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http://mathoverflow.net/questions/97827?sort=oldest
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## A Parameter related to fractional chromatic number and Kneser Graphs
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Let $t \gt 0$ be an integer and $G$ is a simple graph with $\chi_f(G) = t$. Then $t$= inf `$\{ \frac{n}{k}| G \rightarrow KG(n,k)\}$` where $KG(n,k)$ is the Kneser graph.
Does there exist a $k$ such that $G \rightarrow KG(tk,k)$? If so, what is the complexity of finding such a $k$?
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## 1 Answer
Bondy and Hell, in their paper on the star chromatic number, prove that $$\chi_f(G) = \min \left\{ \frac{n}{k}: G\to KG(n,k) \right\}$$ Note that we have $\min$ here, not $\inf$. So the answer to your first question is yes. The proof of this result in my favorite book on Algebraic Graph Theory (which I am pretty sure follows that of Bandy and Hell) constructs the required homomorphism from the fractional coloring. Hence I suspect that the answer to your second question is that it is hard to determine the minimum value of $n$ needed. (But I do not know, and I do not recall seeing anything on this point.)
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http://stats.stackexchange.com/questions/7515/what-are-some-techniques-for-sampling-two-correlated-random-variables
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# What are some techniques for sampling two correlated random variables?
What are some techniques for sampling two correlated random variables:
• if their probability distributions are parameterized (e.g., log-normal)
• if they have non-parametric distributions.
The data are two time series for which we can compute non-zero correlation coefficients. We wish to simulate these data in the future, assuming the historical correlation and time series CDF is constant.
For case (2), the 1-D analogue would be to construct the CDF and sample from it. So I guess, I could construct a 2-D CDF and do the same thing. However, I wonder if there is a way to come close by using the individual 1-D CDFs and somehow linking the picks.
Thanks!
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2
– onestop Feb 22 '11 at 20:59
## 1 Answer
I think what you're looking for is a copula. You've got two marginal distributions (specified by either parametric or empirical cdfs) and now you want to specify the dependence between the two. For the bivariate case there are all kinds of choices, but the basic recipe is the same. I'll use a Gaussian copula for ease of interpretation.
To draw from the Gaussian copula with correlation matrix $C$
1) Draw $(Z=(Z_1, Z_2)\sim N(0, C)$
2) Set $U_i = \Phi(Z_i)$ for $i=1, 2$ (with $\Phi$ the standard normal cdf). Now $U_1, U_2\sim U[0,1]$, but they're dependent.
3) Set $Y_i = F_i^{-1}(U_i)$ where $F_i^{-1}$ is the (pseudo) inverse of the marginal cdf for variable $i$. This implies that $Y_i$ follow the desired distribution (this step is just inverse transform sampling).
Voila! Try it for some simple cases, and look at marginal histograms and scatterpolots, it's fun.
No guarantee that this is appropriate for your particular application though (in particular, you might need to replace the Gaussian copula with a t copula) but this should get you started. A good reference on copula modeling is Nelsen (1999), An Introduction to Copulas, but there are some pretty good introductions online too.
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+1 Nelsen is quite readable. I bought a copy a few years ago even after going through a lot of the online materials. – whuber♦ Feb 22 '11 at 21:36
I found a great tutorial paper and accompanying spreadsheet: behan.ws/copula.pdf and soa.org/files/xls/rsrch-copula-ex.xls – Pete Mar 7 '11 at 15:33
@Pete, the paper is indeed nice. The spreadsheet link, on the other hand, is dead – bgbg Sep 17 '12 at 12:59
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http://mathhelpforum.com/discrete-math/124768-determine-truth-values-following-statements.html
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# Thread:
1. ## Determine the truth values of the following statements
Let S(a,b) be the statement "ab=a/b".
Given that the domain (or universe of discourse) for both variables consists of all positive, non-zero integers, what are the truth values of these two statements?
a) ∀a∃b S(a,b)
b) ∀b∃a S(a,b)
Explain your reasoning when you answer.
2. Originally Posted by Runty
Let S(a,b) be the statement "ab=a/b".
Given that the domain (or universe of discourse) for both variables consists of all positive, non-zero integers, what are the truth values of these two statements?
a) ∀a∃b S(a,b)
b) ∀b∃a S(a,b)
Explain your reasoning when you answer.
Well, we can put them into plain English:
(a) For every integer $a\geq 1$ there is an integer $b\geq 1$ with $ab=\frac{a}{b}$.
This is clearly true by choosing $b=1$ for any $a$.
(b) For every integer $b\geq 1$, there is an integer $a\geq 1$ with $ab=\frac{a}{b}$.
Now, regardless of $a$, it is clear that $b^2=1$. So if $b=2$ then (b) implies $4=1$, which is a contradiction. So (b) is false.
3. Originally Posted by hatsoff
Well, we can put them into plain English:
(a) For every integer $a\geq 1$ there is an integer $b\geq 1$ with $ab=\frac{a}{b}$.
This is clearly true by choosing $b=1$ for any $a$.
(b) For every integer $b\geq 1$, there is an integer $a\geq 1$ with $ab=\frac{a}{b}$.
Now, regardless of $a$, it is clear that $b^2=1$. So if $b=2$ then (b) implies $4=1$, which is a contradiction. So (b) is false.
Thank you very much. Glad to get that one out of the way.
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http://mathoverflow.net/questions/83877?sort=votes
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Are Kazhdan-Lusztig $R$-polynomials the Poincare polynomials of the corresponding affine varieties
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Let $x$ and $y$ be two elements of $S_n$. Let $U(x,y)$ be the intersection $(BxB \cap B_{-} y B)/B$ inside the flag variety. Here $B$ and $B_{-}$ are the groups of upper and lower-triangular matrices respectively. Kazhdan and Lusztig define the $R$ polynomial $R_{x,y}(q)$. As explained in Theorem 1.3 of Deodhar, one such description is that `$$R_{x,y}(q) = \# U(x,y)(\mathbb{F}_q).$$`
Now, $U(x,y)$ is an affine variety, so I don't get to know that the weight filtration matches the cohomological filtration. But, looking at examples, it looks like the coefficient of $(-1)^{\ell(x) - \ell(y) - i } q^i$ in $R_{x,y}(q)$ is the Betti number $\dim H^i(U(x,y))$. For example, with $n=2$, $x=(21)$ and $y$ being the identity, then $U(x,y)$ is the torus `$\mathbb{A}^1 \setminus \{ 0 \}$`, it has $q-1$ points, and its betti numbers are $\dim H^0 = \dim H^1 = 1$.
This is probably something very standard, but I didn't find it the first few places I tried, and it seemed faster to ask here than to keep looking. Thanks!
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As you probably know, $\\#U(x,y)(\mathbb{F}_q)$ the supertrace of the Frobenius acting on $H^*_c(U(x,y))$; one possibility is that the eigenvalue of Frobenius on $H^{2\ell(x)-2\ell(y)-i}_c(U(x,y))$ is $q^{\ell(x)-\ell(y)-i}$. This would happen if the pullback to the torus in the Deodhar decomposition is injective in cohomology. – Ben Webster♦ Dec 19 2011 at 17:51
Unfortunately, that pullback is not injective. I have an example in $GL_5$ with $R$-polynomial $q^6 - 4q^5 + 7q^4 - 8q^3 + 7q^2 - 4q +1$. If these are also betti numbers, then $\dim H^1(U)=4$ and $\dim H^2(U) = 7$. The image of $H^2(U)$ must land in the part of $H^2(T)$ generated by $H^1(U)$ (because pullback is a map of rings) and that part has dimension $\binom{4}{2} = 6$. – David Speyer Dec 19 2011 at 18:03
I seem to recall seeing some paper by someone somewhere in Europe whose name I didn't recognize giving a geometric interpretation of R-polynomials, possibly in connection with Bott-Samelson resolutions or buildings or both. Unfortunately I don't remember enough to track it down. – Alexander Woo Dec 19 2011 at 21:03
@Ben: Dear Ben, could you please let me knowthe fulln name of the Mathematician associated with $\text{Deodhar Decomposition}$. – Chandrasekhar Dec 26 2011 at 15:08
Vinay Vithal Deodhar, according to genealogy.math.ndsu.nodak.edu/id.php?id=9927 . The decomposition Ben is referring to is from ams.org/mathscinet-getitem?mr=782232 – David Speyer Dec 26 2011 at 16:07
1 Answer
I think that this is one of these things that looks plausible in small examples but is false. For example, this would imply that the coefficients of R polynomials are alternating in $q$. This is implied by another conjecture called the Gabber-Joseph conjecture (roughly: coefficients of R-poynomials give dimensions of Ext groups between Verma modules), which is false. See "A counterexample to the Gabber-Joseph conjecture" by Brian Boe.
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2
Besides correcting the spelling of Brian Boe's first name, I'll add a specific MathSciNet listing (it didn't get into arXiv): Boe, Brian D. (1-GA). A counterexample to the Gabber-Joseph conjecture. Kazhdan-Lusztig theory and related topics (Chicago, IL, 1989), 1–3, Contemp. Math., 139, Amer. Math. Soc., Providence, RI, 1992. I discussed this in Remark 8.11 of my 2008 book on the BGG category, hedging my bets at the end on whether there will be a nice interpretation of the R-polynomials. It still seems to be an open question. – Jim Humphreys Dec 19 2011 at 21:22
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http://mathoverflow.net/questions/17997/deriving-symmetries-of-a-gauge-theory/18082
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## Deriving symmetries of a Gauge theory
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Hello, I don't know if this is a good place for exposing my problem but I'll try...
I have a gauge theory with action:
`$S=\int\;dt L=\int d^4 x \;\epsilon^{\mu\nu\rho\sigma} B_{\mu\nu\;IJ} F_{\mu\nu}^{\;\;IJ} $`
Where `$B$` is an antisymmetric tensor of rank two and `$F$` is the curvature of a connection `$A$` i.e: `$F=dA+A\wedge A$`, `$\mu,\nu...$` are space-time indices and `$I,J...$` are Lie Algebra indices (internal indices) I would like to find its symmetries. So I rewrite the Lagrangian by splitting time and space indices `$\{\mu,\nu...=0..3\}\equiv \{O; i,j,...=1..3\}$` I find:
`$L = \int d^3 x\;(P^i_{\;IJ}\dot{A}_i+B_i^{\,IJ}\Pi^i_{\,IJ}+A_0^{\;IJ}\Pi_{IJ})$`
Where `$\dot{A}_i = \partial_0 A_i$`, `$P^i_{\;IJ} = 2\epsilon^{ijk}B_{jk\,IJ}$` is hence the conjugate momentum of `$A_i^{\,IJ}$`
`$B_i^{\,IJ}$` and `$A_0^{\;IJ}$` being Lagrange multipliers we obtain respectively two primary and two secondary constraints:
`$\Phi_{IJ} = P^0_{\;IJ} \approx0$`
`$\Phi_{\;\;IJ}^{\mu\nu} = P^{\mu\nu}_{\;\;IJ} \approx0$`
`$\Pi^i_{\,IJ} = 2\epsilon^{ijk}F_{jk\,IJ} \approx0$`
`$\Pi_{IJ}=(D_i P^i)_{IJ} \approx0$`
Where `$P^0_{\;IJ}$` are the conjugate momentums of `$A_0^{\,IJ}$` and `$P^{\mu\nu}_{\;\;IJ}$` those of `$B_{\mu\nu}^{\;\;IJ}$`. Making these constraints constant in time produces no further constraints.
Whiche gives us a general constraint:
`$\Phi = \int d^3 x \;(\epsilon^{IJ}P^0_{\,IJ}+\epsilon_{\mu\nu}^{IJ}\;P^{\mu\nu}_{\;\;IJ}+\eta^{IJ}\Pi_{IJ}+\eta_i^{IJ}\Pi^i_{\;IJ})$`
Each quantity `$F$` have thus a Gauge transformation `$\delta F = \{F,\Phi\}$` where `$\{...\}$` denotes the Poisson bracket.
Knowing that this theory have the following Gauge symmetry:
`$\delta A = D\omega$`
`$\delta B = [B,\omega]$`
Where `$\omega$` is a 0-form, I would like to retrieve these transformations using the relation below. (where `$\Phi$` is considered as the generator of the Gauge symmetry) but my problem is that I don't know how to proceed, I already did this with a Yang-Mills theory and it worked... but for this theory it seems to le intractable! Someone to guide me?
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## 2 Answers
Unless for some reason you absolutely must work within the Hamiltonian approach, you can just directly look for the complete set of (infinitesimal Lie point) symmetries of the Euler--Lagrange equations or of the action itself. The procedure is standard and described in many good books. For instance, you can look into those by Olver (more math-y) or Stephani (somewhat closer to physics). Using the theory from these books you can also verify whether the transformation at the end of your question is indeed a symmetry.
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Although the second book looks very interesting, I hesitate to start reading for the following reason: - Symmetries in physics are related to the indetermination generated by the fact that the Hamiltonian contains constraints times arbitrary parameters, if one tries to write down two possible future states corresponding to two different parameters (using the Poisson bracket with the Hamiltonian), these two states will be physically equivalent whereas different by a term consisting of constraints times arbitrary parameters. Suggesting that constraints are the generators of these Gauge transf. – Pedro Mar 14 2010 at 11:19
So I don't thing that a pure algebraic method will exhibit these "Gauge transformations"!! On the other hand, I confess I misunderstand the fact that a Lagrangian which did not ask anyone find itself invariant under symmetries exhibited by the passage to the Hamiltonian formalism!! I mean, a priori the form of a Lagrangian has no relation with will be done in the Hamiltonian framework. – Pedro Mar 14 2010 at 11:28
PS: I don't want to verify that this is a symmetry, it's already done. I want to derive this symmetry, actually I have a little more complicated action which I need to analyse... – Pedro Mar 14 2010 at 12:26
First of all, the method outlined above is systematic and it should yield, inter alia, the gauge symmetries (and also the obvious symmetries like, say, Poincare or conformal ones). Moreover, while the notion of symmetry can be considered within the Hamiltonian framework, the notion of symmetry (including but not restricted to the gauge symmetry) does make perfect sense without reference to this framework, see e.g. Olver's book I linked above. I hope that perhaps some other people can better understand the menaing of your comments and be of more help. – mathphysicist Mar 15 2010 at 23:29
### 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.
Although this question is over a year old, some readers may be interested in the following comments.
1. The symmetries of the Lagrangian and the action are not identical.
2. I don't know why you cannot recover the gauge transformations from the constraints, perhaps it has to do with not including the $B_{0i}$ fields in the second expression for the Lagrangian?
3. The constraints give infinitesimal transformations leading to the usual gauge transformations, but there seem to be symmetries which are non-trivial extensions of these.
4. For example, for a gauge group SU(n), the transformations $$A' = UAU^\dagger + \phi$$
B' = UBU^\dagger + UAU^\dagger\wedge\phi + \phi\wedge UAU^\dagger + \phi\wedge\phi
where $U(x) \in$ SU(n), and $\phi = -dUU^\dagger$, leaves the action invariant.
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Thank you for these notes, I just want to say that I finally found how to derive these Gauge transformations (more or less because I arrived to derive them for spatial components only), the demonstration s here docs.google.com/… until page 10 (in french) – Pedro Apr 30 2011 at 8:38
You are welcome. Getting the transformations for the spatial components should be sufficient for the dynamics, and you can extend them to all the components. – Amitabha Lahiri May 2 2011 at 7:32
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http://physics.stackexchange.com/tags/velocity/hot
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Tag Info
Hot answers tagged velocity
28
Double light speed
One of the results of special relativity is that a particle moving at the speed of light does not experience time, and thus is unable to make any measurements. In particular, it cannot measure the velocity of another particle passing it. So, strictly speaking, your question is undefined. Particle #1 does not have a "point of view," so to speak. (More ...
21
Why does the mass of an object increase when its speed approaches that of light?
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Is the escape velocity at Geosynchronous Earth Orbit 0km/hr?
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How to deduce the theorem of addition of velocities?
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Why does the mass of an object increase when its speed approaches that of light?
The complete relevant text in the book is The de Broglie wave equation relates the velocity of the electron with its wavelength, λ = h/mv ... However, the equation breaks down when the electron velocity approaches the speed of light as mass increases. ... Actually, the de Broglie wavelength should be $$\lambda = \frac hp,$$ where p is the ...
9
Double light speed
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Can you completely explain acceleration to me?
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Is instantaneous velocity an abstraction?
Instantaneous velocity can never be measured since there is no way in the real world to do anything instantaneously. All measurements take some amount of time to peform. For example the comment to the question mentioned using the Doppler effect to measure instantaneous velocity. That is not possible since to measure the frequency of a wave you have to ...
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How long must escape velocity be maintained?
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7
What is terminal velocity?
Terminal velocity is the maximum velocity that you can reach during free-fall. If you imagine yourself falling in gravity, and ignore air resistance, you would fall with acceleration $g$, and your velocity would grow unbounded (well, until special relativity takes over). This effect is independent of your mass, since $F = ma = mg \Rightarrow a = g$ Where ...
7
How much does it cost every time your car stops for a traffic light?
Here is a very basic estimation: The kinetic energy of a 1000 kg car moving at 60 km/h is $$E=\frac{mv^{2}}{2}=\frac{1000kg(16.7m/s)^{2}}{2}=138.9 kJ$$ The heat of gasoline combustion is 47 MJ/kg = 35000 kJ/litre. Assuming 10% efficiency of the car's engine, you would need to burn $$\frac{138.9 kJ}{0.1\cdot35000kJ/l}=0.04 litre$$ of gasoline to accelerate ...
7
How to deduce the theorem of addition of velocities?
I endorse Ron's answer – it's the systematic way to proceed. The velocity $v/c$ may be written as $\tanh \eta$ where $\eta$, the rapidity or whatever, is the hyperbolic (Minkowski) counterpart of the (Euclidean) angle. The addition of velocities then boils down to an addition formula for $\tanh(\eta_1+\eta_2)$ because the rapidities just add additively. Let ...
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From how high could have Felix Baumgartner jumped without disintegrating like a shooting star?
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Is it possible to calculate weight of person with sensors found in today's smartphone?
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Are circularly defined {velocity, distance, and time} a problem in physics?
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Why are coordinates and velocities sufficient to completely determine the state and determine the subsequent motion of a mechanical system?
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6
How can an object's instantaneous speed be zero and it's instantaneous acceleration be nonzero?
Suppose you throw a ball upwards at some velocity $v$. When you catch it again it's traveling downwards at (ignoring air resistance) a velocity of $-v$. So somewhere in between throwing and catching the ball it must have been stationary for a moment i.e. it's instaneous velocity was zero. Obviously this was at the top of it's travel. When you throw the ball ...
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How fast would a spaceship have to go to reach Alpha Centauri within a person's lifetime?
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6
How fast would a spaceship have to go to reach Alpha Centauri within a person's lifetime?
If you are not interested in relativistic effects, the answer to your question is easy to workout. According to Wikipedia, Alpha Centauri is 4.24 ly away (4.0114x$10^{16}\mathrm{m}$). So to get there in 60 years ($1892160000\mathrm{s}$). So your non-relativistic answer is \$v = \frac{d}{t} = \frac{4.0114 \times 10^{16}}{1892160000} = 21200000 ...
6
How to brake 'beautifully'?
We notice sudden changes in anything. We don't notice gradual change, whether in time or spread over space. If the car is moving uniformly along straight flat road, its acceleration a=0. Its velocity v is constant. When the brake pedal is pushed, friction causes the car to decelarate. a = some negative number. You can't avoid that. You want to ...
6
When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it?
Short, short version: It's complicated. Slightly longer version: Internal combustion engines have at least two relevant performance characteristics: power and torque. Furthermore the maximum attainable values for both are functions of the current engine speed (RPM). Acceleration will cease if the current requirement for either power or torque equal the ...
6
When driving uphill why can't I reach a velocity that I would have been able to maintain if I started with it?
Here's my guess: As you know, internal combustion engines burn fuel. The power output of the engine is a function of both the current RPM and the amount of fuel you inject. But there's a catch: the engine can burn a limited amount of fuel per cycle, and therefore the higher the RPM, the more fuel can be combusted. So (as @dmckee stated), at a given gear the ...
6
What's the physical difference between the quantities $\langle v_{i}v_{j}\rangle$ and $\langle v_{i}\rangle\langle v_{j}\rangle$?
think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result. When the functions are arbitrary, the average of the product quantifies statistical ...
5
What does this observation of instantaneous velocity in Brownian particles mean?
The article makes no sense. Einstein realized that matter was composed out of atoms, so the number of collisions of a Brownian particle with the surrounding molecule is finite in a finite period of time. However, for times $t$ much longer than the typical scale between the collisions, the particle moves by a distance scaling like $\sqrt{t}$. It follows that ...
5
How fast is earth moving through the universe
Velocity does indeed have to be measured relative to something. We can measure our radial velocity relative to any other astronomical object we care to, by measuring Doppler shifts. But if you want to know our velocity "relative to the Universe as a whole" rather than relative to any one object, we have to be a bit careful to define our terms. Because the ...
5
A Basic Question about Gravity, Inertia or Momentum or something along those lines
The thing you throw in the air is also traveling at the same speed you are, in the same direction. When you throw it up, it doesn't matter that the earth below is moving backwards at speed, nor that the moon is moving past even more quickly, nor that the earth itself is spinning and moving relative to the sun. The ball has a speed and direction and ...
5
Thought experiment that seems to involve something growing at twice the speed of light. Is anything wrong?
The theory of relativity is often expressed as "Nothing can be faster than the speed of light". This is wrong. You can always define certain things that seem to "move" faster than the speed of light. For example, if you shine a laser pointer at the moon and wiggle it around, the little red dot on the moon's surface can "move" faster than the speed of ...
5
What will be the relative speed of the fly? [duplicate]
Have a look at http://en.wikipedia.org/wiki/Galilean_invariance. This is not too mathematical and explains what's going on. The basic idea is that there is no such thing as absolute motion. For example, because the earth is rotating as I sit here typing I'm moving at about 800 miles per hour. Why am I not splattered against my computer screen? It's because ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://math.stackexchange.com/questions/tagged/brownian-motion+control-theory
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Tagged Questions
0answers
51 views
Controlling auto-correlated 1D Brownian motion
I have 1D Brownian motion process $x(t)$, and ability to control it. The control allows to shift the $x$ by $D$ at any time. I need the controlled process to be zero-mean, and to use the control ...
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http://math.stackexchange.com/questions/76708/how-to-determine-a-shape-is-convex-by-giving-polar-form-polynomial-equation
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# How to determine a shape is convex by giving polar form polynomial equation?
It is easy to determine concave, convex curve in xy coordinate. But I am placing a question that I only have a polar polynomial equation like r(ang) = a4*ang^4 + a3*ang^3 + .... + a0;
How I can tell whether it is convex shape by given a equation like this? ** Convert back to xy coordinate is NOT an allowable option.
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Polar equations that don't have trigonometric functions in them often look spiral-like... – J. M. Oct 28 '11 at 16:51
You are right, but assume I only care 1 complete turn (360 deg) and the discontinuous at 0deg/360deg can be ignored. Now how I can find whether the shape is convex? – Marco Oct 28 '11 at 17:21
## 1 Answer
In general for a simple smooth closed curve ${\bf R} = {\bf R}(t)$ traversed counterclockwise, let ${\bf v} = \frac{d{\bf R}}{dt}$ be the velocity vector and ${\bf a} = \frac{d{\bf v}}{dt}$ the acceleration. The curve is convex if $({\bf v} \times {\bf a})_3 \ge 0$ everywhere. For the polar curve $r = r(\theta)$ parametrized by $\theta$, if ${\bf u}_r$ and ${\bf u}_\theta$ are the unit vectors in the radial and counterclockwise directions, we have ${\bf v} = r' {\bf u}_r + r {\bf u}_\theta$ and ${\bf a} = (r'' - r) {\bf u}_r + 2 r' {\bf u}_\theta$, so ${\bf v} \times {\bf a} = (r^2 + 2 (r')^2 - r r'') {\bf u}_r \times {\bf u}_\theta$, and the condition is $r^2 + 2 (r')^2 - r r'' \ge 0$ for all $\theta$. Note that we can't just "ignore" a discontinuity in $r'$, which could correspond to a sharp corner that makes the curve non-convex.
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http://mathhelpforum.com/discrete-math/83698-tell-me-sequence-below-question.html
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# Thread:
1. ## tell me the sequence in the below question;
Find the 15th term of the following sequence
1/2,1/4,1/8,... ( Use sequence formula)
2. $a_n = \frac{1}{2^n}$
3. ## sequence formula
sir please tell me more easily
i cant understand that how can i solve my question according to your given formula..
please kindly again send me the proper solution
4. Originally Posted by bc080402276
sir please tell me more easily
i cant understand that how can i solve my question according to your given formula..
please kindly again send me the proper solution
What Plato gave you should be enough to help you solve the problem. He gave you the pattern of the sequence for an arbitrary n value. Since you're looking for the 15th term of the sequence, you want to find $a_{15}$. And by the formula provided, what would that be?
5. Originally Posted by bc080402276
sir please tell me more easily
i cant understand that how can i solve my question according to your given formula..
please kindly again send me the proper solution
You substitute $n = 15$ into the given formula.
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http://mathhelpforum.com/number-theory/86539-sum-certain-consecutive-odd-numbers-57-2-13-2-find-numbers-print.html
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# The sum of certain consecutive odd numbers is 57^2 – 13^2. Find the numbers.
Printable View
• April 29th 2009, 06:52 PM
chakravarthiponmudi
The sum of certain consecutive odd numbers is 57^2 – 13^2. Find the numbers.
Problem.
The sum of certain consecutive odd numbers is $57^2$– $13^2.$ Find the numbers.
solution
$\sum 1+3+5+....+(2n-1) = n^2 [1]$
$57^2=n^2$
$n=57$
also from equ 1
$n=\frac{l+1}{2}$
therefore
$\frac{l_1+1}{2} = 57$
$l_1=113$
similarly
$13^2=n^2$
$l_2=25$
$57^2 - 13 ^2 = 27^2+29^2+....113^2$
a = 27 , d = 2
l= 113
a+nd = 113
27 + 2n = 113
n = 43
this is where I am facing the problem. To satisfy the condition I should have n = 44 and not 43.
because 57 - 13 = 44. other way to find the n.
• April 30th 2009, 03:12 AM
Opalg
Quote:
Originally Posted by chakravarthiponmudi
a+nd = 113
The formula for the n'th term of an arithmetic progression is a + (n–1)d, not a + nd.
• April 30th 2009, 06:28 AM
Soroban
Hello, chakravarthiponmudi!
You already have all the information you need.
Quote:
The sum of certain consecutive odd numbers is $57^2 - 13^2$. . Find the numbers.
Solution
$\sum^n_{k=1}(2k-1) \:=\:1+3+5+ \hdots+(2n-1) \:=\: n^2$
That formula says:
. . the sum of the first $n$ odd numbers is $n^2.$
So $57^2$ is the sum of the first 57 odd numbers.
. . and $13^2$ is the sum of the first 13 odd numbers.
Hence: . $57^2 - 13^2 \:=\:\text{(first 57 odd numbers)} - \text{(first 13 odd numbers})$
. . . . . . . . . . . . $= \;(1 + 3 + 5 + \hdots + 113) - (1 + 3 + 5 + \hdots + 25)$
. . . . . . . . . . . . $= \;27 + 29 + 31 + \hdots + 113$
All times are GMT -8. The time now is 06:18 AM.
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http://cstheory.stackexchange.com/questions/10169/from-edge-disjoint-paths-to-independent-paths
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# From edge-disjoint paths to independent paths
Let $\mathcal{G}_k$ denote the set of all graphs that contain two vertices $x,y$ and $k$ edge-disjoint $x-y$ paths.
Define $f(k)$ to be the maximum such that for every graph $G\in \mathcal{G}_k$ there are two vertices $x',y'$ with $f(k)$ independent $x'-y'$ paths in $G$.
Here, a set of paths is independent if none contains an internal vertex of another.
Are any lower bounds for $f(k)$ known in the literature?
In particular, I need $f(3)=3$ in an algorithm where I need to find some obstructions in a graph. This is not hard to prove, but I'm wondering whether it is known in the literature, maybe as a special case of a more general theorem.
As a different formulation, how large can you make $f(k)$ in the following statement?
If $G$ is an undirected graph containing two vertices $x,y$ and $k$ edge-disjoint $x-y$ paths, then $G$ contains two vertices $x',y'$ with $f(k)$ independent $x'-y'$ paths.
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3
$f(k)$ is a maximum of what? What quantity does it denote? – Shir Feb 10 '12 at 22:30
Some doubts: in a graph $G \in \mathcal{G}_k$, are other non $k$ edge-disjoint paths allowed between $x$ and $y$? A graph $G \in \mathcal{G}_k$ must contain at least one pair of vertices $x$ and $y$ connected with $k$ disjoint edges? or exactly one pair? What about a graph that contais a pair of vertices connected with $k$ disjoint edges and also a pair of vertices connected with $k+1$ disjoint edges? – Marzio De Biasi Feb 10 '12 at 23:16
Apologies, I corrected the definition of $f(k)$. – Serge Gaspers Feb 10 '12 at 23:51
2
I could be mistaken but it seems that the well-known series parallel graph (the recursive diamond graph, see cseweb.ucsd.edu/~dasgupta/254-embeddings/claire.pdf) can be used to show that $f(k) = 2$ for any value of $k$. – Chandra Chekuri Feb 12 '12 at 15:00
@ChandraChekuri: Actually, for the diamond graph $G_2$, we have two vertices $s,t$ with $4$ edge-disjoint $s-t$ paths, but we have 3 independent $x'-y'$ paths, where $x'=s$ and $y'$ is one of the 2 degree-4 vertices besides $s$ and $t$. I believe that these graphs can be used, though, to show that $f(k)=3$ for any $k\ge 3$. – Serge Gaspers Feb 12 '12 at 16:07
show 1 more comment
## 1 Answer
It turns out that $f(k)=3$ for all $k\ge 3$. Recursive diamond graphs are extremal. See arXiv:1203.4483 for a proof.
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2
yippee. a cstheory.SE citation :) – Suresh Venkat♦ Mar 21 '12 at 16:49
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http://cstheory.stackexchange.com/questions/3692/dijkstra-parallelization/3693
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# Dijkstra parallelization
I'd like to know what is the best method to parallelize the Dijkstra algorithm.
Thanks.
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Try a literature search for parallel shortest path. You could start with e.g. "A Randomized Parallel Algorithm for Single-Source Shortest Paths" by Philip N. Klein and Sairam Subramanian in J. of Algorithms and papers citing it. – Warren Schudy Dec 23 '10 at 20:12
thanks for the link. I'll take a look at that paper. – jutky Dec 26 '10 at 22:12
## 2 Answers
There's a parallel algorithm for shortest paths from Carla Savage's 1977 Ph.D. thesis that consists of forming a square matrix with 0 on the diagonal entries, the length of the edges on the off-diagonal entries corresponding to edges, and a suitably large number for the remaining off-diagonal entries, and then repeatedly squaring this matrix in the (min,+) algebra.
After $\lceil\log_2 n\rceil$ squaring steps, the numbers in the resulting matrix are the distances between each pair of vertices. Each squaring step is easy to parallelize with logarithmic time and cubic work. So overall this algorithm takes $O(\log^2 n)$ time and $O(n^3\log n)$ work. With a little care (using a slightly more complicated algebra) this can be modified so that it also provides the first step of each shortest path in the same time and work bounds.
However, if you only want single source shortest paths rather than all pairs shortest paths, I don't know of anything that comes close to the total work of the sequential Dijkstra algorithm and that provides much in the way of a parallel speedup.
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Thanks for the idea. But I use a slight variation of the single source shortest path Dijkstra. – jutky Dec 9 '10 at 22:55
Is what you are describing above the same as what was based on an idea of Steve Hedetniemi as described here?: highered.mcgraw-hill.com/sites/dl/free/0072880082/299355/… and also here: deepblue.lib.umich.edu/handle/2027.42/59763 – Joseph Malkevitch Dec 10 '10 at 3:04
The matrix algebra part is the same, but the references you give don't seem to have the squaring trick for getting high powers of the matrix much more quickly than multiplying together n copies of the matrix. So they end up doing O(n^4) work, instead of closer to O(n^3) in Savage's version, and they're not as parallel. – David Eppstein Dec 10 '10 at 3:20
– JɛffE Dec 10 '10 at 4:36
All-pairs shortest paths by repeated (min,+) matrix squaring is closely related to Floyd-Warshall, but they are not the same. In this respect, it is useful to think of Floyd-Warshall as (min,+) Gaussian elimination. Both approaches lend themselves to coarse-grained parallelisation, as discussed in http://dx.doi.org/10.1007/3-540-48224-5_15
The work required for repeated squaring can be improved from O(n^3 log n) to O(n^3) by the technique of selective path doubling, introduced by Alon, Galil and Margalit, and also discussed in the above paper.
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http://sumidiot.wordpress.com/2009/11/05/neighbors-in-farey-sequences/
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# ∑idiot's Blog
The math fork of sumidiot.blogspot.com
## Neighbors in Farey Sequences
I was reading about Ford circles today, and was planning on writing about them. But then I noticed that one of the things I wanted to say was based on some things I didn’t say about Farey sequences when I talked about them a few days ago. So I thought I’d go back and fill in that gap, and save Ford circles for tomorrow.
So what did I skip? Well, I mentioned that to get from one Farey sequence to the next, you throw in mediants whose denominators aren’t too big. Is there a way, given a fraction $h/n$ in $F_n$ to figure out what terms in $F_{n-1}$ it was the mediant of? Well, we could, of course, list the terms in $F_{n-1}$, in order, and then pick the two that surround the value $h/n$. I’m after another way.
Since $h/n$ is supposed to be reduced, $h$ and $n$ are relatively prime, and so there are integers $s$ and $t$ such that $hs-nt=1$ (this usually (in my mind) gets written with a “+”, not a “-”, but just change the sign of your $t$). Once you’ve found such a pair $(s,t)$, all of the other pairs that satisfy the same equation are of the form $(s+mn,t+mh)$, where $m$ is any integer. Among the $s+mn$, there is only one in the interval $[0,n)$. Abusing notation a bit, or assuming you picked well to start, let’s call this value $s$, and its partner $t$. I claim that $t/s$ is the term directly before $h/n$ in $F_n$.
Before talking about $t/s$, I should have checked that $s\neq 0$. If this is not the case, then $hs-nt=1$ means $n=1$, in which case we are in $F_1$ which is just 0/1 and 1/1. Not much interesting going on there. So let’s go ahead and assume $s>0$.
Now, the linear equation $hs-nt=1$ gets re-written $h/n-t/s=1/(ns)$ to show that $h/n>t/s$ (so the fractions are at least in the correct order). Messing about with inequalities some more, you can check that $0\leq t<s$, and since $s$ and $t$ are relatively prime (they satisfy the linear equation), the fraction $t/s$ is actually in $F_n$. Let $a/b$ be the predecessor of $h/n$ in $F_n$ (we want to show $a=t,b=s$). Since it is a neighbor, we showed last time that then $hb-na=1$. But now we’re saying that all solutions of the equation $hx-ny=1$ have a fixed form. That is, $b=s+mn$ and $a=t+mh$ for some integer $m$. But if $m\neq 0$ then $b$ is not in $[0,n)$, and so $a/b$ is not in $F_n$. So $m=0$ and $a=t,b=s$, and we’re done.
Man, ok, so, after all that, we now know how to find the term before $h/n$ in $F_n$. I’ll call it $t/s$, sticking with our notation above. What about the term after $h/n$? You could play mostly the same games as above, I suppose. Or, if the successor is $T/S$, then we know that $h=t+T$ and $n=s+S$, since we made $h/n$ as a mediant. So $T=h-t$ and $S=n-s$, easy enough. [Update 20091109: You have to be careful with this. The above method for finding the successor to $h/n$ in $F_n$ only works because you are in $F_n$. If you look for successors in $h/k$ in $F_n$, where $k<n$, this method, using the predecessor, is wrong. For example, 1/4 is the predecessor of 1/3 in $F_5$, but the fraction (1-1)/(3-4) suggested by the previous method is clearly NOT the sucessor of 1/3 in $F_5$]
So we know the neighbors of $h/n$ in $F_n$. What is the next Farey sequence when $h/n$ gets a new neighbor? Let’s stick to the next time it gets a new neighbor “on the left” (less than itself). We know that when that happens, it had to be because we made the mediant with the old neighbor, our friend $t/s$. So the next neighbor is $(t+h)/(s+n)$, which doesn’t show up until $F_{s+n}$. The newest neighbor after that will be the mediant with this new neighbor, i.e., $(t+2h)/(s+2n)$, in $F_{s+2n}$. Continuing on, we see that all of the neighbors are of the form $(t+mh)/(s+mn)$, for some positive integer $m$. I called these $(1,m)$-weighted mediants (of $t/s$ and $h/n$) yesterday.
The story goes pretty much the same on the right side. I called the neighbor on the right, in $F_n$, $T/S$. The next neighbors will be $(h+T)/(n+S)$, $(2h+T)/(2n+S)$, etc. These are the $(m,1)$-weighted mediants of $h/n$ and $T/S$.
We can put both of these into the same statement by noticing that the $(1,-m)$ mediant of $t/s$ with $h/n$ is the same as the $(m-1,1)$ mediant of $h/n$ with $T/S$. In other words, all of the neighbors of $h/n$, in all of the Farey sequences, can be written as $(1,m)$ weighted mediants of $t/s$ with $h/n$. They will have numerators of the form $a=t+mh$ and denominators of the form $b=s+mn$. Guess what? These are exactly all of the solutions to $hx-ny=1$. Go back a few paragraphs, it’s the same thing I said there.
In summary, the neighbors to $h/n$, in any Farey sequence, are in one-to-one correspondence with solutions to $hx-ny=1$.
If I knew what I were doing, I could probably have gotten to this statement a bit more quickly. But I don’t remember the last time I knew what I was doing, so there you have it.
### Like this:
Tags: farey, mablowrimo
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### 4 Responses to “Neighbors in Farey Sequences”
1. Jaime Says:
November 5, 2009 at 9:51 am | Reply
Farey sequences are a cool subject indeed, I’m thoroughly enjoying your series on them. When you find time, go back to Euler’s Flying Circus and give 192 and 198 a try…
198 is arguably the hardest problem posted at Project Euler if you look at how long its been posted and how many people have got it right. Yet it’s quite straightforward to solve once you know all you now know…
• sumidiot Says:
November 5, 2009 at 7:18 pm | Reply
Thanks for the encouragement! And for the pointer to fun Project Euler problems. I wasn’t to get to those two for a while, working my way through in order. By the time I got there, I’d probably forget all of the math I’d need to solve them. It might be a few days, but I’ll attack those two next.
• sumidiot Says:
November 9, 2009 at 10:07 am | Reply
Curse you Jaime! I’ve been working on 198, and spending far too much time on it. I’ve got 3 “solutions” going, but all of them are too slow to ramp up to the bound 10^8. Got an idea how to improve things… but not sure when I should let myself get back to the problem, versus doing “real work”.
2. Problem 198 – Ambiguous Numbers « Leonhard Euler’s Flying Circus Says:
December 9, 2009 at 11:32 pm | Reply
[...] I’m getting to this problem a little out of order. Jaime told me to. I was talking about Neighbors in Farey Sequences, on another blog, and he said this problem would fit well with what I was doing. In fact, if you [...]
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http://mathhelpforum.com/statistics/57651-continuous-random-variables-probability-density-functions.html
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# Thread:
1. ## Continuous random variables – probability density functions
The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.
$<br /> f(x) = \begin{cases} <br /> k(x + x^2), <br /> & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}<br />$
How could I know it represent a p.d.f or not?
2. evaluate if:
$\int_{-\infty}^{\infty} f(x)dx=1$
3. Well, my problem is not to finding k. I’ve confusion to recognize is it p.d.f or not.
If I try on this way,
f(-0.5) = k(0.75)
f(1) = 2k
I didn’t understand how the above info/working or else will show
f(x) < 0. So it could not be a p.d.f. (As my book’s answer.)
4. If $\int_{-1}^{1} f(x)dx=1$
So: k=3/2.
5. $\int_{-\infty}^{\infty} f(x)dx=1$
you gotta "chop" the integral in three parts:
$\int_{-\infty}^{-1} 0dx+\int_{-1}^{1} k(x + x^2)dx+\int_{1}^{\infty} 0dx=1$
continue...
6. Originally Posted by geton
If $\int_{-1}^{1} f(x)dx=1$
So: k=3/2.
that's the right way can't say if that's the answer gotta do it but im gonna sleep atm
7. Answer:
f(x) < 0.
Therefore f(x) cannot be a probability density function.
8. Originally Posted by geton
The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.
$<br /> f(x) = \begin{cases} <br /> k(x + x^2), <br /> & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}<br />$
How could I know it represent a p.d.f or not?
$f(x) = k(x + x^2)$ can only represent a pdf if it's non-zero for all values of x. This is clearly NOT the case therefore it's cannot represent a pdf.
9. Originally Posted by mr fantastic
$f(x) = k(x + x^2)$ can only represent a pdf if it's non-zero for all values of x. This is clearly NOT the case therefore it's cannot represent a pdf.
Why I'm trying to prove f(x) < 0? Totally time wasting...
10. Originally Posted by geton
The following could represent a probability density function (p.d.f.) or not. If could, find the value of k.
$<br /> f(x) = \begin{cases} <br /> k(x + x^2), <br /> & \text {$-1 \le x \le 1,$} \\ 0, & \text {otherwise.} \end{cases}<br />$
How could I know it represent a p.d.f or not?
Hi geton,
One of the requirements for a pdf is that $f(x) \geq 0$ for all values of x. Now
$x + x^2 < 0 \text{ for } -1 < x < 0$
and
$x + x^2 > 0 \text{ for } 0 < x < 1$,
so if $k > 0$ then $f(x) \geq 0$ fails for $-1 < x < 0$, and if $k < 0$ then $f(x) \geq 0$ fails for $0 < x < 1$, so no value of $k \neq 0$ will make f suitable for a pdf.
The only remaining possibility is $k = 0$ but then $\int f(x) dx = 0$, so that won't work either.
11. Originally Posted by awkward
Hi geton,
One of the requirements for a pdf is that $f(x) \geq 0$ for all values of x. Now
$x + x^2 < 0 \text{ for } -1 < x < 0$
and
$x + x^2 > 0 \text{ for } 0 < x < 1$,
so if $k > 0$ then $f(x) \geq 0$ fails for $-1 < x < 0$, and if $k < 0$ then $f(x) \geq 0$ fails for $0 < x < 1$, so no value of $k \neq 0$ will make f suitable for a pdf.
The only remaining possibility is $k = 0$ but then $\int f(x) dx = 0$, so that won't work either.
Thank you so much. I really appreciate it.
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http://physics.stackexchange.com/questions/34784/how-would-be-tv-reception-in-a-spaceship-travelling-close-to-speed-of-light
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# How would be tv reception in a spaceship travelling close to speed of light?
My little brother has made me a tough question (specially for a computer science engineer).
Imagine that there is a spaceship orbiting earth close at nearly speed of light (say 99%). Someone in earth transmits a tv signal to spaceship.
What will the astronauts see in their television?
• Nothing?
• Accelerated version of what was recorded? (My best guess)
• Normal speed of what was recorded long time ago?
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## 1 Answer
Relativistic motion in a circle is a surprisingly subtle problem to treat. You might want to have a look at Invariant spacetime - distance - Circular Motion, which addresses a related problem. That answer contains a link to a paper that goes into the problem in more detail.
The end result is that the observer on the Earth sees time moving slowly for the astronaut, while the astronaut sees time moving fast for the people on Earth. The factor by which time runs fast or slow is $\gamma$ where:
$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$
The factor $\gamma$ is greater than 1 for any non-zero velocity $v$ and it tends to infinity as $v$ tends to the speed of light. So if the Earth broadcasts a programme lasting 1 hour the spaceship will receive the whole broadcast in 1/$\gamma$ hours so the programme will appear to the astronauts to be speeded up.
NB the TV transmission from the Earth would be blue shifted, so a normal TV couldn't receive it. However I'm assuming the astronaut's TV is sophisticated enough to receive the changed frequency and cope with the timing changes needed to decode the picture.
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2
– John Rennie Aug 23 '12 at 14:46
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http://mathhelpforum.com/algebra/148181-algebra-2-graphing-help.html
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Thread:
1. Algebra 2 graphing help
Hey guys, what's up? I know this is going to sound dumb, but I forgot how to do it.
xy=-3. How would you graph that? Im drawing a blank right now. Thanks for the help!
2. Originally Posted by robot214
Hey guys, what's up? I know this is going to sound dumb, but I forgot how to do it.
xy=-3. How would you graph that? Im drawing a blank right now. Thanks for the help!
recall that the form $y=\frac{1}{x}$ is called an equilateral hyperbola
first rewrite the equation as $y=-\frac{3}{x}$ the - sign will flip the graph and the 3 will expand it.
here is graph of $y=\frac{1}{x}$ and $y = -\frac{3}{x}$
3. Thanks so much!
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http://mathhelpforum.com/trigonometry/67832-trigonometric-concepts.html
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# Thread:
1. ## trigonometric concepts
Reading this, something rings in my mind, but I cant find the room the ring comes from I did this last year, but since I took an online class and therefor have no teacher,it's difficult to get the infos. The only thing I can remember is how to calculate sin,cos,tan using a scientific calculator
Mrs. Imconfused is having difficulty understanding the trigonometric concepts. She doesn't know when to use the Primary Trigonometric Ratios (sin, cos and tan), the Sine Law or the Cosine Law. She also needs clarification on what each one (primary trig ratios, sine law, cosine law) can be used to calculate. Provide an explanation that will help to solidify her understanding. Be sure to discuss what each one will calculate and how to decide which one to use. Explanations, diagrams, examples, formulas and mathematical terminology should all be included in your solution. Be thorough!
