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http://physics.stackexchange.com/questions/23316/how-fast-is-heat-transferred-by-conduction
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# How fast is heat transferred by conduction?
How fast is heat transferred by conduction? Is there some simple, but quantitative way that starts from some properties of the material (e.g. its thermal conductivity) and makes rough predictions, for example about how much time is needed for temperature to change at one end of the body when it is placed, at the other end, in thermal contact with another?
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The conduction coefficient for slow temperature gradients can be calculated in many materials using a Kubo type formula, but I am not sure if the result is reliable in metals with a large thermal gradient, in the case that electrons are delocalized over a region which is big enough for the temperature gradient to be non-infinitesimal. – Ron Maimon Apr 5 '12 at 19:01
## 2 Answers
The heat equation for these kind of problems (assume 1D), reads
$$\frac{\partial T}{\partial t} = a \frac{\partial^2 T}{\partial x^2}$$
Where of course $T$ is temperature, $t$ is time, $x$ is position and $a$ the thermal diffusivity: $a=\frac{\lambda}{\rho c_p}$ (respectively thermal conductivity, density and heat capacity. This equation can be derived from Fourier's law.
Suppose you have a block at temperature $T_0$ which you put in contact with a block of temperature $T_1$ at $x=0$. Then you have a set of boundary conditions
$$T(x,0) = T_0 \\ T(0,t) = T_1 \\ T(x\to\infty,t)=T_0$$
It is not easy, but it has been derived that the solution to this equation is
$$\frac{T-T_0}{T_1-T_0}=1-\frac{2}{\sqrt{\pi}}\displaystyle\int_0^{\frac{x}{2\sqrt{at}}} e^{-s^2}ds$$ Where the solution to this integral is referred to as error function
This solution describes the transient and spatial profile of the heated piece of material.
For short times, only a certain amount of the material is heated. Using the error function, one can define the penetration depth, which is $x_p=\sqrt{\pi a t}$, which is obviously a measure for how far the increased temperature ranges into the material.
Suppose you domain has a finite length and is isolated at the other end. The heat transfer coefficient from the wall at $x=0$ is nearly constant, and the average temperature of the block will converge with an exponential decay to the boundary temperature (see Vladimir's answer). This can be derived from the heat equation by assuming that $\frac{T-T_0}{T_1-T_0}=1 - f(t) g(x/L)$
Note: This book was used as a reference for some of the equations.
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One can easily estimate the time of reaching a steady state. If you have a layer of a certain thickness and you know the thermal properties of the material, the problem is solved in any textbook on heat conduction. The temperature at the other end varies with time and the final stage is very simple (called a regular regime): $T(t)\approx T(\infty)+Ae^{-\lambda_0 t}$. The lowest eigenvalue $\lambda_0$ is determined with the layer thickness $L$, its heat conductivity $\kappa$, specific heat capacity $c$ and the material density $\rho$.
$$\lambda_0 = \frac{\pi^2 \cdot\kappa}{4\rho \cdot c \cdot L^2}$$
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http://mathoverflow.net/questions/122308/when-a-riemannian-manifold-is-of-hessian-typ
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## When a Riemannian manifold is of Hessian Typ
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
When a Riemannian manifold is of Hessian Type (i.e., a Riemannian manifold which its metric is Hessian)
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This doesn't seem to be a question. – HW Feb 19 at 14:26
This is very important queston because for studying local geometry of the moduli space of special lagrangian submanifolds we need to Riemannian manifolds of Hessian type which are useful for the study of monge ampere equations – Hassan Jolany Feb 19 at 14:36
You can see some works of Hichin related to my previous comment – Hassan Jolany Feb 19 at 14:37
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I think HW means the question is too board. – Ralph Feb 19 at 14:41
I meant that, grammatically, it was not a question. I don't want to penalise non-native English speakers, but I didn't understand what the question was. Since the edit, I think I do, viz, 'When is a Riemannian manifold of Hessian type?' – HW Feb 19 at 14:52
## 2 Answers
First, the definition: A Riemannian $n$-manifold $(M^n,g)$ is of Hessian type if there exist $(n{+}1)$ functions $x^1,\ldots,x^n, u$ on $M$ such that $dx^1\wedge\cdots\wedge dx^n\not=0$ and such that $$g = \frac{\partial^2u}{\partial x^i\partial x^j} dx^idx^j$$ (of course, the summation convention is in force for this formula). Note that the independence of the differentials $dx^i$ is needed in order to define the 'partial derivatives' in the formula. One says that $(M^n,g)$ is locally of Hessian type if each point of $M$ has an open neighborhood $U\subset M$ on which there exists a coordinate chart $x:U\to\mathbb{R}^n$ and a function $u\in C^\infty(U)$ such that the above formula holds on $U$.
Since metrics in dimension $n$ depend on $\tfrac12n(n{+}1)$ functions of $n$ variables and the data of a Hessian representation depends only on $(n{+}1)$ functions of $n$ variables, it is clear that, when $n>2$, not every metric is locally of Hessian type, and, in principle, such a set of criterion can be developed, but it's not trivial. Of course, as $n$ increases, the condition of being locally of Hessian type becomes more and more restrictive, even implying algebraic conditions on the Riemann curvature tensor once $n$ is sufficiently large.
However, when $n=2$, this is a determined problem. However, it is never elliptic, so one never gets elliptic regularity. The characteristic variety consists of $3$ points, so at least one of them has to be real. Depending on the sign of the Gauss curvature of the metric, one can sometimes formulate the problem as having $3$ real characteristics and sometimes one can't. Of course, in the real-analytic case, the problem is always solvable locally, so every real-analytic metric in dimension $2$ is locally of Hessian type.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I might be misunderstanding the question, but I believe that if there are functions $h$ and $k$ so that $$\text{Hess}_h = kg,$$ then $(M,g)$ must be (at least locally) a warped product $(a,b) \times_f N^{n-1}$. This follows from integrating along flowlines of $\nabla h$, to compare the induced metrics on different level sets of $h$.
I don't know if this is was the first proof of this result, but the result I've stated above is proven (and discussed a bit more than I have here) in Cheeger-Colding's paper "Lower Bounds on Ricci Curvature and Almost Rigidity of Warped Products" on p 192-194 in this copy of the paper.
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I think that you are misunderstanding the question. I believe that your $\mathrm{Hess}_h$ is actually $\nabla^2 h$, where $\nabla$ is the Levi-Civita connection of $g$. That is not the Hessian with respect to a coordinate system, as I described it above. In your equation, $g$ actually appears on both sides of the equation, whereas, in mine, it only appears on $1$ side and the $(n{+}1)$ unknowns all appear on the other side. – Robert Bryant Feb 19 at 17:18
Yes, exactly I am agree with @ Robert – Hassan Jolany Feb 19 at 19:22
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http://math.stackexchange.com/questions/213607/a-mapping-function-is-requested
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# a mapping function is requested
I have $n$ data points in $d$-dimensional space, i.e $x_1,x_2,...x_n$ where $x_i\in R^d$. I want an interpolation function $f:R^d-> R$ on these data points to map them to $y_1,y_2,...,y_n$, where $y_i=f(x_i)$. In other words, I require an extension to Lagrange interpolation method for $d$ dimensional space.
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## 1 Answer
Let $l_{i,j}:\mathbb{R}\rightarrow\mathbb{R}$ for $j=1,\dots,d$ be the Lagrange basis polynomials for the data $(x_i^{(j)})_i$, i.e. $l_{i,j}(x^{(j)}_k)=1$ if $k=i$ and $0$ otherwise. Then the polynomial in $d$ variables given by
$$L(x):=\sum^n_{i=1} y_i\left(\prod^d_{j=1}l_{i,j}(x^{(j)})\right)$$
is what you are looking for (just plug in the $x_i$): $L(x_i)=y_i$. Note however, that the Lagrange polynomials has some undesirable properties (fast oscillation in between the points fixed by the data, doesn't extrapolate well, ..). So depending on your problem, you might also want to look into other interpolation methods (Bezier curves, etc.).
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Thank you for your answer. – remo Oct 14 '12 at 15:01
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http://math.stackexchange.com/questions/199530/probability-limit-of-a-decreasing-uncountable-family-of-sets/199541
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# Probability limit of a decreasing (uncountable) family of sets.
Let $t^*\in(0,1)$ and $(\Omega,\mathcal{F},P)$ be a probability space. Suppose I have set $A\in\mathcal{F}$ and an uncountable family of sets $(B_t : t\in[0,1])\subset\mathcal{F}$ with the following properties:
1. $P(A)>0$ and $P(B_t)=p$ where $p>0$ is a constant.
2. For each $\omega\in A$ there exists a $\delta>0$ such that $\omega\notin B_t$ for all $t\in (t^*-\delta, t^*+\delta)$.
Is it true that $\limsup_{t\rightarrow t^*}P(A\cap B_t) = 0$? Is the limit well defined?
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## 1 Answer
Let $t_n$ be any sequence converging to $t^*$, and let $C_n = \bigcup_{i=n}^\infty B_{t_n}$. Then $C_n$ is decreasing, so we have $$\lim_{n\to\infty}P(A\cap C_n)= P(A\cap\bigcap_{n=1}^\infty C_n)$$ because finite measures are continuous from above. But $\bigcap_n C_n$ is disjoint from $A$: for any $\omega\in A$, there exists $\delta$ such that $\omega\notin B_t$ if $|t-t^*|<\delta$, so there exists $N$ such that $\omega\notin B_{t_n}$ if $n> N$, so there exists $N$ such that $\omega\notin C_N$. Therefore $$\limsup_{n\to\infty}P(A\cap B_{t_n})\leq\lim_{n\to\infty}P(A\cap C_n)=P(A\cap\bigcap_{n=1}^\infty C_n)=0.$$
So $\limsup_{t\to t^*}P(A\cap B_t)=0$.
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Just for clarity, shouldn't we say that $\limsup_{n\rightarrow\infty} P(A\cap B_{t_n}) \leq \lim_{n\rightarrow\infty} P(A\cap C_n)$, since, a priori, we do not know that the limit of $P(A\cap B_{t_n})$ exists? – Carl Morris Sep 20 '12 at 17:33
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That's true, I'll edit to reflect that. Thanks! – Owen Biesel Sep 21 '12 at 2:16
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http://physics.stackexchange.com/questions/tagged/internal-energy?sort=active&pagesize=15
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# Tagged Questions
The internal-energy tag has no wiki summary.
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http://mathhelpforum.com/differential-geometry/161797-uniform-continuity-proof.html
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# Thread:
1. ## Uniform Continuity Proof
Hi,
I'm not sure how to prove this statement:
Suppose f is uniformly continuous on the set S and let T be the image of S under f. Also assume that the function g is uniformly continuous on T. Prove that g o f is uniformly continuous on S?
Thanks a lot!
2. Originally Posted by AKTilted
Hi,
I'm not sure how to prove this statement:
Suppose f is uniformly continuous on the set S and let T be the image of S under f. Also assume that the function g is uniformly continuous on T. Prove that g o f is uniformly continuous on S?
Thanks a lot!
So, you're trying to show that given $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in S$ $|x-y|<\delta\implies |f(g(x))-f(g(y))|<\varepsilon$, right?
Ok, so what's the first step? You're lcearly going to want to consider two deltas one for the $S$ and one for $T$. Which should you pick first?
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http://mathoverflow.net/questions/18770?sort=newest
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## Is there a name for this property of a topology?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This property seems like it should have a nice name, but I can't find one anywhere. Does anyone know a name for this?
For each non-empty open set $U$, there exist proper open subsets $\{U_i\}_{i\in I}$ such that $U=\cup_i U_i$.
I suppose this could also be formulated as each nonempty open set having an open cover of proper subsets, or being the colimit of its open subsets.
(Also, apologies if this is something obvious I should have thought of.)
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Each nonempty open set? – unknown (google) Mar 19 2010 at 16:41
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As stated its just not a good question. If the space is perfect, that is every point is a limit point, and points are closed, then a space will have this property. Its not a research level question. – Charlie Frohman Mar 19 2010 at 17:11
4
@ Ketil Tveiten, re: Charlie Frohman's comment. For what it's worth, I think this is a fine question. I don't think the answer would be obvious to every "research-level" mathematician, although I'm only a first-year graduate student myself... – Vectornaut Mar 19 2010 at 19:52
5
@Charlie: Are you saying that serious researchers only study T1 spaces?!? – François G. Dorais♦ Mar 19 2010 at 23:25
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I agree with Charlie, and here's why: I don't see off hand any good reason for caring what the name of such a space is. I mean, if you have examples of some of these spaces, and some result that says that this precise property is what you need for some application, then by all means, it should have a name, and knowing the conventional name will help you look up the appropriate literature. But as it is, I'd like some motivation before I'll like the question. – Theo Johnson-Freyd Mar 20 2010 at 1:43
show 3 more comments
## 4 Answers
Willie, this is ,clearly, saying that every local base can not be finite.
I can not write comments that's why i am writing this like answer. Can I?
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Thanks for verifying my question (I am a bit rusty here). You'll get to comment after you earn a bit more reputation points. See the FAQ for more details. – Willie Wong Mar 20 2010 at 15:44
### 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.
Isn't this just the Base of the topology?
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Huh? Take the set {0,1} with the discrete topology. The subsets { {0}. {1}} form a base of the topology, and they don't satisfy the conditions given in the question. Perhaps you are thinking of something else? – Willie Wong Mar 20 2010 at 13:39
Stupid me. I missed the word 'Proper'. – Undergrad Mar 20 2010 at 13:59
Are you just saying that the topology is an atomless lattice? I'd call it "a space with atomless topology".
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1
Consider the integers Z, with open sets (-infty,n). This is a topology and is an atomless lattice, but it doesn't have the desired property. – Joel David Hamkins Mar 20 2010 at 11:08
Hmm, good point. I would like to hear where these spaces come from. – Andrej Bauer Mar 20 2010 at 20:00
In spaces where singleton points are closed, your property is equivalent to saying that the space has no isolated points. Or in other words, that it is perfect.
Clearly, no space with an isolated point can have your property. Conversely, when singletons are closed, then you can subtract one point from any open set and thereby have a proper open subset. So if U has at least 2 points x,y, then U = U-{x} union U-{y}, giving an instance with I of size 2.
However, your property does not imply that points are closed, since the space on reals R, where open sets have the form (-infty, a), has your property, but points are not closed in this space.
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Note that in the example space I give, no finite I would suffice, in contrast to the T1 perfect space situation, where I of size 2 suffices. – Joel David Hamkins Mar 20 2010 at 2:26
Sort of going the other way (you gave a condition that implies the OP's condition), the condition implies that any local system, ordered by set inclusion, cannot have a smallest element. I think this is saying that local base cannot be finite? (This distinguishes your two examples of R with (-infty,a) and Z with (-infty,n). ) – Willie Wong Mar 20 2010 at 14:01
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http://math.stackexchange.com/questions/98272/on-the-vector-product/98279
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# on the vector product
Can the product(element to element) of two vector $\vec{a}$ and $\vec{b}$ with the same dimension be written as the usual vector(or matrix) product?
Say $\vec{a}=[2,3],\vec{b}=[4,5]$ and $\vec{a}~\text{product(element to element)}~\vec{b}=[8,15]$
Thank you!
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if you mean dot product ,then it is just scalar not vector,in your case a*b=23 – dato Jan 11 '12 at 19:56
Yes, I know. I mean I can add some other constant matrix like identity matrix, or other constant vector like $e_j$, anyway, I just want to get rid off that kind of unusual product. – breezeintopl Jan 11 '12 at 20:01
## 2 Answers
No, not if you mean standard matrix multiplication. The matrix product of two $n\times 1$ or $1\times n$ matrices does not exist. Even if you allow taking the transpose, the product of a $1\times n$ matrix and a $n\times 1$ matrix is $1\times 1$ while the product of a $n\times 1$ matrix and a $1\times n$ matrix is $n\times n$. None of these are vectors, so this can't be what you are looking for.
However, as Andreas pointed out, there is a name for the operation you are talking about, the Hadamard product, and it generalizes to all matrices.
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Ok, thank you so much! I understand. – breezeintopl Jan 11 '12 at 20:08
For matrices there is the Hadamard product with $(A \circ B)_{i,j} = (A)_{i,j} \cdot (B)_{i,j}.$ So think of vectors as $1-\text{by}-2$ matrices.
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http://gowers.wordpress.com/2012/11/
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# Gowers's Weblog
Mathematics related discussions
## Archive for November, 2012
### What maths A-level doesn’t necessarily give you
November 20, 2012
I had a mathematical conversation yesterday with a 17-year-old boy who is in his second year of doing maths A-level. Although a sample of size 1 should be treated with caution, I’m pretty sure that the boy in question, who is very intelligent and is expected to get at least an A grade, has been taught as well as the vast majority of A-level mathematicians. If this is right, then what I discovered from talking to him was quite worrying.
The purpose of the conversation was to help him catch up with some work that he had missed through illness. The particular topics he wanted me to cover were integrating $\log x$, or $\ln x$ as he called it, and integration by parts. (Actually, after I had explained integration by parts to him, he told me that that hadn’t been what he had meant, but I don’t think any harm was done.) But as we were starting, he asked me why the derivative of $e^x$ was $e^x$, and what was special about $e$.
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Posted in Mathematical pedagogy | 116 Comments »
### What actually happened
November 9, 2012
The short version is that I’ve had the ablation (see previous post) and the surgeon who did it says that he has a good feeling about it. It’s taken till now to write this because, unlike most people who have ablations, I felt terrible for two days after it — with a headache (normal) and a fever (less normal but not unheard of). The fever was not very high, but high enough to be unpleasant, and meant that the only thing I could bear to do was go to bed, except that on the second night after the operation I had to spend part of the night sitting up on a sofa because my chest hurt too much when I was horizontal. (That was normal, and nothing to worry about.) So today is the first day that I am well enough to do anything as strenuous as writing a blog post.
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Posted in General | 28 Comments »
### Mathematics meets real life
November 5, 2012
I’ve been in two minds about whether to post this. On the one hand, I try to keep personal matters out of this blog — though there has been the occasional exception — but on the other hand I have a topic that fits quite nicely with some of what I’ve been writing about recently, since it concerns a fairly important medical decision that I have had to make based on what felt like inadequate information. Since that is quite an interesting situation from a mathematical point of view, and even a philosophical point of view, and since most people have to make similar decisions at some point in their lives, I have opted to write the post.
The background is that over the last fifteen years or so I have had occasional bouts of atrial fibrillation, a condition that causes the heart to beat irregularly and not as strongly as it should. It is quite a common condition: I’ve just read that 2.3% of people over the age of 40 have it, and 5.9% of people over 65. Some people have no symptoms. I myself have mild symptoms — I can feel a slightly strange, and instantly recognisable, feeling in my chest, and I experience a few seconds of dizziness almost every time I stand up from a relaxed seated position — otherwise known as orthostatic hypotension, which I often used to get anyway (as do many people).
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Posted in General | 46 Comments »
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http://math.stackexchange.com/questions/234832/show-that-e-x2-is-the-solution-to-the-initial-value-problme-y2-4x2y
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# Show that $e^{-x^2}$ is the solution to the initial value problme $y''+(2-4x^2)y=0$ $y(o)=1$ $y'(0)=0$
I have already worked out a lot of stuff already from previous parts of the questions and I think I've got to work from the fact that $a_{2k+1}=0$ and $a_{2k}= (-1)^k/k!$ for k>0 satisfies the recurrence relation of the power series as it states this and then says hence deduce that $y(x)=e^{-x^2}$
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## 2 Answers
$$e^x = \sum^{+\infty}_{k=0}\cfrac {x^k}{k!} \implies e^{-x^2}=\sum^{+\infty}_{k=0}\cfrac {(-x^2)^k}{k!}=\sum^{+\infty}_{k=0}\cfrac {(-1)^k}{k!}x^{2k}$$ The coefficient of $x^{2k}$ satisfies the sequence you found.
Working forward
$$\sum^{+\infty}_{k=0}a_k\ x^k =\sum^{+\infty}_{k=0}a_{2k}\ x^{2k}+\sum^{+\infty}_{k=0}a_{2k+1}\ x^{2k+1}=\sum^{+\infty}_{k=0}\cfrac {(-1)^k}{k!}\ x^{2k}=\sum^{+\infty}_{k=0}\cfrac {(-x^2)^k}{k!}=e^{-x^2}$$
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The only thing you have to do is to show that $f(x)=e^{-x^2}$ satisfies the equation with the initial values. By the existence and uniqueness theorem for ode, you will know that this is the solution.
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http://math.stackexchange.com/questions/298543/how-to-show-ln-lnk1-ln-lnk-frac1k-lnk-forall-k-in-mathbbn
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How to show $\ln(\ln(k+1))-\ln(\ln(k))<\frac{1}{k \ln(k)},\forall K\in \mathbb{N}, K\geq 2.$
How can I show the inequality $\ln(\ln(k+1))-\ln(\ln(k))<\dfrac{1}{k\ln(k)},\forall K\in \mathbb{N}, K\geq 2.$
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1
Welcome to MSE! If this is homework, please tag it as such, people will help you anyway. And please try and tell us what you have tried so far, and where you got stuck. Also, please take a look at the revision of the formatting I did on your post. – Andreas Caranti Feb 9 at 6:57
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Actually, it is not homework question. The question is taken from one of the previous mathematics entrance examination. Thank you – ftolessa Feb 9 at 7:39
2 Answers
Hint
What's the value of the derivative of $\ln(\ln(x))$ at $x = k$?
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From Mean value theorem we have, $$\displaystyle\frac{\ln(\ln(k+1))-\ln(\ln(k))}{k+1-k}=\frac{1}{\ln c}.\frac{1}{c},c\in(k,k+1)$$ $$k<c\Rightarrow\frac{1}{\ln c}.\frac{1}{c}<\frac{1}{\ln k}.\frac{1}{k}$$ $$\Rightarrow\displaystyle{\ln(\ln(k+1))-\ln(\ln(k))}=\frac{1}{\ln c}.\frac{1}{c}<\frac{1}{\ln k}.\frac{1}{k}$$ $$\Rightarrow\displaystyle{\ln(\ln(k+1))-\ln(\ln(k))}<\frac{1}{\ln k}.\frac{1}{k}$$ We are done.
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Thank for your you valuable time. Now it is okay. – ftolessa Feb 9 at 8:02
By the way is it possible to show $\sum_{k=2}^{\infty}(ln(ln(k+1))-ln(lnk))=\infty$ so that by comparison test $\sum_{K=2}^{\infty}\frac{1}{klnk}$ also divergent? – ftolessa Feb 9 at 8:13
I guess yes@user61455 – Abhra Abir Kundu Feb 9 at 8:57
Another test which you can use $\sum a_i$ is convergent if and only if $\sum 2^ka_{2^k}$ is convergent. – Abhra Abir Kundu Feb 9 at 9:02
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http://math.stackexchange.com/questions/111541/nth-roots-and-e
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# $n^{th}$ roots and $e$.
In Churchill's book on complex variables, the $n^{th}$ root of $e$ is defined to be $e^{1/n}$. A comment is made that in this respect $e$ is treated differently than the $n^{th}$ roots of other complex numbers (in the sense that there are typically n roots of the nth root of a number in complex analysis rather than just one as in the case of $e$).
I am curious why $e$ is treated so differently. Is there an obvious reason/motivation why?
Edit: The section from Churchill is,
As anticipated earlier, we define here the exponential function $e^z$ by writing $$e^z = e^xe^{iy}\ \ \ \ \ \ (z = x + iy)\ \ \ \ \ \ \ \ \ (1)$$ where Euler's formula $$e^{iy} = \cos y + i\sin y$$ is used and $y$ is to be taken in radians. We see from this definition that $e^z$ reduces to the usual exponential function in calculus when $y=0$; and, following the convention used in calculus, we often write $\exp z$ for $e^z$.
Note that since the positive $n$th root $\sqrt[n]{e}$ of $e$ is assigned to $e^x$ when $x = 1/n$ ($n = 2,3,\ldots$), expression (1) tells us that the complex exponential function $e^z$ is also $\sqrt[n]{e}$ when $z = 1/n$ ($n = 2,3,\ldots$). This is an exception to the convention that would ordinarily require us to interpret $e^{1/n}$ as the set of $n$th roots of $e$.
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Hmmm, that doesn't sound right. What page of Churchill's book are you looking at? – Álvaro Lozano-Robledo Feb 21 '12 at 2:32
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Apart from the Routine $\TeX$ edit, I have made vars "variables". Please do not make this place into a SMS chat. This is a dedicated forum for Math Q&A. Further, any signature or tagline for the posts is advised against, by the faq. So, I have removed it. – user21436 Feb 21 '12 at 2:33
The constant $e$ actually does have $n$ distinct $n$-th roots, for all $n=1,2,3,\cdots$. The only number in all of $\mathbb{C}$ without such a property is $0$. I'm not familiar with Churchill - is your parenthetical your own understanding or something specifically stated by the author? Generally in complex analysis the function $z\to z^{1/n}$ is chosen with a particular branch in mind rather than assumed multi-valued. // Could you transcribe the comment in full, if possible? – anon Feb 21 '12 at 2:34
The only explanation I can think of is that $f(z)=z^{1/n}$ is taken multi-valued while $e^w$ is defined to be the exponential, given by the usual power series, and hence takes on only one value. – anon Feb 21 '12 at 2:37
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@MattBrenneman and everyone else! I've added the passage in the book. Hopefully it's everything you mentioned, and if not please feel free to add (or emphasize) any other parts! – Alex Feb 21 '12 at 23:49
show 2 more comments
## 2 Answers
The natural exponential function is defined by $$\exp(z) = \sum_{n=1}^\infty {z^n\over n!}.$$ This is an entire function. It is not hard to show it has all of the expected properties.
To define $z^w$ you must define something like $$z^w = \exp(z\log(w)).$$
Unfortunately, the exponential function is $2\pi i$-periodic. Therefore it is not 1-1, so the business of defining a logarithm function becomes tricky. You must choose a domain to restrict the exponential function to so it is 1-1. And there the trouble begins. But where the trouble begins, complex analysis begins in all of its beauty and elegance.
I quote one of my grad school professors, Sidney Graham, who said, "There are those who say that the study of complex variables is the study of the logarithm function."
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The point is that we want "the" exponential function to be single-valued. If you want to write $\exp(1/n)$ as ${\rm e}^{1/n}$, that singles out one "$n$'th root of e". There are still $n$ $n$'th roots of e, it's just that only one of them is written as ${\rm e}^{1/n}$.
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http://mathoverflow.net/questions/38738/a-question-on-the-prime-divisors-of-p-1/49562
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## A question on the prime divisors of p-1
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For each positive integer n we may define the convergent sum $$s(n)=\sum_{p}\frac{(n,p-1)}{p^2}$$ where the summation is over primes p and $(a,b)$ denotes the greatest common divisor of a,b.
It is immediate to deduce that s(n) is bounded on average:
Using $\sum \limits_{d|a, d|b}\phi(d)=(a,b)$ and inverting the order of summation we get
$s(n)=\sum_{d|n}\phi(d) a_d$ where $a_d=\sum_{p \equiv 1 (mod \ d)}p^{-2}$
Ignoring the fact that we sum over primes we get the bound $a_d \ll \frac{1}{d^2}$ which leads to $$\sum_{n \leq x}s(n) \ll x \sum_{d \geq 1}\frac{\phi(d)a_d}{d}=O(x)$$ and $$s(n) \ll \exp(\sum_{p|n}1)$$ The last inequality means that $s(n)$ stays bounded if $\omega(n)$ is bounded. Towards the other direction, it seems fair to expect that $s(n)$ grows to infinity if $\omega(n)$ is large in some quantitative sense, say $\omega(n) \geq (1+\epsilon) \log \log n$. Taking into account that the contribution to the sum $s(n)$ of the primes $p$ that satisfy $(p-1,n) \leq \frac{p}{\log p}$ is bounded, since $\sum_{p}\frac{1}{p \log p}$ converges, we see that $s(n)=s'(n)+O(1)$ where $s'(n)=\sum_{ (p-1,n)>\frac{p}{\log p}} \frac{(n,p-1)}{p^2}$ We are therefore led to the question as to whether a condition of the form $\frac{\omega(n)}{\log \log n}-1 \gg 1$ can guarantee that $s'(n) \to +\infty$ Are there any non-trivial techniques that can be used to answer this question ?
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Is the last $\log n$ in the question supposed to be $\log\log n$? – Gerry Myerson Sep 15 2010 at 6:52
## 3 Answers
Are you just trying to show $s(n)$ is unbounded? and do you insist on a non-trivial technique? Let $n=m!$; then $s(n)>\sum_{p\lt m}{p-1\over p^2}=\sum_{p\lt m}{1\over p}+O(1)$ and of course the sum diverges.
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(corrected the typo log log n, thanx !) your choice of n shows that $s(n) \gg \log \log \log n$ infinitely often. Choosing $n=\prod_{p \leq m}(p-1)$ we deduce that $s(n) \gg \log \log n$ infinitely often. But my question was of another nature : Is it true that $s(n) \to \infty$ for each sequence of integers n that have a large number of prime divisors (not for just a particular sequence of the form $m!$ or $\prod_{p \leq y}(p-1))$ ? – Captain Darling Sep 15 2010 at 12:13
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
A few more ideas: using the chebyshev upper bound, by partial summation we have $\sum_{p>y}p^{-2}=O(\frac{1}{y \log y})$ and therefore we see that $s(n)=\sum_{p \leq \frac{n}{\log n}}\frac{(p-1,n)}{p^2}+O(1).$ Furthermore by the equality $\sum_{p \leq x}p^{-1}=\log \log x +A +(\frac{1}{\log x})$ we get $$s(n)=\sum_{p \leq n^{1/3}}\frac{(p-1,n)}{p^2}+O(1)$$ and one can make this more accurate. There is something in the expression $s(n)=\sum_{d|n}\phi(d)a_d$ that is linked to Linnik's constant(or the Elliot-Halberstam Conjecture). In particular, using $a_d=\sum_{p=1(mod d), p>y} p^{-2} \leq d^{-2}\sum_{m>y/d}m^{-2}$ one can deduce that $$s(n)=\sum_{d|n}\phi(d) a'(d)+O(1)$$ where $$a'(d)=\sum_{p \leq d \log d {\log \log d}^2, p=1(mod d)}\frac{1}{p^2}$$ That seems to suggest that for each n for which $s(n)$ is quite large then for many divisors $d|n$ there might be many primes $p=1 (mod d)$ in the interval $[d,d \log d {\log \log d}^{2}]$ and conversely, but I haven't been able to establish a clear connection between these two facts. To this end we may compute the mean values $\sum_{n \leq x} s^{2k}(n), k \geq 0$ which is quite straightforward. Does all this set-up reminds you anything I could look up?
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Sorry I posted something wrong and can't delete.
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http://mathhelpforum.com/pre-calculus/101780-absolute-value-inequality.html
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# Thread:
1. ## Absolute value inequality
Solve the inequality |x^2 - x| - |x| < 1 analytically.
2. $\left| x \right|\left| x-1 \right|-\left| x \right|-1<0.$
Three cases to solve this:
• $x\le0.$
• $0\le x\le1.$
• $x\ge1.$
3. Hmm how did you come up with the second case, $0 \leq x \leq 1$? In the past for all equalities I had set the overall expressions to be either $x < 0$ or $x \geq 0$
4. Originally Posted by xxlvh
Hmm how did you come up with the second case?
Because you can't rule anything out. You must assume that x can be anything. Only upon inspection can you begin to narrow the solutions to specified intervals.
5. Originally Posted by xxlvh
Hmm how did you come up with the second case, $0 \leq x \leq 1$? In the past for all equalities I had set the overall expressions to be either $x < 0$ or $x \geq 0$
Because you have |x- 1|, not just |x|. The formula for absolute value "changes" when whatever is inside the absolute value equals 0. For |x|, that is x= 0. For |x-1| it is x-1= 0 or x= 1.
6. If it is for either x = 0 or x = 1, why is the case 0 < x < 1 since that involves all the values between those numbers as well?
Sorry for my slowness but I have received no prior instruction in inequalities.
So far here is my attempt to solve, nearly positive there's a few mistakes:
Case 1: x < 0
$-x(-x-1) + x - 1 < 0$
$x^2 + x + x - 1 < 0$
$x^2 + 2x - 1 < 0$
(used quadratic formula to solve x's)
$(x + 2.4142)(x - 0.4142) < 0$
solution: 0.24142 < x < 0.4142
If I pay attention to the earlier restriction on x, would that be changed to 0.24142 < x < 0 ?
Case 2:
I don't even know where to begin to solve this one
Case 3: x > 1
$x(x-1) - x - 1 < 0$
$x^2 - x - x - 1 < 0$
$x^2 - 2x - 1 < 0$
$(x + 0.4142)(x - 2.4142) < 0$
solution: -0.4142 < x < 2.4142
which then becomes 1 < x < 2.4142
7. Originally Posted by xxlvh
Hmm how did you come up with the second case, $0 \leq x \leq 1$? In the past for all equalities I had set the overall expressions to be either $x < 0$ or $x \geq 0$
Put $x=0$ in that inequality, is it true? In the same fashion put $x=1,$ is it true?
In both cases, those values satisfy the inequality, so that's why I picked $x\le0$ instead of $x<0.$
As for the sake of the problem, you solved the first case wrong, 'cause for $x\le0$ the inequality becomes $x^{2}-x+x-1<0\implies -1<x<1.$ Since $x\le0,$ the first solution set is $(-1,0].$
For the second case, the inequality becomes $x(1-x)-x-1<0\implies x^{2}+1>0,$ which is obviously true, so the second solution set it's just $[0,1].$
Finally, having $x\ge1,$ the inequality becomes $x^{2}-x-x-1<0\implies (x-1)^2<2,$ and then $1-\sqrt2<x<1+\sqrt2,$ so the third solution set is $\Big[1,1+\sqrt2\Big).$
Put these solutions set together and the original inequality verifies for each $-1<x<1+\sqrt2.$
8. Thank you, it's much clearer now. I'm just curious, when solving for the second case would it be possible to go about it without first factoring the original equation out into |x||x-1|-|x|-1 < 0? Under normal circumstances I doubt I could've thought of that myself
9. if you don't factorice, then you can't realize what are the critical points, we need them to solve the initial inequality.
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http://mathoverflow.net/questions/39923/can-a-self-adjoint-operator-have-a-continuous-set-of-eigenvalues/77406
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## Can a self-adjoint operator have a continuous set of eigenvalues?
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This should be a trivial question for mathematicians but not for typical physicists.
I know that the spectrum of a linear operator on a Banach space splits into the so-called "point," "continuous" and "residual" parts [I gather that no boundedness assumption is needed but I could be wrong]. I further know that the point spectrum coincides with the set of eigenvalues of the operator. It seems from the terminology that the point spectrum is a discrete set of isolated point and that the eigenvalues cannot form a continuum. But I haven't been able to find a clear statement in a math reference about this.
Actually, I'm mostly interested in self-adjoint operators on a Hilbert space; so a simpler version of my question would be: Can a self-adjoint operator have a continuous set of eigenvalues? And if yes, under what conditions do the eigenvalues have to be discrete?
I appreciate any help.
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The answer is yes for certain (easy) self-adjoint operators on a suitable non-seperable Hilbert space. I assume you want to restrict your attention to separable ones? – Yemon Choi Sep 25 2010 at 1:50
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While this doesn't answer your question about self-adjoint operators, the unilateral backwards shift $S:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$, given by $$S(x_1,x_2,\dots) = (x_2,x_3,\dots)$$ has lots of eigenvalues -- namely the whole open unit disc. – Yemon Choi Sep 25 2010 at 1:54
Thanks for both of your good comments. Yes I do assume separable Hilbert spaces. The example is also pretty illuminating. – Mahdiyar Sep 26 2010 at 6:16
## 8 Answers
Eigenvectors for different eigenvalues of a self-adjoint operator are orthogonal. In a separable Hilbert space, any orthogonal set is countable. So a self-adjoint operator on separable Hilbert space has only countably many eigenvalues. (As noted, this does not mean the spectrum is countable.)
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Such a simple answer! Thanks. This is exactly what I expected for the "simpler version" of my question. Actually Piero D'Ancona also provided the same answer to my question but this one was more to the point and simpler to understand (for someone like me!). – Mahdiyar Sep 26 2010 at 7:03
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
You are confusing two notions. First of all, the point spectrum just means eigenvalues; there is no assumption that these form a discrete set. The shift operator is a simple example where the spectrum is "continuous".
The condition for the eigenvalues to be discrete is precicsely that the operator $A:H \to H$ is compact. It is however possible for non-compact self adjoint operators to have a discrete spectrum. The simplest example of this is the orthogonal projection operator $P:H \to Y$ where $Y$ is a closed subspace of the Hilbert space $H$. Here the spectrum is $0$ and $1$.
Bounded self adjoint operators have no residual spectrum but they do indeed have a continuous spectrum. Take any compact operator $A:H \to H$ where dim$H=+\infty$. Then $0$ belongs to the continuous spectrum because otherwise $A:H \to H$ would be invertible, implying that dim$H <\infty$. Continuous = "exists a set of approximate eigenvectors".
If you want a continuous range of spectrum take $Af = x f(x)$ on $L^2([0,1])$. Then the range of the spectrum is just $[0,1]$. There are no eigenvalues for this operator and moreoever since the residual spectrum is empty for self adjoint operator, $[0,1]$ is the spectrum and it is equivalent to the continuous spectrum.
So a final point, continuous just means that $R(A-\lambda I)$ is not dense but that $\lambda$ is not an eigenvalue. It has nothing to do with the actual spectrum being discrete or continuous.
So I think you were mixing up two notions but hopefully I've provided examples for both.
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"take Af=xf(x) on L2([0,1]). Then the range of the spectrum is just [0,1]. These are all eigenvalues." -- Are you sure? – Yemon Choi Sep 25 2010 at 2:58
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Also, the eigenvalues of a compact operator do not form a discrete set, since they accumulate at zero. – Yemon Choi Sep 25 2010 at 2:59
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If you take multiplication by the function $f(n) = n$ on the integers with counting measure, then the spectrum coincides with the point spectrum and is the integers, so discrete, but of course the operator is not bounded and so not compact. – Dick Palais Sep 25 2010 at 3:50
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I thought "discrete" meant "discrete as a subset of the complex plane" but perhaps I've misremembered the terminology – Yemon Choi Sep 25 2010 at 4:19
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I wouldn't call a set with an accumulation point, "discrete". How would it be different from a set with many accumulation points? – Martin Argerami Sep 25 2010 at 10:09
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One version of the spectral theorem states that if $A$ is a selfadjoint operator on a separable Hilbert space $H$, then we can find a set $X$, a $\sigma$-finite measure on $X$, a unitary operator $U:H\to L^2(X)$, and a measurable function $a:X\to \mathbb{R}$, such that $A=U^*aU$. In other words, any selfadjoint operator, in essence, is nothing but 'multiplication by a real valued function', in suitable 'coordinates'. The spectrum of $A$ coincides with the essential range of the function $a$.
Now, an eigenvalue must be a real number $\lambda$ such that the set $a^{-1}(\lambda)$ has a positive measure. Since $X$ is $\sigma$-finite, the eigenvalues can be at most a countable set. But you can have a continuous spectrum of course, only the points will not be eigenvalues. You can play with real valued functions (and with measures on $X$; any $\sigma$-finite measure is ok) to train intuition.
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Recall that the spectrum of an operator $A$ on a Hilbert space is the set of vales $\lambda$ such that $A-\lambda I$ does not have a bounded inverse . So if $A$ is multiplication by a function in $L^2$ of a measure space, then any point of the essential range of the function is in the spectrum. So, for example (and this is the classic example) the spectrum of multiplication by $x$ on the real line is the whole line. However, a point of the spectrum is not necessarily an eigenvalue. In fact $\lambda$ is an eigenvalue (or in the point spectrum) iff $A- \lambda I$ has a non-trivial null space. (And as others have pointed out, on a separable Hilbert space, the point spectrum is countable.)
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I add this remark because it may be part of what the OP wants. Note that, as to the spectrum of a bounded linear self-adjoint operator on $\ell^2$, of course, it can be any compact set $K$ of $\mathbb{R}$. Just take a diagonal operator where the set of the diagonal elements (eigenvalues) is dense in $K$.
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Just so we have the silly example here. Consider $\ell^2([0,1])$ meaning the set of SEQUENCES $u_{x}$ indexed by a number $x\in [0,1]$. So the scalar product is $$\langle u, v \rangle = \sum_{x \in [0,1]} \overline{u_x} v_x.$$ In order for $u \in \ell^2([0,1])$, we have that $u_x \neq 0$ for at most countably many $x$. So this is very different from $L^2([0,1])$.
Now consider the operator $$(Au)_x = u_x.$$ This operator is diagonal and it's eigenvalues are $[0,1]$. The eigenvector corresponding to $x \in [0,1]$ is $$u_y = \begin{cases} 1, & x=y\\ 0, & \text{otherwise}.\end{cases}$$ Of course the key to this example and the question is that $\ell^2([0,1])$ is a NON-separable Hilbert space.
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The hyperbolic Laplacian is self adjoint and its spectrum has a continuous component.
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What is a hyperbolic laplacian? Isn't the laplacian elliptic? – Dorian Sep 25 2010 at 3:45
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The spectrum of the standard Laplace operator on $R^n$ is $\sigma(-\Delta)=[0,+\infty)$. – Piero D'Ancona Sep 25 2010 at 11:31
I would guess that hyperbolic Laplacian means Laplacian on hyperbolic space. – Emerton Sep 26 2010 at 1:19
The resolvent set is the set of all $\zeta \in \mathbb{C}$ for which $T-\zeta$ is invertible (which means especially that the Range is all of $H$). The spectrum $\Sigma$ is the complement of the resolvent.
Normally, a $\lambda \in \Sigma$ is only called eigenvalue, if a $x \in H$ exists such that $Tx = \lambda x$. The set of those eigenvalues is countable as already argued but not necessarily discrete. For example: A compact operator on a Hilbert space has a countable spectrum, but not necessarily a discrete one; the spectrum has an accumulation point at zero, which may be an eigenvalue.
A noncompact bounded operator may have a countable spectrum consisting only of "true" eigenvalues, but then the eigenvalues -- counted with multiplicity -- have a nonzero accumulation point.
An unbounded operator whose resolvent is compact (for some element of the resolvent set) has a discrete spectrum; this criterion works on Laplacians on compact manifolds, for example. The Laplacian on $L^2(\mathbb{R}^n)$ with usual Lebesgue measure -- which is self-adjoint if $\mathrm{dom}(\nabla) = H^2$ -- however has as spectrum the whole positive real line, but no eigenvalue.
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http://physics.stackexchange.com/questions/45451/spring-coupled-platforms-conservation-of-momentum-can-it-be-solved-with-fres/45468
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Spring coupled platforms & conservation of momentum - can it be solved with freshman physics?
This question came up as an exercise in a first year undergraduate course I was a TA for. It turned out to be a lot more difficult (impossible?) than anticipated...
Two platforms of mass $M_1$ and $M_2$ ($M_1\neq M_2$) are connected by a spring of constant $k$, and are initially at rest with the spring unstretched, sitting on a frictionless surface. A man of mass $m$ stands on one platform and begins to run, always with a constant speed $v$ measured relative to the platform he is running on. What is the maximum speed reached by the other platform, relative to the ground?
My intuition is telling me that I need to know something about how the man gets from rest to his constant speed (instantaneously? very slowly? with some smooth acceleration?) to solve this, but I haven't been able to prove to myself that this is required. If this is a requirement, I think the most reasonable assumption would be that he reaches his full speed instantaneously.
What I'm most interested in is whether this problem can be tackled with a typical freshman's toolbox - simple arguments around conservation of energy/momentum, no/very limited differential equations, basic calculus. I can see an easy way to get an upper bound on the maximum speed from energy/momentum considerations, but I don't see a way to check if this speed is ever reached.
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1 Answer
You can ignore the spring to get the stead state solution using conservation of linear momentum.
With the spring, you will need the natural frequency of the system and its mode shapes to find the maximum velocity. If the man starts with an impulse to $M_1$ to get him from $0$ to $v$ in a small amount of time then the problem is someone hitting the two mass system with a hammer and it can be solved (albeit not on freshman level).
I was able to solve it by fitting the following general solution to the two differential equations
$$x_1(t) = X_1 \sin(\omega t+\varphi)-X_1 \omega t \cos\varphi-X_1 \sin\varphi-\frac{m v}{M_1} t \\ x_2(t) = X_2 \sin(\omega t+\varphi)-X_2 \omega t \cos\varphi-X_2 \sin\varphi$$
and solving for $X_1$, $X_2$, $\varphi$, $\omega$. It helps to start with $X_1=-\frac{M_2}{M_1+m} X_2$.
The above fits the initial conditions at rest, but with $v_1 =-\frac{m v}{M_1}$.
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+1 for providing a full solution, but what I was really hoping for was either a proof or disproof of whether this can be solved with a limited tool set. – Kyle Dec 4 '12 at 16:02
Sure for MIT students! – ja72 Dec 4 '12 at 18:10
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http://www.physicsforums.com/showthread.php?t=148386
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Physics Forums
Recognitions:
Homework Help
## An introduction to the analysis of statically indeterminate systems
It is assumed that the reader is familiar with the basic concepts of structural analysis. Further on, I am aware of the possible inconsistencies and errors in this writing, and hereby invite all who have comments and objections to contribute and help to make this small tutorial useful for anybody interested in the analysis of statically indeterminate systems.
1. THE FORCE METHOD.
1.1. INTRODUCTION THROUGH AN EXAMPLE.
Perhaps the most standard example of the force method application is the system in Figure 1. It is obvious that the system's degree of statical indeterminacy equals one, since the number of constraints equals 4, and the number of equations of equilibrium we can use equals 3. Note that, if we removed the roller support at point B, we would be left with a statically determinate console with a support at point A, which (the support) represents three kinematic constraints. Hence, we can refer to the support at point B as a statical constraint, since the system, after having this support removed, still remains geometrically stable (i.e. it does NOT become a mechanism).
Figure 1.
The geometrical and material characteristics are all displayed in Figure 1. The module of elasticity equals E, and it is constant along the whole length L, as is the moment of inertia of the cross section, denoted by I (with respect to the y-axis, which is pointing 'towards us').
Let us remove the roller support at point B, and 'replace it' with its reactive force $R_{B}$. We shall explicitely use the equations of equilibrium for the z-direction (1), and we shall set the sum of the torques with respect to point A to equal zero (2):
(1)
$$\sum F_{z} = 0 \rightarrow R_{A} + R_{B} = qL \ \text{,}$$
(2)
$$\sum M_{A} = 0 \rightarrow M_{A} + R_{B}L = \frac{qL^2}{2} \ \text{,}$$
where $M_{A}$ is the reaction torque at point A, $R_{A}$ the reaction force in the z-direction at the same point, and $R_{B}$ the reaction force in the same direction at point B. (It is assumed that the positive direction of the torques is counter-clockwise, and the positive direction of the vertical forces is in the positive direction of the z-axis).
It is obvious that the system of two equations (1) and (2), which represents the conditions of equilibrium of the system, has no unique solution, since we are left with two equations and three unknowns, $M_{A}$, $R_{A}$ and $R_{B}$. Further on, we can write the system of equations in the form:
(1')
$$R_{A} = qL - R_{B}$$
(2')
$$M_{A} = \frac{qL^2}{2} - R_{B}L \ \text{.}$$
Now we can see the the equations of equilibrium will be satisfied for any value of the reaction $R_{B}$, i.e. the system has a one-parameter solution, where the parameter is $R_{B}$. This means that there is an infinite number of equilibrium states of the system. But, the one we are trying to find is the one which satisfies the displacement conditions of the system - i.e. the vertical displacement at point B equals zero, since, in the original system there exists a support at point B which prevents a vertical displacement at that point.
The vertical displacement of the point B has two contributions: one comes from the load q, and the other from the reactive force $R_{B}$ acting at that point. So, it can be expressed as the superposition of these two contributions, respectively:
$$w_{B} = \frac{qL^4}{8EI} - R_{B}\frac{L^3}{3EI} \ \text{.}$$
(Note that the displacement of the reactive force $R_{B}$ has a negative sign, since the force is assumed to point in the negative direction of the z-axis.) As mentioned, the displacement must equal zero, which represents the additional condition we need to describe the desired 'real' state of equilibrium of the system, so $w_{B} = 0$ (3) implies:
$$R_{B} = \frac{3}{8}qL \ \text{.}$$
Now we can finally solve equations (1') and (2'), i.e. calculate the reactive forces $R_{A}$ and $M_{A}$, and, of course, solve standard statics problems as before - find bending moment and shear force diagrams, displacements, etc.
The provided example was supposed to serve as a brief introduction to the force method, whose summary can easily be distinguished from the very same example. First, a support, which represents a statical constraint is removed and replaced with a reactive force in the 'position and direction' of the support. Further on, the equations of equilibrium are set up. Finally,a kinematic condition (a condition of displacement, i.e. a geometric condition) is set up to determine which state of equilibrium is the real one. As a final illustraion, one can find deflection curves for some values of $R_{B}$ which do not satisfy the condition (3) in Figure 2, while Figure 3 represents the actual deflection curve of the bar, which is in consistence with the condition (3). (The console is placed on the horizontal axis, and the diagrams are plotted for L = 5. The vertical axis represents the vertical displacement w.)
Figure 2.
In the end, it is of great importance to mention that this rough illustration of the force method concept applies to systems with more than one degree of statical indeterminacy - in general, for a statically indeterminate system with the degree of statical indeterminacy n, one merely has to pick n different supports (i.e. statical constraints) and replace them with forces acting in the points and the directions of the supports. Of course, this system, which is referred to as the primary system, is not allowed to be a mechanism, which means we can not remove just any support as we wish. In our example, we could as well add a hinge to point A, which physically means we have removed the part of the rigid support which transfers torques. After that, we would have to add a torque M to that point, and that torque should satisfy condition equivalent to (3), which would state that the rotation at point A should equal zero. In general, when dealing with statically indeterminate systems with a degree of statical indeterminacy equal to n, we have n displacement conditions (in our case we have only one - equation (3) ).
Figure 3.
A more detailed and formal outline of the force method shall be introduced in section 1.2.
