to Set$. The second part $Sh(X) \to Set$ is no problem, but the first part $Set \to Sh(X)$ usually doesn't exist. So we expand the universe of sets a bit by adding a generic point to $X$. Note that the double-negation locale $X$ gets destroyed in the final process of generating the generic extension. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 73, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9418069124221802, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/38511/flow-of-momentum-is-pressure?answertab=oldest	# Flow of momentum is pressure In the diagonal terms of the energy-momentum tensor, the flow of $x$-momentum in the $x$-direction is the $x$-pressure. Why the flow of momentum is pressure? - ## 1 Answer Consider some basic unit analysis. Pressure is defined as force/area which is the same as momentum/area/time since F=dp/dt. Momentum flow would be the momentum passing through a unit area per unit time so it's the same units. More physically, think of a gas at constant pressure in a box. If you popped a little hole of unit area in the side of the box, the pressure would be the amount of momementum escaping per unit time. Hope that helps! -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9179148077964783, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2009/09/22/hard-choices/?like=1&source=post_flair&_wpnonce=55e0e6033e	# The Unapologetic Mathematician ## Hard Choices So yesterday we noted that the big conceptual problem with partial derivatives is that they’re highly dependent on a choice of basis. Before we generalize away from this, let’s note a few choices that we are going to make. First of all, we’re going to assume our space $\mathbb{R}^n$ comes with a positive-definite inner product, but it doesn’t really matter which one. We’re choosing a positive-definite form with signature $(n,0)$ instead of a form with some negative-definite or even degenerate portion — where we’d get $-1$s or ${0}$s along the diagonal in an orthonormal basis — because we want every direction to behave the same as every other direction. More general signatures will come up when we talk about more general spaces. But we do want to be able to talk in terms of lengths and angles. Now this doesn’t mean we’ve chosen a basis. We can choose one, but there’s a whole family of other equally valid choices related by orthogonal transformations. Ideally, we should define things which don’t depend on this choice at all. If we must make a choice in our definitions, the results should be independent of the choice. Often, this will amount to the existence of some action of the orthogonal group on our structure, and the invariance of the results under this action. Definitions which don’t depend on the choice are related to what physicists mean when they say something is “manifestly coordinate-free”, since we don’t even have to mention coordinates to make our definitions. Those which depend on a choice, but are later shown to be independent of that choice are a lesser, but acceptable, alternative. Notice also that this avoidance of choices echoes the exact same motives when we preferred the language of linear transformations on vector spaces to the language of matrices acting on ordered tuples of numbers. But, again, we have made a choice of some inner product. But this doesn’t matter, because all positive-definite inner products “look the same”, in the sense that if we pick an orthonormal basis for each of two distinct inner products, there’s going to be a general linear transformation which takes the one basis to the other, and which thus takes the one form to the other. That is, the forms are congruent. So as long as we have some inner product, any inner product, to talk about lengths and angles, and to translate between vectors and covectors, we’re fine. ### Like this: Posted by John Armstrong \| Analysis, Calculus ## 1 Comment » 1. [...] point indeed. And why should we have any such special point, from a geometric perspective? We already insisted that we didn’t want to choose a basis for our space that would make some directions more [...] Pingback by \| September 28, 2009 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 4, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9280853867530823, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/181727-induction-help.html	Thread: 1. Induction help Hi I need help in regards with proving this problem. If the sequence is 1,13,37,73,121 then it should follow like this 1+12+24+36......12n+(12n+12) For the formula 6n^2 - 6n + 1. How would I go on to prove this with induction exactly? Thanks 2. Originally Posted by monkbear Hi I need help in regards with proving this problem. If the sequence is 1,13,37,73,121 then