2. Originally Posted by Serialkisser
Reading this, something rings in my mind, but I cant find the room the ring comes from I did this last year, but since I took an online class and therefor have no teacher,it's difficult to get the infos. The only thing I can remember is how to calculate sin,cos,tan using a scientific calculator
Mrs. Imconfused is having difficulty understanding the trigonometric concepts. She doesn't know when to use the Primary Trigonometric Ratios (sin, cos and tan), the Sine Law or the Cosine Law. She also needs clarification on what each one (primary trig ratios, sine law, cosine law) can be used to calculate. Provide an explanation that will help to solidify her understanding. Be sure to discuss what each one will calculate and how to decide which one to use. Explanations, diagrams, examples, formulas and mathematical terminology should all be included in your solution. Be thorough!
When in doubt, check Wikipedia
Trigonometry - Wikipedia, the free encyclopedia
3. ## Trigonometry
Hello Serialkisser
Originally Posted by Serialkisser
Mrs. Imconfused is having difficulty understanding the trigonometric concepts. She doesn't know when to use the Primary Trigonometric Ratios (sin, cos and tan), the Sine Law or the Cosine Law. She also needs clarification on what each one (primary trig ratios, sine law, cosine law) can be used to calculate. Provide an explanation that will help to solidify her understanding. Be sure to discuss what each one will calculate and how to decide which one to use. Explanations, diagrams, examples, formulas and mathematical terminology should all be included in your solution. Be thorough!
Here is a summary: you will need to think of examples and draw some diagrams.
The Primary Trig Ratios will be used where you have a right-angled triangle. Draw a diagram showing a right-angled triangle, marking an angle of $90^o$ and labelling one of the other angles $x$. Label the three sides $H$, $O$ and $A$, and write down the three ratios:
$\sin x = \frac{O}{H}$, $\cos x = \frac{A}{H}$ and $\tan x = \frac{O}{A}$
Now explain (with a simple example of each) when you would use each of these formulae. For example, you'd use sine if you knew $x$ and $H$, and wanted to calculate $O$; or you'd use tan if you knew $O$ and $A$, and wanted to calculate $x$. The question asks you to be thorough, so make sure you include all three ratios in as many ways as you can think of.
The Sine and Cosine Laws will be used where you have a non-right-angled triangle. Draw a diagram showing a triangle ABC, and write down the laws:
Sine Law: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
And the three Cosine Laws: $a^2 = b^2 +c^2 -2bc\cos A, b^2 = \dots, c^2 = \dots$ (I'll leave those to you)
Then:
Use the Sine Law if you know:
• 2 angles and a side, and want a second side. (You may need to find the third angle first, using the fact that the three angles add up to $180^o$.)
• 2 sides and a non-included angle, and want a second angle. (This is called the Ambiguous Case because there may be two possible angles, one of which is the supplement of the second.)
Use one of the Cosine Laws if you know:
• 2 sides and the included angle, and want the third side.
• 3 sides and want an angle.
If you have 2 sides and the included angle, and want a second angle, you'll have to find the third side first, using the Cosine Law, and then the angle using the Sine Law.
To be thorough in your answer, you could again think up some problems where you need to use these rules, and work out the answers showing how the formulae are used.
I hope that gives you something to work on.
Grandad
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http://physics.stackexchange.com/questions/21425/what-meaning-do-changes-in-the-absolute-value-of-gibbs-free-energy-have-in-a-sim/21644
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# What meaning do changes in the absolute value of Gibbs free energy have in a simple expansion process?
Below is a simple representation of the thermodynamics of a steam turbine. Stream kinetic and potential energy changes are neglected and no other type of non-PV work is done besides shaft work. Enthalpy and entropy are straight from the steam tables. Gibbs free energy is calculated using $G=H-TS$. Outlet pressure was arbitrarily chosen as 400 psig. Steam quality is 1.0 in both inlet and outlet conditions.
What physical significance, if any, does the change in Gibbs free energy have? In this example, note that Gibbs free energy increases which seems counterintuitive for a supposedly spontaneous process. Indeed, if I use an isentropic efficiency of 0 in this model to represent a simple throttle valve, I get a decrease in Gibbs free energy to -798 Btu/lb which makes sense as a spontaneous process.
I am familiar with the availability approach. I would just like to understand what significance the change in absolute values of Gibbs free energy have, if any.
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## 2 Answers
The short answer is, it doesn't really mean much of anything, because it's not meaningful to compare the Gibbs free energy of two systems unless they're at the same temperature. To show you why, I'll quickly run you through the derivation of the Gibbs free energy and point out where the constant temperature assumption comes in.
The second law of thermodynamics tells us that the total entropy of a system, plus the entropy of its surroundings, must be non-decreasing. I'll write this as $\Delta S_\text{total} = \Delta S + \Delta S_\text{surroundings} \ge 0$.
Suppose we have a system (of any kind) in contact with an environment that remains at constant temperature and pressure. This system undergoes a process (of any kind) that changes its internal energy by $\Delta U$ (with positive sign meaning the system's energy increases) and its volume by $\Delta V$ (positive sign meaning it does work on the environment). We know that the system has done an amount of work $p\Delta V$ on the environment, so we can say that $\Delta U = Q - p\Delta V$, with $Q$ an amount of heat transferred from the surroundings to the system (or in the opposite direction if it's negative).
Since the surroundings are at a constant temperature $T$, they must have lost an amount of entropy equal to $Q/T = \frac{\Delta U}{T} + \frac{p\Delta V}{T}$. So we have $$\Delta S_\text{total} = \Delta S - \frac{\Delta U}{T} - \frac{p\Delta V}{T}.\qquad\qquad(i)$$ This quantity, as we know from the second law, must always seek a maximum, as long as the temperature and pressure remain constant.
Now, up to this point the assumption of constant temperature hasn't really been necessary. If we assume that $\Delta U$ etc. are small we can replace them with differentials to get $dS_\text{total} = dS - pdV/T - dU/T$, which we can integrate if we know how $p$ and $T$ depend on $U$ and $V$.
However, for some completely weird and arbitrary reason, the tradition in physics is to multiply equation $(i)$ by $-T$ to get $$\Delta G \stackrel{\textit{def}}{=} -T\Delta S_\text{total} = \Delta U + p\Delta V - T\Delta S,$$ which must now be minimised because of the change of sign. I suppose this is done in order to put the quantity into energy units, which for some reason physicists and chemists are more comfortable with than entropy units. But it comes at a price - the transformation from maximising $\Delta S_\text{total}$ to minimising $-T\Delta S_\text{total}$ only works if $T$ is a constant, and if you violate this assumption then $\Delta G$ can either be positive or negative for a spontaneous process, so its sign no longer really means anything.
To work out $\Delta S_\text{total}$ for your system you would have to integrate $dS_\text{total}$ as described above, and to do that you would have to know the heat capacity and thermal expansion coefficient of steam (as functions of $T$ and $p$, though I'd guess they're pretty constant over that range) in addition to the data you posted. This would certainly come out positive, and a lower value would mean a more efficient turbine, with $\Delta S_\text{total}$ approaching zero in the reversible limit.
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The absolute value of the Gibbs free energy decreases as the spontaneous process happens. The negative value represents the spontaneity of expansion. The more negative (smaller) the G, the more work you can obtain from the system. So as work is done on the surroundings, G becomes closer to zero, and in this case increases.
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I don't think this is correct. A negative value of Gibbs free energy is just an artifact of what datum basis was chosen for enthalpy and entropy (in this case it is the triple point of water in the liquid phase). Moreover, the Gibbs free energy decreases which is the counterintuitive part. Approaching zero again doesn't hold any special meaning. – Jason Waldrop Feb 27 '12 at 18:12
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http://math.stackexchange.com/questions/307011/how-many-integer-solutions-of-a-exist-for-this-equation
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How many integer solutions of “a” exist for this equation?
$a\times b = 8 \times (a + b)$
I have used Wolfram Alpha and it has given me 14 integer solutions.
But how can we find those solutions ? Which method ?
edit: integer solutions are as following:
$a = [-56, -24, -8, 0, 4, 6, 7, 9, 10, 12, 16, 24, 40, 72]$
$b = [7, 6, 4, 0, -8, -24, -56, 72, 40, 24, 16, 12, 10, 9]$
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2 Answers
Hint: $$ab-8(a+b)=0 \iff (a-8)(b-8)=64$$
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the amount of integers can be found this way but finding the exact values of those integers is still bugging me. – Selim Arikan Feb 18 at 13:56
@SelimArikan Use factorization of 64. For example, $$(a-8)(b-8)=32\cdot 2$$ So one proposed solution of this equation is $a-8=32$, $b-8=2$, so $a=40, b=10$. – tetori Feb 19 at 1:51
thank you for your explanation – Selim Arikan Feb 19 at 7:25
So, $$a=\frac{8b}{b-8}=\frac{8(b-8)+64}{b-8}=8+\frac{64}{b-8}\iff (b-8)\mid64$$
Now, $64=2^6$ has $6+1=7$ positive divisors
So, we have $7$ solutions if we consider the only the positive values of $b-8$.
So, if we include all the non-zero integers, we shall have $2\cdot7=14$ divisors.
Clearly, $b-8\ne0$ else $a$ will be infinite.
So, the values of $b-8$ are $\pm 2^i$ for integer $i\ge0$ and $\le6$
Hence,
if $i=0,b-8=\pm1,b=9,7;$
if $i=1,b-8=\pm2,b=10,6;$
if $i=2,b-8=\pm4,b=12,4;$
if $i=3,b-8=\pm8,b=16,0;$
if $i=4,b-8=\pm16,b=24,-20;$
if $i=5,b-8=\pm32,b=40,-24;$
if $i=6,b-8=\pm64,b=72,-56;$
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But there are only three negative solutions for a, [-56, -24, -8] and how to find those values ? – Selim Arikan Feb 18 at 12:02
@SelimArikan, observe that we have $7$ negative and $7$ positive values for $b-8$. Sorry for the delay. – lab bhattacharjee Feb 18 at 15:00
thank you for your explanation – Selim Arikan Feb 19 at 7:24
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http://mathoverflow.net/questions/37010?sort=newest
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## Product Measure Only Possible Measure?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $X$ be a separable complete metric space and $Z$ be the set of all integers. Let $\nu$ be a Borel probability measure on $X^Z$ invariant under the shift function $S:X^Z \to X^Z$. Is it necessarily the case that $\nu = \mu^Z$ for some Borel probability measure $\mu$ on X?
Thanks
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## 4 Answers
No (unless X is a one-point space). The mean of two distinct shift-invariant product probability measures is a shift-invariant probability measure, though not a product.
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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.
Every such a $\nu$ is the law of some stationary process on $X^{\mathbb{Z}}$. Of course not every stationary process is i.i.d.
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Measures with the property you describe are called Bernoulli measures. There are many, many invariant measures that are not Bernoulli: one class of examples is given by the measures concentrated on periodic orbits (as rpotrie points out in another answer); another important class is the Markov measures. These are given by a measure $\mu$ on $X$ (which if $X$ is finite is simply a probability vector) together with a function $p\colon X \to \mathcal M(X)$ that represents transition probabilities, where $\mathcal{M}(X)$ is the space of Borel probability measures on $X$. Then one defines a measure $\nu$ on $X^\mathbb{Z}$ by \begin{multline} \nu(X_1 \times X_2 \times \cdots \times X_n) = \\ \int_{X_1} \int_{X_2} \cdots \int_{X_{n-1}} p(x_{n-1},X_n) dp(x_{n-2},x_{n-1}) \cdots dp(x_1,x_2) d\mu(x_1), \end{multline} where $\int dp(x,\cdot)$ represents integration with respect to $p(x)$. Note that this simplifies quite a bit if $X$ is finite, in which case $\mu$ is a probability vector, $p$ turns into a stochastic matrix, and you just need to write down the measure of an arbitrary cylinder. In any case, these give you a broad class of invariant measures that are not Bernoulli, but are very important for many applications.
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You need $\mu$ to be the stationary distribution of kernel $p$, right? – mr.gondolier Aug 29 2010 at 7:19
Yes, sorry, I left that out. $p$ induces a map $\mathcal M(X) \to \mathcal M(X)$, and you need $\mu$ to be a fixed point for this map. (Or in the finite case, you need the probability vector $\mu$ to be an eigenvector for the stochastic matrix $p$ with eigenvalue $1$.) – Vaughn Climenhaga Aug 29 2010 at 12:49
The answer is no. A trivial example is to concentrate the measure in a "periodic orbit", this will give an invariant measure for the shift.
But there are a whole lot of invariant measures (including full support measures which probably are more interesting).
The measure which is a product measure, has though some important features. For example, if you look at its "entropy".
(See K. Sigmund, Generic properties for Axiom A diffeomorphisms, Inventiones Math 11 (1970) for the case of the space X being finite)
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This also works for N in place of Z. – Carl Mummert Aug 29 2010 at 1:03
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http://simple.wikipedia.org/wiki/Partial_derivative
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# Partial derivative
In calculus, an advanced branch of mathematics, the partial derivative of a function is basically the derivative of one named variable, and the unnamed variable of the function is held constant. In other words, the partial derivative takes the derivative of certain indicated variables of a function and doesn't differentiate the other variable(s). The notation
$\frac{\partial f}{\partial x}$
is usually used, although other notations are not invalid. Usually, although not in all cases, the partial derivative is taken in a multivariable function, (i.e., the function has three or more variables, whether independent or dependent variables).
## Examples
If we have a function $f(x, y)=x^2+y$, then there are several partial derivatives of f(x, y) that are all equally valid. For example,
$\frac{\partial x}{\partial y}[f(x, y)]=2x+1$
Or, we can do the following
$\frac{\partial}{\partial x}[f(x, y)]=2x$
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http://micromath.wordpress.com/2008/05/26/calculus-without-limits/?like=1&source=post_flair&_wpnonce=b467bcc1c8
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# Mathematics under the Microscope
Atomic objects, structures and concepts of mathematics
Posted by: Alexandre Borovik | May 26, 2008
## Calculus without limits
I re-publish my old post from my old blog, with comments (some of them new) included in the body.
My post:
Thanks to Michael Livshits, I became aware of an alternative approach to calculus based on eliminating the concept of a limit and replacing it by uniform Lipschitz bounds. For example, definition of derivative becomes
$|f(x)-f(a)-f'(a)(x-a)| < K(x-a)^2$
I would love to learn more. Any comments?
“Calculus without limits” T-shirt is available from Michael’s e-shop, take a look.
Comments from the old blog:
Torus said…
The book of Marsden and Weinstein “Calculus Unlimited” is available here. It has a similar approach.
Anonymous said…
Dear Alexandre, I appreciate your interest in my efforts to simplify Calculus. In addition to many links from my home page at mathfoolery.org, there is a book by a Chinese mathematician Qun Lin called “Free Calculus: A Liberation from Concepts and Proofs,” ISBN 9789812704597, published in 2007 by World Scientific, which is based on similar ideas, summarized in his paper at http://www.math.ohiou.edu/~shen/calculus/linqun.pdf
There are also some articles by Mark Bridger at http://www.math.neu.edu/~bridger/LBC/lbcswp.pdf
as well as “The Lost Calculus: Tangency and Optimization without Limits” by Jeff Suzuki, Mathematics Magazine, December 2005, http://www.maa.org/pubs/mag_dec05_toc.html and very recent one by Harold Edwards, dedicated to Euler’s tricentenial, at http://www.ams.org/bull/2007-44-04/S0273-0979-07-01174-3/home.html
In reply to torus: while “Calculus Unlimited” is a fine attempt to introduce differentiation without using limits, it is still based on differentiability at a given point, and needs some heavy tools like completeness and compactness to get any practical results.
For everybody else: the correct formula in Alexandre’s original post is |f(x)-f(a)-f’(a)(x-a)| is less or equal to K|x-a|^2
I am Миша Лившиц and I approve this message. The views expressed are my own and not of any organization (I hate them all).
Anonymous said…
“Free Calculus: A Liberation from Concepts and Proofs”
Why would anyone want to be “liberated” from proofs? Surely, mathematical proof is one of the supreme and most beautiful achievements of human culture!
Anonymous said…
Not all of them, silly, only the superfluous ones, the ones that just clog up your brain.
Zen Harper said…
This definition is definitely incorrect; it does not work for all differentiable functions f.
For a simple example, let f(x)=|x|^p with p strictly between 1 and 2, and a=0=f(a)=f’(a). Then f’ exists (and is continuous) on the whole real line, but there is no fixed K such that |f(x)| is less than Kx^2 for all sufficiently small x.
Everyone has their own opinion on calculus teaching, but I personally feel that the classical methods and proofs of standard Analysis are already about as close to perfection as undergraduate maths gets, in terms of beauty and simplicity. If students can’t get it, or aren’t interested, they should switch to Physics or Engineering…
(Of course, you may have the perfectly valid opinion that Analysis and Calculus are different subjects; but I disagree, at least as far as mathematics students are concerned. University is the time to start learning real maths properly, and to undo the terrible damage caused by GCSE and A-level, i.e. tedious memorisation of methods without proof or understanding).
Alexandre Borovik said…
Zen Harper: it is a not wrong definition, it is an alternative definition. Yes, it sacrifices functions like the one sugested by you. But it allows to develop a consistent theory with full proofs, even if concerned with narrower classes of funcions.
misha said…
Zen Harper: Take your favorite modulus of continuity m, then replace K(x-a)^2 in the definition by Km(|x-a|)(|x-a|). Enjoy the simplicity and directness of all the proofs. See more discussion on the next reincarnation of this blog.
You also said: “… I personally feel that the classical methods and proofs of standard Analysis are already about as close to perfection as undergraduate maths gets, in terms of beauty and simplicity. If students can’t get it, or aren’t interested, they should switch to Physics or Engineering…”
This is the most arrogant piece of nonsense that I’ve seen in a while. First of all, Calculus and Analysis are not carved in stone, there is a lot of room for improvement, both conceptual and pedagogical. Just compare the proofs in both approaches and see the difference. Second, most of the students who are forced to take Calculus, will use it as a tool for applications, and they are not interested in mathematical hair-splitting, they want to understand the practical aspects of the subject. It is the indiscriminate shoving of the mostly irrelevant mathematical formalism down the students’ throats that breeds the “tedious memorisation of methods without proof or understanding,” as you have aptly put it. And finally, the approach based on explicit uniform estimates, that I and other people suggest, brings Calculus into the realm of modern mathematics, puts Classical Analysis in its proper context and makes it easier to understand even for the math majors. I rest my case.
Zen Harper said…
I was surprised to see such a strong backlash against my comments. I am not trying to start a war here, I am just stating my opinion. Maybe we will just have to agree to disagree on this one.
I’m sorry if you think I sound arrogant. Maybe it came across slightly wrong because I am thinking in terms of the English university system, where students don’t have “majors” and “minors” – they make a choice (before university) whether to study a degree in mathematics, physics, engineering, or something else.
My point is that Mathematics students (in the English system I am talking about) are there to learn Mathematics, and that means proper, undiluted rigorous proofs in great generality; the “pathological” differentiable functions, for which your proposed alternatives fail, are admittedly not needed for most purposes in physics, engineering, and other applications. But they are useful sometimes (even in applications! e.g. random walks, Brownian motion, probability, functional analysis, etc. etc. etc.) and mathematicians definitely do need them; the longer students delay in seeing them, the harder it will be for them to get used to the Analytic way of thinking, and understanding of “pathological” functions.
If instead you’re talking about Calculus courses for non-mathematics students in the American system, then that’s an entirely different set of students, and not the one I was talking about. I suppose I should have made this clear.
Another very important point, which I forgot to say precisely because it normally goes without saying, is that “differentiable function” is a totally standard phrase, with a standard meaning, agreed upon by over 99% of all mathematicians. You really shouldn’t use standard terminology to refer to something else; this will only confuse students more when they read other books (and is totally contrary to accepted practice).
It’s fine to use alternative approaches, but then you must use alternative terminology also; you should tell students the classical definitions (even if you are not planning to use them); and you should point out the negative aspects of your alternative approach.
Otherwise, you are simply being dishonest to your students. You have a duty to teach students mathematics; if you want to use your own non-standard variants, you must make it clear to them.
In reply to Misha, about choosing a modulus of continuity m: I agree completely that it simplifies some proofs. But it still doesn’t give you all differentiable functions! As soon as you make a choice about which m to use, and fix it, you are then restricting to a subclass of differentiable functions. As you make m tend to zero more slowly, you make the class of functions larger; but you will never be able to deal with all differentiable functions SIMULTANEOUSLY, whatever fixed choice of m you make. If instead you allow m to change depending on the function f, which has already been in common use in standard textbooks for decades, then (as is well-known and standard) you get simply a trivial rephrasing of the classical definition.
In summary: really, the only thing I object to is the use of the word “differentiable”. Just make up your own alternative word instead, and no-one will complain.
Zen Harper said…
P.S. to my previous comment, with a stupid mistake at the end:
Of course, the modulus m must depend not just on f, but also on the point a, in order to recover the full class of differentiable functions.
misha said…
To Zen Harper: Brownian motion trajectories are (locally) uniformly Holder continuous with any exponent less than 1/2. When you get to functional analysis, you can use distributions and flush your classical differentiation theory down the toilet, it’s a mess anyway.Talking about terminology, I call the functions differentiable in the stronger sense uniformly Lipschitz differentiable, or uniformly Holder differentiable, accorrding to the modulus of continuity, so nobody is misled. Since any continuous function on a closed finite interval has a uniform modulus of continuity, we can capture any continuously differentiable function with this approach, and these are the most important. Peter Lax used uniform differentiability in his calculus book of 1976, and Mark Bridger has recently published a text book on introductory analysis, based on uniform notions.
And who needs to deal with ALL the diferentiable functions SIMULTANEOUSLY? Can you give an example? Similarly, in most analytic applications, you need more than continuity to get any meaningful results, “general” continuous functions are highly pathological.
Finally, I don’t suggest to totally throw away the classical definitions of pointwise continuity and pointwise differentiability (althought they are mostly inadequate and need compactness and other crutches to get to anything useful). I’m just saying that starting with them in calculus and introductory analysis is not very clever, especially in calculus, where these lofty notions are almost never used and never properly explained, and you just end up looking as a fool, paying a lip service to these notions and giving complicated or vague explanations, that only confuse your students, while much simpler and clear explanations are available.
Look, when you study or teach a new subject, it’s a bad idea to start with the most general definitions and pathologies, you start with the simple stuff that is immediately useful in solving problems and applicable to the other subjects, you can get to generalities and pathologies later, and they will make more sense after you or your students get a good grasp of simple examples and applications.
Zen Harper said…
To Misha again…
On this particular subject, I sympathise with the saying “if it ain’t broke, don’t fix it…” Really, what’s wrong with the traditional limits approach?
Just a quick example: the Lebesgue Differentiation Theorem, as far as I know, gives almost no information beyond mere pointwise differentiability (only almost everywhere). I only mentioned Brownian motion as an example of “natural” but non-differentiable continuous functions; I wasn’t thinking in terms of additional continuity properties.
It seems to me that the question “what is the best way to teach calculus/analysis?” is both subjective and highly dependent upon the target set of students. We still disagree on a number of issues; I cannot see why you seem to be so reluctant to use limits in the first course, and why you think the modulus of continuity approach is so much easier for students. I think we both agree, however, that understanding of limits is essential, and definitely needs to be done at some point; so when, exactly, do you propose to teach it? The second year? The third year? That doesn’t leave very much time for more advanced courses.
Also, I would say (although this is quite subjective) that “continuity” is more “mathematically natural” than continuity plus Holder or Lipschitz estimates; and so, similarly, pointwise differentiability is more “natural” than differentiability plus additional estimates. Your approach emphasises “unnatural” estimates, which are not “typical” of “most” functions.
There is a strong danger that students will mistake your special Holder and Lipschitz estimates (which are still, of course, very useful for some purposes) for being “typical”.
I don’t agree at all with your final paragraph (although, of course, it is a matter of taste). I think students will not appreciate your approach, in the long run, precisely BECAUSE you delay too long in showing them the nasty examples! A similar thing happens with the famous exp(-1/x^2) function, which confuses most students who have been so strongly conditioned to think in terms of Taylor series. In the long term, I think it is better for students to see such things as soon as possible. Just try doing a quick survey of third year undergraduates in any good UK university, to find out how many of them are aware of such nasties (even without knowing the detailed examples); I bet you will find a surprisingly large number who aren’t.
Of course, you don’t necessarily have to spend a large amount of time on the “pathologies” (many of which, like exp(-1/x^2), are not really pathological at all); mere knowledge of their existence is already enough for many purposes.
Here are some of my axioms:
AXIOM 1: Mathematics is very hard, and unavoidably needs serious work.
I think most of us would agree with this, if we are being honest.
AXIOM 2: The main purpose of a Mathematics B.A. or B.Sc. (or M.Math.) degree SHOULD be to teach students enough mathematics to prepare them for further study or research (either in mathematics, or in a highly mathematical “applied” field).
It is NOT to train future computer programmers, bankers, accountants, airline pilots, etc. etc. how to do their jobs (for which a mathematics degree is almost a total waste of time, if we are being honest). Such people only study mathematics for fun; and it is DEFINITELY not to provide basic numeracy skills (for which GCSE or A-level maths is already more than adequate). University lecturers (I am one myself) are about the worst possible people to give vocational training, and we should stop trying to pretend otherwise.
AXIOM 3: a university degree, whatever the subject, SHOULD be very difficult and demanding. A secondary purpose is to “toughen up” students, similarly to hard physical training for athletes. Another purpose is to actually show students what MATHEMATICS is; unfortunately, GCSE and A-level are totally watered down and inadequate nowadays, and do not really contain any serious maths at all.
AXIOM 4: three or four years is, in fact, only a very short amount of time; far less than the optimal pre-Ph.D. time. There is no time to be wasted!
…anyway, now that I’ve stated my axioms I can defend my position some more: first, limits undeniably must be covered (Axiom 2). You can’t avoid them for too long (Axiom 4), so they really should be taught no later than the first undergraduate year. The fact that students find limits difficult is completely irrelevant (by Axioms 1 and 3). By Axiom 4, it would not be completely unreasonable if you only had enough time for one approach: EITHER the standard limits approach, OR your modulus of continuity approach. Which should be omitted? (Of course the optimal course is to cover both; but then, the world we live in is far from optimal).
I still think (coming back to Axioms 1,2,3) that any student who understands limits properly will then have no difficulty understanding your alternative approach; but any student who can’t understand limits will also not understand your approach (they themselves might THINK they do, but REALLY they don’t; how many students REALLY understand uniform continuity or term-by-term differentiation of power series, for example?)
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## Responses
1. To Zen Harper:
If you think it’s not broke, especially outside of the pure math department, you are in denial. Everybody complains about it, it’s an elephant in the room that you choose to ignore.
As for Lebesque differentiation theorem (that says that any monotonic function is differentiable almost evrywhere), that certainly does not belong to the first semester analysis course, so I’m not sure why you are bringing it up, it’s showcase example of how classical differentiation doesn’t quite work with Lebesque integral, because if you integrate the derivative, you will recover only the absolutely continuous part of your monotonic function. To reconstruct any pointwise differentiable function from its derivative you either need the Henstock-Kurzweil integral, or you have to look at your function as a distribution. The classical (19th century) analysis works well only with (piecewise) continuously differentiable bunctions, i.e., piecewise uniformly differentiable functions.
Limits of functions and pointwise continuity can be treated after the theory based on uniform estimates is developed, and it will not take less time than in the traditional approach since the ground is well prepared. So the time you spend on the streamlined theory is not wasted, as you may argue, but serves as a good preparation for the more advanced topics. See an essay by Hermann Karcher for more details. By the way, he said that the students who learned it this way, reported less problems with more advanced topics, such as numerical analysis.
Why do you say “continuity plus Holder or Lipschitz estimates?” It’s like saying “continuity plus differentiability,” I would not expect it from a mathematician, but it’s a minor point. As for being “typical,” it depends on typicality. In my opinion, non-Holder continuous functions are on the fringe of mainstream mathematics. There is no danger at all that students will think that all continuous functions are Holder or Lipschitz, especially if you point it out and give a few examples.
Now about your AXIOMS and the implications that you derive.
AXIOM 1. Yes, mathematics is hard but it doesn’t mean you have to make it even harder by starting with the hard topics. And it doesnt mean that making the ideas of calculus more accessible is not worth the effort. By articially making mathematics less accessible than it has to be, you really shoot yourself in the foot, as a mathematician.
AXIOM 2 Doesn’t mean the ideas of calculus should be taught only in the orthodox 19th century way that you advocate. And be not taught at all to non-specialists.
AXIOM 3 Doesn’t mean that the university degree should be made even harder by rather thoughtless and rigid arrangement of the material in undergraduate analysis.
AXIOM 4 Doesn’t mean that starting with uniform estimates is a waste of time, most likely, it will save some time, even during the first year, and lead to better results at that.
Your position seems to be that everything is hunky-dory the way it is, and any objections are irrelevant. I’m not buying it. Your claim that the students who have difficulties with the canonical approach will have no less difficulties with a simplified approach doesn’t hold water, axioms or no axioms.
By: misha on June 3, 2008
at 12:28 am
2. In the third paragraph of my previous post “…and it will not take less time…” should be “…and it will take less time…,” sorry for sloppy editing. I’ll post later on yet another way to look at differentiation, as division of f(x)-f(a) by x-a in a certain ring, the way that may appeal to math majors.
By: misha on June 3, 2008
at 12:39 am
3. Here we go. Nobody needs limits to differentiate polynomials, since we can explicitly divide $p(x)-p(a)$ by $x-a$, as polynomials in $x$, and stick $x=a$ into the result of the division to get $p'(a)$. Similarly, the Cauchy’s definition of differentiability of $f$ at $a$ is divisibility of $f(x)-f(a)$ by $x-a$ in the ring of functions of $x$ continuous at a ($a$ is kept constant).
Now, we can work with the other rings (and modules) to get the other flavors of differentiability. If we view $p(x)-p(a)$ and $x-a$ as functions of 2 variables and our ring is the ring of unifomly continuous functions of 2 variables, we get the uniform differentiability. Taking Lipschittz functions of 2 variables will gives us the uniform Lipschitz differentiability with the inequality $|f(x)-f(a)-f'(a)(x-a)|\leq K(x-a)^2$ that was the starting point of this discussion. Instead of Lipschitz, we can use Holder, or the class related to some other modulus of continuity. It even works for distributions.
This view of differentiation gives us a broader perspective, shows us how to simplify calculus, and also extends to many variables, complex variables and even more general situations. Some people say it is a crime against Analysis, but I don’t see it this way.
By: misha on June 4, 2008
at 7:10 am
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http://en.m.wikipedia.org/wiki/Montonen%E2%80%93Olive_duality
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# Montonen–Olive duality
In theoretical physics, Montonen–Olive duality is the oldest known example of S-duality or a strong-weak duality. It generalizes the electro-magnetic symmetry of Maxwell's equations. It is named after Finnish Claus Montonen and British David Olive.
## Overview
In a four-dimensional Yang-Mills theory with N=4 supersymmetry, which is the case where the Montonen–Olive duality applies, one obtains a physically equivalent theory if one replaces the gauge coupling constant g by 1/g. This also involves an interchange of the electrically charged particles and magnetic monopoles. See also Seiberg duality.
In fact, there exists a larger SL(2,Z) symmetry where both g as well as theta-angle are transformed non-trivially.
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## Mathematical formalism
The gauge coupling and theta-angle can be combined together to form one complex coupling
$\tau = \frac{\theta}{2\pi}+\frac{4\pi i}{g^2}.$
Since the theta-angle is periodic, there is a symmetry
$\tau \mapsto \tau + 1.$
The quantum mechanical theory with gauge group G (but not the classical theory, except in the case when the G is abelian) is also invariant under the symmetry
$\tau \mapsto \frac{-1}{n_G\tau}$
while the gauge group G is simultaneously replaced by its Langlands dual group LG and $n_G$ is an integer depending on the choice of gauge group. In the case the theta-angle is 0, this reduces to the simple form of Montonen–Olive duality stated above.
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## References
• Edward Witten, Notes from the 2006 Bowen Lectures, an overview of Electric-Magnetic duality in gauge theory and its relation to the Langlands program
This quantum mechanics-related article is a stub. You can help Wikipedia by expanding it.
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http://www.physicsforums.com/showthread.php?p=4128573
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Physics Forums
## From what I understand, there s some sort of computer calculating pi...
... and for a long time, I took this for granted, but now I'm starting to wonder how it is doing this. From my knowledge, the only way one could calculate pi is to measure the diameter and radius of a circle, and find the ratio, but to calculate an irrational decimal as thoroughly as we have would require an unrealistically accurate measurement. There may be some sort of equation that can be used to calculate pi, but I have no idea what it is.
So, how are mathematicians managing to calculate pi?
You mean arclength right? Other ways to calculate pi is to use infinite series that we know converge to pi. One example, though not the best, is the arctan(1) series. Also called leibniz series. Edit: You may understand this provided you have taken at least calculus 2 or learned it prior.
But from what I understand, pi is needed in order to calculate arctan... I may be wrong. I really don't know much about this.
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## From what I understand, there s some sort of computer calculating pi...
Not quite.
There are lots of nice formulas for calculating pi, most of them fall under the category of infinite series, meaning there are ways to represent pi as the sum of an infinite amount of numbers (with each successive number being added becoming smaller so it approaches pi as the amount of numbers in the series you add up goes to infinity - something mathematicians call convergence).
This is one of many ways to calculate pi:
http://en.wikipedia.org/wiki/Leibniz_formula_for_π
(This is actually not a very efficient method, but it looks nice :)
All a computer does is, using formulas like these, add up millions and billions of fractions to get more and more accurate values.
Recognitions: Homework Help I like $$\frac{1}{\pi}=\frac{2\sqrt{2}}{9801} \sum_{n=0}^\infty \frac{(4n)!}{(n!)^4} \cdot \frac{1103+26390n}{396^{4n}}$$ but there are many ways see http://mathworld.wolfram.com/PiFormulas.html http://www.numberworld.org/misc_runs/pi-5t/details.html http://en.wikipedia.org/wiki/Pi http://en.wikipedia.org/wiki/Machin-like_formula http://mathworld.wolfram.com/Machin-LikeFormulas.html for Arctan use a series for Arctan and pi/4=Actan(1) along with a Machin-like formula such as Arctan(1)=4Arctan(1/5)-Arctan(1/239) to speed things up
Special mention goes to the BBP formula, which allows one to calculate the nth hexadecimal digit of ∏ without calculating the first (n-1)th digits.
Recognitions:
Homework Help
Quote by lurflurf Arctan(1)=4Arctan(1/5)-Arctan(1/239)
Note arctan(1) = π/4
another formula, note that dividing by (1/8)^2 can be done with a right shift of 6 bits:
arctan(1) = 6 arctan(1/8) + 2 arctan(1/57) + arctan(1/239)
a bunch of arctan formulas here:
http://www.jjj.de/arctan/arctanpage.html
Programs like pifast:
pifast.html
and the pi demo program for apfloat:
http://apfloat.org
use highly optimized extended precision math and normally use the Chudnovsky formula mentioned on the wiki page linked to above (other methods are offered as options). Since division is relatively slow. it's my understanding that the denominator is generated by multiplication into a huge number then only a single division is done, usually by inverting the denominator and multiplying.
All of this is way beyond anything practical in terms of physics, other than sub-sets of a large value for pi or e could be used for random numbers.
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http://mathoverflow.net/questions/5568/categorifying-the-group-representations
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## Categorifying the group representations
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Given a Lie algebra $g$, one constructs $\mathcal A$, a category whose $K_0$ is the universal enveloping algebra of $g$. Conjecture: any $\mathcal A$-acted triangulated category $\mathcal V$ (with its $K$ locally finite) decomposes to $\oplus \mathcal V_\lambda$ with braid action; and there is bijection between $g$-representations and minimal such categories.
Is there a good — if possible, non-$sl_2$ — example of such a category $\mathcal A$, minimal categories $V_\lambda$ and braid action which explains why one would have such a construction?
Update: Found the notes of the talk that has two $sl_n$ examples, one from quivers, another from sheaves on the grassmannian, $\mathcal V :=\oplus^n_i D^b\mathop{\rm constr}/\mathop{\rm Gr}(i,n)$.
A more accessible text for either example would be welcome! Because if the best way to understand these is to "get" quantum groups, that's quite a big topic. My idea was more like "maybe this is a good place to start".
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## 1 Answer
The $sl_n$ version of this shouldn't so bad. I think it's just self-dual objects in parabolic category O and shuffling functors, though I'll admit, I haven't checked this myself, and doubt it's written properly somewhere. Probably the best reference is the papers of Brundan and Kleshchev (for example "Schur-Weyl duality for higher levels").
I suspect the inspiration for such a conjecture isn't really particular examples so much a philosophy about what sort of structures on a quantum group should be categorifiable.
By the way, I think your conjecture might be a bit too strong (at least as I interpret it). The 2-representations of a 2-Kac-Moody algebra aren't semi-simple (I've got a huge supply of non-semisimple examples categorifying tensor products).
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According to the notes, the conjecture doesn't say that any V decomposes as a sum numbered by irreps of g, merely that V decomposes as a sum according to weights and that there's a filtration numbered by irreps. For the reference, the talk in question has notes at math.utexas.edu/users/benzvi/GRASP/lectures/IAS/… – Ilya Nikokoshev Nov 15 2009 at 0:53
Ah, well, this is what happens when you don't explain your notation. The decomposition corresponding to weight spaces is no conjecture; any reasonable version of categorifying Kac-Moody algebras will have that property. The braid action is more interesting though. Rouquier constructs a candidate, but doesn't prove that it satisfies the braid relations. – Ben Webster♦ Nov 15 2009 at 2:50
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http://mathhelpforum.com/pre-calculus/111504-please-help-i-kind-know-how-do-but-i-dont-know-if-my-answer-right.html
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# Thread:
1. ## Please help! I kind of know how to do this, but I dont know if my answer is right.
y=f(x)=6x^2-x-2/3x^2-11x-20
What are the intervals, test point k, f(k)?
y=f(x)=x^3+x^2-12x/x^2-2x-8
This is an oblique asymptote
What are the intervals, test point k, f(k)?
2. Originally Posted by Jluse7
y=f(x)=6x^2-x-2/3x^2-11x-20
What are the intervals, test point k, f(k)?
y=f(x)=x^3+x^2-12x/x^2-2x-8
This is an oblique asymptote
What are the intervals, test point k, f(k)?
what do you mean by "test point k" and f(k) ?
what, exactly, are you supposed to do with these rational functions?
3. The test point is just a point you randomly chose like if it was negative infinity, -5 you could choose -6 and plug it in to the equation and whatever you get is the f(k). I'm just not sure what the x-int., y-int, horizontal and vertical asymptotes are.
4. Originally Posted by Jluse7
The test point is just a point you randomly chose like if it was negative infinity, -5 you could choose -6 and plug it in to the equation and whatever you get is the f(k). I'm just not sure what the x-int., y-int, horizontal and vertical asymptotes are.
x-intercepts ... factor the numerator and set it equal to 0, then solve
y-intercepts ... f(0)
horizontal asymptote for the first expression is just y = the ratio of the leading coefficients.
vertical asymptotes ... factor the denominator and set equal to 0, then solve for x.
5. Sounds like the problem is to determine where that fraction is positive and negative. A fraction can change sign only where the numerator or denominator is 0. Set the numerator equal to 0 and solve for x. Set the denominator equal to 0 and solve for x. Those values of x divide the number line into intervals. Choose one "test point" in each interval.
6. Okay im just not sure if my answers are right and was double checking to see if they right. But I have no idea if I did them right
7. Originally Posted by Jluse7
Okay im just not sure if my answers are right and was double checking to see if they right. But I have no idea if I did them right
post your solutions ... be clear what each represents.
8. x-int= I need help figuring this out because I'm not used to x being bigger than 1
y-int= 1/10
horizontal asymptote= y=2
vertical asymptote= I need help figuring this I'm not used to x being bigger than 1
9. Originally Posted by Jluse7
x-int= I need help figuring this out because I'm not used to x being bigger than 1
y-int= 1/10 ok
horizontal asymptote= y=2 ok
vertical asymptote= I need help figuring this I'm not used to x being bigger than 1
$6x^2-x-2 = (3x-2)(2x+1)$
$3x^2-11x-20 = (3x+4)(x-5)$
you need to start getting used to "x being bigger than 1" when factoring.
here's a link you should visit ...
Factoring Quadratics: The Hard Case
10. Okay I got that. Now for finding the x-int and vertical asymptote how would you write that so you could plot it on a graph?
11. Originally Posted by skeeter
x-intercepts ... factor the numerator and set it equal to 0, then solve
vertical asymptotes ... factor the denominator and set equal to 0, then solve for x.
...
12. Okay here's what I got tell me if anything is wrong.
x-int= (-1/2,0) and (2/3,0)
y-int= y= 1/10
horizontal asymptote= y=2
vertical asymptote= x=5 x=-4/3
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http://gilkalai.wordpress.com/2008/07/03/helly-cayley-hypertrees-and-weighted-enumeration-iii/?like=1&source=post_flair&_wpnonce=d42062ed33
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Gil Kalai’s blog
## Helly, Cayley, Hypertrees, and Weighted Enumeration III
Posted on July 3, 2008 by
This is the third and last part of the journey from a Helly type conjecture of Katchalski and Perles to a Cayley’s type formula for “hypertrees”. (On second thought I decided to divide it into two devoting the second to probabilistic questions.) This part will include several diversions, open problems, and speculations.
### 11. How to make it work – the matrix tree theorem
Our high dimensional extension to Cayley’s theorem reads:
$\sum |H_{d-1}(K,{\bf Z})|^2 = n^{{n-2} \choose {d}},$
where the sum is over all d-dimensional simplicial complexes K on n labelled vertices, with a complete (d-1)-dimensional skeleton, and which are Q-acyclic, namely all their (reduced) homology groups with rational coefficients vanish.
Looking at the various proofs of Cayley’s formula (there are many many many beautiful proofs), the proof that I know to apply is the one based on the matrix tree theorem.
Consider the signed incidence matrix A’ between all (d+1)-subsets and all d-subsets of {1,2,…,n} that represents the boundary operator of simplicial homology. The rank of this matrix is ${n-1} \choose {d}$, and just like in the ordinary matrix tree theorem you delete rows to be left with linearly independent rows. Here you delete all rows corresponding to sets containing ‘n’ and you are left with a matrix A. Now we compute the determinant of $det (A \cdot A^{tr})$ directly, and compare the result to a computation based on the Cauchy-Binet Formula.
The eigenvalues of $A \cdot A^{tr}$ are the eigenvalues of the d-th Laplacian of the complete d-dimensional simplicial complex with n vertices. It is easy to inspect what they are and the determinant of $A \cdot A^{tr}$ is indeed $n^{{n-2} \choose {d}}$.
The many square determinants correspond to d-dimensional simplicial complexes K on our labelled set of vertices, which satisfy $f_{d-1}(K) = {{n} \choose {d}}$ and $f_d(K) = {{n-1} \choose {d}}$. Now if K has non-vanishing d-th homology, the determinant is zero. If K is a Q-acyclic simplicial complex (i.e., its (reduced) homology groups with rational coefficients are trivial) then it has a non zero determinant. So far, it is like trees, but next comes a surprise. The contribution of K is the square of the number of elements in $H_{d-1}(K)$, the (d-1)th homology group of K. This torsion group is finite, but for d >1 it need not be trivial.
Emily Peters presents the matrix-tree theorem. From “The Sarong Theorem Archive” – an electronic archive of images of people proving theorems while wearing sarongs.
### 12. An even simpler use of Cauchy-Binet worth knowing
Consider the $n \times 2^n$ matrix A, whose columns are all +1 -1 vectors of length n. Computing $det A A^{tr}$, via Cauchy-Binet Formula (or by other easy methods) asserts that the expected value $(det (B))^2$ for all $n \times n$ +1/-1 matrices behaves roughly like $(n!)$. This was observed by Turan and Szekeres who also found a formula for the sum of the fourth powers of all 0-1 n by n matrices. See a leter paper by Turan. (I am not aware of a similar formula for the fourth power of the size of the homology groups for hypertrees.) Much is known about the determinant and related properties of random 0-1 matrices and the analogy between torsion in the homology groups of random complexes and determinants of random matrices looks like a good analogy.
### 13. Torsion
One consequence of the formula compared to the total number of available simplicial complexes is that the torsion group is typically huge. (For d>1, the expected value of $|H_{d-1}(K,{\bf Z})|^2$ for Q-acyclic complexes counted by the formula, is asymptotically larger than $((d+1)/e)^{{n-2} \choose {d}}$; I am not aware of explicit examples with such a huge torsion group.)
How should we think about torsion in homology? It seems that thinking about the size of the torsion as a behaving like the determinant of a random matrix, may give a good intuition for many cases.
### 14. Extending other proofs for Cayley’s theorem?
Cayley’s counting trees theorem has many wonderful proofs. Can any other proof extend to the case of Q-acyclic simplicial complexes? For example, one proof relies on the exponential theorem that relates the exponential generating functions for connected and general graphs with a certain property P. (Followed by the Lagrange inversion formula.) Is there an analog of the exponential formula when connectivity is replaced by higher homology? Is there any analog of Prüfer sequences? I am not aware of any other proof that works.