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Recognitions: Homework Help 1.2. A FORMAL OUTLINE (I). Under the assumption that the idea of the force method was successfully illustrated in Section 1.1., we shall now try to generalize and formalize that idea. It first has to be mentioned that the statical constraints that are to be removed from the system do not have to be external constraints, such as supports - one can add internal hinges or roller connections to the system too, which also minimizes the degree of statical indeterminacy. This is shown in Figure 4. However, one has to add a pair of genenralized forces (forces or moments) which are to prevent a relative generalized displacement (translation or rotation) at that point. The pair of generalized forces consists of two forces with an equal magnitude and direction, but of opposite orientation. Figure 4. As mentioned before, for a system with a degree of statical indeterminacy which equals n, one has to remove (and/or add) a total number of n supports (and/or internal hinges). Of course, the resulting primary system must not be a mechanism, i.e. it has to have zero degrees of freedom. We can accomplish this in a lot of ways - it will be seen later that there are smarter ways to do that, since we shall need to sketch different internal force diagrams, which can become fairly complicated, depending on the supports we choose to remove (or the hinges we choose to add). After creating the primary system, we place the set of generalized forces $\left\{X_{1}, \cdots, X_{n}\right\}$ on the system in such a manner that each force acts at the point and in the direction of the removed support. In the example in Section 1.1., the generalized force was the reaction $R_{B}$, which was placed at the point of the removed support. (Note that the generalized forces may represent pairs of forces too, as shown in Figure 4.) After doing so, it is intuitively clear that one has to solve a system of n equations, each of which expresses a displacement condition at one of the points where the supports were removed (or where internal hinges were added - we shall, for practical reasons, from now on refer to this procedure as support removal only). These equations are often referred to as compatibility equations, since they guarantee that the primary system with the generalized forces acting on it must have the same displacements as the original system. We shall now analyse the structure of the i-th equation, which has the form: $$\sum_{k=1}^n X_{k} \delta_{ik} + \delta_{i0} = 0 \ \ \text{(1),}$$ where $X_{k}$ represents the force acting at point k, $\delta_{ik}$ represents the generalized displacement at the point i, caused by the force k, and $\delta_{i0}$ represents the generalized displacement at point i, caused by the original loads applied to the system (not the added generalized forces). In words, the total displacement at every point i has two contributions: one is from the generalized forces $\left\{X_{1}, \cdots, X_{n}\right\}$, and the other one comes from the loads applied to the original system. This total displacement must vanish at every point, which is expressed by (1). (It is important to understand that every generalized force has a contribution to the displacement at point i, i.e. at every point of the system.) We shall now present some theorems and principles which justify and explain the formulation of the n equations of compatibility, each of which has the form (1). Perhaps the most important theorem is Castigliano's (second) theorem, which states that the partial derivative of the potential energy of deformation of a linearly elastic body with respect to the i-th force acting on the body equals the displacement at the point and direction of the force, i.e. $$\delta_{i} = \frac{\partial U}{\partial F_{i}} \ \ \text{(2).}$$ The proof of this theorem shall not be presented here. We shall only mention that it relies on Betti's theorem (which can be found here), which is derived under the assumption that the principle of superposition holds. Without going more in-depth, one can intuitively assume that expression (2) will be useful in the force method, since the most important issue is to find displacements at certain points, while the generalized forces acting at those points remain the unknowns - which is why the method is called the force method. Indeed, if we set $F_{i} = X_{i}$, it can easily be seen that, for a system with a degree of statical determinacy which equals n, we shall have (in consistence with the idea of the force method laid out at the beginning of this section) n equations of the form $$\delta_{i} = 0, i = 1, \cdots, n \ \ \text{(3),}$$ which, after applying Castigliano's theorem, become: $$\frac{\partial U}{\partial X_{i}}=0, i = 1, \cdots, n \ \ \text{(3').}$$ The solution of the system is the ordered n-touple $(X_{1}, \cdots, X_{n})$, which represents the reaction forces in all the removed supports. The primary system is now in the same state of equilibrium as the original system - the equations of equilibrium can easily be solved to retrieve the values of all the other reactive forces (and internal forces, automatically), and it can easily be seen that the displacement of the system equals the displacement of the original system at every point, which almost directly follows from condition (3). The only thing left to do now is to show how the deformational potential energy of the system, referred to as U, is constructed, and how the system of equations (3') is equivalent to the system (1).
Recognitions: Homework Help 1.3. A FORMAL OUTLINE (II). If we consider a statical system consisting of m uniform bars, every of which is straight, it can be shown that the potential energy of a single bar j equals: $$U_{j} = \int_{0}^{L_{j}}\left(\frac{N^2}{2EA}+\frac{M^2}{2EI}\right)dx \ \text{.}$$ (Note that the notation is inconsistent due to practical reasons - the j-th bar has a local coordinate system associated with it, so it is not correct for the variables of indegration to be dx for every bar of the system.) The upper expression for deformational potential energy is derived by defining the work of internal forces of the system on differential elements, since that work must equal the potential energy of deformation stored in that element. The original expression consists of six terms, since the resulting force and torque can, in space, be decomposed to three components each. Since we assume all the loads act in the xz plane, we are left with the normal force N and the bending torque M only (the bars bend in the xz plane). The influence of the shear force (perpendicular to N) on the potential energy of deformation is neglible, so the expression is left with two terms only. It is also assumed that the module of elasticity E, the area of the cross section A, and the moment of inertia of the cross section I are all constant for every bar of the system. Further on, the potential energy of the whole system consisting of m bars, is: $$U=\sum_{j=1}^m \int_{0}^{L_{j}}\left(\frac{N^2}{2EA}+\frac{M^2}{2EI}\right)dx \ \ \text{(1).}$$ Since the principle of superposition holds (the material behaviour is linearly elastic), the internal forces N and M can be expressed in the form: $$N(x) = N^0(x)+\sum_{k=1}^n X_{k}n_{k}(x)$$ $$M(x) = M^0(x)+\sum_{k=1}^n X_{k}m_{k}(x) \ \ \text{(2).}$$ In words, the forces at every point are represented as the sum of forces due to the original load applied to the bar structure ($N^0(x)$ and $M^0(x)$) and the forces due to the n generalized forces $X_{k}$ ($\sum_{k=1}^n X_{k}n_{k}(x)$ and $\sum_{k=1}^n X_{k}m_{k}(x)$). The functions $n_{k}(x)$ and $m_{k}(x)$ represent the internal forces at some point due to a unit load applied to the structure at the point and in the direction of the generalized force $X_{k}$ (which arises from the principle of superposition, too). As considered in the previous section, we first apply Castigliano's theorem, which states, once again: $$\delta_{i} = \frac{\partial U}{\partial X_{i}} \ \text{.}$$ After applying this to expression (1), we have: $$\delta_{i} = \frac{\partial U}{\partial X_{i}} =\sum_{j=1}^m \int_{0}^{L_{j}}\left(\frac{N}{EA}\frac{\partial N}{\partial X_{i}} +\frac{M}{EI} \frac{\partial N}{\partial X_{i}} \right)dx \ \text{.}$$ After plugging equations (2) into the expression above and rearranging, we obtain: $$\delta_{i} = \sum_{k=1}^n X_{k} \sum_{j=1}^m\int_{0}^{L_{j}}\left( \frac{n_{k}(x)n_{i}(x)}{EA} + \frac{m_{k}(x)m_{i}(x)}{EI} \right) dx + \sum_{j=1}^m \int_{0}^{L_{j}} \left( \frac{N^0(x)n_{i}(x)}{EA}+\frac{M^0(x)m_{i}(x)}{EI} \right) dx \ \text{.}$$ The expression above represents the displacement at the point i where the generalized force $X_{i}$ acts. The displacement can be a translation, rotation, or a relative translation or rotation (depends on how we choose the primary system). Further on, we set: $$\delta_{ik}=\sum_{j=1}^m\int_{0}^{L_{j}} \left( \frac{n_{k}(x)n_{i}(x)}{EA}+\frac{m_{k}(x)m_{i}(x)}{EI} \right) dx \ \text{, and}$$ $$\delta_{i0} = \sum_{j=1}^m \int_{0}^{L_{j}} \left( \frac{N^0(x)n_{i}(x)}{EA} + \frac{M^0(x)m_{i}(x)}{EI} \right) dx \ \ \text{(3).}$$ Since the displacement at every point must vanish (see previous section), we obtain the system of equations: $$\delta_{i} = 0, i = 1, \cdots, n \ \text{, i.e.}$$ $$\sum_{k=1}^n X_{k} \delta_{ik} + \delta_{i0} = 0, i = 1, \cdots, n \ \ \text{(4),}$$ which represents the system of equations of compatibility of the force method. We shall once again repeat what has already been said in the previous section; $X_{k}$ represents the generalized force acting at point k, $\delta_{ik}$ represents the generalized displacement at the point i, caused by the force k, and $\delta_{i0}$ represents the generalized displacement at point i, caused by the original loads applied to the system. By solving the system of equations (4) we obtain the values of the unknown forces $X_{1}, \cdots, X_{n}$ which represent the reaction forces in the supports we have removed (or the forces acting on the added hinges which are preventing relative displacement). After doing so, one merely has to use equations (2) to obtain the functions of the internal forces of the structure. Since the system of equations (4) is, as stated before, equivalent to the system of equations (3') in the previous section, there is another important conclusion which arises. It is easily seen that the system of equations represents a necessary condition for the function U to have an extremum. Further on, it can be seen that this extremum is a minimum, hence we can conclude that, in a statically indeterminate system, the unknowns, i.e. the n generalized forces $X_{k}$ must have such a value for which the potential energy of deformation U of the system is minimized. This is referred to as the principle of minimum potential energy. The theory behind the force method is now slightly brightened up, but every reader has the right to ask himself how to actually calculate the values of the integrals (3), which are needed to solve the system of equations (4). The answer to that question is given in the next section.
Recognitions:
Homework Help
## An introduction to the analysis of statically indeterminate systems
1.3. A FEW COMMENTS.
As mentioned, calculation of the intergals (3) from the previous section is the most important operative tool which one has to develop in order to use the force method.
A formal proof shall not be presented here, but it can be seen that the summands in the expressions (3) are of the form
$$k \int_{0}^{L_{j}} f(x)\cdot g(x) dx \ \text{,}$$
where k stands for $\frac{1}{EI}$ or $\frac{1}{EA}$, and it is assumed that the module of elasticy E, the moment of inertia of the cross section I, and the area of the cross section A are all constant. Further on, since at least one of the functions f or g presents the internal force (bending moment or normal force) caused by a generalized unit force acting at a point where a support has been removed (see previous section), at least one of the functions shall be a linear function (i.e. the internal force diagram shall be linear). This is the fact from which Vereshchagin derived the method of integration of the expressions (3) - one merely has to find the area of the diagram (between 0 and Lj, of course) which is the diagram of the internal force caused by the original loads acting on the primary system (not the diagram of the generalized unit loads), and multiply it (that area) with the ordinate of the unit load diagram at the point of the centroid of area of the non unit load diagram. This will be shown in an example in the next section, and the reader shouldn't be worried if the description of the method of integration was confusing.
Another useful fact considering the equations (3) is that $\delta_{ik} = \delta_{ki}$, which follows directly from Maxwell's theorem, which states that the displacement at the point and in the direction of the first unit load caused by the second unit load equals the displacement at the point and in the direction of the second unit load, caused by the first unit load. This theorem holds for a linearly elastic body.
After determining the values of (3), i.e. of $\delta_{ik}$ and $\delta_{i0}$, we plug them back into the system of equations of compatibility (4). The system can be represented in matrix form:
d X + do = 0.
(Note that the matrix d is symmetric.)
Solving the system is a standard linear algebra procedure, and it is left to the reader which method to choose.
As mentioned before, after solving the system, we can, using the equations of equilibrium, find all the reactions in the system - we merely have to solve a statically determinate system with the original loads acting on it, along with the loads $X_{1}, \cdots, X_{n}$. Further on, by using equations (2) from the previous section, we can find the values of the internal forces N and M at any point of the system, and sketch normal force and bending moment diagrams, which is one of the standard assignments in structural statics.
In the next (last) section the reader will be provided with an example of a simple statically indeterminate system treated with the force method.
Recognitions: Homework Help 1.4. 'OUTRODUCTION' THROUGH AN EXAMPLE. As a final example, we shall analyze the system shown in Figure 5. Let:F = 50 [kN], q = 25 [kN/m], and L = 5 [m].The material and cross-sectional characteristics are: E = 3*107 [kN/m2], b/h = 0.5/0.5, I = 0.54/12 [m4], and A = 0.52 = 0.25 [m2]. (Missing figure) Figure 5 The total number of reactions is 5, and the number of independent equations of equilibrium we can use is 3, which implies that the system's degree of statical indeterminacy equals 2. The chosen primary system is displayed in Figure 6. (Missing figure) Figure 6 As it can be seen in the figure above, the primary system is chosen so that a a hinge is added at the support at point C, and the hinge support at point A is turned into a roller hinge support. Further on, at point C there is now a moment $X_{2}$ added, and at point A there is a horizontal force $X_{1}$. Note that the (generalized) forces $X_{1}$ and $X_{2}$ act in the direction of the released support parts (or added hinges), i.e. the forces act in the direction of the new added degree of freedom. The system is now statically determinate - the number of reactions equals the number of equations of equilibrium we can use. The equations of compatibility (eq. (4) in Section 1.3.) are now: $$X_{1}\delta_{11}+X_{2}\delta_{12}+\delta_{10} = 0$$ $$X_{1}\delta_{21}+X_{2}\delta_{22}+\delta_{20} = 0 \ \text{(1).}$$ In order to solve the equations for $X_{1}$ and $X_{2}$ we have to calculate the coefficients $\delta_{ik}$ and $\delta_{i0}$, where i, k = 1, 2. In order to do that, we shall use internal force diagrams referred to as $M^0$, $N^0$, $m_{1}$, $m_{2}$, $n_{1}$ and $n_{2}$. The 'capital letter' diagrams $M^0$ and $N^0$ represent the bending moment and normal force in the system due to the original loads applied to the system (the uniform load q and the concentrated force F). The other diagrams represent bending moments and normal forces due to the unit generalized displacement at a point i - for example, the diagram $m_{2}$ represents the bending moment in the system caused by the unit moment acting at point C in the same direction as the moment $X_{2}$; the diagram $n_{1}$ represents the normal force in the system due to a unit concentrated force at point A acting in the same direction as the force $X_{1}$, etc. The diagrams are nothing but sketches of the functions $M^0(x)$, $N^0(x)$, $m_{1}(x)$, etc., which are calculated on the local coordinate systems of each member of the system, AB, and BC. The functions can, of course, easily be set up after calculating the reactions. The diagrams are shown in Figure 7 (Note that every diagram has its own scale, due to practical reasons.) (Missing figure) Figure 7 We shall only calculate the value of the coefficient $\delta_{10}$, which represents the displacement at the point where the force $X_{1}$ is acting, due to the loads F and q. By definition, we have: $$\delta_{10} = \sum_{j=1}^2 \int_{0}^{L_{j}} \left( \frac{N^0(x)n_{1}(x)}{EA} + \frac{M^0(x)m_{1}(x)}{EI} \right) dx$$ $$= \int_{0}^{L_{1}} \left( \frac{N^0(x)n_{1}(x)}{EA} + \frac{M^0(x)m_{1}(x)}{EI} \right) dx + \int_{0}^{L_{2}}\left( \frac{N^0(x)n_{1}(x)}{EA} + \frac{M^0(x)m_{1}(x)}{EI} \right) dx$$ $$= \frac{1}{EI} \int_{0}^{L_{1}} M^0(x)m_{1}(x) dx + \frac{1}{EI} \int_{0}^{L_{2}} M^0(x)m_{1}(x) dx + \frac{1}{EA} \int_{0}^{L_{1}} N^0(x)n_{1}(x) dx + \frac{1}{EA} \int_{0}^{L_{2}} N^0(x)n_{1}(x) dx \ \ \text{(2).}$$ In our case $L_{1} = L_{2} = L$. We shall now use Vereshchagin's theorem to integrate the expressions above. We have: $$\delta_{10} = \frac{1}{EI} \left[ 2.5\cdot 125\cdot\frac{1}{2}\cdot(2.5+\frac{2}{3}\cdot 2.5)\right] + \frac{1}{EI} \left[ 125\cdot 5\cdot\frac{1}{2} \cdot ( \frac{2}{3}\cdot 5) + \frac{2}{3} \cdot \frac{25\cdot 5^2}{8} \cdot 5 \cdot (-\frac{1}{2}\cdot 5) \right]$$ $$\ \ \ + \frac{1}{EA} \left[ 87.5\cdot 5 \cdot (1) \right] + \frac{1}{EA} \left[ 50 \cdot 5 \cdot (1) \right]$$ $$= 6.7583333\cdot 10^{-3} \ \text{.}$$ It is useful to compare the calculated term with the expression (2) to see how Vereshchagin's integration method is used. Every term multiplied by $\frac{1}{EI}$ or $\frac{1}{EA}$ consists of two factors. The first one represents the area of the diagram $M^0$ or $N^0$, and the second one (the bracketed one) represents the value of the functions $m_{1}$ or $n_{1}$ at the abscissa of the centroid of area of the $M^0$ or $N^0$ diagram. For example, the first term in the expression above consists of two factors: the first one, $2.5\cdot 125\cdot\frac{1}{2}$ is the area of the diagram $M^0$ along the length $L_{1}$, i.e. on the element AB. The second factor, $2.5+\frac{2}{3}\cdot 2.5$ equals the value of the $m_{1}$ diagram at the (local) abscissa of the centroid of area of the $M^0$ diagram. (This can be calculated easily - it is suggested for the reader to make a sketch and 'follow' the calculations laid out in order to avoid unnecessary confusion.) The value of the coefficient $\delta_{11}$ will be calculated, $$\delta_{11} = \frac{1}{EI} \left[ 5\cdot 5 \cdot \frac{1}{2} \cdot(\frac{2}{3}\cdot 5)\cdot 2 \right] + \frac{1}{EA} \left[ 1\cdot 5 \cdot (1) \cdot 2 \right] = 5.346666667 \cdot 10^{-4} \ \text{,}$$ and the other coefficients are left for the reader to calculate as an exercise. The values one should obtain after doing that are: $$\delta_{20} = -1.78333355\cdot 10^{-4} \ \text{,}$$ $$\delta_{22} = 1.069333333\cdot 10^{-5} \ \text{,}$$ $$\delta_{12} = \delta_{21} = 2.653333333\cdot 10^{-5} \ \text{.}$$ After plugging the coefficients in the system of equations (1), and solving, we have: $$X_{1} = -15.3986[/tex \ \text{, and}$$ $$X_{2} = 54.8855 \ \text{.}$$ All we have to do now is use the equations (2) from Section 1.3, i.e. $$N(x) = N^0(x)+X_{1}n_{1}(x) + X_{2}n_{2}(x)$$ $$M(x) = M^0(x)+X_{1}m_{1}(x) + X_{2}m_{2}(x)$$ to obtain the internal forces N and M at any point of the system. This is left as an exercise too. The final diagrams of the internal force are shown in Figure 8. (The reactions can be calculated too, of course.) (Missing figure) Figure 8 As a final exercise, the reader can solve the example from Section 1.1. with the same primary system as suggested in that section, and then try out a different primary system.
Recognitions: Homework Help Noticed figures are missing, will re-upload them in a day or two.
Recognitions: Homework Help Figure 1. Figure 2. Figure 3. Figure 4.
Recognitions: Homework Help Figures 1-4 uploaded, having trouble finding Figures 5-8, though. Will look into it. Edit: had to upload them separately, since I can't edit the old posts, I hope this isn't a problem.
Mentor Blog Entries: 10 If you can upload figures 5-8, I can edit the old posts to include them.
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http://physics.stackexchange.com/questions/28108/calculating-mass-of-object-by-lifting-edge
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# Calculating mass of object by lifting edge
Assuming a solid rectangular plate, hinged along one edge. How does one calculate the mass of the plate if the force necessary to lift the opposite edge is known?
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Something's missing in your problem... – Pygmalion May 10 '12 at 12:35
@Pygmalion What are you thinking of? It is ok to make generic assumptions regarding mass and dimensons. – Arik Raffael Funke May 10 '12 at 13:48
As described, plate is the vertical position at the start. If you wish to lift it, you have exert some force. What is the direction of the force? And don't forget that after applying force, the plate will start accelerating... Problem is hugely underdefined. – Pygmalion May 10 '12 at 13:55
Have you ever heard of a Free Body Diagram ? – ja72 May 10 '12 at 18:35
Does the problem assume slow (static) lifting, or a sudden lift with appreciable vertical acceleration? The answer is different for those two cases. – ja72 May 10 '12 at 19:29
## 1 Answer
This is blatantly a homework question, so we're only allowed to discuss methods, and not give you the answer. With any problem like this the very first thing to do is draw a diagram. From the limited information in your question I think the situation loks like this:
You know the force $F$ that you're using to lift the end of the plate, and you want to know the force $mg$, where $m$ is the mass of the plate and $g$ is the acceleration due to gravity.
You find $mg$ by taking moments about the hinge. The distance from the hinge to the end of the plater where you're applying the force is $l$, and if the plate is a rectangle the force $mg$ acts from the centre of mass, which is $l/2$ away from the hinge.
The rest is up to you!
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Thanks for the answer. It was actually not a homework question but I had an engineering problem where I needed to find the weight of an installed lid. Many thanks. – Arik Raffael Funke May 25 '12 at 16:12
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http://mathoverflow.net/questions/39317/maximum-principle-fails-when-uc-cant-find-example/39327
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Maximum Principle fails when u∉C²(Ω)? Can’t find example.
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I would like an example where the maximum principle fails in a bounded smooth domain $\Omega$ where one has a solution which is not $C^2(\Omega)$ to $Lu=0$ where $L$ is elliptic and linear. This obviously must rely on the coefficients being discontiuous for the elliptic operator since otherwise one can do interior regularity estimates. All of the examples I have tried to come up with turn out to not actually be weak solutions so I'm stuck on this. Perhaps maximum principles extend to the non-smooth setting? Any insight on this would be appreciated.
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2 Answers
Yes indeed, the maximum principle extends to the non-smooth setting. I am not sure that there is a complete theory, because so many situations can occur. But at least let me mention the following situations.
1. In Hopf's maximum principle, one only needs that $u$ admits second order derivatives (this is less than ${\mathcal C}^2$) and the coefficients of $L$ be bounded.
2. In the variational case, that if $L$ is self-adjoint, the $H^1$ solution is unique and minimizes a functional $\int (Lu,u)dx$. Using the fact that if $v$ is in the Sobolev space $H^1$, then the absolute value $|v|$ is in this space too, one again proves a maximum principle.
3. The previous item covers the one-dimensional case, because then second order operator can always be multiplied by an appropriate weight in such a way that they become self-adjoint. This is the reason why the spectrum of a one-D second-order operator is real (Liouville's theory).
4. If $L$ is elliptic with smooth coefficients, and if $u$ is a distribution such that $Lu\in{\mathcal C}^\infty$, then $u\in{\mathcal C}^\infty$. Therefore the maximum principle is valid.
5. The modern theory of second-order elliptic (not necessarily linear) equations is based upon the maximum principle. One declares that $u\in{\mathcal C}^0$ is a solution if, given an arbitrary point, there exists a super-solution (resp. a sub-solution) $\phi_\pm$ such that $\phi_\pm(x)=u(x)$, and $\phi_+\le u\le \phi_-$ otherwise.
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Thanks for all of the nice examples and facts Denis! I'll use them to think a bit more about this. – Dorian Sep 19 2010 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.
Maximum principle, in general, can be applied in the viscosity solutions setting (in this case, the viscosity solution is only continuous). You can have a look at the excellent paper by Crandall-Ishii-Lions as following: http://arxiv.org/abs/math/9207212
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http://math.stackexchange.com/questions/17522/are-there-rules-of-logic-dealing-with-the-implies-operator
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# Are there rules of Logic dealing with the implies operator?
I'm just beginning a course in discrete mathematics and I'm learning a few of the basic laws to prove propositions. I understand how to use propositions that use the logical connectives, AND , OR, NOT. However I'm not sure how to prove a proposition that has an implies operator. For example prove the following is a tautology.
$(p\land (p\implies q))\implies q$
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## 4 Answers
There are two possible answers. One, your language doesn't really contain the $\implies$ operator, and it is thought of as a shorthand (as in user3123's answer): $$a \implies b \text{ is a shorthand for } \lnot a \lor b.$$ Two, your language does include it, and then you have some axioms expressing its "meaning". For example, you could have a system with only $\lnot$ and $\implies$, having the following axioms
$$A \implies (B \implies A)$$
$$(A \implies B) \implies ((B \implies C) \implies (A \implies C))$$
$$(A \implies B) \implies (\lnot B \implies \lnot A)$$
and Modus Ponens as the only inference rule: given $A$ and $A \implies B$, deduce $B$. This system is complete (can prove every true proposition). The other connectives are then shorthands:
$$A \lor B \text{ stands for } \lnot A \implies B$$
$$A \land B \text{ stands for } \lnot(\lnot A \lor \lnot B)$$
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There's a nice shortcut to proving that something with an implies operator is a tautology. Assume it's not a tautology: the only way $a \Rightarrow b$ can be false is if $a$ is true and $b$ is false, so you can assume that, etc. and this is often a quick way to reach a contradiction.
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Funny that you can use the logical operators and ($\land$), or ($\lor$) and not ($\lnot$) but not the imply operator as its nothing else:
$(a \implies b) \Leftrightarrow (\lnot a \lor b)$
Maybe this helps? Just replace those imply operators in your equation.
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Is what you wrote equivalent to my proposition, just rewritten? – lampShade Jan 14 '11 at 19:56
No, I just gave you a way to translate the imply-operator into something you already know (a combination of the not and or operator). – Listing Jan 14 '11 at 19:58
You may draw a truth table, in order to investigate the alternatives (see Implication at wikipedia). In words, if $A$ and $B$ are some propositions then $A\Rightarrow B$ is a third proposition which is true if and only if $A$ is false, or both $A$ and $B$ are true.
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http://physics.stackexchange.com/questions/3005/derivation-of-maxwells-equations-from-field-tensor-lagrangian/3006
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Derivation of Maxwell's equations from field tensor lagrangian
I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density $L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ (where $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is the field tensor).
Substituting in for the definition of the field tensor yields $L = -\frac{1}{2}[(\partial_\mu A_\nu)(\partial^\mu A^\nu) - (\partial_\mu A_\nu)(\partial^\nu A^\mu)]$. I know I should be using $A^\mu$ as the dynamical variable in the Euler-Lagrange equations, which become $\frac{\partial L}{\partial A_\mu} - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)} = - \partial_\mu\frac{\partial L}{\partial(\partial_\mu A_\nu)}$, but I'm confused about how to proceed from here.
I know I should end up with $\partial_\mu F^{\mu\nu} = 0$, but I don't quite see why. Since $\mu$ and $\nu$ are dummy indices, I should be able to change them: how do the indices in the lagrangian relate to the indices in the derivatives in the Euler-Lagrange equations?
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3 Answers
Well, you are almost there. Use the fact that $${\partial (\partial_{\mu} A_{\nu}) \over \partial(\partial_{\rho} A_{\sigma})} = \delta_{\mu}^{\rho} \delta_{\nu}^{\sigma}$$ which is valid because $\partial_{\mu} A_{\nu}$ are $d^2$ independent components.
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Dear amc, first, write your Lagrangian density as $$L = -\frac{1}{4} F_{\mu\nu}F^{\mu\nu} = -\frac{1}{2} (\partial_\mu A_\nu) F^{\mu\nu}$$ Is that fine so far? The $F_{\mu\nu}$ contains two terms that make it antisymmetric in the two indices. However, it's multiplied by another $F^{\mu\nu}$ that is already antisymmetric, so I don't need to antisymmetrize it again. Instead, both terms give me the same thing, so the coefficient $-1/4$ simply changes to $-1/2$.
Now, the field equations force you to compute the derivatives of the Lagrangian with respect to $A_\mu$ and its derivatives. First of all, the derivative of the Lagrangian $L$ with respect to $A_\mu$ components themselves vanishes because the Lagrangian only depends on the partiial derivatives of $A_\mu$. Is that clear so far?
So the equations of motion will be $$0 = -\partial_\mu [\partial L / \partial(\partial_\mu A_\nu)] = \dots$$ Whoops, you already got to this point. But now, look at my form of the Lagrangian above. The derivative of the Lagrangian with respect to $\partial_\mu A_\nu$ is simply $$-\frac{1}{2} F^{\mu\nu}$$ because $\partial_\mu A_\nu$ simply appears as a factor so the equations of motion will simply be $$0 = +\frac{1}{2} \partial_\mu F^{\mu\nu}$$ However, I have deliberately made one mistake. I have only differentiated the Lagrangian with respect to $\partial_\mu A_\nu$ included in the first factor of $F_{\mu\nu}$, with the lower indices. However, $\partial_\mu A_\nu$ components also appear in $F^{\mu\nu}$, the second factor in the Lagrangian, one with the upper indices. If you add the corresponding terms from the Leibniz rule, the result is simply that the whole contribution will double. So the right equation of motion, including the natural coefficient, will be $$0 = \partial_\mu F^{\mu\nu}$$ The overall normalization is important because this equation may get extra terms, like the current, whose coefficient is obvious, and you don't want to get a relative error of two between the derivative of $F$ and the current $j$.
Cheers Lubos
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+1 Elegant! I like it. – nervxxx May 12 at 2:22
We vary the action $$\delta \int {Ldt} = \delta \int {\int {\Lambda \left( {A_\nu ,\partial _\mu A_\nu } \right)d^3 xdt = 0} }$$ ${\Lambda \left( {A^\nu ,\partial _\mu A^\nu } \right)}$ is the density of lagrangian of the system.\ So: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A^\nu }}\delta A_\nu + \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}\delta \left( {\partial _\mu A_\nu } \right)} \right)d^3 xdt = 0} }$$ By integrating by parts we obtain: $$\int {\int {\left( {\frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right)\delta A_\nu d^3 xdt = 0} }$$ $$\frac{{\partial \Lambda }}{{\partial A_\nu }} - \partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = 0$$ W e have to determine the density of lagrangian. One terms deals with the interaction of the charges with the electromagnetic field. This term is $J^\nu A_\nu$. One term is the density of energy of the electromagnetic field: this term is the difference of the magnetic field and the electris field .So we have: $$\Lambda = J^\mu A_\mu + \frac{1}{{4\mu _0 }}F^{\mu \nu } F_{\mu \nu }$$ We have: $$\frac{{\partial \Lambda }}{{\partial A_\nu }} = J^\nu$$ so: $$\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}F^{\kappa \lambda } F_{\kappa \lambda } } \right) = \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\left( {\partial ^\kappa A^\lambda - \partial ^\lambda A^\kappa } \right)\left( {\partial _\kappa A_\lambda - \partial _\lambda A_\kappa } \right)} \right)} \right) =$$ $$= \frac{1}{{4\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa - \partial ^\lambda A^\kappa \partial _\kappa A_\lambda + \partial ^\lambda A^\kappa \partial _\lambda A_\kappa } \right)} \right) =$$ The third and the fourth are the same of first and the second terms. You can do $k \leftrightarrow \lambda$: $$\partial _\mu \frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}} = \frac{1}{{2\mu _0 }}\partial _\mu \left( {\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda - \partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right)} \right)$$ but: $$\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\kappa A_\lambda } \right) = \partial ^\kappa A^\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\kappa A_\lambda } \right) + \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda } \right) =$$ $$= \partial ^\kappa A^\lambda \delta _\kappa ^\mu \delta _\lambda ^\nu + g^{\kappa \alpha } g^{\lambda \beta } \partial _\kappa A_\lambda \frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial _\alpha A_\beta } \right) = 2\partial ^\mu A^\nu$$ We have: $$\frac{\partial }{{\partial \left( {\partial _\mu A_\nu } \right)}}\left( {\partial ^\kappa A^\lambda \partial _\lambda A_\kappa } \right) = 2\partial ^\nu A^\mu$$ so: $$\partial _\mu \left( {\frac{{\partial \Lambda }}{{\partial \left( {\partial _\mu A_\nu } \right)}}} \right) = \frac{1}{{\mu _0 }}\partial _\mu \left( {\partial ^\mu A^\nu - \partial ^\nu A^\mu } \right) = \frac{1}{{\mu _0 }}\partial _\mu F^{\mu \nu }$$ The lagrangian equations provide the non homogeneus maxwell equations:
$$\partial _\mu F^{\mu \nu } = \mu _0 J^\nu$$
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Thanks for this detailed solution. – Oaoa Jan 25 at 14:59
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http://crypto.stackexchange.com/questions/5694/public-keys-and-their-protocols/5850
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# Public keys and their protocols
I'm having difficulties understanding exactly what this protocol means:
$S \to D : \{N_S , S\}K_D$
$D \to S : \{N_S , N_D \}K_S$
$S \to D : \{N_D \}K_D$
"where $S$ represents the supervisor’s console and $D$ represents the door controller and the other symbols have their usual security protocol meanings."
I understand $N$ to be a nonce, a randomly generated number/hash that will only be used once.
I'm assuming $K$ is the public key?
Could someone explain this protocol for me, I’ve looked everywhere and cannot find anything like this.
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## 3 Answers
It seems like an authentication protocol. In the first message, $S$ identifies herself. In the 2nd message $D$ proves knowledge of her private key (otherwise she could not return $N_s$. In the third message, $S$ proves knowledge of her private key.
You asked in a comment about replay attacks. There are two possibilities, the attacker is impersonating $S$ or $D$.
For $S$, the attacker might replay a captured 1st message. $D$ would see it, decrypt, return $N_S$ with a randomly generated $N_D$. The attacker could not possibly know what $N_D$ is and could not send the last message.
For impersonating $D$, $S$ would send the first message, but the attacker couldn't decrypt it and therefore could not send the 2nd message.
So, it seems the protocol is secure against replay attacks.
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But if there are two different doors $D_1$, $D_2$, then one of the doors is able to impersonate the supervisor to the other door. – CodesInChaos Dec 14 '12 at 13:30
Under the assumption that $K_X$ represents $X$'s public key, this is the three-message version of the Needham-Schroeder public-key protocol. Many mutual authentication protocols derive from this design.
Wikipedia has the seven-message version of the protocol. The essential difference between that version and the three-message version is that the seven-message version explicitly includes the certificate exchanges for the public keys. The three-message version simply assumes that the correct public keys are known.
The protocol was proven secure using a formalism called BAN logic, which is hardly used nowadays. Almost 20 years after the protocol was proposed it was shown that it is vulnerable to a man-in-the-middle attack if the adversary has his own public/private key pair, see this paper by Gavin Lowe.
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This looks like a symmetric-key protocol: $S$ first sends a nonce $N_S$ along with its name encrypted under the key $K_D$. Then $D$ responds with an encryption of $N_S$ along with another nonce $N_D$, this time encrypted under $K_S$. Finally $S$ sends $N_D$ to $D$, this time encrypted under $K_D$.
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Is it possible for a message to be retransmitted from an attacker? – George Orwell Dec 14 '12 at 1:48
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http://math.stackexchange.com/questions/200423/show-that-the-set-w-is-a-countable-set/200426
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# Show that the set W is a countable set.
Q: Let the alphabet A be a finite set of $n$ letters such that A=${a_1, a_2,...,a_n}$ and let W be the set of all possible words written from this finite alphabet (here a word is a finite concatenation of letters). Show that W is countable.
I don't know how to do this...?
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2
Do you know that the union of countably many countable sets is countable? If so, you could partition $W$ into sets of words of a fixed length, and show that this makes $W$ a countable union of countable sets. – Manny Reyes Sep 21 '12 at 18:38
2
Can you show that the union of countably many finite sets is countable? – Mustafa Gokhan Benli Sep 21 '12 at 18:38
1
@Manny you were 19 seconds faster :) – Mustafa Gokhan Benli Sep 21 '12 at 18:39
That is genius! Thank you both! – Allison Cameron Sep 21 '12 at 18:41
That question has nothing to do with real analysis, and has a lot more to do with elementary set theory. – Asaf Karagila Sep 21 '12 at 19:47
## 3 Answers
@William has already given a construction using the standard proof that the countable unions of countable sets are countable, but I would like to also present a direct proof for completeness.
Given $A=\{a_1,\dots,a_n\}$ and $W=\{\text{Words of finite length}\}$, where a word is a concatenation of elements from $A$, let $W_m$ denote the set of words of length $m$. The number of elements of $W_m$ is precisely $n^m$, hence finite. We can now rewrite $W$ in the following way.
$$W=\bigcup_{m=1}^\infty W_m$$
We can form a bijection (one-to-one mapping) between the natural numbers in the following way. Assign to the elements of $W_1$ the numbers $\{1,2,\dots,n\}$ in a non-repeating way. To the elements of $W_2$ assign the numbers $\{n+1,n+2,\dots,n+n^2\}$ in a non-repeating way. Now supposing that we have assigned to sets $W_1,\dots,W_k$ the numbers $\{1,2,\dots,\sum_{i=1}^k n^i\}$, assign to the elements of $W_{k+1}$ the numbers $\{1+\sum_{i=1}^k n^i,2+\sum_{i=1}^k n^i,\dots,n^{k+1}+\sum_{i=1}^k n^i\}$.
The above construction shows the existence of a bijection between $W$ and the natural numbers, which are countable. Therefore, $W$ is countable.
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Hint: The number of words of length $n$ is finite. Now consider the set of words of all possible lengths $n \in \mathbb{N}$.
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So the set of words of all possible lengths n must also be finite? – Allison Cameron Sep 21 '12 at 18:40
@AllisonCameron No. The set of all possible lengths is countably infinite. It's $1, 2, \ldots$ (or $\mathbb{N}$). So the set of all words of all possible lengths is the union of countably many finite sets. – Ayman Hourieh Sep 21 '12 at 18:41
If $A$ is a finite alphabet of size $k$. Let $B_n$ denote the words of length $n$. Then the size of $B_n$ is $k^n$. The set of all words in $A$ is $\bigcup_{n \in \mathbb{N}} B_n$. Countable union of countable sets (finite sets) are countable. Hence the set of all words are countable.
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http://mathoverflow.net/questions/22295?sort=newest
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## (∞,1) vs Category weakly enriched over spaces
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
What is the difference between:
($\infty,1$) categories - in which have for two objects you have an ($\infty,0$) category of morphisms (i.e. a space of morphisms)
and
categories weakly enriched over spaces - by that I mean categories such that hom(x,y) is always a space and composition is defined only up to (coherent) homotopy
?
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## 2 Answers
If "space" means the same thing in the two cases, as seems to be implied, then there is no difference, at least not at that level of precision.
There are many different models of $(\infty,1)$-categories. Many of these, like $A_\infty$-categories, Segal categories, complete Segal spaces, and simplicial categories, try to make explicit the intuition that they are "categories weakly enriched over spaces," though in various different ways. Others, like quasicategories, are less directly related to the enrichment point of view, although they yield a provably equivalent theory (in a certain appropriate sense).
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
In the quasi-category formalism of $(\infty,1)$-categories, it is hard to make sense of whether they are strictly enriched, or weakly enriched in infinity-groupoids. This is mostly due to the ambiguity (only up to homotopy) of what is meant by $Hom(X,Y)$. (There are various models for this, for example Hom^R(X,Y), Hom^L(X,Y), or the actual Hom in its associated simplicial category (apply the left-adjoint to the homotopy coherence nerve)). To avoid this, lets work with simplicial categories (which is justified as Bergner's model structure is Quillen equivalent to Joyal's). In this model for infinity-categories, the enrichment is STRICT. The fibrant cofibrant simplicial categories have each $Hom(X,Y)$ a Kan-complex = an infinity groupoid. The homotopy theory of simplicial categories models categories STRICTLY enriched in infinity-groupoids and WEAK functors between them. However, when people say Kan-complexes=infinity groupoids, they really mean they are WEAK infinity groupoids. (In a very concrete case, the tricategory of weak 2-groupoids is equivalent to the tricategory with CW-complexes as objects, continuous functions as morphisms, homotopies as 2-cells, and homotopy classes of homotopies between homotopies as 3-cells). Just as Mike said, the quasi-category approach (and hence the simplicial category approach) are provably equivalent to other approaches which take infinity-categories explicitly to be categories WEAKLY enriched in infinity-groupoids. Therefore, you can interpret this as a sort of strictification result. But, be careful, this is really a SEMI-strictification result; it is important that we are still enriching in WEAK infinity-groupoids. This could be thought of as a higher-dimensional analogue of the fact that, although not every tricategory is equivalent to a strict-3 category=category enriched in STRICT 2-categories, every tricategory is equivalent to a category enriched in weak 2-categories (to make this precise, every tricategory is equivalent to a category enriched in Gray see: http://ncatlab.org/nlab/show/Gray-category). However, in any rate, we still have to consider WEAK functors between such STRICTLY enriched guys. This is important to note (e.g. although every bicategory is equivalent to strict 2-category, there are more weak functors between them than strict ones). So the moral of the story is we can model $(\infty,1)$ categories as categories strictly enriched in weak infinitiy-groupoids with weak functors between them.
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http://mathoverflow.net/revisions/60398/list
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## Return to Answer
2 notation error.
I'm pretty certain the answer is no provided $n \geq 3$. Let $X$ be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold
It's a contractible open 3-manifold which is not homeomorphic to $\mathbb R^3$.
$X^2$ I claim is homeomorphic to $\mathbb R^6$. I don't have a slick proof of this. The idea is, ask yourself if you can put a boundary on $X^2$ to make $X^2$ the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if $X^2$ is simply-connected at infinity, you're okay. But the fundamental-group of the end of $X^2$ has a presentation of the form $(\pi_1 X Y * \pi_1 XY) / <\pi_1 X^2>$Y^2>$, where$*$denotes free product and angle brackets "normal closure". This , and$\pi_1 Y$is "the fundamental group at infinity for$X$". So$\pi_1 X^2\$ is the trivial group.
Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is $D^6$.
1
I'm pretty certain the answer is no provided $n \geq 3$. Let $X$ be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold
It's a contractible open 3-manifold which is not homeomorphic to $\mathbb R^3$.
$X^2$ I claim is homeomorphic to $\mathbb R^6$. I don't have a slick proof of this. The idea is, ask yourself if you can put a boundary on $X^2$ to make $X^2$ the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if $X^2$ is simply-connected at infinity, you're okay. But the fundamental-group of the end of $X^2$ has a presentation of the form $(\pi_1 X * \pi_1 X) / <\pi_1 X^2>$, where $*$ denotes free product and angle brackets "normal closure". This is the trivial group.
Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is $D^6$.
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http://physics.stackexchange.com/questions/31142/does-the-fundamental-frequency-in-a-vibrating-string-not-necessarily-have-the-st/31157
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# Does the Fundamental Frequency in a Vibrating String NOT Necessarily Have the Strongest Amplitude?
I am doing some experiments on musical strings (guitar, piano, etc.). After performing a Fourier Transform on the sound recorded from those string vibrations, I find that the fundamental frequency is not absolutely the component with the largest amplitude (or energy).
I learnt from introductory physics courses that fundamental frequency is the "major" component of a vibrating string. So why does it sometimes does not possess the highest amplitude?
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## 3 Answers
I did a bit of discussion on this subject in this thread on Music.SE.
The fundamental doesn't necessarily have the strongest amplitude. As said by Alfred Centauri, it depends on the initial configuration: ideally, the string returns to exactly that configuration after each $\tfrac1{\nu_1}$, and the amplitude of each harmonic in frequency space is proportional to its amplitude in the initial configuration in wavenumber space. Now the initial configuration for plucked instruments happens to be roughly what I depicted in the first figure in the Music.SE answer: something between a triangle and a sawtooth, and as we know both have a monotonically descreasing1 sequence of Fourier coefficients ($\mathcal{O}(\tfrac1{n^2})$ for triangle, $\mathcal{O}(\tfrac1{n})$ for sawtooth), so the fundamental does tend to have the strongest amplitude on the string in the beginning. But this may not hold true for an actual instrument sound for a variety of reasons: on the guitar, if you strike the string quickly with a pick, the initial configuration may be more like a smoothened Dirac peak, for which all of the lower frequencies have the same amplitude. Similarly for a loud piano note. Then, resonances in the instruments' bodies may lead to the fundamental decaying faster than some of the other harmonics, so after a while they are stronger, though instrument builders classically try to prevent this. Similarly, as said by John Rennie, the bodies may be more efficient at conducting some of the harmonics to the air than the fundamental and likewise the microphone at measuring the sound.
Finally, you should assure yourself of what this Fourier spectrum of yours actually shows: in audio production, spectrum analysers do not display something proportional to the amplitude $A(\omega)$ at frequency $\omega$, but in fact to $A(\omega)\cdot\sqrt{\omega}$ so that pink noise appears to have constant amplitude. (For both historic / technical and convenience reasons.) So if in such a spectrum the fundamental turns up slightly weaker than e.g. the second harmonic, it may in actuality still be stronger.
1Actually, the sequence of Fourier coefficients of a triangular wave itself is not monotonic, but alternates between a monotonic sequence (for odd $n$) and zero. Since the fundamental has odd $n=1$, that amounts to the same result.
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I would expect that if you pluck a guitar string the mode with the highest amplitude is the fundamental, though this isn't necessarily the mode with the highest energy since the energy of a vibrating string is proportional to the square of the frequency (and the square of the amplitude).
However you're not measuring the amplitude of the string vibration. You're measuring the EMF generated by a microphone that is recording the air vibrations produced by the string. You need to consider how energy is transferred from the string to the air, and from the air to the microphone. In the case of a guitar a lot of the sound is generated by the body of the guitar, so you also need to factor in the efficiency of transfer of energy from the string to the body of the guitar.
All this seems complicated to me, and I don't know of any simple rules for how much of the energy of each mode eventually ends up at the microphone. I did do similar experiments at school, but that was a long time ago and I honestly can't remember whether the fundamental was the biggest mode or not. I remember that the higher harmonics were pretty big, i.e. the waveform looked nothing like a sine wave, so if you tell me the fundamental sometimes isn't the biggest component I can believe it.
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It all depends on the initial configuration. If the string's initial configuration (shape of string at $t=0$) includes a node at the center, the string will not vibrate at the fundamental frequency.
This YouTube video is a demonstration of this phenomenon.
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http://mathoverflow.net/questions/50693/stable-motivic-cohomology-with-finite-coefficients
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## Stable motivic cohomology with finite coefficients?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In this question, which attracted no responces so far, I've asked whether it is possible to extend the Beilinson-Lichtenbaum etale descent rule for motivic cohomology to singular varieties, in particular, by choosing the Grothendieck topologies appropriately.
Subsequently I've looked into Voevodsky's paper "Motives over simplicial schemes" (Journ. K-theory 2010) and found that he seems to explain, in the introduction, that there are two kinds of motivic cohomology for singular varieties: the effective and the stable one. The effective motivic cohomology of a variety $X$ over a field $F$ with coefficients in an abelian group $A$ is given by $$H^i_M(X,A(j))=Hom_{DM(F)}(M(X),A(j)[i])=Hom_{DM(X)}(\mathbb Z,A(j)[i]),$$ (see also section 4 of the same paper), while the stable motivic cohomology is $$H^i_{stable}(X,A(j))=\varinjlim\nolimits_n Hom_{DM(X)}(\mathbb Z(n),A(n+j)[i]).$$ The reason for the two theories being different is that the cancellation theorem (claiming that the Tate twist is fully faithful) does not hold for motives over a singular variety.
As I tried to explain in my previous question, it seems that the Beilinson-Lichtenbaum etale descent rule for motivic cohomology with finite coefficients $$H_M^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}),$$ where $\pi\colon Et\to Zar$, breaks down when $X$ is no longer smooth. Can it be true that an isomorphism like $$H_{stable}^i(X,\mathbb Z/m(j)) = H_{Zar}^i(X,\tau_{\le j}R\pi_*\mu_m^{\otimes j}),$$ holds for arbitrary singular varieties $X$? Or can it be made to hold by replacing the pair of topologies (Zariski, etale) with a different pair, with one or both of the topologies including resolutions of singularities as covers, perhaps?
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http://stats.stackexchange.com/questions/tagged/probability?sort=votes&pagesize=15
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I am hoping that I can ask this question the correct way. I have access to play-by-play data, so it's more of an issue with best approach and constructing the data properly. What I am looking to do ...
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### Confidence interval and probability - where is the error in this statement?
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### How should I mentally deal with Borel's paradox?
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http://mathoverflow.net/revisions/69321/list
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## Return to Answer
4 added 184 characters in body
I don't understand the questions for the following reason: If the image of $X$ is complemented in $E_1$ then the extension is split. Indeed, if $P$ is a projection of $E_1$ onto the image of $X$ then $1-P$ is a projection onto an isomorph of $Y$ by the open mapping theorem (see e.g. Nicolas Monod's thesis Corollary 4.2.4 for a detailed proof).
First question: If you're asking about a pair of extensions of $Y$ by $X$ with $E_1$ split and $E_2$ non-split, take $X = c_0$ and $Y = \ell^{\infty}/c_0$. Then $E_2: 0 \to X \to \ell^{\infty} \to Y \to 0$ is not split by Phillips' lemma (see Whitley's note in the Monthly for a simple proof), and $E_1: 0 \to X \to X \oplus Y \to Y \to 0$ is split by definition.
Second question: Yes, $(x, y) \mapsto \left(\frac{1}{2}(x+y), \frac{1}{2}(x+y)\right)$ is a projection of $X \oplus X$ onto $\Delta$. I recommend you to prove that this sequence is isomorphic to the obvious extension $0 \to X \to X \oplus X \to X \to 0$ (inclusion into the first summand, projection onto the second).
Two final remarks:
• A very interesting procedure for producing non-split extensions of Banach spaces is the twisted sum construction due to Kalton-Peck (I recently learned about this from Bill Johnson in this thread).
• Basically, you're asking about the Yoneda Exts in the exact category of Banach spaces with the exact structure consisting of all kernel-cokernel pairs. If you're interested in such abstract nonsense, please allow me a bit of self-advertisement.
3 added 27 characters in body
I don't understand the questions for the following reason: If the image of $X$ is complemented in $E_1$ then the extension is split. Indeed, if $P$ is a projection of $E_1$ onto the image of $X$ then $1-P$ is a projection onto an isomorph of $Y$ by the open mapping theorem (see e.g. Nicolas Monod's thesis Corollary 4.2.4 for a detailed proof).
First question: If you're asking about a pair of extensions of $Y$ by $X$ with $E_1$ split and $E_2$ non-split, take $X = c_0$ and $Y = \ell^{\infty}/c_0$. Then $E_2: 0 \to X \to \ell^{\infty} \to Y \to 0$ is not split by Phillips' lemma (see Whitley's note in the Monthly for a simple proof), and $E_1: 0 \to X \to X \times oplus Y \to Y \to 0$ is split by definition.
Second question: yesYes, $(x, y) \to mapsto \left(\frac{1}{2}(x+y), \frac{1}{2}(x+y)\right)$ is a projection of $X \times oplus X$ onto $\Delta$.
Two final remarks:
• a
• A very interesting procedure for producing non-split extensions of Banach spaces is the twisted sum construction due to Kalton-Peck (I recently learned about this from Bill Johnson in this thread).
• Basically, you're asking about the Yoneda Exts in the exact category of Banach spaces with the exact structure consisting of all kernel-cokernel pairs. If you're interested in such abstract nonsense, please allow me a bit of self-advertisement.
Post Undeleted by Theo Buehler
2 added 1013 characters in body; added 1 characters in body
I don't understand the questions for the following reason: if If $X$ is complemented in $E_1$ then the extension is splitfor algebraic reasons: . Indeed, if $P$ is a projection of $E_1$ onto the image of $X$ then $1-P$ is a projection onto an isomorph of $Y$.Y\$ by the open mapping theorem (see e.g. Nicolas Monod's thesis Corollary 4.2.4 for a detailed proof).
First question: If you're asking about a pair of extensions of $Y$ by $X$ with $E_1$ split and $E_2$ non-split, take $X = c_0$ and $Y = \ell^{\infty}/c_0$. Then $0 \to X \to \ell^{\infty} \to Y \to 0$ is not split by Phillips' lemma (see Whitley's note in the Monthly for a simple proof), and $0 \to X \to X \times Y \to Y \to 0$ is split by definition.