it should follow like this 1+12+24+36......12n+(12n+12) For the formula 6n^2 - 6n + 1. How would I go on to prove this with induction exactly? Thanks 3. Originally Posted by monkbear How would I go on to prove this with induction exactly? Thanks Induction works in two steps. First you prove it true for some base case. In this case, if P(n) reads $1 + 12 + 24 + ... + 12n + (12n + 12) = 6n^2 - 6n + 1$ show that P(0) is true. That should be straight-forward computation. The inductive step is to assume it is true for some $k \geq 0$. Assuming P(k), demonstrate P(k+1). By establishing those two facts regarding P(n), you can prove it is true for every $n\geq 0$. Do you understand why this two-step process works? In other words, do you know why the principle of mathematical induction (PMI) is a valid rule of inference? Originally Posted by Plato That link didn't work for me Plato (using FF2). Broken? 4. Originally Posted by bryangoodrich that link didn't work for me plato (using ff2). Broken? fixed. 5. I've never used Wolfram Alpha before, but I have to say ... that's awesome! 6. Originally Posted by monkbear Hi I need help in regards with proving this problem. If the sequence is 1,13,37,73,121 then it should follow like this 1+12+24+36......12n+(12n+12) For the formula 6n^2 - 6n + 1. How would I go on to prove this with induction exactly? Thanks You should have a closer look at the pattern. $u_1=1=1+12(0)$ $u_2=13=1+12(1)$ $u_3=37=1+12(1)+12(2)$ $u_n=1+12(1)+12(2)+....+12(n-1)$ To prove using induction that $u_n=1+12(1)+12(2)+....+12(n-1)=6n^2-6n+1$ P(k) $1+12(1)+12(2)+....+12(k-1)=6k^2-6k+1$ P(k+1) $1+12(1)+12(2)+.....+12(k-1)+12k=6(k+1)^2-6(k+1)+1\;\;\;?$ Try to show that IF P(k) is true THEN P(k+1) will also be true. Proof If P(k) is true, then P(k+1) is $1+12(1)+12(2)+...+12(k-1)+12k=6k^2-6k+1+12k=6k^2+6k+1$ which compares to $6(k+1)^2-6(k+1)+1=6\left(k^2+2k+1\right)-6k-6+1$ $=6k^2+12k+6-6k-6+1$ $=6k^2+6k+1$ It remains to test the base case for n = 1, which is clearly true.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9254982471466064, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/56182-please-check-me.html	# Thread: 1. ## please check for me I have a theorem states that: Let $G=H\times K$, and let $H_{1}\triangleleft H$ and $K_{1}\triangleleft K$. Then $H_{1}\times K_{1}\triangleleft G$ and $G/(H_{1}\times K_{1})\cong(H/H_{1})\times(K/K_{1})$. Then I have this corollary: If $G=H\times K$, then $G/(H\times\{{1}\})\cong K$. Please check for me if my proof for the corollary is correct. Proof: We consider $H_{1}=H$, i.e. H has no proper normal subgroup. Also, $K_{1}=\{1\}$. By last theorem, $G/(H_{1}\times K_{1})\cong (H/H_{1})\times(K/K_{1})$. We have $G/(H\times\{1\})\cong (H/H)\times(K/\{1\})$. Since H has no proper normal subgroup, so quotient group of H is not defined. Also, $(K/\{1\})=K$. Hence, $G/(H\times\{1\})\cong K$ 2. Originally Posted by deniselim17 Proof: We consider $H_{1}=H$, i.e. H has no proper normal subgroup. Also, $K_{1}=\{1\}$. By last theorem, $G/(H_{1}\times K_{1})\cong (H/H_{1})\times(K/K_{1})$. We have $G/(H\times\{1\})\cong (H/H)\times(K/\{1\})$. Since H has no proper normal subgroup, so quotient group of H is not defined. Also, $(K/\{1\})=K$. Hence, $G/(H\times\{1\})\cong K$ It is correct except for a phrase you use "H has no proper normal subgroup", why not? It does not matter. Just choose $H_1 = H$. We really do not care if it has proper normal subgroup (by the way it does, it has $\{ 1 \}$) and then apply the theorem and get your result. But then you say something that is again not necessary "so the quotient group of H is not defined", yes it is! Certainly $H\triangleleft H$ with $H/H \simeq \mathbb{Z}_1$. It is defined, it is just not a very interesting group.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9333363771438599, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/19000/how-do-i-find-the-tension-in-additional-strings-in-this-problem?answertab=votes	# How do I find the tension in additional strings in this problem? A mass of 5.00 kg hangs attached to three strings as shown in the figure (see image below). Find the tension in each string. Hint: Consider the equilibrium of the point where the strings join. So finding $T_3$ was relative quick, knowing $W = mg$, $m = 5.00\text{ kg}$, $g = 10\ \mathrm{ms^{-2}}$ $$W = 50\text{ N} = T_3$$ It is the other two ($T_1$ and $T_2$) I'm not sure. In fact I'm at lost with what I learned in class and applying. I noticed the problem states a hint (equilibrium where strings join) but I'm not seeing how this information can be used. - Hint: think about components of vectors. – David Zaslavsky♦ Jan 2 '12 at 1:44 You were right, thank you. the rest unfolded easily, I found the