### 15. Weights to the rescue of other conjectures?
Can we use subtle weights to save other promising but false enumerative conjectures? The farthest reaching fantasy in this direction would be to try to save MacMahon’s conjecture regarding space partitions of the number n. This conjecture is about enumerating spacial arrays of numbers that sum up to n. The conjecture is true for small values of n but fails for larger values. Can subtle weights come to the rescue? (MacMahon’s conjecture extends the formulas for ordinary partitions and for plane partitions.)
### 16. Incidence matrices
Perles’ observation in the beginning of this story was about the rank of the incidence matrices modulo 2 of k-subsets versus (k+1) subsets of an n element set. This was the starting point for a work by Linial and Rothschild. They asked: What is the rank of the incidence matrix of $N \choose r$ versus $N \choose k$ modulo p? and gave a complete answer for p=2. Richard Wilson gave a complete answer for general values of p and came quite close to presenting the “Smith form” of these matrices. Frumkin and Yakir gave representation-theoretic interpretation of Wilson’s result and proved “q-analogs”. Namely they replaced k-subsets of an n-element set by k-dimensional linear (and in a later work affine) subspaces of an n dimensional linear space over a field with q elements. They gave a complete formula when p and q are prime.
### 17. Duality and Self-Dual trees
Here is a very nice notion of duality that occurs in many places. Start with a simplicial complex K on a set X of vertices. Take the family F of all the complements to all sets in K. (This is not a simplicial complex, it is closed under supersets and not under subsets.) Now, take the family K* of all subsets of X not in F. Formally, $K^*=2^X \backslash$ {$X \backslash S: S \in K$}.
K* is a simplicial complex again. It is the Alexander dual of algebraic topology, and the “blocker” of polyhedral combinatorics.
If n=2d+2 the duals of our hypertrees are also d-dimensional. Molly Maxwell counted self-dual hypertrees with the same weights we used, and for odd dimensions the count gives precisely the square root of $n^{{n-2} \choose {d}}$. She deduced it from a more general theorem on matroids duality. For even values of d this is not the case but something may still work.
For d=1 the number of self-dual trees (simply stars) on 4 vertices is four the square root of 16 the total number of labelled trees. There is a theorem of Tutte extending this to self dual trees inside self dual planar maps. For d=3 and 8 vertices, the weighted number of all hypertrees is $8^{20}$ and by Maxwell’s theorem the weighted number of self dual ones is $8^{10}$. For d=2, n=6 - the weighted number of hypertrees is $6^6$ and if we exclude the triangulations of the real projective plane we get $6^3$. For d=4 the total weighted sum of hypertrees with 10 vertices is $10^{70}$, and somehow, a clever weighted sum of the self dual ones should give you $10^{35}$.
### 18. The Perles-Katchalski conjecture and associated eumeration problem
The assertion of the Perles-Katchalski conjecture holds for general classes of simplicial complexes described by homological properties, and we can ask again if the extremal examples enumerate nicely.
Let $\cal K$ be the class of (d+r)-dimensional simplicial complexes with the properties that
(L) For every induced subcomplex K’, $H_i(K',Q)=0, i \ge d$
Now, the homological extension of the Perles-Katchalski Theorem asserts that a (d+r)-dimensional simplicial complex K with n vertices satisfying condition (L) has at most ${{n} \choose {d} } - {{n-r-1} \choose {d}}$ d-dimensional faces.(For r=0 we need not worry about induced subcomplexes since, in this case, non trivial d-th homology for a subcomplex immediately extends to the whole complex.
We can try to “enumerate” simplicial complexes with n labelled vertices satisfying property (L) with precisely ${{n} \choose {d} } - {{n-r-1} \choose {d}}$ d-dimensional faces. (All these complexes will have the same number of i-faces for every i.)
We can expect that an appropriate enumeration of these objects (probably those containing a specific r-face), will give us a formula of the form $m^{{n-2-r} \choose {d}}$, where m is the number of r-dimensional faces of such a simplicial complex K. For the case d=1, no weights are needed and this speculation reduces to a formula of Beineke and Pippert for counting “k-trees”. (We may even expect finer enumeration formulas according to degree-sequences of r-faces; This is known for “k-trees”. See the following paper by C. Rényi and A. Rényi.)
### 19. Adin’s colorful extension.
Ron Adin extended the weighted enumeration of hypertrees to “colored complexes”, thus confirming (with extra weights added) another conjecture of Bolker.
### 20. Gelfand’s question.
Ron Adin gave a lecture about his work in a Stockholm ’89 meeting in combinatorics which was one of the earliest meetings with many participans from Russia, among them Gelfand, Vershik, Zelevinskii, Serganova, and others. Gelfand was excited about combinatorics (or what he regarded as combinatorics) at the time and was quite interested in Adin’s result. One question he asked me was: why is it that in combinatorics there is so much emphasis on graphs compared to higher dimensional objects.
I personally like the combinatorics of high dimensional objects but I could think of three answers. (Gelfand was quite satisfied with them).
a) For many purposes moving from sets to graphs represents a major conceptual jump, more than moving up from graphs to higher dimensional objects.
b) Higher dimensional objects can often be represented by graphs.
c) Many of the miracles of graph theory fail at higher dimensions.
(And here is a post by David Epstein titled “why graphs?”)
Another memory from the 89 conference is this: Israel Gelfand has a somewhat wide-spanned competative nature. Gelfand looked at Ron Adin and asked me: “He is orthodox isn’t he?”, “yes” I replied. Gelfand thought a little and then said: “But not as orthodox as my Dima.”
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### 5 Responses to Helly, Cayley, Hypertrees, and Weighted Enumeration III
1. Richard Stanley says:
Gil asks about saving MacMahon’s conjecture on space (or solid) partitions using subtle weights. If the weights are allowed to be subtle enough then there is an answer of sorts. The enumeration of plane partitions follows from the Cauchy identity
$\sum s_\lambda(x)s_\lambda(y) = \prod_{i,j\geq 1} (1 - x_i y_j)^{-1}$
for Schur functions. We substitute $x_i=q^{i-1}$ and $y_i=q^i$ to get MacMahon’s generating function on the right. On the left we get pairs of weighted tableaux that can be merged into a plane partition. (This proof is due to Bender and Knuth.) It is not unreasonable to replace the right-hand side by
$\prod_{i,j,k\geq 1}(1 - x_i y_j z_k)^{-1}.$
If we set $x_i=y_i=q^{i-1}$ and $z_i=q^i$ then we get MacMahon’s incorrect conjectured generating function for solid partitions. What about the left-hand side? If we expand in terms of Schur functions we get
$\sum_{\lambda,\mu,\nu} g_{\lambda,\mu,\nu} s_\lambda(x) s_\mu(y) s_\nu(z),$
where $g_{\lambda,\mu,\nu}$ is the notoriously intractable “Kronecker coefficient,” i.e., the multiplicity of the irreducible character $\chi^\lambda$ of $S_n$ in the tensor (or pointwise) product of $\chi^\mu$ and $\chi^\nu$. Thus solid partitions are replaced by certain triples of tableaux weighted by $g_{\lambda,\mu,\nu}$. This is probably too subtle for most people’s taste, especially since we don’t end up with any kind of solid partition. However, it is a rather natural extension of the elegant proof of Bender and Knuth and suggests why solid partitions may remain forever intractable.
2. Gil Kalai says:
Dear Richard, many thanks!
3. Pingback: Plans and Updates « Combinatorics and more
4. Pingback: A Beautiful Garden of Hypertrees « Combinatorics and more
5. Gil Kalai says:
I think that the “correct” class of simplicial complexes that extends both our hypertrees and Beineke and Pippert’s “k-trees” is the class of Bi-Cohen-Macauly simplicial complexes K. Namely simplicial complexes K such that both K and its Alexander dual are Cohen Macaulay. See, e.g., the paper of Gunnar Floystad, Jon Eivind Vatne
http://arxiv.org/abs/math/0209061
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http://math.stackexchange.com/questions/91319/work-problems-a-deeper-look?answertab=oldest
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# Work problems: A Deeper Look
We all know that work problems are some applications of algebra in reality. But the equation that corresponds to work problems is not as clearly stated as investment, mixture and uniform-motion problems.
Now, suppose we have the set of workers $A$ and each worker $a_i \in A$ where $n$ is the number of workers. Each worker $a_i$ can do a certain job $J$ in $h_i$ hours. Now if all the workers do the job $J$ altogether and started at the same time, we come into an equation which is $$x\sum _{i=1} ^n \frac{1}{h_i} = 1$$
Now, we solve for $x$ which is the number of hours the whole workers can do the job altogether. Considering the fact that they started at the same time. What if there exists a worker who started earlier by an hour from the rest or what if the situation is more complicated.
The question now is, what could be possible explanations of the formula that could give as clarity of the formula?
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## 1 Answer
Looked at properly, this kind of work problem is just a kind of rate problem: since $h_i$ is the number of hours that it takes $a_i$ to do the job by himself, his rate of work is $$r_i=\frac1{h_i}\text{ jobs/hour}\;,$$ and $$r=\sum_{i=1}^n\frac1{h_i}=\sum_{i=1}^nr_i\text{ jobs/hour}$$ is the work rate of the entire crew of $n$ workers. Then, just as in uniform motion problems, if $j$ similar jobs are to be done, and $x$ is the amount of time required to do them, we must have $$rx=j\tag{1}$$ and therefore $$x=\frac{j}{r}\;.\tag{2}$$ This is just $$\begin{cases}\text{rate}\cdot\text{time}=\text{distance}\\\\\text{time}=\frac{\text{distance}}{\text{rate}}\end{cases}$$ in a slightly different guise.
People are used to thinking of rates in miles per hour or the like, but the reciprocal notion isn’t completely unfamiliar. Runners, for example, are very familiar with it: one normally speaks of running six-minute miles, for instance, not of running at a speed of ten miles per hour.
Variations in which one or more workers start early are easily handled in this framework. If worker $a_i$ starts $k_i$ hours early, he does $r_ik_i$ jobs before the official starting time, and the team of $n$ workers does $$\sum_{i=1}^nr_ik_i$$ jobs before the official starting time. Thus, only $$j-\sum_{i=1}^nr_ik_i$$ jobs remain to be done when work officially starts, so if $x$ now represents the number of hours needed after the official starting time, $(1)$ becomes $$rx=j-\sum_{i=1}^nr_ik_i\;,$$ and $(2)$ becomes $$x=\frac{j}r-\sum_{i=1}^n\frac{r_i}rk_i\;.$$
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http://math.stackexchange.com/questions/241629/sum-of-two-sine-curves
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# Sum of two sine curves
How can we compute the sum $$\sin(f_1) + \sin(f_2)$$ I know it is $$2\sin\left(\frac{f_2 + f_1}{2}\right) \cos\left(\frac{f_2 - f_1}{2}\right)$$ but how can it be derived with elementary trigonometric identites?
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## 3 Answers
The sine angle addition rule reads:
$$\sin(u\pm v)=\sin(u)\cos(v)\pm\sin(v)\cos(u)$$
You can prove your identity by rewriting $f_1=\frac{f_2+f_1}{2}-\frac{f_2-f_1}{2}$ and similarily for $f_2$ and applying the sine addition formula twice.
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$$f_1= \frac{f_1+f_2}{2}+\frac{f_1-f_2}{2}$$ $$f_2= \frac{f_1+f_2}{2}-\frac{f_1-f_2}{2}$$
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$$\sin(f_1) + \sin(f_2)=\sin \cfrac {2f_1}{2}+ \sin\cfrac {2f_2}{2}$$
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http://stats.stackexchange.com/questions/50156/are-standardized-betas-in-multiple-linear-regression-partial-correlations
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# Are standardized betas in multiple linear regression partial correlations?
Since standardized betas are correlation coefficients in bivariate regression, is it the case that standardized betas in multiple regression are partial correlations?
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Short answer: No. – Glen_b Feb 17 at 2:51
Because the value of a beta can be anything (including of absolute value greater than $1$), it cannot generally be interpreted as a partial correlation. – whuber♦ Feb 17 at 3:00
## 2 Answers
Longer answer.
If I have this right --
Partial correlation:
$$r_{y1.2} = \frac{r_{y1}-r_{y2}r_{12}}{\sqrt{(1-r^2_{y2})(1-r^2_{12})}}$$
equivalent standardized beta:
$$\beta_1 = \frac{r_{y1}-r_{y2}r_{12}}{\sqrt{(1-r^2_{12})}}$$
As you see, the denominator is different. Indeed, since the additional term in the denominator of the first thing is between 0 and 1 (inclusive), it looks like $\beta_1$ will almost always be smaller than $r_{y1.2}$ though they could be equal if the two "independent variables" (and here's why I dislike that term) are ... well, independent - or at least uncorrelated.
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+1. But not least fact is that the numerator is the same. This implies that both coefficients are just different ways to standardize the raw regression coefficient b. – ttnphns Feb 17 at 8:05
I've in another question the following covariance matrix C for the three variables X,Y,Z given:
$$\text{ C =} \small \left[ \begin{array} {rrr} 1&-0.286122&-0.448535\\ -0.286122&1&0.928251\\ -0.448535&0.928251&1 \end{array} \right]$$ From its cholesky-decomposition L $$\text{ L =} \small \left[ \begin{array} {rrr} X\\Y\\Z \end{array} \right] = \left[ \begin{array} {rrr} 1&.&.\\ -0.286122&0.958193&.\\ -0.448535&0.834816&0.319215 \end{array} \right]$$ we can directly retrieve the partial correlation between Y,Z wrt. X as $\small corr(Y,Z)_{\cdot X} = 0.958193 \cdot 0.834816$ Now if we have the variables ordered such that the dependent variable is Z then the betas are computed by inverting the square-submatrix of the range in L which is populated by the independent variables X,Y: $$L_{X,Y} = \small \left[ \begin{array} {rrr} 1&.\\ -0.286122&0.958193 \end{array} \right]$$ and its inverse, which is inserted into a 3x3 identity-matrix to form the matrix $M$: $$M = \small \left[ \begin{array} {rrr} 1&.&.\\ 0.298605&1.043631&.\\ .&.&1 \end{array} \right]$$ Then the betas occur by the matrix-multiplication $\beta = L \cdot M$
$$\beta =\small \left[ \begin{array} {rrr} X\\Y\\Z \end{array} \right] = \small \left[ \begin{array} {rrr} 1&.&.\\ .&1&.\\ -0.199254&0.871240&0.319215 \end{array} \right]$$
which indicates, that the $\beta_X$ contribution for $Z$ is $\small \beta_X=-0.199254$ and the $\beta_Y$ contribution for $Z$ is $\small \beta_Y=0.871240$ . The unexplained variance in Z is the bottom-right entry squared: $\small resid^2= (0.319215)^2$
We see in $M$ that -being an inverse of a partial cholesky-matrix- it can contain values bigger than $1$ - and as well the Beta-matrix can then have entries bigger than 1.
So - to come back to your question- the partial correlation between $Y$ and $Z$ were the product of the entries in the second column of the L-matrix. The $\small \beta_Y$ however is the product of the entry in the second column of the Z-row with the inverse of that in the Y-row and the relation between the concepts of partial correlation and $\small \beta$ can be described by this observation.
Additional comment: I find it a nice feature, that we get by this also the compositions of $X$ and $Y$ in terms of $X$ and $Y$ - which of course are trivially 1. It is also obvious, how we would proceed, if we had a second dependent variable, say $W$, and even that scheme can smoothly be extended to compute/show the coefficients of the generalization to the canonical correlation - but that's another story....
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http://mathhelpforum.com/differential-geometry/139574-logarithmic-function.html
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Thread:
1. logarithmic function
how do i show that $\frac{d^2}{dx^2}\int_{-1} ^1 {log|x-t|}dt$= $\int_{-1}^1 \frac{-1}{(x-t)^2}dt$ is INCORRECT?
i know that to interchange limit and integral, a function must be uniformly convergent to its limit. but i dont know exactly how to show it.
2. Originally Posted by Kat-M
how do i show that $\frac{d^2}{dx^2}\int_{-1} ^1 {log|x-t|}dt$= $\int_{-1}^1 \frac{-1}{(x-t)^2}dt$ is INCORRECT?
i know that to interchange limit and integral, a function must be uniformly convergent to its limit. but i dont know exactly how to show it.
Can you differentiate an absolute value function at every point?
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http://mathhelpforum.com/calculus/94894-looking-integration-pointer-print.html
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# Looking for an integration pointer!
Printable View
• July 11th 2009, 12:29 PM
superduper
Looking for an integration pointer!
Hey guys, I got two problems here, that are probably extremely basic; but i'm having a hard time with these. When I finally find a $u$-substitution path I can generally work my way through, but I'm hitting a block on these, even just a pointer or what $u$ should be would be great!
1) $\int x^2 \sqrt{x+1}dx$
For this one, I'm trying to find a $u$ sub but I can't seem to make it work, I keep wishing real hard that the $x^2$ was under the radical and the $x$ was outside.(Speechless)
2) $\int\sin^2{(3x)}dx$
$= \int(\sin{(3x)})^2dx$
Again, no idea what to set $u$ equal to because there is no $2(\sin{(3x)})(\cos{(3x)})(3)$ with the $dx$
Any toss as to where to go next would be great, thanks for taking a look!
• July 11th 2009, 12:38 PM
Random Variable
$\sin^{2}u = \frac {1}{2} - \frac {\cos 2u}{2}$
• July 11th 2009, 12:41 PM
Krizalid
Quote:
Originally Posted by superduper
1) $\int x^2 \sqrt{x+1}dx$
For this one, I'm trying to find a $u$ sub but I can't seem to make it work, I keep wishing real hard that the $x^2$ was under the radical and the $x$ was outside.(Speechless)
Put $t=\sqrt{x+1}$ written backwards as $t^2=x+1.$
• July 12th 2009, 02:32 PM
superduper
Hey, so I got an answer for the radical integrand, but I'm still having trouble with the Trig question.
I set $u=3x$, and to integrate my textbook has a formula that states $\int\sin^2{u}du=\frac{1}{2}u-\frac{1}{4}\sin{2u}+C$.
So I applied this as follows,
$<br /> \int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$
Where $u=3x$ and $du=3dx$
Therefore,
$<br /> \int_{0}^{\frac{\pi}{3}}\sin^2{(3x)}dx$
= $3\int_{0}^{\frac{\pi}{3}}\sin^2{(u)}du$
= $3(\frac{1}{2}u+\frac{1}{4}\sin{(2u)})|^{\pi/3}_{0}$
= $[\frac{3}{2}\sin{((3)(\frac{\pi}{3}))}+\frac{3}{4}\ sin{((2)(3)(\frac{\pi}{3})}]-[\frac{3}{2}\sin{(0)}+\frac{3}{4}\sin{(0)}]$
= $[\frac{3}{2}\sin{(\pi)}+\frac{3}{4}\sin{(2\pi)}]-[\frac{3}{2}\sin{(0)}+\frac{3}{4}\sin{(0)}]$
= $[0+0+-[0+0]$?
I'm not sure where I'm going wrong here.
I was going to sub in the post that RandomVariable made, but I was just wondering if you could clarify that for me, I thought that identity was supposed to look like
$sin^2{\frac{u}{2}}=\frac{1+cos(u)}{2}$
EDIT- I'm mistaken, what you posted was $cos2u$ but I'm still lost as to where that identity comes from? I can't find it anywhere in my textbook.
Thanks again for the help guys.
• July 12th 2009, 02:44 PM
Plato
Go to this website W|A
Type in this exact phrase: integrate x^2 sqrt[x+1].
Click the equals bar at the end if the input window.
Be sure you click ‘show steps’ to see the solution.
• July 12th 2009, 03:02 PM
superduper
Wow Plato, that website is amazing, thanks for showing me that!
That solution showed me that I was wrong, but I was on the right track, messed up the u's somewhere.
Right off the bat, WolframAlpha says
$\int x^2\sqrt{x+1}dx$
Where $u=\sqrt{x+1}$ and $du=\frac{1}{2\sqrt{x+1}}$
So they say
$\int x^2\sqrt{x+1}dx$
= $2\int u^2 (u^2-1)^2 du$
= $2\int (u^6-2u^4+u^2)du$
Could you elaborate on this for me please? From the above relation shouldn't $u^2=x+1$ not $u^2=x^2$ I have no idea how they got this, I came up with
$\int_{0}^3 (u-1)^2\sqrt{u}du$ where $u=x+1$
= $\int_{0}^3 (x^{\frac{5}{2}}-2u^{\frac{3}{2}}+u^{\frac{1}{2}})du$
• July 12th 2009, 03:25 PM
Random Variable
Quote:
I was going to sub in the post that RandomVariable made, but I was just wondering if you could clarify that for me, I thought that identity was supposed to look like
http://www.mathhelpforum.com/math-he...58cacba3-1.gif
EDIT- I'm mistaken, what you posted was http://www.mathhelpforum.com/math-he...e194f001-1.gif but I'm still lost as to where that identity comes from? I can't find it anywhere in my textbook.
Just let u= 2x. And it should be $\sin^{2} \Big(\frac{u}{2}\Big) = \frac {1-\cos u}{2}$
$\cos^{2} \Big(\frac {u}{2} \Big) = \frac {1+\cos u}{2}$
• July 12th 2009, 03:32 PM
superduper
But the original equation was $sin(3x)$ so if we make $u=2x$ don't we have to turn that into $\frac{3}{2}u=3x$ or something to make that equation usable? Or actually $\frac{2}{3}u=\frac{2}{3}(3x)$? I'm getting lost now.
• July 12th 2009, 04:25 PM
Random Variable
Quote:
Originally Posted by superduper
But the original equation was $sin(3x)$ so if we make $u=2x$ don't we have to turn that into $\frac{3}{2}u=3x$ or something to make that equation usable? Or actually $\frac{2}{3}u=\frac{2}{3}(3x)$? I'm getting lost now.
$\sin^{2}(3x) = \frac {1 - \cos2(3x)}{2} = \frac {1-\cos(6x)}{2}$
The u=2x thing was to show that the identity you remember is the same identity.
• July 12th 2009, 04:51 PM
superduper
Oooh, that makes sense. Thanks for the clarification!
So here's what I have so far, would you mind poking over it and letting me know what you think?
$\int_{0}^{\frac{\pi}{3}} sin^2(3x)dx$
Where $u=3x$ and $3du=dx$
Therefore,
$\int_{0}^{\frac{\pi}{3}} sin^2(3x)dx$
= $3\int_{0}^{\frac{\pi}{3}} sin^2u du$
= $3\int_{0}^{\frac{\pi}{3}} \frac{1-cos(2u)}{2} du$
= $\frac{3}{2} \int_{0}^{\frac{\pi}{3}} 1-cos(2u) du$
= $\frac{3}{2}(u-sin(2u))|_{0}^{\frac{\pi}{3}}$
Looking good so far? Thanks again for your time.
• July 12th 2009, 05:16 PM
Random Variable
if you let $u = 3x$, then $du=3dx$ which implies that $\frac {du}{3} = dx$
so the integral becomes $\frac {1}{3} \int \sin^2(u) \ du$
Now you need to change the limits of integration.
The upper limit, which was $x=\frac{\pi}{3}$, becomes $u =3*\frac{\pi}{3} = \pi$
The lower limt, which was $x=0$, becomes $u =3*0=0$
so the integral now becomes $\frac{1}{3}\int^{\pi}_{0} \sin^{2}(u) \ du$
now using the identity
$= \frac{1}{3} \int^{\pi}_{0} \big(\frac {1}{2} - \frac{\cos(2u)}{2} \Big) du$ $= \frac{1}{6}\int^{\pi}_{0} du - \frac{1}{6} \int^{\pi}_{0} \cos(2u) \ du$
$\frac {u}{6} \Big|^{\pi}_{0} - \frac{\sin(2u)}{12} \Big|^{\pi}_{0}$
• July 12th 2009, 05:28 PM
superduper
Perfect, I can finish with that. Just one quick question on the last step.
Why is $\int cos(2u)du=\frac{sin(2u)}{2}$? I don't see that formula anywhere either, I thought it was;
$\int cos (u) du= sin (u)$
Thanks!
• July 12th 2009, 05:36 PM
Random Variable
Quote:
Originally Posted by superduper
Perfect, I can finish with that. Just one quick question on the last step.
Why is $\int cos(2u)du=\frac{sin(2u)}{2}$? I don't see that formula anywhere either, I thought it was;
$\int cos (u) du= sin (u)$
Thanks!
You have to make a substituion.
Let $v = 2u$, then $dv=2du$ which implies $\frac {dv}{2}=du$
so the intergal becomes $\frac {1}{2} \int \cos v \ dv = \frac {1}{2} \sin v = \frac {1}{2} \sin(2u)$
• July 12th 2009, 05:37 PM
superduper
Oh I see, that makes sense, I've never done more than 1 substitution in the same problem before. Thanks for sticking with me here, really appreciate it.
All times are GMT -8. The time now is 03:54 AM.
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http://stats.stackexchange.com/questions/tagged/centering+resampling
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# Tagged Questions
1answer
259 views
### Is centering needed when bootstrapping the sample mean?
When reading about how to approximate the distribution of the sample mean I came across the nonparametric bootstrap method. Apparently one can approximate the distribution of $\bar{X}_n-\mu$ by the ...
1answer
163 views
### Shifting bootstrap confidence interval to be centered around original parameter
I've been doing a bit of research into bootstrapping as I've been told one method of performing it, and this seems to differ from what I can find in other sources. I have a sample, and want to ...
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http://mathoverflow.net/questions/5995/positive-systems-in-n-the-n-1-simplex/6040
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## “Positive systems” in n * the (n-1)-simplex
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let S := the nonnegative integer solutions to {$a_1 + ... + a_n = n$}, and center := (1,1,1,...,1). Call a vector v generic if v.s = v.center <=> s = center. Then each generic v defines a positive system in S, the subset { s in S : v.s > v.center }.
Already at n=3 it is possible for one positive system to contain another.
Up to permutation, we may as well take v strictly decreasing. Having done so, is there a reasonable way to classify the maximal positive systems?
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## 2 Answers
To make the obvious comment, $v$ and $w$ define the same positive system if they lie on the same side of $(s-c)^{\perp}$, where $s$ ranges through the noncentral points of the simplex and $c$ is the center. So your question is whether we can characterize the regions in the complement of this hyperplane arrangement.
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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.
If you replace S by the lattice points in the positive orthant, and forget the condition that the hyperplane passes through a particular point, and require that the hyperplane only cuts off finitely many points, then you are looking at things called "corner cuts" by Onn and Sturmfels in Cutting corners.
Anyway, I'm not sure if this will help, but maybe it'll suggest something.
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http://en.wikibooks.org/wiki/Vectors
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# Vectors
A Wikibookian believes this page should be split into smaller pages with a narrower subtopic. You can help by splitting this big page into smaller ones. Please make sure to follow the naming policy. Dividing books into smaller sections can provide more focus and allow each one to do one thing well, which benefits everyone. You can ask for help in dividing this book in the assistance reading room.
A reader requests expansion of this book to include more material. You can help by adding new material (learn how) or ask for assistance in the reading room.
## The basic idea
A vector is a mathematical concept that has both magnitude and direction. Detailed explanation of vectors may be found at the Wikibooks module Linear Algebra/Vectors in Space. In physics, vectors are used to describe things happening in space by giving a series of quantities which relate to the problem's coordinate system.
A vector is often expressed as a series of numbers. For example, in the two-dimensional space of real numbers, the notation (1, 1) represents a vector that is pointed 45 degrees from the x-axis towards the y-axis with a magnitude of $\sqrt 2$.
Commonly in physics, we use position vectors to describe where something is in the space we are considering, or how its position is changing at that moment in time. Position vectors are written as summations of scalars multiplied by unit vectors. For example:
$a \hat{i} + b \hat{j} + c \hat{k}$
where a, b and c are scalars and $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors of the Cartesian (René Descartes) coordinate system. A unit vector is a special vector which has magnitude 1 and points along one of the coordinate frame's axes. This is better illustrated by a diagram.
A vector itself is typically indicated by either an arrow: $\vec{v}$, or just by boldface type: v, so the vector above as a complete equation would be denoted as:
$\vec{v} = a \hat{i} + b \hat{j} + c \hat{k}$
The magnitude of a vector is computed by $|\vec{v}| = \sqrt{\sum_i(x_i^2)}$. For example, in two-dimensional space, this equation reduces to:
$|\vec{v}| = \sqrt{x^2+y^2}$.
For three-dimensional space, this equation becomes:
$|\vec{v}| = \sqrt{x^2+y^2+z^2}$.
# Exercises
Find the magnitude of the following vectors. Answers below.
$\vec{v} = (4, 3)$
$|\vec{v}| = \sqrt{4^2 + 3^2} = 5$
$\vec{v} = (5, 3)$
$|\vec{v}| = \sqrt{5^2 + 3^2} = \sqrt{34}$
$\vec{v} = (1, 0)$
$|\vec{v}| = \sqrt{1^2 + 0^2} = 1$
$\vec{v} = (4, 4)$
$|\vec{v}| = \sqrt{4^2 + 4^2 } = \sqrt{32}$
$\vec{v} = (5, 0, 0)$
$|\vec{v}| = \sqrt{5^2 + 0^2 + 0^2} = 5$
## Using vectors in physics
Many problems, particularly in mechanics, involve the use of two- or three-dimensional space to describe where objects are and what they are doing. Vectors can be used to condense this information into a precise and easily understandable form that is easy to manipulate with mathematics.
Position - or where something is, can be shown using a position vector. Position vectors measure how far something is from the origin of the reference frame and in what direction, and are usually, though not always, given the symbol $\vec{r}$. It is usually good practice to use $\vec{r}$ for position vectors when describing your solution to a problem as most physicists use this notation.
Velocity is defined as the rate of change of position with respect to time. You may be used to writing velocity, v, as a scalar because it was assumed in your solution that v referred to speed in the direction of travel. However, if we take the strict definition and apply it to the position vector - which we have already established is the proper way of representing position - we get:
$\frac{d\vec{r}}{dt} = \frac{da}{dt}\hat{i} + \frac{db}{dt}\hat{j} + \frac{dc}{dt}\hat{k}$
However, we note that the unit vectors are merely notation rather than terms themselves and are in fact not differentated, only the scalars which represent the vector's components in each direction differentiate.
Assuming that each component is not a constant and thus has a non-zero derivative, we get:
$\vec{v} = a' \hat{i} + b' \hat{j} + c' \hat{k}$ where a', b' and c' are simply the first derivatives with respect to time of each original position vector component.
Here it is clear that velocity is also a vector. In the real world this means that each component of the velocity vector indicates how quickly each component of the position vector is changing - that is, how fast the object is moving in each direction.
# Vectors in Mechanics
Vector notation is ubiquitous in the modern literature on solid mechanics, fluid mechanics, biomechanics, nonlinear finite elements and a host of other subjects in mechanics. A student has to be familiar with the notation in order to be able to read the literature. In this section we introduce the notation that is used, common operations in vector algebra, and some ideas from vector calculus.
## Vectors
A vector is an object that has certain properties. What are these properties? We usually say that these properties are:
• a vector has a magnitude (or length)
• a vector has a direction.
To make the definition of the vector object more precise we may also say that vectors are objects that satisfy the properties of a vector space.
The standard notation for a vector is lower case bold type (for example $\mathbf{a}\,$).
In Figure 1(a) you can see a vector $\mathbf{a}$ in red. This vector can be represented in component form with respect to the basis ($\mathbf{e}_1, \mathbf{e}_2\,$) as
$\mathbf{a} = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 \,$
where $\mathbf{e}_1\,$ and $\mathbf{e}_2\,$ are orthonormal unit vectors. Orthonormal means they are at right angles to each other (orthogonal) and are unit vectors. Recall that unit vectors are vectors of length 1. These vectors are also called basis vectors.
You could also represent the same vector $\mathbf{a}\,$ in terms of another set of basis vectors ($\mathbf{g}_1,\mathbf{g}_2\,$) as shown in Figure 1(b). In that case, the components of the vector are $(b_1,b_2)\,$ and we can write
$\mathbf{a} = b_1 \mathbf{g}_1 + b_2 \mathbf{g}_2 \, ~.$
Note that the basis vectors $\mathbf{g}_1\,$ and $\mathbf{g}_2\,$ do not necessarily have to be unit vectors. All we need is that they be linearly independent, that is, it should not be possible for us to represent one solely in terms of the others.
In three dimensions, using an orthonormal basis, we can write the vector $\mathbf{a}\,$ as
$\mathbf{a} = a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + a_2 \mathbf{e}_3 \,$
where $\mathbf{e}_3\,$ is perpendicular to both $\mathbf{e}_1\,$ and $\mathbf{e}_2\,$. This is the usual basis in which we express arbitrary vectors.
## Vector Algebra Operations
Some vector operations are shown in Figure 2.
### Addition and Subtraction
If $\mathbf{a}\,$ and $\mathbf{b}\,$ are vectors, then the sum $\mathbf{c} = \mathbf{a} + \mathbf{b}\,$ is also a vector (see Figure 2(a)).
The two vectors can also be subtracted from one another to give another vector $\mathbf{d} = \mathbf{a} - \mathbf{b}\,$.
### Multiplication by a scalar
Multiplication of a vector $\mathbf{b}\,$ by a scalar $\lambda\,$ has the effect of stretching or shrinking the vector (see Figure 2(b)).
You can form a unit vector $\hat\mathbf{b}\,$ that is parallel to $\mathbf{b}\,$ by dividing by the length of the vector $|\mathbf{b}|\,$. Thus,
$\hat\mathbf{b} = \frac{\mathbf{b}}{|\mathbf{b}|} ~.$
### Scalar product of two vectors
The scalar product or inner product or dot product of two vectors is defined as
$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta)$
where $\theta\,$ is the angle between the two vectors (see Figure 2(b)).
If $\mathbf{a}\,$ and $\mathbf{b}\,$ are perpendicular to each other, $\theta = \pi/2\,$ and $\cos(\theta) = 0\,$. Therefore, ${\mathbf{a}}\cdot{\mathbf{b}} = 0$.
The dot product therefore has the geometric interpretation as the length of the projection of $\mathbf{a}\,$ onto the unit vector $\hat\mathbf{b}\,$ when the two vectors are placed so that they start from the same point.
The scalar product leads to a scalar quantity and can also be written in component form (with respect to a given basis) as
${\mathbf{a}}\cdot{\mathbf{b}} = a_1 b_1 + a_2 b_2 + a_3 b_3 = \sum_{i=1..3} a_i b_i~.$
If the vector is $n$ dimensional, the dot product is written as
${\mathbf{a}}\cdot{\mathbf{b}} = \sum_{i=1..n} a_i b_i~.$
Using the Einstein summation convention, we can also write the scalar product as
${\mathbf{a}}\cdot{\mathbf{b}} = a_i b_i~.$
Also notice that the following also hold for the scalar product
1. ${\mathbf{a}}\cdot{\mathbf{b}} = {\mathbf{b}}\cdot{\mathbf{a}}$ (commutative law).
2. ${\mathbf{a}}\cdot{(\mathbf{b}+\mathbf{c})} = {\mathbf{a}}\cdot{\mathbf{b}} + {\mathbf{a}}\cdot{\mathbf{c}}$ (distributive law).
### Vector product of two vectors
The vector product (or cross product) of two vectors $\mathbf{a}\,$ and $\mathbf{b}\,$ is another vector $\mathbf{c}\,$ defined as
$\mathbf{c} = {\mathbf{a}}\times{\mathbf{b}} = |\mathbf{a}||\mathbf{b}|\sin(\theta) \hat{\mathbf{c}}$
where $\theta\,$ is the angle between $\mathbf{a}\,$ and $\mathbf{b}\,$, and $\hat{\mathbf{c}}\,$ is a unit vector perpendicular to the plane containing $\mathbf{a}\,$ and $\mathbf{b}\,$ in the right-handed sense (see Figure 3 for a geometric interpretation)
In terms of the orthonormal basis $(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\,$, the cross product can be written in the form of a determinant
${\mathbf{a}}\times{\mathbf{b}} = \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}~.$
In index notation, the cross product can be written as
${\mathbf{a}}\times{\mathbf{b}} \equiv e_{ijk} a_j b_k ~.$
where $e_{ijk}$ is the Levi-Civita symbol (also called the permutation symbol, alternating tensor).
## Identities from Vector Algebra
Some useful vector identities are given below.
1. ${\mathbf{a}}\times{\mathbf{b}} = - {\mathbf{b}}\times{\mathbf{a}}$.
2. ${\mathbf{a}}\times{\mathbf{b}+\mathbf{c}} = {\mathbf{a}}\times{\mathbf{b}} + {\mathbf{a}}\times{\mathbf{c}}$.
3. ${\mathbf{a}}\times{({\mathbf{b}}\times{\mathbf{c}})} = \mathbf{b}({\mathbf{a}}\cdot{\mathbf{c}}) - \mathbf{c}({\mathbf{a}}\cdot{\mathbf{b}})$ ~.
4. ${({\mathbf{a}}\times{\mathbf{b}})}\times{\mathbf{c}} = \mathbf{b}({\mathbf{a}}\cdot{\mathbf{c}}) - \mathbf{a}({\mathbf{b}}\cdot{\mathbf{c}})$ ~.
5. ${\mathbf{a}}\times{\mathbf{a}} = \mathbf{0}$~.
6. ${\mathbf{a}}\cdot{({\mathbf{a}}\times{\mathbf{b}})} = {\mathbf{b}}\cdot{({\mathbf{a}}\times{\mathbf{b}})} = \mathbf{0}$~.
7. ${({\mathbf{a}}\times{\mathbf{b}})}\cdot{\mathbf{c}} = {\mathbf{a}}\cdot{({\mathbf{b}}\times{\mathbf{c}})}$~.
## Vector Calculus
So far we have dealt with constant vectors. It also helps if the vectors are allowed to vary in space. Then we can define derivatives and integrals and deal with vector fields. Some basic ideas of vector calculus are discussed below.
## Derivative of a vector valued function
Let $\mathbf{a}(x)\,$ be a vector function that can be represented as
$\mathbf{a}(x) = a_1(x) \mathbf{e}_1 + a_2(x) \mathbf{e}_2 + a_3(x) \mathbf{e}_3\,$
where $x\,$ is a scalar.
Then the derivative of $\mathbf{a}(x)\,$ with respect to $x\,$ is
$\cfrac{d\mathbf{a}(x)}{dx} = \lim_{\Delta x\rightarrow 0} \cfrac{\mathbf{a}(x+\Delta x) - \mathbf{a}(x)}{\Delta x} = \cfrac{da_1(x)}{dx} \mathbf{e}_1 + \cfrac{da_2(x)}{dx} \mathbf{e}_2 + \cfrac{da_3(x)}{dx} \mathbf{e}_3~.$
Note: In the above equation, the unit vectors $\mathbf{e}_i$ (i=1,2,3) are assumed constant.
If $\mathbf{a}(x)\,$ and $\mathbf{b}(x)\,$ are two vector functions, then from the chain rule we get
$\begin{align} \cfrac{d({\mathbf{a}}\cdot{\mathbf{b}})}{x} & = {\mathbf{a}}\cdot{\cfrac{d\mathbf{b}}{dx}} + {\cfrac{d\mathbf{a}}{dx}}\cdot{\mathbf{b}} \\ \cfrac{d({\mathbf{a}}\times{\mathbf{b}})}{dx} & = {\mathbf{a}}\times{\cfrac{d\mathbf{b}}{dx}} + {\cfrac{d\mathbf{a}}{dx}}\times{\mathbf{b}} \\ \cfrac{d[{\mathbf{a}}\cdot{({\mathbf{a}}\times{\mathbf{b}})}]}{dt} & = {\cfrac{d\mathbf{a}}{dt}}\cdot{({\mathbf{b}}\times{\mathbf{c}})} + {\mathbf{a}}\cdot{\left({\cfrac{d\mathbf{b}}{dt}}\times{\mathbf{c}}\right)} + {\mathbf{a}}\cdot{\left({\mathbf{b}}\times{\cfrac{d\mathbf{c}}{dt}}\right)} \end{align}$
## Scalar and vector fields
Let $\mathbf{x}\,$ be the position vector of any point in space. Suppose that there is a scalar function ($g\,$) that assigns a value to each point in space. Then
$g = g(\mathbf{x})\,$
represents a scalar field. An example of a scalar field is the temperature. See Figure4(a).
If there is a vector function ($\mathbf{a}\,$) that assigns a vector to each point in space, then
$\mathbf{a} = \mathbf{a}(\mathbf{x})\,$
represents a vector field. An example is the displacement field. See Figure 4(b).
## Gradient of a scalar field
Let $\varphi(\mathbf{x})\,$ be a scalar function. Assume that the partial derivatives of the function are continuous in some region of space. If the point $\mathbf{x}\,$ has coordinates ($x_1, x_2, x_3\,$) with respect to the basis ($\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\,$), the gradient of $\varphi\,$ is defined as
$\boldsymbol{\nabla}{\varphi} = \frac{\partial \varphi}{\partial x_1}~\mathbf{e}_1 + \frac{\partial \varphi}{\partial x_2}~\mathbf{e}_2 + \frac{\partial \varphi}{\partial x_3}~\mathbf{e}_3 ~.$
In index notation,
$\boldsymbol{\nabla}{\varphi} \equiv \varphi_{,i}~\mathbf{e}_i ~.$
The gradient is obviously a vector and has a direction. We can think of the gradient at a point being the vector perpendicular to the level contour at that point.
It is often useful to think of the symbol $\boldsymbol{\nabla}{}$ as an operator of the form
$\boldsymbol{\nabla}{} = \frac{\partial }{\partial x_1} ~\mathbf{e}_1 + \frac{\partial }{\partial x_2} ~\mathbf{e}_2 + \frac{\partial }{\partial x_3} ~\mathbf{e}_3 ~.$
## Divergence of a vector field
If we form a scalar product of a vector field $\mathbf{u}(\mathbf{x})\,$ with the $\boldsymbol{\nabla}{}$ operator, we get a scalar quantity called the divergence of the vector field. Thus,
$\boldsymbol{\nabla}\cdot{\mathbf{u}} = \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3} ~.$
In index notation,
$\boldsymbol{\nabla}\cdot{\mathbf{u}} \equiv u_{i,i} ~.$
If $\boldsymbol{\nabla}\cdot{\mathbf{u}} = 0$, then $\mathbf{u}\,$ is called a divergence-free field.
The physical significance of the divergence of a vector field is the rate at which some density exits a given region of space. In the absence of the creation or destruction of matter, the density within a region of space can change only by having it flow into or out of the region.
## Curl of a vector field
The curl of a vector field $\mathbf{u}(\mathbf{x})\,$ is a vector defined as
$\boldsymbol{\nabla}\times{\mathbf{u}} = \det \begin{vmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ \frac{\partial }{\partial x_1} & \frac{\partial }{\partial x_2} & \frac{\partial }{\partial x_3} \\ u_1 & u_2 & u_3 \\ \end{vmatrix}$
The physical significance of the curl of a vector field is the amount of rotation or angular momentum of the contents of a region of space.
## Laplacian of a scalar or vector field
The Laplacian of a scalar field $\varphi(\mathbf{x})\,$ is a scalar defined as
$\nabla^2{\varphi} := \boldsymbol{\nabla}\cdot{\boldsymbol{\nabla}{\varphi}} = \frac{\partial^2 \varphi}{\partial x_1} + \frac{\partial^2 \varphi}{\partial x_2} + \frac{\partial^2 \varphi}{\partial x_3}~.$
The Laplacian of a vector field $\mathbf{u}(\mathbf{x})\,$ is a vector defined as
$\nabla^2{\mathbf{u}} := (\nabla^2{u_1}) \mathbf{e}_1 + (\nabla^2{u_2}) \mathbf{e}_2 + (\nabla^2{u_3}) \mathbf{e}_3 ~.$
## Identities in vector calculus
Some frequently used identities from vector calculus are listed below.
1. $\boldsymbol{\nabla}\cdot{(\mathbf{a} + \mathbf{b})} = \boldsymbol{\nabla}\cdot{\mathbf{a}} + \boldsymbol{\nabla}\cdot{\mathbf{b}}$~.
2. $\boldsymbol{\nabla}\times{(\mathbf{a} + \mathbf{b})} = \boldsymbol{\nabla}\times{\mathbf{a}} + \boldsymbol{\nabla}\times{\mathbf{b}}$~.
3. $\boldsymbol{\nabla}\cdot{(\varphi\mathbf{a})} = {(\boldsymbol{\nabla}{\varphi})}\cdot{\mathbf{a}} + \varphi (\boldsymbol{\nabla}\cdot{\mathbf{a}})$~.
4. $\boldsymbol{\nabla}\times{(\varphi\mathbf{a})} = {(\boldsymbol{\nabla}{\varphi})}\times{\mathbf{a}} + \varphi (\boldsymbol{\nabla}\times{\mathbf{a}})$~.
5. $\boldsymbol{\nabla}\cdot{({\mathbf{a}}\times{\mathbf{b}})} = {\mathbf{b}}\cdot{(\boldsymbol{\nabla}\times{\mathbf{a}})} - {\mathbf{a}}\cdot{(\boldsymbol{\nabla}\times{\mathbf{b}})}$~.
## Green-Gauss Divergence Theorem
Let $\mathbf{u}(\mathbf{x})\,$ be a continuous and differentiable vector field on a body $\Omega\,$ with boundary $\Gamma\,$. The divergence theorem states that
${ \int_{\Omega} \boldsymbol{\nabla}\cdot{\mathbf{u}}~dV = \int_{\Gamma} {\mathbf{n}}\cdot{\mathbf{u}}~dA }$
where $\mathbf{n}\,$ is the outward unit normal to the surface (see Figure 5).