Second question: yes, $(x, y) \to \left(\frac{1}{2}(x+y), \frac{1}{2}(x+y)\right)$ is a projection of $X \times X$ onto $\Delta$.
Two final remarks:
• a very interesting procedure for producing non-split extensions of Banach spaces is the twisted sum construction due to Kalton-Peck (I recently learned about this from Bill Johnson in this thread).
• Basically, you're asking about the Yoneda Exts in the exact category of Banach spaces with the exact structure consisting of all kernel-cokernel pairs. If you're interested in such abstract nonsense, please allow me a bit of self-advertisement.
Post Deleted by Theo Buehler
1
I don't understand the questions for the following reason: if $X$ is complemented in $E_1$ then the extension is split for algebraic reasons: if $P$ is a projection of $E_1$ onto the image of $X$ then $1-P$ is a projection onto an isomorph of $Y$.
First question: If you're asking about a pair of extensions of $Y$ by $X$ with $E_1$ split and $E_2$ non-split, take $X = c_0$ and $Y = \ell^{\infty}/c_0$. Then $0 \to X \to \ell^{\infty} \to Y \to 0$ is not split by Phillips' lemma, and $0 \to X \to X \times Y \to Y \to 0$ is split by definition.
Second question: yes, $(x, y) \to \left(\frac{1}{2}(x+y), \frac{1}{2}(x+y)\right)$ is a projection of $X \times X$ onto $\Delta$.
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http://math.stackexchange.com/questions/144332/interesting-recurrence-relation-tn-t-sqrtn-tn-sqrtn-n
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# Interesting Recurrence Relation $T(n) = T(\sqrt{n}) + T(n-\sqrt{n}) + n$
I found an interesting recurrence that I do not know how to solve. I think this has to do with quicksort with pivots at rank $\sqrt{n}$. I do not know how to approach this problem nor found any helpful resources about it.
Here is the recurrence:
$$T(n)=T(\sqrt{n})+T(n−\sqrt{n})+n$$ Any help would be much appreciated. Thanks!
Let's say the base case is T(N) where N < 2 is $O(1)$
-
8
I think you have to specify what $n$ is, and give some context, Doing what comes naturally I find the contradiction: T(0)=2T(0). T(0)=0. T(1)=T(1)+T(0)+1. T(0)=-1. – Mark Bennet May 12 '12 at 20:10
maybe it is recurence for exact square integers(for integers,from which we can take always square root – dato May 12 '12 at 20:13
@dato: Mark used the recurrence for n=0 and for n=1, which are both exact square integers. – Did May 12 '12 at 20:15
1
@dato 0 and 1 are squares. If I plug $n=1$ into the equation I need the value of T(0), so I can't exlude 0. If I put in $n=4$, I need T(2) etc – Mark Bennet May 12 '12 at 20:17
It's really the asker's responsibility to clarify the question, but given the context of analysis of quicksort, I would guess that the actual recurrence ought to be $T(0) = T(1) = 1$ and $T(n) = T(\lfloor\sqrt n\rfloor) + T(n-\lfloor\sqrt n\rfloor) + n$ for $n \ge 2$. – Rahul Narain May 12 '12 at 22:16
show 1 more comment
## 1 Answer
If I just worry about asymptotics and ignore the issue of whether the square root is an integer, I would guess $T(n)\approx bn^a$. Substituting that in gives $bn^a= bn^{a/2}+b(n-\sqrt n)^a+n\approx bn^{a/2}+bn^a-ban^{a-\frac 12} +n$ which is correct to the first two orders if $a=\frac 32, b=\frac 23$
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http://mathoverflow.net/questions/37202?sort=votes
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## Derived algebraic geometry via dg rings?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Jacob Lurie's stuff seems to develop derived algebraic geometry via $E_\infty$ rings and/or maybe something like simplicial commutative rings. Ben Wieland's comment in this question indicates that Lurie never deals with commutative dg algebras. However, it is supposed to be true that all of these different things are the same (meaning more precisely that their model categories are Quillen equivalent) in characteristic zero.
So my question is:
Is the theory of derived algebraic geometry via dg rings or dg algebras in characteristic zero developed anywhere? If not, why not?
My motivations:
1. I feel like there must be a good reason why Lurie does not use dg rings/algebras, other than the fact that they apparently don't work well in positive characteristic. So I wonder what the reasons are.
2. I don't know very much about homotopy theory, so I find the $E_\infty$ rings approach to DAG a bit daunting. I am personally more comfortable with dg algebras.
3. I am personally more interested in things involving "sheaves of dg algebras" than things involving "sheaves of $E_\infty$ rings" (such as elliptic cohomology (and TMF), which I understand is one of Lurie's motivations).
-
4
You may want to take a look at the work of Toen and Vezzosi, e.g. arxiv.org/abs/math/0210407 – Greg Stevenson Aug 31 2010 at 1:37
Thanks. That (and the papers of Ciocan-Fontanine and Kapranov) seems to be precisely what I am looking for. – Kevin Lin Aug 31 2010 at 1:58
3
You may find that many homotopy theorists really will not think twice about casually conflating "sheaves of DG-algebras" and "sheaves of `$E_\infty$` rings" in characteristic zero, precisely because of the strong way in which they're known to have equivalent homotopy theories. I can't put words in Lurie's mouth, but I think that the answer to (1) really is that dg-objects don't cover many of the contexts homotopy theorists are interested in. You may just need to wait a little for references to come out that specialize Lurie's work to dg-algebras. – Tyler Lawson Aug 31 2010 at 2:19
This equivalence is a Quillen equivalence of model categories, or ...? – Kevin Lin Aug 31 2010 at 2:26
@Kevin: Yes, exactly. – Tyler Lawson Aug 31 2010 at 4:51
show 2 more comments
## 1 Answer
Dear Kevin,
This is more or less an amplification of Tyler's comment. You shouldn't take it too seriously, since I am certainly talking outside my area of expertise, but maybe it will be helpful.
My understanding is that homotopy theorists are extremely (perhaps primarily) interested in torsion phenomena. (After all, homotopy groups are often non-trivial but finite.) TMF, for example, involves quite subtle torsion phenomena. Coupled with Tyler's remark that homotopy theorists have no fear of $E_{\infty}$ rings, and so are (a) happy to identify them with dg-algebras in char. zero, and (b) don't feel any psychological need to fall back on the crutch of dg-algebras, this makes me suspect that your assumption (1) is likely to be wrong. (I share your motivation (2), but this is a psychological weakness of algebraists that homotopy theorists seem to have overcome!)
In particular, one of Lurie's achievements is (I believe) constructing equivariant versions of TMF, which (as I understand it) involves (among other things) studying deformations of $p$-divisible groups of derived elliptic curves. It seems hard to do this kind of thing without having a theory that can cope with torsion phenomena.
Also, when Lurie thinks about elliptic cohomology, he surely includes under this umbrella TMF and its associated torsion phenomena. (So your (3) may not include all the aspects of elliptic cohomology that Lurie's theory is aimed at encompassing.)
-
It's not quite true that homotopy theorists are only interested in torsion (and integrality) phenomena; rational homotopy theory is a thriving field (though with techniques that are rather different from the rest of homotopy theory). However, <em>stable</em> rational homotopy theory is completely trivial (or as trivial as the theory of graded rational vector spaces is). – Torsten Ekedahl Aug 31 2010 at 4:06
I think this is a fairly good characterization. I can't help but comment on two things. First, as in ordinary characteristic zero algebraic geometry, the derived elliptic curves and their p-divisible groups are not themselves torsion objects, but are merely representing objects for torsion, and so this part of the story still doesn't strictly require `$E_\infty$`-algebras. – Tyler Lawson Aug 31 2010 at 4:56
1
Second, `$E_\infty$`-algebras are often studied in models where they are strictly commutative monoids - we use the same terminology to reflect that there is an underlying structure independent of which model is chosen. So far as "fear" of `$E_\infty$`-algebras, I might add an analogy with little mathematical content. Injective modules might be initially less intuitive than projective ones. However, this does not mean that algebraic geometers are necessarily brave when basing homological algebra on injective sheaves; the alternative is simply not always available to them! – Tyler Lawson Aug 31 2010 at 5:02
Dear Torsten and Tyler, Thank you for these clarifying remarks. – Emerton Aug 31 2010 at 11:48
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http://mathoverflow.net/revisions/29359/list
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## Return to Answer
1 [made Community Wiki]
Of course, the Cayley Hamilton Theorem is not really hard, and there are many many proofs of it. But you have to admit that, at least when you first step into linear algebra, it's rather surprising that it's enough to proof the theorem for diagonal matrices (which is a very short calculation). Because then you can derive it for diagonalizable matrices, which are dense with respect to the Zariski Topology (assuming w.l.o.g. that the ground field is algebraically closed). The latter is because every non-empty open subset is dense, a rather strange but here very useful property.
The same procedure applies to other polynomial identites in linear algebra, for example that the characteristic polynomials of $AB$ and $BA$ coincide.
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http://mathhelpforum.com/calculus/70645-integrals.html
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# Thread:
1. ## Integrals
find the following indefinite integral (using u substitution):
$\int x \sqrt[3]{x+1} dx$
evaluate the following definite integrals using the Fundamental Theorem of the Calculus (u sub involved):
$\int_{0}^{\frac{\pi}{4}} (1 + tan^2 x)sec^2 x dx$
$\int_{0}^{2|b|} \frac{x}{\sqrt{x^2 + b^2}} dx$
2. Originally Posted by qzno
find the following indefinite integral (using u substitution):
$\int x \sqrt[3]{x+1} dx$
evaluate the following definite integrals using the Fundamental Theorem of the Calculus (u sub involved):
$\int_{0}^{\frac{\pi}{4}} (1 + tan^2 x)sec^2 x dx$
$\int_{0}^{2|b|} \frac{x}{\sqrt{x^2 + b^2}} dx$
1) Let $u=x+1$
2) Let $u=\tan x$
3) Let $u=x^2+b^2$
3. I already tried this for them all haha
4. for the second one I got the evaluated answer to be:
$\frac{(tan 1)^3}{3}$
is this correct?
and i still cant do the last question : (
thanks to anyone who helps!!
5. For the first one when i let u = x + 1, i got the answer to be: $\frac{3x(x+1)^{\frac{4}{3}}}{4} + c$
is this correct?
i still havnt figured out the third one : (
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http://math.stackexchange.com/questions/122201/is-this-function-on-the-surface-smooth?answertab=votes
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# Is this function on the surface smooth?
Consider the following functions $F_{ij}:S\subset{\mathbb R}^3\to{\mathbb R}$, $$F_{ij}(y) = \begin{cases} \frac{(y_i-x_i)(y_j-x_j)(y-x)\cdot n(y)}{|y-x|^3},&y\neq x; \\ 0,& y=x.\end{cases} \quad i,j = 1,2,3$$ where $S$ is a surface which has a continuously varying normal vector, $x=(x_1,x_2,x_3)\in S$ is given, $y=(y_1,y_2,y_3)\in S$, $n(y)$ is the normal vector at point $y$ [EDITED: which is assumed to be smooth]. Here $(y-x)\cdot n(y)$ is the dot product. Using the method in the answer to the question on MO, I conclude that given $i,j$ $$\lim_{y\to x}F_{ij}(y)=0$$ which implies that $F_{ij}(y)$ is continuous at $y=x$.
Here is my question:
• Is $F_{ij}(y)$ smooth at $x$? If it is not, what would be the key properties to fail the smoothness?
An immediate idea is that I should test the smoothness of $F_{ij}$ by definition. The difficulty is that with a parameterization $y=y(\alpha,\beta)$, it is not trivial to find the high order partial derivatives for $F_{ij}(y(\alpha,\beta))$. I am not even able to determine if $F_{ij}$ is $C^1(S)$.
[EDITED:] Using the parameterization in this answer, I am able to get something like $$F_{12}(\alpha,\beta)=\frac{-\alpha\beta (\alpha h_{\alpha}+\beta h_{\beta}-h)}{(\alpha^2+\beta^2+h^2)^{3/2}}$$ where $y(\alpha,\beta)=(\alpha,\beta,h(\alpha,\beta))$. Due to the root term, I guess $F_{12}$ can not be $C^1$.
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You need that $n(x)$ is smooth in this case. – azarel Mar 19 '12 at 19:30
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http://mathoverflow.net/questions/81789/is-this-a-subcase-of-the-fundamental-lemma
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Is this a subcase of the fundamental lemma?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $F$ be a local field and $G= GL(n,F)$.
Assume that $\gamma$ is an element of $G$ and $G_\gamma$ is its centralizer. The orbital integral is defined as $$O_\gamma^G( \phi) = \int\limits_{G_\gamma \backslash G} \phi( g^{-1} \gamma g) d g.$$
We can assume wlog that $\gamma$ is elliptic. Can we lift the elliptic orbital integral on $G$ to an elliptic orbital integral on $G' =GL(n,F')$, where $F' = F[X] / det(X-\gamma)$?
More precisely, can we find an $T:C_c^{\infty}(G) \rightarrow C_c^\infty(G')$ such that $$O_\gamma^G(\phi) = O_\gamma^{G'} ( T\phi).$$
I think for explicit computations, it is much more convenient to work on $G'$, where $\gamma$ is diagonalizable nad $G'_\gamma$ can be chosen to be the diagonal matrices.
What is the Reference? What is the Buzzword for this?
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1 Answer
I will offer some words on this, but only because no-one else has; I was holding out hoping that one of the more automorphic people would chip in. It might be worth taking much of the below with a pinch of salt.
So I've been trying to penetrate the "fundamental lemma" literature myself, and let me begin by showing my hand and saying that my current impression, that may be wrong, is that "the fundamental lemma" is not a well-defined mathematical statement, it is a principle that applies in many situations, and I think that mathematics is open to the situation where there will be precise mathematical statements formulated in the future, when people are studying situation $X$, and that people will call these statements "the fundamental lemma for $X$". For example, I think that perhaps in 50 years' time when people are proving general base change (rather than cyclic base change) there may be a "general base change fundamental lemma".
So this of course immediately raises the question -- what did Ngo Bao Chau do? Well, my perception is that he has proved the version of the fundamental lemma that shows up in the theory of endoscopy.
As I've already tried to make clear, I am not 100% sure on all of this. But let me muddle on regardless. My impression is that what the game looks like is this. There is the general functoriality principle, which says that if I have two connected reductive groups $G$ and $H$ over a global field, and a map between the $L$-group ${}^LH\to{}^LG$, then there should be some sort of way of transferring automorphic representations on $H$ to automorphic representations on $G$. Already this is a "principle" rather than a precisely-formulated statement, because I think there are lots of issues with packets and multiplicities, that certainly I don't understand, when trying to make a precise statement, and I am not sure I've seen a precise statement in the literature that ties up all the loose ends that I want to see tied up, other than some very weak ones that just demand compatibility at the unramified places and don't care about multiplicities at all.
But it turns out, when attempting to e.g. study the zeta functions of some Shimura varieties, that Langlands needed some very special cases of the functoriality principle, where $H$ was an endoscopic subgroup for $G$, and here he postulated a strategy for proving functoriality using the trace formula. The idea is that you try and stabilise the trace formula for $G$, whatever that means, and this involves, amongst other things, figuring out some way of transferring functions locally and matching the orbital integrals that come up. The upshot is that you relate the trace formula for $G$ to the trace formula for all the endoscopic subgroups for $G$.
So my perception is that, in its initial state, the situation was this: $G$ was a connected reductive group, now over a local field, $H$ an endoscopic subgroup (and part of the data of this endoscopic situation is that you're given a map ${}^LH\to{}^LG$) and there is supposed to be a "transfer" map, that maps functions on $G$ to functions on $H$. Even the existence of the transfer map is not at all clear in general, I don't think. For example Labesse-Langlands have to do some calculations (only a few pages, but some work) to prove that one can transfer functions from $SL(2)$ to a subtorus (this is the simplest example of endoscopy, I think).
So my impression is that the general notion of moving from functions on one group, locally, to functions on another, is called "transfer of functions". My understanding is that the transfer map is not at all well-defined, that sometimes one can characterise the image (as being functions whose orbital integrals vanish on some certain subgroup), and that at the end of the day the precise relationship you want between the function and its transfer can be quite complicated. I think that in general you want the orbital integral of one function to be the orbital integral of the other multiplied by a "fudge factor" whose definition is the key point of a 100-page paper by Langlands and Shelstad. One can already see these fudge factors in the $SL(2)$ case with Langlands-Labesse.
My understanding of what the fundamental lemma is, is the following: in the situation where $G$ and $H$ are unramified, one extra condition you could put on the "transfer of functions" map is that the identity element for the unramified Hecke algebra for $G$ gets sent to the identity element for the unramified Hecke algebra for $H$. Hence in this situation, "the fundamental Lemma", I think, boils down to the assertion that a certain volume equals a fudge factor that it takes 100 pages to define, multiplied by another volume.
I'm slowly getting to the point :-) I think I can answer one of your questions at least -- the "buzzword" you're looking for is not "fundamental lemma" but "transfer of functions" or perhaps "local transfer of functions". I think.
However, my understanding is that the situation you are looking at is not an "endoscopic situation". In particular I don't think that reading the complete works of Ngo Bao Chau will give you an answer to your question. You have groups $G$ and $G'$ and they're both $GL(n)$ but over different fields, so if I were trying to prove a hard global theorem and I needed the type of transfer that you're looking for, I would probably not be trying to stabilise the trace formula (indeed I think the trace formula for $GL(n)$ is already stable and that "$GL(n)$ has no endoscopy" in some sense) -- I would probably be trying to prove base change.
Now here are, for me, some BIG problems I would fear when attempting to try and get an answer to your question from the literature that I know about.
The first is that, when trying to prove global base change for $GL(n)$ for a global cyclic extension $L/K$, I attempt to find a relation between the trace formulas for $GL(n)$ over $L$ and over $K$ and then I attempt to start matching up terms etc etc, and the problem is that one trace formula is a sum over conj classes in $GL(n,L)$ and the other is a sum over $GL(n,K)$. As far as I know, people don't know how to relate these two sets in a natural way. So what they do is they relate conjugacy classes in $GL(n,K)$ to twisted conjugacy classes in $GL(n,L)$. I think the local story looks like this: say $E/F$ is now local and has cyclic Galois group. Given $\gamma$ in $GL(n,E)$ they take its "norm" in the most naive way (multiply it by its Galois conjugates) and get an element of $GL(n,F)$ and in this nice cyclic situation they can inject twisted conj classes ($x\sim \sigma(g)xg^{-1}$ with $\sigma$ a generator of the Galois group) in $GL(n,E)$ with conj classes in $GL(n,F)$. Using this trick they want to relate the usual trace formula for $GL(n,K)$ with a twisted trace formula for $GL(n,L)$, and to do this the transfer of functions they require is a twisted transfer! In particular, they need a machine which, given a function $a$ on $GL(n,E)$ spits out a function $b$ on $GL(n,F)$ such that the orbital integrals of $b$ equal certain twisted orbital integrals for $a$, possibly again multiplied by some fudge factors, but I believe that in this base change situation the fudge factors (which are I think not covered by the Langlands-Shelstad monster because this is not an endoscopic situation) are all 1 anyway.
The upshot is that in the Arthur-Clozel book, proving cyclic base change for $GL(n)$, you will see a definition of transfer of functions, which looks formally a bit similar to the thing you write above, but there are certain crucial differences:
1) if $a$ transfers to $b$ then an orbital integral for $b$ will equal a twisted orbital integral for $a$, rather than an orbital integral.
2) The twisted orbital integral for $a$ will be attached to an element $\gamma$ and the orbital integral for $b$ that it equals will be attached not to $\gamma$ but to its naive norm (which is a well-defined conj class, I believe, in $GL(n,F)$)
3) $E/F$ will ALWAYS be cyclic (or perhaps more generallu they will allow $E=F^n$ so $E$ is not actually a field but it's still etale over $F$).
Now I look at the question you're asking, and there are of course two ways of approaching it: the first is to try and get your hands dirty and write down the map between functions yourself. But it sounds to me like you're hoping that you can take another approach -- to get what you want from the literature. And what I am really scared by is that although what you write looks to me superficially like transfer of functions, you are in a situation where $F'$ is not in general a cyclic Galois extension of $F$, so you can throw away Arthur-Clozel, and you are not I think in an endoscopic situation either, so you can also throw away Ngo Bao Chau, and unfortunately I personally do not know what is left. Of course this will largely be my own ignorance and probably there are people who have thought about "transferring" in its own right, independent of links to functoriality. But I am now not sure where to point you.
Aah, the joyful tones of my daughter, who has apparently just woken up. What timing she has! I've gotta go, but I've said all I can say anyway.
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Thanks. You are only adressing in the last part of your answer my original question. The rest of your ellaboration actually adresses a question, which was implicit or even only in the title;) So thanks for reading inbetween the lines. After your explanation, I think the answer is contained in Laumon "Cohomology of Drinfeld modules" Part 1, Proposition 4.7.1, page 114, at least for unramfied extensions. So you claim it is sufficient to prove every thing for the unit? This uses invariance here? I probably have to also include weights in my orbital integrals.... – Marc Palm Nov 26 2011 at 10:19
so I will have to reprove most of the chapter 4 in Laumon. Also my field extensions will be ramified in general. Are there some standard tools, which allow unramified extensions to be treated as ramified ones? – Marc Palm Nov 26 2011 at 10:25
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I am not an expert so I can only offer you a superficial overview rather than practical help with details. My general impression is that in certain situations (e.g. an unramified endoscopic situation), the transfer map is staring at you in the face -- it's coming from the theory of the Satake isomorphism. The problem is proving that the "obvious" transfer does the job you want, i.e. that one orbital integral equals another. I think that if you can check it for the identity functions then you can check it for all the functions, and the fundamental lemma is the assertion that... – Kevin Buzzard Nov 26 2011 at 10:49
1
...it's OK for the identity function. So in some simple situations the problem of transferring functions is equivalent to the fundamental lemma. The moment one moves away from these simple situations, I know nothing and I'm afraid I'm not the person to ask. I'll try to remember to take a look at Laumon's book on Monday when I'm back at work, but I am a little worried that it will still mean very little to me on anything other than a superficial level. Ask me the same questions again in 5 years and we'll see if I've learnt any more! I'm still struggling through Labesse-Langlands! – Kevin Buzzard Nov 26 2011 at 10:52
Ok I finally remembered to look. It seems to me that Laumon is just doing what Arthur-Clozel do (but perhaps in char $p$ rather than char 0): proving the base change fundamental lemma. In particular he is proving that an orbital integral downstairs is equal to a twisted orbital integral upstairs. You do not seem to have the twist in your question, so it seems to me that Laumon's calculations are not directly relevant. – Kevin Buzzard Nov 29 2011 at 10:45
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http://physics.stackexchange.com/questions/tagged/action?sort=faq&pagesize=15
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# Tagged Questions
The action tag has no wiki summary.
5answers
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### Hamilton's Principle
Hamilton's principle states that a dynamic system always follows a path such that its action integral is stationary (that is, maximum or minimum). Why should the action integral be stationary? On ...
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### Deriving the Lagrangian for a free particle
I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading "Mechanics" by Landau and Lifshitz recently and hit a few roadblocks right away. Proving that a free particle ...
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### What is the relativistic action of a massive particle?
all Lorentz observers watching a particle move will compute the same value for the quantity $$ds^2 = -(c \, dt)^2 + dx^2 + dy^2 + dz^2,$$ $$ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu},$$ and ''ds/c'' is then ...
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### What variables does the action $S$ depend on?
Action is defined as, $$S ~=~ \int L(q, q', t) dt,$$ but my question is what variables does $S$ depend on? Is $S = S(q, t)$ or $S = S(q, q', t)$ where $q' := \frac{dq}{dt}$? In ...
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### Why the Principle of Least Action?
I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$. Okay, fine. You write down the action ...
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### Is the principle of least action a boundary value or initial condition problem?
Here is a question that's been bothering me since I was a sophomore in university, and should have probably asked before graduating: In analytic (Lagrangian) mechanics, the derivation of the ...
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### Why can't any term which is added to the Lagrangian be written as a total derivative (or divergence)?
All right, I know there must be an elementary proof of this, but I am not sure why I never came across it before. Adding a total time derivative to the Lagrangian (or a 4D divergence of some 4 ...
2answers
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### Why lagrangian is negative number?
In the special relativistic action for a massive point particle, $$\int_{t_i}^{t_f}\mathcal {L}dt,$$ why is the Lagrangian $$\mathcal {L}=-E_o\gamma^{-1}$$ a negative number?
2answers
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### How do I show that there exists variational/action principle for a given classical system?
We see variational principles coming into play in different places such as Classical Mechanics (Hamilton's principle which gives rise to the Euler-Lagrange equations), Optics (in the form of Fermat's ...
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### Derivation of Maxwell's equations from field tensor lagrangian
I've started reading Peskin and Schroeder on my own time, and I'm a bit confused about how to obtain Maxwell's equations from the (source-free) lagrangian density \$L = ...
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### Do an action and its Euler-Lagrange equations have the same symmetries?
Assume a certain action $S$ with certain symmetries, from which according to the Lagrangian formalism, the equations of motion (EOM) of the system are the corresponding Euler-Lagrange equations. Can ...
2answers
75 views
### More general invariance of the action functional
I will formulate my question in the classical case, where things are simplest. Usually when one discusses a continuous symmetry of a theory, one means a one-parameter group of diffeomorphisms of the ...
1answer
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### What is the action for an electromagnetic field if including magnetic charge
Recently, I try to write an action of an electromagnetic field with magnetic charge and quantize it. But it seems not as easy as it seems to be. Does anyone know anything or think of anything like ...
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### Gauge fixing and equations of motion
Consider an action that is gauge invariant. Do we obtain the same information from the following: Find the equations of motion, and then fix the gauge? Fix the gauge in the action, and then find the ...
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### Conversion of the Nambo-Goto action into the Polyakov action?
I`ve read that the Nambo-Goto action containing the induced metric $\gamma_{\alpha\beta}$ $$\tag{1} S_{NG} ~=~ -T\int_{\tau_i}^{\tau_f} d\tau \int_0^{\ell} d\sigma \sqrt{-\gamma}$$ can be converted ...
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### $\hbar$, the angular momentum and the action
Is there anything interesting to say about the fact that $\hbar$, the angular momentum and the action have the same units or is it a pure coincidence?
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### Varying an action (cosmological perturbation theory)
I am stuck varying an action, trying to get an equation of motion. (Going from eq. 91 to eq. 92 in the image.) This is the action $$S~=~\int d^{4}x \frac{a^{2}(t)}{2}(\dot{h}^{2}-(\nabla h)^2).$$ ...
3answers
285 views
### Calculating lagrangian density from first principle
In most of the field theory text they will start with lagrangian density for spin 1 and spin 1/2 particles. But i could find any text where this lagrangian density is derived from first principle.
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484 views
### Does Action in Classical Mechanics have a Interpretation? [duplicate]
Possible Duplicate: Hamilton's Principle The Lagrangian formulation of Classical Mechanics seem to suggest strongly that "action" is more than a mathematical trick. I suspect strongly ...
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51 views
### Why vary the action with respect to the inverse metric?
Whenever I have read texts which employ actions that contain metric tensors, such as the Nambu-Goto, Polyakov or Einstein-Hilbert action, the equations of motion are derived by varying with respect to ...
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http://www.maths.usyd.edu.au/u/pubs/publist/preprints/2010/thomas-26.html
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## Infinite generation of non-cocompact lattices on right-angled buildings
### Anne Thomas and Kevin Wortman
#### Abstract
Let $\Gamma$ be a non-cocompact lattice on a locally finite regular right-angled building $X$. We prove that if $\Gamma$ has a strict fundamental domain then $\Gamma$ is not finitely generated. We use the separation properties of subcomplexes of $X$ called tree-walls.
This paper is available as a pdf (108kB) file.
Wednesday, September 22, 2010
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http://math.stackexchange.com/questions/167752/bipartite-simple-graph-matching-proof/270259
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# bipartite simple graph matching-proof
Prove that if $G$ is a bipartite simple graph and every vertex has the same degree $k$, then the edges of $G$ can be partitioned into $k$ matchings. Can we then prove the same if every vertex has degree $\le k$?
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It's not clear in your statement of the problem whether or not every vertex is to still have the same degree. – Tim Duff Jul 7 '12 at 8:27
The second bit asks whether the same can be proven if every vertex has degree AT MOST $k$ rather than $=k$. Does that answer your question? – Aria Fitzpatrick Jul 7 '12 at 8:31
No. Does every vertex have the same degree or may vertices have different degrees? – Tim Duff Jul 7 '12 at 8:48
The first proof requires that all vertices have deg=k, while the second doesn't(but has a max cap on deg=k). – Aria Fitzpatrick Jul 7 '12 at 8:59
1
@TimDuff Note that the question does not require the $k$ matchings to be perfect. – Erick Wong Jul 7 '12 at 9:07
## 3 Answers
The "big stick" route is via iterative application of Hall's Marriage Theorem, which for our purposes tells us that a regular bipartite graph has a perfect matching (restricted to regular bipartite graphs, the proof of this is not too hard either).
Then given $k$-regular bipartite $G=(V_{1}\uplus V_{2},E)$ we can find a perfect matching $M \subseteq E$. Let $G'=(V_{1}\uplus V_{2}, E\setminus M)$. $G'$ is clearly a $(k-1)$-regular bipartite graph, hence we can apply Hall's theorem again, et cetera.
You could also go the other way, and give an inductive proof. The base case of $k=1$ is obvious, then with the assumption that $k=p-1$ gives us a $(p-1)$-matching it's relatively straight forward to show that for $k=p$ you can find a perfect matching, then apply the inductive assumption for the graph without that matching, though you're really just showing the special case of Hall's theorem again to prove that you can get the first perfect matching.
For the second part you should be able to do basically the same with Hall's theorem. First discard any degree zero vertices, then take a matching which covers the smaller side, remove this and iterate. A simple contradiction argument shows that you can do this at most $k$ times. (Assume you do it $k$ times and there's some edge left over, but it can't be part of any of the $k$ previous matchings, then at least one of its end points must have an edge in all the matchings, and hence must have degree greater than $k$)
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Could you maybe do this more rigorously? – Aria Fitzpatrick Jul 7 '12 at 15:17
A bipartite graph $G$ with degree bounded above by $k$ can be extended (as necessary by adding both vertices and edges) to a regular bipartite graph $H$ of degree $k$. Partitioning the edges of $H$ into $k$ perfect matchings then induces a partition of edges of $G$ into $k$ (not necessarily perfect) matchings.
Added: It would probably be in keeping with the problem to phrase the conclusion as partitioning $G$ into at most $k$ matchings, since the above construction may produce matchings in $H$ that miss $G$.
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I'll give a self-contained proof; I "borrowed" this proof from Theorem 7.6.3 in A Textbook of Graph Theory by Balakrishnan and Ranganathan.
Theorem (König): Let $G$ be a bipartite graph with maximum degree $\Delta$. Then $G$ can be assigned a proper edge colouring using $\Delta$ colours.
Proof: We proceed by induction on the number of edges $m$ (having fixed the vertex set, and vertex bipartition). The theorem is true for $m=1$. Now assume $m>1$.
Let $uv$ be an edge in $G$. By the inductive hypothesis, $G \setminus \{uv\}$ can be properly edge coloured using $\Delta$ colours. An example is given below for $\Delta=3$.
In this colouring of $G \setminus \{uv\}$:
• Let $i$ be a colour not used for an edge with an endpoint $u$ in $G \setminus \{uv\}$.
• Let $j$ be a colour not used for an edge with an endpoint $v$ in $G \setminus \{uv\}$.
Note: both $i$ and $j$ exist, since the degrees of $u$ and $v$ in $G \setminus \{uv\}$ are at most $\Delta-1$.
We can assume $u$ is the endpoint of an $j$-coloured edge, otherwise, we can assign $uv$ the colour $j$, and we are done.
Let $P$ be the maximum-length path starting at $u$ following edges of colours $j$ and $i$ alternatingly. In the above example, we take $i=\text{blue}$ and $j=\text{red}$, so $P$ is the path depicted below.
Importantly, $P$ cannot have $v$ as a vertex; this would imply one of the following:
• If $v$ were an internal vertex of $P$, then $v$ is an endpoint of an $j$-coloured edge, giving a contradiction.
• If $v$ were a terminal vertex of $P$, then either $u$ and $v$ belong to the same part (contradicting that $G$ is bipartite) or $v$ is an endpoint of an $j$-coloured edge, giving a contradiction.
If we swap the colours $i$ and $j$ in $P$, we obtain a new proper edge colouring of $G \setminus \{uv\}$ using $\Delta$ colours. (It is indeed proper, since $P$ is of maximum-length.) In the above example, we obtain the following modified colouring.
However, this time, both $u$ and $v$ are not the endpoint of an $j$-coloured edge, so we can add back in the edge $uv$ and colour it $j$, thereby obtaining a proper edge colouring of $G$ using $\Delta$ colours. In our running example, the final result is the following.
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http://mathoverflow.net/questions/35408?sort=oldest
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## Naturally occuring groups with cardinality greater than the reals.
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In group theory, the single most important piece of information about a group is its cardinality, which is of course either finite, countably infinite, or uncountably infinite. Usually, however, uncountably infinite simply means a cardinality of $\aleph_{1}$, the same as $\mathbb{R}$. My question is: is there anywhere that groups with cardinality strictly greater than $\aleph_{1}$ arise naturally? Of course, it is easy enough to construct groups with arbitrarily large cardinality, but I cannot recall ever seeing them used.
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19
The cardinality of $\mathbb{R}$ is not necessarily $\aleph_1$. – François G. Dorais♦ Aug 13 2010 at 0:45
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What gave you the impression that the single most important piece of information about a group is its cardinality? It misses nearly all of the richness of group theory. For instance, many interesting classes of finite simple groups can be studied using a lot of the same tools as are used in the study of simple complex Lie algebras and semisimple linear algebraic groups. – BCnrd Aug 13 2010 at 0:54
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Turning this remark around, can one give a "natural" example of a group which has cardinality $\aleph_1$, independent of CH? – Pete L. Clark Aug 13 2010 at 0:55
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Pete, very good question! Here is one answer: it is consistent with ZFC that there are no Borel sets, and even no analytic sets, of size $\aleph_1$. In such a model of set theory, the only groups (built out of real numbers) of size $\aleph_1$ would have high descriptive set-theoretic complexity. This could be taken as a negative answer to your question. But a positive answer could still arise by builiding a group directly out of the countable ordinals, rather than by using reals. – Joel David Hamkins Aug 13 2010 at 1:05
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You can push it up to $\Sigma^1_2$ using the Mansfield Solovay theorem, and if you assume PD, then there are no projective sets of size $\aleph_1$. In those models, there are arguably no natural examples of sets of reals of size exactly $\aleph_1$. – Joel David Hamkins Aug 13 2010 at 3:48
show 2 more comments
## 5 Answers
In line with Joel's answer, my favorite "outrageously large group" is the group $G = \operatorname{Aut}(\mathbb{C})$ of field automorphisms of the complex numbers. It has cardinality $2^{2^{\aleph_0}}$, which is pretty scary. But that's just the beginning of how large it is. For instance, from the study of real-closed fields, one can deduce that the number of conjugacy classes of order $2$ elements of $G$ is also $2^{2^{\aleph_0}}$. It is also an extension of the absolute Galois group of $\mathbb{Q}$ (a profinite group which is conjectured to have among its quotients every finite group, up to isomorphism) by the huge simple group $\operatorname{Aut}(\mathbb{C}/\overline{\mathbb{Q}})$.
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A nice reference for some of the basic facts on Aut(C) is the 1957 Lester R. Ford prize-winning article: Paul B. Yale, Automorphisms of the complex numbers, Math. Mag. 39 (1966), 135-141. jstor.org/stable/2689301 – Skip Aug 13 2010 at 0:59
Why is $\operatorname{Aut}(\mathbb C/\overline{\mathbb Q})$ simple? – Emil Jeřábek Mar 1 2011 at 17:26
2
@Emil: That is a special case of a theorem of Daniel Lascar: see the end of math.uga.edu/~pete/galois.pdf for a precise reference. – Pete L. Clark Mar 1 2011 at 17:30
Thanks ! – Emil Jeřábek Mar 1 2011 at 18:58
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I would expect that automorphism groups of natural structures would count as natural groups in your sense. But automorphism groups of uncountable structures often have size larger than the continuum. In general, the size of the automorphism group of a structure of size $\kappa$ is bounded above by $2^\kappa$, which is strictly larger than $\kappa$, and this upper bound is often reached, when the structure is insufficient to restrict the general nature of automorphisms. For example, the number of bijections of an infinite set of size $\kappa$ with itself is $2^\kappa$.
I am sure that you will be able to construct many other natural structures of uncountable size $\kappa$, whose automorphism groups have size $2^\kappa$, and these would seem to the sort of examples you seek.
P.S. Let me also note that your remark that the reals have size $\aleph_1$ is only correct when the Continuum Hypothesis holds. In general, the size of the reals, also known as the continuum, is $2^{\aleph_0}$, which is also denoted $\beth_1$, whereas $\aleph_1$ is simply the first uncountable cardinal.
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Does a group showing up in a College Algebra (pre-calculus) course count as arising naturally? I'm pretty sure we teach students to add two functions (from the reals to the reals) pointwise to get a new function, even there. Of course, on the one hand really we only ask them to deal with the countable subset of functions with a finite description, and on the other hand Abelian groups are not as interesting, but technically that defines a group with cardinality greater than the continuum. (We also define inverse functions and composition, but at first glance it seems that strictly monotone functions must have only continuum cardinality.)
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But if you are truly in pre-calculus, you likely only find piece-wise continuous functions, and even if you allow countably many pieces, you still have only continuum many such functions, the same size as $\mathbb{R}$, since on each piece they are determined by their values on the rationals. – Joel David Hamkins Aug 13 2010 at 1:38
@Joel: Hence my caveat about only asking them to grok the countable subgroup of finitely described functions. – Tracy Hall Aug 13 2010 at 1:53
Yes, that is a nice way to say it. (+1). – Joel David Hamkins Aug 13 2010 at 1:55
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Every Banach space is an Abelian group, and many of the usual Banach spaces we study have larger cardinality than the reals. – Carl Mummert Aug 13 2010 at 12:55
@CM: Really? (Meaning: "I know very little about Banach spaces", not "I doubt that very much.") What is such an example? – Pete L. Clark Aug 13 2010 at 23:24
See Canter's Theorem . Cardinality of the power set of R is strictly greater than that of R.
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That should be "Cantor". There is a question of which group structure on P(R) you mean; one possibility would be the one given in Henno's answer. – Todd Trimble Mar 1 2011 at 18:40
Any product group like $\{0, 1\}^I$ for index sets $I$, using mod 2 addition coordinatewise. It's just (isomorphic to) the power set of $I$ using symmetric difference as the addition. It's of course also a ring (pointwise multiplication / intersection ). These Boolean groups often come up in general topology.
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http://mathoverflow.net/questions/25513?sort=oldest
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## Zariski closed sets in C^n are of measure 0
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This is related to another question in which it is proved that Zariski open sets are dense in analytic topology.
But it is intuitive that something more is true. Namely, that they are the sets where some polynomials vanish, and consideration of a few examples in $\mathbb R^n$ where they are of Lebesgue measure $0$, suggest strongly that the Zariski-closed sets(except the whole affine space) are of measure $0$ in $\mathbb C^n$ as well. This should be quite simple; but I am unable to prove it due to inexperience in measure theory.
The nice thing about proving this is that once this is done, then we are able to claim safely that so-and-so statement is true almost everywhere, if it is true on a Zariski-open set.
So, in a more measure theoretic formulation:
Let $X$ be a set in $\mathbb C^n$ contained in the zero locus of some collection of polynomials. How to show that $X$ is of measure $0$?
In fact my feeling is that more should be true, ie, we can replace polynomials by analytic functions at least, and get the same result.
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Three out of four ain't bad! (Hint: an s is missing in the title.) I've batted worse with someone else's name in a paper. – Willie Wong May 21 2010 at 20:45
Thanks Willie Wong! I am sorry to have messed it up like this. – Akela May 21 2010 at 21:33
Induct on $n$. When $n = 1$, this is true since zeros of a non trivial polynomial is a finite set. For inductive step use Fubini's theorem. – Ashutosh Nov 27 2010 at 0:19
## 4 Answers
If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.
Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Clearly, it is sufficient too show it for a closed set given by $f=0$ where $f$ is analytic. (write your set as included in a countable union of such described sets). Then, using the normal form of analytic germs as finite ramified coverings, you're done.
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There is a very naive argument for this. As Henri says, it reduces to a zero set of a polynomial $f$. Write $$f(z_1,\ldots,z_n)=\sum_{j=0}^d g_j(z_1,\ldots,z_{n-1})z_n^j.$$ where the polynomial $g_d$ is not identically zero. For each $(z_1,\ldots,z_{n-1})\in\mathbb{C}^n$, there are only finitely many $z_n$ with $f(z_1,\ldots,z_n)=0$ unless $g_d(z_1,\ldots,z_{n-1})=0$. Inductively these exceptional $(n-1)$-tuples form a set of measure zero in $\mathbb{C}^{n-1}$ and now the result follows from Fubini's theorem (regarding $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ and going down two real dimensions).
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And if you want to take this to an extreme... for a function on a domain in ${\bf R}^n$, it's enough to assume that at every $x$ there's a ball $B_x$ centered at $x$ and a multiindex $\alpha$ for which $\partial^{\alpha} f$ is nonzero and continuous on $B_x$.
To see this, first note that it suffices to show that the zeroes of $f$ in a given $B_x$ have measure zero. This is proven by induction on $|\alpha|$. If $\alpha = 0$ it's trivial, and if $\partial^{\alpha'}f(x) \neq 0$ for any $\alpha '$ with $|\alpha '| < |\alpha|$ it follows by the induction hypothesis, shrinking $B_x$ if necessary. Otherwise we can write $\partial^{\alpha} f = \partial_{x_i}\partial^{\beta} f$ for some $\beta$, where we can assume $\partial^{\beta} f(x) = 0$. By the implicit function theorem, if $B_x$ is small enough the zeroes of $\partial^{\beta} f$ in $B_x$ form a $C^1$ hypersurface with measure zero. For each $y$ off this surface, $\partial^{\beta} f$ is nonzero and then you can apply the inductive hypothesis to find an appropriate $B_y$. A simple compactness argument shows you need only countably many $B_y$.. so you're done.
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http://nrich.maths.org/6009/solution
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### Folium of Descartes
Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.
### Witch of Agnesi
Sketch the members of the family of graphs given by y = a^3/(x^2+a^2) for a=1, 2 and 3.
### Quick Route
What is the quickest route across a ploughed field when your speed around the edge is greater?
# Least of All
##### Stage: 5 Challenge Level:
Thank you Ruth from Manchester High School for Girls for your solution to this problem.
Most people's first reaction to this question is exactly the same as Ruth's:
"My initial conjecture was that the minimum value of $f(x)$ is when $x=0$ for any value of $a$, because the function is even and increases as $|x|$ increases."
Using calculus, we shall see that this is not so and that the minimum value of $f(x)$ does depend on the value of $a$ .
The given expression is \begin{eqnarray} f(x) &=& (1 + (a+x)^2)(1 + (a-x)^2)\\ &=&(1 + x^2 + a^2 +2ax)(1 + x^2 +a^2 -2ax)\\ &=& (1 + x^2 + a^2)^2 - 4a^2x^2 \\ &=& x^4 + 2x^2(1 + a^2) + (1 +a^2)^2 - 4a^2x^2 \\ &=& x^4 + 2x^2(1 - a^2) + (1+a^2)^2. \end{eqnarray}
As this is a quartic in $x$ there will be one or three turning points.
Differentiating $f$ to find the minima:
$$f'(x) = 4x^3 + 4x(1-a^2) = 4x(x^2 + (1 - a^2))$$
Case 1 : $(1 - a^2) < 0$.
The derivative $f'(x) = 0$ for $x = 0$ and $x = \pm \sqrt (a^2 - 1)$. The second derivative $f''(x) = 12x^2 + 4(1 - a^2) < 0$ at $x = 0$ which gives a maximum value but $f''(x) = 12x^2 + 4(1 - a^2) > 0$ for $x = \pm \sqrt (a^2 - 1)$ giving two minimum points on the quartic where $x = \pm \sqrt (a^2 - 1)$. The minimum value at each point is $f(x) = 4a^2$ where the position of X for these minimum values is clearly dependent of $a$.
Case 2 : $(1 - a^2) > 0$.
The derivative $f'(x) = 0$ if and only if $x = 0$. The second derivative
$f''(x) = 12x^2 + 4(1 - a^2) > 0$ so there is one minimum value $f(x)= (1+a^2)^2$ where the position of X, at $x=0$, is independent of $a$ agreeing with the conjecture.
Case 3 : $(1 - a^2) = 0$.
Note that where $a^2 = 1$ there is a single minimum $f(x) = 4$ at $x=0$ giving continuity between Case 1 and Case 2.
The most likely first conjecture agrees with Case 2 but does not allow the possibility of Case 1. Realising that the function we are minimizing is a quartic we should have taken into account the possibility of two minimum values.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://physicspages.com/2011/09/26/gausss-law/
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Notes on topics in science
## Gauss’s Law
Required math: vectors, calculus
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Sec 2.2.
Gauss’s law in electrostatics is a relation between the charge contained by a closed surface and the electric field that crosses that surface. The easiest way to see how it works is to begin with a point charge at the origin and a spherical surface centred at the origin. By symmetry the electric field due to the point charge is
$\displaystyle \mathbf{E}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{\mathbf{r}}$
The flux of this field through the sphere is defined as the surface integral of the component of the field that is normal to the surface. That is the flux ${\Phi}$ is defined as
$\displaystyle \Phi=\int\mathbf{E}\cdot d\mathbf{a}$
where the integral extends over the surface, and ${d\mathbf{a}}$ is a differential vector whose magnitude is a differential area element and whose direction is normal to the surface at each point.
In the case of a sphere, it is not surprisingly easiest to use spherical coordinates, and in that case
$\displaystyle d\mathbf{a}=r^{2}\sin\theta d\theta d\phi\hat{\mathbf{r}}$
That is, the area element points radially outwards at each point on the sphere.
Combining these results, we see that for a point charge
| | | |
|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|-----------------------------------------------------------------------------|
| $\displaystyle \Phi$ | $\displaystyle =$ | $\displaystyle \int\mathbf{E}\cdot d\mathbf{a}$ |
| $\displaystyle$ | $\displaystyle =$ | $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\left(\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{\mathbf{r}}\right)\cdot\left(r^{2}\sin\theta d\theta d\phi\hat{\mathbf{r}}\right)$ |
| $\displaystyle$ | $\displaystyle =$ | $\displaystyle \frac{q}{4\pi\epsilon_{0}}\int_{0}^{2\pi}\int_{0}^{\pi}\sin\theta d\theta d\phi$ |
| $\displaystyle$ | $\displaystyle =$ | $\displaystyle \frac{q}{\epsilon_{0}}$ |
That is, the flux due to a point charge depends only on the magnitude of the charge and not on the radius of the sphere that contains it. This makes intuitive sense, since if we imagine a point charge ‘emitting’ the electric field, then as long as we provide a surface that wraps up the charge completely, the flux through that surface will be the same regardless of the size of the surface. In fact, it shouldn’t depend on the shape of the surface either, so long as that surface completely encloses the charge. A more general derivation using an arbitrary closed surface is possible (using the solid angle subtended by the elemental area), but it adds nothing to the general idea.
From here, we can generalize the idea to a collection of point charges using the principle of superposition, and get, for a collection of ${n}$ point charges, all enclosed by the surface:
$\displaystyle \int\mathbf{E}\cdot d\mathbf{a}=\frac{1}{\epsilon_{0}}\sum_{i=1}^{n}q_{i}$
For a continuous charge distribution, where the charge density is ${\rho(\mathbf{r})}$, we get
$\displaystyle \int\mathbf{E}\cdot d\mathbf{a}=\frac{1}{\epsilon_{0}}\int\rho(\mathbf{r})d^{3}\mathbf{r}$
where it is important to note that the integral on the left is over the enclosing surface, while that on the right is over the volume enclosed by that surface. This is the integral form of Gauss’s law for electrostatics.
Using the divergence theorem, we can equate the charge density with the divergence of the electric field:
$\displaystyle \nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_{0}}$
This is the differential form of Gauss’s law. Both these forms are very powerful in solving various types of problems since they allow electric fields to be calculated, often without requiring complicated integrals.
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By , on Monday, 26 September 2011 at 15:44, under Electrodynamics, Physics. Tags: divergence theorem, Gauss's Law; electric flux. 15 Comments
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http://mathoverflow.net/questions/26402?sort=oldest
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## {transcendental numbers} \ {computable transcendental numbers}
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I know Chaitin's constant Ω is not computable (and therefore transcendental). Are there other specific, known noncomputable numbers? I am trying to understand what distinguishes a computable transcendental number, such as π, from a noncomputable transcendental number, such as Ω. Is there anything revealing that can be said about the set difference {transcendental numbers} \ {computable transcendental numbers}?
I ask this as a novice. I am re-visiting a wonderful book that sadly can no longer be updated by Victor Klee, in which he and Wagon pose this as an open problem: If an irrational number is real-time computable, is it then necessarily transcendent? [Problem 23]
Update (19Jun12). There is an illuminating discussion under the title "Why The Hartmanis-Stearns Conjecture Is Still Open" at the Lipton-Regan blog. The Hartmanis-Stearns Conjecture is the open problem mentioned above: If a number is real-time computable, it is either rational or transcendental. If true, this has what strikes me as a counterintuitive consequence: that algebraic irrationals like $\sqrt{3}$ are in some sense "more complicated" than transcendentals.
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1
I would expect any sensible definition of "computable" makes $\sqrt2$ a computable number... yet it is not trascendent. – Mariano Suárez-Alvarez May 30 2010 at 0:42
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What does real-time computable mean? – Pete L. Clark May 30 2010 at 1:52
7
Chaitin's Ω is not a specific number. It is only defined relative to a universal prefix-free machine, and there are many choices of such a machine with no single "correct" choice. Any sense of "specific" that applies to Ω also applies to the Turing jump of the empty set, 0'. – Carl Mummert May 30 2010 at 2:47
1
A number is real-time computable, if there is an algorithm that computes its digit sequence for which there is a constant c, such that for every n, the algorithm uses at most cn steps to compute the first n digits. – Halfdan Faber May 30 2010 at 5:08
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In the context of Turing machines, a real-time algorithm always inputs a symbol, then outputs a symbol within a constant c number of steps. – Halfdan Faber May 30 2010 at 5:15
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## 4 Answers
Note: Answer is pending update per attached comments.
The difference, stated informally, is that that the non-computable transcendentals in their k-base digit representation are entirely random and non compressible. A computable transcendental, such as e, can be represented by a finite algorithmic description, such as a series expansion, which is a form of compression. For the non-computable numbers no such shorter representation exist. Their shortest computational description is their own infinite digit sequence. You can read more about computational complexity here: http://en.wikipedia.org/wiki/Kolmogorov_complexity.
There is a wealth of similar numbers to the Ω class of numbers. In general it is "easy" to come up with new definitions for such numbers. These all belong to the countably infinite set of non-computable definable numbers.
To make matters worse, what is left are the non-definable (and therefore also non-computable) numbers. They are the numbers that cannot be described in any way what-so-ever, other than by just iterating through their infinite non-compressible digit sequence. The set of all non-definable numbers is uncountable.