correct solution. – Cloud Jan 2 '12 at 2:05 4 Good to hear it. Perhaps you could write up the solution you found and post it as an answer. (Unfortunately the system requires you to wait 8 hours before answering your own question, but I'd encourage you to come back and do it tomorrow.) – David Zaslavsky♦ Jan 2 '12 at 2:13 (By the way, it's fantastic that such a small hint was enough to get you through the problem. I wish every homework question was like that.) – David Zaslavsky♦ Jan 2 '12 at 6:31 Anybody care to fix the image link? – Qmechanic♦ Feb 2 '12 at 19:43 ## 2 Answers Here is the formal methodology for this type of problems. When things are in equilibrium, then the sum of the forces acting in any direction equal to zero. So add up the vertical forces on the knot and make them equal to zero, as well as the horizontal forces equal to zero. From those two equations solve for the two unknowns $T_1$ and $T_2$. - $\newcommand{\t}[1]{\frac{T_#1}{\sin\theta_#1}}$ The correct way is with vectors. The shortcut way is to use Lami's theorem. Basically, if the tension in each string is $T_1$, and the angle opposite each string is $\theta_1$, then $$\t 1=\t 2=\t 3$$ Referring to this diagram, $\frac{A}{\sin \alpha}=\frac{B}{\sin \beta}=\frac{C}{\sin \gamma}$, where $A,B,C$ are forces. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9516063332557678, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/46929/schrodingers-equation-explanation-to-non-physicist/46972	# Schrodinger's equation (explanation to non physicist) For a report I'm writing on Quantum Computing, I'm interested in understanding a little about this famous equation. I'm an undergraduate student of math, so I can bear some formalism in the explanation. However I'm not so stupid to think I can understand this landmark without some years of physics. I'll just be happy to be able to read the equation and recognize it in its various forms. To be more precise, here are my questions. Hyperphysics tell me that Shrodinger's equation "is a wave equation in terms of the wavefunction". 1. Where is the wave equation in the most general form of the equation? $$\mathrm{i}\hbar\frac{\partial}{\partial t}\Psi=H\Psi$$ I thought wave equation should be of the type $$\frac{\partial^2}{\partial^2t}u=c^2\nabla^2u$$ It's the difference in order of of derivation that is bugging me. From Wikipedia "The equation is derived by partially differentiating the standard wave equation and substituting the relation between the momentum of the particle and the wavelength of the wave associated with the particle in De Broglie's hypothesis." 2. Can somebody show me the passages in a simple (or better general) case? 3. I think this questions is the most difficult to answer to a newbie. What is the Hamiltonian of a state? How much, generally speaking, does the Hamiltonian have to do do with the energy of a state? 4. What assumptions did Schrödinger make about the wave function of a state, to be able to write the equation? Or what are the important things I should note in a wave function that are basilar to proof the equation? With both questions I mean, what are the passages between de Broglie (yes there are these waves) and Schrödinger (the wave function is characterized by)? 5. It's often said "The equation helps finds the form of the wave function" as often as "The equation helps us predict the evolution of a wave function" Which of the two? When one, when the other? - 3 Philisophically I always find requests to explain an equation for the laymen to be a little strange. The point of writing it in math is to have a precise, and complete representation of the theory... – dmckee♦ Dec 15 '12 at 16:13 You're right. That's why I tried to make it clear I'm not asking an explanation of the "equation" as you mean it, rather the meaning of the "symbols in it". In particulart question number 1 is the most important for me now. – Temitope.A Dec 15 '12 at 17:04 – Qmechanic♦ Dec 15 '12 at 18:21 ## 3 Answers You should not think of the Schrödinger equation as a true wave equation. In electricity and magnetism, the wave equation is typically written as $$\frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = \frac{\partial^2 u}{\partial x^2}$$ with two temporal and two spatial derivatives. In particular, it puts time and space on 'equal footing', in other words, the equation is invariant under the Lorentz transformations of special relativity. The one-dimensional time-dependent Schrödinger equation for a free particle is $$\mathrm{i} \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$ which has one temporal derivative but two spatial derivatives, and so it is not Lorentz invariant (but it is Galilean invariant). For a conservative potential, we