In index notation,
$\int_{\Omega} u_{i,i}~dV = \int_{\Gamma} n_i u_i~dA$
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http://mathhelpforum.com/number-theory/1589-convergence-primes.html
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# Thread:
1. ## Convergence of primes
Let $p_n$ represent the n-th prime number. Determine what $\lim_{n\rightarrow \infty}\frac{p_{n+1}}{p_n}=?$. Does it even have a convergence?
Notice the following
$p_n<p_{n+1}<2p_n$ (Betrand's Conjecure)
thus,
$1<\frac{p_{n+1}}{p_n}<2$
Thus, by the limit limitation theorem,
$1\leq \lim_{n\rightarrow \infty}\frac{p_{n+1}}{p_n} \leq 2$ (I, of course, assumed the limit exists).
My guess is that the limit is one.
2. Nice problem. To my knowledge, the spacing of primes would increase as x approaches infinity, correct? If this is true it would seem that the limit would not converge.
Another cool prime problem is the sum of the reciprocals. Does $\sum_{n=1}^{\infty}\frac{1}{p_n}$ converge?
3. [QUOTE=Jameson]Nice problem. To my knowledge, the spacing of primes would increase as x approaches infinity, correct? If this is true it would seem that the limit would not converge. [qUOTE]
$p_n\sim n.\ln(n)$, so we might expect that:
$\frac{p_{n+1}}{p_n}\sim \frac{(n+1). \ln(n+1)}{n.\ln(n)}$.
Now the RHS of this relation goes to $1$ as $n \rightarrow \infty$.
This is not a proof because of the nature of the $\sim$
relation used here, but it is indicative that its not
implausible that the limit might exist and be equal to $1$.
Also, computing this ratio for all pairs of consecutive
primes $<10,000,000$ shows that this could
well be true.
RonL
4. Jameson, the problem you posted determine, $\sum^\infty_{k=1}\frac{1}{p_k}$ it diverges. It was demonstrated by Euler.
However my problem still remains unanswered. I fully believe that it converges to one. Do you are agree CaptainBlack, Jameson?
5. Oh nice post CaptainBlack. I'll have to do some research on prime number theory. At first glance the limit just doesn't seem to approach 1.
6. It is also possible that the limit does not exist at all. That the limit of this quotient fluctuates for values close to one.
The reason why I am doing this problem is to have an elegant proof that a 'polynomial prime producing phunction' (I just could not resist using 4 p's ) does not exist. The books on number theory give an elementary ugly proof. Thus, since it is true that the ratio of terms in any polynomial progression always converges to one. If we assumed that a 'polynomial prime producing phunction' does not converge to one then we have a contradiction.
Notice we can use the above concept to prove that a polynomial function for the Fibonacci sequence does not exist. Because in the Fibonacci sequence the ratio of consecutive terms must converge to 1 but in reality they converge to the Divine Proportion.
7. Originally Posted by ThePerfectHacker
However my problem still remains unanswered. I fully believe that it converges to one. Do you are agree CaptainBlack, Jameson?
Further research reveals that the required result was proven (or is a special
case of what was proven) by Erdös in a paper in 1949. As yet I haven't
found a proof online.
However it does look to me quite easy to show that if the limit exists it
must be one, as otherwise it would contradict the prime number theorem.
RonL
8. The standard reference for precise results of this kind is J.B. Rosser and L. Schoenfeld, "Approximate formulas for some functions of prime numbers", Illinois J. Maths, vol.6 (1962) 64--94.
Their (3.12) and (3.13) are $n \log n < p_n < n\log n + n\log\log n$ for all $n \ge 6$ which is enough to make CaptainBlack's argument precise.
9. Originally Posted by rgep
The standard reference for precise results of this kind is J.B. Rosser and L. Schoenfeld, "Approximate formulas for some functions of prime numbers", Illinois J. Maths, vol.6 (1962) 64--94.
Their (3.12) and (3.13) are $n \log n < p_n < n\log n + n\log\log n$ for all $n \ge 6$ which is enough to make CaptainBlack's argument precise.
Too lazy to check did you apply the squeeze theorem?
10. ## Are you sure of that?
Originally Posted by rgep
The standard reference for precise results of this kind is J.B. Rosser and L. Schoenfeld, "Approximate formulas for some functions of prime numbers", Illinois J. Maths, vol.6 (1962) 64--94.
Their (3.12) and (3.13) are $n \log n < p_n < n\log n + n\log\log n$ for all $n \ge 6$ which is enough to make CaptainBlack's argument precise.
I think that if n*log(n)<Pn where Pn is the n°ème prime number if i have understand correctly would involved that there would be no twin prime number (ie Pn=(Pn+1)+2 ) you must have forgoten a little something in your formula but indeed your formula gives sense to CapitainBlack "tilda" notation
hargh..
11. Originally Posted by SkyWatcher
I think that if n*log(n)<Pn where Pn is the n°ème prime number if i have understand correctly would involved that there would be no twin prime number (ie Pn=(Pn+1)+2 ) you must have forgoten a little something in your formula but indeed your formula gives sense to CapitainBlack "tilda" notation
hargh..
I'm not sure I understand your statement that
$n \log(n)<p_n$,
would imply that there are no twin primes.
Now if for all $n$ sufficiently large:
$(n+1)\log(n+1)>n \log(n)+\log \log(n)+2$
then we would have but a finite number of twin prime
pairs, but I don't think this is the case.
Excercise: show that as $n \rightarrow \infty$:
$(n \log (n)+\log \log(n)) - (n+1) \log (n+1) \rightarrow \infty$
(if it does).
RonL
12. Originally Posted by CaptainBlack
then we would have but a finite number of twin prime
pairs, but I don't think this is the case.
Excercise: show that as $n \rightarrow \infty$:
$(n \log (n)+\log \log(n)) - (n+1) \log (n+1) \rightarrow \infty$
(if it does).
RonL
of course i was a litle fast on that one (milk was boiling)
but i know that i have read somewere there are an infinite number of twins primes i have a litle idea (i will check and maybe start a thread on this)
now i will try to reply your exercise : i love limits problem and its a long time i didn't did one:
here it goes (the forth the sixth...)
We have to proove that for any positive integer M we can find an integer noted here A for wich when replacing n by A in your formula we find a result greater than M (by definition of +infinity limit)
We know that log is a strictly growing up function so is her reciprocal
x->10^x keep that in mind to understand the rest (i cannot do a strictly correct demonstration on keyboard)
So we have log(n+1)>log(n) for every positive number
So for A and replacing n by A in the formula developing the formula and making a majoration will lead us to proove that for every M we can find A
for wich log(A)-log(log(A))>M
wich is equivalent to show that
A/log(A)>10^M <=> A>10^M *log(A)
we know that
log(A) < (number of digits in base 10 of A)+1
So give me an M i will give you A=10^(M^2) wich satisfy the inequality
because 10^(M^2) is greater than 10^M*((M^2)+1) ( for M>1 indeed)
Hush!
13. I do not know if there are infinite number of twin primes,
SkyWatcher.
An attempt was made by Viggo Brun which tried to show that the series:
$\sum^{\infty}_{k=1}\frac{1}{t_k}$
diverges, (where $t_k$ is k-th twin prime).
Thus, if there are finitely many twin primes then this series must for definetly converge. Thus, by contropositive, if this series diverges then there are infinitely many twin primes. But Brun demonstrated that it actually converges. Thus, the results are inconclusive. This is called Brun's Constant.
Thus, the question still remains, are there infinitely many twin prime? I once tried to prove it by assuming there are finitely many and constructing a number which would also be a prime twin, (like in Euclid's prove of infinitely many prime numbers) but no success.
14. I have been looking to paper encyclopedies and it seems that the problem of "is there an infinity of twin primes" was not conclued at the times those encyclopedies were writen.
so my memory had made two mistakes
1 i thought i had seen somewhere (in internet) that it was proven that there are an infinity of those (twin primes)
2 i had made a mistake about "is there an infinity of primes" (see jameson thread on that) i had understood (long time ago) that the product of the N first prime plus one was prime (it walks when you consider what you can calculate whithout using a computer (for the first prime)).
anyway if the formula i am doubtful about was right you would'nt had to squeeze your brain trying to find if there is an infinity of that stuff because it's an evident coloraly of such a théorème that's there is not!
what made me looking about the thrueness of this formula is that it's using decimal logarithm (as far as we have the same notation in english and in french) and i could'nt see why the number ten would had to do with the distribution of prime number.
15. Originally Posted by SkyWatcher
I have been looking to paper encyclopedies and it seems that the problem of "is there an infinity of twin primes" was not conclued at the times those encyclopedies were writen.
so my memory had made two mistakes
1 i thought i had seen somewhere (in internet) that it was proven that there are an infinity of those (twin primes)
2 i had made a mistake about "is there an infinity of primes" (see jameson thread on that) i had understood (long time ago) that the product of the N first prime plus one was prime (it walks when you consider what you can calculate whithout using a computer (for the first prime)).
anyway if the formula i am doubtful about was right you would'nt had to squeeze your brain trying to find if there is an infinity of that stuff because it's an evident coloraly of such a théorème that's there is not!
what made me looking about the thrueness of this formula is that it's using decimal logarithm (as far as we have the same notation in english and in french) and i could'nt see why the number ten would had to do with the distribution of prime number.
Just because your found on the internet it said that they had a proof does not mean anything. I once searched for Fermat's Last Theorem, then I get hunders of links that promise they have "proofs". Thus, I enter some on them and I see that the people that made the site do not know what they are talking about. For example, someone gives a false proof on one of these website and in the end he writes, "...Thus, it does not work for integral,rational and irrational numbers." WHAT! it certainly does work for irrational numbers. If it said that on Wikipedia, or Mathworld then you can trust it otherwise people try to make themselves look smart by tricking people in making them think they got proofs.
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http://www.physicsforums.com/showthread.php?t=246182
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Physics Forums
## Ideal Gas Law and Isobaric Processes
Hi,
So let's take the standard example of a gas in a container with a piston at the top. Charles' Law states that at constant pressure, an increase in temperature (kinetic energy of gas molecules) will increase the volume. This makes sense both conceptually and mathematically (per PV = nRT). However, in an isobaric process (pressure is constant again), the kinetic energy of the gas molecules is what is moving the piston so it must have lost some energy in doing so. Therefore, although initially Esystem = q that was added, it's net energy change would be Esystem = q - Pext*V.
So this is my problem, wouldn't this isobaric process be inconsistent with the ideal gas law? That is, using strictly the ideal gas law, woudn't the ending temperature not take into account the work done on the piston? Or are we assuming in using the ideal gas law that the work to keep the pressure constant is from an external source?
Thanks.
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Quote by nothing123 Hi, So let's take the standard example of a gas in a container with a piston at the top. Charles' Law states that at constant pressure, an increase in temperature (kinetic energy of gas molecules) will increase the volume. This makes sense both conceptually and mathematically (per PV = nRT). However, in an isobaric process (pressure is constant again), the kinetic energy of the gas molecules is what is moving the piston so it must have lost some energy in doing so. Therefore, although initially Esystem = q that was added, it's net energy change would be Esystem = q - Pext*V. So this is my problem, wouldn't this isobaric process be inconsistent with the ideal gas law? That is, using strictly the ideal gas law, woudn't the ending temperature not take into account the work done on the piston? Or are we assuming in using the ideal gas law that the work to keep the pressure constant is from an external source? Thanks.
You're mixing two different concepts. The ideal gas equation of state shows the relationship between P, V and T at a specified state. An isobaric process (or any process) shows how a substance changed from an initial state to a final state.
Does that help?
CS
Could you clarify exactly what you mean by states? I mean, since the heat added only changes the kinetic energy of the system (which is proportional to temperature), wouldn't we be able to find the same change in temperature whether we used Echange = q + w or whether we used T = PV/nR? Thanks for your help so far.
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## Ideal Gas Law and Isobaric Processes
Quote by nothing123 Could you clarify exactly what you mean by states? I mean, since the heat added only changes the kinetic energy of the system (which is proportional to temperature), wouldn't we be able to find the same change in temperature whether we used Echange = q + w or whether we used T = PV/nR? Thanks for your help so far.
By state I mean a set of properties that completely describe the condition of the system. If the system is not changing, it is in equilibrium. If the system undergoes a process, something has changed the system and it is now in an alternate state. The ideal gas law describes the state of an ideal gas. If the ideal gas undergoes a process, the process path from state 1 to state 2 will describe how the system changed. Once at state 2 the ideal gas relation can describe the system at that state.
Also, the properties of an ideal gas at two different states for a fixed mass are related by the ideal gas law as well and is called the combined gas law IIRC. However, you would need to know some of the properties at both states in order to solve for the unknown. Hence, the process path must be known.
BTW,
$$\Delta U = Q_{net,in} - W_{net,out}$$
Shows that for a closed system the internal energy of a substance decreases if it does work on it's surrounds. Hence it's temperature would decrease. So for an isobaric process like you described, the change in internal energy and thus temperature would depend on how much heat was added, and how much work was done by the piston on its surroundings.
This is typically stated using the enthalpy for simplicity:
$$Q - W_{other} = H_2 - H_1$$
Does that help?
CS
Thank you very much. Clearing up that state definition really helped.
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Quote by nothing123 Could you clarify exactly what you mean by states? I mean, since the heat added only changes the kinetic energy of the system (which is proportional to temperature), wouldn't we be able to find the same change in temperature whether we used Echange = q + w or whether we used T = PV/nR? Thanks for your help so far.
T = PV/nR does not tell you the what P is and what V is. It only tells you what their product is. If T increases, PV must increase at the same rate. If it happens at constant volume, there is no work done by the gas. If it occurs at constant pressure, work is done so the heat flow is greater. PV=nRT always applies, as does dQ = dU + PdV.
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http://physics.stackexchange.com/questions/19904/generalisation-of-reversible-equation-to-non-reversible-situations-because-it-on?answertab=votes
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# Generalisation of Reversible Equation to Non-Reversible Situations Because it Only Contains 'Properties of the System'
I've just been reading through Van Ness' Understanding Thermodynamics, and I'm having a little trouble following his argument at one point. He is deriving the useful differential equation:
$$dU=T \ dS -P\ dV$$
By noting that:
$$dU=dQ-dW$$ for all processes, and then also noting that, only for reversible processes,
$$dQ_{rev}=T\ dS$$ and $$dW_{rev}=P\ dV$$
and substituting these last 2 equation into the first equation. As long as we're talking about reversible processes, I'm fine with this. But then he then goes on to say:
Now we derived this equation for a reversible process, but once derived we see that it contains just properties of the system, and so it must not depend on the kind of process considered. What we have really done is derive an equation for a special case, and then conclude that it must be general.
Maybe this is blindingly obvious, but I'm really struggling to follow his logic through. He's derived an equation for one particular case, and then, just because this equation doesn't explictly refer to anything other than properties of the system, he concludes that it must hold for all cases. I mean, for the reversible case, sure, but I'm struggling to see why, logically, $dU=TdS-PdV$ would hold for a given irreversible case.
To give a very rough analogy of my thinking, I feel live I've found an equation like Boyle's law:
$$P\propto\frac{1}{V}$$ and can then conclude that, just because this equation (a) holds for one particular system (e.g. low pressure, high volume gas) and (b) only contains 'properties of the system' $(P, V)$, then it must therefore hold for all situations, which is of course nonsense.
Anyway, I'd appreciate any help, thanks
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## 1 Answer
Some properties of systems are special because they don't depend on the path taken to get there. The are called "functions of state". To take a trivial example, the volume of a system is a state function. A given system in a given state will always have the same volume regardless of how you prepared that system.
The equation for $dU$ only contains properties that are functions of state. That means it doesn't depend on the path taken for a change, and therefore it applies to changes in irreversible systems as well as reversible systems.
For more details (and I must admit I had to look it up to check) see http://en.wikipedia.org/wiki/Fundamental_thermodynamic_relation
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Thanks for your help, I do really appreciate it, but I'm still not 110% satisfied with your logic. Perhaps could you point out what's wrong with this 'counter-example'. There's a class of reversible processes (in particular, adiabatic ones) which, on top of obeying the complex equation above, also obey the simpler $dS=0$. Now we know that $S$ is a state function, so why can't we use the same logic and say that, just because $dS=0$ holds for all adiabatic, reversible processes, it must hold for all adiabatic processes, including irreversible ones (obviously ridiculous)? Thanks – tom Jan 25 '12 at 2:27
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http://mathoverflow.net/questions/45704?sort=votes
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## Shimura datum of family of fake elliptic curves
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Suppose we have a PEL type $(H,\phi ,*;T,O,V)$ where H is a rational nonsplit quaternion algebra, $\phi$is an embedding of Q-algebra $\phi : H-->M(2,R)$, and * is a positive anti involution of H; O is the maximal order of H , and V level structure . Associate these datum a shimura curve parametric fake elliptic curves in a standar way. My question is :what is the group for the shimura datum ,is it the group H*, the invertible element of H ? And since this family parametirc abelian two folds , what is the map from this group to GSP(4,Q)?
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You should look at the paper of Goresky and Tai, "Anti-holomorphic multiplication and a real algebraic modular variety", J. Differential Geom. vol. 65 no. 3. – David Hansen Nov 12 2010 at 16:30
## 2 Answers
Actually, fake elliptic curves are discussed in Chapter 9 of Lang's Introduction to algebraic and abelian functions.
In order to describe the map to $GSP(4,Q)$, recall that there is the standard antiinvolution $x\to x'$ on the quaternion $Q$-algebra $H$ such that both $tr(x)=x+x'$ and $Norm(x)=xx'=x'x$ are rational numbers for all $x\in H$. Every positive antiinvolution of $H$ is of the form $x \to \gamma^{-1}x'\gamma$ where $\gamma$ is a fixed nonzero (invertible) element of $H$ such that $\gamma^2$ is a negative rational number and therefore $\gamma'=-\gamma$. This gives rise to the alternating nondegenerate $Q$-bilinear form $$E: H \times H \to Q, x,y \mapsto tr(\gamma x y').$$ Now let us consider the following faithful action of the multiplicative group $G$ of $H$ on the $Q$-vector space $V=H$: $$u(x)=x u'$$ for $x \in V=H$ and $u \in G$. Then
$$E(u(x),u(y))=tr(\gamma x u'u y')=(u'u)tr(\gamma x y'),$$
which means that $$E(u(x),u(y))=Norm(u) E(x,y).$$
This gives us the embedding $G \to GSP(V,E)\cong GSP(4.Q)$.
The same construction over arbitrary commutative $Q$-algebras $R$ gives us the desired embedding of the corresponding $Q$-algebraic groups.
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Thank you very much!This is exactly what I need! – TOM Nov 12 2010 at 23:29
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Question 1(what is the group for the Shimura datum):
Well, remember that $H^\times$ is just a bare group. A Shimura datum requires an algebraic group over $\mathbf{Q}$: that is, a functor from $\mathbf{Q}$-schemes to groups. Assuming you mean the group whose $\mathbf{Q}$ points, yes and you can see this in example 5.24 of http://www.jmilne.org/math/xnotes/svi.pdf although you can also use the algebraic group whose $\mathbf{Q}$ points are the norm one units of $H$ if you were interested in the connected Shimura datum (which is another example in milne's notes).
Question 2(what is the map from this group to the symplectic group):
I don't know. Is it even clear that a forgetful map of coarse moduli spaces which happen to be Shimura varieties induces a morphism of Shimura data? Either way your question sounds strongly related to the work of Victor Rotger's thesis which asks about the irreducibility of the quaternionic locus in $\mathcal{A}_2$.
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I think that if you use the norm one units of $H$ than you get a "connected Shimura variety" rather than a Shimura variety proper, since the norm one units don't have enough room to accept a map from all of $Res^{\mathbb C}_{\mathbb R} \mathbb G_m$ (as is necessary for a Shimura datum); they can only accept a map from the norm one part, which gives a connected Shimura datum. – Emerton Nov 12 2010 at 14:21
Thank you, I've edited to include your correction. – stankewicz Nov 12 2010 at 15:08
@stankewicz: Regarding 2: as you can see from Professor Zarhin's answer, this is true in this case (and the use of additional endomorphisms to construct a Riemann form works in other situations as well, as he says). As a matter of general philosophy, I would say that yes, naturally defined morphisms between Shimura varieties do tend to be induced by homomorphisms of the Shimura data. – Pete L. Clark Nov 12 2010 at 22:55
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http://math.stackexchange.com/questions/5259/approximating-pi-using-monte-carlo-integration/5287
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# Approximating $\pi$ using Monte Carlo integration
I'm trying to approximate $\pi$ using Monte Carlo integration; I am approximating the integral
$$\int\limits_0^1\!\frac{4}{1+x^2}\;\mathrm{d}x=\pi$$
This is working fine, and so is estimating the error (variance), $\sigma$. However, when I then try to use importance sampling with a Cauchy(0,1) distribution, things start to go wrong:
$$\frac{1}{n}\sum\limits_{i=0}^n\frac{f(x_i)}{p(x_i)}=\frac{1}{n}\sum\limits_{i=0}^n\frac{\frac{4}{1+x^2}}{\frac{1}{\pi(1+x^2)}}=\frac{1}{n}\sum\limits_{i=0}^n\frac{4\pi(1+x^2)}{1+x^2}=\frac{1}{n}\sum\limits_{i=0}^n4\pi=4\pi$$
Obviously something's wrong, since the mean is computed independently of the random variables I generate. Where is this going wrong? Is the distribution too close to $f$?
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I think there's some confusion over the range of your integration and thesupport of the Cauchy distribution; it's supported on (-\infty, \infty), not on (0,1). Your sum (presuming that x_i = i and the x's in the second formula are supposed to be i's) covers (0, \infty), so you're not computing the integral from 0..1 that you think you are. Can you be a bit more explicit about your sampling process? – Steven Stadnicki Sep 23 '10 at 0:25
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"Monte Carlo is an extremely bad method; it should be used only when all alternative methods are worse." ~ Alan Sokal – J. M. Sep 23 '10 at 14:14
## 1 Answer
This is quite a common error when doing Monte Carlo integration. The support of the random variable you choose must be equal to the range of integration. Though the Cauchy distribution has support on $\mathbb{R}$ we can adapt it slightly to make it work here.
Note that: $\int_0^1 \frac{1}{\pi(1+x^2)} dx = \frac{1}{\pi}\tan^{-1}(1)=\frac{1}{4}$
Thus $g(x) = \frac{4}{\pi(1+x^2)}$ for $x\in(0,1)$ is a probability density with support on $(0,1)$.
using this we have
$\frac{1}{n}\sum_{i=0}^n \frac{f(x_i)}{g(x_i)} = \pi$
This is not a problem! Since $g(x) = \frac{1}{\pi}f(x)$ you have found exactly the right probability distribution to use to evaluate $\int_0^1f(x)dx$! The error is zero for any sample, regardless of the size.
Note however, I needed to know how to integrate $\int_0^1\frac{1}{\pi(1+x^2)}$ in the first place to form the probability distribution. Thus for arbitrary functions it is impossible to get this situation.
Also note the Monte Carlo is not a very good way of approximating integrals in general. Far better deterministic methods are quadrature rules.
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I've been told that Monte Carlo methods get more useful the higher the dimension. – Qiaochu Yuan Sep 23 '10 at 7:53
Here are my thoughts on the multi-dimensional setting: Monte Carlo methods in practice are going to give a larger error than their quadrature counterparts. However, in high dimensions quadrature rules require more evaluations of the integrand and this may be unacceptable (if evaluations are expensive). – alext87 Sep 23 '10 at 8:43
My take on it is: Monte Carlo methods only get more useful the higher the dimension because quadrature rules become less so. (As you can tell, I think Monte Carlo method's popularity are more down to their simplicity rather than their usefulness.) – alext87 Sep 23 '10 at 8:53
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(Quasi-)Monte Carlo methods are not as affected by the "curse of dimensionality" as deterministic quadrature rules. That's the only reason we're even interested in Monte Carlo in practical work, in the absence of any exploitable structure (e.g. symmetry) in the integrand. – J. M. Sep 23 '10 at 9:57
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The other place where Monte Carlo reigns supreme is in integration over complicated regions; imagine trying to properly formulate an integral over, e.g., the intersection of three tori, or even something as simple as the interior of a convex polyhedron. As long as you have some mechanism for deciding point-in-region, and some reasonable bounding sphere/box/'easy' shape for your region, though, you can just rejection-sample to limit your Monte Carlo samples to the interior of the region and then normalize the result by dividing out by the sum of sample weights. – Steven Stadnicki Sep 23 '10 at 22:17
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What you need is this. You'll find its derivation in any standard thermodynamics text. $\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V} - P.$ I agree, though, with Studiot. It is impolite not to acknowledge attempts to help. It's perfectly in order to add that you didn't understand the post, but better if you can say exactly what you didn't understand.
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http://mathoverflow.net/questions/17182?sort=votes
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Is an irreducible holomorphic symplectic manifold a simple Lie algebra?
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The tangent bundle of a hyper-Kahler manifold gives a quadratic Lie algebra in the derived category. Can this be regarded as a simple Lie algebra according to Vogel's definition?
A point of view that came from studying Rozansky-Witten invariants is that the tangent bundle of a holomorphic symplectic manifold or hyper-Kahler manifold is a Lie algebra with a non-degenerate invariant symmetric bilinear form. Here the tangent bundle is taken as an object in the derived category and then shifted. The Atiyah class is interpreted as a Lie bracket and the Bianchi identity as the Jacobi identity. The symplectic form is interpreted as a symmetric form since we shifted. Some references are (and please add or request any reference I have omitted)
MR2024627 (2004m:57026) Roberts, Justin . Rozansky-Witten theory. Topology and geometry of manifolds (Athens, GA, 2001), 1--17, Proc. Sympos. Pure Math., 71, Amer. Math. Soc., Providence, RI, 2003.
MR2110899 (2005h:53070) Nieper-Wißkirchen, Marc . Chern numbers and Rozansky-Witten invariants of compact hyper-Kähler manifolds. World Scientific Publishing Co., Inc., River Edge, NJ, 2004. xxii+150 pp. ISBN: 981-238-851-6
MR2472137 (2010d:14020) Markarian, Nikita . The Atiyah class, Hochschild cohomology and the Riemann-Roch theorem. J. Lond. Math. Soc. (2) 79 (2009), no. 1, 129--143.
Now Vogel has constructed a universal simple Lie algebra. The question is whether the tangent bundle of an irreducible holomorphic symplectic manifold meets Vogel's criteria for a simple Lie algebra. This question is for algebraic geometers so I will expand on this. The first condition is that End(L)=End(I) where I is the trivial representation so End(I) is the commutative ring of scalars. In this example Ext(O). This obviously fails for the product of two manifolds so I have naively excluded this by imposing the irreducible condition. The second condition is that $\mathrm{Hom}(\bigwedge^2L,L)$ is a free End(I)-module with basis the Lie bracket.
One reason I find this confusing is that End(I) has nilpotent elements whereas I am used to a field.
If the answer to both questions is Yes then we get a character of Vogel's universal ring. I would expect this to be of interest to both subjects.
Edit The paper http://arxiv.org/abs/1205.3705 has now been posted on the arxiv and this proves that $K3$-surfaces do give a character of Vogel's ring.
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Vogel's paper: math.jussieu.fr/~vogel/A299.ps.gz – Allen Knutson Mar 5 2010 at 19:45
I have a question. In his paper Vogel does not require that $Hom(\wedge^2 L,L)$ is one dimensional (I am refering to the definition of simple on the bottom of page 11 in his paper "The universal Lie algebra"). Why do you need it here? – DamienC Aug 24 2011 at 13:44
Vogel has two conditions. He defines simple by $End(L)$ is one dimensional. He then has a second condition in 3.2 Theorem that a certain commutative square is Cartesian. My intention is that the two second conditions are equivalent. The second condition is needed to define a character of the universal ring. – Bruce Westbury Aug 24 2011 at 16:31
If I'm not mistaken this point of view on RW invariants is due to Kontsevich (arXiv:dg-ga/9704009) and Kapranov (arXiv:alg-geom/9704009). – David Ben-Zvi Dec 2 2011 at 17:14
1 Answer
I believe this is a very interesting question, that I have been asking myself for quite a long time.
Nevertheless, I have been told by Prof. Beauville that even in the irreducible case one does not have that $$Ext_X(\mathcal O_X,\mathcal O_X)=Ext_X(T_X,T_X)$$
Namely, consider $X$ being the Hilbert scheme of two points on a $K3$ surface.
Then $Ext_X(\mathcal O_X,\mathcal O_X)=\mathbb{C}\oplus\mathbb{C}[-2]\oplus\mathbb{C}[-4]$.
But `$Ext_X(T_X,T_X)=Ext_X(\mathcal O_X,(T^*_X)^{\otimes 2})$` contains $Ext_X(\mathcal O_X,\Omega^2_X)$, which is huge ($h^{2,2}=232$).
Anyway, I must say that this does not kill the question (this just tells we have to reformulate it). I hope to be able to write more about it soon.
EDIT: it seems that the answer to the question is NO. The point is that 232 is also the dimension of $H^1(X,S^3(T_X))$ ($X$ is again a $K3$), therefore $Ext_X^1(S^2(T_X),T_X)=RHom_X(\wedge^2(T_X[-1]),T_X[-1]))$ has dimension $\geq232$.
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Physics Forums
## Cannot tell when a probability generating function converges for |s|<1
Hi, I have a problem that is already solved... I thought 3 of the 4 functions were probability generating functions, but I got one wrong and don't know why.
The solution says $g(s)=1+s-s^2$ is not a probability generating function. However, g(1)=1 and I think g(s) converges to 1 for |s|<1. Isn't that correct? If so, what is it that invalidates this function as a probability generating function?
The solution says that $g(s)=(1/3)*(1+s+s^4)$ and $g(s)=(2-s^2)^{-1}$ are prob. gen. functions and that $g(s)=1+s-s^2$ and $g(s)=(1/2)(1+s+s^3)$ are not (I know the last one is not because g(1)=3/2).
Thanks a lot for your help! :) And if you could also give me an explanation of other things I need to look for in a function to tell if it is a p.g.f., I would really appreciate that.
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If $1 + s - s^2$ were a probability generating function, then we would have $P(2) = -1$.
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http://mathoverflow.net/questions/100620/good-reduction-of-abelian-varieties-s-t-why-is-this-ring-henselian/100621
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## Good reduction of abelian varieties [S-T] — Why is this ring henselian?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
First of all, I find it hard to formulate a good title for this question. Sorry that it is so vague.
Let's move on te the question itself. Lately I have been studying the article "Good reduction of abelian varieties" by Serre and Tate.
At a certain point (in the proof of Lemma 2) they claim that a ring is henselian, and I don't see why. I will introduce the notation, so that I can specify my question.
Let $K$ be a field, $v$ a discrete valuation of $K$, $K_{s}$ a seperable closure of $K$ and $\bar{v}$ an extension of $v$ to $K_{s}$. Let $I$ and $D$ denote the inertia group and the decomposition group of $\bar{v}$.
Let $L$ be the fixed field of the inertia group $I$, and $O_{L}$ the ring of $\bar{v}$-integers in $L$.
As far as I can see, no other assumptions are made.
Why is the ring $O_{L}$ henselian?
If I am not mistaken $L$ is the maximal unramified extension of $K$. I have searched Serre's "Local fields" for reasons why $O_{L}$ might be complete (hence henselian) but I could not find them.
Does anyone know a reference for this question? Or a direct answer? (Thanks in advance.)
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1
Even if $K=\mathbf{Q}_p$, the field $L$ is not complete. When learning local fields, it is a good exercise to prove that an algebraic extension of $\mathbf{Q}_p$ is complete if and only if it is finite. – François Brunault Jun 26 at 15:50
Ok. I knew the if part, but not the only if. Thanks. – Johan Commelin Jun 28 at 17:28
## 2 Answers
$O_L$ is not complete. The completion is usually uncountable, but if $K$ is countable then $K_s$ is countable.
I think the easiest way is just to prove it. Let $f$ be an irreducible polynomial with a simple root mod $\bar{v}$. Then the derivative of $f$ is nonzero mod $\bar{v}$, so it's nonzero, so $f$ is separable, so its roots are in $K_S$. Every root that doesn't disappear mod $\bar{v}$ is a $\bar{v}$-integer. Look at the Galois action on those roots. The inertia group preserves each root's residue mod $\bar{v}$, so it fixes that root, so that root lies in $L$.
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Thanks. Apperently the straightforward approach was the way I should have taken. I had not expected that. But then, I don't have very much experience with henselian rings (-; – Johan Commelin Jun 25 at 19:31
@Will: you mean "if $K$ is countable then $K_s$ is countable"! – Laurent Moret-Bailly Jun 26 at 6:28
@Will: Actually I have onee question about your last sentence of the proof. I agree that the inertia group preserves residues of the roots. But how does it follow that therefore the roots are fixed? – Johan Commelin Jun 26 at 13:45
Since the root under consideration is a simple root modulo $\bar{v}$ by assumption. Its residue doesn't change, and it he has to remain a root of the polynomial (since the polynomial is defined over $L$) so the root itself must be fixed. – Will Sawin Jun 26 at 17:43
Ah. Because the polynomial is monic, the degree of the reduction does not go down. So there is a "kind of" 1-1 correspondence between the roots (if we take multiplicities into account). And thus roots reducing to simple roots are fixed by I. Got it. – Johan Commelin Jun 28 at 17:30
### 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 fact, $O_L$ is the strict henselianization of $O_K$ (with residue field a fixed algebraic closure of the residue field of $K$). It is not complete, but it is Henselian, and it is "the minimum" of all the Henselian rings containing $O_K$ and with residue field algebraically closed.
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Thanks for your answer. I read about henselinizations and strict henselinizations. One question remains: How do you prove your claim? Probably by the same straightforward approach as above, right? (I did not try that much yet, on proving your claim, I should add.) – Johan Commelin Jun 25 at 19:33
1
The straightforward approach should work. The idea is that if $\alpha \in O_L$ is a root of an irreducible polynomial $f$, then $\bar{v}$ restricted to $K(\alpha)$ is unramified, so $\alpha$ modulo $\bar{v}$ is a simple root of $f$ modulo $\bar{v}$ and lies in the algebraic closure of the residue field of $O_k$ mod $v$, so any strictly henselian ring containing $O_K$ has an algebraically closed residue field, so $\alpha$ mod $\bar{v}$ is a simple root of $f$, and is Henselian, so that extends to a root, so it contains $\alpha$. It contains every element of $O_L$, so it contains $L$. – Will Sawin Jun 25 at 19:55
Will Sawin explained exactly the proof I had in mind. Thanks! – Xarles Jun 25 at 20:32
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http://math.stackexchange.com/questions/134838/finding-a-vector-in-a-n-l-s?answertab=votes
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# Finding a vector in a n.l.s.
Let $X$ be a normed linear space and $Y$ a closed proper subspace. Prove that for all $\varepsilon > 0$, there is an $x \in X$ with $\|x\| = 1$ and such that $\|x − y\| ≥ 1 − \varepsilon$ for all $y \in Y$ .
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## 1 Answer
This is a basic result in functional analysis sometimes called Riesz's Lemma. You should be able to find the statement and proof in most functional analysis texts. For instance this is the Lemma on page two in Joseph Diestel's Sequences and Series in Banach Spaces. A Google search reveals a standard (it seems) proof of this result on page 10 here.
If you just want a (big) hint:
Assume $\epsilon\in(0,1)$ (the result is trivial for $\epsilon\ge1$). Pick any $x$ in $X\setminus Y$. Then the distance $d$ from $x$ to $Y$ is positive (we use the hypothesis that $Y$ is closed here). Thus, you may choose a $z\in Y$ with $0\lt\Vert x-z\Vert <{d\over 1-\epsilon}$. Consider the vector $x-z\over\Vert x-z\Vert$.
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http://en.wikipedia.org/wiki/Resolution_(logic)
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# Resolution (logic)
In mathematical logic and automated theorem proving, resolution is a rule of inference leading to a refutation theorem-proving technique for sentences in propositional logic and first-order logic. In other words, iteratively applying the resolution rule in a suitable way allows for telling whether a propositional formula is satisfiable and for proving that a first-order formula is unsatisfiable; this method may prove the satisfiability of a first-order satisfiable formula, but not always, as it is the case for all methods for first-order logic (see Gödel's incompleteness theorems and Halting problem). Resolution was introduced by John Alan Robinson in 1965.
The clause produced by a resolution rule is sometimes called a resolvent.
## Resolution in propositional logic
### Resolution rule
The resolution rule in propositional logic is a single valid inference rule that produces a new clause implied by two clauses containing complementary literals. A literal is a propositional variable or the negation of a propositional variable. Two literals are said to be complements if one is the negation of the other (in the following, $a_i$ is taken to be the complement to $b_j$). The resulting clause contains all the literals that do not have complements. Formally:
$\frac{ a_1 \lor \ldots \vee a_i \vee \ldots \lor a_n, \quad b_1 \lor \ldots \vee b_j \vee \ldots \lor b_m} {a_1 \lor \ldots \lor a_{i-1} \lor a_{i+1} \lor \ldots \lor a_n \lor b_1 \lor \ldots \lor b_{j-1} \lor b_{j+1} \lor \ldots \lor b_m}$
where
all $a$s and $b$s are literals,
$a_i$ is the complement to $b_j$, and
the dividing line stands for entails
The clause produced by the resolution rule is called the resolvent of the two input clauses.
When the two clauses contain more than one pair of complementary literals, the resolution rule can be applied (independently) for each such pair; however, the result is always a tautology.
Modus ponens can be seen as a special case of resolution of a one-literal clause and a two-literal clause.
### A resolution technique
When coupled with a complete search algorithm, the resolution rule yields a sound and complete algorithm for deciding the satisfiability of a propositional formula, and, by extension, the validity of a sentence under a set of axioms.
This resolution technique uses proof by contradiction and is based on the fact that any sentence in propositional logic can be transformed into an equivalent sentence in conjunctive normal form.[citation needed] The steps are as follows.
• All sentences in the knowledge base and the negation of the sentence to be proved (the conjecture) are conjunctively connected.
• The resulting sentence is transformed into a conjunctive normal form with the conjuncts viewed as elements in a set, S, of clauses.
• For example $(A_1 \lor A_2) \land (B_1 \lor B_2 \lor B_3) \land (C_1)$ gives rise to the set $S=\{A_1 \lor A_2, B_1 \lor B_2 \lor B_3, C_1\}$.
• The resolution rule is applied to all possible pairs of clauses that contain complementary literals. After each application of the resolution rule, the resulting sentence is simplified by removing repeated literals. If the sentence contains complementary literals, it is discarded (as a tautology). If not, and if it is not yet present in the clause set S, it is added to S, and is considered for further resolution inferences.
• If after applying a resolution rule the empty clause is derived, the original formula is unsatisfiable (or contradictory), and hence it can be concluded that the initial conjecture follows from the axioms.
• If, on the other hand, the empty clause cannot be derived, and the resolution rule cannot be applied to derive any more new clauses, the conjecture is not a theorem of the original knowledge base.
One instance of this algorithm is the original Davis–Putnam algorithm that was later refined into the DPLL algorithm that removed the need for explicit representation of the resolvents.
This description of the resolution technique uses a set S as the underlying data-structure to represent resolution derivations. Lists, Trees and Directed Acyclic Graphs are other possible and common alternatives. Tree representations are more faithful to the fact that the resolution rule is binary. Together with a sequent notation for clauses, a tree representation also makes it clear to see how the resolution rule is related to a special case of the cut-rule, restricted to atomic cut-formulas. However, tree representations are not as compact as set or list representations, because they explicitly show redundant subderivations of clauses that are used more than once in the derivation of the empty clause. Graph representations can be as compact in the number of clauses as list representations and they also store structural information regarding which clauses were resolved to derive each resolvent.
## A simple example
$\frac{a \vee b, \quad \neg a \vee c} {b \vee c}$
In plain language: Suppose $a$ is false. In order for the premise $a \vee b$ to be true, $b$ must be true. Alternatively, suppose $a$ is true. In order for the premise $\neg a \vee c$ to be true, $c$ must be true. Therefore regardless of falsehood or veracity of $a$, if both premises hold, then the conclusion $b \vee c$ is true.
## Resolution in first order logic
In first order logic, resolution condenses the traditional syllogisms of logical inference down to a single rule.
To understand how resolution works, consider the following example syllogism of term logic:
All Greeks are Europeans.
Homer is a Greek.
Therefore, Homer is a European.
Or, more generally:
$\forall x. P(x) \Rightarrow Q(x)$
$P(a)$
Therefore, $Q(a)$
To recast the reasoning using the resolution technique, first the clauses must be converted to conjunctive normal form. In this form, all quantification becomes implicit: universal quantifiers on variables (X, Y, …) are simply omitted as understood, while existentially-quantified variables are replaced by Skolem functions.
$\neg P(x) \vee Q(x)$
$P(a)$
Therefore, $Q(a)$
So the question is, how does the resolution technique derive the last clause from the first two? The rule is simple:
• Find two clauses containing the same predicate, where it is negated in one clause but not in the other.
• Perform a unification on the two predicates. (If the unification fails, you made a bad choice of predicates. Go back to the previous step and try again.)
• If any unbound variables which were bound in the unified predicates also occur in other predicates in the two clauses, replace them with their bound values (terms) there as well.
• Discard the unified predicates, and combine the remaining ones from the two clauses into a new clause, also joined by the "∨" operator.
To apply this rule to the above example, we find the predicate P occurs in negated form
¬P(X)
in the first clause, and in non-negated form
P(a)
in the second clause. X is an unbound variable, while a is a bound value (term). Unifying the two produces the substitution
X ↦ a
Discarding the unified predicates, and applying this substitution to the remaining predicates (just Q(X), in this case), produces the conclusion:
Q(a)
For another example, consider the syllogistic form
All Cretans are islanders.
All islanders are liars.
Therefore all Cretans are liars.
Or more generally,
∀X P(X) → Q(X)
∀X Q(X) → R(X)
Therefore, ∀X P(X) → R(X)
In CNF, the antecedents become:
¬P(X) ∨ Q(X)
¬Q(Y) ∨ R(Y)
(Note that the variable in the second clause was renamed to make it clear that variables in different clauses are distinct.)
Now, unifying Q(X) in the first clause with ¬Q(Y) in the second clause means that X and Y become the same variable anyway. Substituting this into the remaining clauses and combining them gives the conclusion:
¬P(X) ∨ R(X)
The resolution rule, as defined by Robinson, also incorporated factoring, which unifies two literals in the same clause, before or during the application of resolution as defined above. The resulting inference rule is refutation complete[citation needed], in that a set of clauses is unsatisfiable if and only if there exists a derivation of the empty clause using resolution alone.
## References
• Robinson, J. Alan (1965). "A Machine-Oriented Logic Based on the Resolution Principle". 12 (1): 23–41.
• Leitsch, Alexander (1997). The Resolution Calculus. Springer-Verlag.
• Gallier, Jean H. (1986). Logic for Computer Science: Foundations of Automatic Theorem Proving. Harper & Row Publishers.
• Lee, Chin-Liang Chang, Richard Char-Tung (1987). Symbolic logic and mechanical theorem proving ([Nachdr.] ed.). San Diego: Academic Press. ISBN 0-12-170350-9.
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|
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http://physics.stackexchange.com/questions/tagged/harmonic-oscillator
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# Tagged Questions
The term "harmonic oscillator" is used to describe any system with a "linear" restoring force that tends to return the system to a equilibrium state. There is both a classical harmonic oscillator and a quantum harmonic oscillator. Both are used to as toy problems that describe many physical systems.
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### Why is the damping force on a spring oscillator linearly dependent on velocity?
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### Frequency with the spring scale [closed]
Grocery stores often have spring scales in their produce department to weigh fruits and vegetables. The pan of one particular scale has a mass of $0.5 kg$, and when you place a $0.5 kg$ sack of ...
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### Does spatial coupling prohibit resonances due to an external source field?
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While reading up on statistical physics, I am going through the calculation of the partition function of the harmonic oscillator in the microcanonical ensemble. The result for the partition function ...
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### Quantum harmonic oscillator. Finding operators
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### Hyperbolic, parabolic, elliptical PDE related to under-, critical- and overdamped in harmonic osciallation
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### Coordinate representation of quantum ladder operator?
I can't seem to figure out how to derive the coordinate representation of the $a_+$ ladder operator in quantum mechanics. I know that $a_-$ is $\sqrt{\frac{1}{2mwh}} (mwx + i\dot{p})$ in which where ...
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### FWHM in resonance amplitude square derivation
Consider a linear harmonic oscillator subject to a periodic force: $$\ddot x + 2 \beta \dot x + \omega _0 ^2x = f_0\cos \omega t$$ The solution tends to: $$A \cos (\omega t - \delta)$$ where: ...
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### Metronome synchronisation applied to swings
The movement of several metronomes can be synchronised when a movable floor is utilised which couples the movement of the different metronomes. Is it possible to apply this sort of synchronisation to ...
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### Can someone please derive $T=2\pi\sqrt{l/g}$ or prove it without using calculus?
I don't know much calculus, but I want to know that how one derives the formula to find time period $T$ of a simple pendulum.
2answers
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### Why is simple harmonic motion called so?
Is the motion of a simple pendulum, a simple harmonic motion? It stops vibrating after sometime.