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Haldfan, I give up with my attempts to understand the question. Maybe, not a motivation, but just the question itself. The only thing I do understand that the question has nothing to do with nt.number-theory. – Wadim Zudilin May 30 2010 at 7:05
yes, agreed; the tag 'number theory' is misleading. – Halfdan Faber May 30 2010 at 7:27
What is an "uncountable [real] number"? – Pete L. Clark May 30 2010 at 9:21
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The claims about non-definable numbers in this answer are problematic, as I explain in this MO answer: mathoverflow.net/questions/44102/… and also in my article: jdh.hamkins.org/…. The basic situation is that the notion of "definable" is simply not expressible, and if there are any models of ZFC at all, then there are models in which every real number is definable. – Joel David Hamkins Apr 18 2012 at 8:07
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I look forward to your update. Although a naive treatment of definability may be common, I think we can aspire here at mathoverflow to treat the topic with precision and correctness. The naive treatment is indeed incorrect, since using it one could argue like this: there are only countably many definable ordinals, but uncountably many ordinals; thus, there is a least ordinal $\alpha$ that is not definable. But this defines $\alpha$, a contradiction if this usage of "definability" is legitimate. – Joel David Hamkins Apr 20 2012 at 6:08
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The other standard example is to order the Turing machines and take the binary number with the nth decimal being 1 if the nth Turing machine stops. The computability of this number is (obviously) equivalent to the Halting problem.
Here computable means that the digits are literally computable by a Turing machine. Thus, Sqrt(2) is certainly computable. The book you're referring to defines a notion of "real-time computable" which also puts restrictions on how many steps it takes to compute the digits.
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I think the same sort of trick (sticking a 1 or 0 in each decimal place according to some rule) can be played with several variants of the "Turing machine trick."
Here's one of a somewhat different flavor. Choose an enumeration of the Diophantine equations (over $\mathbb{Z}$), and define a number with decimal expansion $0.a_1a_2a_3\ldots$ where
$$a_i=\begin{cases} 1&\text{ if the $i$-the Diophantine equation has a solution}\\ 0&\text{ if not.} \end{cases}$$
This is non-computable by the negative solution to Hilbert's 10th problem.
(Though to be fair, by that same solution, this is probably tantamount to permuting the digits in one of the Turing machine examples.)
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Let me consider your question: Are there other specific, known non-computable numbers?
There are indeed numerous specific, known definable real numbers that we know are not computable. Many of these real numbers arise when considering various notions of definability in various stronger-than-computable systems, such as the arithmetic hierarchy, the projective hierarchy and other hierarchies of definability and complexity, some of which I describe in my answer to I. J. Kennedy's question Are some numbers more irrational than others? In my answer to Paul Budnik's question Is there a well-defined subset of the integers that cannot be defined as a property of a recursive process or Turing machine?, I mention several specific numbers that are definable, but not computable. In this information-theoretic context, the difference between a real number and a set of natural numbers is often not so great, and every set of natural numbers (or sentences in a given language) naturally corresponds to a real number, whose digits indicate membership or non-membership in that set. Let me list several definable non-computable reals by their names:
• The real $0'$, pronounced "$0$-jump". This is just the halting problem, which you can think of as the real number whose $n^{th}$ digit is $1$ if the $n^{th}$ Turing machine program halts on trivial input, and was mentioned already in some of the other answers.
• Kleene's $O$.
• $0''$, $0'''$, the double jump, triple jump and so on, which relativizes the halting problem to an oracle.
• $x'$ for any definable $x$, we have the halting problem relative to Turing machines with oracle $x$, which is strictly harder than $x$ to compute.
• Tot, the set of programs computing total functions. This has complexity $\Pi^0_2$.
• Fin, the set of programs computing a finite function. This has complexity $\Sigma^0_2$.
• TA, or "true arithmetic", is the set of arithmetic sentences true in $\langle\mathbb{N},+,\cdot,0,1,\lt\rangle$. It is Turing equivalent to $0^{(\omega)}$, the $\omega^{th}$ jump.
• WO, the set of programs computing a well-ordered relation on $\mathbb{N}$. This has complexity complete $\Pi^1_1$, just beyond the hyperarithmetic hierarchy.
• Th(HC), the set of statements true for hereditarily countable sets. This is an analogue of TA for the projective hiearchy.
• One can define other specific real numbers, which are not computable, such as the real number whose $n^{th}$ binary digit records whether or not $2^{\aleph_n}=\aleph_{n+1}$. This real number is definable, but not necessarily computable. In fact, for any real number, there is a forcing extension of the universe in which this definition defines that number. So in this sense, any number you like can be made to be a specific definable number in some set-theoretic universe.
• $0^\sharp$, pronounced "$0$-sharp", is a real number whose existence is a kind of large cardinal assertion, equivalent to the existence of a proper class of order-indiscernible ordinals for the constructible hierarchy. The real $0^\sharp$ is the theory of these indiscernibles.
• One can iterate this with $0^{\sharp\sharp}$ and $x^\sharp$ for any $x$.
• $0^\dagger$, pronounced "$0$-dagger", is a similar theory for indiscernibles over a richer inner model $L[\mu]$, with a measurable cardinal. This is a specific real number whose existence has consistency strength greater than the existence of a measurable cardinal.
• There are other similar reals, such as $0$-hand-grenade and others, which carry a stronger large cardinal strength.
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http://mathoverflow.net/questions/7687?sort=oldest
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## Clifford algebra as an adjunction?
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Background
For definiteness (even though this is a categorical question!) let's agree that a vector space is a finite-dimensional real vector space and that an associative algebra is a finite-dimensional real unital associative algebra.
Let $V$ be a vector space with a nondegenerate symmetric bilinear form $B$ and let $Q(x) = B(x,x)$ be the associated quadratic form. Let's call the pair $(V,Q)$ a quadratic vector space.
Let $A$ be an associative algebra and let's say that a linear map $\phi:V \to A$ is Clifford if $$\phi(x)^2 = - Q(x) 1_A,$$ where $1_A$ is the unit in $A$.
One way to define the Clifford algebra associated to $(V,Q)$ is to say that it is universal for Clifford maps from $(V,Q)$. Categorically, one defines a category whose objects are pairs $(\phi,A)$ consisting of an associative algebra $A$ and a Clifford map $\phi: V \to A$ and whose arrows $$h:(\phi,A)\to (\phi',A')$$ are morphisms $h: A \to A'$ of associative algebras such that the obvious triangle commutes: $$h \circ \phi = \phi'.$$ Then the Clifford algebra of $(V,Q)$ is the universal initial object in this category. In other words, it is a pair $(i,Cl(V,Q))$ where $Cl(V,Q)$ is an associative algebra and $i:V \to Cl(V,Q)$ is a Clifford map, such that for every Clifford map $\phi:V \to A$, there is a unique morphism $$\Phi: Cl(V,Q) \to A$$ extending $\phi$; that is, such that $\Phi \circ i = \phi$.
(This is the usual definition one can find, say, in the nLab.)
Question
I would like to view the construction of the Clifford algebra as a functor from the category of quadratic vector spaces to the category of associative algebras. The universal property says that if $(V,Q)$ is a quadratic vector space and $A$ is an associative algebra, then there is a bijection of hom-sets
$$\mathrm{hom}_{\mathbf{Assoc}}(Cl(V,Q), A) \cong \mathrm{cl-hom}(V,A)$$
where the left-hand side are the associative algebra morphisms and the right-hand side are the Clifford morphisms.
My question is whether I can view $Cl$ as an adjoint functor in some way. In other words, is there some category $\mathbf{C}$ such that the right-side is $$\mathrm{hom}_{\mathbf{C}}((V,Q), F(A))$$ for some functor $F$ from associative algebras to $\mathbf{C}$. Naively I'd say $\mathbf{C}$ ought to be the category of quadratic vector spaces, but I cannot think of a suitable $F$.
I apologise if this question is a little vague. I'm not a very categorical person, but I'm preparing some notes for a graduate course on spin geometry next semester and the question arose in my mind.
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I was just wondering this a week ago! I'd also be very interested in the answer to this question. – Qiaochu Yuan Dec 3 2009 at 18:04
Are categorical questions not definite? :P – Mariano Suárez-Alvarez Dec 3 2009 at 18:14
I(and likely others) would appreciate it if you posted these notes for download. There are some references that I know of that address relationships between Clifford algebras and other branches of modern physics(Girard, Iordănescu), but I would like to see what perspective you are coming from on this. Thanks for the question though, it is interesting! – B. Bischof Dec 3 2009 at 22:17
Certainly. I put all my notes online as a matter of principle. I've only just started preparing them, though... so it may be a while. Thanks for your interest! – José Figueroa-O'Farrill Dec 4 2009 at 7:06
## 5 Answers
Disqualifier: this isn't a complete answer.
There's a basic "chalk and cheese" problem here. The "categories" that you are comparing are of two different types, although they do seem similar on the surface. On the one hand you have an honest algebraic category: that of associative algebras. But the other category (which, admittedly, is not precisely defined) is "vector spaces plus quadratic forms". This is not algebraic (over Set). There's no "free vector space with a non-degenerate quadratic form" and there'll (probably) be lots of other things that don't quite work in the way one would expect for algebraic categories. For example, as you require non-degeneracy, all morphisms have to be injective linear maps which severely limits them. You could add degenerate quadratic forms (which means, as AGR hints, that you regard exterior algebras as a sort of degenerate Clifford algebra - not a bad idea, though!) but this still doesn't get algebraicity: the problem is that the quadratic form goes out of the vector space, not into it, so isn't an "operation".
However, you may get some mileage if you work with pointed objects. I'm not sure of my terminology here, but I mean that we have a category $\mathcal{C}$ and some distinguished object $C_0$ and consider the category $(C,\eta,\epsilon)$ where $\eta : C_0 \to C$, $\epsilon : C \to C_0$ are such that $\epsilon \eta = I_{C_0}$. In Set, we take $C_0$ as a one-point set. In an algebraic category, we take $C_0$ as the free thing on one object. Then the corresponding pointed algebraic category is algebraic over the category of pointed sets (I think!).
The point (ha ha) of this is that in the category of pointed associative algebras one does have a "trace" map: $\operatorname{tr} : A \to \mathbb{R}$ given by $(a,b) \mapsto \epsilon(a \cdot b)$. Thus one should work in the category of pointed associative $\mathbb{Z}/2$-graded algebras whose trace map is graded symmetric.
In the category of pointed vector spaces, one can similarly define quadratic forms as operations. You need a binary operation $b : |V| \times |V| \to |V|$ (only these products are of pointed sets) and the identity $\eta \epsilon b = b$ to ensure that $b$ really lands up in the $\mathbb{R}$-component of $V$ (plus symmetry).
Whilst adding the pointed condition is non-trivial for algebras, it is effectively trivial for vector spaces since there's an obvious functor from vector spaces to pointed vector spaces, $V \mapsto V \oplus \mathbb{R}$ that is an equivalence of categories.
Assuming that all the $\imath$s can be crossed and all the $l$s dotted, the functor that you want is now the forgetful functor from pointed associative algebras to pointed quadratic vector spaces.
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I'm accepting this answer; although to be honest I'm still not close to understanding the "pointed" point of view. I'm learning category theory in earnest, so perhaps I can come back to this a little later. Thanks! – José Figueroa-O'Farrill Mar 3 2010 at 0:02
Gosh, I'd forgotten about this completely! Well done for going back through your questions and sorting them out. If, at some point, you want to work through the details then I'd like to see the workings (even help if that's allowed). You should feel free to do this on the nlab if you wanted! – Andrew Stacey Mar 3 2010 at 9:20
### 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.
If I understand the definitions correctly:
Let $C$ be the category of pairs (V,q) where V is a vector space on a fixed field and q is a quadratic form. A morphism $f: (V,q) \rightarrow (V',q')$ is a linear map $V \to V'$ preserving the quadratic form.
Let $D$ be the category of unital algebras over the field. Morphisms are linear maps preserving multiplication and identity.
We've got a forgetful functor $D \rightarrow C$ that maps an algebra V to the quadratic vector space $(V,q)$ where $q(x)=(x \cdot x) \cdot 1$. This functor has as left adjoint the Clifford algebra construction.
(I'm inexperienced, so this might be plain wrong. But surely an adjoint functor is hiding here.)
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q(x) isn't a quadratic form. But I think if e denotes the identity and e* denotes its dual then defining q(x) = e* x^2 works. – Qiaochu Yuan Dec 3 2009 at 18:11
Also, I'm not sure if it matters or not whether you want algebra homomorphisms to preserve the identity. – Qiaochu Yuan Dec 3 2009 at 18:19
Thanks, corrected. – sdcvvc Dec 3 2009 at 18:45
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"its dual"? What is the dual to the identity? – Theo Johnson-Freyd Dec 3 2009 at 19:39
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I am not sure if this is whole answer, but it seems to be in the right direction. First, $q(x) = - e^* x^2$ seems better, the way I have defined Clifford maps. My main concern is that $e^*$ is not canonically defined. Perhaps one has to add more structure to the algebras... – José Figueroa-O'Farrill Dec 3 2009 at 19:41
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UPDATE: the following argument is wrong, see the comments.
If $\mathcal{C}l$ admits a right adjoint then it preserves colimits, and coproducts in particular. Now, in your category of quadratic vector spaces, the coproduct of $(V, Q)$ and $(V', Q')$ is $(V \oplus V', Q \oplus Q')$; for associative algebras $A$ and $A'$, its coproduct is given by tensor product over $\mathbb{R}$. Hence, it is necessary that $$\mathcal{C}l(V \oplus V', Q \oplus Q') \cong \mathcal{C}l(V, Q) \otimes_{\mathbb{R}} \mathcal{C}l(V', Q')$$
Here's a counterexample: take $V = V' = \mathbb{R}$ with $Q = Q' = -1$. By the classification of Clifford algebras, we know that $\mathcal{C}l(\mathbb{R}, -1) \cong \mathbb{C}$ and $\mathcal{C}l(\mathbb{R}^2, \mathrm{diag}(-1,-1)) \cong \mathbb{H}$. It is now enough to observe that $$\mathbb{H} \not\cong \mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$$
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The coproduct in the category of associative algebras is not the tensor product: two maps $f:A\to C$ and $f:B\to C$ give a map $A\otimes B\to C$ only if the images of $f$ and of $g$ commute. – Mariano Suárez-Alvarez Dec 3 2009 at 18:42
Curiously, since Clifford algebras are $\mathbb{Z}_2$-graded, if you were to take the $\mathbb{Z}_2$-graded tensor product in your first displayed equation, then this would be a true statement. This perhaps suggestst that I have to consider the category of $\mathbb{Z}_2$-graded associative algebras. – José Figueroa-O'Farrill Dec 3 2009 at 18:53
My bad: Clifford algebras are Z_2-graded, and hence the coproduct is the Z_2-graded tensor product. Proposition 1.5 on page 11 of Lawson and Michelsohn's "Spin Geometry" asserts that Cl indeed preserves coproducts. – Alberto García-Raboso Dec 3 2009 at 18:54
In any case, the Clifford algebra of an orthogonal direct sum of quadratic spaces is isomorphic to the twisted tensor product of the corresponding tensor algebras: this is the "super" tensor product, the one which introduces signs by the parity graduation of the Clifford algebras. – Mariano Suárez-Alvarez Dec 3 2009 at 18:55
Yes, but there are useful ways of being wrong :) – José Figueroa-O'Farrill Dec 3 2009 at 19:44
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This answer builds on sdcvvc's answer and the comments below it, and in particular concerns the (non)existence of a canonical quadratic form $q$ (in sdcvvc's notation).
Let me denote by $\mathcal{Q}$ the category of quadratic real vector spaces (where the symmetric bilinear form is not necessarily nondegenerate), and by $\mathcal{A}$ some subcategory of the category $\mathcal{A}ss$ of finite-dimensional real unital associative algebras that contains the image of the Clifford functor $\mathcal{C}l: \mathcal{Q} \to \mathcal{A}ss$.
Notice that $\mathcal{Q}$ contains $\mathrm{\mathbf{Vect}}_\mathbb{R}$ as the full subcategory whose objects of the form $(V, 0)$, and that the restriction of the functor $\mathcal{C}l: \mathcal{Q} \to \mathcal{A}$ to this subcategory is the exterior algebra functor $V \mapsto \Lambda^{\ast}V$. Then,
$$\mathrm{Hom}_{\mathcal{A}}(\Lambda^\ast V, A) \cong \lbrace \phi: V \to A \; | \; \phi(v)^2 = 0 \rbrace$$
You can make $\Lambda^{\ast}(-)$ into a left adjoint by restricting $\mathcal{A}$ to be the category of $\mathbb{Z}_2$-graded supercommutative algebras (maybe you can take a bigger subcategory?). The right adjoint should then be the functor taking such an algebra to its odd-degree part considered as a vector space. This makes the Clifford condition $\phi(v)^2 = 0$ trivially true.
It is the latter observation the one that allows us to cook up such an $\mathcal{A}$. However, in the general case the Clifford condition does involve the quadratic form on the vector space that is the domain, and so it doesn't seem possible to me that we could do something like the above universally.
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I'm hoping for a second opinion on this question. The same question occurred to me, and google led me to this thread. At first glance, the consensus answer here (there is no right-adjoint to $Cl$) seems a plausibly argued. But after some thought, I'm not convinced.
We know that a universal construction, if it exists for every object in the source category, always gives an adjunction between categories.
An object satisfying the universal property for a Clifford algebra can be explicitly constructed from any vector space with quadratic form as a quotient of the tensor algebra. So an object satisfying the universal property always exists, therefore it is a left-adjoint. And what should the right-adjoint functor to the Clifford functor? Why nothing other than the underlying map from associative algebras to quadratic spaces, with quadratic form $q(x)=x^2$. This is the only possible quadratic form on the underlying vector spaces which will make the stipulation in the universal construction about the linear maps into morphisms in the category of quadratic vector spaces.
I should conclude that the right-adjoint of $Cl$ is a forgetful functor $k\text{-Alg}\to k\text{-Quad}$ which takes an associative algebra and forgets multiplication but remembers how to square vectors. The unit of this adjunction is the Clifford algebra structure map, and the counit is the map from the Clifford algebra on the quadratic vector space underlying any algebra $A$ to $A$ which takes $a_1\cdot a_2\mapsto a_1a_2$.
This is of course exactly the unaccepted answer that sdcvvc gives above, though without much detail. Qiaochu Yuan says that the claimed quadratic form $q(x)=x^2$ on the underlying vector space of an associative algebra is not actually quadratic. I cannot see why not. Why is sdcvvc's answer incorrect?
Alberto García-Raboso gives an answer as well, where in the discussion it is settled that $Cl$ preserves finite coproducts. If we can also show that it preserves cokernels then we know that it must have a right-adjoint, by Freyd adjoint functor theorem, right?
And have I misunderstood the relationships between universal morphisms and adjunctions? Is it not the case that we can simply read off the adjoint functor out of the universal property?
And do we really need to consider, as Andrew Stacy suggests, some kind of pointed vector spaces? If so, why?
I wanted to post my questions as comments, not an answer, but I guess I don't have enough rep. Please forgive me.
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$q(x) = x^2$ doesn't take values in the underlying field! – Qiaochu Yuan Jul 3 at 5:44
Right, duh. Thank you, Qiaochu, for explaining to the slow kid. So how do we reconcile the fact that the Clifford functor isn't a left-adjoint with the fact that every universal property determines an adjunction? I guess we conclude that the universal property which characterizes the Clifford algebra doesn't actually meet the technical definition of a universal morphism, in the sense that there is no functor for which the Clifford construction is initial in the slice over it? Are there other examples of universal properties which are not universal morphisms like this? – Joe Hannon Jul 3 at 14:32
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http://mathhelpforum.com/discrete-math/73672-binomial-expansion.html
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# Thread:
1. ## Binomial expansion
In each of the following expansions , find the term as stated .
(1) $(1+x)^{10}$ , 5th term
(2) $(2a+b)^{12}$ , 10th term
(3) $(x-\frac{1}{x})^6$ , constant term
2. Originally Posted by thereddevils
In each of the following expansions , find the term as stated .
(1) $(1+x)^{10}$ , 5th term
(2) $(2a+b)^{12}$ , 10th term
(3) $(x-\frac{1}{x})^6$ , constant term
Use the formula:
$(a+b)^n=\sum_{k=0}^n {n \choose k} a^{n-k} \cdot b^k$
Keep in mind that you start counting by zero! Thus the 5th term correspond with k = 4.
Therefore the 5th term of $(1+x)^{10} \rightarrow {10\choose 4}1^4\cdot x^6 = 210x^6<br />$
For the last example: A constant term only occurs if the exponents of x and $\frac1x$ are equal, thus you have to calculate the 4th term with k = 3. Since k is an odd number you get as the constant summand -20.
3. Originally Posted by earboth
Therefore the 5th term of $(1+x)^{10} \rightarrow {10\choose 4}1^4\cdot x^6 = 210x^6<br />$
I guess there is a slight typo , should be 210x^4
4. Originally Posted by mathaddict
I guess there is a slight typo , should be 210x^4
With $(1+x)^{10}$ the 5th term is:
${10\choose 4}1^4\cdot x^6 = 210x^6$ and the 7th term is ${10\choose 6}1^6\cdot x^4 = 210x^4<br />$
Mostly the binoms are given as $(x+1)^{10}$ and then you would have been right.
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http://stats.stackexchange.com/questions/32701/comparing-observed-data-to-simulated-data
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# Comparing observed data to simulated data?
I have a set of observed values (shown with black dots in the figure) that I would like to compare to some simulated data (100 simulated datasets shown as box plots with quartiles, extremes (excluding outliers) as whiskers, and outliers as white dots).
The observed values are outside of the 95% confidence intervals of the simulated dataset so it seems pretty obvious that there is a significant difference between the observed and the simulated; but in some cases, the difference might not be so obvious. Which statistical test (preferably in R) can I do on this data to get a P-value for the significant difference between the observed and the simulated sets?
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It is unusual for boxplots to show confidence intervals. Don't these show quartiles, fences, and any outliers? It also seems worthwhile asking why you would want a p-value when the difference is so blatantly obvious: clearly the simulation does not reproduce the observed values. – whuber♦ Jul 20 '12 at 15:34
Hi whuber, Sorry I am mistaken, the whiskers represent the maximum and minimum (excluding outliers). The reason I want to do this, is that in some of my datasets it will be a much closer fit, so then a p-value become very useful. – blJOg Jul 20 '12 at 16:09
Could you comment on the nature of the simulation? In principle, it is merely drawing data from an underlying distribution. We could probably bypass the simulation altogether by knowing the distribution. This leads to very simple tests, such as the Chi-squared--and obviates the need to run the simulation, too! – whuber♦ Jul 20 '12 at 18:13
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So I have observed a number of mutations in 48 horses and determined how many mutations are shared between pairs of horses. 800 pairs of horses have no mutations in common, approx 200 pairs of horses share 1 mutation etc... For each observed mutation, the mutation is randomly assigned to the same number of horses it was found in. I then determine based on this random assignment, how many pairs of horses share 0...8 mutations if mutations were spread randomly in the horse population. – blJOg Jul 22 '12 at 9:51
## 3 Answers
Building on the comment by @whuber, a straightforward test that would deliver reasonable results would be a Chi square test that compares your observed counts in each bin of "number of shared mutations" with the expected counts in that bin. This is particularly the obvious way to do it if you know how the data were simulated and hence you can directly use the expected counts in the bin; but even if you don't it is a reasonable pragmatic approach.
In the current case the expected number with zero shared mutations is zero, which will make a Chi square statistic infinite (because its calculation involves dividing by the expected number in each bin), and give a p-value of zero, which is to be expected. Basically this reflects that it is literally impossible for the observed data to have been generated from a distribution that gives zero probability to zero mutations.
````> horses <- data.frame(
+ expected = c(0,270,410,230,80,10),
+ observed = c(800,230,40,10,5,1),
+ mutations =c(0,1,2,3,4,"lots")
+ )
>
> horses$expected.scaled <- horses$expected *
+ sum(horses$observed) / sum(horses$expected)
>
> horses
expected observed mutations expected.scaled
1 0 800 0 0.00
2 270 230 1 293.22
3 410 40 2 445.26
4 230 10 3 249.78
5 80 5 4 86.88
6 10 1 lots 10.86
>
> X <- with(horses,
+ sum((observed-expected.scaled)^2/expected.scaled)
+ )
> X
[1] Inf
> 1-pchisq(X,5)
[1] 0
````
-
Consider the frequency distribution from each simulation run (and the observed data) as a 9-dimensional vector with $j$th element being the number of observations with $j$ shared mutations. Now you have an observed vector $z$, and a set of simulated vectors $x_i$, $i=1,\ldots,100$. Your goal is to determine if the vector $z$ could have come from the distribution defined by $x_i$'s.
A possible approach is to look at deviations from the mean of the simulations. You could use Euclidean distance, or Euclidean distance on the square-roots (which has variance-stabilizing properties), or Mahalanobis distance, or anything else you come up with. Now your simulated data will give you a random sample of 100 of such distances $||x_i-\bar{x}||$, which you can compare to the observed distance $||z-\bar{x}||$ to get a p-value.
If you are braver, you can even make multivariate joint normality assumptions, fit a multivariate normal to the simulations (i.e. get its mean and variance matrix) and calculate a p-value based on that. This would be the parametric equivalent of using the Mahalanobis distance.
-
One way to test and get a p-value by comparing actual data to simulated data (though in a way very different from what you show above) is discussed in:
```` Buja, A., Cook, D. Hofmann, H., Lawrence, M. Lee, E.-K., Swayne,
D.F and Wickham, H. (2009) Statistical Inference for exploratory
data analysis and model diagnostics Phil. Trans. R. Soc. A 2009
367, 4361-4383 doi: 10.1098/rsta.2009.0120
````
And is implemented in the `vis.test` function in the `TeachingDemos` package for R.
One approach more along the lines that you are showing would be to compute a p-value for each x-value above as the proportion of simulated points that are as extreme or more extreme than the observed point. Then use meta-analysis techniques to combine the p-values (9 in the case of your plot above), one option is that the negative log of a p-value (under the null) follows a chi-square distribution with 2 degrees of freedom and you can add chi-square statistics (though the degrees of freedom should probably be discounted since I am guessing that the parts above are not independent). Or if you can define "more extreme" in terms of the whole set of 9 points from each simulation then you could just do the simple p-value instead of combining 9.
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default
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http://mathoverflow.net/questions/49979?sort=votes
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## Statistics of a simple Markov chain
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Imagine a two-state Markov chain which hops between the states $\pm 1$ with probability $p<1/2$, so that the autocorrelation function after $k$ steps is
$\rho_k = (2p-1)^k$
If I take an exponential moving average of this series with weighting parameter $\lambda$, what does the distribution of values of the new series look like?
Probably the answer is "gaussian, centered on 0" but what is the variance? Is there a known result that makes this computation trivial?
-
All you need to find the variance is to be able to compute the expectation of the product $X_iX_j$ and that is what the autocorrelation function is for (which, if $p$ is the probability of jumping, not staying, is $(1-2p)^k$, by the way). Expanding the product of the sums into the sum of the products is trivial and so is the summation of the resulting series. As it has been mentioned by maxdev, this will be approximately Gaussian for fixed $p$ only if $\lambda$ is close to $1$. – fedja Dec 20 2010 at 21:51
## 3 Answers
The answer is certainly not "Gaussian". What you describe is often called Bernoulli convolutions and, even in the independent case (in your setting, $p=1/2$), the limiting object is quite complicated and interesting since it involves some deep number theoretic properties of $\lambda$.
To begin with, let $(X_n)_{n\in \mathbb{Z}}$ denote the $\pm1$-valued Markov chain with probability $p$ of switching states. Let $(Y_n)_{n\in \mathbb{Z}}$ denote the exponential moving average of parameter $\lambda$ with $0<\lambda < 1$ you are interested in, that is, $$Y_n=\sum_{k=0}^{+\infty}\lambda(1-\lambda)^kX_{n-k}.$$ The Markov chain is centered and has correlation $E(X_nX_{n+k})=(1-2p)^k$ for every integers $n$ and $k\ge0$. (Hence you should check your formula.) From there, one sees that $Y_n$ is centered and one can compute its variance. If I am not mistaken, one finds something like $$E(Y_n^2)=\frac{1-2p(1-\lambda)/(2-\lambda)}{1+2p(1-\lambda)/\lambda}.$$ The stationary distribution of the moving average is a different story. It is often best described as a measure-valued fixed point problem, as follows. First, $Y_n=X_nY_+$ where $Y_+$ and $X_n$ are independent, and $Y_+$ is distributed like $Y_n$ conditioned on $[X_n=+1]$. Second, $Y_+$ is distributed like $\lambda+(1-\lambda)ZY_+$, where $Z=\pm1$, $P(Z=+1)=1-p$, $P(Z=-1)=p$, and $Z$ independent of $Y_+$.
This indirect description of the stationary distribution is often the most useful tool to get some information on it.
As regards your original "Gaussian" hint, note that conditioning on $(X_{n-k})_{0\le k\le N-1}$ for a given $N$ yields that $Y_n$ is in one of $2^N$ intervals of length $2(1-\lambda)^N$. If $2(1-\lambda)<1$, this simple remark shows that the distribution of $Y_n$ is concentrated on a Cantor set of Lebesgue measure zero (hence this probability distribution does not even have a density with respect to the Lebesgue measure, and it has no atom either). The argument uses only the fact that each $X_n=\pm1$ almost surely and not the structure of the process $(X_n)_n$.
Another easy case is when $\lambda=1/2$. Then, if $(X_n)_n$ is in fact independent ($p=1/2$), one recognises the usual binary expansion of a random number hence $Y_n$ is uniformly distributed on $[-1,1]$, but for every other value of $p$, the distribution of $Y_n$ is concentrated on a subset of $[-1,1]$ of Lebesgue measure zero.
For much more on the stationary distributions of moving averages like the ones which interests you, some starting points could be the paper Sixty years of Bernoulli convolutions by Peres, Schlag and Solomyak, and the book Some random series of functions by Kahane.
-
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The measure $\mu_\lambda$ on an interval whose distribution is given by the random variable $$\sum_{n=1}^\infty \epsilon_n\lambda^n,$$ where the $\epsilon_n$ assume the values 0 and 1 (or $\pm1$) independently with probabilities $(p,1-p)$ is called a biased Bernoulli convolution.
If one assumes $\lambda\in(0,1/2)$, then $\mu_\lambda$ is supported by a Cantor set and is consequently singular. If $\lambda=1/2$, then it is a well known singular measure on $[0,1]$. (It is invariant and ergodic under the doubling map $\tau x=2x\ \bmod 1$ and so is the Lebesgue measure.)
The most interesting case is $\lambda\in(1/2,1)$. Here if $p\in[1/3,2/3]$, then for a.e. $\lambda$ the measure $\mu_\lambda$ is known to be equivalent to the Lebesgue measure. If $\lambda^{-1}$ is a Pisot number, then it is singular. (Which was essentially proved by Erdős in 1939.)This is almost all that is known about these measures.
For more detail see, e.g.,
B. Solomyak, Notes on Bernoulli convolutions
-
By exponential moving average you mean something like $b_k = C \sum_{i=k}^{\infty} {\frac{a_i}{\lambda^i}}$ ? If $\lambda$ is small enough then this would depend hugely on $a_k$ and so it wouldn't be gaussian right?
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http://mathhelpforum.com/advanced-algebra/118893-jordan-form-help.html
|
# Thread:
1. ## Jordan form help...
I stumbled across this question, and I'm not quite sure how to go about and find the Jordan form of this kind of question. Any help on how to get started?
Let
N be an element of L(P4(C)), defined by N(p(z)) = zp'''(z)+p'(z).Show that N is nilpotent and find the Jordan form of N.
Any help is greatly appreciated!
2. I am also curious as to the solution of this problem
3. Originally Posted by Majialin
I stumbled across this question, and I'm not quite sure how to go about and find the Jordan form of this kind of question. Any help on how to get started?
Let
N be an element of L(P4(C)), defined by N(p(z)) = zp'''(z)+p'(z).Show that N is nilpotent and find the Jordan form of N.
Any help is greatly appreciated!
Find the matricial representation of N wrt the basis $\{1,z,z^2,z^3,z^4\}$ of $P_4(\mathbb{C})[z]$ . From here it follows at once that N is nilpotent, and check that $N^5=0$ but $N^4\ne 0$, so that its Jordan Form is...
Tonio
|
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http://mathhelpforum.com/math-topics/162723-physics-multiple-choices-question.html
|
# Thread:
1. ## Physics ( Multiple choices question)
How do i find which force creates the largest moment about the pivot?
What is the Pole on the point of the solenoid with an arrow pointing towards it drawn by me as induced by the other circuit?
Any help is appreciated. I am looking for the reasoning behind not only the answer!! thx
2. The torque is $rF\sin\theta$ where r is the distance from the hinge to P, F is the magnitude of the force and $\theta$ is the angle between the beam and the force.
In the second problem, you know the initial air volume, pressure (atmospheric plus the excess mercury) and temperature. Using the Ideal gas law, you can find the amount of air. Then, using the new volume, you can find the new pressure.
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http://physics.stackexchange.com/questions/13832/integrals-of-motion?answertab=votes
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# Integrals of Motion
Landau writes in chapter 2 of his Mechanics book "The number of independent integrals of motion for a closed mechanical system with s degrees of freedom is 2s-1". Then he goes on to show how -- "Since the equations of motion for a closed system do not involve time explicitly, the choice of the origin of time is entirely arbitrary, and one of the arbitrary constants in the solution of the equations can always be taken as an additive constant t0 in time. Eliminating t + t0 from the 2s functions, we can express the 2s-1 arbitrary constants as functions of q and q’ (generalized co-ordinates and velocities) and these functions will be the integrals of the motions." Could someone elaborate?
-
1
Elaborate on what? The dynamics is second-order so there will always be 2N parameters (I don't like that Landau calls them integrals of motion since they are not what is understood by that term in modern physics, i.e. dynamically conserved quantities) for the system with N degrees of freedom. Picking initial time eliminates one of the parameters and leaves you with 2N -1. – Marek Aug 22 '11 at 5:40
2
Why people often say "Landau writes", "Landau calls"? It is Landau and Lifshitz who prepared this physics course. Moreover, there is no single sentence that Landau wrote in his famous course. All "paperwork" was due to Lifshitz, Landau was the inspirer, the scientific adviser and the editor of the course. Sometimes people even joking like: "In the physics course by L&L there is no single word of Landau and no single thought of Lifshitz" (c) – Physicsworks Aug 22 '11 at 9:21
@Physicsworks: so in view of Landau's supposedly horrible case of writer's block in his later life, should we say "Landau calls" and "Lifshitz writes"? `:-)` – Willie Wong Aug 22 '11 at 16:27
@Willie Wong: yeah :) BTW: Since early 30's Lifshtiz even wrote scientific papers for Landau (I heard this from one of Landau's coworkers). – Physicsworks Aug 22 '11 at 18:19
@Marek: I am an beginning undergraduate so 'naturally' am not familiar with the mathematical formalism (or maybe it is just because am a dumb ass). Whatever be the reason, I am thinking of the 2n parameters as the n co-ordinates and n velocities(varying in time). I agree with the fact that the origin of time can be chosen arbitrarily. However the following argument escapes me. I have tried to made sense of it, but have little belief in my ideas. So, HELP ME OUT!!! – Sourav Aug 23 '11 at 1:04
## 1 Answer
I think the source of your confusion is mathematics, not physics. It is important here (and L&L did mention this) that the system of differential equations is autonomous. If this is the case, than along with solution $q_i=q_i(t,C_1,\dots,C_{2s}),$ it has a solution $q_i=q_i(t-t_0,C_1,\dots,C_{2s}).$ Because the former is the general solution, the later must reduced to it, that is, it should be $$q_i(t-t_0,C_1,\dots,C_{2s})=q_i(t,C_1'(C_1,\dots,C_{2s},t_0),\dots,C_{2s}'(C_1,\dots,C_{2s},t_0)).$$ Putting $C_{2s}=0$, yields
$$q_i(t-t_0,C_1,\dots,C_{2s-1})=q_i(t,C_1'(C_1,\dots,C_{2s-1},t_0),\dots,C_{2s}'(C_1,\dots,C_{2s-1},t_0)).$$
For example, in the case of a free 1D motion $x=C_1t+C_2$. Making the shift in time one has: $$x=C_1t+(C_2-C_1t_0)$$ (expression in brackets is $C_2'$). And letting $C_2=0$ we find $$x=C_1(t-t_0).$$
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http://physics.stackexchange.com/questions/tagged/representation-theory
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# Tagged Questions
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This is a question that has been posted at many different forums, I thought maybe someone here would have a better or more conceptual answer than I have seen before: Why do physicists care about ...
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http://mathhelpforum.com/algebra/56641-got-class-30-mins-3-problems-i-dont-know-how-solve.html
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Thread:
1. Got class in 30 mins, 3 problems I dont know how to solve
These are incredibly difficult
2) A fourth degree polynomial f(x) with real coefficients and leading coefficient of 1 has zeroes of -1, 2,1 -i. Write the polynomial as a product of linear and quadratic factors with real coefficients that are irreducible over R (this R represents the real number like I think)
2) Find values a,b, and c for exponential function f(x) = cb^-x + a. Given that the horizontal asymptote is y = 72. Y intercept is 425 and point P (1,248.5) lies on the graph.
I did y=mx + b and got M=176.5 but what about the rest (c,b,a,x)?
3) Determine the domain and range of f^1 for the function f(x) = (4x+5)/(3x-8), without actually finding f^1
2. Originally Posted by mwok
These are incredibly difficult
2) A fourth degree polynomial f(x) with real coefficients and leading coefficient of 1 has zeroes of -1, 2,1 -i. Write the polynomial as a product of linear and quadratic factors with real coefficients that are irreducible over R (this R represents the real number like I think)
Mr F says: If the coefficients are real then 1 + i is also a root. Then the polynomial can be written f(x) = (x + 1)(x - 2)(x - 1 + i)(x - 1 - i). The last two factors expand to give (x - 1)^2 + 1 = ......
2) Find values a,b, and c for exponential function f(x) = cb^-x + a. Given that the horizontal asymptote is y = 72. Y intercept is 425 and point P (1,248.5) lies on the graph.
Mr F says: You should know that according to the model the horizontal asymptote is y = a. Therefore a = 72.
Substitute (0, 425) and (1, 248.5) into ${\color{red}f(x) = c b^{-x} + 72}$ to get two equations in c and b. Solve these two equations simultaneously.
I did y=mx + b and got M=176.5 but what about the rest (c,b,a,x)?
3) Determine the domain and range of f^1 for the function f(x) = (4x+5)/(3x-8), without actually finding f^1
Mr F says: dom f = ran f^-1 and ran f = dom f^-1. Note also that ${\color{red}f(x) = \frac{47}{3(3x-8)} + \frac{4}{3}}$
..
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http://math.stackexchange.com/questions/13838/what-kind-of-vector-space-does-the-set-of-complex-real-solutions-of-the-follow/13858
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# What kind of vector space does the set of (complex/real) solutions of the following differential equation form?
I'm doing a multiple choice exercise and I'm stuck at a question, where there are two possible answers (may be both or none correct).
Consider the differential equation
$$y''+y'+y=0$$
for $y(x)$. Which of the following statements are true?
• The set of complex solutions forms a two-dimensional complex vector space.
• The set of real solutions forms a two-dimensional real vector space.
I'm really not comfortable with vector spaces and the differentiation of real and complex vector space makes me even more unconfident. I solved the differential equation with the ansatz $y(x) = e^{\lambda x}$ and got to $\lambda^2 + \lambda + 1 = 0$, an equation which does not possess any real solutions. However, when I proceeded, I figured that I could rewrite the exponential function (with complex exponent) with the help of the sine and the cosine, so in the end I got a general solution with no complex coefficients.
What kind of vector space is this?
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Hint: How many parameters do you need to describe the solutions in each case? Beware, in the complex case, we're talking about complex parameters and in the real case, we're talking about real parameters. – Raskolnikov Dec 10 '10 at 19:05
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@Huy: Those constants are the parameters. Your real solutions are all linear combinations of $e^{-x/2}\sin(\sqrt{3}x/2)$ and $e^{-x/2}\cos(\sqrt{3}x/2)$, so the dimension is at most $2$ (over $\mathbb{R}$); since the two are actually linearly independent... In the complex case, you have a similar situation but with your solutions being (complex)-linear combinations of $e^{(-1+\sqrt{3}i)x/2}$ and $e^{(-1-\sqrt{3}i)x/2}$. – Arturo Magidin Dec 10 '10 at 19:25
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@Huy: Surely you can figure it out and be confident in your answer... If $ae^{(-1+\sqrt{3}i)x/2} + be^{(-1-\sqrt{3}i)x/2} = 0$ (the zero function), must $a=b=0$? If so, yes, they are linearly independent. If not, then that gives you a dependence. – Arturo Magidin Dec 10 '10 at 20:15
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@Huy: Of course it has to be the zero function (notice I even said so). We are in a vector space of functions, so the zero vector is the zero function. We are checking linear independence of the functions. That means that for every $x$ you must get $0$ as the result, because we are assuming that the expression you have equals the zero vector/zero function. – Arturo Magidin Dec 10 '10 at 20:23
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An important point that: the zero vector $\neq$ zero number. That's why I like the use of the notation $\vec{0}$ in this context. It makes things immediately clearer. – Raskolnikov Dec 10 '10 at 20:33
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## 2 Answers
As Huy says, both statements are true.
Well, if you are a hammer, all the problems seem nails to you, but since this is a linear differential equation, Laplace transform again gives us the general solution in both, real and complex, cases.
$${\cal L}[y''] + {\cal L}[y'] + {\cal L}[y] = 0 \ \Longleftrightarrow \ (s^2 F(s) -sy(0) -y'(0) ) + (sF(s) -y(0)) + F(s) = 0 \ ,$$
where $F(s) = {\cal L}[y]$. Now, put $a= y(0)$ and $b= y'(0)$ and solve this equation for $F(s)$:
$$F(s) = \dfrac{as + (a+b)}{s^2+s+1} \ .$$
Since the polynomial $s^2+s+1$ has no real roots, the distinction between the real and complex cases arrives at this point.
In the complex case, you write
$$s^2 + s + 1 = (s-\alpha)(s-\beta) \ ,$$
where $\alpha = \dfrac{-1+i\sqrt{3}}{2}$ and $\beta = \dfrac{-1-i\sqrt{3}}{2}$ are the roots of your characteristic polynomial. Hence the solution of your differential equation is
$$y(t) = A{\cal L}^{-1} \left[\dfrac{1}{s-\alpha} \right] + B {\cal L}^{-1} \left[\dfrac{1}{s-\beta} \right] = A e^{\alpha t} + B e^{\beta t} \ ,$$
where $A$ and $B$ are complex numbers, depending on $a , b$, which again I let you the pleasure to compute :-) .
Anyway, the fact is that the solution of your differential equation is the $\mathbb{C}$-linear span of two linearly independent complex functions
$$e^{\alpha t} \qquad \text{and} \qquad e^{\beta t} \ .$$
Thus a $\mathbb{C}$-vector space of dimension two.
In the real case, you write
$$s^2 + s + 1 = \left(s + \frac{1}{2} \right)^2 + \frac{3}{4}$$
and this time, the solution set of your differential equation is
$$y(t) = A {\cal L}^{-1} \left[ \frac{s + \frac{1}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} } \right] + B {\cal L}^{-1} \left[ \frac{ \frac{\sqrt{3}}{2}}{ \left( s+ \frac{1}{2} \right)^2+ \frac{3}{4} } \right] = A e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2}t\right) + B e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2}t\right) \ ,$$
where $A$ and $B$ are real numbers, depending on $a,b$, which... Anyway again: the solution set is the $\mathbb{R}$-linear span of two linearly independent real functions
$$e^{-\frac{t}{2}}\cos \left( \frac{\sqrt{3}}{2} t \right) \qquad \text{and} \qquad e^{-\frac{t}{2}}\sin \left( \frac{\sqrt{3}}{2} t \right) \ .$$
Thus, a real vector space of dimension two.
Alternatively, you could read what Wikipedia says about linear differential equations.
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Both statements are true.
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You could write up your answer. Then you may even get some feedback on cleaning it up. – Arturo Magidin Dec 11 '10 at 7:09
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http://crypto.stackexchange.com/questions/5553/bent-combining-functions/5554
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# Bent Combining Functions
Bent functions are boolean functions with maximum nonlinearity and are widely studied for their potential applications in cryptography. Lets say you wanted to use a simple two input AND gate as a nonlinear combiner for the outputs of two Linear Feedback Shift Registers . Because a two input AND gate is a bent function, is it technically correct to call the output sequence of this combination generator a bent sequence ( I need to know this for a paper I am writing).Thank you for commenting.
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## 2 Answers
No. As far as I can tell, this does not appear to produce a bent sequence. (Or, if it does, that fact requires proof/justification.)
According to Wikipedia, a "bent sequence" is a sequence of the form $(-1)^{f(x_0)}, \dots, (-1)^{f(x_{2^n-1})}$ where $x_0,\dots,x_{2^n-1}$ is the sequence of $n$-bit values in lexicographic order and $f$ is some bent function. That's different from what you are mentioning. If what you're mentioning happens to fit this definition, that fact is far from obvious.
I don't know if this is the same definition of "bent sequence" you are using. If it isn't, I suggest editing your question to include the definition you are using.
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@DW: nice answer but it does matter what I call the sequence if I am writing a paper that will be seen by experts . I want to be technically correct. Maybe there is no name designation for the sequence I described, why don't you invent a name for it, I will cite you in the paper ;-) – William Hird Dec 3 '12 at 9:26
@WilliamHird, OK, you are right! It does make sense to use standard terminology. I withdraw my comment about names. (A suggestion for future questions: maybe including the definition of the term you are asking about in the question would help others answer?) – D.W. Dec 3 '12 at 19:05
@DW: You are right , I did not define what I meant by a bent sequence , I incorrectly assumed that a bent boolean function like the AND function when used as a combining function for psuedorandom sequences will produce a bent sequence. A novice error to be sure. – William Hird Dec 4 '12 at 7:56
Bent functions can be obtained from linear feedback shift registers but not by the method that you propose. In fact, the method you propose (using an AND gate to combine the output of two linear feedback shift registers) will not give you a bent function.
Bent functions (though not by that name) were studied for use as signature sequences in direct-sequence spread-spectrum communications. In the literature on this topic, such sequences are called the small set of Kasami sequences. The sequences are of length (or period) $2^{2m}-1$ and can be obtained as the (XOR) sum of the outputs of two maximal-length linear feedback shift registers (LFSRs) of lengths $2m$ and $m$ respectively. The feedback polynomials of the LFSRs thus are primitive polynomials of degrees $2m$ and $m$ respectively, and they produce maximal-length LFSR sequences ($m$-sequences) of periods $2^{2m-1}$ and $2^m-1$ respectively. Thus, $2^m+1$ copies of one period of the shorter sequence get added into one period of the longer sequence, and the result is of period $2^{2m}-1$. A total of $2^m-1$ distinct sequences can generated by using a fixed nonzero initial loading of the longer LFSR and $2^m-1$ nonzero initial loadings of the shorter LFSR. Note that one cannot choose both feedback polynomials arbitrarily. If the longer polynomial is chosen as a specific primitive polynomial of degree $2m$, then only one primitive polynomial of degree $m$ will give you a small set of Kasami sequences. If the shorter polynomial is chosen first, then the longer one must be chosen from a subset of the set of all primitive polynomial of degree $2m$. For more details, see the paper D. V. Sarwate and M. B. Pursley, "Cross-correlation properties of pseudorandom and related sets of sequences," Proc. IEEE, May 1980.
The set of sequences thus generated are of period or length $2^{2m-1}$ whereas the description of bent functions says they are of length $2^{2m}$. To relate this to Wikipedia's version, consider the Boolean function $$f(\mathbf x) = f(x_1, x_2, \ldots, x_2m) = x_1x_2 \oplus x_3x_4 \oplus \cdots \oplus x_{2m-1}x_{2m}.$$ Wikipedia'a definition says that the bent function is the sequence of values of $(-1)^{f(\mathbf x)}$ as we cycle through the $2^{2m}$ values of $\mathbf x$ in lexicographic order. A sequence from the small set of Kasami sequences cycles through all the nonzero values of $\mathbf x$ in a different order, usually in what might be called shift-register order (successive contents of a length-$2m$ maximal-length LFSR) or Galois-field order (polynomial representation of successive powers of a primitive element of $\mathbb F_{2^{2m}}$. So, if you need the symbols in Wikipedia order, you are going to have to do a permutation of the $2^{2m}-1$ bits from the output of the XOR gate that adds the two seqeuences and prepend a $+1$ to get the missing $(-1)^{f(\mathbf 0)}$.
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I'm not sure how your answer relates to my question. The AND function for two variables is bent according to Claude Carlet(he should know!) so if you use it to combine two m-sequence LFRS"s the output is a nonlinear representation of the input. Looking at the truth table for the AND function we see that three out of four lines of the table resolve to 0. So an attacker does not know if the 0 was caused by inputs {0,0} {0,1} or {1,0}. Of course the output will not be balanced but for my circuit I don't care about balance . Sorry if mis-named this sequence a "bent sequence". – William Hird Dec 3 '12 at 16:47
You propose taking the AND of two LFSR outputs which is not a bent function in this particular context (where the number of variables is the LFSR length $n$, usually $n \gg 2$), and I doubt very much that Claude would say the AND function is a bent function of $n$ variables. I told you how to get a bent function (of $n = 2m$ variables) from LFSRs with a total of $3n/2$ stages instead of your proposed idea using $2n$ stages. Incidentally, the Kasami sequences that you get this way are slightly imbalanced, but you say you don't care so that is OK. – Dilip Sarwate Dec 3 '12 at 18:13
How does what Claude said contradict what I said, that the AND function is not a bent function of $n$ variables if $n\gg 2$? You need not snidely and suggestively put Claude's name in quotation marks. For the record, I am an Associate Editor of the journal Cryptography and Communications - Discrete Structures, Boolean Functions and Sequences of which Claude is the Editor, and do know him quite well. I am flagging your comment. – Dilip Sarwate Dec 4 '12 at 11:01
I meant no disrespect to Mr. Carlet, I was suggesting that I have no way to verify that any emails I have received are actually from the real Mr. Carlet. You have misunderstood my comment. – William Hird Dec 4 '12 at 12:47
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http://mathoverflow.net/revisions/101062/list
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## Return to Answer
1 [made Community Wiki]
It is very common in set theory to prove that a particular model or structure is well-founded by mapping it into a fixed well-founded structure. The point is that if $j:\langle M,{\in^M}\rangle\to \langle N,{\in}\rangle$ is $\in$-preserving, then any instance of ill-foundedness in $M$ would carrry over to $N$, which has none; and so $M$ is well-founded. This method is often applied in the context of iterated ultrapower constructions.
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http://math.stackexchange.com/questions/17152/given-an-infinite-number-of-monkeys-and-an-infinite-amount-of-time-would-one-of/17156
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# Given an infinite number of monkeys and an infinite amount of time, would one of them write Hamlet?