usually add $V(x) \psi$ to the right hand side. Now, you can solve the Schrödinger equation is various situations, with potentials and boundary conditions, just like any other differential equation. You in general will solve for a complex (analytic) solution $\psi(\vec r)$: quantum mechanics demands complex functions, whereas in the (classical, E&M) wave equation complex solutions are simply shorthand for real ones. Moreover, due to the probabilistic interpretation of $\psi(\vec r)$, we make the demand that all solutions must be normalized such that $\int \|\psi(\vec r)\|^2 dr = 1$. We're allowed to do that because it's linear (think 'linear' as in linear algebra), it just restricts the number of solutions you can have. This requirements, plus linearity, gives you the following properties: 1. You can put any $\psi(\vec r)$ into Schrödinger's equation (as long as it is normalized and 'nice'), and the time-dependence in the equation will predict how that state evolves. 2. If $\psi$ is a solution to a linear equation, $a \psi$ is also a solution for some (complex) $a$. However, we say all such states are 'the same', and anyway we only accept normalized solutions ($\int \|a\psi(\vec r)\|^2 dr = 1$). We say that solutions like $-\psi$, and more generally $e^{i\theta}\psi$, represent the same physical state. 3. Some special solutions $\psi_E$ are eigenstates of the right-hand-side of the time-dependent Schrödinger equation, and therefore they can be written as $$-\frac{\hbar^2}{2m} \frac{\partial^2 \psi_E}{\partial x^2} = E \psi_E$$ and it can be shown that these solutions have the particular time dependence $\psi_E(\vec r, t) = \psi_E(\vec r) e^{-i E t/\hbar}$. As you may know from linear algebra, the eigenstates decomposition is very useful. Physically, these solutions are 'energy eigenstates' and represent states of constant energy. 4. If $\psi$ and $\phi$ are solutions, so is $a \psi + b \phi$, as long as $\|a\|^2 + \|b\|^2 = 1$ to keep the solution normalized. This is what we call a 'superposition'. A very important component here is that there are many ways to 'add' two solutions with equal weights: $\frac{1}{\sqrt 2}(\psi + e^{i \theta} \phi)$ are solutions for all angles $\theta$, hence we can combine states with plus or minus signs. This turns out to be critical in many quantum phenomena, especially interference phenomena such as Rabi and Ramsey oscillations that you'll surely learn about in a quantum computing class. Now, the connection to physics. 1. If $\psi(\vec r, t)$ is a solution to the Schrödinger's equation at position $\vec r$ and time $t$, then the probability of finding the particle in a specific region can be found by integrating $\|\psi^2\|$ around that region. For that reason, we identify $\|\psi\|^2$ as the probability solution for the particle. • We expect the probability of finding a particle somewhere at any particular time $t$. The Schrödinger equation has the (essential) property that if $\int \|\psi(\vec r, t)\|^2 dr = 1$ at a given time, then the property holds at all times. In other words, the Schrödinger equation conserves probability. This implies that there exists a continuity equation. 2. If you want to know the mean value of an observable $A$ at a given time just integrate $$<A> = \int \psi(\vec r, t)^* \hat A \psi(\vec r, t) d\vec r$$ where $\hat A$ is the linear operator associated to the observable. In the position representation, the position operator is $\hat A = x$, and the momentum operator, $\hat p = - i\hbar \partial / \partial x$, which is a differential operator. The connection to de Broglie is best thought of as historical. It's related to how Schrödinger figured out the equation, but don't look for a rigorous connection. As for the Hamiltonian, that's a very useful concept from classical mechanics. In this case, the Hamiltonian is a measure of the total energy of the system and is defined classically as $H = \frac{p^2}{2m} + V(\vec r)$. In many classical systems it's a conserved quantity. $H$ also lets you calculate classical equations of motion in terms of position and momentum. One big jump to quantum mechanics is that position and momentum are linked, so knowing 'everything' about the position (the wavefunction $\psi(\vec r))$ at one point in time tells you 'everything' about momentum and evolution. In classical mechanics, that's not enough information, you must know both a particle's position and momentum to predict its future motion. - Thank you! One last question. How do somebody relate the measurment principle to the equations, that an act of measurment will cause the state to collapse to an eigenstate? Or is time a concept indipendent of the equation? – Temitope.A Dec 16 '12 at 11:37 Can states of