1answer
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### Sitting on the bob of a pendulum
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### From the local Hooke's law to the global one
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### Constant magnetic field applied to a quantum harmonic oscillator
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4answers
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### Why is linear independence of harmonic oscillator solutions important?
The equation of motion for the harmonic oscillator (mass on spring model) $$\frac{d^2x}{dt^2} + \omega_0^2 x = 0$$ with $\omega_0^2 = D/m$, $D$ and $m$ being the force constant of the spring and the ...
1answer
135 views
### Shift operator (integral calculus involving Hermite polynomials) [closed]
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### Spring with mass [closed]
A block of mass $M$ is attached to a spring that has mass $m$ and the force constant $k$. The block is placed on a horizontal frictionless surface. Find the period of small-amplitude oscillations ...
0answers
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### Mass spring system, increase mass [closed]
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3answers
469 views
### Partition function for quantum harmonic oscillator
Hi guys I'm currently trying to solve a mock exam for an exam in a few days and am a bit confused by the solutions they gave us for this exercise: Exercise: A solid is composed of N atoms which ...
1answer
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### Why don't tuning forks have three prongs?
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3answers
758 views
### How to derive the period of spring pendulum?
So I wanted to find out how to (simply, if that's possible) derive the formula for a period of spring pendulum: $T=2\pi \sqrt{\frac{m}{k}}$. However, Google doesn't help me here as all I see is the ...
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http://mathhelpforum.com/differential-geometry/207056-function-distance-point-outside-sphere-any-point-its-surface-print.html
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# function for distance from a point outside a sphere to any point on its surface
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• November 8th 2012, 10:28 AM
mosibat
1 Attachment(s)
function for distance from a point outside a sphere to any point on its surface
Hi
I have a point m outside a sphere. The sphere center is o and r is the radius of sphere. Distance from point m to o is l.
If we draw a line from m to any point on the surface of sphere, this line has a length. Minimum length is l - r and maximum length would be l + r.
I want the function for distance from m to any point on the surface of sphere. Also how to draw the graph of this function (all possible distances).
http://mathhelpforum.com/attachment....1&d=1352402766
• November 8th 2012, 10:42 AM
HallsofIvy
Re: function for distance from a point outside a sphere to any point on its surface
We can set up a coordinate system in which the sphere has center at the origin and the given point, m, is on the z-axis. The equation of the sphere in that coordinate system is $x^2+ y^2+ z^2= r^2$ and m is (0, 0, l). Given a point (x, y, z) on the sphere, above the xy-plane, then $z= \sqrt{r^2- x^2- y^2}$ then the distance from m to that point is $\sqrt{x^2+ y^2+ (l- \sqrt{r^2- x^2- y^2})^2}$. If (x, y, z) is below the xy-plane, $z= -\sqrt{r^2- x^2 y^2}$ then the distance is $\sqrt{x^2+ y^2+ (l+ \sqrt{r^2- x^2- y^2})^2}$.
• November 8th 2012, 12:40 PM
mosibat
Re: function for distance from a point outside a sphere to any point on its surface
Thanks. That was exactly what I was looking for.
All times are GMT -8. The time now is 02:21 PM.
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http://mathoverflow.net/questions/32538/finite-axiom-of-choice-how-do-you-prove-it-from-just-zf/32545
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## Finite axiom of choice: how do you prove it from just ZF?
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The axiom of choice asserts the existence of a choice function for any family of sets F. Suppose, however, that F is finite, or even that F just has one set. Then how do we prove the existence of a choice function?
The usual answer is that we just go from set to set, picking an element from each set. Since F is finite, this process will terminate. What I'm really wondering is how we can always choose from a single set. The informal answer seems to be just that it's possible... but this isn't an axiom, so it must be justified some other way.
So: how do you prove from the axioms of just ZF without choice, that for any nonempty x there exists a function f:{x}->x?
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By definition of "nonempty", x contains an element y, so let $f$ be given by the set of ordered pairs {({x}, y)}. – Charles Staats Jul 19 2010 at 21:29
Sorry, that should be {(x, y)}. – Charles Staats Jul 19 2010 at 21:33
Well, saying the product is non-empty is equivalent to the axiom of choice. I assume the OP means, how do we show this claim from the axioms? – Daniel Litt Jul 19 2010 at 21:37
The product is, by definition, the set of all ordered pairs. In this case, we can explicitly write down an ordered pair, so clearly the product is nonempty. But I'm not entirely sure I understand your comment, so my apologies if this comment does not address it. – Charles Staats Jul 19 2010 at 21:41
Ah sorry; I think the OP's question boils down to: how, from the axioms of ZF, do we "explicitly write down an ordered pair"? Since if we can write down ordered pairs, we're obviously done. – Daniel Litt Jul 19 2010 at 21:48
show 1 more comment
## 4 Answers
Although the answers already given are correct, let me add some information (essentially just rephrasing the bracketed part of Thomas Scanlon's answer) that I've found useful for students who raised this question. Consider the problem, at the end of the original question, of "choosing" from a single set $x$. As several people have pointed out, we are given the existential statement, "There is an element in $x$." What should be noticed in addition is that what we want to prove is also an existential statement, "There is a choice function." We have an explicit construction, which I'll call $C$, that will convert any element of $x$ into a choice function, namely sending any $y$ to `$\{(x,y)\}$` (as in Charles Staats's comment on the original question). If we can't explicitly define any particular $y$, then we won't be able to define any particular choice function either, but the problem doesn't require us to explicitly define a choice function; we need only prove that one exists. And that follows, thanks to $C$, from the existence of elements in $x$.
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I was in the process of writing this myself, I'll just add one more thing to clarify: As Andreas said, when dealing with a finite number of sets you can explicitly write the function in the form of a formula. However, in the infinite case you cannot write this sort of formula (it would require infinitely many quantifiers to mention which element is chosen) and therefore you're in need of a stronger tool that would guarantee the existence of such function. – Asaf Karagila Jul 19 2010 at 23:47
Thank you, this is very helpful. As I indicated in my comments to Thomas Scanlon's answer, my confusion arose because I was looking for an "explicit choice function". – unknown (google) Jul 20 2010 at 0:03
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@Asaf Karagila: You can only write that formula in the case when the product is externally finite. As the other answers indicate, it is enough that the product is internally finite, in which case an induction argument shows the choice function exists. Moreover, the standard reading of "a finite number of sets" is the internal one. Andreas Blass's comment is safe because he only refers to choosing from a single set. – Carl Mummert Jul 20 2010 at 0:04
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
To say that $x \neq \varnothing$ is exactly to assert $(\exists y) y \in x$ and the truth of this existential formula will be witnessed by some set $a$ for which $a \in x$. Thus, the set $f := \lbrace \langle x, a \rangle \rbrace$ is a function from $\lbrace x \rbrace$ to $x$.
[I think that this point, namely that the Axiom of Choice is not used to instantiate the truth of an existential formula by naming a witness is the heart of your question. I am not claiming that there will be a constructive choice of the witness. It is just that if an existential formula is true, there must be a witness.]
More generally, to prove that if $F$ is a finite set of nonempty sets, then $\prod F := \lbrace f : f \text{ a function with dom}(f) = F \text{ such that } (\forall x \in F) f(x) \in x \rbrace$ is non-empty one argues by induction.
If $F = \varnothing$, then the empty function is an element of $\prod F$. If $\text{card}(F) = n+1$, then we may express $F = F' \cup \lbrace x \rbrace$ as disjoint union where $\text{card}(F') = n$. As $x \neq \varnothing$, from the calculation above we find some $f:\lbrace x \rbrace \to x$. By induction, there is some $g \in \prod F'$. I will leave it to you to exhibit a bijection between $\prod F$ and $(\prod F') \times \prod \lbrace x \rbrace$ and thereby complete the argument.
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I think your first point is the main one for the question. It's alluding to the "existential elimination" inference rule, which says that from the assumption $\exists x A(x)$, you can assert $A(c)$ for some new constant $c$. The soundness of the rule is a direct consequence of the way semantics for first-order logic are defined. I have previously encountered confusion from non-logicians about this issue. – Carl Mummert Jul 19 2010 at 22:45
@Thomas Scanlon Thanks. Is it true, then, that there isn't a constructive way of choosing the element in general? I think my confusion arose because sometimes people motivate the axiom of choice by saying that for infinite families, there's no "definite procedure" or "explicit method" of choosing your function, so we need to create an axiom. (E.g. en.wikipedia.org/wiki/Axiom_of_choice#Usage) But it seems that even in the finite case, there's no way to explicitly build your function, in one sense of the word "explicit". – unknown (google) Jul 20 2010 at 0:01
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If you have constructive verification that the set is not empty, you can constructively convert this to a choice function. For example, perhaps you have a constructive proof that the set is nonempty. But if you have no constructive information about the set, of course there's no way to explicitly construct a member of the set, regardless of the axiom of choice. – Carl Mummert Jul 20 2010 at 0:13
@Carl: This is true, but it applies to infinite choice just the same (that’s why the axiom of choice is sound under variants of recursive realizability for various constructive theories). – Emil Jeřábek Jan 23 at 11:30
There are two finite choice theorems, the internal one and the external one, both are true in ZF.
As Charles Staats pointed out, the external version is a tautology (modulo some finite combinatorics): if $a_1,\dots,a_n$ are all nonempty, then there are $z_1 \in a_1$,...,$z_n \in a_n$ and then $\lbrace (a_1,z_1),\ldots,(a_n,z_n)\rbrace$ is the desired choice function for the family $X = \lbrace a_1,\dots,a_n \rbrace$ of nonempty sets.
The internal version "every finite family of nonempty sets has a choice function" is stronger since a model of ZF may have nonstandard finite cardinals. The proof in this case is by induction on the cardinality of the family.
The empty family has a trivial choice function — the empty function. Suppose we know the theorem to be true for families of size $n$. Let $X$ be a family of nonempty sets with size $n+1$. Let $g:n+1\to X$ be a bijection. Let $X' = g[n]$ and $a = g(n)$. Then $X'$ is a family of nonempty sets of size $n$, which therefore has a choice function $f':X' \to \bigcup X'$. Since $a$ is nonempty, we can find $z \in a$ and hence $f = f' \cup \lbrace (a,z) \rbrace$ is a choice function for the original family $X$.
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The axiom of pairing is not a tautology... – Emil Jeřábek Jan 23 at 11:26
Right! That was poorly formulated... – François G. Dorais♦ Jan 23 at 14:01
The Axiom of Choice does not allow to "choose" a choice function, it only says that a choice function exists. To show that a choice function for a single nonempty set exists, you do not need to "choose" an element in the set, it is enough to show that at least one element exists (i.e. the set is nonempty). Every element in the set will give a different choice function.
What did you mean by "choosing" an element from a set? ;)
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http://mathhelpforum.com/calculators/82873-implicit-differentiation-ti-nspire-cas.html
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# Thread:
1. ## Implicit Differentiation on TI-Nspire CAS
Hi there, I'm trying to do the following implicit differentiation on my TI-Nspire CAS, and I can't figure it out.
$v = \frac{1}{3} \pi r^2 h$
I want to get: $\frac{dv}{dt} = \frac{2}{3} \pi r \frac{dr}{dt} \frac{dh}{dt}$
Is there any way to get the Nspire to produce that sort of output?
I know that ImpDif of $x^2 + y^2 = 36$ can be implicitly derived by the Nspire to yield $\frac{-x}{y}$ but how about cases where a function is being derived with respect to a variable not found in the function itself (like $t$ as in the case above)? (if this makes sense)
Warm regards,
Abe
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http://programmingpraxis.com/2010/07/20/solving-systems-of-linear-equations/?like=1&_wpnonce=ef2a6945bd
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# Programming Praxis
A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer
## Solving Systems Of Linear Equations
### July 20, 2010
In today’s exercise we continue the examination of matrix operations that we began previously. Our goal is to be able to solve a system of equations; along the way we will see two methods for decomposing matrices.
We begin with some terminology. All the matrices we will be looking at are square, meaning that they have the same number of rows and columns. A lower-triangular matrix L has all entries Lij = 0 for i < j; thus, all entries above the main northwest-to-southeast diagonal are zero. An upper-triangular matrix U has all entries Uij = 0 for i > j; thus, all entries below the main northwest-to-southeast diagonal are zero. A lower- or upper-triangular matrix is unit lower- or upper-triangular if all the entries along the main diagonal are 1. A permutation matrix P has exactly one 1 in each row and column and 0 elsewhere; it is called a permutation matrix because multiplying a vector X by a permutation matrix has the effect of permuting the elements of X. The identity matrix I has 1 in each entry along the main diagonal and 0 elsewhere. A matrix M is singular if it has no inverse, that is, there is no matrix M-1 such that M M-1 = I.
An LU decomposition of a matrix A finds two matrices L, which is unit lower-triangular, and U, which is upper-triangular, such that A = L U. The algorithm is called Gaussian elimination, and works from top to bottom. First, multiples of the first equation are subtracted from the other equations so that the first variable is removed from those equations. Then multiples of the second equation are subtracted from the remaining equations so that the second variable is removed from those equations. Then the third equation, and the fourth, and so on, until all the equations have been processed and the matrix is in upper-triangular form. Here is an example of the LU decomposition of matrix A into its factors L × U:
$\begin{pmatrix} 2 & 3 & 1 & 5 \\ 6 & 13 & 5 & 19 \\ 2 & 19 & 10 & 23 \\ 4 & 10 & 11 & 31 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 3 & 1 & 0 & 0 \\ 1 & 4 & 1 & 0 \\ 2 & 1 & 7 & 1 \end{pmatrix} \times \begin{pmatrix} 2 & 3 & 1 & 5 \\ 0 & 4 & 2 & 4 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 3 \end{pmatrix}$
There are two problems with LU decomposition: First, the algorithm leads to a divide-by-zero error on singular matrices. Second, it is prone to numerical instability for small divisors. The solution is to rearrange, or permute, the equations so that the pivot element is always the largest remaining element, greatly reducing the likelihood of numerical instability.
An improved decomposition is the LUP decomposition, which finds for an input matrix A three matrices L, U, and a permutation matrix P such that P A = L U. Rather than actually moving equations, the permutation matrix records the rearrangements. For example, here is the LUP decomposition of the matrix A given by P × A = L × U:
$\begin{pmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{pmatrix} \times \begin{pmatrix} 2 & 0 & 2 & 3/5 \\ 3 & 3 & 4 & -2 \\ 5 & 5 & 4 & 2 \\ -1 & -2 & 17/5 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 2/5 & 1 & 0 & 0 \\ -1/5 & 1/2 & 1 & 0 \\ 3/5 & 0 & 2/5 & 1 \end{pmatrix} \times \begin{pmatrix} 5 & 5 & 4 & 2 \\ 0 & -2 & 2/5 & -1/5 \\ 0 & 0 & 4 & -1/2 \\ 0 & 0 & 0 & -3 \end{pmatrix}$
Given the LUP decomposition, it is simple to solve a system of linear equations. Forward substitution solves the lower-triangular system by calculating the first variable, which is part of an equation with one unknown, then substitutes that into the second equation, reducing it from two unknowns to one unknown, and so on. Then back substitution runs backward, calculating the final values of the variables in the original matrix. Here’s an example, where we wish to solve for the vector X given A X = B:
$\begin{pmatrix} 1 & 2 & 0 \\ 3 & 5 & 4 \\ 5 & 6 & 3 \end{pmatrix} X = \begin{pmatrix} 1/10 \\ 25/2 \\ 103/10 \end{pmatrix}$
The LUP decomposition P A = L U is
$\begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \times \begin{pmatrix} 1 & 2 & 0 \\ 3 & 5 & 4 \\ 5 & 6 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 3/5 & 1 & 0 \\ 1/5 & 4/7 & 1 \end{pmatrix} \times \begin{pmatrix} 5 & 6 & 3 \\ 0 & 7/5 & 11/5 \\ 0 & 0 & -13/7 \end{pmatrix}$,
the result of forward substitution L Y = P B is
$\begin{pmatrix} 1 & 0 & 0 \\ 3/5 & 1 & 0 \\ 1/5 & 4/7 & 1 \end{pmatrix} \times Y = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \times \begin{pmatrix} 1/10 \\ 25/2 \\ 103/10 \end{pmatrix} = \begin{pmatrix} 103/10 \\ 25/2 \\ 1/10 \end{pmatrix}$, giving $Y = \begin{pmatrix} 103/10 \\ 158/25 \\ -39/7 \end{pmatrix}$,
and the result of the back substitution U X = Y is
$\begin{pmatrix} 5 & 6 & 3 \\ 0 & 7/5 & 11/5 \\ 0 & 0 & -13/7 \end{pmatrix} \times X = \begin{pmatrix} 103/10 \\ 158/25 \\ -39/7 \end{pmatrix}$, giving $X = \begin{pmatrix} 1/2 \\ -1/5 \\ 3 \end{pmatrix}$,
which is the solution.
Your task is to write functions that perform LU-decomposition and LUP-decomposition and solve systems of linear equations. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
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Posted by programmingpraxis
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### 4 Responses to “Solving Systems Of Linear Equations”
1. Remco Niemeijer said
July 20, 2010 at 10:45 PM
Since I’m currently at a conference I don’t have as much time to devote to this as I’d like, but here’s my implementation of LU decomposition:
```lu :: Fractional a => [[a]] -> ([[a]], [[a]])
lu = unzip . map unzip . elim where
elim [] = []
elim ~((r:rs):xs) = zip (1 : repeat 0) (r:rs) :
zipWith (:) (map (\(y:_) -> (y / r, 0)) xs)
(elim $ map sub xs) where
sub ~(y:ys) = zipWith (-) ys $ map (y / r *) rs
```
If I have some time before Friday I’ll see if I can work on the rest.
Also, you might want to fix the examples in the exercise, since they contain some errors (e.g. in the first example: 1×3 + 3×4 + 1×0 + 0×0 = 15, not 19). The versions used in the solution code are correct.
2. programmingpraxis said
July 21, 2010 at 1:18 PM
Fixed four errors. I didn’t do that very well, did I?
3. July 21, 2010 at 8:25 PM
[...] Praxis – Solving Systems Of Linear Equations By Remco Niemeijer In yesterday’s Programming Praxis exercise our task is to implement some more matrix-related functions, [...]
4. Remco Niemeijer said
July 21, 2010 at 8:26 PM
My Haskell solution (see http://bonsaicode.wordpress.com/2010/07/21/programming-praxis-solving-systems-of-linear-equations/ for a version with comments):
```import Control.Arrow
import Data.List
import qualified Data.List.Key as K
mult :: Num a => [[a]] -> [[a]] -> [[a]]
mult a b = [map (sum . zipWith (*) r) $ transpose b | r <- a]
elim :: Fractional a => [a] -> [a] -> [a]
elim ~(x:xs) ~(y:ys) = zipWith (-) ys $ map (y / x *) xs
identity :: Num a => [[a]] -> [[a]]
identity = zipWith (zipWith const) (iterate (0:) (1 : repeat 0))
lu :: Fractional a => [[a]] -> ([[a]], [[a]])
lu = unzip . map unzip . f where
f [] = []
f ~(x:xs) = zip (1 : repeat 0) x :
zipWith (:) (map (\(y:_) -> (y / head x, 0)) xs)
(f $ map (elim x) xs)
perm :: (Fractional a, Ord a) => [[a]] -> [[a]]
perm m = f $ zip (identity m) m where
f [] = []
f xs = a : f (map (second $ elim b) $ delete (a,b) xs)
where (a,b) = K.maximum (abs . head . snd) xs
lup :: (Fractional a, Ord a) => [[a]] -> ([[a]], [[a]], [[a]])
lup xs = (perm xs, l, u) where (l,u) = lu $ mult (perm xs) xs
lupsolve :: (Fractional a, Ord a) => [[a]] -> [a] -> [a]
lupsolve a b = f y u where
(p,l,u) = lup a
y = foldl (\x (l', pb') -> x ++ [pb' - sum (zipWith (*) x l')])
[] (zip l (concat . mult p $ map return b))
f _ [] = []
f ~(y':ys) ~((r:rs):us) = (y' - sum (zipWith (*) rs z)) / r : z
where z = (f ys $ map tail us)
```
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# Tagged Questions
1answer
71 views
### Definite integral including the ratio and power functions of a single variable
I find trouble in calculating the following integral: $$\int_0^R \frac{m\cdot x}{m+s\cdot x^a} \,dx$$ Mathematica does not provide an output for this function, however, there seems to be an output ...
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### Why does the power rule work?
If $$f(x)=x^u$$ then the derivative function will always be $$f'(x)=u*x^{u-1}$$ I've been trying to figure out why that makes sense and I can't quite get there. I know it can be proven with limits, ...
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### Proof for power functions
Which is greater? $\sqrt{n}^{\sqrt{n+1}}$ or $\sqrt{n+1}^\sqrt{n}$ I know that $\sqrt{n}^{\sqrt{n+1}}$ is greater but I tried using induction and I couldn't figure it out. Thanks for the help.
1answer
63 views
### Summation of powers inequality
Can anyone provide a slick proof of the following? Let $0 < x \le 1$. Then $\displaystyle \sum_{k=0}^{n-1} x^k \ge \frac {1} {1 - (1 - 1/n)x}$.
1answer
294 views
### Comparing Powers with Different Bases Using Logarithms?
I looked all over to see if a question like this had already been answered, but I couldn't find it. So here goes: I need a general formula for comparing two (insanely huge) powers. I'm pretty sure ...
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### Power function inequality
Let $x$ and $p$ be real numbers with $x \ge 1$ and $p \ge 2$ . Show that $(x - 1)(x + 1)^{p - 1} \ge x^p - 1$ . I recently discovered this result. I am sure it is known, but it is new to me. It is ...
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### $2^x - a$ touches $\log_2(x)$
I was playing around with the functions $2^x$ and $\log_2(x)$. As they are the inversions of each other, I thought there was a simple number $a$ for which $2^x - a$ touches $\log_2(x)$. Using ...
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http://math.stackexchange.com/questions/4435/finding-alpha-such-that-f-alphaxy-fxfy?answertab=votes
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# Finding $\alpha$ such that $f(\alpha(x+y))=f(x)+f(y)$
Problem taken from the link: http://web.mit.edu/rwbarton/Public/func-eq.pdf I am stating the question here
For which $\alpha$ does there exists a nonconstant function $f: \mathbb{R} \to \mathbb{R}$ such that $f(\alpha(x+y))=f(x)+f(y)$ for all $x,y \in \mathbb{R}$.
Clearly for $\alpha=1$ we see that this case is satisfied, by taking the identity function. But are there other values of $\alpha$ for which this condition is satisfied.
-
## 1 Answer
$f(0+0)=f(0) + f(0)$ implies $f(0)=0$. Set $y=0$ to get $f(\alpha x)=f(x)$ so that $f(\alpha (x+y))=f(x+y)$ and the function is additive. If $\alpha \neq 1$ this means $f((\alpha - 1)x)=0$ and hence $f=0$.
This is too easy to be on rwbarton's list, and in fact I cannot find this problem in the file that you linked. Which problem number is it in the list?
-
@T: I am sorry! the file which i had downloaded is outdated. Seems, Reid has updated it. The one which i have was dated on June 2005, but the file present now is September 2006. – anonymous Sep 12 '10 at 2:30
Good to know, thanks. – T.. Sep 12 '10 at 5:07
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# Thread:
1. ## Poincare's Theorem
Let H1 & H2 be subgroups of G. How can it be shown that if H1 & H2 have finite index in G then so the intersection of H1 & H2.
2. Originally Posted by thomas_donald
Let H1 & H2 be subgroups of G. How can it be shown that if H1 & H2 have finite index in G then so the intersection of H1 & H2.
Hint (acually the solution!!): $(H_1 \cap H_2)g=H_1g \cap H_2g,$ for all $g \in G.$
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Physics Forums
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## Bosons and Fermions in a rigorous QFT
I'm wondering, is there still a sharp distinction between Bosons and Fermions in a rigorous QFT, if exsits?
My question is motivated by the following, consider one of the equations of motion of QED:
$$\partial_\nu F^{\nu \mu} = e \bar{\psi} \gamma^\mu \psi$$
In our familiar perturbative QED (Here I'm not 100% sure if I use the word "perturbative" correctly. I simply mean fields are quantized as free fields, and we introduce an interaction built from free fields operators, like an iteration method), LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.
I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Science Advisor One of the greater successes of axiomatic QFT was the proof that at least in some quantum field theories bosonic and fermionic behaviour results quite naturally for localized charges. See e.g. the book by R. Haag, Local quantum physics.
Recognitions: Science Advisor The above mentioned field equation cannot be quantized directly b/c it has to be gauge fixed. In A°=0 at least for the time-indep. constraint (the Gauss law) this equaton is implemented on the physical Hilbert space. This does not require any "free field approach".
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## Bosons and Fermions in a rigorous QFT
Quote by kof9595995 [...] LHS is made of Bosonic operators and RHS is made of Fermionic operators, and since the Bosonic sector and Fermionic sector are independent in the total Fock space, perturbative QED fails to satisfy this equation of motion.
The product of two spin-1/2 operators is bosonic (being a superposition of spin-0 and spin-1 in general). (You might want to review Clebsch-Gordan coefficients and associated angular momentum decomposition theory in ordinary QM if you're not already familiar with it.)
[...] then is there still a sharp distinction between Bosons and Fermions?
Yes, in the sense that there's a superselection rule between them. But when you start forming products, things get more complicated.
Recognitions: Science Advisor Perhaps it makes sense to consider a simple qm example; the problem for the two-dim. harmonic oscillator as a toy model would be $$H_i = \frac{1}{2}p_i^2 + \frac{1}{2}x_i^2 = a_i^\dagger a_i + \frac{1}{2}$$ $$H = H_1 + H_2$$ Now we can easily solve equations like $$(H - N)|N\rangle = 0$$ for some eigenvalues N, but structurally the equation "bosonic operator = fermionic operator" would be something like $$(H - N) = 0$$ and this is obviously not allowed as an operator equation b/c $$(H - N) = 0 \;\Rightarrow\; (H - N)|m,n\rangle = 0 \;\forall m,n \;\Rightarrow\; (m+n+1-N) = 0\;\forall m,n$$ Gauge fixing introduces some additional structures like resolving "unphysical bosonic operators" in terms of fermionic operators via Gauss law, but I still don't see how this is sufficient to resolve the problems for the remaining operator equations. Neverthelesse there should be some solution e.g. for QE´D and QCD where these problems have been treated w/o using perturbation theory.
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Quote by kof9595995 I suppose if a rigourous QED exists, this equation of motion should be satisfied, but this in turn means the fermion operator and bosonic operator must act on a Hilbert space they share together, then is there still a sharp distinction between Bosons and Fermions?
Is there anybody who believes that a rigorous QED exists?
Recognitions: Science Advisor I don't believe in a rigorous QED ... ... but this problem seems to be trivial and there must be a solution in "standard textbook QED with canonical quantization using Fockspace". I think the subtlety is the regularization of the operator product on the r.h.s. which requires e.g. point splitting with gauge-field insertion (in order not to destroy gauge invariance); this would introduce a gauge field dependency on the r.h.s. whereas the fermionic contribution on the l.h.s. comes from the solution of the Gauss law constraint i.e. A° expressed in terms of fermionic charge density. That means that quantization, gauge fixing and regularization translates the equation "bosonic operator = fermionic operator" into something like "bosonic + fermionic operator = fermionic + bosonic operator" which has a chance to hold as operator equation in terms of standard Fock space creation and annihilation operators. This is of course no rigorous proof but is indispensable already for standard textbook QED.
Quote by DrDu One of the greater successes of axiomatic QFT was the proof that at least in some quantum field theories bosonic and fermionic behaviour results quite naturally for localized charges. See e.g. the book by R. Haag, Local quantum physics.
That sounds quite a long way to go, is there a layman explanation of it?
Quote by tom.stoer I don't believe in a rigorous QED ...
Then what's the underlying theory that makes perturbative QED plausible?
Quote by tom.stoer Neverthelesse there should be some solution e.g. for QE´D and QCD where these problems have been treated w/o using perturbation theory.
Can you give a concrete example? Better a QED example since I'm not familiar with QCD.
Recognitions: Science Advisor My example is described in #7 We must not mix two issues - perturbative QED - Fock space with creation and annihilation operators The latter does not imply perturbation theory
Quote by tom.stoer My example is described in #7 We must not mix two issues - perturbative QED - Fock space with creation and annihilation operators The latter does not imply perturbation theory
Could you elaborate? I have always thought a Fock space only makes sense for perturbation theory, since I would imagine an interaction should destroy the simple direct product structure of different sectors.
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Quote by kof9595995 That sounds quite a long way to go, is there a layman explanation of it?
It was formulated about 1969 but there are still people working on it. Its an extremely mathematical approach.
See the article by Roberts:
http://kolxo3.tiera.ru/M_Mathematics..._.pdf#page=274
Recognitions: Homework Help Science Advisor A useful perspective on this question is provided by lattice gauge theory. For example, Gauss' law, which is the 0th component of the equation you wrote, is satisfied as an identity on the physical hilbert space. In other words, in the true hilbert space electric field lines can only end where charges are located. Nevertheless, there are many ways to distinguish bosons and fermions. In this model the fermions carry charge while the gauge bosons do not. There are composite operators made of fermions that are bosonic and carry charge, but there is still a fermion number that remains sensible. Of course, this is not to say that there is no blurring of the lines. Bosonization in one dimension is a procedure for exchanging bosons and fermions (and is even relatively rigorous). In higher dimensions one can have solitons built from gauge and bosonic matter degrees of freedom that can carry weird charges and even be fermions.
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Quote by kof9595995 Could you elaborate? I have always thought a Fock space only makes sense for perturbation theory, since I would imagine an interaction should destroy the simple direct product structure of different sectors.
Of course going beyond perturbation theory is still difficult, but neither impossible nor excluded in principle
Think about a Hamiltonian
$$H = \frac{p^2}{2} + V(x)$$
You can rewrite this as
$$H = \frac{p^2}{2} + \frac{x^2}{2} + \left[V(x) - \frac{x^2}{2}\right] = \frac{p^2}{2} + \frac{x^2}{2} + \tilde{v}(x) = a^\dagger a + \frac{1}{2} + \tilde{u}$$
Now it's up to you to solve this problem exactly or to use perturbation theory in u.
I think it's possible to rewrite a QFT in terms of creation and annihilation operators acting on Fock spaces, but I don't see why a perturbative treatment is mandatory.
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Quote by tom.stoer Of course going beyond perturbation theory is still difficult, but neither impossible nor excluded in principle Think about a Hamiltonian ...
I don't consider this problem as too representative for the situation in QFT. The main problem in field theory is that the perturbed and free fields live in different Hilbert spaces -- a problem usually absent in ordinary QM.
Superconductivity can be taken as a toy model on how to solve this problem:
The BCS Hamiltonian can be diagonalized introducing new field operators by the Bogoliubov Valatin transformation. The new field operators are also fermionic but describe the interacting particles.
They cannot be obtained perturbationally from the free electron gas.
Recognitions:
Science Advisor
Quote by DrDu The main problem in field theory is that the perturbed and free fields live in different Hilbert spaces ... ... Superconductivity can be taken as a toy model on how to solve this problem: The BCS Hamiltonian can be diagonalized introducing new field operators by the Bogoliubov Valatin transformation. The new field operators are also fermionic but describe the interacting particles. They cannot be obtained perturbationally from the free electron gas.
I never said that.
What I am saying is that one may start with Fock space operators and then use some non-perturbative techniques. Bogoljubov transformation, bosonization, ... are examples. That is more than just a solution, it's a kind of formal re-definition of the theory.
There is no need to use perturbation theory only b/c of Fock space states, neither before nor after Bogoljubov transformation.
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http://mathoverflow.net/questions/24594/survey-articles-on-homotopy-groups-of-spheres/24598
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## Survey articles on homotopy groups of spheres
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Are there general surveys or introductions to the homotopy groups of spheres? I'm interested especially in connections to low-dimensional geometry and topology.
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Retagged reference request in lieu of big list and soft-question. Seemed more specific, but feel free to undo. – Cam McLeman May 14 2010 at 16:13
Cam, "reference-request" is a more precise tag. Thanks. – Colin Tan May 15 2010 at 4:26
## 3 Answers
While my Algebraic Topology book and my unfinished book on spectral sequences (referred to in other answers to this question) contain some information about homotopy groups of spheres, they don't really qualify as a general survey or introduction. One source that fits this bill more closely is Chapter 1 of Doug Ravenel's "green book" Complex Cobordism and Stable Homotopy Groups of Spheres, from 1986. This introductory chapter starts at a reasonably accessible level, with increasing prerequisites in the later sections of the chapter. More recent surveys ought to exist, although at the moment I can't recall any. With the recent solution of the Kervaire invariant problem by Hill-Hopkins-Ravenel, this would be a good time for an updated survey.
Connections between homotopy groups of spheres and low-dimensional geometry and topology have traditionally been somewhat limited, with the Hopf bundle being the thing that comes most immediately to mind. A fairly recent connection is Soren Galatius' theorem that the homology groups of $Aut(F_n)$, the automorphism group of a free group, are isomorphic in a stable range of dimensions to the homology groups of "loop-infinity S-infinity", the space whose homotopy groups are the stable homotopy groups of spheres.
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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.
You can find something in Allen Hatcher, "Spectral Sequences" .
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Hatcher's "Algebraic Topology".
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http://mathhelpforum.com/calculus/156500-triple-integral.html
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# Thread:
1. ## Triple integral
Find the volume of the ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ by solving the triple integral after making the transformation $x = au, y = bv$ and $z = cw$
I'm just a bit confused here because this does not seem like a Jacobian transformation. I know what the triple integral would be in cartesian terms, ie, $\displaystyle{\int_{-a}^a \int_{-\sqrt{b^2-\frac{b^2x^2}{a^2}}}^{\sqrt{b^2-\frac{b^2x^2}{a^2}}} \int_{-\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}}^ {\sqrt{c^2-c^2\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)}} 1 dzdydx}$
But clearly, we don't want to integrate that, so what does it mean by "after making the transformation $x = au, y = bv$ and $z = cw$"? Do I just simply replace all the x with au and y with bv etc, then what happens to the dz dy and dx?
Thanks heaps!
2. Do I just simply replace all the x with au and y with bv etc
Yes.
then what happens to the dz dy and dx?
You get the same thing you'd normally get in a single integration: dx = a du, dy = b dv, and dz = c dw. Your limits change, your integrand changes, and your differentials change. So what's your new integral, and how do you think you should go about computing it?
3. Ahhh okay, thanks heaps, yeah because I was just evaluating triple integrals with transformations involving the Jacobian so I got confused when I saw this question (I am studying ahead of class and this question was on one of my classes exercise sheets).
I understand it now, thanks Ackbeet with all your help! I really appreciate it!!
4. Perhaps the simplest thing to do is NOT to make the transformation in the integral or, if fact, to do the integral at all. If you let x= au, y= bv, and z= cw, then the equation of the ellipse becomes
$\frac{a^2u^2}{a^2}+ \frac[b^2v^2}{b^2}+ \frac{c^2w^2}{c^2}= u^2+ v^2+ w^2= 1$ which is a sphere, of radius 1, in the uvw-coordinate system. The Jacobian (I'm not sure what you mean by a "Jacobian transformation" but every transformation has a Jacobian) is
$\left|\begin{array}{ccc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w}\end{array}\right|= \left|\begin{array}{ccc}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right|= abc$ so that volume of the ellipse is the volume of the sphere, $\frac{4}{3}\pi 1^3= \frac{4}{3}$, times abc.
(Since you have been "evaluating triple integrals with transformations involving the Jacobian" that's exactly what happens here: dxdydz= abc dudvdw.)
5. You're welcome. Have a good one!
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http://mathoverflow.net/questions/10512/theories-of-noncommutative-geometry/10533
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Theories of Noncommutative Geometry
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
[I have rewritten this post in a way which I hope will remain faithful to the questioner and make it seem more acceptable to the community. I have also voted to reopen it. -- PLC]
There are many ways to approach noncommutative geometry.
What are some of the most important currently known approaches? Who are the principal creators of each of these approaches, and where are they coming from? E.g., what more established mathematical fields are they using as jumping off points? What problems are they trying to solve?
Two examples:
1) The Connes school, with an approach from C$^*$-algebras/mathematical physics.
2) The Kontsevich school, with an approach from algebraic geometry.
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Sorry, I think this question is far too broad (see mathoverflow.net/faq#encyclopedia). For such broad questions, it's better to look at the nLab page (ncatlab.org/nlab/show/…), the wikipedia page (en.wikipedia.org/wiki/Noncommutative_geometry), and to use Google. Though this is a real question, it's not a specific enough question to be a good fit for Math Overflow. – Anton Geraschenko♦ Jan 2 2010 at 23:06
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There is a saying, "there are more noncommutative geometry theories than there are noncommutative geometers". This had the potential to generate a lot of interesting answers. You should have let some more answers come, before hastily closing it. See the answers and then decide whether this was a real question or not. Let noncommutative geometers answer. This is the hottest question in that subject, I would say. – Anweshi Jan 3 2010 at 2:40
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So please reopen it for 2-3 days and see what happens. I suppose you have nothing much to lose by that. You can always close it later. Is it not fair enough? In an hour or so, already 4 positive votes came for the question. – Anweshi Jan 3 2010 at 2:42
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I tend to think that the current division of the two approaches is not correct. In addition to Kontsevich being associated to at least two approaches to NC Geom(His formally smooth spaces work with Soibelman, and his work with A. Rosenberg), but there are other perspectives in the algebraic geometry realm beyond the things Kontsevich works on. In the summary article I posted, he outlines at least 5, only one of which has anything to do with Kontsevich. While I agree that it is convenient to delineate fields into those worked on by Fields Medalists, I think this is probably a mistake. – B. Bischof Feb 6 2010 at 21:46
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I agree with the comment of B. Bischof, especially the last sentence (No one talks of NCG a la Higson -Moscovici - Nistor, it seems;) and for a more rounded picture of noncommutative algebraic geometry, Lieven Le Bruyn's website has pointers to a much richer literature.) – Yemon Choi Feb 6 2010 at 22:05
show 6 more comments
5 Answers
In accordance with the suggestion of Yemon Choi, I am going to suggest some further delineation of the approaches to "Non-commutative Algebraic Geometry". I know very little about "Non-commutative Differential Geometry", or what often falls under the heading "à la Connes". This will be completely underrepresented in this summary. For that I trust Yemon's summary to be satisfactory. (edit by YC: BB is kind to say this, but my attempted summary is woefully incomplete and may be inaccurate in details; I would encourage anyone reading to investigate further, keeping in mind that the NCG philosophy and toolkit in analysis did not originate and does not end with Connes.)
Also note that much of what I know about these approaches comes from two sources:
1. The paper by Mahanta
2. My advisor A. Rosenberg.
Additionally, much useful discussion took place at Kevin Lin's question (as Ilya stated in his answer).
I think a better break down for the NCAG side would be:
A. Rosenberg/Gabriel/Kontsevich approach
Following the philosophy of Grothendieck: "to do geometry, one needs only the category of quasi-coherent sheaves on the would-be space" (edit by KL: Where does this quote come from?)
In the famous dissertation of Gabriel, he introduced the injective spectrum of an abelian category, and then reconstructed the commutative noetherian scheme, which is a starting point of noncommutative algebraic geometry. Later, A. Rosenberg introduced the left spectrum of a noncommutative ring as an analogue of the prime spectrum in commutative algebraic geometry, and generalized it to any abelian category. He used one of the spectra to reconstruct any quasi-separated (not necessarily quasi-compact), commutative scheme. (Gabriel-Rosenberg reconstruction theorem.)
In addition, Rosenberg has described the NC-localization (first observed also by Gabriel) which has been used by him and Kontsevich to build NC analogs of more classical spaces (like the NC Grassmannian) and more generally, noncommutative stacks. Rosenberg has also developed the homological algebra associated to these 'spaces'. Applications of this approach include representation theory (D-module theory in particular), quantum algebra, and physics.
References in this area are best found through the MPIM Preprint Series, and a large collection is linked here. Additionally, a book is being written by Rosenberg and Kontsevich furthering the work of their previous paper. Some applications of these methods are used here, here, here, and here. The first two are focusing on representation theory, the second two on non-commutative localization.
Kontsevich/Soibelman approach
They might refer to their approach as "formal deformation theory", and quoting directly from their book
The subject of deformation theory can be defined as the "study of moduli spaces of structures...The subject of this book is formal deformation theory. This means $\mathcal{M}$ will be a formal space(e.g. a formal scheme), and a typical category $\mathcal{W}$ will be the category of affine schemes..."
Their approach is related to $A_{\infty}$ algebras and homological mirror symmetry. References that might help are the papers of Soibelman. Also, I think this is related to the question here. (Note: I know hardly anything beyond that this approach exists. If you know more, feel free to edit this answer! Thanks for your understanding!)
(Some comments by KL: I am not sure whether it is appropriate to include Kontsevich-Soibelman's deformation theory here. This kind of deformation theory is a very general thing, which intersects some of the "noncommutative algebraic geometry" described here, but I think that it is neither a subset nor a superset thereof. In any case, I've asked some questions related to this on MO in the past, see this and this.
However, there is the approach of noncommutative geometry via categories, as elucidated in, for instance, Katzarkov-Kontsevich-Pantev. Here the idea is to think of a category as a category of sheaves on a (hypothetical) non-commutative space. The basic "non-commutative spaces" that we should have in mind are the "Spec" of a (not necessarily commutative) associative algebra, or dg associative algebra, or A-infinity algebra. Such a "space" is an "affine non-commutative scheme". The appropriate category is then the category of modules over such an algebra. Definitively commutative spaces, for instance quasi-projective schemes, are affine non-commutative schemes in this sense: It is a theorem of van den Bergh and Bondal that the derived category of quasicoherent sheaves on a quasi-projective scheme is equivalent to a category of modules over a dg algebra. (I should note that in my world everything is over the complex field; I have no idea what happens over more general fields.) Lots of other categories are or should be affine non-commutative in this sense: Matrix factorization categories (see in particular Dyckerhoff), and probably various kinds of Fukaya categories are conjectured to be so as well.
Anyway I have no idea how this kind of "noncommutative algebraic geometry" interacts with the other kinds explained here, and would really like to hear about it if anybody knows.)
Lieven Le Bruyn's approach
As I know nearly nothing about this approach and the author is a visitor to this site himself, I wouldn't dare attempt to summarize this work.
As mentioned in a comment, his website contains a plethora of links related to non-commutative geometry. I recommend you check it out yourself.
Approach of Artin, Van den Berg school
Artin and Schelter gave a regularity condition on algebras to serve as the algebras of functions on non-commutative schemes. They arise from abstract triples which are understood for commutative algebraic geometry. (Again edits are welcome!)
Here is a nice report on Interactions between noncommutative algebra and algebraic geometry. There are several people who are very active in this field: Michel Van den Berg, James Zhang, Paul Smith, Toby Stafford, I. Gordon, A. Yekutieli. There is also a very nice page of Paul Smith: noncommutative geometry and noncommutative algebra, where you can find almost all the people who are currently working in the noncommutative world.
References: This paper introduced the need for the regularity condition and showed the usefulness. Again I defer to Mahanta for details. Serre's FAC is the starting point of noncommutative projective geometry. But the real framework is built by Artin and James Zhang in their famous paper Noncommutative Projective scheme.
Non-commutative Deformation Theory by Laudal
Olav Laudal has approached NCAG using NC-deformation theory. He also applies his method to invariant theory and moduli theory. (Please edit!)
References are on his page here and this paper seems to be a introductory article.
Apologies
Without a doubt, I have made several errors, given bias, offended the authors, and embarrassed myself in this post. Please don't hold this against me, just edit/comment on this post until it is satisfactory. As it was said before, the nlab article on noncommutative geometry is great, you should defer to it rather than this post.
Thanks!
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Wow, that's sterling stuff (and thanks for the kind words on my answer, although I'm really not sure it deserves them!) – Yemon Choi Feb 6 2010 at 23:28
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I just described the work of Pierre-Gabriel – Shizhuo Zhang Feb 7 2010 at 5:15
Thanks Shizhuo! I did some minor typesetting changes to what you wrote btw. – B. Bischof Feb 7 2010 at 5:47
I just made some small edits to the phrasing (ironing out some minor points of language). Hope this is fine – Yemon Choi Feb 7 2010 at 10:10
Yes, thanks Yemon – B. Bischof Feb 7 2010 at 22:57
show 1 more comment
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The nLab's page on noncommutative geometry provides a lot of information. (It's actually a good idea to look into nLab before posting any general question: I did post some when I didn't know about nLab existence and was often sent to nLab anyway, usually politely.)
Also, you should take a look at Kevin Lin's question Non-commutative algebraic geometry.
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The nLab page is just one long page. I want the answers and viewpoints of various people. Isn't MO better for that? I also did see Kevin Lin's question. But he is asking from the point of view of an algebraic geometer looking to get noncommutative. I believe, there is more to NCG than just that. – Anweshi Jan 2 2010 at 21:32
Some thoughts and links on the analysts' NCG, from someone who doesn't practice it. Caveat lector. (Some edits made to erroneous history.)
NCG a la Connes was originally non-commutative differential geometry (which is why extra structure is needed in, say, the definition of spectral triple). Having only recently looked at Connes' original long, two-part paper in the Publications of the IHES, I think this is a better place to start than the later Big-Picture-Which-Is-Really Fundable works of various people thereafter. Work of Connes & Moscovici and others on trying to generalize the Atiyah-Singer index theorem also give some indication of the original motivation. (This is where someone more expert than me should really step in and say something about the work of C & M Mischenko, Kasparov and their co-authors on the Novikov conjecture, or indeed the work of Higson et al. on the Baum-Connes conjecture.)