Of course, we've all heard the colloquialism "If a bunch of monkeys pound on a typewriter, eventually one of them will write Hamlet."
I have a (not very mathematically intelligent) friend who presented it as if it were a mathematical fact. Which got me thinking... Is this really true? Of course, I've learned that dealing with infinity can be tricky, but my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite. Therefore, it isn't necessarily given that the monkeys would write Hamlet.
Could someone who's better at this kind of math than me tell me if this is correct? Or is there more to it than I'm thinking?
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@Jason: +1 for asking a question that elicits such golden answers as are found below! – Sean Tilson Jan 12 '11 at 2:14
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Note that the set of finite-length strings of characters (i.e. "works") is countably infinite, not uncountable. – Nate Eldredge Jan 12 '11 at 3:23
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Doesn't one monkey suffice? – Rasmus Jan 12 '11 at 9:34
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– GWLlosa Jan 12 '11 at 12:34
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“We’ve all heard that a million monkeys banging on a million typewriters will eventually reproduce the entire works of Shakespeare. Now, thanks to the Internet, we know this is not true.” (Robert Silensky) – Ed Guiness Jan 12 '11 at 16:53
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## 14 Answers
Some references (I am mildly surprised that no one has done this yet). This is called the infinite monkey theorem in the literature. It follows from the second Borel-Cantelli lemma and is related to Kolmogorov's zero-one law, which is the result that provides the intuition behind general statements like this. (The zero-one law tells you that the probability of getting Hamlet is either zero or one, but doesn't tell you which. This is usually the hard part of applying the zero-one law.) Since others have addressed the practical side, I am telling you what the mathematical idealization looks like.
my intuition says that time is countably infinite while the number of works the monkeys could produce is uncountably infinite.
This is a good idea! Unfortunately, the number of finite strings from a finite alphabet is countable. This is a good exercise and worth working out yourself.
Edit: also, regarding some ideas which have come up in the discussions on other answers, Jorge Luis Borges' short story The Library of Babel is an interesting read.
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Hmmm... I must admit to being a bit surprised to find out that there's an "infinite monkey" theorem. – Jason Jan 13 '11 at 1:32
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@Jason: of course, the theorem isn't really about monkeys. It's about a particular model of monkeys. (Mathematics cannot prove anything about the world: it can only prove things about models of the world.) One can argue, as so many are doing in this thread, for or against this model, but when people quote this result I am assuming that they are referring to the model. – Qiaochu Yuan Jan 13 '11 at 1:53
@Qiaochu - I noticed that. It doesn't make the idea of an "infinite monkey theory" any less amusing though. ;-) – Jason Jan 13 '11 at 2:19
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@Trufa "¿De qué otra forma se puede amenazar que no sea de muerte? Lo interesante, lo original, sería que alguien lo amenace a uno con la inmortalidad." – belisarius Feb 3 '11 at 2:25
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I imagine the reason is no, and for relatively simple reasons. Monkeys won't see the use of typing. After at most a few keypresses they'll either get bored or destroy the typewriter in a fit. I doubt an infinite number of monkeys could even put together a full page full of nonsense but reasonable-length words with punctuation. You could ask the same question about spiders. Put an infinite number of spiders on typewriters and they won't produce Hamlet either, mostly because most spiders lack the strength to type. – Ryan Budney Dec 2 '11 at 18:11
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I found online the claim (which we may as well accept for this purpose) that there are $32241$ words in Hamlet. Figuring $5$ characters and one space per word, this is $193446$ characters. If the character set is $60$ including capitals and punctuation, a random string of $193446$ characters has a chance of $1$ in $60^{193446}$ (roughly $1$ in $10^{344000}$) of being Hamlet. While very small, this is greater than zero. So if you try enough times, and infinity times is certainly enough, you will probably produce Hamlet. But don't hold your breath. It doesn't even take an infinite number of monkeys or an infinite number of tries. Only a product of $10^{344001}$ makes it very likely. True, this is a very large number, but most numbers are larger.
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In fact one monkey alone would write Hamlet an infinite number of times given unlimited time. – jericson Jan 12 '11 at 1:54
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+1 for: "True, this is a very large number, but most numbers are larger." – Sean Tilson Jan 12 '11 at 2:08
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But what is even more alarming, is that the same monkeys - and probably (in the mathematical sense of the word) within the same time limit mentioned above by Ross Millikan - will write Hamlet with every possible typo imaginable. – Fredrik Meyer Jan 12 '11 at 2:50
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@Fredrik Meyer: ...and there was I thinking the decline of English was down to textspeak and all forms of youthful disregard for conventions; when actually the inaccurate spelling movement is being led by a band of thespian-tendencied research monkeys. – Orbling Jan 12 '11 at 3:38
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"but most numbers are larger". Not really. There's this infinite set of negative numbers to contend with. :) – Haacked Jan 12 '11 at 8:34
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[Note that in my answer I am actually assuming that there are only a finite number of monkeys. I don't see what is gained by having both the number of monkeys and the time frame be infinite: mathematically speaking $\aleph_0 \times \aleph_0 = \aleph_0$, and it is somewhat confusing to contemplate infinitely many monkeys typing simultaneously: too much is happening at once. In fact, there might as well be only one monkey, or at any rate only one typewriter.]
Let me take the unusual (for me) step of considering the practical aspects of this question as well.
As Ross Millikan has explained, there is a simple mathematical model of monkey keyboard pounding under which it is easy to see that the claim is true: the probability that at least one of the monkeys will type out Hamlet approaches $1$ as the time $n$ approaches infinity.
However there is an assumption here: namely, that the pounding on the typewriter is random or sufficiently close to random. One way to formalize this is to say that after typing any $n$ characters, the probability of hitting any given key as the $n+1$st character is at least $P$, where $P$ is positive and independent of $n$.
The problem is that for actual typewriter banging, this is a very unlikely assumption. The issue is similar here to what happens if you ask someone to produce a random sequence of digits, say from $0$ to $9$, or even a random sequence of $H$'s and $T$'s (for "heads" and "tails"). Just closing your eyes and banging away will produce something very far from being random.
If the question is meant to apply to actual monkeys with their nonrandom motor behavior, then it is something else entirely. I would be tempted to say that the probability of producing Hamlet does not approach $1$ as time approaches infinity, but I'm not sure off the top of my head how to justify this.
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Pete: I think this might be my favorite answer to any question ever. I know you meant it seriously, and it is certainly a well thought answer, but I find some of the phrases hilarious. Only a mathematician could write the phrase "there is a simple mathematical model of monkey keyboard pounding". – Sean Tilson Jan 12 '11 at 2:11
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– Matt E Jan 12 '11 at 3:25
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– KCd Jan 12 '11 at 6:39
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I believe there was an actual experiment done with monkeys in a room with typewriters to see what exactly they would produce. If I recall correctly the monkeys destroyed the typewriters, science. – JSchlather Jan 12 '11 at 8:19
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@Jacob Schlather. Not only did they destroy them, they treat them like toilets, too. – Atømix Jan 12 '11 at 20:24
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If you have an infinite number of monkeys, then an infinite subset of them will just sit and type out Hamlet, letter-for-letter, straight away. So after a few hours you will have an infinite number of copies of Hamlet.
If you have a finite number of monkeys then you may have to wait. But given in infinite amount of time (and immortal monkeys, etc.), you'll get your Hamlet. Eventually.
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So infinite monkeys, no problem; finite monkeys, request infinite improbability drive for Christmas present. This poses the question, how do you count your monkeys? – Orbling Jan 12 '11 at 3:41
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Fortunately, Hamlet is out of copyright. But they'll also write copies of everything that is in copyright, plus hexadecimal representations of binary files for every copyrighted movie, TV show, recording, etc., so they're still in big trouble. – Mike Scott Jan 12 '11 at 8:46
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The problem is distinguishing the correct copies of Hamlet from the infinite number of copies with significant typographic errors --- for example, the one in which Hamlet and Ophillia create a time machine and run off together to America. – vy32 Jan 12 '11 at 16:04
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@Orbling With an infinite improbability drive you get whales, not monkeys :D – belisarius Jan 14 '11 at 3:44
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@Orbling And the quotes are mostly harmless – belisarius Jan 14 '11 at 15:52
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Hamlet was indeed written after a finite number of monkeys (Shakespeare was a a primate from the family Hominidae!)
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Which makes it all the more remarkable, don't you think? – Atømix Jan 12 '11 at 20:29
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Entertaining, but does not really answer the question. – timur Jan 13 '11 at 5:03
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The important difference here is that the monkeys you are referring to are not randomly selected, nor do they type randomly. – Konrad Rudolph Jan 13 '11 at 14:51
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hsdiljcmksl ndewjknxcjkw;jkei80900 – Simon Jan 13 '11 at 16:27
@muntoo: your comment was removed mostly for its containing a non-broken string of letters that is too long for the display (and causing inconvenience for the rest of the page). – Willie Wong♦ Jan 14 '11 at 18:30
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It would probably be faster to apply selective pressure to breed intelligent, literary monkeys. Our common ancestor with chimps lived only 4 million years ago, and by design I'm sure we could make something similar happen faster starting with chimps, bonobos, or whatever.
We could probably get orcas and other whales up to speed very quickly, too!
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What is the probability that I now read the word bonobo for the SECOND time today after a lifetime of never having heard it - ahh the internets... – mplungjan Jan 12 '11 at 12:54
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Even if you do that, why would they produce an exact copy of Shakespeare? I mean almost surely their history will be different than ours. – timur Jan 13 '11 at 5:11
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@timur True; you would have to reproduce the exact conditions that lead to Hamlet it produce the same exact text at the same exact time. Or, as Carl Sagan put it, "If you want to make an apple pie from scratch, you must first create the universe." – WCWedin Jan 13 '11 at 23:49
No, just pay them minimum wage to write it out. Before they become smarter than us and destroy the Statue of Liberty! – Kevin Cantu Jan 14 '11 at 10:19
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Apparently Hollywood is already halfway there, judging by their product. – cdjaco Jan 14 '11 at 23:23
You don’t even need an infinite number of monkeys! For any $\epsilon > 0$ and $k \geq l(\text{Hamlet})$, there is some number $N$ such that $N$ monkeys at typewriters, each typing for $k$ keystrokes, will produce a copy of Hamlet with probability greater than $1-\epsilon$. (This holds under some quite weak conditions on our model of monkey typing.)
This is an example of the general “soft analysis to hard analysis” principle, championed by Terry Tao among others: most any proof in analysis may be transformed into a proof of a quantitative statement such as the one above.
This can be made precise in some generality using various rather beautiful proof-theoretic methods, such as variants of Gödel’s Dialectica translation; lovely results along these lines have been obtained by e.g. Avigad, Gerhardy and Towsner. In this particular case, the bounds we get will of course depend on the model of monkey typing used.
For instance, if we assume that the keystrokes are independently uniformly distributed, if our Hamlet-recognition criterion is case-, punctuation- and whitespace-insensitive, then for the case $k = l(\text{Hamlet})$,
$$N = \lceil \frac{\log \epsilon}{\log (1 - \frac{1}{26^{l(\text{Hamlet})}})}\rceil$$
will work. (The proof is an exercise for the reader.)
Project Gutenberg’s copy of Hamlet (first folio) weighs in at 117,496 alphanumeric characters. So if we want to produce Hamlet (first folio) with probability 1/2, in the minimal number of keystrokes, then by some quick slapdash estimating (rounding up a little to be on the safe side), something like $10^{170,000}$ monkeys should certainly suffice!
I guess empirical testing is out — ethical controls are so tricky. Anyone want to run some simulations?
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Oops — I see now that my answer overlaps somewhat with Ross Millikan’s excellent one above. However, there is a fair amount that is different, so I will leave this nonetheless… – Peter LeFanu Lumsdaine Jan 13 '11 at 5:40
For some reason I want to attribute this reasoning to Douglas Hofstadter, though I couldn't tell you which of his books it's from. Here goes:
If you could get sufficient randomness from an infinite number of monkeys (this is trivial if you assume that by mere chance, a infinite subset will fit the bill – or you could follow Arjang's approach and aggregate across monkeys for more entropy, which has the benefit of getting results much faster), you already have every variation of every story ever told. You'll even have every story that ever could be told. Just get an infinite number of monkeys (or a slightly smaller number of computers) and opening a publishing business. Make a million bucks and retire.
But this rings false, especially since modern computing power (relative to the difficulty of the task) is practically infinite, putting the practice of this philosophy within reach. Just imagine trying it yourself. It's not the monkeys or the computers or the printers doing all of the work. Suddenly, you are wading through millions of pages of gibberish text looking for the book that will make you rich. Good luck. (It is the fact that the filtering process is much slower than the production process that makes me say that computer power is practically infinite.)
It's not the characters on the page that make Hamlet. Hamlet is a synthesis of information, a composition that can only be guided by intelligence. In a sea of random characters, the sequence that maps isomorphically to Hamlet is just more noise.
This may sound like a qualified "yes" in response to the question, but in reality it is an empathic "no." It is not the act of producing the character sequence of Hamlet in a random string that writes Hamlet – it is the act of finding Hamlet in a random string that writes Hamlet. The distinction may sound subtle, but the two tasks are profoundly different.
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"modern computing power […] is practically infinite" — I think you misunderstand what "infinite" means. "A very large number" is not "practically infinite". In fact, the world's total computing power isn't even, "relative to the difficulty of the task", close to 10^344000, which is roughly how big you need things to be for Hamlet to be produced. It's a minuscule insignificant fraction of that, which means that we have no chance of generating Hamlet today by an actual uniform random process. Yes, if it were generated, we'd have trouble finding it, but that's a far out concern. – ShreevatsaR Jan 13 '11 at 8:03
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@ShreevatsaR A very large number" is not "practically infinite". I think the stress should be the word "practical" rather than "infinite." If the value could be either its current value or infinity without changing the bottleneck task (qualitatively or quantitatively), how is that anything other than practically infinite? It would be been better if I'd clarified that point more explicitly and earlier in the paragraph, though. – WCWedin Jan 13 '11 at 20:53
@ShreevatsaR Yes, if it were generated, we'd have trouble finding it, but that's a far out concern. On second thought, though, I think this is where our opinions diverge. In my opinion, this is a more fundamental point than physical limitations, or even the theoretical implication of infinity. You could say I played a sly trick in redefining "generating Hamlet," but I was really trying to make a point about the nature of information. To sum it up differently, I would say, "The question is irrelevant, because data is not information without context." Or, in Hofstadter's words, "Mu." – WCWedin Jan 13 '11 at 21:00
As far as I can see, the question is not about whether we can sift through the generated text to find intelligent text ("opening a publishing business. Make a million bucks…"), but whether a specific long text (Hamlet) may be generated by a random process at all. (That is, it's not the question here, though it may have been Hofstadter's, knowing his concerns.) So that part is simply irrelevant. And if you're looking for Hamlet specifically (letter-for-letter), it's a rather simple task that can be done with almost the same amount of resources as stepping through the generated text. – ShreevatsaR Jan 15 '11 at 18:31
@ShreevatsaR: Given the context, you could fairly say that I answered "the wrong question." I would even agree. – WCWedin Jan 17 '11 at 6:30
The answer is yes, With infinite time and all of the infinite monkeys will produce hamlet and every other works infinitely many times.
After the first keystroke of the infinitely many monkeys, there will be infinitely many hamlets if you just grabbed the first letter from each one of them.
Also at the first keystroke, infinitely of monkeys would have types the first letter of the entire hamlet already, this shows that if you have infinite number of monkeys you only each one of them to type just as many letters as the number of letters in hamlet to have already infinitely many copies of hamlet, so there is no need for infinite time to have one of them produced hamlet, infinitely many of the monkeys would produce infinitely many hamlets in finite amount of time as long as the finite amount of time is equal or greater than the time to type a single copy of hamlet. (this was already mentioned in Bennett McElwee's answer)
If you rephrase the question to : Would an infinite random sequence of letters in 2 dimensions contain at least one hamlet in every and each one of it's columns/rows? then the answer is yes, but there would be infinite number of hamlets contained in each row/column, As an infinite random sequence will contain all it's possible finite sub sequences, infinitely many times. ( Reference needed ).
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I suspect this is incorrect.. Consider a random sequence using only 1's and 2's. No reason why it contains all possible numbers especially 123 inside it. – picakhu Jan 12 '11 at 4:16
@picakhu : edited, of course a sequence of 1's and 2's wont have a 3 in it! :) added the finite subsequence into edit – Arjang Jan 12 '11 at 4:32
I think this is still incorrect.. Assume that a 11 is banned. so we can have a random sequence like 121222212221222122222222212121212221212 which has no sub-sequence 11 in it. – picakhu Jan 12 '11 at 4:37
@picakhu : Why random sequence of 1's and 2's is not allowed to have 11? of course if a subsequence is always take out from the sequence then it will not occur. – Arjang Jan 12 '11 at 5:13
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@picakhu I could then just as easily say that the sequence 212 contains the sub-sequence 11 and the sequence 2122 does not contain the sub-sequence 22. If the interpretations are isomorphic, you are free to choose either one. That said, monkeys are unlikely to produce output that can be universally interpreted in more than one meaningful, non-trivial way. I.e., interpreting 'aa' as 'b' is meaningless in the general case, and the choice of computer character encoding (ASCII vs. UTF-8, say) is trivial. My point, I guess, is that random letter generators won't perform high-level encoding. – WCWedin Jan 12 '11 at 22:19
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Let me be the one who says NO. It is a bit hard to handle an infinite number of monkeys (because you will need an infinite number of bananas to feed them), but if you agree to have a finite number of monkeys and infinite time, I would guess that considering the time it should take for this to happen, it is likely that the universe would die before it actually does.
To quote a chat of one of my friends with my first year "intro to CS" professor:
professor: "So there is no chance that a fair coin would fall with the head side up 50 times in a row."
my friend: "No!, there is such a chance it's just very small...."
professor: "Well yes, its like the chance for all the nitrogen molecules in the air to gather around your head and have you die from lack of oxygen."
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Ultrafinitism takes out all the fun from mathematics. Especially from the parts dealing with infinite processes. – Asaf Karagila Jan 12 '11 at 8:41
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From a [ultra]finitistic perspective, I don't think you can say "NO"; you can only say that the question is meaningless as there's no such thing as infinite time or infinite monkeys. – ShreevatsaR Jan 12 '11 at 14:42
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And there would need to be an infinite amount of space to contain the monkeys... not to mention infinite typewriters... better ramp up typewriter production in China. – Atømix Jan 12 '11 at 20:33
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@ShreevatsaR Talk about practical thinking. – muntoo Jan 14 '11 at 6:10
Also, I doubt the monkeys will specifically twist their hands for each letter. The QWERTY keyboard was designed to make it slower/harder to type, after all, so I doubt the mindless monkeys will go to all that trouble to contort their 'inferior' hands. – muntoo Nov 21 '11 at 6:59
NOTE: By probability, I mean the chance of it happening per iteration, and starting with a new page each time.
Let's take a step back, shall we? (Not too many, because there's a cliff behind you.) Let's think of what the probability is of producing the following randomly:
A
Assuming there are only 26 characters (A-Z, uppercase), the probability would be $\frac{1}{26}$.
AA
It'd be $(\frac{1}{26})^2$. This:
AAA
It'd be $(\frac{1}{26})^3$. And this:
XKCD
It'd be $(\frac{0}{26})^4$. [Just kidding, it's: $(\frac{1}{26})^4$].
So, for every character we add to the quote, it will be: $(1/26)^c)$, where $c$ represents the number of characters.
Basically, it would be a probability of $(\frac{1}{26*2+12})^c$ since the characters used could be: A-Z, a-z, .!?,;: "'/() Of course, there could be more characters, but that's just an example. :)
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Could I please have the ability to downvote myself? 10 months and 6 days later, I find out that I am an idiot. – muntoo Nov 21 '11 at 7:02
You can delete the answer, right? – The Chaz 2.0 Feb 27 '12 at 14:50
@TheChaz I will leave it up here as proof that even a genius can be an idiot at times. – muntoo Feb 28 '12 at 1:37
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Don't be so hard on yourself! Maybe this is proof that given a finite amount of time, a human with a keyboard will type... ________ ??? – The Chaz 2.0 Feb 28 '12 at 3:13
Upvote from me. This is a genius answer, you proved the opposite, muntoo. (kidding, no offense :) – Rock Jun 6 '12 at 5:09
Here's a rather interesting article discussing the probability of a monkey producing Shakespeare's works and uses a random letter generator to demonstrate some results.
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If there truly was an infinite amount of time and monkey, yes, it could happen. However, we know that time and monkeys are both limited. Let's say that the universe will be gone when there are no more neutrons. According to this article, that will be about 10^40 years. There are approximately 4*10^78 atoms. And let's just say that an atom monkey can type at 10^15 keys per second. Let's also assume that there are 40 keys on a typewritter (26 A-Z, numbers, period, comma, semicolon, and space). That'll give the following:
There will be 4e78*365*24*3600*1e15 key strokes per atom, giving a total of 1.26e101. There would then be 5.05e177 key strokes. 40^108=1.0531e+173. That means that there would be segments of around 108 characters of Hamlet around, but certainly not the whole work.
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+1 The answer is no because the question is impossible. Given that the mass and age of the universe are both finite, it is impossible to have an infinite number of monkeys typing for an infinite amount of time. – Qwerky Jan 12 '11 at 16:26
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@Qwerky: I think you are interpreting the question too literally. That is one interpretation, but there are other interpretations less tied to physical reality. – Qiaochu Yuan Jan 12 '11 at 17:38
Monkeys don't produce a proper random distribution on keystrokes. Not even a Markov-chain of keystrokes.
Given infinite time (and an undying support of monkeys) or just infinite monkeys, they will produce infinite text. But that does not need to imply that Hamlet will be part of this infinite long text.
There is empirical evidence from such an experiment, reported by the BBC.
The actual text produced is obviously not that random at all.
(@Henry: Thx for these links.)
[/edit]
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Did you mean "perhaps" instead of "eventually"? – Rasmus Jan 16 '11 at 10:50
daah... yeah, of course. silly me. – The german word "eventuell" means "possibly" in english. It's one of those strange words... The inventor of the english language must have written some bugs in one of the first english dictionaries, which is now hardwired in the standard library. – comonad Jan 27 '11 at 21:22
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Monkeys don't produce a proper random distribution on keystrokes. Not even a Markov-chain of keystrokes. How do you know? – Did Nov 11 '11 at 20:57
Just look at the monkey, dude! Not better than a baby on the piano, nearly repeating the same wham from before. (Assumed that it's not Markov's child playing.) – comonad Nov 12 '11 at 7:44
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@Didier: There is empirical evidence from an experiment reported by the BBC and I suspect that the actual text produced would fail most test of randomness. – Henry Nov 14 '11 at 0:34
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## protected by t.b.Nov 9 '11 at 18:53
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.
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http://mathhelpforum.com/differential-geometry/129610-banach-algebras-question-4-a.html
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# Thread:
1. ## Banach algebras question 4
I've the lats three questions about maximal ideals in Banach algebras.
How can I show the following (in red shape) part of proof?
2. Originally Posted by Arczi1984
I've the lats three questions about maximal ideals in Banach algebras.
How can I show the following (in red shape) part of proof?
I'm not sure what the difficulty is here. The first sentence of the proof shows that if x is invertible then $T(x^{-1})$ is the inverse of $Tx$. In other words, $(Tx)^{-1} = T(x^{-1})$. The comment about the spectrum then follows immediately from the fact that $\lambda\notin\sigma(x)\;\Leftrightarrow\;x-\lambda e$ is invertible.
3. Here is my mistake. When I read it I saw non-invertible I spent to much time over this paper and now I see problems where everything is clear. I'm sorry for this question and thanks for answer.
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http://mathhelpforum.com/advanced-statistics/194916-asymptotic-variance-print.html
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Asymptotic Variance
Printable View
• January 4th 2012, 07:41 AM
Duke
Asymptotic Variance
What is the asymptotic variance of an estimator? There seems to be several definitions on the web but I know the one I'm after includes the expectation of the second derivative of the log-likelihood.
• January 4th 2012, 10:01 PM
matheagle
Re: Asymptotic Variance
an estimator, by definition is a statistic, hence its a function of the data, which is the sample
$X_1,\ldots, X_n$
so it's a random variable and it has a distribution.
Which means it has a mean and a variance.
If its mean, its expectation is equal to the unknown parameter(which is a constant)
then that estimator is said to be unbiased.
Next we look at its variance.
And we like to see its long term behaviour.
That limit is the asymptotic variance.
What you seem to want is Fischer's...
http://ocw.mit.edu/courses/mathemati...s/lecture3.pdf
All times are GMT -8. The time now is 04:32 PM.
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http://math.stackexchange.com/questions/225697/is-this-the-correct-way-to-prove-exists-y-forall-x-px-vee-qy-equiv-f?answertab=oldest
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Is this the correct way to prove $\exists y \forall x (P(x) \vee Q(y)) \equiv \forall xP(x) \vee \exists x Q(x)$?
$\exists y \forall x (P(x) \vee Q(y)) \equiv \forall xP(x) \vee \exists x Q(x)$
If the LHS is true, then there are two cases: P(x) is true, in which case $\forall$ P(x) is true and the RHS is true, and Q(y) is true, in which case $\exists$ x Q(x) is true and the RHS is true.
If the RHS is true, then there are two cases $\forall$ P(x) is true, in which case P(x) is true and the LHS is true, and $\exists$ x Q(x) is true, in which case Q(y) is true and the LHS is true.
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– MJD Oct 31 '12 at 3:49
1 Answer
I think the easiest way to prove this is to use the rules for manipulating quantifiers. Starting with the LHS, we have
$$\begin{align*} \exists y[\forall x(P(x)\vee Q(y))] &\equiv \exists y [\forall x P(x) \vee \forall x Q(y)]\\ &\equiv \exists y [\forall x P(x) \vee Q(y)] \\ &\equiv \exists y\forall x P(x) \vee \exists y Q(y) \\ &\equiv \forall xP(x) \vee \exists yQ(y) \\ &\equiv \forall xP(x) \vee \exists xQ(x). \end{align*}$$
This is because $\mathbf{Q}\mathbf{v}(\mathbf{P}(\mathbf{v})\ \mathbf{B}\ \phi)\equiv (\mathbf{Q}\mathbf{v}\mathbf{P}(\mathbf{v}))\ \mathbf{B}\ \phi$ is valid where $\mathbf{Q}$ is a quantifier, $\mathbf{v}$ is a variable, $\mathbf{P}$ is a predicate, $\mathbf{B}$ is either the conjunction or disjunction operator, and $\phi$ is a formula where $\mathbf{v}$ isn't free.
For your proof, when assuming the LHS, splitting directly into cases isn't possible because the disjunction is "trapped" by the quantifiers. We must say: let $c$ be an object such that $\forall x(P(x)\vee Q(c))$ is true (we know such a $c$ exists because of the assumption). Now let $a$ be an arbitrary object. We know $P(a)\vee Q(c)$ is true. Suppose now that $Q(c)$ is true. Then certainly $\exists xQ(x)$. Now suppose $P(a)$ is true. Since $a$ was arbitrary, $\forall xP(x)$ is true.
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shouldn't it be $P(c) \vee Q(c)$? – Gladstone Asder Oct 31 '12 at 17:29
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http://physics.stackexchange.com/questions/12201/why-dont-waves-with-different-wavelengths-interfere-with-each-other-in-white-li/12204
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# Why don't waves with different wavelengths interfere with each other in white-light?
The book I'm reading about optics says at some point that "each color (wavelength) contained in the white light interferes only with itself". But why is this so?
Edit: I moved the rest of the question elsewhere.
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Closely related: . – dmckee♦ Jul 13 '11 at 2:25
## 3 Answers
Add any instant in time, light of different wavelengths can be said to interfere. However, because of the extreme frequencies of optical light, any cross interference will get time-averaged away very quickly unless the two waves are very close in frequency.
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Very simple and clear. Just one thing: did you mean "At any instant"? Also, I was wondering if I should strip down the question and retain only the first sentence; the rest might be a bit messy or slightly off-topic. What do you think? – Waldir Jul 12 '11 at 17:09
I think it might be better if I moved the rest of it to another question. I'll do so now. – Waldir Jul 12 '11 at 17:19
Interference is not an instant phenomenon but a time-lasting one. The time should be sufficient to speak of certain frequencies. Two different frequencies can be easily distnguished with the corresponding resonators but this process takes time. It is in a long term condition that two frequencies are separated (resonators pumped). Instantly we cannot even speak of a frequency.
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Imagine two light beam each of its own wavelength $L_1=720\text{ nm}$ and $L_2=719\text{ nm}$. One can show I suppose that every $720$ cycles of the $L_2$ beam the total constructive interference shall repeat.
In air the $L_1$ beam light frequency is $$F_1=\frac{C}{L_1}=417\times 10^{14}\text{ Hz}$$
and therefore the frequency of the interference effect is $$F_\text{int}=\frac{F_1}{720}=57.9*10^{12}\text{ Hz}$$
This frequency is rather high to be detectable. When two light beams are of identical wavelength however, the interference effect is time independent and therefore observed as an interference pattern. It is obvious that there is the interference between light beams of a different "color" albeit difficult to observe. Please read about the white light interferometry.
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http://www.physicsforums.com/showthread.php?t=190540
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Physics Forums
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## Wavefunction collapse: is that really an axiom
Can the wavefunction collapse not be derived or is it really an axiom?
How can the answer to this question (yes or no) be proven?
If it is an axiom, is it the best formulation, is it not a dangerous wording?
Let's enjoy this endless discussion !!!
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Blog Entries: 19 Recognitions: Science Advisor There are several different interpretations of QM. In some of them, there is no need for a collapse postulate.
Recognitions: Gold Member Science Advisor Staff Emeritus I think the thing we can sensibly say is that wavefunction collapse cannot follow from the unitary time evolution, which is easy to establish.
## Wavefunction collapse: is that really an axiom
Quote by vanesch I think the thing we can sensibly say is that wavefunction collapse cannot follow from the unitary time evolution, which is easy to establish.
But it is if you include the interaction with the measuring device into the QM model.
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Quote by lalbatros But it is if you include the interaction with the measuring device into the QM model.
Only if you choose some model other than unitary evolution for describing the measurement process.
Do you mean that the "measurement axiom" is contradictory to my/the postulate that the (unitary) equation of evolution governs all interactions? (including measurement sytems)
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Quote by lalbatros Do you mean that the "measurement axiom" is contradictory to my/the postulate that the (unitary) equation of evolution governs all interactions? (including measurement sytems)
yes, of course! That's the whole issue (or better, half the issue) in the "measurement problem". It is (to me at least) one of the reasons to consider seriously MWI.
There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.
Quote by vanesch yes, of course! That's the whole issue (or better, half the issue) in the "measurement problem". It is (to me at least) one of the reasons to consider seriously MWI. There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.
The algebra is simple and true.
The problem is that the wavefunction collapse doesn't really exist.
Just like microreversibility doesn't contradict the second law: microreversibility doesn't necessarily imply the existence of a chaos demon.
Quote by vanesch yes, of course! That's the whole issue (or better, half the issue) in the "measurement problem". It is (to me at least) one of the reasons to consider seriously MWI. There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.
The algebra is simple and true.
(even simple inspection of collapse algebra is enough for that, specially on the density matrix)
The problem is that the wavefunction collapse doesn't really exist.
Just like microreversibility doesn't contradict the second law: microreversibility doesn't necessarily imply the existence of a chaos demon.
In addition, I am quite sure that the collapse axiom can be derived from the Schrödinger equation. But the understanding is missing, to my knowledge.
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Quote by lalbatros In addition, I am quite sure that the collapse axiom can be derived from the Schrödinger equation. But the understanding is missing, to my knowledge.
There is actually a proof that it cannot. The proof is based on the fact that the Schrodinger equation involves only local interactions, while the collapse, including the cases with two or more entangled particles, requires nonlocal interactions.
There is, however, something that contains some elements of a collapse but can be obtained from the Schrodinger equation. This is the environment-induced decoherence. And it is closely related to the second law emerging from time-symmetric laws of a large number of degrees of freedom. See e.g.
http://xxx.lanl.gov/abs/quant-ph/0312059 (Rev. Mod. Phys. 76, 1267-1305 (2004))
Quote by vanesch This can be shown in 5 lines of algebra.
could you please show it then? i don't know about you guys, but i am so sick of qualitative arguments involving wavefunction collapse, etc.
to address the OP, it is my understanding is that if you postulate the Born interpretation then wavefunction collapse follows from that; in other words, physicists weren't just sitting around and postulated "wave collapse" as some popular books/shows would like one to believe.
more concretely,
$$\langle \Omega \rangle_\alpha = \sum_i \omega_i | \langle \omega_i | \alpha \rangle|^2$$
where the Born interpretation is that the quantity$$| \langle \omega_i | \alpha \rangle|^2$$ is to be interpreted as the probability amplitude of measuring a value $$\omega_i$$
$$\langle \Omega \rangle_\alpha = \sum_i \omega_i | \langle \omega_i | \alpha \rangle|^2$$
$$= \sum_i \omega_i \langle \omega_i | \alpha \rangle ^* \langle \omega_i | \alpha \rangle$$
$$= \sum_i \omega_i \langle \alpha | \omega_i \rangle \langle \omega_i | \alpha \rangle$$
$$= \sum_i \langle \alpha | \omega_i \rangle \omega_i \langle \omega_i | \alpha \rangle$$
$$= \sum_i \langle \alpha | \omega_i \rangle \langle \omega_i | \Omega | \omega_i \rangle \langle \omega_i | \alpha \rangle$$
$$= \langle \alpha |\left(\sum_i | \omega_i \rangle \langle \omega_i |\right) |\Omega |\left(\sum_i | \omega_i \rangle \langle \omega_i | \right)\alpha \rangle$$
$$=\langle \alpha | \Omega | \alpha \rangle$$
so for some state $$\phi = \sum_i c_i \psi_i$$ that is NOT an eigenket of the operator (but can always be formed from a linear combination of eigenkets), we have:
$$\langle \phi | \Omega | \phi \rangle = \langle \phi | \Omega | \sum_j c_j \psi_j \rangle$$
$$=\langle \phi | \sum_j c_j \omega_j \psi_j \rangle$$
$$=\sum_j \langle \sum_i c_i \psi_i | c_j \omega_j \psi_j \rangle$$
$$=\sum_i \sum_j c_i^* c_j \omega_j \langle \psi_i | \psi_j \rangle$$
and by orthogonality of states, we have:
$$=\sum_i |c_i|^2 \omega_i$$
which shows that the average value of our experiments will be a weighted average of the eigenstates, i.e. the "wavefunction collapse" s.t. any individual measurement will be a particular eigenvalue. in the classical limit, the spectra of eigenvalues is nearly continuous and so the effect is unnoticeable.
so whats the big deal?? can someone explain to me why this is, for some people, such a big damn mystery??
Quote by vanesch There's no unitary evolution (no matter how complicated) that can result in a collapsed wavefunction. This can be shown in 5 lines of algebra.
There is no need for any algebra to show that the wavefunction collapse is not described by a unitary evolution. This follows simply from the definition of the wavefunction. By definition, the wavefunction is a probability amplitude. This means that the measurements described by the wavefunction is a random probabilistic unpredictable process, which cannot be described by deterministic "unitary evolution". That's the whole point of quantum mechanics. In my opinion, looking for a unitary description of the collapse is equivalent to looking for "hidden variables".
Eugene.
I agree, meopemuk. The collapse is not an unitary transformation, and it is even not a transformation at all. After the collapse, there is no wave function anymore, but a statistical mixture. That's the axiom. My view is that after the interaction of a small system with a measuring device, the state of the small system loses its meaning, and only the combined wavefunction has a meaning. The problem that remains is how does the axiom emerge from the "complex" evolution. This is a challenge, and I am confident that it will or can be explained trivially. I am also sure that solving this problem is not really useful for the progress of QM, that is it of the kind of problem that time and generations solves.
Recognitions: Gold Member Science Advisor Staff Emeritus Ok, here goes the "proof". Axiom 1: every state of a system is a ray in Hilbert space. Now, consider the system "measurement device + SUT" (SUT = system under test, say, an electron spin). This is quantum-mechanically described by a ray in hilbert space. As we have degrees of freedom belonging to the SUT and other degrees of freedom belonging to the measurement device, the hilbert space of the overall system is the tensor product of the hilbert spaces of the individual systems H = H_m x H_sut Now, consider that before the measurement, the SUT is in a certain state, say |a> + |b> and the measurement system is in a classically-looking state |M0>. As we have now individually assigned states for each of the subsystems, the overall state is given by the tensor product of both substates: |psi0> = |M0> x ( |a> + |b> ) Now we do a measurement. That comes down to having an interaction hamiltonian between both subsystems, and from that interaction hamiltonian follows a unitary evolution operator over a certain time, say time T. We write this operator as U(0,T), it evolves the entire system from time 0 to time T. Now, let us first consider that our SUT was in state |a> and our measurement system was in (classically looking) state |M0>, which is its state before a measurement was done. "doing a measurement" would result in our measurement device get into a classically looking state |Ma> for sure, assuming that |a> was an eigenvector of the measurement. As such, our interaction between our system and our measurement apparatus, described by U(0,T) is given by: U(0,T) { |M0> x |a> } = |Ma> x |a> Indeed, the state of the measurement device is now for sure the classically-looking state Ma, and (property of a measurement on a system in an eigenstate) the SUT didn't change. We can tell now the same story if the system was in state b: U(0,T) { |M0> x |b> } = |Mb> x |b> where Mb is the classically looking state of the measurement apparatus with the pointer on "b". Now from linearity of U follows that: U(0,T) { |M0> x (|a> + |b>) } = |Ma> x |a> + |Mb> x |b> We didn't find "sometimes |Ma> x |a> and sometimes |Mb> x |b>" ; the unitary evolution gave us an entangled superposition. Now, you can say "yes, but |u> + |v> means: sometimes |v> and sometimes |u>" but that's not true of course. Consider the other measurement apparatus N which does the following: U(0,T) { |N0> x (|a> + |b>) } = |Nu> x (|a> + |b> ) U(0,T) { |N0> x (|a> - |b>) } = |Nd> x (|a> - |b>) From this, we can deduce that U(0,T) {|N0> x |a> } = 1/2 (|Nu> x { |a> + |b> } + |Nd> x { |a> - |b> }) Clearly if |u> + |v> means "sometimes u and sometimes v" then we could never have that |a> + |b> (which is then "sometimes a and sometimes b") always gives rise to Nu and never gives rise to Nd, because |a> gives rise to sometimes Nu and sometimes Nd.
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Quote by lalbatros I agree, meopemuk. The collapse is not an unitary transformation, and it is even not a transformation at all. After the collapse, there is no wave function anymore, but a statistical mixture. That's the axiom.
The problem is that the transition for a density matrix to go from "superposition" to "statistical mixture" is the following transformation:
take the densitymatrix of the "superposition", and write it in the matrix form in the *correct basis*. Now put all non-diagonal elements to 0. You now have the statistical mixture.
But again, that is a point-wise state change (if you take the density matrix to define the state) which is not described by the normal evolution equation of the density matrix. In other words, the transformation "superposition" -> "mixture" for the density matrix is again a state change which is not described by a physical interaction (which is normally described by the usual evolution equation of the density matrix).
In other words, that's nothing else but another way of writing down a non-unitary evolution, which is not the result of a known physical interaction.
My guess is that 99% of all experiments involve a single measurement of the system's state. (The counterexample is the bubble chamber, where we repeatedly measure particle's position and obtain a continuous track) In these cases we do not care what is the state of the systems and its wavefunction after the measurement (collapse). It is important to realize that one needs to consider the abrupt change of the wavefunction after measurement only in (not very common) experiments with repeated measurements performed on the same system. Eugene.
Quote by vanesch The problem is that the transition for a density matrix to go from "superposition" to "statistical mixture" is the following transformation: take the densitymatrix of the "superposition", and write it in the matrix form in the *correct basis*. Now put all non-diagonal elements to 0. You now have the statistical mixture. But again, that is a point-wise state change (if you take the density matrix to define the state) which is not described by the normal evolution equation of the density matrix. In other words, the transformation "superposition" -> "mixture" for the density matrix is again a state change which is not described by a physical interaction (which is normally described by the usual evolution equation of the density matrix). In other words, that's nothing else but another way of writing down a non-unitary evolution, which is not the result of a known physical interaction.
I would rather say:
which is more conveniently described by a non-unitary transformation
Decoherence can already easily wipe off non-diagonal elements.
I also remember my master thesis 25+ years ago.
I worked on the Stark effect in beam-foil spectroscopy: the time-dependence with the quantum beats and the atomic decay.
(by the way the off-diagonal elements of the H-atoms exiting the foil were crucial in the simulation)
The hamiltonian had to simulate also the decays of the atomic levels.
Looks-like a non-unitary transformation too, isn't it?
I also didn't want to embarass myself with full QED stuff.
Guess how I modelled that:
- adding a non-hermitian term to the hamiltonian (related to the decay rates)
- and calculating the resulting non-unitary evolution operator
(the density matrix was therefore decaying, which is quite natural)
This is nothing strange or surprising and it shows clearly how non-unitary evolution can simply occur as a limit case of an unitary transformation.
With such a simple approach the Stark effect is very well calculated, for the energy levels, for the perturbed lifetimes and for the time-dependence and polarisations of light emission.
In somewhat pedantic words (I am not a mathematician): the limit of a series of unitary transformations may not be an unitary transformation, looks like that to me. And this can just mean that in some situations the unitary evolution may just be an academic vision: coarse graining makes it practical and non-unitary.
Why should I go for more science-fiction?
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# Computation of Laplace-Beltrami operator in a conformally equivalent metric
Could anyone tell me where I'm wrong with the following elementary calculation? Given a smooth Riemannian manifold $(M, g)$, I'd prove that if $\tilde{g}$ is conformally equivalent to $g$ (that is, $\tilde{g} = e^{2w}g$ for some smooth function $w$), then $\Delta_{\tilde{g}} = e^{-2w}\Delta_g$. Now, recalling that the Laplace-Beltrami operator $\Delta_g$ is defined as (or better, this is a possible definition)
$$\Delta_g:=g^{ij}\left(\frac{\partial^2}{\partial x_i \partial x_j} - \Gamma^k_{i j}\frac{\partial }{\partial x_k}\right).$$
With elementary calculations, I obtained (and I'm pretty sure that, at least this, is right...)
$$\tilde{\Gamma}^k_{ij} = \Gamma^k_{ij} - \left(\delta_{ik}\frac{\partial w}{\partial x_j} + \delta_{kj}\frac{\partial w}{\partial x_i} - g^{k\ell}g_{ij}\frac{\partial w}{\partial x_{\ell}}\right),$$
where $\tilde{\Gamma}^k_{ij}$ are the Christoffel's symbols of the Levi-Civita's connection associated to the metric $\tilde{g}$ and $\delta_{ij}$ is the Kronecker's $\delta$. At this point, I have
$$\Delta_{\tilde{g}} = e^{-2w}\Delta_g + e^{-2w}g^{ij}\left(\delta_{ik}\frac{\partial w}{\partial x_j} + \delta_{kj}\frac{\partial w}{\partial x_i} - g^{k\ell}g_{ij}\frac{\partial w}{\partial x_{\ell}}\right)\frac{\partial}{\partial x_k}$$
But
$$g^{ij}\delta_{ki}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_k} = g^{ij}\frac{\partial w}{x_j}\frac{\partial }{\partial x_i},$$
$$g^{ij}\delta_{kj}\frac{\partial w}{\partial x_i} \frac{\partial }{\partial x_k} = g^{ij}\frac{\partial w}{\partial x_i}\frac{\partial}{\partial x_j} \stackrel{i \leftrightarrow j}{=} g^{ji}\frac{\partial w}{\partial x_j}\frac{\partial}{\partial x_i} \stackrel{g^{ji} = g^{ij}}{=} g^{ij}\frac{\partial w}{\partial x_j}\frac{\partial}{\partial x_i},$$
$$\underbrace{g^{ij}g_{ij}}_{= g^{ij}g_{ji} = 1} g^{k\ell} \frac{\partial w}{\partial x_{\ell}}\frac{\partial }{\partial x_k} = g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i}.$$
That is, one the last terms seems survive. Where am I wrong?
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## 2 Answers
The proposed relationship $$\Delta_{\tilde{g}} = e^{-2w} \Delta_g$$ only holds for surfaces. In your final calculation, assuming $M$ is a surface, you should have $g^{ij} g_{ij} = 2$, so that the final term is \begin{align*} e^{-2w}g^{ij}\left(\delta_{ik}\frac{\partial w}{\partial x_j} + \delta_{kj}\frac{\partial w}{\partial x_i} - g^{k\ell}g_{ij}\frac{\partial w}{\partial x_{\ell}}\right)\frac{\partial}{\partial x_k} & \\ = g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i} + g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i} - 2g^{i j}\frac{\partial w}{\partial x_j}\frac{\partial }{\partial x_i} & = 0. \end{align*}
An easier approach which avoids using Christoffel symbols or messing around with indices is to use the formula $$\Delta_g = - \frac{1}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \right).$$ With this formula, you can compute \begin{align*} \Delta_{\tilde{g}} & = - \frac{1}{e^{dw}\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( e^{dw} \sqrt{|\det g|} e^{-2w} g^{ij} \frac{\partial}{\partial x^i} \right) \\ & = - \frac{e^{-dw}}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( e^{(d-2)w} \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \right) \\ & = - \frac{(d-2)e^{-2w}}{\sqrt{|\det g|}} \frac{\partial w}{\partial x_j} \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} - \frac{e^{-2w}}{\sqrt{|\det g|}} \frac{\partial}{\partial x_j} \left( \sqrt{|\det g|} g^{ij} \frac{\partial}{\partial x^i} \right) \\ & = \Delta_g - (d-2)e^{-2w} g^{ij} \frac{\partial w}{\partial x_j} \frac{\partial}{\partial x^i}, \end{align*} so $\Delta_{\tilde{g}} = e^{-2w} \Delta_g$ only holds when $d = 2$. (You'll get the same formula when you correctly state $g^{ij}g_{ij} = d$ in your work above).
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For Riemannian manifolds of dimension $d$, $g^{ij}g_{ji} = d$. This means your formula works, but only in dimension two.
There is an alternative expression for the Laplace-Beltrami operator involving $\sqrt{\det g}$ which gives you the factor $d-2$ directly.
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Is there a coordinate-free way to arrive at this expression for $\Delta$? – Guillermo Jan 25 at 23:43
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## Having trouble understanding/accepting the definition of work. Thought exercise...
One concept in physics that has never set well with me is the way work and energy are defined.
According to all the physics sources I've looked at, work is defined as:
$$W = \vec{F} \cdot \vec{d}$$
(for a constant force over a distance)
However, intuitively the notion of taking the dot product of F with the displacement doesn't seem quite right to me. I am well familiar with the properties and interpretation of the dot product, but I still feel like something is missing.
Consider the following thought exercise:
Imagine we have two separate objects in space, object A and object B. Object A is at rest and object B is moving at a constant velocity through space. In other words, there is no net force acting on either object.
Suppose that we now apply a force $F_1$ to object A and a separate force $F_2$ to object B for an equal amount of time and that $F_1 = F_2$ and that both forces point in the same direction as the velocity of object B does. Object A will cover a distance $d_A$ and object B will cover a distance $d_B$ in this time interval.
Because A started at rest it will cover a smaller distance over the time interval than object B will cover. Therefore according to the definition of work the work done on object B over the time interval will be greater than the work done on object A, even though equal force was applied to both in equal time.
This in turn implies that the amount of kinetic energy applied to A to cause the resulting change in momentum would not be sufficient to apply the same change in momentum to object B, even though we already know that the same amount of force over time was applied to achieve it. This seems like a contradiction to me.
It seems to me that a more intuitively accurate formula for work might be something like this:
$$W = \vec{F} \cdot \frac{\vec{d}}{ \lVert \vec{d} \rVert}\, t$$
(where t is the amount of time over which the force is applied to the object)
Notice that this alternate formula gives the same value for both objects A and B, whereas the standard formula does not.
It seems like you can't really cheat by just changing frames of reference, because in this example we're trying to calculate energy totals for our current frame of reference as the observer. The standard formula seems to imply that more kinetic energy must be put into changing the momentum of already moving objects than of objects at rest, which seems to violate the basic laws of physics.
Furthermore, I can't help but notice that the energy laws, such as conservation of energy are very similar to the laws of conservation of momentum. Perhaps they are in fact one and the same when proper analysis is applied. Perhaps change in momentum in a direction (as given by my alternate formulation of work above) is in fact the real unit of mechanical work. Granted, one would need to reformulate other equations in physics that depended on it because of the change, but maybe it could in fact work.
Energy is basically supposed to measure a systems "ability" to enact a change in momentum right (assuming we're considering only mechanical energy), and isn't it true that if a particle collides with a particle at rest then it transfers momentum into the other, and by so doing enacts a change in momentum? Thus doesn't this further support the argument that energy may in fact be a measure of momentum in disguise (in the purely mechanical cases at least).
Can you disprove my reasoning and clarify what work actually is and why the standard definition is what it is? Or perhaps, could my alternate definition be more correct somehow?
All the sources I've found on work and energy don't really discuss why it's defined to be the way it is; they instead just repeatedly reference other definitions and terms without really linking any of those terms to any real reason. How were these things originally thought up and why are the formulas what they are?
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It seems to me you want to look at impulse instead of work. You're talking about a situation where a force acts between two objects for a constant time, and in that case, the impulse--the change in momentum imparted--is relatively simple to calculate: $$\Delta p = F \Delta t$$ Saying that the two objects have a different amount of work done on them gets a, "So what?" response from me. You want something to be the same between the two of them, impulse is it. Your situation of a force that goes on and off based on time is problematic to analyze because these two particles would have potential energy based on their positions that just goes poof when the force is turned off. From a Newtonian mechanics perspective, energy is just passed around between objects, either stored as potential energy due to outside forces or converted into kinetic energy. You can say that energy is conserved because there is a symmetry of the laws of physics with respect to time. Any symmetry leads to a corresponding conserved quantity, and time symmetry leads to energy. To understand why time and energy are related in this way, I suggest you read up on the quantity called action. The action has the units of Energy x Time, Distance x Momentum, or Angle x Angular Momentum, and the symmetries of the laws of physics with respect to shifts in time, translations, and rotations (time, distance, and angle) generate the symmetries that conserve energy, momentum, and angular momentum. You could say that action is the most fundamental quantity, and we just had to give names for "action / distance (that is, momentum)" and so on because the conservation laws made them useful. Most higher level physics doesn't deal directly with Newton's laws anymore, only the equivalent equations of motion based on an action principle (even for classical physics). The work done on a particle by a force is just the amount of energy converted from potential energy of the particle to kinetic energy.
Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi WraithGlade! Force is rate of increase of momentum. Force is not rate of increase of energy.
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## Having trouble understanding/accepting the definition of work. Thought exercise...
You can test for yourself what impact speed has on work using a bike and a hill. Ride a bike partway up a hill for a certain amount of time. Then do it again for the same amount of time, but faster. See if you are more tired the second time.