entanglement be seen in the equation to? – Temitope.A Dec 16 '12 at 11:47 Note that user10347 talks of a potential added to the differential equation. To get real world solutions that predict the result of a measurement one has to apply the boundary conditions of the problem. The "collapse" vocabulary is misleading. A measurement has a specific probability of existing in the space coordinates or with the fourvectors measured. The measurement itself disturbs the potential and the boundary conditions change, so that after the measurement different solutions/psi functions will apply. – anna v Dec 16 '12 at 13:23 One type of measurement is strong measurement, where we the experimentalists, measure some differential operator $A$, and find some particular (real) number $a_i$, which is one of the eigenvalues of $A$. (Important detail: for $A$ to be measureable, it must have all real eigenvalues.) Then, we know the wavefunction "suddenly" turns into $\psi_i$, which is the eigenfunction of $A$ whose eigenvalue was that number $a_i$ we measured. The system has lost of knowledge of the original wavefunction $\psi$. The probability of measuring $a_i$ is $\|<\psi_i \| \psi>\|^2$. – emarti Dec 18 '12 at 7:12 @Temitope.A: Entanglement isn't obvious in anything here because I've only written single-particle wavefunctions. A two-particle wavefunction $\Psi(\vec r_1, \vec r_2)$ gives a probability $\int_{V_1}\int_{V_2}\|\Psi\|^2 d \vec r_1 d \vec r_2$ of detecting one particle in a region $V_1$ and a second particle in a region $V_2$. A simple solution for distinguishable particles is $\Psi(\vec r_1, \vec r_2) = \psi_1(\vec r_1) \psi_2(\vec r_2)$, and it can be shown that this satisfies all our conditions. An entangled state cannot be written so simply. (Indistinguishable particles take more care.) – emarti Dec 18 '12 at 9:32 show 2 more comments What you write is the time-dependent Schrödinger equation. This is not the equation of a true wave. He postulated the equation using a heuristic approach and some ideas/analogies from optics, and he believed on the existence of a true wave. However, the correct interpretation of $\Psi$ was given by Born: $\Psi$ is an unobservable function, whose complex square $\|\Psi\|^2$ gives probabilities. In older literature $\Psi$ is still named the wavefunction, In modern literature the term state function is preferred. The terms "wave equation" and "wave formulation" are legacy terms. In fact, part of the confusion that had Schrödinger, when he believed that his equation described a physical wave, is due to the fact he worked with single particles. In that case $\Psi$ is defined in an abstract space which is isomorphic to the tri-dimensional space. However, when you consider a second particle and write $\Psi$ for a two-body system, the isomorphism is broken and the superficial analogy with a physical wave is completely lost. A good discussion of this is given in Ballentine textbook on quantum mechanics (section 4.2). The Schrödinger equation cannot be derived from wave theory. This is why the equation is postulated in quantum mechanics. There is no Hamiltonian for one state; the Hamiltonian is characteristic of a given system with independence of its state. Energy is a possible physical property of a system, one of the possible observables of a system; it is more correct to say that the Hamiltonian gives the energy of a system in the cases when the system is in a certain state. A quantum system always has a Hamiltonian, but not always has a defined energy. Only certain states $\Psi_E$ that satisfy the time-independent Schrödinger equation $H\Psi_E = E \Psi_E$ are associated to a value $E$ of energy. The quantum system can be in a superposition of the $\Psi_E$ states or can be in more general states for which energy is not defined. Wavefunctions $\Psi$ have to satisfy a number of basic requirements such as continuity, differentiability, finiteness, normalization... Some texts emphasize that the wavefunctions would be single-valued, but I already take this in the definition of function. The Schrödinger equation gives both "the form of the wave function" and "the evolution of a wave function". If you know $\Psi$ at some initial time and integrate the time-dependent Schrödinger equation you obtain the form of the wavefunction to some other instant: e.g. the integration is direct and gives $\Psi(t) = \mathrm{Texp}(-\mathrm{i}/\hbar \int_0^t H(t') dt') \Psi(0)$, where $\mathrm{Texp}$ denotes a time-ordered exponential. This equation also gives the evolution of the initial wavefunction $\Psi(0)$. When the Hamiltonian is time-independent, the solution simplifies to $\Psi(t) = \exp(-\mathrm{i}Ht/\hbar) \Psi(0)$. For stationary states, the time-dependent