It was only afterwards that Connes started championing an NCG perspective on The Standard Model. (Although, if you want real connections with mathematical physics, there was some work of Jean Belissard on identifying gaps in spectra of certain operators in a quantum-mechanical model with K-theoretic invariants of associated C*-algebras. See this paper of Kaminker and Putnam and the references therein for more details.)
Personally, I am bit leery of the Big Picture motivation for noncommutative geometry, at least of this variety. The most useful variant of such motivation that I can think of, is that degenerate group actions on topological spaces give rise to better-behaved homotopy groupoids; thanks to a theorem of Rieffel (IIRC), when the group action on the space is nice, the commutative C*-algebra of the quotient space is Morita equivalent -- i.e. has "the same module theory" -- as the noncommutative C*-algebra of the homotopy groupoid.
I apologize for the rambly nature of this answer, but with all due respect to Anweshi I think his question, or at least the version of it which I can currently see, is so broad (as per Anton's original comments) that the only responses are either encyclopaedic - I haven't even had space to mention the historical role played by Brown-Douglas-Fillmore theory, for instance - or sales pitches. Nevertheless, if someone can persuade Nigel Higson to drop by, I'm sure he could give a much better answer ;)
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I really like this answer. I know little about A.Connes geometry. But I am interested in the work of Higgsen and J.Block. – Shizhuo Zhang Feb 7 2010 at 7:04
The physics part of the picture comes from a classic article
• String Theory and Noncommutative Geometry by Seiberg and Witten
I can't say for sure, but I think it still isn't completely transplanted to math by either Connes or Kontsevich.
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There is a beautiful three-part series of lectures by Jonathan Block that introduces both the Connes and Kontsevich schools of non-commutative geometry:
http://www.math.upenn.edu/~tpantev/rtg09bc/lecnotes/block-ncg.pdf
One of the main motivating examples throughout the lectures is Lusztig's proof proof of the Novikov conjecture.
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That looks really nice! I should have mentioned something more about the Novikov conjecture and its role in the development of NCG, but couldn't remember what little I'd seen of the details as opposed to the "adverts" – Yemon Choi Feb 7 2010 at 5:29
Thanks for this link! – B. Bischof Feb 7 2010 at 5:55
Thanks for posting this, Richard. I had forgotten about this. Btw, these notes were taken by our friend Aaron Smith! – Kevin Lin Apr 18 2010 at 9:50
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http://math.stackexchange.com/questions/153097/flattening-a-2d-normal-distribution
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# “Flattening” a 2D Normal Distribution
I would like to model the probability of a point being at a certain place on a 2D grid. The X coordinate of the point varies according to a normal distribution with mean $0$ and standard deviation $\sigma$. The Y coordinate varies according to a normal distribution with the same mean and same standard deviation.
I know that the probability of being a certain distance from $(0, 0)$ is the same in all directions. I would therefore like to "flatten" my 2 distributions into a single distribution where the random variable is the distance from $(0, 0)$. In other words, if I know the distributions of x and y, what is the distribution of $\sqrt{x^2+y^2}$?
Am I right in my intuition that this will also be a normal distribution? (Or, half of a normal distibution, since the distance cannot be below zero). If so, how do I calculate the standard deviation of this distribution? If not, what type of distribution would it be, and what are the pdf and cdf functions of it?
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## 1 Answer
The distribution you are after is called the Rayleigh distribution function. The pdf is given by $$f_R(r) = \dfrac{r}{\sigma^2} \exp \left(-\dfrac{r^2}{2 \sigma^2} \right) \text{ where } r \geq 0$$
$$F_R(r; \sigma) = \mathbb{P}(R \leq r) = \mathbb{P}(X^2 + Y^2 \leq r^2) = \int_{-r}^{r} \mathbb{P} \left(Y \in \left[ \sqrt{r^2-x^2},\sqrt{r^2-x^2} \right] \right) f_X(x) dx\\ = \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} f_Y(y)f_X(x) dx = \dfrac1{2 \pi \sigma^2} \int_{-r}^{r} \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \exp \left( -\left( \dfrac{y^2 + x^2}{2 \sigma^2} \right) \right) dy dx$$ Changing variable from $(x,y) \rightarrow (\rho,\theta)$, we get that $$F_R(r; \sigma) = \dfrac1{2 \pi \sigma^2} \int_{\rho=0}^{r} \int_{\theta=0}^{2 \pi} \exp \left( - \dfrac{\rho^2}{2 \sigma^2} \right) \rho d \rho d \theta\\ =\dfrac1{\sigma^2} \int_{\rho=0}^{r} \rho \exp \left(- \dfrac{\rho^2}{2 \sigma^2} \right) d \rho \\ = -\left. \exp \left( - \dfrac{\rho^2}{2 \sigma^2}\right) \right \rvert_{\rho=0}^{\rho=r} \\ = 1 - \exp \left( - \dfrac{r^2}{2 \sigma^2} \right)$$ $$f_R(r) = \dfrac{d F_R(r)}{dr} = \dfrac{r}{\sigma^2} \exp \left( - \dfrac{r^2}{2 \sigma^2}\right)$$
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– leonbloy Jun 3 '12 at 1:17
@leonbloy True. – user17762 Jun 3 '12 at 1:19
So, just to confirm, the $\sigma$ in your Rayleigh pdf function is the same variable as the $\sigma$ that represents the standard deviation of the initial normal distributions, right? – Ord Jun 3 '12 at 15:24
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http://unapologetic.wordpress.com/2007/08/11/internal-categories/?like=1&_wpnonce=34114a6dfa
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# The Unapologetic Mathematician
## Internal Categories
Just like we have monid objects, we can actually define something we could sensibly call a “category object”. In this case, however, it will be a little more accurate to use the term “internal category”.
This is because a (small) category isn’t just a set with extra structure. It’s two sets with extra structure. We have a set $O$ of objects, a set $M$ of morphisms, a function $i:O\rightarrow M$ assigning the identity morphism to each object, functions $s:M\rightarrow O$ and $t:M\rightarrow O$ assigning the source and target objects to each morphism, and an arrow $\gamma:M{}_s\times_tM\rightarrow M$ telling us how to compose certain pairs of morphisms. This involves a “fibered product”, which is just the pullback in $\mathbf{Set}$. We take the arrows $s$ and $t$ from $M$ to $O$ and pull back the square to get the set of all pairs of morphisms so that the source object of one is the target of the other.
Then there are a bunch of relations which hold:
• The source of the identity arrow on an object is the object itself.
• The target of the identity arrow on an object is the object itself.
• The identity arrow on an object acts as a left and right identity for the composition.
• The source of a composition is the source of the second member of the pair.
• The target of a composition is the target of the first member of the pair.
• The composition is associative.
I’ll leave you to write these out purely in terms of the functions $m$, $i$, $s$, and $t$.
Now we can take this whole setup and drop it into any other category, as long as that category has pairwise pullbacks. If $\mathcal{C}$ does have these pullbacks, then a category internal to $\mathcal{C}$ (or a “category object”) consists of a pair of objects and four morphisms of $\mathcal{C}$, which must satisfy the above relations. Then a category internal to $\mathbf{Set}$ is a small category.
When we’re talking about categorifying something like a group, we want to replace the underlying set of a group with a small category. That is, we want to have a group object in $\mathbf{Cal}$. But we know that internalizations commute, so this is the same thing as a “category object” in groups! That is, instead of looking for a category with a multiplication functor and so on, we can instead look for a pair of groups with source, target, composition, and identity homomorphisms between them.
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://mathoverflow.net/questions/50441?sort=oldest
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## Forbidden mirror sequences
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\cal{M}$ be a finite collection of two-sided mirrors, each an open unit-length segment in $\mathbb{R^2}$, and such that the segments when closed are disjoint. A ray of light that reflects off the mirrors determines a mirror sequence or mirror string consisting of the mirror indices in order of reflection. Obviously no mirror string can contain $a a$ as a substring. I am wondering if mirror sequences may be characterized by a list of forbidden substrings.
For example, consider the special case of parallel mirrors, none collinear. Orient them all vertical, and label them sorted left to right. Then for labels $a < b < c$, these are forbidden substrings:
$$b * a c * b$$ $$b * c a * b$$
with $*$ being any string (including the empty string). So in the above example, after 415 appears, 4 cannot occur again. Perhaps these are the only forbidden patterns for parallel mirrors?
Edit. Apologies for not initially phrasing a clear question.
Are there strings of mirror indices that cannot be realized by some ray reflecting among some collection $\cal{M}$ of mirrors? Is there a list of such strings that characterize all the realizable sequences?
In contrast to the parallel-mirrors example above, I am primarily interested in mirrors without constraints on their placements or orientations.
Perhaps analogous sequences have been studied before, maybe in another context? Pointers appreciated!
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If you have three mirrors forming a triangle with an obtuse angle, then aba is a forbidden configuration (where ab is the obtuse angle). – Peter Shor Dec 26 2010 at 21:12
@Peter: Yes; thanks. But of course $aba$ can be realized by other configurations of mirrors. Ideally I am seeking sequences that are unrealizable by any collection of mirrors. – Joseph O'Rourke Dec 26 2010 at 23:07
@Joseph: You should make your question clearer, then. Although that's a much nicer question than the one I thought you were asking. – Peter Shor Dec 26 2010 at 23:31
1
Some mildly related questions (also related to Bill Thurston's commentary) are considered in: Quas, Anthony N.(4-CAMB-S) Infinite paths in a Lorentz lattice gas model. (English summary) Probab. Theory Related Fields 114 (1999), no. 2, 229–244. 60K35 (28D05 82B40) – Igor Rivin Dec 27 2010 at 4:00
1
@Igor: Nice connection! "We consider infinite paths in an illumination problem on the lattice $\mathbb{Z}^2$, where at each vertex, there is either a two-sided mirror (with probability $p \ge 0$) or no mirror (with probability $1-p$)." – Joseph O'Rourke Dec 27 2010 at 13:47
show 1 more comment
## 3 Answers
Maybe count how many canonical such sequences are possible for a few small numbers of mirrors and check OEIS?
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also with a max sequence length – aster Dec 26 2010 at 20:37
### 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.
Edited 12/27 to try to clarify, and to correct regular expression formatting (\* is needed so * doesn't turn into italization)
On some level its obvious that the set of mirror sequences are characterized by the list of finite substrings that can never occur, if you allow lists to be infinite, and if you count the closure of the set of mirror trajectories as mirror trajectories. (There might be subtle issues of what happens when a light beam hits exactly at the edge of a mirror: in this case, if you allow both outcomes that are the limits of its perturbation, it gives a closed set). But I don't think this is what you meant. It reminds me of having to use a word such as "industrious" in a sentence in 2nd grade. My teacher didn't like my sentences, "'Industrious' is a word."
Perhaps a better formulation of the question is to ask whether the set of mirror sequences is a regular language, and if so, to describe it. For some presumably easier cases, the set of arrangements of mirrors can be specialized to range over some particular subset, such as parallel mirrors, or mirrors all at angles of the form $k \pi / n$ for some $n$.
Thus, for two mirrors a and b, the language is b?(ab)*a?, interpreted with the usual regular expression convention where ? means 0 or 1 occurrence of the preceding term and * means 0 or more occurrences. This is equivalent to not(.*aa.*) and not(.*bb.*) where . matches any symbol.
For three parallel mirrors, you can have a sequence such as (ab)+(cb)+(ab)+, where + denotes 1 or more occurrence of the previous term, but this sequence implies that c is sandwiched between a and b, so you can't have any more c's after that. I won't try to write out a complete regular expression, but
Claim: For any number of parallel mirrors, the set of possible sequences (as the positions of mirrors and the initial beam are vaired) is a regular language. Furthermore, the number of possible sequences of length $n$ grows only as a polynomial in \$n.
Proof: One may assume that the beam is coming from the left and angling upward. (The cases that it is vertical or horizontal are trivial).
To make a finite state automaton (FSA) to recognize valid mirror sequences, first make a nondeterministic finite state automaton (NDFSA) whose states are possible arrangements of the mirrors and the current position of the beam in relation to them: the relevant data is the horizontal linear order, the question of whether the beam is going to the left or right, and the vertical partial order of the tops and bottoms of the mirrors and the beam, where the partial order $x < y$ on points in the plane is whether $y$ is in the cone bounded by beams at $x$ at the beam angle and its mirror image. Each time the beam hits a mirror, the point where it hits has a one of a finite number of positions in the new partial order, inserting the new point and removing the previous position of the beam; this gives an NDFSA. The (deterministic) FSA has states which are sets of states of the NDFSA, and the transition takes a set of states to the set of all possible states obtained by an NDFSA transition from one of them.
There are probably nice methods to actually implement this. Instead of a set of linear orders, we can keep track of a partially ordered set of mirrors that we have seen; each new letter pair xy tells us either that x < y or x > y. A letter can either occur at only odd positions, or only even positions --- when the mirrors are parallel, it's impossible for the beam to reflect from the front, then get around to the other side and hit the back. Whenever we see such as yx[^y]*xy, where [^y] means a character that is not y, it implies that the mirrors other than x in the [^y]* substring are sandwiched between x and y, and the beam has gone above them: they can never occur again.
Perhaps you can see or intuit from the way the finite-state mirror model evolves that it's impossible to get branching recurrence, so the number of sequences grows only as a polynomial. Instead of trying to make this precise, there's a proof-by-authority for this last point: pass to a surface obtained by taking two copies of the plane, cutting slits where the mirrors are, and gluing them together along the slits so that the light rays follow geodesics on the resulting surface (with exceptional points where the trajectories fork at the ends of the mirrors.) The trajectories now become simple curves on the resulting surface, generically going out to the point at infinity at both ends. It is known that the set of all homotopy classes of simple curves on any surface grows as a polynomial of degree 6g-6, where $g$ is the "genus" or number of holes of the surface, since simple closed curves are lattice points in the space of measured laminations on a surface.
The same proof works when mirrors are confined to angles of the form $2\pi/n$, where you take $2n$ copies of the plane, one for each reflected image of the plane, mod translations. Cf. "A rational billiard flow is uniquely ergodic in almost every direction" by Kerckhoff, Masur and Smillie for a masterful use of this technique of passing to a surface obtained by gluing multiple copies.
Even though the number of mirror sequences has polynomial growth in the case of mirrors at angles $\pi/n$, nonetheless, even when $n= 2$ i.e. for mirrors that are vertical or horizontal, mirror sequences are not a regular language. The classicla special case of a rectangular billiard table illustrates this: for a rectangle you can glue four copies together to get a torus, where the trajectories become lines of constant slope. These sequences have been well studied, for instance in computer graphics, where they give sequences of pixels used to depict a thin straight line. They have a simple recursive characterization (essentially the Euclidean algorithm), but they are not determined by a finite state machine. For an arbitrary arrangement of four mirrors, the richest sequences are those that come from the billiard table --- otherwise, mirror sequences degenerate to a case of 2 or 3 parallel mirrors at the start and/or finish. Thus, there is no finite set of excluded sequences for this case.
For more less restricted arrangements, the conditons will only become more complicated.
In the general case, there are some constraints having to do with inequalities that can be deduced on angles of the mirrors.
First, if there is a sequence of length $n$ alternating between two mirrors x and y, it implies that one of the two angles between the two mirrors is less than $\pi/(n-1)$. You can see this by starting with mirrors that form an angle $\alpha$, and reflecting it around. A light ray hitting mirror $x$ unfolds as a straight line crossing this pattern, so if it crosses $k$ edges, the $k-1$ angles between them cannot add to more than $\pi$.
Using this, for a complicated enough sequence involving just 3 letters, you get inequalities concerning the three angles of the triangle formed by extending the mirrors.
I don't know whether there is a good general algorithm to decide if a sequence is a mirror sequence. I suspect that there should be an algorithm that in practice is fairly efficient, generalizing the idea from the case of parallel mirrors to describe increasingly restrictive geometric inequalities about the arrangement of mirrors, perhaps as a first approximation using floating point inequalities.
I suspect that the number of mirror sequences probably has polynomial growth even in the general case when no conditions are imposed on the angles of the mirrors, but I haven't thought through a proof. If there are $m$ mirrors, then when the angles are restricted to have the form $\pi k / n$, the covering surface where trajectories are simple proper arcs (tending to infinity at both ends) has Euler characteristic $-2 n ( m-1)$. The dimension of the space of measured geodesic laminations on the covering surface grows linearly with $n$, but the light trajectories are integral curves of a quadratic differential that has an equivariance property with respect to deck transformations, which limits its dimension. The dimension of the moduli space of mirror arrangements with any given constant angle depends only on the number of mirrors: we can normalize the first mirror to be the unit interval in the $x$-axis, and then each additional mirror has 3 degrees of freedom (when its angle is fixed), so the dimension is $3m-3$. This should equal the degree of the polynomial growth for mirror sequences with such mirrors. However, since mirror sequences are not determined by an FSA, the number of valid mirror sequences is probably not the set of values of an actual polynomial except when the mirrors are parallel, so one would need estimates of initial behavior or particular constants to conclude that the limit has polynomial growth.
A weaker question is whether the shift map on mirror sequences has topological entropy 0 (i.e, it can be recoded into a sequence with bit rate 0). I think this follows. If we add the extra information of the angles of the mirrors to the dynamical system (usually this is probably redundant anyway), then each particular set of angles defines a closed invariant subset. The angle of the beam at any time is in the subgroup of $S^1$ generated by the angles of the mirrors, which is an abelian group of rank at most $m$. There is an infinite-sheeted covering surface where the trajectories become simple, and the covering surface can be thought of as equivariantly embedded in $\mathbb R^m$ (provided $m \ge 3$; for \$m = 3, make it an immersion if it's hard to get an embedding). This setup seems to imply that the topological entropy is 0. Perhaps some expert can confirm or deny.
I think it would probably be worth expanding the notation to distinguish between hitting the front and back of a given mirror, perhaps using capitalization, x vs X.
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1
I appreciate these illuminating :-) insights!! Because the complement of a regular language is regular, your theorem could also be approached via forbidden sequences. It is remarkable that with just horizontal and vertical mirrors, there is no finite set of forbidden sequences! The connection to graphics line-pixels is both startling and (now) obvious. And the argument that growth is polynomial by "passing to a surface" is beautiful! – Joseph O'Rourke Dec 27 2010 at 13:43
@Joseph O'Rourke: Yes, the set of forbidden sequences is a regular language whenever the list of good sequences is a regular language. Note however that for 3 or more parallel mirrors, no finite list of forbidden sequences is sufficient: for instance, any string of the form yx(zx)*yx*z is forbidden and arbitrarily long, but any proper substring is valid. Various people have studied the symbolic dynamics of quadratic differentials, interval exchange maps, and billiards, subjects that are all closely related to this ... I'm hoping some of the experts will post. – Bill Thurston Dec 27 2010 at 16:48
Here is a possible way of producing such forbidden configurations. Suppose you have $(ab)^k$ for some large $k$. Then I'd like to claim that $a$ and $b$ must be nearly parallel (see Thurston's answer). So produce a sequence with three mirrors containing $(ab)^k$, $(bc)^k$, $(ac)^k$ and a forbidden configuration for parallel mirrors such as the one you gave above (which you then have to prove also is forbidden for nearly parallel mirrors).
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@Peter: Ingenious way to constrain the geometry! That seems to provide what I couldn't find: a forbidden string. Nice! – Joseph O'Rourke Dec 27 2010 at 21:04
@Joseph: You still need to show that a forbidden configuration for parallel mirrors is also forbidden with near-parallel mirrors. I think you might be able to do this by compactness, but it's not immediate. – Peter Shor Dec 27 2010 at 22:36
Actually, thinking about it, there are forbidden sequences for parallel mirrors which are achievable by near-parallel mirrors arbitrarily close to parallel, so the compactness argument discussed in the previous comment is not only not immediate, but also unachievable. – Peter Shor Dec 28 2010 at 13:00
@Peter: Yes, I see that completing the argument is not straightforward. – Joseph O'Rourke Dec 28 2010 at 14:39
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http://en.wikipedia.org/wiki/Uniqueness
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Uniqueness quantification
From Wikipedia, the free encyclopedia
(Redirected from Uniqueness)
Jump to: navigation, search
"Unique" redirects here. For other uses, see Unique (disambiguation).
This article includes a list of references, related reading or external links, but . (January 2013)
In mathematics and logic, the phrase "there is one and only one" is used to indicate that exactly one object with a certain property exists. In mathematical logic, this sort of quantification is known as uniqueness quantification or unique existential quantification.
Uniqueness quantification is often denoted with the symbols "∃!" or ∃=1". For example, the formal statement
$\exists! n \in \mathbb{N}\,(n - 2 = 4)$
may be read aloud as "there is exactly one natural number n such that n - 2 = 4".
Proving uniqueness
Proving uniqueness turns out to be mostly easier than that of existence or expressibility. The most common technique to proving uniqueness is to assume there exists two quantities (say, a and b) that satisfies the condition given, and then logically deducing their equality, i.e. a = b.
As a simple high school example, to show x + 2 = 5 has only one solution, we assume there are two solutions first, namely, a and b, satisfying x + 2 = 5. Thus
$a + 2 = 5\text{ and }b + 2 = 5. \,$
By transitivity of equality,
$a + 2 = b + 2. \,$
By cancellation,
$a = b. \,$
This simple example shows how a proof of uniqueness is done, the end result being the equality of the two quantities that satisfy the condition. We must say, however, that existence/expressibility must be proven before uniqueness, or else we cannot even assume the existence of those two quantities to begin with.
Reduction to ordinary existential and universal quantification
Uniqueness quantification can be expressed in terms of the existential and universal quantifiers of predicate logic by defining the formula ∃!x P(x) to mean
$\exists x\, P(x) \wedge \neg \exists x,y \, ( P(x) \wedge P(y) \wedge (x \neq y)),$
or equivalently,
$\exists x \, ( P(x) \wedge \forall y\,(P(y) \to x = y)).$
An equivalent definition that has the virtue of separating the notions of existence and uniqueness into two clauses, at the expense of brevity, is
$\exists x\, P(x) \wedge \forall y\, \forall z\,((P(y) \wedge P(z)) \to y = z).$
Another equivalent definition with the advantage of brevity is
$\exists x\,\forall y\,(P(y) \leftrightarrow x = y).$
Generalizations
One generalization of uniqueness quantification is counting quantification. This includes both quantification of the form "exactly k objects exist such that …" as well as "infinitely many objects exist such that …" and "only finitely many objects exist such that…". The first of these forms is expressible using ordinary quantifiers, but the latter two cannot be expressed in ordinary first-order logic.
References
[1] [2]
1. Kleene, Stephen (1952). Introduction to Metamathematics. Ishi Press International. p. 199.
2. Andrews, Peter B. (2002). An introduction to mathematical logic and type theory to truth through proof (2. ed. ed.). Dordrecht: Kluwer Acad. Publ. p. 233. ISBN 1-4020-0763-9.
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http://mathhelpforum.com/advanced-algebra/92473-finding-eigenvector-print.html
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# Finding the eigenvector
Printable View
• June 10th 2009, 01:56 PM
Nerdfighter
Finding the eigenvector
Given a specific Eigenvalue $\lambda$, how do I find an eigenvector of a Matrix $A$?
Specifically, if $\lambda = 4$ and $A =$ $\left[\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right]$
By the way, is there a way to make my Matrix look a little neater/lined up?
• June 10th 2009, 02:07 PM
TheAbstractionist
If $\begin{pmatrix}x\\y\\z\end{pmatrix}$ is an eigenvector, then $\left(\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right)\begin{pmatrix}x\\y\\z\end{ pmatrix}\ =\ \lambda\begin{pmatrix}x\\y\\z\end{pmatrix}\ =\ 4\begin{pmatrix}x\\y\\z\end{pmatrix}.$
There will be infinitely many eigenvectors corresponding the the eigenvalue 4, so just pick one of them.
Quote:
Originally Posted by Nerdfighter
By the way, is there a way to make my Matrix look a little neater/lined up?
Instead of centre-aligning them, right-align them (as I have done).
• June 10th 2009, 02:17 PM
Nerdfighter
How Do I pick one? I obviously can't pick any numbers for x, y, and z.
• June 10th 2009, 02:44 PM
TheAbstractionist
I did not say pick any numbers for $x,y,z.$ Pick $x,y,z$ so that they satisfy the eigenvalue equation.
• June 10th 2009, 02:46 PM
Nerdfighter
Quote:
Originally Posted by TheAbstractionist
I did not say pick any numbers for $x,y,z.$ Pick $x,y,z$ so that they satisfy the eigenvalue equation.
What is a good method to find $x,y,z$ such that $x,y,z$ satisfies the eigenvalue equation?
Could I simply pick the zero vector to be an eigenvector?
• June 10th 2009, 03:02 PM
TheEmptySet
Quote:
Originally Posted by Nerdfighter
Given a specific Eigenvalue $\lambda$, how do I find an eigenvector of a Matrix $A$?
Specifically, if $\lambda = 4$ and $A =$ $\left[\begin{array}{rrr}3&0&-1\\2&3&1\\-3&4&5\end{array}\right]$
By the way, is there a way to make my Matrix look a little neater/lined up?
Solve the matrix equation
$A\vec v = \lambda \vec v \iff (A-\lambda I)\vec v =0$
Since your eigen value is 4 we get
$(A-4 I)\vec v =0$
$\begin{bmatrix} <br /> 3-4 && 0&& -1\\<br /> 2 && 3-4 && 1 \\<br /> -3 && 4 && 5-4 \\<br /> <br /> \end{bmatrix}$
$\begin{bmatrix} <br /> -1 && 0&& -1\\<br /> 2 && -1 && 1 \\<br /> -3 && 4 && 1 \\<br /> <br /> \end{bmatrix}$
putting this in reduced row form we get
$\begin{bmatrix} <br /> 1 && 0&& 1\\<br /> 0 && 1 && 1 \\<br /> 0 && 0 && 0 \\<br /> <br /> \end{bmatrix}$
Solving the above homogenious equation gives
$z=t, y=-t,x=-t$ or
$\begin{bmatrix}<br /> -t \\ -t \\ t <br /> \end{bmatrix}=t \begin{bmatrix}<br /> -1 \\ -1 \\ 1 <br /> \end{bmatrix}$
So that is the eigen vector
p.s if you use the \begin{bmatrix} \end{bmatrix} tags it will auto format all of your columns
• June 10th 2009, 03:06 PM
Nerdfighter
Thank you, that's what I needed!
• June 10th 2009, 10:41 PM
Swlabr
Quote:
Originally Posted by Nerdfighter
Could I simply pick the zero vector to be an eigenvector?
No, the zero vector is never an Eigenvector, as if it was every scalar would be an Eigenvalue. An Eigenvector for the corresponding Eigenvalue $\lambda$ is defined to be a "non-zero vector such that $Mv = v \lambda$".
• June 10th 2009, 10:42 PM
Nerdfighter
Quote:
Originally Posted by Swlabr
No, the zero vector is never an Eigenvector, as if it was every scalar would be an Eigenvalue. An Eigenvector for the corresponding Eigenvalue $\lambda$ is defined to be a "non-zero vector such that $Mv = v \lambda$".
Ooh, very good to know. Thank you.
All times are GMT -8. The time now is 10:35 AM.
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http://www.physicsforums.com/showthread.php?p=4240256
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Physics Forums
## Series and Future Stock Prices
1. The problem statement, all variables and given/known data
I have a question here that seems so simple, yet I can't seem to wrap my head around what they're asking.
It says that one method of pricing a stock is to set the share price equal to the sum of all future dividends for infinitely many years, with dividends discounted to their present value. We assume that we are always able to invest money at a guaranteed rate of 5% per year. This implies that $1.05 received next year is worth only$1 today. What is \$1 received n years from now worth today?
The second part says that if a corporation promises to pay a dividend of \$1 per share every year for all years in the future, what is the total value of all future dividends for one share for infinitely many years discounted to their value today.
2. Relevant equations
Possibly the compound interest formula?
A = P( 1 + r )n
3. The attempt at a solution
I am assuming the equation for the first part will have something to do with the compound interest formula, but I'm not sure how to apply it here. I was thinking of something like 1 - .05n, but that doesn't seem quite right to me.
As for the second part of the problem, it is dependent on what I find for the first part, but I think that it has to do with finding the limit of whatever I find to be the answer for the first part of the question.
Any suggestions that will point me in the right direction here?
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Quote by Illania 1. The problem statement, all variables and given/known data I have a question here that seems so simple, yet I can't seem to wrap my head around what they're asking. It says that one method of pricing a stock is to set the share price equal to the sum of all future dividends for infinitely many years, with dividends discounted to their present value. We assume that we are always able to invest money at a guaranteed rate of 5% per year. This implies that $1.05 received next year is worth only$1 today. What is $1 received n years from now worth today? The second part says that if a corporation promises to pay a dividend of$1 per share every year for all years in the future, what is the total value of all future dividends for one share for infinitely many years discounted to their value today. 2. Relevant equations Possibly the compound interest formula? A = P( 1 + r )n
Yes, with "P" the future value and A= \$1. Assuming r= 0.05, that gives 1= P(1.05n) so that P= 1/(1.05n).
3. The attempt at a solution I am assuming the equation for the first part will have something to do with the compound interest formula, but I'm not sure how to apply it here. I was thinking of something like 1 - .05n, but that doesn't seem quite right to me. As for the second part of the problem, it is dependent on what I find for the first part, but I think that it has to do with finding the limit of whatever I find to be the answer for the first part of the question. Any suggestions that will point me in the right direction here?
For the second part, sum over all n:
$$\sum_{n=0}^\infty \frac{1}{1.05^n}$$
That is a "geometric series" of the form $\sum r^n$ with r= 1/1.05.
Hm, I have only seen geometric series of the form $\Sigma$ Arn-1. I understand how to manipulate a term to make a power of n a power of n-1, but the exponent here would be a negative n. Should I instead to be looking at the series as $\sum\left(\frac{1}{1.05}\right)^{n}$ and then manipulate the series to get that power to n-1?
Recognitions:
Homework Help
## Series and Future Stock Prices
Quote by Illania Hm, I have only seen geometric series of the form $\Sigma$ Arn-1. I understand how to manipulate a term to make a power of n a power of n-1, but the exponent here would be a negative n. Should I instead to be looking at the series as $\sum\left(\frac{1}{1.05}\right)^{n}$ and then manipulate the series to get that power to n-1?
What do YOU think?
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http://programmingpraxis.com/2011/04/12/
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# Programming Praxis
A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer
## House Of Representatives
### April 12, 2011
The United States counts its citizens every ten years, and the result of that census is used to allocate the 435 congressional seats in the House of Representatives to the 50 States. Since 1940, that allocation has been done using a method devised by Edward Huntington and Joseph Hill that minimizes percentage differences in the sizes of the congressional districts.
The Huntington-Hill method begins by assigning one representative to each State. Then each of the remaining representatives is assigned to a State in a succession of rounds by computing $g(n, p) = \frac{p}{\sqrt{n(n+1)}}$ for each State, where n is the current number of representatives (initially 1), p is the population of the State, and g(n, p) is the State’s population divided by the geometric mean of the current number of representatives and the number of representatives that the State would have if it was assigned the next representative. The geometric mean g(n, p) is calculated for each State at each round and the representative assigned to the State with the highest geometric mean g(n, p).
For instance, once each State has been assigned one representative, the geometric mean g(n, p) for each State is its population divided by the square root of 2. Since California has the biggest population, it gets the 51st representative. Then its geometric mean is recalculated as its population divided by the square root of 2 × 3 = 6, and in the second round the 52nd representative is assigned to Texas, which has the second-highest population, since it now has the largest geometric mean g(n, p). This continues for 435 − 50 = 385 rounds until all the representatives have been assigned.
Your task is to compute the apportionment of seats in the House of Representatives; the population data is given on the next page. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
Posted by programmingpraxis
Filed in Exercises
13 Comments »
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http://en.m.wikibooks.org/wiki/Sound_Synthesis_Theory/Sound_in_the_Digital_Domain
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# Sound in the Digital Domain
## Introduction
Digital systems (e.g. computers) and formats (e.g. CD) are clearly the most popular and commonplace methods of storing and manipulating audio. Since the introduction of the compact disc in the early 1980s, the digital format has provided increasingly greater storage capacity and the ability to store audio information at an acceptable quality. Although analogue formats still exist (vinyl, tape), they typically serve a niche audience. Digital systems are ubiquitous in modern music technology. It must be stressed that there is no argument as to whether one domain, be it analogue or digital is superior, but the following provides some desirable features of working with audio in the digital domain.
• Storage. The amount of digital audio data capable of being stored on a modern hard drive is far greater than a tape system. Furthermore, we can choose the quality of the captured audio data, which relates directly to file size and other factors.
• Control. By storing audio information in digital, we can perform powerful and complex operations on the data that would be extremely difficult to realise otherwise.
• Durability. Digital audio can be copied across devices without any loss of information. Furthermore, many systems employ error correction codes to compensate for wear and tear on a physical digital format such as a compact disc.
↑Jump back a section
## Digital <-> Analogue Conversion
Acoustic information (sound waves) are treated as signals. As demonstrated in the previous chapter, we traditionally view these signals as varying amplitude over time. In analogue systems, this generally means that the amplitude is represented by a continuous voltage; but inside a digital system, the signal must be stored as a stream of discrete values.
Figure 2.1. An overview of the digital <-> analogue conversion process.
Digital data stored in this way has no real physical meaning; one could describe a song on a computer as just an array of numbers; these numbers are meaningless unless there exists within the system a process that can interpret each number in sequence appropriately. Fig. 2.1 shows an overview of the process of capturing analogue sound and converting it into a digital stream of numbers for storage and manipulation in such a system. The steps are as follows:
1. An input such as a microphone converts acoustic air pressure variations (sound waves) into variations in voltage.
2. An analogue to digital converter (ADC) converts the varying voltage into a stream of digital values by taking a 'snapshot' of the voltage at a point in time and assigning it a value depending on its amplitude. It typically takes these 'snapshots' thousands of times a second, the rate at which is known as the sample rate.
3. The numerical data is stored on the digital system and then subsequently manipulated or analysed by the user.
4. The numerical data is re-read and streamed out of the digital system.
5. A digital to analogue converter (DAC) converts the stream of digital values back to a varying voltage.
6. A loudspeaker converts the voltage to variations in air pressure (sound).
Although the signal at each stage comes in a different form (sound energy, digital values etc.), the information is analogous. However, due to the nature of the conversion process, this data may become manipulated and distorted. For instance, low values for sample rates or other factors at the ADC might mean that the continuous analogue signal is not represented with enough detail and subsequently the information will be distorted. There are also imperfections in physical devices such as microphones which further "colour" the signal in some way. It is for this reason that musicians and engineers aim to use the most high-quality equipment and processes in order to preserve the integrity of the original sound throughout the process. Musicians and engineers must consider what other processes their music will go through before consumption, too (radio transmission etc.).
↑Jump back a section
## Sampling
Sound waves in their natural acoustic form can be considered continuous; that is, their time-domain graphs are smooth lines on all zoom factors without any breaks or jumps. We cannot have these breaks, or discontinuities because sound cannot switch instantaneously between two values. An example of this may be an idealised waveform like a square wave - on paper, it switches between 1 and -1 amplitude at a point instantaneously; however a loudspeaker cannot, by the laws of physics, jump between two points in no time at all, the cone has to travel through a continuous path from one point to the next.
Figure 2.2. Discrete samples (red) of a continuous waveform (grey).
Sampling is the process of taking a continuous, acoustic waveform and converting it into a digital stream of discrete numbers. An ADC measures the amplitude of the input at a regular rate creating a stream of values which represent the waveform in digital. The output is then created by passing these values to the DAC, which drives a loudspeaker appropriately. By measuring the amplitude many thousands of times a second, we create a "picture" of the sound which is of sufficient quality to human ears. The more and more we increase this sample rate, the more accurately a waveform is represented and reproduced.
### Nyquist-Shannon sampling theorem
The frequency of a signal has implications for its representation, especially at very high frequencies. As discussed in the previous chapter, the frequency of a sine wave is the number of cycles per second. If we have a sample rate of 20000 samples per second (20 kHz), it is clear that a high frequency sinusoid such as 9000 Hz is going to have fewer "snapshots" than a sinusoid at 150 Hz. Eventually there reaches a point where there are not enough sample points to be able to record the cycle of a waveform, which leads us to the following important requirement:
The sample rate must be greater than twice the maximum frequency represented.
Why is this? The minimum number of sample points required to represent a sine wave is two, but we need at least slightly more than this so that we're not dependent phase (samples at exactly twice the sine wave frequency, the samples may fall on the peaks of the sine wave, or on the zero crossings). It may seem apparent at this time that using just two points to represent a continuous curve such as a sinusoid would result in a crude approximation - a square wave. And, inside the digital system, this is true. However, both ADCs and DACs have low-pass filters set at half the sample rate (the highest representable frequency). What this means for input and output is that any frequency above the cut-off point is removed and it follows from this that the crude sine representation - a square wave in theory - becomes filtered down to a single frequency (i.e. a sine wave). From this, we have two mathematical results:
$F_{s} > 2 f_{max}$ and $F_{N} = \frac{F_{s}}{2}$
Where $F_{s}$ is the sample rate, $f_{max}$ is the highest frequency in the signal. $F_{N}$ is the Nyquist frequency. Frequencies over the Nyquist frequency are normally blocked by filters before conversion to the digital domain when recording; without such processes there would be frequency component foldover, otherwise known as aliasing.
### Sampling accuracy and bit depth
It has been established that the higher the sample rate, the more accurate the representation of a waveform in a digital system. However, although there are many reasons and arguments for higher sample rates, there are two general standards: 44100 samples per second and 48000 samples per second, with the former being most commonplace. The main consideration for this is the fact that the human hearing range extends, at maximum, to an approximate limit (that varies from person to person) of 20000 Hz. Frequencies above this are inaudible. Considering the example of 44.1 kHz, we find that the Nyquist frequency evaluates to 22050 Hz, which is more than the human hearing system is capable of perceiving. There are other reasons for this particular sample rate, but that is beyond the scope of this book.
Figure 2.3. Effects of increased sample rate and bit depth on representing a continuous analogue signal.
There is one more important factor to consider when considering the sampling process: bit depth. Bit depth represents the precision with which the amplitude is measured. In the same way that there are a limited amount of samples per second in a conversion process, there are also a limited amount of amplitude values for a sample point, and the greater the number, the greater the accuracy. A common bit resolution found in most standard digital audio systems (Hi-Fi, Compact Disc) is 16 binary bits which allows for a range of 65536 ($2^{16}$) individual amplitude values at a point in time. Lower bit values result in a greater distortion of the sound - a two bit system ($2^2$) only allows for four different amplitudes, which results in a massively inaccurate approximation of the input signal.
↑Jump back a section
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http://math.stackexchange.com/questions/tagged/game-theory+decision-theory
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http://math.stackexchange.com/questions/15710/why-is-it-that-mathbbq-cannot-be-homeomorphic-to-any-complete-metric-spac
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# Why is it that $\mathbb{Q}$ cannot be homeomorphic to _any_ complete metric space?
Why is it that $\mathbb{Q}$ cannot be homeomorphic to any complete metric space?
Certainly $\mathbb{Q}$ is not a complete metric space. But completeness is not a topological invariant, so why is the above statement true?
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Don't forget to accept the answer you find most useful (if there is one); click on the checkmark on the left of the answer. If you think none of the answers are enough, then you might want to ask further questions so they can be clarified. – Arturo Magidin Dec 29 '10 at 22:01
## 5 Answers
By the Baire Category Theorem, a space that is homeomorphic to a complete metric space must be a Baire Space: the intersection of a countable family of open dense sets must be dense. But $\mathbb{Q}$ does not have this property, because it is countable and no point is isolated: for each $q\in\mathbb{Q}$, let $\mathscr{O}_q = \mathbb{Q}\setminus\{q\}$. This is open (since $\{q\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{Q}$) and dense, since every open ball with center in $\mathbb{q}$ intersects $\mathscr{O}_q$. But $$\bigcap_{q\in\mathbb{Q}} \mathscr{O}_q = \emptyset$$ is not dense. Hence $\mathbb{Q}$ is not a Baire space, hence cannot be homeomorphic to a complete metric space (or even to an open subset of a complete pseudometric space).
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I had an intuitive argument using the Baire Category Theorem; thank you for writing it out explicitly. – Tyler Dec 28 '10 at 3:34
Thank you. Any hint how to show Baire is a topological property? – user Dec 28 '10 at 3:50
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@user: What exactly are you having trouble with in proving that being a Baire space is invariant under homeomorphisms? Being an open subset and being a dense subset are both invariant under homeomorphisms, and since homeomorphisms are bijections, the image of the intersection is the intersection of the images. – Arturo Magidin Dec 28 '10 at 3:53
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Note that just countable is not enough: you need every singleton to be nowhere dense. If a space has isolated points, it can be complete, like $\mathbf{N}$ or a convergent sequence. – Henno Brandsma Dec 28 '10 at 4:14
@Henno: Yes, good point; I did not express that very well. For the particular choice I made I need the complement of every singleton to be open and dense, and for the space to be countable. – Arturo Magidin Dec 28 '10 at 4:56
A complete metric space is not a countable union of nowhere dense sets, but $\mathbb{Q}$ is (each singleton is nowhere dense). Since nowhere denseness is a topological property, $\mathbb{Q}$ cannot be homeomorphic to a complete metric space.
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There is another way to test whether a metric space is homeomorphic to a complete metric space (that is, whether the space is completely metrizable: there is a complete metric that induces the same topology). This is known as the Choquet game or the strong Choquet game. It was introduced by Choquet in his Lectures on analysis in 1969; a modern treatment can be found in Classical descriptive set theory by Kechris.
The idea is that each metric space is associated with a two-player game of perfect information, of countable length, such that the metric space is homeomorphic to a complete separable metric space if and only if the second player has a winning strategy in the game. All the details and proofs can be found in Kechris' book.
It is easy to show, once you have all the definitions, that the first player has a winning strategy on this game when it is played on $\mathbb{Q}$, which means the second player does not have a winning strategy, so $\mathbb{Q}$ is not completely metrizable. This gives an alternate way to prove that the rationals are not $G_\delta$ in the reals (cf. the answer by Jonas Meyer). First we prove that the rationals are not completely metrizable, by Choquet's method. Then we prove that a metric space is completely metrizable if and only if it is $G_\delta$ in its completion (actually, this is used in the proof of Choquet's result, so we had to prove it already). Since the completion of $\mathbb{Q}$ is $\mathbb{R}$, this shows that $\mathbb{Q}$ is not $G_\delta$ in $\mathbb{R}$.
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A subspace of a separable complete metric space is homeomorphic to a complete metric space if and only if it is a $G_\delta$. (I have a reference that says the direction relevant to your question is proved on page 197 of Bourbaki's General topology, volume 2, but I don't currently have access to that page. The other direction is proved on page 196.) The set of rational numbers is not a $G_\delta$ in the real numbers. I must admit that the only way I know how to prove that $\mathbb{Q}$ is not a $G_\delta$ is to appeal to Baire's theorem. (Every open set containing the rationals is dense, and the set of irrationals is a countable intersection of dense open sets, so if the set of rationals were a $G_\delta$, then the empty set (obtained by intersecting the sets of rational and irrational numbers) would be a countable intersection of dense open sets, contradicting Baire's theorem.)
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Separability isn't required, although the fact that $G_\delta$ means a countable intersection of open sets is required. – Carl Mummert Dec 28 '10 at 16:11
@Carl: Thanks. I actually don't know how to prove that if a subspace of a completely metrizable space is completely metrizable then it is a $G_\delta$, and I included the separability hypothesis because I had only seen the statement for the separable case (Polish spaces), and that applies here. I better try to prove it or look it up. – Jonas Meyer Dec 29 '10 at 4:57
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The Polish case is definitely the more common one. There is a proof of the general result in Kechris' Classical descriptive set theory (Theorem 3.11). It goes via an extension lemma attributed to Kuratowski: a continuous function from a subset $A$ of a complete metric space into a complete metric space can be extended to a continuous function on a $G_\delta$ set $G \supseteq A$. – Carl Mummert Dec 29 '10 at 12:12
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I'm very late to the party here, but as I stumbled over this... I gave an argument for Kuratowski's characterization for completely metrizable subspaces in this answer here via Lavrentiev's theorem (the extension lemma mentioned by Carl). It is quite similar to what Kechris does but expressed in somewhat different language. I'm not aware of a substantially different argument. – t.b. Aug 15 '11 at 18:32
There is a MISTAKE in the "proof" below the the line. That was the original post. See the discussion that follows. I'm leaving it here as it may be instructive for someone, like it was for me. I'm happy to delete this post if that is considered the best course of action in this community. Please, let me know. Thanks to Asaf Karagila and Nate Eldredge for pointing out the mistake.
My mistake was when I wrote "and more importantly, that $\lim z_{m}=0$", because, for example the numbers $d\left(h(y_{m+p}),h(y_{m}\right)$ may be constant (for all positive integers $p$). Counterexamples have already been sketched in the comments below (thanks, Asaf, Nate). Let me provide a counterexample in more detail: in my setup, if $Y=(−\pi/2,\pi/2)$, $X=ℝ$ and $h(y) = \tan y$, then setting $x_{n} = n$ and $y_{n} = h^{−1}(x_{n})$, we obtain $y_{n}$ that increases monotonically towards $\pi/2$, and is thus Cauchy. But $x_{n}$ is not Cauchy, by construction.