Same force does not mean same energy. You can see this in everyday examples. Accelerating a car from 60-120 is at least 3x harder than accelerating from 0-60, and not just because of air resistance, but because W=Fd. Even with the same force, you will still cover 3x the distance due to the higher initial speed. Indeed you will also burn 3x more fuel. What actually happens in a real car is that the force is NOT the same at higher speeds, because you have to be in a higher gear. You had to trade some of the force in into distance, and so your force (and acceleration) will drop....but the engine ends up working just as hard. Likewise, stopping a car from 120 will heat up the brakes 4x more than stopping from 60. Interestingly, you also need to jump from the 4th floor to hit the ground 2x as fast as jumping from the 1st floor (ground floor being 0). But you were talking about objects in space. F=wd still applies in space, but it's a bit more complicated because you are probably using rockets, and a rocket takes it's mass with it. So, while it might be just as easy to accelerate a rocket moving at 10 as it is to accelerate an identical rocket moving at 20, remember that the faster moving rocket had to have something bringing it up to its speed in the first place. After all, they both started on earth from 0. which means that when standing on the launchpad, the faster rocket still needs to be much larger (4x, although probably more, I don't understand the rocket equation enough) than the slower one. You then might be tempted to ignore the fact that both rockets started at 0, and just look at their starting speeds as being "0". But then you'd be looking at two different frames of reference, and work is indeed different depending on which frame of reference you're looking at. That's essentially the problem with rockets, they need to bring their frame of reference with them wherever they go, and that gets heavy.
Thanks for the replies everyone! All of the replies were quite useful in regards to different aspects of the intuition. I'm going to spend some more time thinking about this and trying practice problems from my physics books and thought exercises until I've got a solid understanding. Action sounds interesting and I was even thinking earlier if Fd exists then maybe pd (momentum*distance) did too. Good to know. One more question: We have energy which is (force * distance) and action which is (momentum * distance), so does that means there's a third quantity (mass*position, or maybe mass*position*distance)? If it exists, what is it's name? It seems like it should be conserved also perhaps.
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http://mathhelpforum.com/pre-calculus/50135-domain-problem.html
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# Thread:
1. ## Domain Problem
What's the domain of G(t) = ln(t^4-1)?
How did you get the answer, the answer in the book says that it's (-infinity, -1) union (1, infinity) but I don't really understand why.
2. For $x \in \left[ { - 1,1} \right]\quad \Rightarrow \quad \ln \left( {x^4 - 1} \right) \mbox{ is meaningless!}$
3. Originally Posted by Plato
For $x \in \left[ { - 1,1} \right]\quad \Rightarrow \quad \ln \left( {x^4 - 1} \right) \mbox{ is meaningless!}$
What?
4. Originally Posted by dm10
What?
dn10, you have a very hard road of learning ahead of you if you truly want to understand this material.
You should pay more attention to basic definitions.
Of course, if you could care less then forget it.
5. $t^{4}-1=\left( t^{2}+1 \right)\left( t^{2}-1 \right)>0,$ this last 'cause logarithm is defined for numbers greater than zero. Hence, it remains to solve $t^2-1>0$ which gives the expected answer.
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http://math.stackexchange.com/questions/123910/let-p-be-a-prime-prove-that-p-divides-abpbap-for-all-integers-a-and/123914
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# Let $p$ be a prime. Prove that $p$ divides $ab^p−ba^p$ for all integers $a$ and $b$.
Let $p$ be a prime. Prove that $p$ divides $ab^p−ba^p$ for all integers $a$ and $b$.
(This is not a homework)
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Hint: It helps to think of it as $ab^p \equiv ba^p \mod p$. – Tara B Mar 24 '12 at 12:47
Oh, I see that while I was laboriously typing my comment on an iPad, people provided actual answers. Oh well. – Tara B Mar 24 '12 at 12:49
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The variant of Fermat's Theorem that says $x^p\equiv x\pmod{p}$ makes this automatic. – André Nicolas Mar 24 '12 at 16:54
## 4 Answers
KV Raman's answer is quite right, but I'll write my answer anyway because I find it tidier.
$\mathbb{Z}_p^* = \mathbb{Z}_p\setminus \{0\}$ is a cyclic group of order $p-1$ under multiplication, so $a^{p-1} \equiv 1 \mod p$ for all integers not divisible by $p$. If $a$ is divisible by $p$, then $a^n \equiv 0 \mod p$ for all $n$. So we have $a^p \equiv a \mod p$ for $a\in \mathbb{Z}$. (This is Fermat's Little Theorem.)
Now for any integers $a$ and $b$, $$ab^p \equiv ab \equiv a^pb \mod p$$ and so $ab^p - ba^p \equiv 0 \mod p$, which means $p$ divides $ab^p - ba^p$.
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You got my vote +1 for your proof Tara. – Kirthi Raman Mar 24 '12 at 13:19
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+1. That's much easier than arranging things such that the $a^{p-1}$ form of Fermat's little theorem applies directly. – Henning Makholm Mar 24 '12 at 13:21
$$ab^p-ba^p = ab(b^{p-1}-a^{p-1})$$
If $p|ab$, then $p|(ab^p-ba^p)$ and also if $p \nmid ab$, then gcd$(p,a)=$gcd$(p,b)=1, \Rightarrow b^{p-1} \equiv a^{p-1} \equiv 1\pmod{p}$ (by Fermat's little theorem).
This further implies that $\displaystyle{p|(b^{p-1}-a^{p-1}) \Rightarrow p|(ab^p-ba^p)}$.
Q.E.D.
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Hint $\$ little Fermat $\rm\Rightarrow mod\ p\!:\ n^p \equiv n\ \Rightarrow\ f(a,b^p)\equiv f(a,b)\equiv f(a^p,b)\$ for all $\rm\:f\in \mathbb Z[x,y]$
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@Downvoter Will the downvoter please explain? – user21436 Mar 30 '12 at 15:02
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http://mathoverflow.net/questions/82781/hodge-decomposition-of-a-symplectic-form
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## Hodge decomposition of a symplectic form.
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Can anyone explain to me what the Hodge decomposition form of a symplectic form in a special symplectic manifold looks like?
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What exactly are you asking? What does a special symplectic manifold mean? Do you mean a Kaehler manifold? The standard Hodge decomposition does not make sense unless you have a Riemannian metric. Do you mean some other, less common, Hodge decomposition? If you mean the usual Hodge decomposition on a Kaehler manifold, then the symplectic (Kaehler) form is already harmonic, so it does not decompose further. – Spiro Karigiannis Dec 6 2011 at 12:24
Mirjana, I guess you are referring to the notion of "special symplectic manifold" introduced by Alekseevsky et al as a variant of the more popular "special Kaehler" geometry. It would be good to edit your question to include a full definition and references, explaining how you know that such a complex manifold actually has a Hodge decomposition. The need for this level of detail is illustrated by the fact that Spiro, an expert on manifolds with special holonomy, does not understand what you are asking. – Tim Perutz Dec 6 2011 at 15:23
@Tim, "how you know that such a complex manifold actually has a Hodge decomposition" I thought that every symplectic form can be written as a direct sum of a closed, coclosed and harmonic form? – Mirjana Dec 7 2011 at 5:52
Mirjana: Ah, so I misunderstood too. But what metric do you want to use for the Hodge theory? Note that Prop. 4 of the paper of Alekseevsky et al is about the type decomposition w.r.t. $J$, not the Hodge decomposition. [You've been leaving comments, but please edit the question to make it clearer.] – Tim Perutz Dec 7 2011 at 14:00
@Tim In Proposition 4 it is explicitly written that it is the Hodge decomposition... – Mirjana Dec 7 2011 at 15:48
show 1 more comment
## 2 Answers
Using the additional information that the OP provided in the comments to Yael Fregier's answer, I can elaborate as follows:
I still don't know what "special complex manifold" means, but in any case, I will assume the following. If $(M, J, \nabla)$ is a complex manifold with a connection $\nabla$ coming from a metric $g$, (that is, $\nabla$ is the Levi-Civita connection of $g$), then we get an associated symplectic form $\omega(X,Y) = g(JX, Y)$, and $\omega$ is parallel with respect to $\nabla$ if and only if $J$ is parallel with respect to $\nabla$, if and only if $J$ is integrable and $\omega$ is closed. That is, $(M, g, J, \omega)$ is Kaehler. In this case, $\omega$ is harmonic, so its Hodge decomposition is $\omega = \omega \in \Delta_2$, where $\Delta_2$ is the space of harmonic $2$-forms on $M$, using the notation of the OP.
If $\nabla$ does not come from a metric, you still need some metric to define the co-derivative $d^* = \delta$ of $d$, and to define the Laplacian $\Delta$. One can indeed do this with a different connection $\nabla$, as long as you have a metric. But in this case it is not clear to me what the symplectic form $\omega$ is, and how it is related to $J$ and $\nabla$.
Added later: I think I just realized that the OP is not asking about the Hodge decomposition of the form $\omega$ in particular, just the "Hodge decomposition" for a "special symplectic manifold." There is a version of "symplectic Hodge theory." See, for example, these notes by Victor Guillemin: http://www-math.mit.edu/~vwg/shlomo-notes.pdf --- I don't know if this is the same thing mentioned in Yael Fregier's answer. Otherwise, I remain confused by the question.
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Thanks for the reference to Victor Guillemin's notes. I have taken a look, what I talked about corresponds to sections 4,5 and 8 of these notes (modulo taking duals). But he does more, in particular what I described was the linear version, whereas he also treats the global aspects in section 7. – yael fregier Dec 6 2011 at 22:55
I found the definition and the Hodge decomposition in this paper: Special complex manifolds, D.V Alekseevsky, V. Cortes, C. Devchand. And I get confused when I start to read the proof of Proposition 4. – Mirjana Dec 7 2011 at 0:31
### 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 agree with Spiro Karigiannis that the question would need some clarifications. I slightly modify the question by replacing the first "form" by "for". In this case I also agree with Spiro about the fact that a metric would be needed to speak of Hodge decomposition... Unless you refer to the Hodge-Lepage decomposition which has a meaning for a symplectic manifold without metric :
Given a symplectic vector space $(V,\omega)$, one can associate to it two operators $\omega^+$ and $\omega^-$ which act on the exterior algebra on $V^*$ by respectively left multiplying a given form $\alpha$ by $\omega$ (i.e. $\omega^+(\alpha)=\omega\wedge\alpha$) or by contracting $\alpha$ by the Poisson bivector $\pi$ associated to $\omega$ ($\omega^-(\alpha)=i_\pi(\alpha)$). These two operators satisfy the relations of the Lie algebra $sl(2)$ and they cut out the space of differential forms into irreducible $sl(2)$-modules which are also modules over the Lie algebra of symplectomorphisms since the operators $\omega^+$ and $\omega^-$ are invariant under the action of this Lie algebra. This decomposition is called the Hodge-Lepage decomposition, and the highest weight vectors are called effective forms. One can find all the details, and some explicit formulas in Darboux coordinates in chapter 5 of the book "Contact Geometry and Nonlinear Differential Equations" by Kushner, Lychagin and Roubtsov (encyclopedia of Mathematics and its Applications (No. 101)) at Cambridge University Press.
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You can't edit comments, so it's best to write it in a separate TeX editor, check that it looks good, and then paste it in. You can then delete your previous comments. – Spiro Karigiannis Dec 6 2011 at 14:21
Ok, here is the definition of special symplectic manifold: A special symplectic manifold (M,J,∇,ω)is a special complex manifold (M,J,\nabla) together with a \nabla−parallel symplectic structure \omega. And there is a theorem(Hodge decomposition)which asserts that \Omega^{k}=Im d_{k-1} + Im \delta_{k+1} + \Delta_{k},where \Omega^{k}is a set of all k−forms,d_{k}maps \Omega^{k} to Ω^{k+1}, δ^{k}=∗ d ∗ maps Ω^{k} to Ω^{k−1}, where ∗ is Hodge star operator. \Delta_{k} is the space of k-harmonic forms. – Mirjana Dec 6 2011 at 14:53
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http://mathoverflow.net/questions/24426?sort=oldest
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## What do people mean by “subcategory”?
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Mac Lane defines a subcategory as a subset of objects and a subset of morphisms that form a category. But the first rule of category theory is that you do not talk about equality of objects. Up to equivalence, the definition becomes a faithful functor. This is a useful concept, but I don't think it fits the name. I don't want groups to be a subcategory of sets!
This is not a question about aesthetics, but about usage. I don't think people tend to use Mac Lane's definition. Maybe they're just wrong, but I'd like to know if there is another definition which fits the usage better. All the time I see people say things like "we may assume that our subcategory contains every object isomorphic to an object of the subcategory." I guess we can expand to an equivalent subcategory to achieve this, but we probably have to choose how the objects are isomorphic (though it may be easier if to change the ambient category). This is a much more natural thing to do (and safer) if the subcategory is full, or at least contains all the automorphisms of its objects. This leads me to suspect that people are assuming or thinking of some stronger definition than faithful.
Do people tend to mean the official definition? or do they also require full? containing all the automorphisms? Are there other useful intermediate notions?
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"I don't want groups to be a subcategory of sets." Why not? (I do.) – Pete L. Clark May 12 2010 at 19:58
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Haha, it's even crazier than that: <Groups> is a subcategory of <Sets>, but also, <Sets> is a subcategory of <Groups>, via the "free group generated by X"--functor. – Xandi Tuni May 12 2010 at 20:08
The question "should Grp really be a subcategory of Set?" is a problem under either definition: with a bit of hacking, we can find a category equivalent to Grp as a Mac Lane subcat of Set. Define, say, an "ersatz group" to be a set X containing a unique element x (call it the "code" of X) such that x is a pair (X\{x},\mu), where \mu is a multiplication making X\{x} a group; and a map or these is a function preserving the code and giving a group homomorphism on the rest. I like your question, but I don't think the "Grp" example really motivates it as well as one might initially think. – Peter LeFanu Lumsdaine May 12 2010 at 21:34
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@Xandi: That's surely no crazier than the fact that $\mathbb{Q}$ "is" a subset of $\mathbb{N}$ in the categorical sense; it's just another illustration of how "$A$ is a sub-foo of $B$" is almost always a statement not just about $A$ and $B$ but also about an implicit inclusion map. :-P – Peter LeFanu Lumsdaine May 12 2010 at 22:41
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@Ben: I was confused. I don't want groups to be a subcategory of sets either. See my response below for clarification. – Pete L. Clark May 12 2010 at 23:35
show 1 more comment
## 6 Answers
Do you want the notion of "subcategory" to be invariant under categorical equivalence? If so, then "pseudomonic" functors are the right thing: faithful, and full on isomorphisms.
But I don't think one would want this any more than the notion of "inclusion" of topological spaces to be invariant under homotopy equivalence (which would make it meaningless).
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Absolutely agreed on the second paragraph! An example here of an interesting subcategory inclusion that is not full on isomorphisms: "differential manifolds; smooth maps" inside "differential manifolds; continuous maps". On the other hand, "pseudomonic" isn't the only notion of subcategory that's invariant under equivalence --- "faithful" is fine for that too, as are the other candidates discussed on the nlab at Andrew S's link. – Peter LeFanu Lumsdaine May 12 2010 at 22:35
Absolutely disagreed on the second paragraph! Requiring a categorical notion to be invariant under equivalence is really much more like requiring the notion of inclusion of topological spaces to be invariant under homeomorphism. Homotopy equivalence is a weaker notion of equivalence imposed on topological spaces when we use them as models for "homotopy types;" the "natural" notion of sameness for topological spaces is homeomorphism, just as the natural notion of sameness for categories is equivalence. – Mike Shulman May 13 2010 at 6:20
@Mike: we can probably argue endlessly on this one, but if you take the geometric realization of a category, equivalences go to homotopy equivalences and not to homeomorphisms. So it's not just a vague analogy I'm referring to here, it's an honest-to-god functor -- although it's of course true that geometric realization is by no means a subcategory inclusion... – Tilman May 13 2010 at 6:41
It's interesting that you're the only person to defend the official definition (maybe implicitly TJF). The second paragraph helped me: homotopy theorists find subspaces helpful, even if they are "evil." There are two useful notions of subcategory, so it's bad that people use the same word. I think you're right that pseudomonic is what people should mean if they want a categorical notion, but I'm not convinced it is what they mean. – Ben Wieland May 13 2010 at 16:34
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
(This is not my answer (as I didn't have anything to do with this page), hence the community-wiki tag.)
http://ncatlab.org/nlab/show/subcategory
In particular, you'll find some sympathy for the viewpoint that "groups" should not be a subcategory of "sets".
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There is some very nice discussion there, but I don't think in actual mathematical practice "subcategory" is used for any of those definitions besides the ones mentioned in this question. – Peter LeFanu Lumsdaine May 12 2010 at 21:40
Actually, I was probably overstating things in that last comment... most subcategories used in practice do fall under some of the other definitions discussed there, and it's very rare to see quantification over arbitrary subcategories, so I guess this point is often moot! – Peter LeFanu Lumsdaine May 12 2010 at 22:48
That's a great resource and I should have thought of looking there first, but, in the interest of closure, I'm choosing Tilman's definite answer. – Ben Wieland May 13 2010 at 16:24
Do people tend to mean the official definition?
I think "official" belongs in scare-quotes... I tend to think that "subcategory" is an evil notion. I'm not published anywhere, but in my notes I use "subcategory" to mean "a subset of $\operatorname{Hom}$, closed under $1_{-}$ and composition." Then you can suitably mimic Mac Lane's definition by specifying $$\operatorname{SubC}(A,X)=\operatorname{SubC}(X,A) = \left\{\begin{array}{c} \{1_A\} & A=X \\ \emptyset & A\neq X \end{array} \right.$$ for any objects $A$ you might want to ignore; they become isolated and trivial. That does feel a bit kludgy, though.
or do they also require full? containing all the automorphisms?
Probably not, in either case. That just looks weird to me... but what do I know?
Are there other useful intermediate notions?
Definitely.
Mac Lane's definition of subcategory given in the question corresponds to a functor which is strongly faithful in the sense that $F(g)=F(h)\implies g=h$ without hypotheses on the sources and targets of $g,h$; this strong property comes from the fact that functors don't generally have sensible image categories: what do you do if all objects are mapped to the same object!?
The ordinary sense of faithful is a slightly less strict condition: if $f,g:S\to T$ and $F(f)=F(g)$ then $f=g$.
Consider functors of groupoids $F:\mathcal{A}\to\mathcal{B}$. Every such functor factors in an essentially unique way as $$F = G_3 \circ G_2 \circ G_1$$ where $G_i$ omits only property $i$ among
1. Faithful
2. Full
3. Essentially Surjective
The same construction${}^1$ that gives this factorization makes good sense for general categories as well, although it's then complicated by other functor properties you might want to consider (reflects isomorphisms, reflects isomorphy, etc.). $G_3$ might be called "full objectwise-subcategory" (the reference calls it "forgets only property") and $G_2$ ("only structure"), I think, generalizes the notion I describe at the top of this answer.
(To be clear, "$G:\mathcal{X}\to \mathcal{Y}$ is essentially surjective" means that every morphism of $\mathcal{Y}$ factors as $i\circ G(\varphi) \circ j$ for isomorphisms $i,j$ in $\mathcal{Y}$. This implies a property of $G$ relative to objects which isn't worth spelling out here.)
${}^1$take a day to read this page
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I think in practice, most people would (if pressed) describe the definition they're using as something like the Mac Lane definition, plus being allowed to replace either category with another equivalent category [where "equivalent" means "with a chosen equivalence in mind"].
This is clearly not equivalent to the unmodified Mac Lane definition. (E.g. the "walking isomorphism" $I$ has a ML-subobject that is discrete on two objects, but $I$ is equivalent to to the terminal category $1$, which has no subcategory equivalent to a discrete cat. on two objects.)
This is equivalent to the definition "a subobject is a faithful functor". Precisely, given a category $\mathcal{C}$, we can define two 2-categories, $\mathrm{SubCat}(\mathcal{C})$ and $(\mathrm{Cat}/C)_\mathrm{faithful}$, with objects respectively
• an object of $(\mathrm{Cat}/C)_\mathrm{faithful}$ is a faithful functor into $\mathcal{C}$;
• an object of $\mathrm{SubCat}(\mathcal{C})$ is a chain $\mathcal{D} \simeq \mathcal{D}' \subseteq \mathcal{C}' \simeq \mathcal{C}$, where $\subseteq$ denotes the inclusion of a literal Mac Lane subcategory;
and an arrow is a triangle/"ladder-triangle", commutative up to specified natural isomorphism(s); and 2-cells in each case are natural isomorphisms commuting with everything in sight.
Then (if I'm not mistaken) these two 2-categories are 2-equivalent.
I think this usually what "subcategory" is used to mean in practice. "Repletification" --- making your subcategory "closed under isomorphisms" --- is then unproblematic: it just involves composing in an extra equivalence $D'' \simeq D'$ on the front, where an object of $D''$ is an object of $C$ together with a chosen isomorphism to some object of your original subcategory.
(As you point out, for literal Mac Lane subcategories this is problematic except when the subcategory is full on isomorphisms.)
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If I want a "full subcategory", I say simply say so.
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Upon request, I will clarify -- and partially take back! -- my earlier comment.
What I find completely unobjectionable is Mac Lane's definition of a subcategory $\mathcal{D}$ of a category $\mathcal{C}$ as being given by subclasses of objects and morphisms which forms a category under the induced composition. I don't see what else you would want a subcategory to be. I do agree that the notion of "subcategory" is not one of the more useful categorical concepts I know, and it even has some potential to be evil in the sense that modern categorists use the word. (Surely it would be more in the spirit of things to talk about a functor from $\mathcal{C}$ to $\mathcal{D}$ which satisfies certain "injectivity" properties.)
Now let's return to the statement "I don't want Groups to be a subcategory of Sets". In my comment I said that I did want this, but I don't now know why I said that: I think I must simply have been confused. Indeed, it is not obvious to me that this definition makes Groups a subcategory of Sets, at least not in any unique or benign way.
If you asked me to spell out the most evident categorical relationship between sets and groups, I would first of all point to the category NonEmptySets -- now that's a subcategory of Sets! -- and then the "forgetful" functor from Groups to NonEmptySets. This functor is (at least assuming the Axiom of Choice) surjective: every nonempty set is the underlying set of some group. But most sets can be endowed with a group law in multiple (usually nonisomorphic) ways, so this is not an "inclusion functor".
(Even the other way around, namely the free group functor from Sets to Groups seems not to quite make Sets into a subcategory of groups, because the class of sets is not a subclass of the class of groups.)
Maybe you are thinking of doing something tricky: defining a group to be an ordered pair [identifying ordered pairs with sets in one of the usual -- silly! -- ways] $(S,\circ)$ where $S$ is a set and $\circ$ is a subset of $S \times S \times S$ satisfying certain axioms. (Note that this is definitely incompatible with the above way of thinking about groups as having -- but not being -- an "underlying set".) But isn't this especially evil?
Comments more than welcome.
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I was thinking of modifying the ambient category, replacing NonEmptySets with the category whose objects are groups, but whose morphisms are maps of sets, ignoring the groups structure. It should be possible to define groups as special sets, but if you change the cardinality, as you and Peter LeFanu Lumsdaine (in the comments at top) do, this changes the functor from groups to sets. But I think you can stuff the multiplication table into the identity element. You just have to change the name of the identity in the table. But, mainly, I was just thinking thinking that Grps -> Sets is faithful. – Ben Wieland May 13 2010 at 0:14
yes, you're right --- my suggested coding did unintentionally change the functor. but as you say, this can be fixed --- you have to hack a little harder (to get around the fact that the multiplication table wants to talk about itself, which in the standard ZFC implementation it can't "literally" do) but yep, it does still work. – Peter LeFanu Lumsdaine May 13 2010 at 3:32
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http://stats.stackexchange.com/questions/34595/is-there-an-intuitive-interpretation-of-a-negative-variogram-nugget-value
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Is there an intuitive interpretation of a negative variogram “nugget” value?
A variogram plots the variance of the difference between sample pairs on a field (any dimensionality) against spatial separation (the "lag") of those samples.
The extrapolation from observed small-lag variances to a zero-lag variogram value is generally termed the "nugget" value.
Variograms I've come across before (geostatistics) always had positive "nuggets", and these values were generally interpreted as arising from (and being some sort of guide to the magnitude of) a combination of measurement error and sub-observation-scale variation in the observed field.
However, I'm just now looking at variograms of some imaging data and any reasonable extrapolation of the low-lag variogram clearly results in a negative "nugget" value. Of course it's always possible to "force" a non-negative nugget by, say, fitting an exponential model... but this doesn't look very convincing when plotted.
Is there any simple interpretation of negative nugget values (or a simple explanation of how they arise) as there is for positive ones ?
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2
The nugget actually is not the extrapolation to zero: it is the (necessarily nonnegative) jump at zero. When extrapolation of a fitted ("model") variogram based on the data is negative, this suggests that the spatial process is smooth, which for certain kinds of images might be no surprise, and you should be using a smooth model. This behavior also arises from data with finite (rather than point) support, which is usually the case with digital images, whose values are pixel averages. The true underlying variance is "smeared" out, frequently resulting in what you describe. – whuber♦ Sep 6 '12 at 17:11
1 Answer
The empirical variogram plots squared values, so I don't see how they can be negative. In the simple (Geostatistical) model:
$Y_i = S(x_i) + Z_i,$
where $Z_i \sim N(0,\tau^2)$ are assumed to be mutually independent, often assumed to represent measurement error. Here $\tau^2$ is the nugget variance, which is what your estimate from an empirical variogram gives, since for two measurements at the same location:
$Y_{i1} - Y_{i2} \sim N(0,2\tau^2)$
The variogram gives $\frac{1}{2} \text{Var}\{Y_{i1} - Y_{i2} \}$, which is $\tau^2$ (the nugget).
So the nugget is a variance, which means it can't be negative. The empirical variogram plots $\frac{1}{2}(Y_{i1} - Y_{i2})^2$, which is an estimate for the variance (since this is a zero-mean process), so an estimate for the nugget $\tau^2$, and again can't be negative.
So for your problem, if the variogram is showing negative values then it's probably something else masquerading as a variogram. And if it just looks like it might dip below zero, but doesn't actually, then it's a trick of the eye I think. Hope that helps.
(Some of this answer was taken from Section 2.5 of the book Model-based Geostatistics, by Diggle and Ribeiro (2000). Any mistakes though are mine not theirs!)
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http://mathhelpforum.com/advanced-algebra/60127-sysmetric-tridiagonal-matrix-print.html
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# sysmetric tridiagonal matrix
Printable View
• November 17th 2008, 04:57 PM
tinng
sysmetric tridiagonal matrix
For the following symmetric matrix A, find a symmetric tridiagonal matrix T, such that A and T have the same eigenvalues.
A=
[ 2 1 0 0
1 1 2 2
0 2 0 -1
0 2 -1 4]
• November 18th 2008, 10:49 AM
Opalg
Quote:
Originally Posted by tinng
For the following symmetric matrix A, find a symmetric tridiagonal matrix T, such that A and T have the same eigenvalues.
$A=\begin{bmatrix}2&1&0&0\\ 1&1&2&2\\ 0&2&0&-1\\ 0&2&-1&4\end{bmatrix}$
I suspect that you're meant to do this using Householder's method. Notice that the only things that stop A from being already tridiagonal are the 2s at the bottom of the second column and the end of the second row. We somehow have to get rid of them.
Start by splitting A into 2×2 blocks: $A = \begin{bmatrix}B&C^{\,\textsc{t}}\\C&D\end{bmatrix }$, where $B = \begin{bmatrix}2&1\\1&1\end{bmatrix}$, $C = \begin{bmatrix}0&2\\0&2\end{bmatrix}$, $D = \begin{bmatrix}0&-1\\-1&4\end{bmatrix}$.
Let Q be a 2×2 orthogonal matrix, and let P be the 4×4 matrix with 2×2-block decomposition $P = \begin{bmatrix}I&O\\O&Q\end{bmatrix}$, where I is the identity and the Os are zero matrices. Then P is orthogonal, and
$PAP^{\textsc{t}} = \begin{bmatrix}B&C^{\,\textsc{t}}Q^{\textsc{t}}\\ QC&QDQ^{\textsc{t}}\end{bmatrix}.\qquad(*)$
We need to choose Q so that the bottom right-hand element of QC is 0, or in other words $Q\begin{bmatrix}2\\2\end{bmatrix} = \begin{bmatrix}*\\0\end{bmatrix}$. In general, this can be done by Householder's method of using reflection matrices. In this case, there is also a straightforward geometric solution, taking Q to be the matrix representing a clockwise rotation through π/4. Then $Q = \begin{bmatrix}1/\sqrt2&1/\sqrt2\\ -1/\sqrt2&1/\sqrt2\end{bmatrix}$, and $Q\begin{bmatrix}2\\2\end{bmatrix} = \begin{bmatrix}2\sqrt2\\0\end{bmatrix}.$ Plug this into the equation (*), and you get $PAP^{\textsc{t}} = \begin{bmatrix}2&1&0&0\\ 1&1&2\sqrt2&0\\ 0&2\sqrt2&1&2\\ 0&0&2&3\end{bmatrix}.$
That matrix is visibly symmetric and tridiagonal, and it has the same eigenvalues as A because it is orthogonally equivalent to it.
All times are GMT -8. The time now is 09:12 AM.
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http://mathoverflow.net/questions/72800/every-real-function-has-a-dense-set-on-which-its-restriction-is-continuous
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## Every real function has a dense set on which its restriction is continuous
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The title says it all: if $f\colon \mathbb{R} \to \mathbb{R}$ is any real function, there exists a dense subset $D$ of $\mathbb{R}$ such that $f|_D$ is continuous.
Or so I'm told, but this leaves me stumped. Apart from the rather trivial fact that one can find a dense $D$ such that the graph of $f|_D$ has no isolated points (by a variant of Cantor-Bendixson), I don't know how to start. Is this a well-known fact?
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Thanks for asking. I feel more grown up now that I know about this theorem. – Richard Kent Aug 13 2011 at 0:58
@Cole → I don't think that's what you meant (maybe I misunderstood): $f$ itself might be continuous nowhere: Blumberg's theorem only promises continuity of the restriction $f|_D$. And even concerning continuity of $f|_D$, it's not true that one can strengthen the conclusion to $D$ being comeager (=residual): I believe there exists a real function which takes every real value on any $G_\delta$ which is dense in a non-trivial interval (along the lines of [Sierpiński](matwbn.icm.edu.pl/ksiazki/fm/fm1/fm1113.pdf), replacing closed sets of positive measure by dense $G_\delta$ on interval). – Gro-Tsen Aug 17 2011 at 15:20
Gro-Tsen, what I said is not what I thought I said, since I misunderstood the statement of Blumberg's theorem. I'll delete my comment so that it doesn't confuse anyone else. – Cole Leahy Aug 17 2011 at 16:35
## 1 Answer
It is a theorem due to Blumberg (New Properties of All Real Functions - Trans. AMS (1922)) and a topological space $X$ such that every real valued function admits a dense set on which it is continuous is sometimes called a Blumberg space.
Moreover, in Bredford & Goffman, Metric Spaces in which Blumberg's Theorem Holds, Proc. AMS (1960) you can find the proof that a metric space is Blumberg iff it's a Baire space.
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Blumberg's paper is online: ams.org/journals/tran/1922-024-02/… – Richard Kent Aug 13 2011 at 0:18
4
Thanks for the reference! For completeness, there's a self-contained and rather nice proof (or at least nicer than Blumberg's original one) of the statement, plus an extensive discussion, in the more general setting, in chapter 8 of Goffman, Nishiura, and Waterman's book Homeomorphisms in Analysis, available online at ams.org/publications/online-books/surv54-index – Gro-Tsen Aug 13 2011 at 23:09
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http://physics.stackexchange.com/questions/55227/how-many-orders-of-magnitude-in-energy-spans-the-domain-of-validity-of-the-stand
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How many orders of magnitude in energy spans the domain of validity of the Standard Model?
I am wondering if it makes sense to state that the upper limit is roughly 1012 eV (up to know the physics probed by the LHC seems to be pretty consistent with the SM) and the lower one is ... the upper bound for the photon mass (somewhere between 10-14 and 10-26 eV according to http://arxiv.org/pdf/hep-ph/0306245v2.pdf).
If I understand well the argument in this article, one could say the larger value 10-14 eV comes from the Standard Model of particles (the photon would acquire mass by the Higgs mechanism, its large-scale behavior beeing effectively Maxwellian) while the smaller value 10-26 eV would come from the standard model of cosmology (the value of the galactic field today leading to this value from the hypothesis of a "Proca regime for all scales" ??).
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Don't forget about the neutrino sector (!). It is also a popular belief that dark matter is described by some field(s) that interact weakly with the usual SM fields, but I couldn't tell you which mass scales are crucial. – Vibert Feb 26 at 23:22
The neutrino sector is very important of course but as far I understand the issue, and following my line of reasoning (I don't know if it is sound) - looking for upper limits for the unknown masses of very light particles - it seems that for neutrinos the upper limit is much larger than the photon (around 100 meV for an order of magnitude estimate ?). – laboussoleestmonpays Feb 28 at 9:23
You're right. Since we only measure differences in neutrino masses, giving solid bounds is difficult. The latest Particle Data Group booklet gives a bound of $m \lesssim 1$ eV. But this does imply that extrapolating the SM to $10^{-26}$ eV without including the neutrinos is a risky business ;) – Vibert Feb 28 at 12:36
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http://nrich.maths.org/6972
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### The Numbers Give the Design
Make new patterns from simple turning instructions. You can have a go using pencil and paper or with a floor robot.
### Straight to Curves
How can you make a curve from straight strips of paper?
### Colouring Curves Game
In this game, try not to colour two adjacent regions the same colour. Can you work out a strategy?
# Celtic Knot
##### Stage: 2 Challenge Level:
A Celtic knot pattern $4\times4$ looks like this. You can see how it is built up on a grid.
A Celtic knot pattern $8\times8$ looks like this on a grid.
A Celtic knot pattern $6\times6$ looks like this.
Can you build it up? There is a copy of the design to download here.
You can do it by making a set of cards to use on a $6\times6$ grid in a $24$ cm square with four copies of this sheet. There is also a second page with two copies of the design.
Or you could try doing it using this interactivity.
Full screen version
This text is usually replaced by the Flash movie.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://cs.stackexchange.com/questions/9964/how-to-construct-this-generalized-xor-without-needing-an-extra-vector
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# How to construct this generalized xor without needing an extra vector?
## Operator - Generalized Symmetric Difference
If you take binary xor and generalize it to other radices you can do so by the absolute value of the difference of each element in a radix vector. However this doesn't have the same properties as the binary symmetric differene. The reason is that in throwing away the "sign" of the difference we are unable to reconstruct an operaand given the result and the other as we can in binary xor. So we lose the nice property
ABA = B
However we keep other nice properties like
A0 = A
AA = 0
There is a way to maintain this property. However, as far as I can tell it involves emitting 3 vectors for any result. The first vector is the usual symmetric difference, the other two vectors are binary vectors of equal length to the first that record the sign of the result, one such vector for each order of the operands, on being the binary complement of the other. In this way, an original operand can be recovered, given the result and the other opernad, AND that other operands "sign" vector.
For example :
Say we have 2 base 10 vectors corresponding to the numbers 1137 and 9284, what i s the xor of these two numbers in base 10?
```` 7 3 1 1 4 8 2 9
4 8 2 9 7 3 1 1
Signed
Result 3 -5 -1 -8 -3 5 1 8
Sign
Vector 0 1 1 1 1 0 0 0
Symmetric
Difference 3 5 1 8
````
Recover 1137 given 8153 and 9284
````3 5 1 8
+ - - -
4 8 2 9
7 3 1 1
````
My question is : is there a better construction of generalized symmetric difference in any radix > 2 such that we don't need to 'remember the sign' ?
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## 1 Answer
The most common definition of `xor` is $a\oplus b=(a+b)\bmod 2$, on vertices applied to every coordinate seperately of course. In base-10 case, you have to introduce two operations: $a\oplus b=(a+b)\bmod 10$ and $a\ominus b=(a-b)\bmod 10$. (Notice that in base-2 case they coincide). Now you have
$$c=a\ominus b \quad\text{for coding}\quad\text{and}\quad a=c\oplus b \quad\text{for decoding.}$$
Actually, you can use $\ominus$ for both operations, just a bit cleverer:
$$c=b\ominus a \quad\text{for coding}\quad\text{and}\quad a=b\ominus c \quad\text{for decoding.}$$
This works because $a=b\ominus c=(b-c)\bmod 10=(b-(b\ominus a))\bmod 10$ $= (b-(b-a)\bmod 10)\bmod 10=a\bmod 10=a$.
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I like this. This is great. Much better. Can you elaborate a little bit on your thinking for how you derived this combined operator from the two operators, such as "why does 5 work?" Also is there anything possible for odd bases like ternary? – Cris Stringfellow Feb 20 at 13:12
@Cris I just found that any number does work in this case, therefore $0$, too ;) see the edit. – tohecz Feb 20 at 13:15
so what you are saying is that subtraction mod the base works as xor in any base? – Cris Stringfellow Feb 20 at 13:20
Yes. In sense of cryptography/coding, yes. Just be careful that $\ominus$ is not symmetrical. Still, given $b$ is a fixed code and defining $f_b(x)=b\ominus x$, we have that $f_b$ is an involution, i.e., $f_b(f_b(x))=x$ for all $x$. (Well, $f_b$ defined this way will be an involution is any abelian group, so why not in $\mathbb Z_{10}$?) – tohecz Feb 20 at 13:26
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http://mathhelpforum.com/differential-equations/196446-finding-separatrix.html
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# Thread:
1. ## Finding the separatrix
I have a pair of differential equations:
$\frac{dx}{dt} = x (3 - x - 2 y)$
$\frac{dy}{dt} = y (2 - x - y)$
We can see that there are equilibrium points at $(0,0), (0,3), (2,0), (1,1)$ which we can further determine are respectively unstable, stable, stable and a saddle point. Depending on the initial conditions, as $t \to \infty$, we expect the system to settle at one of the stable points (i.e. x or y goes 'extinct'). The separatrix is the curve which will determine the end behavior depending on which side of it the initial conditions lie. It should pass through the origin and the saddle point. I am wondering if it is possible to find the exact curve? I have tried a lot of things at this point and am beginning to think that it might not always be possible. I've attained a good approximation numerically, but would love to be able to find the exact curve so if anyone knows how to do so or if it's not possible I'd greatly appreciate the help!
Thanks
2. ## Re: Finding the separatrix
At this point I'm beginning to feel like my efforts are futile. It may just be the case that the exact curve cannot be found and my numerical approximation will have to cut it. A bit of a letdown in a way, but also learned a lot making the effort
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http://mathoverflow.net/questions/53767/constructions-unique-up-to-non-unique-isomorphism/53781
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## Constructions unique up to non-unique isomorphism
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
1) Fields have algebraic closures unique up to a non-unique isomorphism.
2) Nice spaces (without base point) have universal covering spaces unique up to a non-unique isomorphism.
3) Modules have injective hulls unique up to a non-unique isomorphism.
Such situations can lead to interesting groups - the absolute Galois group, the fundamental group, and the "Galois" groups of modules introduced by Sylvia Wiegand in Can. J. Math., Vol. XXIV, No. 4, 1972, pp. 573-579.
I'd appreciate any insight into the abstract features of situations which give rise to this type of phenomenon. And I'd appreciate as many examples from as many parts of mathematics as possible.
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It is possible to functorially define the universal covering space of a non-pointed space X: take the diagram $\Pi_1(X) \to Top$ sending a point $x\in X$ to the (pretty well unique) covering space one gets from the pointed space $(X,x)$, and a homotopy class of paths to the map between covering spaces this induces. The colimit of this diagram is the universal covering space of $X$, and this time this is functorial wrt maps $X\to Y$. I learned this from Todd Trimble, I think. – David Roberts Jan 30 2011 at 9:59
6
@ David: "the colimit of this diagram is the universal covering space of X", are you sure this is correct? What you described is essentially the groupoid of universal covering spaces and covering space maps inside Top and you are taking the colimit inside Top. This is the same as taking a single covering space and taking the quotient by all covering automorphisms, so you just get X back. – Chris Schommer-Pries Jan 30 2011 at 14:46
2
Doesn't this always happen when the construction has not-trivial automorphisms? – Nick S Jan 30 2011 at 18:36
1
@ David - this coequalizer still just gives B back. Try an explicit example like B = circle. The point is that there are many paths and so the different universal covers are identified with eachother in more than one way. This forces you to take a quotient of the universal covers which is too small (namely B itself) and no longer a universal cover. You can see that this has to be the case because the paths from b to b (up to homotopy) are just $pi_1$ and so this colimit factors through the quotient by the action of $pi_1$. Agreed? Where in the n-lab is this written? – Chris Schommer-Pries Jan 31 2011 at 18:57
5
There cannot be any such construction. The group of homeomorphisms from the circle to itself has no compatible action on the (or should I say "a") universal covering space. – Tom Goodwillie May 17 2011 at 4:10
show 1 more comment
## 12 Answers
The first two examples can be described more or less uniformly. Associated to a field $F$ is the category $C_F$ of algebraic field extensions of $F$ (whose objects are morphisms $F \to E$ and whose morphisms are commutative triangles). This category has a weak terminal object given by any algebraic closure $F \to \bar{F}$. The full subcategory on the algebraic closures is what one might call the absolute Galois groupoid of $F$ (which is a perfectly canonical construction), and choosing an object in this groupoid (which is not) gives the absolute Galois group.
Similarly, associated to a nice space $X$ is the category $C_X$ of connected covers of $X$ (whose objects are covering maps $Y \to X$ and whose morphisms are commutative triangles). This category has a weak initial object given by any universal cover $\bar{X} \to X$. The full subcategory on the universal covers is (equivalent to?) the fundamental groupoid of $X$ (again, a perfectly canonical construction), and choosing an object in this groupoid (which is not) gives the fundamental group.
So you will get this kind of behavior in any situation where you have a weak universal object instead of a universal one. (This partially covers the third example, since injectivity is also a weak universal property.) A general way to engineer a situation similar to the above two might be to look at something like the category of (epi?)morphisms into an object or (mono?)morphisms out of it in your favorite category and see what happens.
In any case, if you are only interested in these constructions because they produce interesting groups, then I think nowadays the modern thing to do is to produce interesting groups using Tannaka-Krein duality.
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Nice observation! In fact, the Galois group of modules alluded to in example 3) is also of this type: it's the automorphism group of the weak initial object given by the injective hull inside the category of embeddings (monomorphisms) $M \to I$ into an injective, so it's analogous to example 1) and dual to 2). – Theo Buehler Jan 30 2011 at 14:18
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Not to mention it's also the weak terminal object in the category of essential extensions of M (with embeddings for the maps). – Harry Altman Jan 30 2011 at 14:27
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"You will get this kind of behavior in any situation where you have a weak universal object instead of a universal one" --- this is so only when the weak universal objects are all isomorphic, no? – Steven Landsburg Jan 30 2011 at 15:00
@Steven: I guess that would give a groupoid which is not connected, but it's still possible to restrict attention to each of its connected components. – Qiaochu Yuan Jan 30 2011 at 15:17
I don't understand your last comment. It seems to me that this is due to the fact that after all I'm not so sure what the good definition of "weak initial/terminal" object is, anyway. – Theo Buehler Jan 30 2011 at 15:58
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Any two injective resolutions (of an object in an abelian category) are homotopy equivalent, but this homotopy equivalence is not unique. This is of course because the lifting property in the definition of "injective" does not require any uniqueness.
The connected sum of oriented manifolds is unique up to homeomorphism, but this homeomorphism is not unique.
A bit silly, but: In a short exact sequence $0 \to A \to B \to C \to 0$ in a semisimple abelian category $B$ is unique up to isomorphism (namely, $B \cong A \oplus C$), but the isomorphism is not unique.
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But the equivalence of injective resolutions is unique in the correct higher sense, i.e. the space of choices is contractible. – Thomas Nikolaus May 17 2011 at 7:31
@Thomas, is that indeed true? I've never thought of it but it looks like an interesting question. What do you mean by 'space of choices' in this case? – Fernando Muro Feb 10 2012 at 12:22
For a field $k$ and a natural number $n$, the vector space of dimension $n$ over $k$ is unique up to a non-unique isomorphism, though this somehow feels "less unique" to me than your other examples. I thought at first that this might be due to its not fitting into the class of examples described by Qiaochu, but I suppose you can force it into that class by considering the category of $n$-dimensional vector spaces over $k$. But that in turn feels considerably more ad hoc (at least to me) than considering the category of algebraic field extensions.
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Except when the field is $\mathbf{F}_2$ and the dimensions is $1$:) – Chandan Singh Dalawat Jan 30 2011 at 15:52
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Or any field and the dimension is 0 ;-)) – Johannes Hahn Jan 31 2011 at 0:15
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I attended a talk about some problem in the classification of—well, of something to do with finite fields, I don't remember—at which a colleague was moved to ask “What happens here if $\lambda \ne \mu$?” “Oh, yes,” said the speaker, as if discounting a trivial special case, “the results don't work if there are more than 2 non-$0$ scalars in the field.” – L Spice May 17 2011 at 5:03
My favourite: the mapping cone of a morphism in a triangulated category is unique up to non-unique isomorphism. This fact has originated a lot of research in this topic, and it still does.
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In recent work in set theory the concept of "canonical structure" has emerged, in connection with combinatorial work on pcf theory. The idea is that there are many constructions that depend on the axiom of choice but, once realized, are actually independent of the specific choices made. Usually, this involves two steps: You construct an object, which is not quite canonical (say, a collection of subsets of a cardinal $\kappa$), but then you recognize that there is a natural ideal (say, the non-stationary ideal on $\kappa$) and the corresponding equivalence classes are canonical. Of course, by switching to a new model of set theory, the "canonical structure" may change, so sometimes one thinks of it as a sort of invariant of the models.
The first papers that explicitly mentioned the name "canonical structure" are by Cummings, Foreman, and Magidor, "Canonical structures in the universe of set theory", Parts I and II, Annals of Pure and Applied Logic 129 (2004), 211-243, and 142 (2006), 55-75.
The following quote is from the beginning of the introduction to Part I:
It is a distinguishing feature of modern set theory that many of the most interesting questions are not decided by ZFC, the theory in which we profess to work; to put it another way, ZFC admits a large variety of models. A natural response to this is to identify invariants which may take different values in different models, and which codify a large amount of information about a model.
Of particular interest are invariants which are canonical, in the sense that the Axiom of Choice is needed to show that they exist, but once shown to exist they are independent of the choices made. For example the uncountable regular cardinals are canonical in this sense.
Shelah discovered a large class of canonical invariants, the study of which he labeled PCF theory. These invariants include two which are central in this paper; Shelah [24, 26] (under some mild cardinal arithmetic assumptions on the singular cardinal $\mu$) defined two stationary subsets of $\mu^+$, the sets of good and approachable points. The definitions of these sets appear to depend on certain arbitrary choices, but (modulo the club filter) are in fact independent of these choices. Other canonical structures we study in this paper include the stationary sets of tight and internally approachable structures, and the collection of good points on a scale.
The two references cited in the quote are S. Shelah, "On successors of singular cardinals", in M. Boffa, D. van Dalen, and K. McAloon, editors, Logic Colloquium ’78, pages 357–380, Amsterdam, 1979. North-Holland; and S. Shelah, "Cardinal Arithmetic". Oxford University Press, Oxford, 1994.
Besides the ongoing work by Cummings-Foreman-Magidor and Shelah, these ideas have been extended by others; Krueger and Ishiu come to mind.
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Sounds like a case for 'the': ncatlab.org/nlab/show/the – David Roberts Jan 30 2011 at 20:47
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Ok, @David, I'm curious. What do you mean? – Andres Caicedo Jan 31 2011 at 0:47
@Andres: What David is referring to is the fact that universal constructions in category theory have this property, and that this is only a new feature of set theory. For instance, suppose we want to make the product with an object $S$ a functor (i.e. the functor $(-)\times S$. However, while the product is unique up to unique isomorphism, we have to choose a representative of each isomorphism class as well as the connecting morphisms between them. This requires choice or global choice, but after we choose a specific representative of the functor usign choice, it is unique up to unique iso. – Harry Gindi Jan 31 2011 at 9:16
So when I said above "the functor $(-)\times S$", I was abusing language, since any given construction of "the" functor is only "a functor $(-)\times S$". This might seem like we're being overly cautious, but when we move from isomorphism of objects to equivalence of objects (in a bicategory), this makes some difference. This can be resolved by using Mac Lane's coherence theorem for bicategories, but, when we move up to tricategories, such a coherence theorem is proven not to exist. – Harry Gindi Jan 31 2011 at 9:21
I see. Thanks, @Harry! – Andres Caicedo Jan 31 2011 at 14:11
The homology of a differential graded algebra has an $A_\infty$-algebra structure which is unique up to non-unique isomorphism.
See Keller's nice expository paper, for instance. In particular, he states this result in Section 3.3 (as a theorem due to Kadeishvili, among others). It is stated there as a result about the homology of an $A_\infty$-algebra, but any differential graded algebra may be viewed as an $A_\infty$-algebra.
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Hi John, Thanks, I wish you would say more people or give a reference at least. – David Feldman Jan 30 2011 at 23:22
@David: I added a reference. – John Palmieri Jan 30 2011 at 23:36
Here are some examples that are less of an algebraic nature (but all seem to be subsumed by Qiaochu's observation in that they are "weakly initial" or "weakly terminal" objects in appropriate categories):
Consider the categories of metric spaces or complete metric spaces and $1$-Lipschitz maps. Isbell has shown that in these categories there are injective hulls, unique up to non-unique isomorphism. A metric space $I$ is injective if for every isometric embedding $A \to B$ and every $1$-Lipschitz map $A \to I$ there exists a $1$-Lipschitz extension $B \to I$. The automorphism groups of the injective hull of a space seems exceedingly hard to determine (even for finite spaces) but there's one case I find interesting. If $M$ happens to be a (real) Banach space and $I(M)$ is its injective hull then $I(M)$ is a Banach space, uniquely determined up to unique linear isometry, and it is of the form $C(K)$ where $K$ is an extremally disconnected Hausdorff space. H. Elton Lacey and co-authors have given a complete (finite!) list of possible injective hulls of separable Banach spaces.
Closely related are projective covers in the category of compact Hausdorff spaces and continuous maps. There, the projectives are precisely the extremally disconnected spaces (Gleason).
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I should add that there is an abstract version of injective hulls/projective covers considered in Adamek-Rosicky, Locally presentable and accessible categories, that seems to subsume all the examples given so far (except maybe the example involving bases). – Theo Buehler Jan 30 2011 at 14:48
A compact connected semisimple Lie Group $G$ has an essentially unique maximal torus $T$, a maximal abelian subgroup of maximum dimension (the rank of $G$ is the dimension of this torus). Although $G$ has lots of such torii (in fact any element of $G$ is contained in at least one), any two are conjugate to one another by some element of $G$.
In a similar vein, one can break a given maximal torus $T$ up into congruent pieces (the images of Weyl chambers under the exponential map applied to the Lie algebra of $T$), any two of which are equivalent to one another by an element of the Weyl Group of $G$. The value of any class function on $G$ is then completely determined on all of $G$ by its values on a single one of these pieces.
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The countable dense linear order is unique up to a non-unique isomorphism. The countable atomless Boolean algebra is unique up to a non-unique isomorphism. The random graph is unique up to a non-unique isomorphism. (Algebraically closed fields of a given characteristic and transcendence degree, and vector spaces of given dimension over a given field, are other examples already mentioned above.)