Schrödinger equation that you write reduces to the time-independent Schrödinger equation $H\Psi_E = E \Psi_E$; the demonstration is given in any textbook. For stationary states there is no evolution of the wavefunction, $\Psi_E$ does not depend on time, and solving the equation only gives the form of the wavefunction. - 1 Good answer. I would only add that regarding the last point, I think the confusion comes from references to the "time-independent" Schrodinger eigenvalue equation $H\psi_E = E\psi_E$ being conflated with the "time-dependent" evolution equation $\mathrm{i}\hbar \dot{\psi} = H\psi$, when of course the two are entirely different beasts. – Chris White Dec 15 '12 at 21:07 @ChrisWhite Good point. Made. – juanrga Dec 16 '12 at 2:33 6 paragraph: maybe you should add that the equation only holds if H is time-independent. – ungerade Dec 16 '12 at 12:19 @ungerade Another good point! Added evolution when H is time-dependent. – juanrga Dec 16 '12 at 12:49 If you take the wave equation $$\nabla^2\phi = \frac{1}{u^2}\frac{d^2\phi}{dt^2}\text{,}$$ and consider a single frequency component of a wave while taking out its time dependence, $\phi = \psi e^{-i\omega t}$, then: $$\nabla^2 \phi = -\frac{4\pi^2}{\lambda^2}\phi\text{,}$$ but that means the wave amplitude should satisfy an equation of the same form: $$\nabla^2 \psi = -\frac{4\pi^2}{\lambda^2}\psi\text{,}$$ and if you know the de Broglie relation $\lambda = h/p$, where for a particle of energy $E$ in a potential $V$ has momentum $p = \sqrt{2m(E-V)}$, so that: $$\underbrace{-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi}_{\hat{H}\psi} = E\psi\text{,}$$ Therefore, the time-independent Schrödinger equation has a connection to the wave equation. The full Schrödinger equation can be recovered by putting time-dependence back in, $\Psi = \psi e^{-i\omega t}$ while respecting the de Broglie $E = \hbar\omega$: $$\hat{H}\Psi = (\hat{H}\psi)e^{-i\omega t} = \hbar\omega \psi e^{-i\omega t} = i\hbar\frac{\partial\Psi}{\partial t}\text{,}$$ and then applying the principle of superposition for the general case. However, in this process the repeated application of the de Broglie relations takes us away from either classical waves or classical particles; to what extent the resulting "wave function" should be considered a wave is mostly a semantic issue, but it's definitely not at all a classical wave. As other answers have delved into, the proper interpretation for this new "wave function" $\Psi$ is inherently probabilistic, with its modulus-squared representing a probability density and the gradient of the complex phase being the probability current (scaled by some constants and the probability density). As for the de Broglie relations themselves, it's possible to "guess" them from making an analogy from waves to particles. Writing $u = c/n$ and looking for solutions close to plane wave in form, $\phi = e^{A+ik_0(S-ct)}$, the wave equation gives: $$\begin{eqnarray} \nabla^2A + (\nabla A)^2 &=& k_0^2[(\nabla S)^2 - n^2]\text{,}\\ \nabla^2 S +2\nabla A\cdot\nabla S &=& 0\text{.} \end{eqnarray}$$ Under the assumption that the index of refraction $n$ changes slowly over distances on the order of the wavelength, then $A$ does not vary extremely, the wavelength is small, and so $k_0^2 \propto \lambda^{-2}$ is large. Therefore the term in the square brackets should be small, and we can make the approximation: $$(\nabla S)^2 = n^2\text{,}$$ which is the eikonal equation that links the wave equation with geometrical optics, in which motion of light of small wavelengths in a medium of well-behaved refractive index can be treated as rays, i.e., as if described by paths of particles/corpuscles. For the particle analogy to work, the eikonal function $S$ must take the role of Hamilton's characteristic function $W$ formed by separation of variables from the classical Hamilton-Jacobi equation into $W - Et$, which forces the latter to be proportional to the total phase of the wave, giving $E = h\nu$ for some unknown constant of proportionality $h$ (physically Planck's constant). The index of refraction $n$ corresponds to $\sqrt{2m(E-V)}$. This is discussed in, e.g., Goldstein's Classical Mechanics, if you're interested in details. - Your first equation is a wave equation, only if you substitute the total time derivatives by partial ones. Moreover, you introduce a $\Psi = \psi e^{-i\omega t} = \phi$, but the wavefunction $\Psi$ does not satisfy the first equation for a wave. – juanrga Dec 18 '12 at 11:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 90, "mathjax_display_tex": 13, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.933938205242157, "perplexity_flag": "head"}

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