If $h$ were an isometry (as pointed out by Asaf) the "proof" below would be correct. But isometry is more than needed; $h$ being Lipschitz-continuous would be sufficient. In the counter-example above, it is easy to see that $h$ is not Lipschitz (with the problem happening near $\pi/2$ and $-\pi/2$)
This proof is more elementary than the ones presented so far (unless I overlooked something).
Let $X$ and $Y$ be metric spaces with distance functions $d$ and $e$, respectively, and let $h: Y \to X$ be a homeomorphism. Suppose $X$ is complete. We will now show that $Y$ is complete, and hence $Y \neq \mathbb{Q}$.
The idea is simple. Pick an arbitrary Cauchy sequence $y_{n}$ in $Y$. Then the sequence $x_{n} = h(y_{n})$ ought to be Cauchy. Now, this image lives in $X$, and thus it converges to some $x \in X$. It must be the case then that $h^{-1}(x)$ is the limit of $(y_{n})$. I strongly recommend you try to prove this without reading the proof below. The key fact here is that the image of a Cauchy sequence by a continuous function is a bounded set. Nothing fancy.
${\bf Claim:}$ Every Cauchy sequence in $Y$ converges to a point in $Y$.
${\bf\it Proof.}$ Let $(y_{n})$ be an arbitrary Cauchy sequence in $Y$. For every positive integer $n$, define $x_{n} = h(y_{n})$. For all $m \in \mathbb{N}$ define
$$z_{m} = \sup \{ d(h(y_{m+p}), h(y_{m})) : p \in \mathbb{N} \}.$$
Because $d$ is a distance function, $z_{m}$ is always nonnegative. As $d$ and $h$ are both continuous and $(y_{n})$ is Cauchy, it follows that for all $m$, $z_{m} < + \infty$, and, more importantly, that $\lim z_m = 0$. (If it's not obvious, try to write down a detailed proof of this). It follows that the sequence $(x_{n})$ is Cauchy.
Because $X$ is complete and the Cauchy sequence $(x_{n})$ is in $X$, it must be then that there exists $x \in X$ such that $x_{n} \to x$. Define $y = h^{-1}(x)$. Then, for all $n \in \mathbb{N}$,
$$d(y_{n}, y) = d(h^{-1}(x_n), h^{-1}(x))$$
and again, as $n \to \infty$, the right-hand side above goes to zero because $x_n \to x$ and $h^{-1}$ is continuous. It follows that $y_{n} \to y$. The proof is completed by observing that we chose the Cauchy sequence $(y_{n})$ arbitrarily.
$\square$
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Homeomorphism is not necessarily an isometry. The irrationals are homeomorphic to a complete metric space; but certainly there are Cauchy sequences of irrationals which converge to a rational number. – Asaf Karagila Oct 23 '12 at 2:13
> Homeomorphism is not necessarily an isometry. – Guilherme Freitas Oct 23 '12 at 2:36
Yes. That is what I said. You, on the other hand, wrote in your answer that a Cauchy sequence is ought to be mapped to a Cauchy sequence, which is false. Furthermore, the definition of a complete metric space is a metric space where every Cauchy sequence converges. So I'm not sure what you are trying to prove in that claim. – Asaf Karagila Oct 23 '12 at 2:45
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A more obvious counterexample is that $(0,1)$, which is not complete in its usual metric, is homemorphic to $\mathbb{R}$ with its usual metric, which is. – Nate Eldredge Oct 23 '12 at 3:04
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Agreed. Bit by the "yeah... this should be obvious". My mistake was when I wrote "and more importantly, that $\lim z_{m} = 0$", because, for example the numbers $d(h(y_{m+p}), h(y_{m})$ may be constant (for all positive integers $p$). Counterexamples have already been mentioned (thanks, Asaf, Nate). More explicitly: in my setup, if $Y = (-1, 1)$, $X = \mathbb{R}$ and $h^{-1}(x) = \frac{x}{1+|x|}$, then setting $x_{n} = n$ and $y_{n} = h^{-1}(x_{n})$, we obtain $y_n$ that increases monotonically towards 1, and is thus Cauchy. But $x_{n}$ is not Cauchy, by construction. Thanks, guys. – Guilherme Freitas Oct 23 '12 at 3:26
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http://math.stackexchange.com/questions/56022/measurability-of-the-inverse-of-a-measurable-function/56061
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# Measurability of the inverse of a measurable function
Let $m$ denote Lebesgue measure, and let $f:[0,1] \to [0,1]$ be a (Lebesgue) measurable and bijective function. In general, it is not true that $f^{-1}$ is measurable. However, suppose that we now have the condition that $\forall A \subset [0,1]$, $m(A) = 0 \Rightarrow m(f(A)) = 0$. Why does this condition guarantee the measurability of $f^{-1}$?
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Since f is a bijection, $(f^{-1})^{-1}(A)=f(A)$. So you need to take A open in [0,1]-domain, and show that f(A) is measurable in [0,1]-codomain. But if A is open, it is the countable union of disjoint open intervals... – gary Aug 6 '11 at 22:39
Sorry, I don't see where in my suggestion we're using the condition that f preserves sets of measure 0; let me think it thru; I think you can approximate your measurable setO by the union $\cup O_i$ of open intervals, so that $m(O-\cup O_i)$=$\varepsilon$. Then play around with $(f^{-1})^{-1}=f$ to send sets (the non-zero measure part) back-and-forth. – gary Aug 6 '11 at 23:07
## 1 Answer
The analogous fact about Borel functions is true without additional hypotheses: if $f: [0,1] \to [0,1]$ is a Borel bijection then its inverse is also Borel. This follows from the observation that images of Borel sets under injective Borel functions remain Borel (this is far from obvious, and takes some descriptive set theory to prove -- see, e.g., 15.A of Kechris' Classical Descriptive Set Theory).
[Caveat: Injectivity of the Borel function in the previous observation is essential! You can generalize slightly to Borel functions which are countable-to-one, but it's simply false for general Borel functions.]
Armed with this observation, we can also understand the result for measurable bijections. Suppose that $f:[0,1] \to [0,1]$ is a measurable bijection which sends measure $0$ sets to measure $0$ sets. Fix a measure $1$ Borel set $A \subseteq [0,1]$ such that the restriction $f|_A : A \to [0,1]$ is Borel (in other words, for all Borel $Y \subseteq [0,1]$, the set $(f|_A)^{-1}(Y) = f^{-1}(Y) \cap A$ is Borel, not just Lebesgue measurable).
Edit: see below for more information about building such a set $A$.
To show that $f^{-1}$ is measurable, it suffices to show that $f(B)$ is measurable for all Borel $B \subseteq [0,1]$. We see $$f(B) = f(B \cap A) \cup f(B \setminus A) = f|_A(B \cap A) \cup f(B \setminus A).$$ Now $f|_A(B \cap A)$ is the image of a Borel set under an injective Borel function, thus Borel. Also, $f(B \setminus A)$ is the image of a measure $0$ set (as $A$ had full measure), and is thus measure $0$ by hypothesis. That means we've written $f(B)$ as the union of a Borel set with a measure $0$ set, establishing its measurability.
It's easy to see how this argument falls apart if $f$ no longer sends measure $0$ sets to measure $0$ sets. The behavior of $f$ on $[0,1] \setminus A$ can be horribly pathological, making the $f(B\setminus A)$ part of the above expression some nonmeasurable set.
For the sake of clarity, here's an explanation of why there's a measure $1$ Borel set $A \subseteq [0,1]$ such that $f|_A : A \to [0,1]$ is Borel. Let $(O_n)$ be an enumeration of a countable base for the topology on $[0,1]$ (for example, intervals with rational endpoints). For each $n$, let $A_n \subseteq [0,1]$ be a measure $1$ Borel set such that $A_n \cap f^{-1}(O_n)$ is Borel (these exist by standard tightness of measure arguments). Then $A = \bigcap_n A_n$ is again a measure $1$ Borel set, which we claim works.
First, note that for each $m$, $A \cap f^{-1}(O_m) = (\bigcap_n A_n) \cap (A_m \cap f^{-1}(O_m))$, which is Borel. This implies that $A \cap f^{-1}(B)$ is Borel for all $B$ in the $\sigma$-algebra generated by $\{O_n\}$, which means that $f|_A$ is Borel.
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1
Well done, ccc. – gary Aug 7 '11 at 5:45
@gary: Thanks! I wonder whether there's a proof of this that avoids using significant descriptive set theory. – ccc Aug 7 '11 at 15:06
@ccc: Thanks for the help! Sorry if this is dumb, but can you explain what it means to fix a Borel set $A$ s.t. $f|_A$ is Borel? If $f$ is a Lebesgue measurable function, then it will still be defined on Lebesgue subsets of $A$ for any $A$ won't it? I think I understand the overall argument that you made though. Also, would you happen to have a reference for the fact that the images of Borel sets under injective Borel functions are Borel? – user1736 Aug 8 '11 at 4:46
@user1736: Don't apologize for my imprecisely written answer! I've added some more details, which I hope answer your questions. – ccc Aug 8 '11 at 15:30
@ccc: Hm, I think I get the new defintion that you're using now, but is it obvious that you can find such an $A$ that restricts $f$ in such a way that $f^{-1}(Y) \cap A$ is always Borel? – user1736 Aug 8 '11 at 20:21
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http://mathhelpforum.com/trigonometry/86338-trig-transformation.html
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# Thread:
1. ## trig transformation
hello, i need help with this particular transformation, I was just wondering if my answer is correct
the transformation is this f(x) = 3 cos 2 (x - π/4) + 1 it asks me to find the period, amplitude, phase shift, which I have as π, 3 and π/4 to the right respectively. It then goes on to ask me to find the first five critical points and this is were I'm slightly confused, i have them as; (π/4, 2), (4π/4, 1), (8π/4, 2), (12π/4, 1), (16π/4, 2). My intuition tells me that these are horribly horribly horribly wrong. Could someone please enlighten me on what I am doing wrong here? I just really want a nudge in the right direction. Thank you very much.
2. Hello nikita007
Welcome to Math Help Forum!
Originally Posted by nikita007
hello, i need help with this particular transformation, I was just wondering if my answer is correct
the transformation is this f(x) = 3 cos 2 (x - π/4) + 1 it asks me to find the period, amplitude, phase shift, which I have as π, 3 and π/4 to the right respectively.
I agree!
It then goes on to ask me to find the first five critical points and this is were I'm slightly confused, i have them as; (π/4, 2), (4π/4, 1), (8π/4, 2), (12π/4, 1), (16π/4, 2). My intuition tells me that these are horribly horribly horribly wrong. Could someone please enlighten me on what I am doing wrong here? I just really want a nudge in the right direction. Thank you very much.
The critical points are where the graph of the function has zero gradient. In the case of sine and cosine functions, these will be where the sine and cosine have their maximum and minimum values of +1 and -1 respectively.
In the function $f(x) = 3\cos2(x-\tfrac{\pi}{4}) + 1$, this will be where:
$\cos2(x-\tfrac{\pi}{4}) = \pm 1$ and $f(x) =\pm 3 + 1 = 4\text{ or }-2$.
Now you know that $\cos\theta = \pm 1$ when $\theta = 0, \pi, 2\pi, ...$ . So work from there, starting with:
$2(x-\tfrac{\pi}{4}) = 0 \Rightarrow x = \tfrac{\pi}{4} \Rightarrow f(x)= 4$
Then $2(x-\tfrac{\pi}{4}) = \pi \Rightarrow x = \tfrac{3\pi}{4}$ and $f(x)= -2$
... and so on.
Grandad
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http://math.stackexchange.com/questions/260436/equation-to-determine-radius-for-a-circle-that-should-intersect-a-given-point
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# Equation to determine radius for a circle that should intersect a given point?
Simple question. I tried Google but I don't know what search keywords to use.
I have two points on a 2d plane. Point 1 = x1 and y1, and Point 2 = x2 and y2.
I'd like to draw a circle around Point 1, and the radius of the circle should be so that it intersects exactly with Point 2.
What is the equation to determine the required radius?
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## 3 Answers
Let's call the center of the circle: $P_1 = (x_1, y_1).\;$
Let $P_2 = (x_2, y_2)$ be a point on circle. Then:
$r$: radius of the circle = distance between points $P_1$ and $P_2$, where
$$r = \sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}$$
Any point $(x_i, y_i)$ satisfying the equation $(x_i - x_1)^2 + (y_i - y_1)^2 = r^2$ also lies on this circle.
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The radius is simply the distance between the two points. So use the standard Euclidean distance which you should have learned.
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If $P_1(x_1,y_1)$ is the center, the radius will be $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
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http://stats.stackexchange.com/questions/tagged/finite-population
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# Tagged Questions
The finite-population tag has no wiki summary.
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http://math.stackexchange.com/questions/47969/determing-x-y-increase-by-angle
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# determing x,y increase by angle?
If I had a position of 0,0 and I had an angle of 45 degrees (or any number) and my velocity was 1 what would be the x,y increase?
For example if I had a 90 degree angle, and I had a velocity of 1. Then because my angle is straight up, my increase would be 1y and 0x.
Not sure if this makes sense, I am programming a small game, that shoots on any angle, and I need to pass it x and y coordinates so it shoots on the same angle.
Thanks!
Just need the equation, or a tutorial
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## 1 Answer
From $(0,0)$ at angle $\theta$ at velocity $1$ you get to $x=\cos\theta$, $y=\sin\theta$. You know trigonometry?
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used to! so if I had an angle of 45 it would just be x=cos45 and y=sin45 seems pretty straight forward. So x=0.707 and y = 1 – Johnny Quest Jun 27 '11 at 12:52
@Johnny, yes. – Gerry Myerson Jun 27 '11 at 12:55
1
@Johnny - $x$ looks fine by $y$ does not, as you should find $\cos 45 = \sin 45 = 1/\sqrt{2} \approx 0.707$ – Henry Jun 27 '11 at 13:05
oops, you are right, I did tan. is Y always going to be sin? If so why is that? – Johnny Quest Jun 27 '11 at 13:10
1
It is the definition of the sine as opposite/hypotenuse. Since your angle is measured from the $X$ axis, $v_x=v \cos(\theta), v_y=v \sin (\theta)$ – Ross Millikan Jun 27 '11 at 13:13
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http://mathhelpforum.com/discrete-math/83809-n-1-prove-number-even-perm-n-equals-number-odd-perm-n.html
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# Thread:
1. ## n > 1, prove the number of even perm. of [n] equals the number of odd perm. of [n]
For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].
2. Originally Posted by qtpipi
For n > 1, prove that the number of even permutations of [n] equals the number of odd permutations of [n].
Let $\pi$ be a permutation of [n]. Define another permutation $\sigma$ by switching $\pi(1)$ and $\pi(2)$; i.e., define
$\sigma(1) = \pi(2)$
$\sigma(2) = \pi(1)$
$\sigma(j) = \pi(j)$ for $j > 2$.
Since $\pi$ and $\sigma$ differ only by a transposition, they have opposite parity, i.e. one is odd and one is even. So we have established a bijection between the even and odd permutations, and there are equal numbers of each.
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http://physics.stackexchange.com/questions/15841/can-someone-explain-the-color-pink-to-me?answertab=active
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# Can someone explain the color Pink to me?
I just finished watching this interesting video:
http://youtu.be/S9dqJRyk0YM
It does a very quick explanation of how pink light doesn't exist, and that the concept of pink is our brain's attempt at filling in a gap in the spectrum.
My question is, how does our brain perceive this gap? What frequency is this pink color, if it truly doesn't exist? Is it a frequency greater than violet? Or is it greater than red?
Sorry if this question seems trivial.
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6
The point is that some colors that we see don't correspond to frequencies, but to mixtures of frequencies. Brown isn't a single frequency, either! Where is the brown part of a rainbow? – Greg P Oct 17 '11 at 16:52
## 4 Answers
Your eye has three types of receptor cells, the sensitivity of each type peaking in different spectral regions. Roughly speaking, there's one that peaks in red, one blue, one green. (It's not quite so clear cut, but your brain is really good at sorting out messes like this!) When you look at a fire engine (assuming it's red) It's mostly the red receptors that are firing. If your fire engine is white (as they are in New Haven, CT), all three receptors are firing. For pink fire engines (Yes, they exist! I saw two last month! If you don't believe me, ask Google.) the red receptor is firing full-on, but the green and blue are on too, but not full-on. Maybe 80% on. Almost white, but biased toward red. Your brain takes that in and calls it pink.
BTW black is no stimulation of any of the receptors. Brown corresponds to very little stimulation of any of the three types, but of what there is, the reds are firing more than the greens or blues. So brown is close to black, but biased toward red.
But it's far from that simple: your brain takes into account the adjacent color before deciding what to labeling the color of the object you are looking at. It is not simply a matter of mixing frequencies!!
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Not all colors are monochromatic (pure). The spectrum of colors is really only a spectrum of monochromatic colors, and that's why you can represent it on a line.
For non-monchromatic colors, you'd represent them with a gamut instead of a spectrum:
Here you'll see pink at x=0.45,y=0.3. The monochromatic colors are along the edge, i.e. the edge is the spectrum. The blue numbers 380-700 nanometers are the wavelengths of the monchromatic colors.
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The 'real' variations of all colours you can use on for example on a computer screen can't be envisioned in any 2D object - there are three primary colours, and so you in general need a 3D object to see them all. That's why the gamut above does not look like it has enough colours in it to satisfy a designer or artist. Most are missing. – Tom Andersen Oct 19 '11 at 13:52
@Tom: It's a slice at constant intensity, which is why e.g. black is missing. However, it does show all hues. – MSalters Oct 20 '11 at 7:51
In the context of this question it is worth while to read a bit about Edwin H.Land the inventor of the Polaroid camera which was the only way to take instant pictures until the electronic revolution.
He really delved into the differences of frequency versus perception by the human eye.
from the link:Also in this decade (nb 1950s), Land first discovered a two-color system for projecting the entire spectrum of hues with only two colors of projecting light (he later found more specifically that one could achieve the same effect using very narrow bands of 500 nm and 557 nm light).
I have seen photos taken with two frequencies but cannot unearth them through google.
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– Yotam Oct 18 '11 at 8:09
No, it is not technicolor which is about dyes.It was illumination with just two frequencies and the color photograph came up with full colors, a bit muted though. – anna v Oct 18 '11 at 16:37
As far as I could understand the video, we don't see the color pink, what we see is white light without the green part. This is so because we can see green light in two receptors in our eyes. The first one place the light we see on the red-green scale, and the second place it on the blue-yellow scale. In this system we measure the green twice (which pose an interesting question of evolutionary advantage).
There are two things which bother me in the video you posted; 1. The video claims that pink is the mix of red and blue light. This is not the same as claiming that we don't see the green light (explanation below). 2. It might be understood from the video that pink is our understanding of the entire non-visible part of the electromagnetic spectrum which is completely wrong. We simply can't see that part of the light.
I have found this tool which allows you to combine colors and see the result. Notice that when you only mix blue and red you get magenta (mentioned in the video), to get a true, hello kitty style of pink you need to add some green. Magenta is a man-made color which is a single type of light. Pink (and red, blue and what ever natural color'' we want) is a range of colors and not a single mix.
One last remark. Playing with that tool, I have notice that we can do the same trick with cyan (and blueish color) and yellow (and yellowish colors). I think that what is special in the pink case is only that it is such a common and cute'' color which is surprising that this is what we actually see.
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– MSalters Oct 18 '11 at 9:16
You are right, what I should have explained is that our mind place each light we see on the appropriate position in it's red-green range and the blue-yellow range and this is why this are two kinds of color blind. The cell class I took was so long ago... – Yotam Oct 18 '11 at 11:58
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http://www.physicsforums.com/showthread.php?p=4238454
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Physics Forums
seeking derivation of real scalar field Lagrangian
Here and there I come across the following formula for the Lagrangian density of a real scalar field, but not a deriviation.
[tex]
\mathcal{L} = \frac {1}{2} [ \dot \phi ^2 - ( \nabla \phi ) ^2 - (m \phi )^2 ]
[/tex]
Could someone show me where this comes from? The m squared term in particular is a mystery.
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The first two terms look like kinetic and potential energy, but the third one . . all that comes to mind is the mass-energy relationship of special relativity, but I thought this formula was from classical mechanics.
In what context are you studying this? Most often this comes up at the beginning of a quantum field theory course. The Lagrangian is more a definition than anything; when we write down the Lagrangian we are saying, "Let's imagine a field whose dynamics are described by this Lagrangian. How does it behave?" And we can write down any Lagrangian we want. However, there are some constraints we tend to impose on what Lagrangians we write down and study. One is Lorentz invariance: the Lagrangian should describe a field whose dynamics are consistent with special relativity. This limits what terms can appear in the Lagrangian, but doesn't determine it fully. If we ask for the simplest Lorentz invariant Lagrangian for a scalar field, we get the one you wrote down above. This is why this Lagrangian comes up often. m is a free parameter. In quantum field theory, a scalar field with this Lagrangian has excitations which turn out to be particles of mass m.
seeking derivation of real scalar field Lagrangian
Yes, I bumped into it studying quantum field theory. It's supposed to have the same form as the Klein-Gordon Lagrangian, yet come about without any quantum assumptions.
I assumed that there was some physical situation that corresponds to it, as
[tex]
L = \frac {1}{2}m \dot {x}^2 - \frac {1}{2}kx^2
[/tex]
corresponds to a mass on a spring obeying Hooke's law.
why not. the first two terms form a kinetic energy part 1/2(∂ψ/∂t)2-1/2(∇ψ)2 can be written as 1/2(∂μψ)2 which corresponds to kinetic energy and rest is of course some potential energy of spring type thing 1/2(mψ)2.
Oh I see — the first two terms are analogous to the Minkowski displacement vector with magnitude $ds^2 = (c dt) ^ 2 - dx ^2$ and the third term is the potential. Yes, that makes some sense. Thanks, Andrien! :)
That's funny, I just noticed that it doesn't have the same form as the Klein-Gordon Lagrangian at all. One has second derivatives and the other second powers. Now I'm wondering where I read that . . Edit: Oops, one's a Lagrangian and one isn't. Moderators, feel free to delete this entry #7.
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http://www.physicsforums.com/showthread.php?p=3781680
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Physics Forums
## Can the existence of nonstandard hyperreal extensions be proved?
I have begun to read about the hyperreals, and am wondering whether the natural extensions of real-valued functions to hyperreal-valued functions is simply a definition of the hyperreals, or can it be proved? Or is it accepted as an axiom?
For example, if f(x) = sin(x), then is the existence of f*(x) = sin*(x) provable, or is it assumed? (f*(x) and sin*(x) represent the natural extensions.)
Thanks!
BiP
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Recognitions: Gold Member Science Advisor Staff Emeritus Well, there are always a variety of ways to about defining things. Which theorems are axioms* and which are not isn't really important, since you can always shuffle the presentation of the theory around to change things. General, nonstandard models of real analysis are usually defined by elementary equivalence: a superstructure** a non-standard model if and only if the transfer principle works. (more or less, anyways) So, ${}^\star\!\!\sin$ exists because of that. The elements of the most commonly used hyperreal field can be explicitly named by sequences of real numbers. Then, ${}^\star\!\!\sin$ has an explicit formula: you just take the sine of each term in the sequence. (aside: to pre-empt any misunderstanding, while the names are explicit, the same hyperreal number has many different such names. You need an ultrafilter to tell whether two names denote the same hyperreal number. The ultrafilter can't be explicitly constructed in ZF -- its existence is usually derived from the axiom of choice) *: If P is an axiom, then P is also a theorem, by the trivial proof: "because P is true, P is true". **: Don't worry about the technical meaning of this word, it's not really important.
Quote by Hurkyl Well, there are always a variety of ways to about defining things. Which theorems are axioms* and which are not isn't really important, since you can always shuffle the presentation of the theory around to change things. General, nonstandard models of real analysis are usually defined by elementary equivalence: a superstructure** a non-standard model if and only if the transfer principle works. (more or less, anyways) So, ${}^\star\!\!\sin$ exists because of that. The elements of the most commonly used hyperreal field can be explicitly named by sequences of real numbers. Then, ${}^\star\!\!\sin$ has an explicit formula: you just take the sine of each term in the sequence. (aside: to pre-empt any misunderstanding, while the names are explicit, the same hyperreal number has many different such names. You need an ultrafilter to tell whether two names denote the same hyperreal number. The ultrafilter can't be explicitly constructed in ZF -- its existence is usually derived from the axiom of choice) *: If P is an axiom, then P is also a theorem, by the trivial proof: "because P is true, P is true". **: Don't worry about the technical meaning of this word, it's not really important.
Thank you! Is non-standard analysis helpful in any way? Could I prove the theorems of calculus without having to define the hyperreal number line? Or are they simply used as a convenience? I don't want to expose myself to anything I don't need... but I do need my rigor.
BiP
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## Can the existence of nonstandard hyperreal extensions be proved?
Quote by Bipolarity Thank you! Is non-standard analysis helpful in any way?
some find it "more intuitive".
Could I prove the theorems of calculus without having to define the hyperreal number line?
yes.
Or are they simply used as a convenience? I don't want to expose myself to anything I don't need... but I do need my rigor. BiP
if one is working with the hyperreals, certain statements in calculus become more "direct", for example:
f'(x) = st((f(x+dx)-f(x))/dx)
and there is no mention of limits (and thus no epsilons or deltas to quantify over).
it is (for some) easier to understand "dx" as a very small (hyperreal) number (an infinitesimal, close to 0, but not equal to it), then to deal with the limiting process of something which appears to approach 0/0, a non-sensical number. because no formal system of hyperreals existed when calculus was invented, concerns about the rigorousness of calculus persisted until the 19th century, when formal definitions of the real numbers, limits and continuity were given that alleviated these concerns.
it was even later (around 1966 i think) that the original methodolgies of Newton and Liebniz were shown to be logically consistent (that is, a suitable definition of infinitesimal given) by Abraham Robinson.
at this point in time, both standard, and non-standard analysis are considered "rigorous". it does not appear that either approach can prove something the other cannot. that said, most courses offered in advanced analysis will nevertheless take the "standard" approach.
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Quote by Bipolarity Thank you! Is non-standard analysis helpful in any way?
It allows you to stream-line a number of various arguments and concepts -- e.g. in an argument where one considers things that are "sufficiently small" or "sufficiently large", one can often replace them with "infinitesimal" and "transfinite" which are often easier to work with.
Could I prove the theorems of calculus without having to define the hyperreal number line?
Yes, and no. Any theorem about standard analysis that can be proven by non-standard methods can be proven using standard methods. But as a practical matter, we have this quote from mathworld:
Crucially, however, the angle at which the nonstandard analyst looks at the axioms of analysis provides for an average case reduction in complexity that provides shorter proofs of various results, and will one day lead to the proof of a result which is not accessible to classical mathematics without nonstandard methods, precisely because its classical proof is too long to write down in the length of time humans will reside on Earth.
Quote by Deveno if one is working with the hyperreals, certain statements in calculus become more "direct", for example: f'(x) = st((f(x+dx)-f(x))/dx)
This deserves a bit of nitpicking, since I see a lot of people get this wrong. If the derivative is defined (and dx is a nonzero infinitesimal), then the equation is true. However, the derivative is only defined if the right-hand side has the same value for all choices of dx.
For the simplest stuff, the difference between standard and non-standard methods is usually quite superficial; they generally have the same issues and concerns and you can translate directly back and forth. The winnings of non-standard analysis don't really show up until you start putting things together to make more complex arguments or start treating more sophisticated notions.
Recognitions: Science Advisor Just curious: What kind of ultrafilter do we use to mod out the ultraproduct? I guess we want to identify some elements that are close to each other, but I am not sure how. Also: is the transfer principle the same as elementary equivalence?
Quote by Bipolarity I have begun to read about the hyperreals, and am wondering whether the natural extensions of real-valued functions to hyperreal-valued functions is simply a definition of the hyperreals, or can it be proved? Or is it accepted as an axiom? For example, if f(x) = sin(x), then is the existence of f*(x) = sin*(x) provable, or is it assumed? (f*(x) and sin*(x) represent the natural extensions.) Thanks! BiP
Terence Tao wrote an absolutely clear and brilliant exposition of ultrafilters and nonstandard analysis that develops the subject from the ground up without much background required.
http://terrytao.wordpress.com/2007/0...on-management/
Recognitions: Science Advisor Never mind my first question; I just found out that we only need an ultrafilter over the natural, and that the independence of the choice of U is equivalent to AC.
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http://physics.stackexchange.com/questions/54917/surely-space-time-curvature-does-not-explain-gravity-it-just-describe-its-effec
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# Surely space-time Curvature does not explain gravity, it just describe its effects?
In special relativity co-moving objects see the other's 4-velocity as being only temporal.
When they move relative to each other they see the other's 4-velocity has rotated so that it points less in the temporal direction but now has a spatial component.
By the equivalence principle two co-moving objects falling toward a planet see each other's 4-velocity as only temporal in their own (falling) rest frame so they must think the space between them is attached to their rest frame over time. It thus seems that space has the same rotation of its own 4-vector over time (up to a constant if the objects started with a fixed velocity before falling) But surely space does not fall. Also if space-time curvature causes objects to fall, how? I'd have thought it's just a map of how objects move. not a cause of that motion, but if it does cause falling , how? Space isn't moving so as to push or rotate mater. Surely it's curvature it's just a map of the rotations in (light and) matter's 4-vector? How does something about the mass energy tensor alter geodesics or 4-velocity vectors? I see no explanation of gravity in GR merely a more detailed description of the motions it effects.
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– Peter Shor Feb 24 at 12:29
## 4 Answers
so they must think the space between them is attached to their rest frame over time.
You seem to be mixing up spacetime, the dynamical (changing over time) arena in which everything takes place, whose ontological status we can debate endlessly, with coordinates, which are a completely physically meaningless way of assigning numbers to spacetime. Just because two comoving observers agree on coordinate systems doesn't say anything whatsoever about what spacetime is doing. Two planes can be flying in formation, and just because they agree on a convenient set of numbers to describe their position, doesn't mean the air between them is moving with them.
I see no explanation of gravity in GR merely a more detailed description of the motions it effects.
Well, the whole point of GR is that gravity is not a force. Ignoring speculated quantum gravity theories (none of which remotely works at the moment, by the way), gravity is nothing at all like another force. It is just a result of curved spacetime. It is not clear what kind of further explanation you might want. And really, if you fully describe the physics of how everything moves relative to everything else, what more could you want in a theory?
The best I can say as to "how" this occurs is that mass/energy/momentum/etc. alter the distances between neighboring points, and this changes the "natural" path for things to move along. Suppose you distort the surface of the Earth to bring London close to New York - then there will be short "great circles" (i.e. geodesics) that connect them, where there weren't any before.
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Thanks for the reply. You've told me nothing I don't already know so I must not have expressed my queation well. – charles stewart Feb 24 at 9:55
Imagine you live in a universe governed by extremely simple rules, like Conway's Game of Life, for example. Once you discovered those rules, you might wonder, "Why do cells come alive if they have three living neighbors? Why do they die if they have one? How does that work?" (By "how" here I am referring to "what underlying mechanism makes it work?", which is my interpretation of "how" in the original question.)
In a simulation of the Game of Life that you run on your computer, there is a good answer to this. You can examine the source code, look at the hardware of the computer, and eventually arrive at a complete description of exactly what goes on such that little squares on your computer monitor light up and go off according to the rules.
But we're imagining that these rules are just how the universe works. In that case, there may be no reason at all. Maybe it just does it, full stop.
As humans, though, I think we might find that very hard to accept. There are many cellular automata very similar to the Game of Life, but their behavior is not nearly so rich. Why did we get the one universe with the interesting laws? And how does the universe know to implement those laws without screwing up? Surely there must be some wheels and gears beneath there!
That sort of curiosity is extremely important for physicists, and it has led to a lot of new understanding. Peter Shor pointed out in the comments that wondering about how quantum mechanics works led to quantum information and computation. Famously, Einstein wondered about how electromagnetism worked, leading to understanding relativity. Frequently, a theory of physics doesn't quite feel right to us. That drives curiosity. We demand an answer, and eventually it leads to breakthroughs and new theories.
Physicists have derived great benefit from this approach of taking the pieces that don't feel right or don't feel well-enough explained and using that as a springboard to go deeper, but sometimes it also leads to complete frustration. It turns out that the universe isn't obliged to be the way we want.
If you lived in the Game of Life universe, once you figured out the rule it was following, you could keep asking forever, "Why does it have that rule? How does it implement it?" without getting anywhere. The rule itself is just a short little description. It just says that there's a grid of cells and that they light up and turn off according to a simple pattern, and that's all it says. If there was nothing deeper going on than that, oh well. We wouldn't have to give up trying to find a deeper explanation, but we aren't owed one.
My argument is that real laws of physics are the same. So in General Relativity, we posit that the Einstein equation is true. The theory of General Relativity itself makes no comment on this, just as the theory of the Game of Life makes no comment on why cells with three living neighbors come alive.
So when you ask, "How does something about the mass energy tensor alter geodesics or 4-velocity vectors? I see no explanation of gravity in GR merely a more detailed description of the motions it effects," you are right. GR doesn't say how it does it.
It could be that there's an explanation, but it doesn't seem likely to me that the fundamental problem will go away. For example, suppose someone tells you that gravitation works by sending particles called gravitons around, and gives a detailed description of the theory of gravitons. Couldn't we then ask the same question? How do the gravitons interact with spacetime? We could describe the precise mathematical rules, but fundamentally, this anthropocentric feeling of dissatisfaction would remain. Why those rules for gravitons? If they're derived from some set of appealing principles, why those principles?
Elsewhere in physics, how do wavefunctions know to obey the Schrodinger equation? What forces them to obey that equation rather than doing something else? Nothing. They just do that. It's purely a description of how the wavefunctions behave. The problem is the same, as far as I can see. (You can recast QM in some new formulation, but I don't think this averts the "problem".)
To answer your question as best I understand it, you are right that GR is just a description, nothing more. That may not always be true for GR in particular, but it seems likely to me it will always be true for something. (I can't say for sure, of course, since I don't know what the "something" will be!) It is the nature of theories of physics to be just descriptions. We don't have to accept that as a final word, and our desire to understand more deeply fuels our greatest communal quest for knowledge, but ultimately the universe will do what it will do, and can't be bullied into explaining itself just because things don't feel mechanistic enough for us.
note: This answer is completely rewritten after reading the helpful comments from Qmechanic, Peter Shor, and dmckee. Thank you for your input. This answer is essentially philosophical, so disagreement on it is inevitable, and it represents only my personal opinion.
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I have to disagree here. Physics is very much in the business of "how"; if you know a large enough body of "what" you start to get an idea of "how" and "why". Once you have a "how", then you can have a "what if", and if you correctly predict "what if", then you have more confidence in your "how". – KDN Feb 24 at 3:38
I find your statement incomprehensible. I have no idea what you think the words you're using mean, and I suspect your objection lies in not knowing what I intend the words I'm using to mean. I am not trying to describe how people think about physics, solve problems, and make predictions. I am claiming that the theories themselves are descriptive, and are not, for example, teleological. I attempted to make this clear, but apparently I did not. – Mark Eichenlaub Feb 24 at 5:56
1
Thamks for the replies. I take your how point, Indeed it is long familiar. I know geodesic motion is unforced nevertheless the curvature of spacetime purportedly acts on bodies to produce motion. I an curious how. – charles stewart Feb 24 at 11:58
2
Your answer seems to be supporting the "shut up and calculate" interpretation of quantum mechanics, which I believe kept physicists from discovering quantum information theory and quantum computation for years. You are definitely not speaking for all physicists here. – Peter Shor Feb 24 at 12:24
2
Physics often offers answers to "how" or "why" questions, but those come when we find a more fundamental underling theory. Following a chain of "Why?"s will eventually lead to a answer in the form of "Because that is the observed behavior of the universe." Later on, a new theory could replace that with a more detailed answer, but then you can ask why the new theory and the answer will be "Because the world works that way." At the lowest level physics is descriptive. Finding those more fundamental theories is what we do, but don't kid yourself about there being a final insight there. – dmckee♦ Feb 24 at 15:12
show 3 more comments
Also if space-time curvature causes objects to fall, how?
Geodesic advance in distorted space-time results in coordinate acceleration towards the mass:
I see no explanation of gravity in GR merely a more detailed description of the motions it effects.
Physics is about describing observed effects quantitatively. If you seek "explanations" and "causes" beyond that, you have to look somewhere else.
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Also if space-time curvature causes objects to fall, how? I'd have thought it's just a map of how objects move. not a cause of that motion, but if it does cause falling , how? Space isn't moving so as to push or rotate mater. Surely it's curvature it's just a map of the rotations in (light and) matter's 4-vector?
Maybe a more "timeless" perspective helps: Try to take the viewpoint of the whole 4-dimensional universe, locally $\mathbb{R}\times\mathbb{R}^3$ at once, such that there is nothing changing.
Withing this 4-dimensional manifold, the worldline is a unchanging 1-dimensional lin. Only if you want, it can be parameretized alla $s\mapsto (s,\gamma(s))=p(s)\in \mathbb{R}\times\mathbb{R}^3$. Then $p(s)$ is a point on that line.
Realize that e.g. the computer monitor you look at in this moment gets identified with the computer monitor you look at 10 seconds later. If you do that, then the computer monitor is an object of infinite (huge at least) temporal length in spacetime and the idea of it moving thorugh time merely emerges from the curiosity that you can only percieve one moment in time at once.
In the 4-dimensional view, the curvature of all of spacetime never changes (because only 3-dimensional submanifolds can be described as evolving) and hence general relativity, if solved perfectly for all matter in the region it's valid, gives one and only one curvature of spacetime. Then, given that curved manifold the "pushing" really only describes how you are always only in the present and being able to remember the past and to compare it with now, makes you perceive the notion of change.
Apart from that, I agree with Mark Eichenlaub and view physics as descriptive.
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http://unapologetic.wordpress.com/2007/02/08/sets-and-functions/?like=1&source=post_flair&_wpnonce=d0199d491e
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# The Unapologetic Mathematician
## Sets and functions
I’m soon going to need to really use the notion of a function, and I want to make sure that I lay the groundwork properly. This is also a good place to mention a few things about sets.
For most of my purposes, a naïve concept of sets will suffice. A set is a collection of objects, called the elements of the set. In formal treatments of set theory, the elements are themselves sets. In fact, everything in sight is “really” a set. This sort of foundationalist approach, though, tends to obscure the real structure of mathematical theories, so I’ll avoid talking about formal set theory unless it’s absolutely necessary. What we’ll need from set theory are a few operations on sets.
• Specification: If we have a set $X$ and some statement $p$ that can be unambiguously determined true or false for each element of $X$, there is a set containing exactly those elements of $X$ so that $p$ is true. We write this set as $\{x\in X|p(x)\}$, read “the set of those elements $x$ of $X$ such that $p$ is true of $x$“.
• Intersection: This is actually a special case of specification. For our statement we use “$x$ is an element of the set $Y$“. This gives us the set of all elements in both $X$ and $Y$, and we write this set $X\cap Y$. In practice, we will allow intersections not just of two sets, but of any number of sets — even infinitely many.
• Union: For any two sets $X$ and $Y$ there is a set containing any element in either $X$ or $Y$. We write this set as $X\cup Y$. As for intersection, we will allow unions of any collection of sets.
• Cartesian product: For any two sets $X$ and $Y$ there is the set of ordered pairs $(x,y)$, where $x$ is an element of $X$ and $y$ is an element of $Y$. We write this set $X\times Y$. Again, we allow Cartesian products of any number of sets, though only a finite number at a time here.
• Empty set: While not a “construction”, per se, the empty set is something important to pay attention to. This is, predictably enough, the set containing no elements at all. We write it $\varnothing$.
• Subset: If every element of $Y$ is also an element of $X$ we say $Y$ is a subset of $X$ — written $Y\subseteq X$. Note that specification gives us a subset of $X$.
• Power set: For any set $X$ we have the set of all subsets of$X$. We write this set ${\rm P}(X)$.
I may have forgotten some, but I will mention those (and add them here) if I realize it later.
Anyhow, a function is basically a rule that assigns to each element of one set an element of another. Formally, we need to specify a “domain” set $X$ and a “codomain” set $Y$ (the codomain is often called the “range”). For every element $x$ of the domain, there is a uniquely specified element $f(x)$ in the codomain. Often there is some sort of calculational method to determine the value of the function, but a simple lookup table will suffice. We write $f:X\rightarrow Y$ to specify a function with domain $X$ and codomain $Y$.
There are a few properties a function may have that are worth mentioning here. Every function assigns a value in its codomain to every element in its domain. If every element in the codomain is the value of the function at some element of the domain we say that the function is “onto” or a “surjection”. Every function assigns only one value to every element in the domain. If no element in the codomain is the value of the function at more than one element of the domain, we say that the function is “one-to-one” or an “injection”. If both of these properties hold, we say the function is a “bijection”.
Let’s consider four sets: $A=\{{\rm a},{\rm b},{\rm c}\}$, $M=\{{\rm foo},{\rm bar}\}$, $N=\{1, 2, 3\}$, $V=\{w,x,y,z\}$. We use these to define a number of examples of functions.
• $f:N\rightarrow M$, with $f(1)={\rm foo}$, $f(2)={\rm bar}$, and $f(3)={\rm foo}$. This function is surjective, but not injective.
• $g:N\rightarrow V$, with $g(1)=y$, $g(2)=x$, $g(3)=z$. This function is injective, but not surjective.
• $h:A\rightarrow N$, with $h({\rm a})=2$, $h({\rm b})=3$, $h({\rm c})=3$. This function is neither injective nor surjective.
• $k:A\rightarrow A$, with $k({\rm a})={\rm c}$, $k({\rm b})={\rm b}$, $k({\rm c})={\rm a}$. This function is bijective.
### Like this:
Posted by John Armstrong | Fundamentals
## 12 Comments »
1. I can’t say how many times I’ve read descriptions such as this. and i’d say you’ve done well. but i’m a bit confused by your layout in your final paragraph.
Would it not make more sense to the casual reader to put it like this:
” . . .Every function assigns a value in its codomain to every element in its domain. Every function assigns only one value to every element in the domain. If every element . . . etc.”
or even: ‘Every function assigns one and only one value in its codomain to every element of its domain’
but if there is a reason for stating it as you did, please share. i was puzzled about it all night.
it was a dull evening.
Comment by | February 9, 2007 | Reply
2. I may have mangled the style a bit there. What I had in mind is this.
If we think of a function as setting up a correspondence between elements of the domain and elements of the codomain, like a tables of input and output values, we see that every element of the domain shows up as an input, and it shows up only once. A surjection is a function where every element of the codomain shows up at least once and an injection is one where every element shows up at most once.
These two properties — “shows up at least once” and “shows up at most once” — must hold for elements on the domain side of the table. They don’t need to hold on the codomain side, but if they do we have special names for those sorts of functions.
Comment by | February 10, 2007 | Reply
3. [...] of homomorphisms. For now, there are a few definitions we will find useful later. Recall from the discussion about functions that a surjection is a function between functions that hits every point in its codomain at least [...]
Pingback by | February 10, 2007 | Reply
4. You wrote:
the codomain is often called the “range”
In my experience, the codomain is called the `range’ only when the function is surjective (onto). In general, the range (also called the `image’) of the function is a subset of the codomain, consisting of those elements y satisfying the property (so we are using the Specification principle here!) that y is assigned to at least one element of the domain.
In other words: The codomain is the set of potential values of the function, while the range is the set of actual values of the function.
In the example g, the range is only {x, y, z}, not all of the codomain V. And in the example h, the range is only {2, 3}, not all of the codomain N. (The other two examples are onto; for them the range is all of the codomain.)
Comment by | February 15, 2007 | Reply
5. Toby, that’s correct in current usage among most working mathematicians, but there are places where “range” shows up, notably many calculus texts.
I agree that “range” should mean “the values that actually show up”, but I don’t need that concept too often as yet, and I’d rather avoid collisions with other authors here than include the term.
Comment by | February 15, 2007 | Reply
6. [...] direct product says that we can take two groups, form the Cartesian product of their sets, and put the structure of a group on that. Given groups and we form the group as [...]
Pingback by | February 27, 2007 | Reply
7. [...] on the left of exactly one pair in . In this case if is in the relation we write . Remember from our discussion of functions that I was saying every element of has to show up both at least once and at most once (that is, [...]
Pingback by | March 2, 2007 | Reply
8. Peace people
We love you
Comment by HelloWorld | April 28, 2007 | Reply
9. [...] The standard example here (which will motivate much of our later definitions) is : the category of sets. This has as objects the class of all sets (which can’t itself be a set). The morphisms are [...]
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10. I certainly once suffered from the ‘foundationalist fallacy’ (if you know how it’s built, you have a clue what it’s for), probably due to a combination of intellectual style defects and osmosis of misguided New Math principles. But, I think the axiom of extensionality for sets and especially functions is worth focussing on a bit, since the standard technique for proving two functions to be identical is proving that the do the same thing to their inputs.
& isn’t what you are calling specification the one that’s more usually called separation (I think ‘culling’ would be better, but I’ve never seen it called that).
Comment by MathOutsider | October 13, 2007 | Reply
11. MO, it’s really more a philosophical point. I tend to take the language of sets as being a very convenient one, but far from the only one that’s available. Compare this situation with that of the natural numbers. Both the Church and the von Neumann numerals provide models, but neither is inherently any better than the other.
And yes, specification, separation, comprehension, and so on are all different names for the same thing. “Separation” has the sense of taking a set and separating those elements which satisfy a predicate from those which don’t. “Specification” (which I prefer) has the sense of taking a set and specifying (by a predicate) which of its elements are in a subset. “Culling” would have the sense of taking a set and removing those of its elements which don’t satisfy a predicate. They’re all slightly different ways of thinking about the same thing.