In general, if $T$ is a $\kappa$-categorical first-order theory (in a countable language), then the model $M$ of $T$ of cardinality $\kappa$ is unique up to a non-unique isomorphism.
Even more generally: if $T$ is any complete theory and $\kappa$ an infinite cardinal, then the saturated model $M$ of $T$ of cardinality $\kappa$—if it exists at all—is unique up to a non-unique isomorphism. ($M$ is unique by a standard back-and-forth argument. Non-uniqueness of the isomorphism amounts to saying that $\operatorname{Aut}(M)$ is nontrivial. By homogeneity, it suffices to exhibit two elements of $M$ with the same type. If there exists a nonprincipal parameter-free $1$-type, we can easily find two elements that realize it. If all $1$-types are principal, there are only finitely many, hence two elements of $M$ have to realize the same type by the pigeonhole principle.)
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Let me mention Sullivan's minimal models.
Every commutative differential graded $\mathbb{Q}$-algebra (cdga) `$A^*$` concentrated in non-negative degrees and such that `$H^0(A^*)=\mathbb{Q}$` admits a minimal Sullivan model `$i:M^*\to A^*$` where `$M^*$` is a free commutative graded algebra obtained from $\mathbb{Q}$ by adding generators of non-negative degrees so that the differential of each generator is a $\mathbb{Q}$-linear combination of products of length $\geq 2$ of the previous generators, and $i$ is a map of cgda's that induces a cohomology isomorphism (i.e., a quasi-isomorphism).
The minimal model is unique up to a non-unique isomorphism. More generally, if `$f:A^*\to B^*$` is a map of cdga's and `$j:N^*\to B^*$` is a minimal model of `$B^*$`, then there is a cdga map `$g:M^*\to N^*$`, defined up to cdga homotopy, such that $fi=gj$ up to cdga homotopy; moreover, if $f$ is a quasi-isomorphism, then $g$ is an isomorphism.
This reduces the classification of non-negative cdga's up to quasi-isomorphism (and as a consequence, the classification of simply connected topological spaces up to rational homotopy) to the classification of algebras of a certain kind up to isomorphism.
Of course, this example is similar to some mentioned before (in a sense it is the commutative analog of the answer of John Palmieri).
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Vector spaces have a basis that is unique up to a non-unique isomorphism.
Hilbert spaces have an orthonormal basis that is unique up to a non-unique unitary.
(At least, if you accept Zorn's lemma, i.e. the axiom of choice)
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Wait, how do vector spaces have a unique basis? If $V = \langle x, y\rangle$, then $x + y, y$ is also a basis of $V$, but it is not the same basis. – Simon Rose Jan 30 2011 at 16:04
@Simon: but there is an linear isomorphism that maps the former basis to the latter. So indeed a basis is not unique, but unique up to an isomorphism. Finally, this isomorphism itself is not unique, as it can for example be composed with any isomorphism that leaves the latter basis invariant. – Chris Heunen Jan 30 2011 at 16:14
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@Chris: We might be splitting hairs, but I think that this misses the spirit of the question. There is some notion of canonicality which defines an algebraic closure, for example, but no such notion that defines a basis of a vector space. – Simon Rose Jan 30 2011 at 16:23
@Simon: I agree this might only answer the letter of the question and perhaps not its spirit. Neverthelesss, OP asked for as many examples from as many parts of mathematics as possible, and this is one. Moreover, these isomorphisms certainly form interesting groups, namely $SL(n)$ and $U(n)$. Anyway, if one is after insight into the abstract features of such situations, isn't it important to also consider examples that fall outside one's initial intuition? – Chris Heunen Jan 30 2011 at 17:29
I do think an example lurks here, but you haven't nailed it. You don't want a basis, you want only that for which you thought you wanted a basis, namely rigidification. So for one given field and a given dimension fix a model vector space $M$ of that dimension. Define a rigidification of any given vectors space (over the same field, with the same dimension) as in isomorphism $i:V\rightarrow M$. An isomorphism of rigidified vector spaces is a map $j:M\rightarrow M$ that makes the triangle commute. Now you get $GL(n)$ (and other groups if you force more structure on $M$.) – David Feldman Jan 31 2011 at 5:00
One example that springs to mind is when you are secretly working with the objects of a higher category, and so the choice is not unique up to a unique isomorphism, but the choice of isomorphism is also subject to higher coherence data. In a 2-category this would mean the isomorphisms are unique up to a unique invertible 2-arrow and so on. In $\omega$-categories, you may have such coherence all the way to infinity, and so end up with no uniqueness after all. One place where this emerges is when your $\omega$-category has all duals - is then an $\omega$-groupoid, because all the duals make everything weakly invertible!
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http://physics.stackexchange.com/questions/16526/can-a-photon-passing-by-an-open-space-barycenter-of-a-system-of-masses-be-model
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Can a photon passing by an open space barycenter of a system of masses be modeled as if all the system's mass is at the barycenter?
To be clear, this example can't apply to the Solar System, since the barycenter is within the Sun, similarly the Earth/Moon system's barycenter is within the Earth.
But, given a system of gravitationally attacting masses revolving about a barycenter that is not contained within any of the masses, i.e. a barycenter in open space, would a photon passing near that point behave in the same manner as a photon passing a point mass of the same mass as the system itself?
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– Keith Thompson Nov 18 '11 at 8:37
I meant apply to the Solar System as a whole, which Wikipedia told me was inside the sun. – user1003469 Nov 21 '11 at 19:39
2 Answers
The barycenter has no mass and therefore no forces emanating. This is evident by your example of the moon earth barycenter which continually moves in the mantle 1700 km down or so. If it had any effect it would be working as a whip in cream, generating from quakes to volcanoes!
It is just a geometrical point whose use is to give an observer outside the system a reference point for calculations at large distances from the system.
The photon, or any other particle, will feel the gravity forces of the individual masses according to the laws of gravitation, the distances it has from the individual masses. BTW the forces felt by the photon will be very short and transient.
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Yes, I certainly understand that. I'm not sure that answers the question about the modeling, however. A photon passing relatively near the earth can be accurately modeled as if the earth was a point source. A photon passing relatively near the outer solar system should, at some sufficient distance, be successfully modeled as if it were passing a point source at the solar system's barycenter. My question was whether the collective influences of the multiple bodies in a system having an open space barycenter would have the same effect as passing a point source (neglecting motion of the system). – user1003469 Nov 17 '11 at 18:22
So just to be clear, your actual answer to the question would be 'No', correct? It would not behave the same? – user1003469 Nov 17 '11 at 18:24
It is no for the case of of the photon passing by the barycenter through the solar system, and yes if you are at a great distance so that individual constituents of the conglomerate mass are not discernible. In the intermediate range, i.e. outside the solar system but not extremely far, it could be used as a first order approximation, the closer, the worse. – anna v Nov 17 '11 at 20:02
I should clarify that "solar system" above means a hypothetical one, when the barycenter is in the vacuum as the question asks. – anna v Nov 18 '11 at 4:44
@user1003469: The Earth is a special case. Any spherically uniform body is gravitationally equivalent, for any object outside the surface of the sphere, to a point source at the sphere's center. – Keith Thompson Nov 18 '11 at 8:27
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The answer is no, but for the weak-field, slow moving, Newtonian limit, the far away light can be thought of as deflected by the mass as concentrated in two dimensions to the Barycenter, when the mass distribution has a spherically symmetric 2d projection.
You can see that The deflection cannot only depend on the Barycenter using two orbiting two black holes. Then the light which is off-center can swing by the black holes and get deflected an enormous amount, while light going through the center mostly goes straight through.
But in the weak field limit, the deflection of light as it passes through a gravitating region is small, and the angular deflection is given by the sum of
$$\Delta\theta = \sum_i {4GM_i\over c^2R_i}$$
Where $M_i$ is the i-th mass, and R_i is the impact parameter--- the distance of closest approach of the light to the mass. All but the factor of 2 in the formula is simple to derive, and a quick run through is contained in this answer, which has a detailed description of the weak-field bending of light by gravity: How does gravitational lensing account for Einstein's Cross? .
The upshot of the weak field limit is that light is deflected by a collection of nearby compact masses as if all the mass were squashed in a two dimensional sheet perpendicular to the direction of motion, and the light felt an impulsive force equal to the two-dimensional gravitational potential that would be produced by this source.
This means that if light is going in the z-direction, and it passes by a blob of matter, the deflection is according to the projection of that blob of matter to the x-y plane. If this blob of matter is spherically symmetric, a light ray passing through the center is undeflected. The precise deflection at any distance goes as the solution to the 2d Laplace equation, and it can be solved exactly in many circumstances. The linked answer works out many cases.
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http://physics.stackexchange.com/questions/7864/what-am-i-doing-wrong-when-i-try-to-make-this-unit-conversion/7892
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# What am I doing wrong when I try to make this unit conversion?
I am trying to reconcile data that I have found in one publication (Allen 1969) with data that I found in another publication (George 2003) that synthesized this data. The data is root respiration rate, it was originally measured at $27\ ^\circ C$.
## Approach
I am trying to convert a rate of oxygen consumed as volume per mass of root per time to carbon dioxide produced as mass per unit mass per time.
In the appendix table, George 2003 reports the range of root respiration rates, converted to $15\ ^\circ C$ and standard units:
$$[11.26, 22.52] \frac{\mathrm{nmol CO}_2}{\mathrm{g}\ \mathrm{s}}$$
In the original publication Allen (1969), root respiration was measured at $27\ ^\circ$C. The values can be found in table 3 and figure 2. The data include a minimum (Group 2 Brunswick, NJ plants) and a maximum (Group 3 Newbery, South Carolina), which I assume are the ones used by George 2003:
$$[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$
## Step 1
Transformed George 2003 measurements back to the measurement temperature using a rearrangement of equation 1 from George, the standardized temperature of $15\ ^\circ$C stated in the Georgeh table legend, and Q$_{10} = 2.075$ from George 2003, and the measurement temperature of $27\ ^\circ$C reported by Allen 1969:
$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$
$$[11.26, 22.52] * exp(log(2.075)*(27 - 15)/10)$$
Now we have the values that we would have expected to find in the Allen paper, except that the units need to be converted back to the original:
$$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$
## Step 2: convert the units
### Required constants:
• inverse density of $\mathrm{O}_2$ at $27^\circ C$: $\frac{7.69 \times 10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$ first assume that Allen converted to sea level pressure (101 kPa), although maybe they were measured at elevation (Allen may have worked at \~{} 900 kPa near Brevard, NC)
• molar mass of $\mathrm{O}_2$: $\frac{32\mathrm{g}\ \mathrm{O}_2}{\mathrm{mol}}$
• treat 10mg, which is in the unit of root mass used by Allen, as a unit of measurement for simplicity
Now convert $$[27.03,54.07] \mathrm{nmol CO}_2\ \mathrm{g}^{-1}\mathrm{s}^{-1}$$ to units of $\frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{root}\ \mathrm{h}}$. The expected result is the original values reported by Allen: $[27.2, 56.2] \frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$
$$[27.03, 54.07]\ \frac{\mathrm{nmol}\ \mathrm{CO}_2}{\mathrm{g}\ \mathrm{root}\ \mathrm{s}} \times \frac{1\ \mathrm{g}}{100\times10\mathrm{mg}} \times \frac{3600\ \mathrm{s}}{\mathrm{h}} \times \frac{3.2 \times 10^{-8}\ \mathrm{g}\ \mathrm{O}_2}{\mathrm{nmol}\ \mathrm{O}_2}\times \frac{7.69\times10^5\ \mu\mathrm{L}\ \mathrm{O}_2}{\mathrm{g}\ \mathrm{O}_2}$$
## Result:
$$[23.8, 47.8] \frac{\mu\mathrm{L}\ \textrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$
These are the units reported in the Allen paper, but they appear to be underestimates . Since the ratio of observed:expected values are different, it is not likely that Q$_{10}$ or the atmospheric pressure at time of measurement would explain this error.
## Question
Am I doing something wrong?
• Reference 1: Allen, 1969, Racial variation in physiological characteristics of shortleaf pine roots., Silvics Genetics 18:40-43
• Reference 2: George et al 2003, Fine-Root Respiration in a Loblolly Pine and Sweetgum Forest Growing in Elevated CO2. New Phytologist, 160:511-522
Footnote 1: The values from reference 2 are adjusted from the $15^\circ C$ reference temperature to the $27^\circ C$ in reference 1 using the Ahhrenius equation, but I am off by an order of magnitude so I do not think that this is relevant:
$$R_T = R_{15}[\exp(\ln(Q_{10})(T- 15))/10]$$
$$[26.9, 54.0] = [11.2, 22.5] * exp(log(2.075)*(27 - 15)/10)$$
note: I have been updating the equation based errors pointed out by Mark and rcollyer, but the problem remains
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Parts of this don't make sense. The $1000g/g$ is probably a typo, but how could the final answer have grams divided by kilograms? – Mark Eichenlaub Apr 1 '11 at 6:41
@Mark the answer is grams of CO$_2$ respired per kg root per second. The 1000g/kg is a typo – David Apr 1 '11 at 14:14
where does "root" come from? – rcollyer Apr 3 '11 at 2:22
@rcollyer it is the plant root that is consuming O2 or producing the CO2. I put it in following Marks question above, but I will take it out since it seems to magically appear in the answer, and the answer can be found here. – David Apr 3 '11 at 15:43
## 3 Answers
As far as I can tell, you went wrong in a couple of places:
1. instead of converting from $\mathrm{g}$ to $\mathrm{mg}$ your converting to $\mathrm{kg}$ in your first term, and
2. in your fourth term, the units should be $\mathrm{g\, O_2}$ and $\mathrm{ nmol\, O_2}$, for consistency. This would've let you know that you've inverted your conversion from $\mathrm{ nmol\, CO_2}$ to $\mathrm{ nmol\, O_2}$ in your third term.
When I'm performing unit conversions, I do a couple of things to make my life easier. First, like it is displayed in your question, I write everything on two lines so that I can easily distinguish between the numerator and denominator, but I doubt that is at issue here. Second, like you, I tend to do them all at once, but occasionally, like this case, it may be worthwhile to do it piecemeal. Last, I convert systematically, working from left to right. For instance, in this case, I'd convert from $\mathrm{ nmol\, CO_2}$ to $\mathrm{\mu L\, O_2}$ first, and then add in the conversions from $\mathrm{g}$ to $\mathrm{mg}$ and $\mathrm{s}$ to $\mathrm{h}$ after that.
Edit: in your most recent revision, you've removed you conversion from $\mathrm{CO_2}$ to $\mathrm{O_2}$, so the units you get are
$$\frac{ \mu L\, \mathrm{CO_2}}{10 g h}$$
If I remember correctly, the conversion was a power of 2, so it will not fix the deviation between the two units.
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– rcollyer Apr 8 '11 at 19:35
After making the corrections suggested by rcollyer, you were very close. Looking back to the original paper, note that the units were reported in the unexpcted units of $$\frac{\mu\mathrm{L}\ \mathrm{O}_2}{10\mathrm{mg}\ \mathrm{h}}$$
There is a 10 in the denominator, so multiply your answer by 10 and your calculations are pretty close.
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There are a few problems with this formula and calculation. I shall present those that I have found below and hopefully this will help.
First we note that the intended conversion ratios from the data: old = [27.2,56.2]; new = [11.26,22.52] give a conversion factor between new/old = [0.414,0.401]. So perhaps because of rounding errors one is seeking a conversion factor between these two values, of perhaps around 0.41.
This conversion is done in two logically distinct steps: Unit Conversion (U say) - your step 2- and Temperature Conversion (R say) - your step 1. Both of these conversions contain issues.
Step 1 - Temperature Conversion.
The problem here is not so much the formula (as far as I know), but the assumption of $27^0$C. The Allen paper does do plant tests at $27^0$C and also $30^0$C, $35^0$C and $40^0$C. So which value to choose? The main issue for me however is figure 2 in that paper where the data (from table 3) are plotted against mean annual temperature in degrees Fahrenheit ie the true natural ground temperature. This range is $53^0$F to $67^0$F ie [11.6,19.4] Centrigrade range with median at $15.5^0$C. So it is possible that George has taken $15.5^0$C - which gives an R value of 1.04 or even just $15^0$C giving an R value of 1. If that were true it would largely cancel out the error you have found.
Step 2 - Unit Conversion.
There are some problems here however which complicates things.
First the density of $O_2$ would need to change to that at (say) $15^0$C. Now the figure quoted there is not the density but the inverse density (and there is a typo in the microunit). Using the ideal gas law the inverse density would change to 7.37 microunits from your value of 7.69 microunits - a decrease in this term.
Another problem are the errors in the "respiration" formula. This formula does not balance on the Oxygen: LHS=4, RHS=6. Furthermore the LHS chemical is Formaldehyde! So there must be at least a typo here - or perhaps the whole idea of using respiration formulae is wrong here? As this is just a unit conversion it may not rely on the details of plant respiration but just be based on atomic weights (8 mol $CO_2$ = 11 mol $O_2$)??
EDIT I have just seen what is wrong with this chemical formula: it is intended to be an "empirical formula" reducing from the glucose respiration formula (in Wikipedia). So it is 2$O_2$ on the LHS to balance. In that case it is 1 mol $O_2$ = 1 mol $CO_2$.
The George paper does not explain how the $O_2$ to $CO_2$ conversion works and so I dont know how to take this any further, but hopefully the above is helpful.
(Incidentally to "reverse engineer" the formula under such circumstances you might also check the conversion factor implied by the other cited papers in Appendix 1 where it was maybe clearer what was being converted.)
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Thank you for these pointers. I have updated my calculation to account for the 1:1 ratio of O$_2$:CO$_2$, and changed density to inverse density (but the calculation remains the same). The results of the experiment with T at $30,35,40$ are reported as response ratio, e.g. normalized to respiration at T=30 and reported as a percent. So that data can't be converted to a rate as far as I can see (without more assumptions). And it is not likely that he reports measuring a rate at 27C but then adjusts it to 15C without stating it (or the equation used) in the manuscript. – David Apr 7 '11 at 7:16
@David, I agree with the last sentence. The question (which I cannot answer) is whether George just took the results from figure 3 (presented as they were against actual ground temperatures of around 15C) as the basis of doing the conversion. – Roy Simpson Apr 12 '11 at 9:11
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http://mathhelpforum.com/statistics/174036-probability-union-independent-events-proof-induction-argument.html
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# Thread:
1. ## Probability of the union of independent events proof (induction argument)
I am trying to prove the following:
If $E_1, E_2, ... , E_n$ are mutually independent events, then
$<br /> P(E_1 \cup E_2 \cup ... \cup E_n) = 1 - \Pi^{n}_{i=1} [1 - P(E_i)|<br />$
This is true for $n=2$
$<br /> P(E_1 \cup E_2) = 1 - P(E_1^{C} \cap E_2^{C})$ by DeMorgan's Law
and since $E_1$ and $E_2$ are independent, $E_1^{C}$ and $E_2^{C}$ are independent
Therefore $P(E_1 \cup E_2) = 1 - P(E_1^{C} \cap E_2^{C}) = 1-P(E_1^{C})P(E_2^{C})$
Assume true for n
$P(E_1 \cup E_2 \cup ... \cup E_n) = 1 - P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^C) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})$
Then $P(E_1 \cup E_2 \cup ... \cup E_n \cup E_{n+1}) = P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1})<br />$
Thus by the theorem for n=2
$<br /> P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) = 1-P((E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C}) \cap E_{n+1}) = 1- P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C})P(E_{n+1}^{C})$
But by induction hypothesis
$1- P(E_1^{C} \cap E_2^{C} \cap ... \cap E_n^{C}) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})$
Thus
$P(E_1 \cup E_2 \cup ... \cup E_n \cup E_{n+1}) = P((E_1 \cup E_2 \cup ... \cup E_n) \cup E_{n+1}) = 1-P(E_1^{C})*P(E_2^{C})*...*P(E_n^{C})*P(E_{n+1}^C})$
Which is what we want.
I was wondering if I proved the base case correctly and applied the induction hypothesis correctly. Thank you for your help.
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http://www.cfd-online.com/W/index.php?title=Turbulence_free-stream_boundary_conditions&diff=5438&oldid=5437
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[Sponsors]
Home > Wiki > Turbulence free-stream boundary conditions
# Turbulence free-stream boundary conditions
### From CFD-Wiki
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| - | In most CFD simulations it is necessary to specify values of the turbulent properties on the inlets. This is often difficult and can be a source of uncertainty, since the incoming turbulence is rarely known exactly. Most often you are forced to make a more or less educated guess of the incoming turbulence. | + | In most CFD simulations it is necessary to specify values of the turbulence variables at the inlets. For example, if you are using a <math>k-epsilon</math> you have to specify values of <math>k</math> and <math>\epsilon</math> on the inlets. This is often difficult and a source of uncertainty, since the incoming turbulence is rarely known exactly. Most often you are forced to make a more or less educated guess of the incoming turbulence. |
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| - | Estimating the incoming turbulence model variables, like turbulent energy, dissipation or even Reynolds stresses, directly is difficult. Instead it is often easier to think of the incoming turbulence in terms of properties like [[turbulence intensity]] and [[turbulent length-scale]] or [[eddy-viscosity ratio]]. These properties are more intuitive and can more easily be related to physical characteristics of the problem. | + | Estimating the incoming turbulence model variables, like turbulent energy, dissipation or even Reynolds stresses, directly is often difficult. Instead it is easier to think of the incoming turbulence in terms of variables like [[turbulence intensity]] and [[turbulent length-scale]] or [[eddy-viscosity ratio]]. These properties are more intuitive and can more easily be related to physical characteristics of the problem. |
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## Revision as of 14:23, 17 April 2006
In most CFD simulations it is necessary to specify values of the turbulence variables at the inlets. For example, if you are using a $k-epsilon$ you have to specify values of $k$ and $\epsilon$ on the inlets. This is often difficult and a source of uncertainty, since the incoming turbulence is rarely known exactly. Most often you are forced to make a more or less educated guess of the incoming turbulence.
Estimating the incoming turbulence model variables, like turbulent energy, dissipation or even Reynolds stresses, directly is often difficult. Instead it is easier to think of the incoming turbulence in terms of variables like turbulence intensity and turbulent length-scale or eddy-viscosity ratio. These properties are more intuitive and can more easily be related to physical characteristics of the problem.
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http://en.wikipedia.org/wiki/Binocular_disparity
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# Binocular disparity
Binocular disparity refers to the difference in image location of an object seen by the left and right eyes, resulting from the eyes' horizontal separation (parallax). The brain uses binocular disparity to extract depth information from the two-dimensional retinal images in stereopsis. In computer vision, binocular disparity refers to the difference in coordinates of similar features within two stereo images.
A similar disparity can be used in rangefinding by a coincidence rangefinder to determine distance and/or altitude to a target. In astronomy, the disparity between different locations on the Earth can be used to determine various celestial parallax, and Earth's orbit can be used for stellar parallax.
## Definition
Figure 1. Definition of binocular disparity (far and near).
Human eyes are horizontally separated by about 50–75 mm (interpupillary distance) depending on each individual. Thus, each eye has a slightly different view of the world. This can be easily seen when alternately closing one eye while looking at a vertical edge. The binocular disparity can be observed from apparent horizontal shift of the vertical edge between both views.
At any given moment, the line of sight of the two eyes meet at a point in space. This point in space projects to the same location (i.e. the center) on the retinae of the two eyes. Because of the different viewpoints observed by the left and right eye however, many other points in space do not fall on corresponding retinal locations. Visual binocular disparity is defined as the difference between the point of projection in the two eyes and is usually expressed in degrees as the visual angle.[1]
The term "binocular disparity" refers to geometric measurements made external to the eye. The disparity of the images on the actual retina depends on factors internal to the eye, especially the location of the nodal points, even if the cross section of the retina is a perfect circle. Disparity on retina conforms to binocular disparity when measured as degrees, while much different if measured as distance due to the complicated structure inside eye.
Figure 1: The full black circle is the point of fixation. The blue object lies nearer to the observer. Therefore it has a "near" disparity dn. Objects lying more far away (green) correspondingly have a "far" disparity df. Binocular disparity is the angle between two lines of projection in one eye(Mathematically,dn-df, with sign, measured counterclockwise). One of which is the real projection from the object to the actual point of projection. The other one is the imaginary projection running through the nodal point of the lens of the one eye to the point corresponding to the actual point of projection in the other eye. For simplicity reasons here both objects lie on the line of fixation for one eye such that the imaginary projection ends directly on the fovea of the other eye, but in general the fovea acts at most as a reference. Note that far disparities are smaller than near disparities for objects having the same distance from the fixation point.
In computer vision, binocular disparity is calculated from stereo images taken from a set of stereo cameras. The variable distance between these cameras, called the baseline, can affect the disparity of a specific point on their respective image plane. As the baseline increases, the disparity increases due to the greater angle needed to align the sight on the point. However, in computer vision, binocular disparity is referenced as coordinate differences of the point between the right and left images instead of a visual angle. The units are usually measured in pixels.
## Tricking neurons with 2D images
Figure 2. Simulation of disparity from depth in the plane. (relates to Figure 1)
Brain cells (neurons) in a part of the brain responsible for processing visual information coming from the retinae (primary visual cortex) can detect the existence of disparity in their input from the eyes. Specifically, these neurons will be active, if an object with "their" special disparity lies within the part of the visual field to which they have access (receptive field).[2]
Researchers investigating precise properties of these neurons with respect to disparity present visual stimuli with different disparities to the cells and look whether they are active or not. One possibility to present stimuli with different disparities is to place objects in varying depth in front of the eyes. However, the drawback to this method may not be precise enough for objects placed further away as they possess smaller disparities while objects closer will have greater disparities. Instead, neuroscientists use an alternate method as schematised in Figure 2.
Figure 2: The disparity of an object with different depth than the fixation point can alternatively be produced by presenting an image of the object to one eye and a laterally shifted version of the same image to the other eye. The full black circle is the point of fixation. Objects in varying depths are placed along the line of fixation of the left eye. The same disparity produced from a shift in depth of an object (filled coloured circles) can also be produced by laterally shifting the object in constant depth in the picture one eye sees (black circles with coloured margin). Note that for near disparities the lateral shift has to be larger to correspond to the same depth compared with far disparities. This is what neuroscientists usually do with random dot stimuli to study disparity selectivity of neurons since the lateral distance required to test disparities is less than the distances required using depth tests. This principle has also been applied in autostereogram illusions.
## Computing disparity using digital stereo images
The disparity of features between two stereo images are usually computed as a shift to the left of an image feature when viewed in the right image.[3] For example, a single point that appears at the x coordinate t (measured in pixels) in the left image may be present at the x coordinate t - 3 in the right image. In this case, the disparity at that location in the right image would be 3 pixels.
Stereo images may not always be correctly aligned to allow for quick disparity calculation. For example, the set of cameras may be slightly rotated off level. Through a process known as image rectification, both images are rotated to allow for disparities in only the horizontal direction (i.e. there is no disparity in the y image coordinates).[3] This is a property that can also be achieved by precise alignment of the stereo cameras before image capture.
### Computer algorithm
After rectification, the correspondence problem can be solved using an algorithm that scans both the left and right images for matching image features. A common approach to this problem is to form a smaller image patch around every pixel in the left image. These image patches are compared to all possible disparities in the right image by comparing their corresponding image patches. For example, for a disparity of 1, the patch in the left image would be compared to a similar-sized patch in the right, shifted to the left by one pixel. The comparison between these two patches can be made by attaining a computational measure from one of the following equations that compares each of the pixels in the patches. For all of the following equations, L and R refer to the right and left columns while r and c refer to the current row and column of either images being examined. "d" refers to the disparity of the right image.
• Normalized correlation: $\frac{\sum{\sum{ L(r,c) \cdot R(r,c) }}}{\sqrt{(\sum{\sum{ L(r,c)^2 }}) \cdot (\sum{\sum{ R(r,c)^2 }})}}$
• Sum of squared differences: $\sum{\sum{ (L(r,c) - R(r,c-d))^2 }}$
• Sum of absolute differences: $\sum{\sum{ \left | L(r,c) - R(r,c-d) \right \vert }}$
The disparity with the lowest computed value using one of the above methods is considered the disparity for the image feature. This lowest score indicates that the algorithm has found the best match of corresponding features in both images.
The method described above is a brute-force search algorithm. With large patch and/or image sizes, this technique can be very time consuming as pixels are constantly being re-examined to find the lowest correlation score. However, this technique also involves unnecessary repetition as many pixels overlap. A more efficient algorithm involves remembering all values from the previous pixel. An even more efficient algorithm involves remembering column sums from the previous row (in addition to remembering all values from the previous pixel). Techniques that save previous information can greatly increase the algorithmic efficiency of this image analyzing process.
## Uses of disparity from images
Knowledge of disparity can be used in further extraction of information from stereo images. One case that disparity is most useful is for depth/distance calculation. Disparity and distance from the cameras are negatively correlated. As the distance from the cameras increases, the disparity decreases. This allows for depth perception in stereo images. Using geometry and algebra, the points that appear in the 2D stereo images can be mapped as coordinates in 3D space.
This concept is particularly useful for navigation. For example, the Mars Exploration Rover uses a similar method for scanning the terrain for obstacles.[4] The rover captures a pair of images with its stereoscopic navigation cameras and disparity calculations are performed in order to detect elevated objects (such as boulders).[5] Additionally, location and speed data can be extracted from subsequent stereo images by measuring the displacement of objects relative to the rover. In some cases, this is the best source of this type of information as the encoder sensors in the wheels may be inaccurate due to tire slippage.
## In popular culture
Binocular disparity forms the premise for a sketch from the film Wayne's World in which Wayne, who is lying in bed as Tia Carrere's character, Cassandra, perches above him, compares the respective images from his left and right eyes while noting which is which by saying "Camera 1 ... Camera 2 ... Camera 1 ... Camera 2."
## References
1. Qian, N., Binocular Disparity and the Perception of Depth, Neuron, 18, 359-368, 1997.
2. Gonzalez, F. and Perez, R., Neural mechanisms underlying stereoscopic vision, Prog Neurobiol, 55(3), 191-224, 1998.
3. ^ a b
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http://math.stackexchange.com/questions/98105/why-are-smooth-manifolds-defined-to-be-paracompact?answertab=oldest
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# Why are smooth manifolds defined to be paracompact?
The way I understand things, roughly speaking, the importance of smooth manifolds is that they form the category of topological spaces on which we can do calculus. The definition of smooth manifolds requires that they be paracompact. I've looked all over, but I haven't found a clean statement for how paracompactness is a necessary condition to do calculus.
I understand that, by a theorem of Stone, every metric space is paracompact, but I'm not sure why we need global metrizability either.
Question: In what sense is paracompactness exactly the right condition to impose on a topological manifold to allow us to do calculus on it? Is there some theorem of the form "X has [some structure we strictly need in calculus] if and only if it is paracompact"?
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What is the definition of a smooth manifold that you're working with? I've learned that smooth manifolds are topological manifolds with a smooth structure by a certain construction. All topological manifolds are metrizable, and as you said thus paracompact. Paracompactness as it self never was implicitly in the definition of a smooth manifold in my textbooks though. – Thomas E. Jan 11 '12 at 9:41
– kahen Jan 11 '12 at 10:01
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Dear @kahen: formally the distinction between paracompact LEH and second countable LEH is correct but it is also completely artificial: paracompact LEH is equivalent to second countable LEH unless the manifold has more than countably many connected components, in which case it obviously cannot be second countable but might be paracompact. – Georges Elencwajg Jan 11 '12 at 14:43
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You do not need paracompactness to have well-defined derivatives, notably Serre develops the theory of analytic manifolds and Lie groups without this requirement. – Alexei Averchenko May 13 '12 at 15:32
## 3 Answers
A) For a differential manifold $X$ the following are equivalent:
a) X is paracompact
b) X has differentiable partitions of unity
c) X is metrizable
d) Each connected component of X is second countable
e) Each connected component of X is $\sigma$-compact
Partitions of unity are a fundamental tool in all of differential geometry (cf. kahen's answer) and would suffice to justify these conditions but the other equivalent properties can also be quite useful .
B) However occasionally non paracompact manifolds have been studied too. For example:
1) In dimension $1$ you have the long line obtained roughly by taking the first uncountable ordinal set and adding open segments $(0,1)$ between its successive points.
2) In dimension $2$ there exist non paracompact differentiable surface ( Prüfer and Radò). However every Riemann surface, that is a holomorphic manifold of complex dimension $1$ and thus real dimension $2$, is automatically paracompact.
3) Calabi and Rosenlicht have introduced a complex manifold of complex dimension $2$ which is not paracompact .
Edit As an answer to Daniel's question in the comments below, here are a few random examples of consequences of the existence of partitions of unity on a differential manifold $M$ of dimension $n$.
$\bullet$ If $M$ is orientable it has an everywhere non-vanishing differential form $\omega\in \Omega^n(M)$ of degree $n$.
$\bullet$ If $M$ is oriented you can define the integral $\int_M\eta$ of any compactly supported differential form $\eta\in \Omega^n_c(M)$ of degree $n$.
$\bullet$ The manifold $M$ can be endowed with a Riemannian metric.
$\bullet$ Every vector bundle on $M$ is isomorphic to its dual bundle.
$\bullet$ Every subbundle of a vector bundle on $M$ is a direct summand.
A sophisticated point of view (very optional !)
All sheaves of $C^\infty_M$-modules (for example locally free ones, which correspond to vector bundles) are acyclic in the presence of partitions of unity.
This has as a consequence that paracompact manifolds behave like affine algebraic varieties or Stein manifolds in that you can apply to them the analogue of Cartan-Serre's theorems A and B.
This is, in my opinion, the deep reason for the usefulness of partitions of unity on a manifold. (The last bullet for example was directly inspired from its analogue on affine varieties or Stein manifolds)
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Just to be clear: In your definition of a differential manifold $X$, you are requiring that $X$ be Hausdorff and locally euclidean, but not necessarily second-countable, is that right? – Jesse Madnick Jan 11 '12 at 13:27
Dear @Jesse: that's exactly right. – Georges Elencwajg Jan 11 '12 at 14:30
This is what I was looking for- thanks! Is there an easily-stated reason that partitions of unity are an essential tool for doing calculus on manifolds? (and couldn't be traded for something weaker?) – Daniel Moskovich Jan 11 '12 at 23:45
Dear @Daniel, I have added an edit giving examples of applications of partitions of unity. – Georges Elencwajg Jan 12 '12 at 0:29
1
Dear @Daniel, the pleasantly compact booklet Calculus on Manifolds by Michael Spivak explains why you have to use differential forms if you want to arrive at Stokes' theorem. You will see there in detail how partitions of unity are used to that end. The crucial point is that differential forms have built into them the change of variable formula for integrals on open subsets of $\mathbb R^n$. – Georges Elencwajg Jan 12 '12 at 1:33
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Because paracompactness is needed to prove the existence of smooth partitions of unity subordinate to any open covering.
For example as stated in Theorem 2.25 (Existence of Partitions of Unity) in Lee's "Introduction to Smooth Manifolds":
If $M$ is a smooth manifold and $\mathcal X = \{X_\alpha\}_{\alpha \in A}$ is any open cover of $X$, there exists a smooth partition of unity subordinate to $\mathcal X$.
EDIT: It occurs to me that I should probably also state the definition of partition of unity (from earlier on the same page):
Now let $M$ be a topological space, and let $\mathcal X = \{X_\alpha\}_{\alpha \in A}$ be an arbitrary open cover of $M$. A partition of unity subordinate to $\mathcal X$ is a collection of continuous functions $\{\psi_\alpha: M \to \mathbb R\}_{\alpha \in A}$ with the following properties:
(i) $0 \leq \psi_\alpha(x) \leq 1$ for all $\alpha \in A$ and all $x\in M$.
(ii) $\operatorname{supp} \psi_\alpha \subset X_\alpha$.
(iii) The set of supports $\{\operatorname{supp} \psi_\alpha\}_{\alpha\in A}$ is locally finite.
(iv)$\sum_{\alpha\in A}\psi_\alpha(x) = 1$ for all $x \in M$.
He then goes on to prove the extension lemma and the existence of bump functions as well as exhaustion functions.
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– Daniel Moskovich Jan 11 '12 at 23:47
I've thought about this a bit myself, and I think that, calculus-wise, a key point might be that a locally Euclidean space has to be paracompact in order to be completely metrizable. I think that a complete metric is indispensible to any attempt to do calculus, because without it, the analogues of the IVT (which is a direct consequence of completeness of the reals) and MVT, and Stokes Theorem (if you could even formulate it!), would fail.
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http://gilkalai.wordpress.com/2008/05/26/natis-influence/?like=1&source=post_flair&_wpnonce=bbf4ba6a50
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Gil Kalai’s blog
Nati’s Influence
Posted on May 26, 2008 by
When do we say that one event causes another? Causality is a topic of great interest in statistics, physics, philosophy, law, economics, and many other places. Now, if causality is not complicated enough, we can ask what is the influence one event has on another one. Michael Ben-Or and Nati Linial wrote a paper in 1985 where they studied the notion of influence in the context of collective coin flipping. The title of the post refers also to Nati’s influence on my work since he got me and Jeff Kahn interested in a conjecture from this paper.
Influence
The word “influence” (dating back, according to Merriam-Webster dictionary, to the 14th century) is close to the word “fluid”. The original definition of influence is: “an ethereal fluid held to flow from the stars and to affect the actions of humans.” The modern meaning (according to Wictionary) is: ”The power to affect, control or manipulate something or someone.”
Ben-Or and Linial’s definition of influence
Collective coin flipping refers to a situation where n processors or agents wish to agree on a common random bit. Ben-Or and Linial considered very general protocols to reach a single random bit, and also studied the simple case where the collective random bit is described by a Boolean function $f(x_1,x_2,\dots,x_n)$ of n bits, one contributed by every agent. If all agents act appropriately the collective bit will be ’1′ with probability 1/2. The purpose of collective coin flipping is to create a random bit R which is immune as much as possible against attempts of one or more agents to bias it towards ’1′ or ’0′.
Given such a protocol, the influence of a set S of agents towards ’0′ is the probability that R=0 if the agents in S try to tilt the outcomes of the coin flipping towards ’0′ as much as possible. The influence towards ’1′ is defined in the same way. And the influence of S is the sum of these two quantities. To make the definition clearer we should explain what the agent in S can do. Here we assume that in case of simultaneous action by all agents, the “bad guys” can wait to the contributions of all other agents before making their move. The bad guys can only change their inputs to the procedure.
Notations: When the protocol is denoted by $f$, for a set $S$ of processors, denote by $I^+_S(f)$ their influence toward ’1′, by $I^-_S(f)$ their influence toward ’0′, and let $I_S(f)= I^+_S(f)+I^-_S(f)$. The influence of a single processor $k$ is denoted by $I_k(f)$ and the sum $I(f) = I_1(f)+ I_2(f)+ \dots +I_n(f)$ is called the total influence of $f$. (In the Boolean case we refer to a “processor” or “agent” simply as a variable, and talk about influence of a variable, and influence of a set of variables.)
Notions of powers
Boolean functions can be regarded as “voting rules”. A Boolean function describes a way to move from the votes of n voters between two candidates to the collective decision of the society. Thinking of Boolean functions as voting rules provides nice names for special kinds of Boolean functions. “Dictatorship” refers to functions of the form $f(x_1,x_2,\dots, x_n)=x_k$. For the “majority function” (when $n$ is odd) the value of $f$ is ’1′ if and only if for more than half the variables $x_k =1$. For monotone Boolean functions, the influence of the kth variable coincides with the “Banzhaf power index” defined in game theory. Another related important notion of power is the Shapley-Shubik power index.
The KKL Theorem
Picture: Muli Safra
After many months of working on the conjecture, Jeff, Nati, and I managed to prove it. Let $f(x_1,x_2,\dots,x_n)$ be a Boolean function.
Theorem 1 (KKL): If the $\mu (f)=t$ then there exists a variable k so that $I_k(f) \ge K t (1-t) \log n/n$.
A repeated application of this theorem shows that:
Theorem 2 (KKL): If the $\mu (f)=1/2$ then there exists a set S of $K(\epsilon) n/\log n$ variables so that $I_S(f) \ge 1-\epsilon$.
Examples, examples
Majority
For the majority function with n variables the influence of every variable is proportional to $\sqrt n$.
The tribes example
This is the basic example of Ben-Or and Linial for Boolean functions with low influence. The society is divided into a large number of tribes, each having $\log n - \log \log n +\log \log e$ members. The value of f is one if and only if there exists a tribe whose members all vote ’1′. For this example the influence of every variable is $\theta (\log n /n)$.
The next two examples represent more complicated protocols for collective coin flipping. (Not just Boolean functions.)
Mike Saks’ “passing the baton” example
We start with some voter who holds the baton. This voter passes the baton to another random voter. Every voter who gets the baton passes it to a random voter who did not yet hold it. The last voter to hold the baton chooses the random collective bit. In this example, $o(n/\log n)$ bad agents cannot tilt the outcome significantly. (The best strategy for the “bad guys” is to pass the baton to a “good guy”.)
Uri Feige’s “two rooms” example
Every agent enters at random one out of two rooms. The room with fewer agents is selected and every agent in this room enters at random one out of two rooms. This process is continued (more or less) and at the end, as before, the last remaining agent contributes the collective random bit.
This process is immune against a constant number of “bad guys”. (The first such example was found by Noga Alon and Moni Naor.) The number of rounds in this protocol (appropriately optimized) goes to infinity extremely slowly. It is not known whether there is a protocol with similar properties with a bounded number of rounds.
Influence and threshold behavior
Consider a monotone Boolean function $f(x_1,x_2,\dots,x_n)$. Let $\mu_p$ be the product probability space where for every bit $\mu_p(x_i=1)=p.$ The definition of influence extends without change to the setting of biased product distribution. The influence of the $k$th variable on the Boolean function f with respect to $\mu_p$ is denoted by $I_k^p(f)$. The probability $\mu_p (f)$ that $f(x_1,x_2,\dots,x_n)=1$ is a monotone function in $p$ and Russo’s lemma asserts that the derivative of $\mu_p (f)$ with respect to $p$ is precisely the total influence $I^p(f)$. Therefore, large influence is related to “sharp threshold behavior”. Namely, to a very short interval between the value of $p$ where $\mu_p(f)$ is very close to 0, and the value of $p$ where $\mu_p(f)$ is very close to 1. Simple consequences of KKL’s theorem to the study of threshold behavior were noted by Ehud Friedgut and me, and Friedgut found an important theorem giving conditions for sharp threshold behavior when p itself is a function of n. Muli Safra and I wrote a survey paper on influences and threshold behavior.
Aggregation of information and Condorcet’s Jury theorem
Four sentences about the connection with Game Theory: The sharp threshold phenomenon is called in economics “asymptotically complete aggregation of information”. This property goes back to an old theorem from the theory of voting called “Condorcet’s Jury Theorem“. The Shapley-Shubik power index, mentioned above, can be defined as the integral $\int_0^1 I_k^p(f)dp$. (This is not the original axiomatic definition but a later Theorem by Owen.) It turns out that for a sequence of monotone Boolean functions, “sharp threshold phenomenon” is equivalent to “diminishing individual Shapley-Shubik power indices”. (But the quantitative aspects of this result are not satisfactory.)
Influence without independence
A major conceptual challenge is to understand the concept of influence (and related notions of “sharp threshold phenomenon”) for distributions which are not product distributions, namely when the probabilities for individual bits to be ’1′ are not statistically independent. Does an observer in a committee meeting have an influence? Can there be a negative influence? Moving away from statistical independence is often very difficult and yet very important for most applications. This issue is addressed in the paper of Graham and Grimmett, and that of Haggstrom, Mossel, and myself.
Two old conjectures about influence
There are quite a few problems regarding influences which remained unsolved. I will mention only two related conjectures both dealing with the Boolean case.
Conjecture 1: (Benny Chor): Suppose that $\mu(f)=1/2$ there is $c>0$, such that there is a set $S, |S|= n/10$ with $I_S(f) =1-\exp (cn)$.
Conjecture 2: Suppose that $\mu(f) = 0.999^n$. Then there is a set $S, |S|=0.499n$, so that $I_S(f)=1-o(1)$.
KKL theorem gives a weak form of Conjecture 1 where $1-\exp (cn)$ is replaced by $1-n^{\beta}$ and a weak form of Conjecture 2 where $0.999^n$ is replaced by $2^n/ n^{\alpha}$. The main difficulty here (and in various other problems in extremal combinatorics) is that arguing about influences of single variables is the only known method toward influences of large sets. Both these conjectures (as KKL’s theorem itself) have natural formulations in terms of traces of families of sets related to the Sauer-Shelah Theorem.
Update (February 2013): Both these conjectures have now been disproved in a paper with Jeff Kahn: Functions without influential coalitions.
More on Influence
The theory of poetic influence was developed by Yale’s literary critic Harold Bloom. His book “The Anxiety of Influence” deals with the process of influence as well as with the psychology of influence in literature. Philosopher Avishai Margalit studied influence in Philosophy in his paper on Wittgenstein. The paper will appear in a Festschrift for Peter Hacker by Blackwell Oxford, 2009. Section 3 opens with a distinction between influence and power: “Naked power is for anyone to see. Influence is not. It works its wonders in ways not readily observable. Influence is inferred from its effects. This is a major reason for the elusive nature of influence.”
Michael Ben-Or
Trivia question:
What do Michael Ben-Or, Uzi Segal (whose calibration theorem was mentioned in the post on the controversy around expected utility theory), Mike Werman, Ehud Lehrer, Yehuda Agnon, myself, and quite a few others, all have in common.
Hint:
Influence and coin flipping little updates (June, 8)
Mike Saks made valuable comments regarding this post. The first protocol for collective coin-flipping immune to a positive proportion of processors was by protocol with $log^*n$ rounds was by Alexander Russell and David Zuckerman. If each processor contributes only one bit per round there is a matching lower bound by Russell, Saks, and Zuckerman. The open problem Section from RSZ’s paper is very interesting and, as far as we know, no further progress on the problems presented there was made.
If we allow every processor to contribute many bits in every round then the situation is not clear even for one round. This is related to conjectures discussed by Ehud Friedgut.
Related problems are of great interest for quantum computation. Carlos Mochon have recently solved one of the most important longstanding open problems in quantum information theory, and found a protocol for weak coin flipping with arbitrarily small bias, using quantum bits. His work strengthens earlier work by Kitaev.
Related post: The Entropy Influence Conjecture. (Terry Tao’s blog)
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This entry was posted in Combinatorics, Computer Science and Optimization, Open problems, Probability and tagged Boolean functions, Collective coin flipping, Computer Science and Optimization, Distributed computing, Game theory, Influence, Sharp thresholds, Voting rules. Bookmark the permalink.
15 Responses to Nati’s Influence
1. alef says:
Can you please the exact ref to Avishai Margalit’s quote?
2. Gil says:
Dear Alef, I will add a link to the paper.
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7. KKL Question says:
Gil,
I have been curious for a long time now how you (three) came up with the idea of using the Bonami-Beckner inequality in the proof of the KKL theorem. Certainly pre-kkl, this wasn’t a very well known inequality. It is one of those great ideas that I have always marveled at, and (discovering your blog) can’t resist asking about!
8. Gil Kalai says:
Thanks a lot. I plan to have a post about the proof of KKL’s theorem and I hope others will also like it.
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15. Gil Kalai says:
The two conjectures on influence of large coalitions mentioned in the post have now been disproved by Jeff Kahn and me. (See the update in the post for a link to the paper.)
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http://mathhelpforum.com/advanced-algebra/158894-show-injective-surjective.html
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# Thread:
1. ## Show Injective/surjective
f: A x b --> A defined by f(x,y) = x
Injective: Here's what I did
Suppose f(x,y) = f(a,b). Then x = a. Thus not injective
Is this the correct approach?
surjective: Not sure how to go about this... Any advice? Thanks
2. For injective, you need to say that $b$ is not equal to $y$ (you start with $(a, b) \neq (x, y)$, but then you concluded that $a=x$ but not that $b=y$).
So, a proof would be: Note that for $b \neq y$, $f(x, b) = f(x, y)$ but $(x, b) \neq (x, y)$.
What you did was write down the working to find the proof, but it wasn't the proof itself per se.
For surjective, you start with something arbitrary in the image, A, and find something which maps to it. So, let $x \in A$. Can you think of some $(a, b) \in A \times B$ such that $f(a, b) = x$?
3. Originally Posted by jzellt
f: A x b --> A defined by f(x,y) = x
Injective: Here's what I did
Suppose f(x,y) = f(a,b). Then x = a. Thus not injective
Well, what causes the "thus"? To show "not injective", you need to assert that two different pairs, $(x_1, y_1)$ and $(x_2, y_2)$, can give the same value- you need to give a counterexample.
Is this the correct approach?
surjective: Not sure how to go about this... Any advice? Thanks
Look at the definition of "surjective"- a function, f, from A to B, is surjective if and only if, for every y in B, there exist x in A such that f(x)= y.
If you are given a number b does there exist a pair (x, y) such that f(x,y)= b? If so, what is it?
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http://mathoverflow.net/questions/91138/field-constructions/91143
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## Field constructions
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
If $F$ is a field of characteristic $p$ prime, how can one create a field $K$ such that $K$ is created from $F$ (either by modding out or by taking a product which includes $F$ or by some other method which involves $F$) such that $K$ has a different characteristic $j \ne p$? Also, does this question depend on $p$? I would also like to consider creations which are field extensions of $F$. Also, how many of each kind of extension are possible?
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How about Witt vectors? – Tom Goodwillie Mar 14 2012 at 0:43
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Since you allow products of fields, consider $A= \prod_{p} {\mathbf Z}/p{\mathbf Z}$. This ring has characteristic 0: for no positive integer $n$ does $n=0$ in A. Let I be the ideal of elements of A which have finitely many nonzero coordinates. By Zorn's lemma, I is contained in a maximal ideal M of A. What's the characteristic of the field A/M? If it were a prime $p$ then $p=0$ in A/M, so $p$ is in M. But as a (diagonal) sequence in A, $p$ has all but one nonzero coordinate. Use this and the fact that I is in M to show 1 is in M, which is a contradiction. So char(A/M) = 0. – KConrad Mar 14 2012 at 1:19
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You really should read about Witt vectors, as Tom suggests. Although the construction in my previous comment leads to a (nonconstructive) field of char. 0 out of infinitely many fields of different positive characteristics, Witt vectors are a mathematically more significant way to create fields (well, domains) of characteristic 0 out of finite fields in a good (= functorial) manner. – KConrad Mar 14 2012 at 1:21
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I think this question is quite vague. – Martin Brandenburg Mar 14 2012 at 10:11
## 3 Answers
It should be noted that the characteristic of a field is either prime or zero. If it is zero, then it contains the rational numbers. These are two statements you can probably prove even if algebra isn't your cup of tea.
You can study valuation rings of mixed characteristic. A classic example is $\mathbb{Z}_p$ the p-adic integers. This is a ring of characteristic 0 and its fraction field $\mathbb{Q}_p$ is a field of characteristic 0. However, $\mathbb{Z}_p$ has a (unique) maximal ideal generated by $(p)\mathbb{Z}_p$ such that $$\mathbb{F}_p \cong \mathbb{Z}_p/(p)\mathbb{Z}_p$$ which is a finite field of characteristic $p$.
There is the famous Ax-Kochen theorem, which is the following
$$\Pi_\mathcal{F} \mathbb{Q}_p \cong \Pi_\mathcal{F} \mathbb{F}_p((t))$$
where $\mathcal{F}$ is a non-principal ultra-filter on $\mathbb{N}$. This result depends on the continuum hypothesis. (J. Ax and S. Kochen, Diophantine problems over local fields I, American Journal of Mathematics,87 (1965), 605–630.)
However, there is also an isomorphism
$$\Pi_\mathcal{F} \mathbb{Z}_p \cong \Pi_\mathcal{F} \mathbb{F}_p[[t]]$$
where $\mathcal{F}$ is a non-principal ultrafilter on $\mathbb{N}$, which does not depend on the Continuum Hypothesis, in "Use of Ultrapoducts in Commutative Algebra" by Hans Schoutens, found here http://www.springer.com/mathematics/algebra/book/978-3-642-13367-1
Witt Vectors allow you to move from pure characteristic (either zero or positive) to mixed characteristic.
Another way of studying different characteristics is through Algebraic Geometry, by thinking of fields of different characteristics as fibers living over primes in $\mathbb{Z}$.
Edit: It occurred to me that an example here might be useful in order to illustrate this point. Given a curve $C$ over a field of characteristic p>0, we know (Winter's theorem) that there exists a discrete valuation ring (for example, $\mathbb{Z}_p$ of mixed characteristic), call it B, and a family of curves over B. Over the generic fiber, we have a GAGA principle, which means that things we can define over $\mathbb{C}$ (e.g., fundamental group), we can define over a field of $char(k)=0$. Now using Winter's theorem we can 'transfer' our definitions from analytic geometry to algebraic geometry over a field of positive characteristic, which gives us new tools when studying number theory.
Finally, there is even an "physical" way to interpret moving between fields. I highly recommend this article of A. Connes on the topic: "Characteristic one, entropy and the absolute point," Connes & Consani http://arxiv.org/abs/0911.3537
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I've fixed it, I hope I haven't destroyed any of the mathematical content. – David Roberts Mar 14 2012 at 2:00
Thanks. No, you haven't. – Andrew Stout Mar 14 2012 at 2:17
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
"By some other method" is quite vague. If, by modding out, you are thinking of taking a subring and then modding that ring by a maximal ideal (i.e., you obtain $\mathbb{Z}/p\mathbb{Z}$ from the rationals in this way), then it cannot be done.
Call the field $F$, the subring $R$, and the maximal ideal $I$, and assume $F$ has finite characteristic $p$. By definition of subring, the unit of the $R$ is also the unit of $F$. If the field has characteristic $p$, then $p \cdot 1 = 0$ is also true in the subring, and hence, in the quotient field $R/I$. Thus, the quotient field must also have characteristic $p$ if the original field does.
By definition, an extension field $K$ of $F$ would then contain $F$ as a subring, so again, they would have to have the same characteristic.
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I see from your answer that you cannot change the characteristic of a field by an inner field, by modding out. – Erin K Carmody Mar 14 2012 at 0:47
How many fields can one construct in this way? – Erin K Carmody Mar 14 2012 at 1:10
Even if you are interested in fields $K$ that are extensions of $F$, you left the door open for something else.
If $K$ is an extension of $F$, then $K$ has the same characteristic $p$ as $F$, because $p$ is the number of terms in the shortest non-trivial identity $1+\cdots+1=0$.
But you can also start from $F=\mathbb Z/p\mathbb Z$ and build $\mathbb Q_p$, the field of $p$-adic numbers. This one is has characteristic zero. See Dwork's book.
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http://mathoverflow.net/revisions/18854/list
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## Return to Question
2 added homology tag
1
# Homology of Surfaces with Holes
The classification theorem for surfaces says that the complete set of homeomorphism classes of surfaces is
{ $S_g : g \geq 0$ } $\cup$ { $N_k : k \geq 1$ },
where $S_g$ is a sphere with $g$ handles, and $N_k$ is a sphere with $k$ crosscaps. The first homology groups are easy to compute. They are $H_1 (S_g) = \mathbb{Z}^{2g}$, and $H_1 (N_k)=\mathbb{Z}^{k-1} \times \mathbb{Z} / 2\mathbb{Z}$. My question concerns how the homology groups change once we start cutting holes in our surface.
In the orientable case, it is easy to see what happens. The first hole that we cut out does not change the homology. Every additional hole then introduces another factor of $\mathbb{Z}$.
In the non-orientable case something peculiar happens. Consider the projective plane with homology group $\mathbb{Z} / 2 \mathbb{Z}$. If I cut out a hole, then I get the Mobius strip, which has homology group $\mathbb{Z}$ (it is homotopic to a circle). In general, if I cut a hole out of $N_k$, then in the homology group I lose a factor of $\mathbb{Z} / 2 \mathbb{Z}$, and introduce a factor of $\mathbb{Z}$. Each additional hole will then just introduce another factor of $\mathbb{Z}$.
My question: In the non-orientable case what happened to the factor of $\mathbb{Z} / 2 \mathbb{Z}$? Is there a nice geometric explanation of why it went away? I'm slightly disturbed because I had the intuition that torsion was supposed to record non-orientability, but I guess this doesn't work for surfaces with holes.
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http://mathoverflow.net/questions/94402?sort=votes
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## Changing the Type of a Module in MAGMA
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am currently working with irreducible $k[G]$-modules in MAGMA for finite fields $k$ and finite groups $G$. To construct these modules, I am using the commands IrreducibleModules(G,k) This results in MAGMA recognising the modules as being of type $ModGrp$. However, I wish to perform calculations such as finding the projective cover of such a module, for which MAGMA can only perform such calculations on modules over algebras (i.e. of type $ModAlg$). Is there an easy way to coerce MAGMA into treating modules of type $ModGrp$ as being of type $ModAlg$?
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## 1 Answer
MAGMA is perfectly capable of calculating the projective cover of a member of `IrreducibleModules(G,k)`, via the `ProjectiveCover()` function. You can also just call `ProjectiveIndecomposables(G,k)`. Perhaps you are running an old version of MAGMA? Version 2.18-3 is certainly capable of this functionality.
(This would have been a comment rather than an answer if reputation allowed).
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Thanks for the comment. I have tried using these commands, but just get the error messages: Runtime error in 'ProjectiveCover': Bad argument types Argument types given: ModGrp. Perhaps it is as you suggested that the version of MAGMA that I am using is too old (it is V2.15-15) – dward1996 Apr 18 2012 at 14:47
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I believe the projective modules functionality was introduced in V2.17, so you will need to update. – Alastair Litterick Apr 18 2012 at 15:20
@AlistariLitterick Thanks for clarifying that. – dward1996 Apr 18 2012 at 15:22
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http://dsp.stackexchange.com/questions/3557/shape-of-structuring-elements-for-morphological-gradients
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# Shape of structuring elements for morphological gradients
I am looking to understand recommended shapes of structuring elements used in calculating morphological gradients. According to Pierre Soille: Morphological Image Analysis:
Only symmetric structuring elements containing their origin are considered. By doing so, we make sure that the arithmetic difference is always nonnegative.
The arithmetic difference mentioned in the quote is referring to three combinations currently used to calculate the discrete gradient:
• arithmetic difference between the dilation and the erosion;
• arithmetic difference between the dilation and the original image;
• arithmetic difference between the original image and its erosion.
But, I think using a SE containing its origin is enough (it ensures anti-extensivity of dilation and extensivity of erosion). In this case, the following holds and ensures nonnegativity in all three cases:
$\varepsilon_B \leq id \leq \delta_B$ (where $id$ is the identity transform)
I am looking for a reason to enforce the symmetry condition. Intuitively, I understand that using a symmetrical SE is better than using a non-symmetrical one (e.g. examining a symmetrical pixel neighborhood). It was also suggested to me that there might be historical reason for this constraint.
However, I would like specific examples, arguments or references that point to desirable properties of symmetrical SEs (or undesirable properties of non-symmetrical ones).
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How can I download the article that you mention? – Andrey Oct 29 '12 at 16:23
@Andrey It's a book, not an article. And, sorry, can't help you there, I have a paper copy. – penelope Oct 29 '12 at 16:37
## 2 Answers
For planar elements (implied by the wording "structuring element") the containment of origin is enough to maintain the properties of anti-extensivity for erosion, and extensivity for dilation as can be found in many texts and you also pointed that out. So, yes, this is enough for the non-negativity for the arithmetic difference (this is directly shown by contradiction). The reason this piece of text is present in Pierre's book might be a simple one: a mistake. This statement is supported by other papers (like "Morphological Gradients" by Rivest, Soille, Beucher; or "An Overview of Morphological Filtering" by Serra, Vincent) on the morphological gradient defined by Beucher on his thesis. Now, I expect the most common situation to be the application of a gradient in an isotropic way, implying a symmetrical structuring element.
Now, to the second part of the question (supposing isotropy is not enough to conclude the answer). The first reason I can give for using symmetrical elements is to eliminate the burden of dealing with the multiple definitions for erosion and dilation present in the literature. It turns out that when you consider symmetrical elements, the distinct definitions become the same, guaranteeing the same behavior among different implementations. Using anisotropic elements will also translate your objects, which might be only useful for some certain applications. Also, some structuring elements are trivially decomposed when they are symmetric, enabling faster applications of morphological operations.
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I looked up in Jaehne, Gonzalez, Soille (the one you've posted as well as `Mathematical Morphology and Its Applications to Image and Signal Processing`) and some other special morphological papers and haven't found neither any design criteria for the structuring element nor any special hints why it has to be symmetrical.
Personally I think that a symmetrical SE is good for the same symmetrical effects on the object you want to modify. With my existing experience I would not use a non-symmetrical SE, because I can't take it for any object or scenario and I don't know how it will react for other cases.
Nevertheless it's an interesting question and I am trying to get an answer.
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Could you please expand on "symmetrical SE is good for the same symmetrical effects on the object you want to modify"? And, just to emphasize (not sure if you got this from your answer): I'm not asking about non-symmetrical SEs for all the morph. operations (e.g. erosion, dilation, opening): the question is about SE shape for morphological gradients. Nevertheless, thank you for your interest. – penelope Oct 29 '12 at 15:19
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http://physics.stackexchange.com/questions/34303/what-criteria-distinguishes-causality-from-retrocausality/34362
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# What criteria distinguishes causality from retrocausality?
The brilliant philosopher David Hume remarked that if two events are always found to be correlated to each other with one event happening prior to the other, we call the earlier event the cause and the latter event the effect. However, it has been pointed out that two events can be correlated, with one happening after the other, but only because they both have the same common cause, and not because of any direct causation. It's not "post hoc, ergo propter hoc".
What stops me from defining a new verb "retrocause"? It works just like Hume's definition, but only in reverse. We note the presence of a broken egg is always correlated perfectly (or very very nearly so up to an exponential degree of accuracy. Loschmidt reversal, anyone?) with the existence of an unbroken egg in its past. So, we say broken eggs retrocause unbroken eggs. In the same manner, unbroken eggs retrocause hens laying eggs. And hens retrocause female chicks. And female chicks retrocause hatching eggs. And so on and so forth.
In delayed choice experiments, what stops us from saying the choice of apparatus settings retrocause the preferred basis of a quantum system in its past?
What distinguishes causality from retrocausality?
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1
I removed the soft-question tag because the mods have a tendency to close questions tagged with that. I think this is a perfectly "hard" question about the definition of a physical concept. – Nathaniel Aug 16 '12 at 11:09
This is why causality in this sense does not belong to physics, but to the useless parts of philosophy. – Ron Maimon Sep 7 '12 at 15:33
## 8 Answers
What distinguishes causality from retrocausality? The laws of physics are pretty much time-symmetric so it seems the two should be not too different. However, we all know that causality goes from past to future. This is a question that many have worked on and none have fully resolved. For an overview, there are a couple of pages on Wikipedia: Arrow of time and Entropy (arrow of time). It seems the best answer so far is "it has something to do with entropy."
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Another brilliant philosopher, Immanuel Kant claimed that causality, like space and time, are not intrinsic properties of things-in-themselves, nature as it "really" is, as much as they are properties of the observing mind. It is the mind which determines causality, and the mind can be easily fooled.
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A set of events are called causes for an effect if they're sufficient for that effect to exist. This doesn't imply that the existence of this effect is sufficient for the same set of causes to exist. So whereas causality is consistent with the world we're living in, your idea of retrocausality isn't.
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The central concept lies in the concept of "independent variables" and "dependent variables". Independent variables are those which can be "freely" varied, while the value of dependent variables are determined by those of the independent variables. What does "can be freely varied mean"? Can't we invert the roles of independent and dependent variables, invert the function, so to speak? And what do we mean by determine? Do we mean determine uniquely, or only determining a probability distribution? Do we need to include all the controlling factors to make it a unique determination? If only a probability distribution, what is the meaning of probability? Maybe what looks like probabilistic randomness is only pseudorandom deterministic or ignorance? Unless our model of causality is fully deterministic in the sense of a unique determination, what looks like a fully determined probability distribution can always be due to some other factors.
Monte Carlo simulations using pseudorandom generators. Do we say the pseudorandom numbers are independent variables?
Bell entangled particles. Measure a spin there, measure a spin here. In rest frame, there first, but still spacelike. Local Copenhagenist will tell you measurement causes the spin outcome, and the probability for its collapse. Nonlocal hidden variable guy says the apparent randomness of the measured spin here is really because we have not accounted for the value of the spin there. Once we do, what looks like randomness is fully determined by superluminal effects which in a boosted frame becomes retrosuperluminal effects. Another hidden variables guy tells you, all was determined causally and nonsuperluminally when the entangled pair was created in the same place, and the Stern-Gerlach orientation was also superdetermined. It appears the orientation is an independent variable which can be freely varied, and is one of the causes, but it's not really. Retrocausalists will tell you the orientation now retrocausally determines the spin basis during creation.
Your example. Maybe in future, 3D printing of female chicks become possible. Then, you can't say female chicks retrocause hatching eggs. Change that to female chicks and no advanced 3D printing technology retrocause hatching eggs? Qualifiers. Maybe the world is a stage and the apparent absence of 3D printers is only an illusion? Backstage, behind the scenes, a director 3D printed a female chick and planted it in your set when you weren't looking.
Problem of induction. Just because event A always led to event B in its future does not mean it will do so again the next time round.
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If it's possible to vary the "cause" "freely" while keeping all the other "controlling" degrees of freedom fixed, i.e. controlling for all the other factors, and such a variation leads to a change in the outcome of the "effect", that is causality. The problem lies in the choice of what to consider the other controlling degrees of freedom.
Suppose P,Q and R are three propositions taking on true/false values. Suppose R = P XOR Q. Pick Q as the controlling variable. Then, P causes R, but also, R causes P! Pick P as the controlling variable. Then, Q causes R and R causes Q. Now try picking R as the controlling variable. Do you now understand why people refer to the tangled wed of causality?
The existence of intelligent lifeforms, and the very long time it takes for such lifeforms to evolve retrocaused the fine-tuning of the cosmological constant up to $10^{-123}$, and also the need for structure formation which retrocaused slow roll inflation. The existence of carbon based lifeforms retrocaused the carbon nuclear resonance needed for stellar nucleosynthesis. The existence of heavy elements retrocaused the difference of the neutron mass from the sum of electron and proton masses needed for neutrinos to blow off the heavier elements in nova explosions.
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Are you the same person as "Combative and confrontational", "Very difficult person", and perhaps dozens of other one-time posting identities? – Mitchell Porter Aug 22 '12 at 13:05
– Mitchell Porter Aug 22 '12 at 13:31
One reason that this is annoying is that you actually have something to say, but since it is scattered across a dozen identities, it's not possible to see how your thoughts connect. – Mitchell Porter Aug 22 '12 at 13:34
Here is the correct definition for causality: something is considered a cause of a given event if for any predictor wanting to predict the outcome of that event, whether by reasoning, calculation, simulation, or some other means, that predictor has to include some sort of encoding of the cause, and the entire causal chain emanating from it right till the event, somewhere in the intermediate states of the predictor.
It might turn out there might not be any correlations between the cause and effect at all. For instance, the outcome of the effect might turn out to be the same whatever the cause is. But if the only way for any predictor to figure that fact out is to actually run many simulations with different values for the cause, and only then note the effect remains the same in each case, it is still safe to say there is a causal relation. The time ordering between the cause and effect does not show up in this definition.
Similarly, there is no need for the cause to be able to be freely varied. Even if the cause can only be the way it is and no other, if any predictor has to include the cause in order to predict, it is still a cause. For instance, suppose all proofs of a mathematical theorem requires invoking a certain crucial lemma. The truth of the lemma is a mathematical fact which can't be varied. Still, it is correct to remark the lemma caused the theorem to be true.
In order to predict that an egg will smash when it falls on the floor at a high enough speed, it isn't necessary to know that a hen laid it. So, that doesn't count as a cause. Similarly, to predict that an egg is going to fall doesn't require any knowledge that it will smash in the future either. To predict the behavior of a chick might require some knowledge of the biology and psychology of chicks, but it doesn't necessarily require any knowledge of the neuroanatomy of its brain, or the detailed anatomy of the chick.
If the best the predictor can do is to give a probability distribution for the event outcomes, that still always leaves room for hidden causes overlooked by the predictor under the guise of apparent randomness.
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You say "The time ordering between the cause and effect does not show up in this definition." and yet you talk about a "predictor". By definition a predictor predicts the future, so your definition is about time ordering. – FrankH Sep 3 '12 at 12:15
The problem with most of the definitions of causality given here is according to them, if I have a certain memory or record that a certain event happened in this particular manner in my past, my memory of it retrocaused that event to be that way.
I would be interested to know if there is any definition of causality not explicitly mentioning time ordering which rules out this as an example of retrocausality.
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I don't see any problem at all with your definition of retrocausality. After all, it's just a definition... So your first question is moot.
The difference between causality and retrocausality is the parameter time. On the quantum scale we do use equations such that time moves backward to describe certain interactions. However, on the larger scale these many interactions form a system where time moves forward (i.e. positive) not backward (negative). Thus we tend to use causality, not retrocausality.
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-1. "However, on the larger scale these many interactions form a system where time moves forward (i.e. positive) not backward (negative)." As far as I can tell, this is exactly what the questioner is asking and this response is a mindless regurgitation without any physical or mathematical content. If you could please elaborate and convince me otherwise I would be more than happy to remove my downvote (and possibly upvote instead!). – Jonathan Aug 16 '12 at 23:21
There isn't much to elaborate here. As far as I know there is no theoretical proof that time has to move forward. Instead, we simply know from experiment and observation of a large number of systems that this is in fact what happens. Thus we say that time does have a direction, this itself being an assumption. We could of course assume the opposite and say that time is reversed, but this simply doesn't predict the measurements we observe. – mcFreid Aug 17 '12 at 0:34
I don't mean to sound contrarian, and am absolutely sincere--if there is nothing to say, then why say it? – Jonathan Aug 17 '12 at 0:42
Based on my knowledge, I answered the question "what is the difference between causality and retrocausality"? I don't know what you mean by "there is nothing to say"... Even if there isn't a clean, mathematical, answer to this question doesn't mean it's not worthwhile to answer it. I think the fact that we do observe time to move in one direction is significant. I think that I did answer the OP's question by using that observation. – mcFreid Aug 17 '12 at 0:55
But, it seems to me like you want to rephrase the OP's question to be: "is there a mathematical proof that time moves in one direction?" (i.e. the universe is based on causality not retrocausality). My answer to that question is that no such proof exists. You can choose to believe that that answer is insignificant, but I think it is very significant indeed. It is itself a proof that the current theories do not fully predict observed events without further assumptions. – mcFreid Aug 17 '12 at 0:57
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http://mathhelpforum.com/differential-geometry/112632-help-proving-some-facts.html
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# Thread:
1. ## Help with proving some facts
Hi,
I have to prove the following facts, but I am not sure if I figured them out correctly:
f : X --> Y and F : P(X) --> P(Y )
1) If A1 is a subset of A2, then F(A1) is a subset of F(A2)
For this one, I have the following:
For all x that are elements of A1 and A2
(there is a y that is an element of F(A1) s.t. F(x)=y) and (y is an element of F(A2) s.t. F(X)=y)
Therefore,
the fact that x is an element of A1 implies that there is a y that is an element of F(A2) s.t. F(x) = y
Hence, F(A1) is a subset of F(A2)
(Here I feel that, I missed a step constructing the proof with quantifiers.)
Also,
2) For every A that is an element of P(X), A is a subset of the inverse function of F(A)
For this one, I have:
(For all x that are elements of A, there is a y that is an element of F(A) s.t. F(x)=y s.t. the inverse of F(y)=x) implies that (for all x belonging to A, x is an element of the inverse of F(F(A))) implies that (A belongs to the inverse of F(F(A))).
I am also not sure about my logic here, do you guys agree with me?
2. Originally Posted by kaka87
Hi,
I have to prove the following facts, but I am not sure if I figured them out correctly:
f : X --> Y and F : P(X) --> P(Y )
1) If A1 is a subset of A2, then F(A1) is a subset of F(A2)
For this one, I have the following:
For all x that are elements of A1 and A2
(there is a y that is an element of F(A1) s.t. F(x)=y) and (y is an element of F(A2) s.t. F(X)=y)
Therefore,
the fact that x is an element of A1 implies that there is a y that is an element of F(A2) s.t. F(x) = y
Hence, F(A1) is a subset of F(A2)
(Here I feel that, I missed a step constructing the proof with quantifiers.)
Also,
2) For every A that is an element of P(X), A is a subset of the inverse function of F(A)
For this one, I have:
(For all x that are elements of A, there is a y that is an element of F(A) s.t. F(x)=y s.t. the inverse of F(y)=x) implies that (for all x belonging to A, x is an element of the inverse of F(F(A))) implies that (A belongs to the inverse of F(F(A))).
I am also not sure about my logic here, do you guys agree with me?
1. I think you have it, or if not, you're on the right track.
Let $y\in F(A_1)$. Then $\exists x\in A_1$ such that $F(x)=y$. Because $A_1\subset A_2$, $x\in A_2$ also. Thus $y=F(x)\in F(A_2)$, so $F(A_1)\subset F(A_2)$.
2. I'm not sure I understand the second question. It can't be asking you to prove that $A\subset F^{-1}(F(A))$, because that's trivial, but I don't know what else it could be asking.
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http://dsp.stackexchange.com/questions/tagged/frequency-response
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# Tagged Questions
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### Impulse Response Question
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### Obtaining the Impulse response from the frequency of a discrete system
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### Impulse Response / Frequency response Question
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### Find the frequency response of a system
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### Evaluating loudspeakers
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### Calculating the system output using frequency response
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### Plotting the magnitude response of a biquad filter
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### Ultrasonic speaker calibration and emitting calibrated signals
I'm trying to calibrate an ultrasonic speaker with the aim of emitting predictable signals, but alas I keep running into trouble, probably due to my lack of DSP-fu. A little background I want to be ...
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### Amplitude vs Frequency Response
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### Complex impulse response functions?
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### How to estimate a transfer function from a magnitude-only frequency response?
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### How to Average Complex Responses (and Justification)?
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### Function in the Fourier domain of a low-pass filter
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### Help me understand this simple code snippet to get the magnitude frequency response
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### Why are there comb-like hills in a periodogram?
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### What is meant by a system's “impulse response” and “frequency response?”
Can anyone state the difference between frequency response and impulse response in simple English?
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http://mathhelpforum.com/advanced-algebra/153019-vector-spaces-problem.html
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# Thread:
1. ## Vector spaces Problem
Let V1 = span {(1 0 2)}, and V2 = span{(0 0 1)} (they are colomn, not row)
Observe that V1 and V2 are both vector spaces. A new set of vectors, S, is constructed by taking any vector from V1 and any vector from V2 and adding these 2 vectors together.
why S will also be a vector space?
How to find a basis for S and the dimension of S?
2. In other words, $S= \{w |w= u+ v, u\in V_1, u\in V_2\}$.
To show that a subset of a vector space is a subspace, you only need to show that the set is closed under addition and scalar multiplication of vectors. That is that if $w_1$ and $w_2$ are in S, so are $w_1+ w_2$ and $\alpha w_1$ where $\alpha$ is any scalar.
That can be done most conveniently, in one step, by showing that $\alpha w_1+ w_2$ is in the same set.
Well, if $w_1$ is in that set then $w_1= u_1+ v_1$ for some $v_1\in V_1$ and some $v_1\in V_2$. If $w_2$ is in that set, similarly, $w_2= u_2+ v_2$.
Now, $\alpha w_1+ w_2= \alpha(u_1+ v_1)+ (u_2+ w_2)= (\alpha u_1+ u_2)+ (\alpha v_1+ v2)$. Do you see why that is again in S?
Any vector in S is of the form $w= u+ v$. Since u is in the span of (1, 0, 2), $u= \alpha(1, 0, 2)$ and since v is in the span of (0, 0, 1), $v= \beta(0, 0, 1)$. That means that any vector in S can be written $\alpha(1, 0, 2)+ \beta(0, 0, 1)$. You should be able to see from that what the dimension is and a basis.
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http://mathoverflow.net/questions/101031?sort=oldest
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## Kervaire invariant: Why dimension 126 especially difficult?
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Is there any resource that might help non-experts gains some understanding of why the Kervaire invariant problem remains open now only in dimension $126$? ($126 =2^7-2=2^{j+1}-2$; whether $\theta_j=\theta_6$ exists in the "$128$-stem"), i.e., why the celebrated Hill-Hopkins-Ravenel proof technique fails in this last remaining case? I read a delightful exposition by Erica Klarreich ("Mathematicians solve 45-year-old Kevaire Invariant Puzzle," The Best Writing on Mathematics, 2010, 373ff, Simons Foundation link), which piqued my interest. But Michael Hopkins' online presentation slides announcing the result in 2009 ("Applications of algebra to a problem in topology") are beyond my ken. Perhaps this an area too abstruse for all but the experts? I'd appreciate pointers to expositions. Thanks!
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The problem is complicated and either you dive into it or you won't understand anything at all, there's no way of posting a short enough answer to your question. I'd recommendo you to read Snaith's book on the topic an then the papers by Hill-Hopkins-Ravanel and Akhmetev. – Fernando Muro Jul 1 at 0:57
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There was also a Bourbaki Seminar talk by Haynes Miller, the report is available on arXiv arxiv.org/abs/1104.4523 – quid Jul 1 at 1:01
Thanks, quid, that is a great resource, exactly what I was seeking. And thanks, Fernando, for the reference to (I guess?): Snaith, Victor P. (2009), "Stable homotopy around the Arf-Kervaire invariant," Progress in Mathematics, 273, Birkhäuser Verlag. – Joseph O'Rourke Jul 1 at 1:31
## 1 Answer
I'll give a shot at an answer. The relevant dimensions are of the form $2^j-2$. For $j\leq 4$, it is easy and classical that we can construct manifolds of Kervaire invariant one. The problem was reduced'' from differential topology to pure stable homotopy theory by Browder in 1969. Direct calculational methods in homotopy theory were used by Barratt, Jones, and Mahowald to construct a cell complex that can be used to solve the homotopy theory problem and prove that such manifolds also exist in dimensions 30 and 62. I believe a construction of such a manifold has been worked out in dimension 30, but that has certainly not been done in dimension 62. Periodicity phenomena play a huge role in modern stable homotopy theory, and a crucial feature of the Hill, Hopkins, Ravenel proof is a periodicity of order $2^8 = 256$. That enables them to solve the stable homotopy problem and prove there is no manifold of Kervaire invariant one for $j\geq 8$. The reasons $j=7$ is so hard are several. Nobody has a really good reason for guessing which way the answer will go. There is no reason to expect a relevant periodicity of order $2^7$. Direct calculation of the Adams spectral sequence through dimension $126$ is just plain hard: the calculations blow up. There is a chance that the methodology of Barratt, Jones, and Mahowald might extend to prove existence (if that is how the answer turns out!), but it will probably be much harder to prove nonexistence (if that is the answer).
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Thank you, Peter, for this succinct and clear summary! – Joseph O'Rourke Jul 1 at 12:11
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I disagree with the statement that there is no reason for guessing which way the answer will go. There is a good reason, which I learned from Atiyah, to guess that the answer in dimension 126 is the same as in the lower dimensions 62, 30, 14,... Namely, there exist remarkable manifolds in dimensions 128, 64, 32, and 16. These are the octooctonionic projective plane (see math.ucr.edu/home/baez/octonions/node19.html), the quateroctonionic projective plane (see math.ucr.edu/home/baez/octonions/node18.html), the bioctonionic projective plane, and finally $OP^2$ itself. – André Henriques Jul 1 at 19:26
Interesting. But I think to give it substance you would have to display a manifold of dimension 62 and Kervaire invariant one. – Peter May Jul 1 at 20:41
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I should probably add that I hope you are right, since I have a student who hopes to prove that you are (homotopically of course). – Peter May Jul 1 at 21:30
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I realize that this is not a complete argument. Atiyah's hope was that there should exist a geometric construction that starts with the quateroctonionic projective plane and constructs a 62-manifold of Kervaire invariant one, and that the same construction, applied to the octooctonionic projective plane would produce a 126-manifold of Kervaire invariant one. – André Henriques Jul 2 at 12:07
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http://physics.stackexchange.com/questions/47043/quantum-field-theories-with-asymptotic-freedom?answertab=oldest
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# Quantum field theories with asymptotic freedom
QCD is the best-known example of theories with negtive beta function, i.e., coupling constant decreases when increasing energy scale. I have two questions about it:
(1) Are there other theories with this property? (non-Abelian gauge theory, principal chiral field, non-linear sigma model, Kondo effect, and ???)
(2) Are there any simple (maybe deep) reason why these theories are different from others? It seems that the non-linear constraint of the non-linear sigma model (and principal chiral model) is important, but I have no idea how to generalize this argument to other theories.
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## 2 Answers
Yes, such examples are generic in 2+1 and 1+1D.
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Could you elaborate on this? You only answer the first part of the question. – Manishearth♦ Dec 17 '12 at 8:50
One example would be $\varphi^3$ theory, which is treated extensively in Srednicki.
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Is it artificial? Since there is no true ground state for $\phi^3$ thoery. The strong coupling property at low energy may indicates the collapse of perturbation theory. – Tengen Jan 24 at 8:23
You are right, the theory is unstable. – Frederic Brünner Jan 24 at 10:21
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http://quant.stackexchange.com/questions/4165/why-does-the-adjusted-closing-price-take-into-account-dividends
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Why does the adjusted closing price take into account dividends?
I'm trying to get an intuition as to why the adjusted closing price includes a dividend adjustment:
\begin{equation} 1 - \frac{dividend}{close} \end{equation}
I understand why the adjusted closing price "undoes" the effect of a split on a stock. A stock split hasn't changed the economics of a company so the adjustment effectively removes the split so that the adjusted closing prices before and after the split can be compared.
A dividend does change the underlying economics however so why does the adjusted price account for it? Lastly, the formula also doesn't make sense.
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– chrisaycock♦ Sep 21 '12 at 3:42
3 Answers
Let's assume there is no adjustment and that a stock's price is the same after a dividend payment as before. Then I could get free money simply by buying a stock the day before the ex-date and then selling the stock right after the dividend distribution. Clearly no such arbitrage opportunity exists. Therefore, the price of the stock after the dividend payment must be the same as the price before the payment minus the dividend amount. That is, the adjusted price is
\begin{equation} adj = close - dividend \end{equation}
The other argument, as you point-out in your question, is that the fundamentals have indeed changed and that a company with less cash on hand must have a lower market cap. That is, the number of shares outstanding hasn't changed (unlike in a split), so the share price must change.
To get the ratio you asked about, divide both sides of the above equation by $close$:
\begin{equation} \frac{adj}{close} = \frac{close}{close} - \frac{dividend}{close} = 1 - \frac{dividend}{close} \end{equation}
Therefore, I can get the adjusted price simply by multiplying the closing price by $1 - \frac{dividend}{close}$.
To conclude: when looking at a stock's returns from one day to the next, the historical share price must be adjusted for all corporate actions (splits, dividends, and name changes) to present a coherent picture of returns. Otherwise, the returns will appear to have unrealistic gaps.
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It's just a convention, a way to account for the total return of the stock including dividends. This method is for simplicity's sake. It'd be "impossible" to try to take into account all of the infinite possibilities of dividend investment elsewhere.
You could think of it as "the dividend is fully reinvested cost-free", but that's just a short-hand way to visualize it.
If the stock increases by X and a dividend paid is Y then today's return is X+Y. It carries on to each new day, so that's why a dividend paying stock has higher adjusted prices than the true price.
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Often you're more interested in what your overall return would have been (or has been) over a certain holding period rather than how much the stock price has changed. In that case, you'd need to have some way to account for the dividends, and calculating adjusted stock price is a convenient way of doing it.
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http://mathoverflow.net/questions/22950/why-isnt-the-mobius-band-an-algebraic-line-bundle
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## why isn’t the mobius band an algebraic line bundle?
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When I hear the phrase "line bundle" the first thing that pops into my head is a mobius band. But this is a bad picture from an algebraic point of view since any line bundle on an affine variety is trivial. Anyway, my question is: is there a way of seeing more concretely what "goes wrong" when you try to construct the mobius band as an algebraic line bundle over ℝ, and what changes when you move to analytic line bundles?
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## 1 Answer
Consider the real algebraic line bundle $\mathcal{O}(-1)$ over the real algebraic variety $\mathbb{R}\mathbb{P}^1$. It is nontrivial hence continuously isomorphic to the "Moebius" line bundle (there are only 2 line bundles on the circle, up to continuous isomorphism), so its total space is homeomorphic o the "Moebius strip".
By the way, it is false that line bundles on affine algebraic varieties are trivial. One example is the above universal line bundle over $\mathbb{R}\mathbb{P}^1$. If you want an example over $\mathbb{C}$, consider the complement of a point in an elliptic curve: it's an affine variety but its Picard group is far from being trivial.
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thanks for that, the thing about line bundles being trivial was told to me today by a topology professor, i guess she was confused. – Max Flander Apr 29 2010 at 8:05
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Your comment about $\mathbb{R} \mathbb{P}^1$ is correct, but may be confusing to some. One of the interesting features of real algebraic geometry is that $\mathbb{R} \mathbb{P}^n$ is (also) the real analytic manifold associated to a smooth affine variety. So there is, in this sense, no distinction between affine and projective varieties over $\mathbb{R}$. – Pete L. Clark Apr 29 2010 at 8:07
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@maxmoo: There is a related true result here: a coherent sheaf on an affine variety $X$ is acyclic for sheaf cohomology: its higher cohomology groups are all zero. One might be tempted to deduce from this that $\operatorname{Pic}(X) = H^1(X,\mathcal{O}_X^{\times}) = 0$. But this is not valid: $\mathcal{O}_X^{\times}$ is not a coherent sheaf. – Pete L. Clark Apr 29 2010 at 8:11
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@Pete: yes, O^* is not a coherent sheaf simply because it is not a sheaf of O_X-modules, but just a sheaf of abelian groups. – Qfwfq Apr 29 2010 at 8:16
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Just to expand a little on Pete's point: $\mathbb{RP}^1$ is isomorphic to $x^2+y^2=1$. Over the complex numbers, the former is a sphere and the latter is a sphere with two punctures, but the punctures are complex conjugates of each other, so you don't see them in the real picture. I think you should be able to write the Mobius band explicitly as `$\{(x,y,u,v) : \ x^2+y^2=1 \ (u^2-v^2)y=2uv x \}$`. In other words, it is the set of pairs of complex numbers $(z,w)$ such that $z$ has norm $1$ and $\mathrm{arg}(z)=2 \mathrm{arg}(w)$. – David Speyer Apr 29 2010 at 10:39
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http://mathoverflow.net/questions/19942/rank-of-aba-where-b-is-positive-definite/19954
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## Rank of ABA where B is positive definite
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I have a n-by-k matrix A and a n-by-n matrix B, where B is positive definite. I can form the matrix $M = A^t B A$. Playing around, I always found $rk(M) = rk(A)$ but I can't prove this.
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Hint: M and A have the same null space. – Mark Meckes Mar 31 2010 at 12:26
Thanks, I actually found that, but missed that A and M have the same number of columns to finish the proof. – Frank Meulenaar Mar 31 2010 at 12:28
## 2 Answers
$A^T BAx = 0 \implies (Ax)^TB(Ax)=0 \implies Ax=0$ by positve definiteness of $B$. So $ker(M)=ker(A)$ and hence $rk(M)=rk(A)$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This is rather tangential, but I can't help mentioning that this is related to the fact that in a hilbert space, for a (compact) linear operator one has: $Ker A^* \perp Im A$.
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http://mathoverflow.net/questions/70692/why-is-the-chebyshev-function-relevant-to-the-prime-number-theorem/70835
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## Why is the Chebyshev function relevant to the Prime Number Theorem
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Why is the Chebyshev function
$\theta(x) = \sum_{p<=x}\log p$
useful in the proof of the prime number theorem. Does anyone have a conceptual argument to motivate why looking at $\sum_{p<=x} \log p$ is relevant and say something random like $\sum_{p<=x}\log\log p$ is not useful or for that matter any other random function $f$ and $\sum_{p<=x} f(p)$ is not relevant.
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The rate of growth of primes suggests the Chebyshev function might be useful: en.wikipedia.org/wiki/Prime_number_theorem – Kimball Jul 19 2011 at 1:24
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Primes are a multiplicative basis, which is a reason to expect that a function on product of primes is likely to be more natural than a function on say the number of primes, or sum of primes etc. Actually I think that the most natural function is the second Chebyshev function (which is the one appearing in both the complex analytic and elementary proofs of PNT, as well as in the "explicit formula"), but it's not hard to show that the two Chebyshev functions approximate each other. – Gjergji Zaimi Jul 19 2011 at 1:35
You might be interested in what T. Tao writes in this connection in his book on blogs from year three. – Franz Lemmermeyer Jul 19 2011 at 7:05
## 3 Answers
There are several ideas here, some mentioned in the other answers:
One: When Gauss was a boy (by the dates found on his notes he was approximately 16) he noticed that the primes appear with density $\frac{1}{\log x}$ around $x$. Then, instead of counting primes and looking at the function $\pi (x)$, lets weight by the natural density and look at $\sum_{p\leq x} \log p$. Since we are weighting by what we think is the density, we expect it to be asymptotic to be $x$.
Two: Differentiation of Dirichlet series. If $$A(s)=\sum_{n=1}^\infty a_n n^{-s}, \text{ then } A^'(s)=-\sum_{n=1}^\infty a_n (\log n) n^{-s}$$ The $\log$ term appears naturally in the differentiation of Dirichlet series. Taking the convolution of $\log n$ with the Mobius function (that is multiplying by $\frac{1}{\zeta(s)}$) then gives the $\Lambda(n)$ mentioned above. The $\mu$ function is really the special thing here, not the logarithm.
Expanding on this, there are other weightings besides $\log p$ which arise naturally from taking derivatives. Instead we can look at $\zeta^{''}(s)=\sum_{n=1}^\infty (\log n)^2 n^{-s}$, and then multiply by $\frac{1}{\zeta(s)}$ as before. This leads us to examine the sum $$\sum_{n\leq x} (\mu*\log^2 )(n)$$ (The $*$ is Dirichlet convolution) By looking at the above sum, Selberg was able to prove his famous identity which was at the center of the first elementary proof of the prime number theorem:
$$\sum_{p\leq x} \log^2 p +\sum_{pq\leq x}(\log p)(\log q) =2x\log x +O(x).$$
Three: The primes are intimately connected to the zeros of $\zeta(s)$, and contour integrals of $\frac{1}{\zeta(s)}$. (Notice it was featured everywhere here so far) We can actually prove that
$$\sum_{p^k\leq x} \log p= x -\sum_{\rho:\ \zeta(\rho)=0} \frac{x^{\rho}}{\rho} +\frac{\zeta^'(0)}{\zeta(0)}.$$ Notice that the above is an equality, which is remarkable since the left hand side is a step function. (Somehow, at prime powers all of the zeros of zeta conspire and make the function jump.)
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The distribution of sums $\sum_{p\leq x}f(p)$ is intimately related to the analytic properties of the Dirichlet series $\sum_{p\leq x}f(p)p^{-s}$. If $f(p)$ equals the von Mangoldt function, perhaps twisted by some automorphic data like values of a Dirichlet character or Hecke eigenvalues of a cusp form, then $\sum_{p\leq x}f(p)p^{-s}$ equals, up to a Dirichlet series supported on higher prime powers, the negative logarithmic derivative of an automorphic $L$-function $L(s)$, as follows from the Euler product decomposition of $L(s)$. The function $L(s)$ itself is nice, in addition to being an Euler product: it has a meromorphic continuation with at most a few poles and good behavior in vertical strips, has a functional equation, and conjecturally all its zeros lie on the axis of symmetry (Generalized Riemann Hypothesis). The logarithmic derivative $L'(s)/L(s)$ is therefore also meromorphic with simple poles at the zeros and poles of $L(s)$.
In the case of $f(p)=\Lambda(p)$, we have $L(s)=\zeta(s)$ so that there is a simple pole at $s=1$ with residue $1$ and also at every zero of $\zeta(s)$ with residue equal to the negative multiplicity of the zero. Conjecturally the zeros all lie on $\Re(s)=1/2$ which makes $\sum_{n\leq x}\Lambda(n)$ change rather smoothly: it is $x+O(x^{1/2+\epsilon})$. Of course $\sum_{p\leq x}\log(p)$ differs from $\sum_{n\leq x}\Lambda(n)$ only by $O(x^{1/2+\epsilon})$.
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I would say that the von Mangoldt function $\Lambda$ is more interesting than other multiplicative functions because of the relation $log = 1\star \Lambda$, where $\star$ is the convolution of arithmetic functions. This relation encodes the fundamental theorem of arithmetic, so "heuristically" it is of great interest, compared to the other propositions of the OP. – Bernikov Jul 21 2011 at 2:23
@Bernikov: The von Mangoldt function is not multiplicative (it is supported on prime powers). – GH Jul 21 2011 at 9:59
It seems to me that the second Chebyshev function $\psi(x) = \sum_{n \le x} \Lambda(n)$ (where $\Lambda$ is the von Mangoldt function) is more natural, but as Gjergji says the two approximate each other. This is because the behavior of the second Chebyshev function is related by certain general theorems that I'm not familiar with to the behavior of the Dirichlet series
$$\sum_{n \ge 1} \frac{\Lambda(n)}{n^s}$$
and this Dirichlet series is precisely $\frac{-\zeta'(s)}{\zeta(s)}$, or the negative logarithmic derivative of $\zeta(s)$. This has the following intuitive interpretation: if we think of $\zeta(s)$ as the partition function
$$\zeta(s) = \sum_{n \ge 1} e^{-s \log n}$$
of the Riemann gas, then the negative logarithmic derivative of a partition function describes the expected value of energy at a given temperature, a fundamental property.
Edit: This observation goes at least as far back as Mackey, Unitary Group Representations in Physics, Probability, and Number Theory (1978), who wrote:
...Our main point here is that one could have been led to the main outline of the proof of the prime number theorem by using the physical interpretation of Laplace transforms provided by statistical mechanics. In particular, the function $-\frac{\zeta'}{\zeta}$ whose representation as a Dirichlet series (Laplace transform with discrete measure) plays a central role in the proof has a direct physical interpretation as the internal energy function.
Regarding why it is natural to assign the state $n$ the energy $\log n$, the point is that for $s > 1$ we get the only probability distributions on the natural numbers which satisfy certain natural properties; see this blog post, for example.
Of course a basic purely mathematical reason to consider the logarithmic derivative is the ideas around the argument principle.
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Partition functions? Riemann gases? Since when do these give "intuitive interpretations" of the zeta function? :) – David Hansen Jul 19 2011 at 4:52
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@David: well, I suppose it is a matter of whether you care about physical intuition or not. The point is that the statistical properties of the Riemann gas should be closely related to statistical properties of the primes. It should be possible to write a complete proof of the PNT in which every step is physically motivated like this but I don't know if anyone's actually done it... – Qiaochu Yuan Jul 19 2011 at 5:11
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I truly don't find an explanation involving "Riemann gasses" to be "physically motivated". Forcing a translation into the language of physics is not a motivation. – Greg Martin Jul 19 2011 at 9:13
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@Greg: "forced" seems a little harsh to me. Some people take ideas like these quite seriously, and this exact observation was pointed out by no less a mathematician than Mackey: empslocal.ex.ac.uk/people/staff/mrwatkin//zeta/… – Qiaochu Yuan Jul 19 2011 at 12:36
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@Qiaochu: Sorry, I'm still not buying it. No number theorist (i.e., the kind of people who prove statistical theorems about the primes) has ever thought like this. – David Hansen Jul 19 2011 at 18:02
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http://cms.math.ca/cjm/v64/n3/
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Canadian Mathematical Society
www.cms.math.ca
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location: Publications → journals → CJM
« Vol. 64 No. 2 Vol. 64 No. 4 »
Volume 64 Number 3 (Jun 2012)
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Page
Contents
481
Chamorro, Diego
In this article we study some Sobolev-type inequalities on polynomial volume growth Lie groups. We show in particular that improved Sobolev inequalities can be extended to this general framework without the use of the Littlewood-Paley decomposition.
497
Li, Wen-Wei
Dans cet article, on énonce une variante du lemme fondamental pondéré d'Arthur pour le groupe métaplectique de Weil, qui sera un ingrédient indispensable de la stabilisation de la formule des traces. Pour un corps de caractéristique résiduelle suffisamment grande, on en donne une démonstration à l'aide de la méthode de descente, qui est conditionnelle: on admet le lemme fondamental pondéré non standard sur les algèbres de Lie. Vu les travaux de Chaudouard et Laumon, on s'attend à ce que cette condition soit ultérieurement vérifiée.
544
Li, Zhiqiang
A K-theoretic classification is given of the simple inductive limits of finite direct sums of the type I $C^*$-algebras known as splitting interval algebras with dimension drops. (These are the subhomogeneous $C^*$-algebras, each having spectrum a finite union of points and an open interval, and torsion $\textrm{K}_1$-group.)
573
Nawata, Norio
We introduce the fundamental group ${\mathcal F}(A)$ of a simple $\sigma$-unital $C^*$-algebra $A$ with unique (up to scalar multiple) densely defined lower semicontinuous trace. This is a generalization of Fundamental Group of Simple $C^*$-algebras with Unique Trace I and II'' by Nawata and Watatani. Our definition in this paper makes sense for stably projectionless $C^*$-algebras. We show that there exist separable stably projectionless $C^*$-algebras such that their fundamental groups are equal to $\mathbb{R}_+^\times$ by using the classification theorem of Razak and Tsang. This is a contrast to the unital case in Nawata and Watatani. This study is motivated by the work of Kishimoto and Kumjian.
588
Nekovář, Jan
In this article we refine the method of Bertolini and Darmon and prove several finiteness results for anticyclotomic Selmer groups of Hilbert modular forms of parallel weight two.
669
Pantano, Alessandra; Paul, Annegret; Salamanca-Riba, Susana A.
We classify all genuine unitary representations of the metaplectic group whose infinitesimal character is real and at least as regular as that of the oscillator representation. In a previous paper we exhibited a certain family of representations satisfying these conditions, obtained by cohomological induction from the tensor product of a one-dimensional representation and an oscillator representation. Our main theorem asserts that this family exhausts the genuine omega-regular unitary dual of the metaplectic group.
705
Thomsen, Klaus
It is shown that simplicity of the crossed product of a unital AH-algebra with slow dimension growth by an endomorphism implies that the algebra is also purely infinite, provided only that the endomorphism leaves no trace state invariant and takes the unit to a full projection.
© Canadian Mathematical Society, 2013
© Canadian Mathematical Society, 2013 : http://www.cms.math.ca/
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http://mathhelpforum.com/calculus/36611-growth-model-word-problem-w-differential-equations.html
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# Thread:
1. ## Growth model word problem w/differential equations
Instructions:
Biology. A population of eight beavers has been introduced into a new wetlands area. Biologists estimate that the maximum population the wetlands can sustain is 60 beavers. After 3 years, the population is 15 beavers. If the population follows a Gompertz growth model, how many beavers will be in the wetlands after 10 years?
Gompertz growth model:
$\frac{dy}{dt}=kyln\frac{60}{y}$
Here's what I did.
$dy=kyln\frac{60}{y}(dt)$
$\frac{dy}{yln\frac{60}{y}}=kydt$
$\int \frac{1}{y(ln60-lny)}dy=\int kdt$
$u=ln60-lny$
$du=-\frac{1}{y}dy$
$-du=\frac{1}{y}dy$
$- \int \frac {1}{u}du = \int kdt$
$-ln|u| = kt + C$
$-|ln60-lny| = e^{kt+c}$
$-ln\frac{60}{y} = Ce^{kt}$
$\frac{60}{y}=e^{-Ce^{kt}}$
$60=e^{-Ce^{kt}}y$
My values for the variables are:
$y=0, t=0<br /> y=15, t=3<br /> y=?, t=10$
However, the solutions manual states the answer for the general soln is
$y=60e^{-Ce^{kt}}$ which is a difference of signs for the constant in the exponential. I'm not sure where I went wrong but I don't want to start plugging in values until I have the right general solution. Help please?
2. Originally Posted by emttim84
Instructions:
Biology. A population of eight beavers has been introduced into a new wetlands area. Biologists estimate that the maximum population the wetlands can sustain is 60 beavers. After 3 years, the population is 15 beavers. If the population follows a Gompertz growth model, how many beavers will be in the wetlands after 10 years?
Gompertz growth model:
$\frac{dy}{dt}=kyln\frac{60}{y}$
Here's what I did.
$dy=kyln\frac{60}{y}(dt)$
$\frac{dy}{yln\frac{60}{y}}=kydt$
$\int \frac{1}{y(ln60-lny)}dy=\int kdt$
$u=ln60-lny$
$du=-\frac{1}{y}dy$
$-du=\frac{1}{y}dy$
$- \int \frac {1}{u}du = \int kdt$
$-ln|u| = kt + C$
$-|ln60-lny| = e^{kt+c}$
$-ln\frac{60}{y} = Ce^{kt}$
$\frac{60}{y}=e^{-Ce^{kt}}$
$60=e^{-Ce^{kt}}y$
My values for the variables are:
$y=0, t=0<br /> y=15, t=3<br /> y=?, t=10$
However, the solutions manual states the answer for the general soln is
$y=60e^{-Ce^{kt}}$ which is a difference of signs for the constant in the exponential. I'm not sure where I went wrong but I don't want to start plugging in values until I have the right general solution. Help please?
3. Isn't $-|ln\frac{60}{y}| =$ whatever the same as - $ln\frac{60}{y}=$whatever since the negative sign is on the outside of the absolute value anyway?
4. Originally Posted by emttim84
Isn't $-|ln\frac{60}{y}| =$ whatever the same as - $ln\frac{60}{y}=$whatever since the negative sign is on the outside of the absolute value anyway?
no...no it doesnt...because say you have $\pm{x}\cdot{-1}$
all this will do is change "when" the negative occurs..not the existence of it...it is still +/-
5. Originally Posted by Mathstud28
no...no it doesnt...because say you have $\pm{x}\cdot{-1}$
all this will do is change "when" the negative occurs..not the existence of it...it is still +/-
Ahh ok...so because of the +/-, since we're talking about population growth, and you can't have a - if it's growth, then it's assumed that the + value of the absolute value should be used?
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