Comment by | October 13, 2007 | Reply
12. [...] finite sums, but about infinite sums. As such, we consider all possible rearrangements — bijections — which make up the “infinity symmetric group . Now we might not be able to effect [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://math.stackexchange.com/questions/38517/in-relatively-simple-words-what-is-an-inverse-limit/38523
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# In (relatively) simple words: What is an inverse limit?
I am a set theorist in my orientation, and while I did take a few courses that brushed upon categorical and algebraic constructions, one has always eluded me.
The inverse limit. I tried to ask one of the guys in my office, and despite a very shady explanation he ended up muttering that "you usually take an already known construction."
The Wikipedia article presents two approaches, the algebraic and the categorical. While the categorical is extremely vague for me, the algebraic one is too general and the intuition remains hidden underneath the text in a place I cannot find it.
Since I am not too familiar with categories, the explanation most people would try to give me which is categorical in nature seems to confuse me - as I keep asking this question over and over every now and then.
Could anyone explain to me in non-categorical terms what is the idea behind an inverse limit? (I am roughly familiar with its friend "direct limit", if that helps)
(While editing, I can say that the answers given so far are very interesting, and I have read them thoroughly, although I need to give it quite some thinking before I can comment on all of them right now.)
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I was reluctant to tag this under category related tags, as I'm really asking to avoid categories. If anyone thinks of something better to retag it with, be my guest. – Asaf Karagila May 11 '11 at 18:11
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If you're going to ask "Please tell me about X in basic terms", and there is a wikipedia article on X, I think it would be a good idea to read and make reference to the article, as what it says is going to be along the lines of how many will try to answer. (In this case you have gotten some very creative answers. If I were to answer it would be much more along the lines of wikipedia, in part because that's the kind of explanation I would have found helpful as a beginner in this area.) – Pete L. Clark May 11 '11 at 19:53
@Pete: I'm very glad to see you back on the site. I thought that I did mention that I read the wikipedia page and it wasn't very clear to me in the sense that the intuition was hidden and the categorical abstract definition was the center of the text. – Asaf Karagila May 11 '11 at 20:06
– Alex B. May 12 '11 at 5:24
@Alex: It looks to me as though this question is a generalization of that question, and the answers here are slightly more general than the ones in the question you linked. I do agree that the topics are closely related - I did search for "inverse limit" but not for "projective limit" prior to posting this question. (All this regardless as being the OP, and as unbiased as I can possibly be on the matter.) – Asaf Karagila May 12 '11 at 6:48
show 2 more comments
## 5 Answers
I like George Bergman's explanation (beginning in section 7.4 of his Invitation to General Algebra and Universal Constructions.
Suppose you are interested in solving $x^2=-1$ in $\mathbb{Z}$. Of course, there are no solutions, but let's ignore that annoying reality for a moment.
The equation has a solution in the ring $\mathbb{Z}_5$ (in fact, two: both $2$ and $3$, which are the same up to sign). So we want to find a solution to $x^2=-1$ in $\mathbb{Z}$ which satisfies $x\equiv 2 \pmod{5}$.
An integer that is congruent to $2$ modulo $5$ is of the form $5y+2$, so we can rewrite our original equation as $(5y+2)^2 = -1$, and expand to get $25y^2 + 20y = -5$.
That means $20y\equiv -5\pmod{25}$, or $4y\equiv -1\pmod{5}$, which has the unique solution $y\equiv 1\pmod{5}$. Substituting back we determine $x$ modulo $25$: $$x = 5y+2 \equiv 5\cdot 1 + 2 = 7 \pmod{25}.$$
Continue this way: putting $x=25z+7$ into $x^2=-1$ we conclude $z\equiv 2 \pmod{5}$, so $x\equiv 57\pmod{125}$.
Using Hensel's Lemma, we can continue this indefinitely. What we deduce is that there is a sequence of residues, $$x_1\in\mathbb{Z}_5,\quad x_2\in\mathbb{Z}_{25},\quad \ldots, x_{i}\in\mathbb{Z}_{5^i},\ldots$$ each of which satisfies $x^2=-1$ in the appropriate ring, and which are "consistent", in the sense that each $x_{i+1}$ is a lifting of $x_i$ under the natural homomorphisms $$\cdots \stackrel{f_{i+1}}{\longrightarrow} \mathbb{Z}_{5^{i+1}} \stackrel{f_i}{\longrightarrow} \mathbb{Z}_{5^i} \stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow} \mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow} \mathbb{Z}_5.$$
Take the set of all strings $(\ldots,x_i,\ldots,x_2,x_1)$ such that $x_i\in\mathbb{Z}_{5^i}$ and $f_i(x_{i+1}) = x_i$, $i=1,2,\ldots$. This is a ring under componentwise operations. What we did above shows that in this ring, you do have a square root of $-1$.
Added. Bergman here inserts the quote, "If the fool will persist in his folly, he will become wise." We obtained the sequence by stubbornly looking for a solution to an equation that has no solution, by looking at putative approximations, first modulo 5, then modulo 25, then modulo 125, etc. We foolishly kept going even though there was no solution to be found. In the end, we get a "full description" of what that object must look like; since we don't have a ready-made object that satisfies this condition, then we simply take this "full description" and use that description as if it were an object itself. By insisting in our folly of looking for a solution, we have become wise by introducing an entirely new object that is a solution.
This is much along the lines of taking a Cauchy sequence of rationals, which "describes" limit point, and using the entire Cauchy sequence to represent this limit point, even if that limit point does not exist in our original set.
This ring is the $5$-adic integers; since an integer is completely determined by its remainders modulo the powers of $5$, this ring contains an isomorphic copy of $\mathbb{Z}$.
Essentially, we are taking successive approximations to a putative answer to the original equation, by first solving it modulo $5$, then solving it modulo $25$ in a way that is consistent with our solution modulo $5$; then solving it modulo $125$ in a way that is consistent with out solution modulo $25$, etc.
The ring of $5$-adic integers projects onto each $\mathbb{Z}_{5^i}$ via the projections; because the elements of the $5$-adic integers are consistent sequences, these projections commute with our original maps $f_i$. So the projections are compatible with the $f_i$ in the sense that for all $i$, $f_i\circ\pi_{i+1} = \pi_{i}$, where $\pi_k$ is the projection onto the $k$th coordinate from he $5$-adics.
Moreover, the ring of $5$-adic integers is universal for these property: given any ring $R$ with homomorphisms $r_i\colon R\to\mathbb{Z}_{5^i}$ such that $f_i\circ r_{i+1} = r_i$, for any $a\in R$ the tuple of images $(\ldots, r_i(a),\ldots, r_2(a),r_1(a))$ defines an element in the $5$-adics. The $5$-adics are the inverse limit of the system of maps $$\cdots\stackrel{f_{i+1}}{\longrightarrow}\mathbb{Z}_{5^{i+1}}\stackrel{f_i}{\longrightarrow}\mathbb{Z}_{5^i}\stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow}\mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow}\mathbb{Z}_5.$$
So the elements of the inverse limit are "consistent sequences" of partial approximations, and the inverse limit is a way of taking all these "partial approximations" and combine them into a "target object."
More generally, if you have a system of, say, rings, $\{R_i\}$, indexed by an directed set $(I,\leq)$ (so that for all $i,j\in I$ there exists $k\in I$ such that $i,j\leq k$, and a system of maps $f_{rs}\colon R_s\to R_r$ whenever $r\leq s$ which are "consistent" (if $r\leq s\leq t$, then $f_{st}\circ f_{rs} = f_{rt}$), and let's assume that the $f_{rs}$ are surjective, as they were in the example of the $5$-adics. Then you can think of the $R_i$ as being "successive approximations" (with a higher indexed $R_i$ as being a "finer" or "better" approximation than the lower indexed one). The directedness of the index set guarantees that given any two approximations, even if they are not directly comparable to one another, you can combine them into an approximation which is finer (better) than each of them (if $i,j$ are incomparable, then find a $k$ with $i,j\leq k$). The inverse limit is a way to combine all of these approximations into an object in a consistent manner.
If you imagine your maps as going right to left, you have a branching tree that is getting "thinner" as you move left, and the inverse limit is the combination of all branches occurring "at infinity".
Added. The example of the $p$-adic integers may be a bit misleading because our directed set is totally ordered and all maps are surjective. In the more general case, you can think of every chain in the directed set as a "line of approximation"; the directed property ensures that any finite number of "lines of approximation" will meet in "finite time", but you may need to go all the way to "infinity" to really put all the lines of approximation together. The inverse limit takes care of this.
If the directed set has no maximal elements, but the structure maps are not surjective, it turns out that no element that is not in the image will matter; essentially, that element never shows up in a net of "successive approximations", so it never forms part of a "consistent system of approximations" (which is what the elements of the inverse limit are).
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@Asaf: For one thing, every nonzero element of the inverse limit has infinite additive order, but the direct limit contains elements of order $5$ (the image of the generator of $\mathbb{Z}_5$ in the direct limit has order $5$); in fact, the direct limit is torsion, the inverse limit is torsionfree. In the direct limit you are forcing global conformity/consistency (by identifying elements that are eventually mapped to the same thing, even if they are different objects to being with), whereas in the inverse limit you are requiring local conformity. – Arturo Magidin May 11 '11 at 19:22
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@Asaf: he direct limit is the Prufer group $\mathbb{Z}_{5^{\infty}}$. The direct limit is a quotient of the direct sum, the inverse limit is a subgroup of the direct product (there's that categorical duality you didn't want...) The examples are possibly misleading in that they have very special structure maps and index sets... – Arturo Magidin May 11 '11 at 19:46
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@Asaf: In the direct limit, it's as if you are "pulling away" and seeing a larger and larger vista, or more generally, you are seeing bigger and bigger pictures, and you glue them together by forcing the overlaps to agree. In the inverse limits, you are "zooming in" and looking at smaller and smaller views; you already have that the different views are consistent on the overlaps; there are details that you can only see once you are "sufficiently close", and the inverse limit is a way of keeping track of all the details you can see at different resolutions. – Arturo Magidin May 11 '11 at 19:47
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I just have to remark that I have to give a general lecture in some seminar about forcing, I was thinking about the description in "other mathematics" and had very similar description such as Cauchy sequences describing a real number, and so on... :-) – Asaf Karagila May 11 '11 at 20:52
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And will the downvoter say why the downvote? – Arturo Magidin May 13 '11 at 16:57
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Well, I can give you my understanding of inverse limits from a topological view. Topologically, I think of inverse limits as "generalized intersections".
Consider a sequence of inclusions
$X_1 \supset X_2 \supset X_3 \supset \dots$
with an inverse limit $X$. Then $X$ is precisely the intersection of the $X_i$. The key thing here is that a point in the intersection gives you a point $x_i$ in each of the $X_i$, such that $x_{i+1}$ gets mapped to $x_i$. This is kind of silly and clear since you're really talking about the same point, just in the context of different containing sets, but it sets the stage for the more general inverse limit.
More generally, you can have maps between the $X_i$ that are not inclusions.
$X_1 \leftarrow X_2 \leftarrow X_3 \leftarrow \dots$
In this case, you can build an intersection-like object by taking a "point" in it to be a set of points, one from each $X_i$, such that each one is mapped to the previous. A better way to describe this is as a subset of the product $\Pi X_i$, namely points $(x_1, x_2, ...)$ such that $x_{i+1}$ maps to $x_i$. This also gives us a handy topology for this set, namely the induced topology from the product topology.
So say, for example, that $X_1$ is a point, $X_2$ is two (discrete) points, $X_3$ is four discrete points, etc., and each map is some two-to-one map. Then since each point has exactly two preimages, a point in the inverse limit is basically a choice of preimage (say, 0 or 1) at each $i$. The product topology makes these choices "close" if they agree for a long time, and so you may be able to then understand why the inverse limit is the Cantor set.
Basically, I think of the inverse limits as "infinite paths of preimages", with a topology that makes two sequences close if they stay close to each other for a long time.
A fun exercise: Let all the $X_i$ be circles, and let the maps from $X_{i+1}$ to $X_i$ be degree 2 maps (say, squaring the complex numbers with modulus 1). Understand the inverse limit of this system (it's often called a "solenoid").
Hopefully that helps you get a different perspective!
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I am grateful for a topological example, in a crazy land of categories and algebra. Feels very homey :-) However two things are unclear here, what does "a topology that makes two sequences close if they stay close to each other for a long time" mean in actual terms? in the fun exercise are we taking any circles (and what is a degree 2 map anyway?) – Asaf Karagila May 12 '11 at 22:51
Sorry about the hand-waveyness. The inverse limit $X$ is a subset of the product of the $X_i$, and its topology is simply the subspace topology induced by the product topology. So when is a sequence $a_1, a_2, a_3 , ...$ in $X$ convergent? Well remember, each $a_k$ is itself a sequence, with a chosen point in each $X_i$. The $a_k$ converge if, for each fixed $i$, the chosen point in $X_i$ is the same for all large enough $k$. To make the fun exercise more concrete: Let each $X_i$ be a copy of the unit circle in $\mathbb{C}$, and let each map be $z\rightarrow z^2$. – MartianInvader May 17 '11 at 21:03
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To even better clarify "paths of preimages", each point in the inverse limit is basically choosing a point in some $X_i$, choosing a preimage of that point in $X_{i+1}$, a preimage of that point in $X_{i+2}$, etc. Two such sequences are "close" if they have the same choices up to some big $i$. Also, I guess I should mention that inverse limits can be pretty horrible in topology. Even something as simple as the squaring example I gave gives an inverse limit with uncountably many path components. – MartianInvader May 17 '11 at 21:32
Here's how I think of inverse versus direct limits. The following should not be taken too seriously, because of course inverse limits and colimits are completely dual. But in many of the categories one tends to work with daily, they have a somewhat different feel.
In practice, inverse limits are a lot like completions: one has a tower of spaces $X_n \to X_{n-1} \to X_{n-2} \to \dots$ and one wishes to consider the space of "Cauchy sequences": in other words, one has a sequence $(x_n)$ such that $x_n \in X_n$ (this is the Cauchy sequence) and the $x_n$ are compatible under the maps (which is the abstract form of Cauchy-ness). For instance, the completion of a ring is the standard example of an inverse limit in commutative algebra. Here the $X_n$'s (which are the quotients of a fixed ring $R$ by a descending sequence of ideals $I_n$) can be thought of as specifying sets of "intervals" that are shrinking with each $n$, so an element of the inverse limit is a descending sequence of intervals. Perhaps the reason that inverse limits feel this way in many categories of interest is that many categories of interest are concrete, and the forgetful functor to sets is corepresentable, so that (categorical) limits look the same in the category and in the category of sets. In this case, inverse limits are given by precisely the construction above: it is a kind of "successive approximation."
Direct limits, on the contrast, are much more like unions. Here the picture that I usually keep in mind is that of a sequence of objects in time that gets wider as time passes, even though this is not necessarily accurate: in practice, one often wishes to take direct limits over non-monomorphisms. But the construction of a direct limit in the category of sets (and in many concrete categories: often, it happens that the forgetful functor also commutes with direct limits, and one deeper reason for that is that the corepresenting object is relatively small, and small objects have a tendency to commute with filtered colimits because of the above union interpretation) is ultimately the quotient of the disjoint union of the sets such that each element of a set $X_n$ is identified with its image in the next one. Since ultimately it feels like taking a union, direct limits tend to behave very nicely homologically: most often, they preserve exactness. Inverse limits, by contrast, do not usually preserve exactness unless one imposes extra conditions (such as the ML condition).
Finally, as a categorical limit, inverse limits are always easy to map into, while direct limits are always easy to map out of.
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I too think of the Cauchy sequences analogy. This was suggested to me by D.N. Yetter. Good answer +1. Especially the last line, it is catchy and mnemonous. – BBischof May 11 '11 at 18:36
I am very glad to see you taking part of the website again. I understand some of your answer, but I want to give it some more thought before asking specific questions. – Asaf Karagila May 11 '11 at 20:12
After reading Arturo's answer in details your answer still seems to elude from my grasp into the realm of algebraic and categorical definitions. I am aware that I am asking to clean out the categoricity from a very categorical construction, I have lost you after the completion example, and again after two lines in the second paragraph. – Asaf Karagila May 12 '11 at 22:46
@Asaf: Dear Asaf, thank you for the kind words, and I'm sorry my answer wasn't more helpful. I'm not quite sure that I know much more to say (especially which the other answers haven't covered quite nicely!). Cheers, – Akhil Mathew May 13 '11 at 14:36
Do not worry, judging by the +15 (at time of commenting) score of your answer, even if it is the lowest score amongst these answers it still signals that it is a very good answer. I am only grateful my question had the luxury of three wonderful answers that will certain help other people someday! – Asaf Karagila May 13 '11 at 17:23
This has puzzled me for a long time, until I came across the following example that should appeal to a set theorist:
Let $S$ be a set and let $(\Pi_n)$ be a sequence of partitions of $S$ such that $\Pi_{n+1}$ is finer than $\Pi_n$ for all $n$. That is, every cell in $\Pi_n$ is the union of cells in $\Pi_{n+1}$. Now for each $n\geq m$, $\Pi_n$ is finer than $\Pi_m$ and there is a canonical projection $\pi_{mn}:\Pi_n\to\Pi_m$ that maps each cell in $\Pi_n$ to the unique cell in $\Pi_m$ that contains the former as a subset.
$$\Pi_0\longleftarrow\Pi_1\longleftarrow\Pi_2\longleftarrow\ldots$$
The underlying directed set is simply $\mathbb{N}$ with the natural order. The partitions together with the projections form a projective system:
1. $\pi_{nn}$ is the identity on $\Pi_n$.
2. $\pi_{ln}=\pi_{lm}\circ\pi_{mn}$ if $l\leq m\leq n$.
So what is the projective limit? By definition, it is $$\Big\{(P_0,P_1,P_2,\ldots)\in\prod_{n=0}^\infty\Pi_n:P_m=\pi_{mn}(P_n)\text{ for }m\leq n\Big\}.$$
Since all (non-identity-)projections can be made up from projections of the form $\pi_{n(n+1)}$, we can rewrite this as
$$\Big\{(P_0,P_1,P_2,\ldots)\in\prod_{n=0}^\infty\Pi_n:P_n=\pi_{n(n+1)}(P_{n+1})\Big\}$$
$$=\Big\{(P_0,P_1,P_2,\ldots)\in\prod_{n=0}^\infty\Pi_n:P_n\supseteq P_{n+1}\Big\}.$$
So the inverse limit is simply the set of infinite paths in the tree formed from this sequence of partitions.
There a related approaches in probability theory and measure theory as approaches to generalize the Daniell-Kolmogorov extension Theorem. Studying projective limits of probability spaces was pioneered by Salomon Bochner. A typical paper in this strand of literature is this paper by Kazimierz Musiał.
A beautiful (and very short) paper that gives an example in which the projective limit is empty even though all projections are surjective is this note by William Waterhouse.
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This is a very clear example for me! Just to be clear: $P_i$ is a "cell" in partition $\Pi_i$, correct? – magma May 14 '12 at 21:50
Yes, $P_i$ is a "cell" or "block" in the partition. – Michael Greinecker May 14 '12 at 21:51
thank you Michael. – magma May 14 '12 at 22:26
I'm currently dabbling around in a category where morphisms are refinements of computer programs, in the sense that they are functions $f : X \rightarrow Y$ from a program refinement (an implementation) $X$ to a program specification $Y$.
In this setting, a sequence (or net) of morphisms can be either describing an ever more elaborate specification $X_0 \rightarrow X_1 \rightarrow X_2 \rightarrow ...$ or it can be an ever more precise refinement $... \rightarrow X_2 \rightarrow X_1 \rightarrow X_0$. The first sequence has as a direct limit the 'ultimate' specification, while the second sequence has as an inverse (also called projective) limit the 'ultimate' refinement.
As an extra: maybe my own question about Directed and projective limit in Rel helps in looking at the notion in a different way. In Rel, where the two types of limit coincide, my intuition is that a net of relations simply results in 'lines' connecting points in a disjoint union, and the 'infinite' lines form a new object, the limit.
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http://math.stackexchange.com/questions/10648/why-is-a-circle-in-a-plane-surrounded-by-6-other-circles?answertab=votes
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# Why is a circle in a plane surrounded by 6 other circles
When you draw a circle in a plane you can perfectly surround it with 6 other circles of the same radius. This works for any radius. What's the significance of 6? Why not some other number?
I'm looking for an answer deeper than "there are 6*60=360 degrees in a circle, so you can picture it".
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## 4 Answers
The short answer is "because they don't work," but that's kind of a copout. This is actually quite a deep question. What you're referring to is sphere packing in two dimensions, specifically the kissing number, and sphere packing is actually quite a sophisticated and active field of mathematical research (in arbitrary dimensions).
Here's one answer, which isn't complete but which tells you why $6$ is a meaningful number in two dimensions. The packing you refer to is a special type of packing called a lattice packing, which means it comes from an arrangement of regularly spaced points; in this case, the hexagonal lattice. The number $6$ appears here because the hexagonal lattice has $6$-fold symmetry. So a natural question might be whether one can find lattices in two dimensions with, say $7$-fold or $8$-fold symmetry, since these might correspond to circle packings with more circles around a given circle. (Intuitively, we expect more symmetric lattices to give rise to denser packings and to packings where each circle has more neighbors.)
The answer is no: $6$-fold symmetry is the best you can do! This is a consequence of the crystallographic restriction theorem. The generalization of the theorem to $n$ dimensions says this: it is possible for a lattice to have $d$-fold symmetry only if $\phi(d) \le n$, where $\phi$ is Euler's totient function.
The generalization implies that you still cannot do better than $6$-fold symmetry in $3$ dimensions. There are two natural lattice packings in $3$ dimensions, which both occur in molecules and crystals in nature and which both have $6$-fold symmetry, and it turns out that these are the densest sphere packings in $3$ dimensions. It also turns out that they give the correct kissing number in $3$ dimensions, which is $12$ (see the wiki article).
In $4$ dimensions, the kissing number is $24$, and I believe the corresponding packing is a lattice packing coming from a lattice with $8$-fold symmetry, which is possible in $4$ dimensions. In higher dimensions, only two other kissing numbers are known: $8$ dimensions, where the $E_8$ lattice gives kissing number $240$, and $24$ dimensions, where the Leech lattice gives kissing number $196560$! These lattices are really mysterious objects and are related to a host of other mysterious objects in mathematics.
A great reference for this stuff, although it is a little dense, is Conway and Sloane's Sphere Packing, Lattices, and Groups. Edit: And for a very accessible and engaging introduction to symmetry in the plane and in general, I highly recommend Conway, Burgiel, and Goodman-Strauss's The Symmetries of Things.
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3
+1, great answer. – Fernando Martin Nov 17 '10 at 3:02
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+1, but... this is more an answer to "why is a sphere surrounded by 12 other spheres", i.e., the kissing number problem without regard to rigidity. For the planar 6-circles problem I think the point of the question was that the inner circle is perfectly surrounded, which means an optimal rigid packing. Disordered sphere packings (or kissing arrangements) are likely to beat lattice packings except in special dimensions, but in the superspecial dimensions (such as 2,8,24) with rigid kissing configurations, rigidity probably implies that the configuration must come from a lattice packing. – T.. Nov 17 '10 at 18:54
@T..: sure. I was just trying to provide some context to make sense of the 2-dimensional case. – Qiaochu Yuan Nov 17 '10 at 19:01
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The answer is excellent (enjoy the stream of upvotes!), but since we don't have multidimensional ratings to separate quality-of-answer from was-question-addressed, I thought it worth adding the posting and comment emphasizing the role of rigidity and flatness. – T.. Nov 17 '10 at 19:27
– friedo Nov 17 '10 at 20:52
show 1 more comment
The question is equivalent to asking why 6 equilateral triangles fit together exactly around a point, with no additional room left over. The answer is "because the Euclidean plane is flat", a condition equivalent to triangles having angle sum of 180 degrees (half the angle around a point), so that each vertex of a symmetrical triangle has 1/3 of half of a full rotation = 1/6. That the six-circle arrangement exists for any radius is also a special feature of Euclidean geometry: scale invariance.
For flat surfaces such as a cylinder (rolled up plane) or flat torus (as in the Asteroids video game) the perfect 6-circle configurations exist only for small enough radius of the circles. These geometries are, in small regions, the same as the Euclidean plane but differ "globally", e.g. there is a maximum distance between points of the torus. So the magic number of 6 is really about local flatness (absence of curvature) and having this configuration for circles of all size is a global question about the space in which the circles live.
In hyperbolic geometry there are tesselations of the plane by equilateral triangles with angle $180/n$ at each vertex, for any $n \geq 7$. In the picture of the $n=7$ triangular tesselation at http://www.plunk.org/~hatch/HyperbolicTesselations/3_7_trunc0_512x512.gif (triangles in white with the dual tesselation by heptagons shown in blue) if at each triangle vertex you draw a circle inscribed in its heptagon, there will be 7 circles exactly surrounding each circle, with all circles of the same radius. The same type of configuration exists for any $n$ and suitable radius of the circles. So the flatness condition is necessary; the theorem is false in negatively curved two-dimensional geometry. Six is also not the correct number for spherical geomety, and both spherical and hyperbolic geometry lack a radius-independent "number of circles that fit around one circle".
In the geometry of surfaces, having 5-or-fewer as the local number of circles that can be fit around a single circle characterizes positive curvature, as in spherical geometry. Having more-than-6 fit, or extra room when surrounding one circle by six, is a characterization of negative curvature, as in hyperbolic geometry. This is a statement about the local geometry of general 2-dimensional surfaces, and does not assume the surface has the same amount of curvature at all points, as would be the case for the spherical and hyperbolic analogues of Euclidean plane geometry, where there is a homogeneity assumption that "geometry is the same at all points". Having exactly 6 circles fit perfectly is a characterization of locally Euclidean (that is, zero curvature) geometry. If you don't know what curvature is in a technical sense, for purposes of this discussion it is (for surfaces) a number that can be associated to any point on the surface, and curvature being zero in a region of surface is equivalent to the ability to make a distance-preserving planar map of that region. Impossibility of doing this for surfaces of nonzero curvature, such as the sphere which is positively curved, was Gauss' Theorema Egregium which ruled out perfect flat cartography of the Earth.
Flatness plays the same role in the higher dimensional, sphere-packing interpretation of the question that Qiaochu suggested. The necessary packings exist only in a limited set of dimensions. The reason is that exactly surrounding a sphere of radius $r$ by $k$ spheres of that radius means more than a kissing number of $k$, the optimal kissing configuration should also be rigid (tight enough that the spheres cannot be moved). Rigidity is false in three dimensions and is believed to be false in most other dimensions but in those cases where it is known or suspected to be true (i.e., dimensions $d = 2, 8, 24$ and possibly a few others) the existence of a configuration with the same set of equalities between various interpoint distances as in the optimal kissing arrangement (in flat Euclidean space $\mathbb{R}^d$)should force a homogeneous $d$-dimensional geometry to be flat. This is because deforming the curvature of the space in the positive direction would reduce, and negative curvature would increase, the freedom to position such spheres around a central sphere.
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I wish I knew the word for the difference between this question and the accepted one. So much easier to keep a model in my head with this one. Thanks. +1, obviously. – Triptych Jul 21 '11 at 11:07
The centres of three circles of radius r just touching one another will form an equilateral triangle of side length 2r. Exactly 6 such non-overlapping equilateral triangles will be formed in this way as you keep adding more circles around the edge of the central circle. Why 6? These equilateral triangles will have a common vertex at the centre, each forming a 60 degree angle there. And as, you pointed out, "there are 6*60=360 degrees in a circle."
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I think this is covered in the first paragraph of the other answer and to some extent in the question. The angle sum for one triangle is half a full rotation. For a symmetrical triangle the angles must be equal, so each angle is a third of the half-rotation, or one sixth of the angle measure around a point (360/6). The harder part is to explain the significance of six and maybe also the rigidity of the six-circle configuration. The OP was insightful to mention the radius-independence as a notable aspect of this. It is an excellent and deceptively simple question. – T.. Nov 17 '10 at 20:06
@Dan, you are simply rephrasing the question. – John Smith Nov 17 '10 at 21:11
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@John: The 6*60=360 argument combines two steps, 360=2*180 and 180=3*60. The first amounts to a rephrasing of "triangles have angle sum of 180 degrees (half of a circle or full rotation)", ie., the locally flat/Euclidean nature of the geometry. The second step, 3*60=180 is trivial and carries almost no information. It is a restatement of the symmetry of an equilateral triangle, which is a very general argument that holds in any geometry I can imagine, and thus says nothing in particular about the geometrical meaning of any special configuration such as the six surrounding circles. – T.. Nov 17 '10 at 21:30
FWIW, I gave this as a comment (now removed), which prompted his edit of "something deeper than 6*60=360". It's fundamentally related to the fact that we assume the parallel postulate (which turns up in the proof that the angles of a triangle sum up to 180 degrees, which means equilateral triangles can tile the plane, which means... you get the idea.) – J. M. Nov 17 '10 at 22:24
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An answer at any level is a contribution. Users with a broad range of backgrounds will read most threads. I think it is more informative if posters feel free to say whatever comes to mind at their actual knowledge level (rather than PhD's "dumbing down" or neophytes uploading from Wikipedia) as this allows everyone to spend the most time at their highest level of competence. This model would become more realistic when the user population becomes larger and there are enough "eyeballs" on each question to sample the whole spectrum of replies. – T.. Nov 18 '10 at 4:11
show 2 more comments
Im sorry guys to inform u all that the question itself is wrong... the number of circles surrounding a circle of same radius in a plane tangent to the central circle and tangent to each other is 2pi... ie, 2*3.14 = 6.28 circles....
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that means 6 circles wont exactly fit around... there will always be a gap corresponding to the 0.28 circle – user74165 Apr 24 at 15:16
I rather believe your interpretation of the question is wrong. The number of circles should be a natural number.. – Mårten W Apr 24 at 15:38
I realize you don't yet have enough reputation, but this is best as a comment. Regards – Amzoti Apr 24 at 15:38
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http://mathhelpforum.com/advanced-statistics/200623-biased-ness-ols-estimates.html
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# Thread:
1. ## Biased-ness of OLS estimates
If the picture is too small, please view attached pic.
Hey guys, so I'm just reading through the proof of the biased-ness of OLS estimates if an important independent variable is left out and I'm just stuck at the very end, basically the mathematical part that confuses me is the part circled in red.
When we take the expectation of $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ how does it simply end as the same thing?
Ie, Why is $E\left[\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}\right] = \frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ ?
Since $\frac{\sum_{i=1}^n (x_{i1} - \bar{x_1})(x_{i2} - \bar{x_2})}{\sum_{i=1}^{n}\left((x_{i1}-\bar{x_1})^2\right)}$ isn't a constant as it depends on $n$ so it's like a random variable which outputs different values as $n$ changes, so why does it follow the rule that $E(c) = c$ where $c$ is a constant?
Thanks
Attached Thumbnails
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http://mathoverflow.net/questions/97175/small-sums-of-entries-in-submatrices-strange-phenomenon/97293
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## small sums of entries in submatrices - strange phenomenon
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose that $x \in \mathbb{R}^{n}$ is a vector of small positive fractions, i.e. $x_{i} \approx \frac{1}{n}$. The exact values are unknown. I form the matrix $M=diag(x)-\frac{xx^{T}}{2}$ which is a Stieltjes matrix and is diagonally dominant (at least in the cases of interest).
Now what I am interested in are the sums of the entries of small rectangular submatrices of $M$. These sums turn out to be very small in absolute value in practice, however my efforts to estimate them yield bounds that are too crude. This makes me suspect that some phenomenon is at work under the hood, about which I do not know, and I am hoping that somebody can point out a direction in which to look. (Does this situation/property has a name?)
More formally, take a set of $d$ rows and $d$ columns and look at the $d \times d$ submatrix of $M$ that they induce. Call this submatrix $M_{0}$. My reasoning goes like this: at the worst case, each row of $M_{0}$ contains at most one diagonal entry of $M$. The rest of the entries in the row push it down but we do not know by how much exactly. Rows of $M_{0}$ that do not contain a diagonal entry of $M$ sum to at most the value of the diagonal entry. Therefore the best bound I come up with is $dC$, where $C=max_{1 \leq i \leq n}{x_{i}}$.
As I said, in practice this bound is off by an order of magnitude and I would very much like to improve it. But how?
Does this sound like a problem of extremal combinatorics type perhaps? Any other places to look for apppropriate tools?
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## 2 Answers
Perhaps there's something I don't understand, but your bound is asymptotically best possible. That is, by taking all $x_i = c$ and the same set for the $d$ rows and the $d$ columns the sum of entries in $M_0$ is $d c - d^2 c^2/2 = d c (1 - d c/2)$. For any $d$ and any $\epsilon > 0$, take $0 < c < 2 \epsilon/d$ and the true value is more than $1-\epsilon$ times your estimate.
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First of all, thanks a lot for devoting thought to this! Now, I'm almost convinced, but please recall that I also have the extra condition that $x_{i} \approx \frac{1}{n}$. Although for $n$ large enough your argument stilll takes the biscuit, what happens when $n$ is fixed and not terribly big and we have little control on $x_{i}$? – Felix Goldberg May 17 2012 at 8:30
What precisely do you mean by $x_i \approx 1/n$ when $n$ is fixed? – Robert Israel May 18 2012 at 5:30
Never mind, I've seen the light already! :) (The question was a fragment of a larger investigation; what I am really interested in is the sum of sums over certain ubmatrices; your first reply got me out of trying to estimate each sum independently, which is rather a dead-end and into estimating the sum of sums; but that is a different story, which doesn't belong here). Thanks again. – Felix Goldberg May 18 2012 at 11:15
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Unless I don't understand the question: Consider the submatrix formed by rows $R$ and columns $C$. Then the sum of this submatrix is $$\sum_{k\in R\cap C} x_k - \tfrac12 \biggl(\sum_{i\in R} x_i\biggr) \biggl(\sum_{j\in C} x_j\biggr).$$ The question is a bit too vague to say more, as this expression is pretty simple.
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Yes, I've got this far. You are right that without further information this is more or less the end of the story... Thanks! – Felix Goldberg May 18 2012 at 11:16
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http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_LawOfLargeNumbers&redirect=no
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# SOCR EduMaterials Activities LawOfLargeNumbers
## Overview
This is part I of a heterogeneous activity that demonstrates the theory and applications of the Law of Large Numbers (LLN). Part II and Part III of this activity contain more examples and diverse experiments. The SOCR LLN applet is available here.
### Goals of the SOCR LLN activity
The goals of this activity are to:
• illustrate the theoretical meaning and practical implications of the LLN;
• present the LLN in varieties of situations;
• provide empirical evidence in support of the LLN-convergence and dispel the common LLN misconceptions.
### Example
The average weight of 10 students from a class of 100 students is most likely closer to the real average weight of all 100 students, compared to the average weight of 3 randomly chosen students from that same class. This is because the sample of 10 is a larger number than the sample of only 3 and better represents the entire class. At the extreme, a sample of 99 of the 100 students will produce a sample average almost exactly the same as the average for all 100 students. On the other extreme, sampling a single student will be an extremely variant estimate of the overall class average weight.
### Statement of the Law of Large Numbers
If an event of probability p is observed repeatedly during independent repetitions, the ratio of the observed frequency of that event to the total number of repetitions converges towards p as the number of repetitions becomes arbitrarily large.
### The theory behind the LLN
Complete details about the weak and strong laws of large numbers may be found here.
## Exercise 1
This exercise illustrates the statement and validity of the LLN in the situation of tossing (biased or fair) coins repeatedly. Suppose we let H and T denote Heads and Tails, the probabilities of observing a Head or a Tail at each trial are 0 < p < 1 and 0 < 1 − p < 1, respectfully. The sample space of this experiment consists of sequences of H's and Ts. For example, an outcome may be {H,H,T,H,H,T,T,T,....}. If we toss a coin n times, the size of the sample-space is 2n, as the coin tosses are independent. Binomial Distribution governs the probability of observing $0\le k\le n$ Heads in n experiments, which is evaluated by the binomial density at k.
In this case we will be interested in two random variables associated with this process. The first variable will be the proportion of Heads and the second will be the differences of the number of Heads and Tails. This will empirically demonstrate the LLN and its most common misconceptions (presented below). Point your browser to the SOCR Experiments and select the Coin Toss LLN Experiment from the drop-down list of experiments in the top-left panel. This applet consists of a control toolbar on the top followed by a graph panel in the middle and a results table at the bottom. Use the toolbar to flip coins one at a time, 10, 100, 1,000 at a time or continuously! The toolbar also allows you to stop or reset an experiment and select the probability of Heads (p) using the slider. The graph panel in the middle will dynamically plot the values of the two variables of interest (proportion of heads and difference of Heads and Tails). The outcome table at the bottom presents the summaries of all trials of this experiment. From this table, you can copy and paste the summary for further processing using other computational resources (e.g., SOCR Modeler or MS Excel).
• Note: We report the normalized differences of the number of Heads minus the number of Tails in the graph and result table. Let H and T are the number of Heads and Tails, up to the current trial (k), respectively. Then we define the normalized difference | H − T | = $p+ ((1-p)H-pT )/(2/3 \times Max_k)$, where $Max_k = \max_{1 \le i \le k}{||H-T||_i}$ and | | H − T | | i is the maximum difference of Heads and Tails up to the ith trial. Observe that the expectation of the normalized difference E( | H − T | ) = p, since E((1 − p)H − pT) = 0. This ensures that the normalized differences oscillate around the chosen p (the LLN limit of the proportion of Heads) and they are visible within the graph window.
Now, select n=100 and p=0.5. The figure below shows a snapshot of the applet. Remember that each time you run the applet the random samples will be different and the figures and results will generally vary. Click on the Run or Step buttons to perform the experiment and observe the proportion of heads and differences evolve over time. Choosing Continuous from the number of experiments drop-down list in the tool bar will run the experiment in a continuous mode (use the Stop button to terminate the experiment in this case). The statement of the LLN in this experiment is simply that as the number of experiments increases the sample proportion of Heads (red curve) will approach the theoretical (user preset) value of p (in this case p=0.5). Try to change the value of p and run the experiment interactively several times. Notice the behavior of the graphs of the two variables we study. Try to pose and answer questions like these:
• If we set p=0.4, how large of a sample-size is needed to ensure that the sample-proportion stays within [0.4; 0.6]?
• What is the behavior of the curve representing the differences of Heads and Tails (red curve)?
• Is the convergence of the sample-proportion to the theoretical proportion (that we preset) dependent on p?
• Remember that the more experiments you run the closer the theoretical and sample proportions will be (by LLN). Go in Continuous run mode and watch the convergence of the sample proportion to p. Can you explain in words, why can't we expect the second variable of interest (the differences of Heads and Tails) to converge?
## Exercise 2
The second SOCR demonstration of the law of large numbers will be quite different and practically useful. Here we show how the LLN implies practical algorithms for estimation of transcendental numbers. The two most popular transcendental numbers are π and e.
### Estimating e using SOCR simulation
The SOCR E-Estimate Experiment provides the complete details of this simulation. In a nutshell, we can estimate the value of the natural number e using random sampling from Uniform distribution. Suppose X1,X2,...,Xn are drawn from uniform distribution on (0, 1) and define $U= {\operatorname{argmin}}_n { \left (X_1+X_2+...+X_n > 1 \right )}$, note that all $X_i \ge 0$.
Now, the expected value $E(U) = e \approx 2.7182$. Therefore, by LLN, taking averages of $\left \{ U_1, U_2, U_3, ..., U_k \right \}$ values, each computed from random samples $X_1, X_2, ..., X_n \sim U(0,1)$ as described above, will provide a more accurate estimate (as $k \rightarrow \infty$) of the natural number e.
The Uniform E-Estimate Experiment, part of SOCR Experiments, provides a hands-on demonstration of how the LLN facilitates stochastic simulation-based estimation of e.
### Estimating π using SOCR simulation
Similarly, one may approximate the transcendental number π, using the SOCR Buffon’s Needle Experiment. Here, the LLN again provides the foundation for a better approximation of π by virtually dropping needles (many times) on a tiled surface and observing if the needle crosses a tile grid-line. For a tile grid of size 1, the odds of a needle-line intersection are ${ 2 \over \pi} \approx 0.63662$. In practice, to estimate π from a number of needle drops (N), we take the reciprocal of the sample odds-of-intersection.
## Experiment 3
Suppose we row 10 loaded hexagonal (6-face) dice 8 times and we are interested in the probability of observing the event A={3 ones, 3 twos, 2 threes, and 2 fours}. Assume the dice are loaded to the small outcomes according to the following probabilities of the 6 outcomes (one is the most likely and six is the least likely outcome).
| | | | | | | |
|--------|-------|-------|------|-------|-------|-------|
| x | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X=x) | 0.286 | 0.238 | 0.19 | 0.143 | 0.095 | 0.048 |
P(A)=?
Of course, we can compute this number exactly as:
$P(A) = {10! \over 3!\times 3! \times 2! \times 2! } \times 0.286^3 \times 0.238^3\times 0.19^2 \times 0.143^2 = 0.00586690138260962656816896.$
However, we can also find a pretty close empirically-driven estimate using the SOCR Dice Experiment.
For instance, running the SOCR Dice Experiment 1,000 times with number of dice n=10, and the loading probabilities listed above, we get an output like the one shown below.
Now, we can actually count how many of these 1,000 trials generated the event A as an outcome. In one such experiment of 1,000 trials, there were 8 outcomes of the type {3 ones, 3 twos, 2 threes and 2 fours}. Therefore, the relative proportion of these outcomes to 1,000 will give us a fairly accurate estimate of the exact probability we computed above
$P(A) \approx {8 \over 1,000}=0.008$.
Note that that this approximation is close to the exact answer above. By the Law of Large Numbers, we know that this SOCR empirical approximation to the exact multinomial probability of interest will significantly improve as we increase the number of trials in this experiment to 10,000.
## Hands-on activities
The following practice problems will help students experiment with the SOCR LLN activity and understand the meaning, ramifications and limitations of the LLN.
• Run the SOCR Coin Toss LLN Experiment twice with stop=100 and p=0.5. This corresponds to flipping a fair coin 100 times and observing the behavior of the proportion of heads across (discrete) time.
• What will be different in the outcomes of the 2 experiments?
• What properties of the 2 outcomes will be very similar?
• If we did this 10 times, what is expected to vary and what may be predicted accurately?
• Use the SOCR Uniform e-Estimate Experiment to obtain stochastic estimates of the natural number $e \approx 2.7182$.
• Try to explain in words, and support your argument with data/results from this simulation, why is the expected value of the variable U (defined above) equal to e, E(U) = e.
• How does the LLN come into play in this experiment?
• How would you go about in practice if you had to estimate $e^2 \approx 7.38861124$ ?
• Similarly, try to estimate $\pi \approx 3.141592$ and $\pi^2 \approx 9.8696044$ using the SOCR Buffon’s Needle Experiment.
• Run the SOCR Roulette Experiment and bet on 1-18 (out of the 38 possible numbers/outcomes).
• What is the probability of success (p)?
• What does the LLN imply about p and repeated runs of this experiment?
• Run this experiment 3 times. What is the sample estimate of p ($\hat{p}$)? What is the difference $p-\hat{p}$? Would this difference change if we ran the experiment 10 or 100 times? How?
• In 100 Roulette experiments, what can you say about the difference of the number of successes (outcome in 1-18) and the number of failures? How about the proportion of successes?
## Common Misconceptions regarding the LLN
• Misconception 1: If we observe a streak of 10 consecutive heads (when p=0.5, say) the odds of the 11th trial being a Head is > p! This is of course, incorrect, as the coin tosses are independent trials (an example of a memoryless process).
• Misconception 2: If run large number of coin tosses, the number of heads and number of tails become more and more equal. This is incorrect, as the LLN only guarantees that the sample proportion of heads will converge to the true population proportion (the p parameter that we selected). In fact, the difference |Heads - Tails| diverges!
## References
• Dinov, ID., Christou, N., Gould, R Law of Large Numbers: the Theory, Applications and Technology-based Education. JSE, Vol. 17, No. 1, 1-15, 2009.
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http://math.stackexchange.com/questions/211265/continuity-of-floor-function-in-mathbb-z
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# continuity of floor function in $\mathbb Z$
Suppose $$f(x)=\left|x-2\left\lfloor\frac{x+1}{2}\right\rfloor\right|$$ To prove the continuity of $f$ in $\mathbb {R}$ we have to prove it first in $\mathbb {Z}$ then at $\mathbb{R} - \mathbb{Z}$. But why do we have to prove the continuity if $x=2n$ and $x=2n-1$ ($n \in \mathbb {N}$ odd or even) when we want to prove the continuity in $\mathbb {Z}$? (I saw this in a solution but I didn't understand it)
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## 2 Answers
Presumably, the solution that you read thought that the best way to show continuity for all integers was to to it in two cases: when we have odd integers and when we have even integers.
To show that the function is continuous at all integers is equivalent to showing that it's continuous at both the even and odd integers.
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If you try to draw a graph of this function, you'll see why it's necessary to deal with even and odd numbers separately. Basically, it's a kind of zigzag - at each integer, it has a sharp tooth. Its continuity away from the integers is never in any doubt; but the reason why it's continuous at the upward-pointing teeth (odd integers) is completely different from the reason why it's continuous at the downward-pointing teeth (even integers). It would be disproportionately difficult to try and give a continuity proof that works for all integers, without splitting them into the two separate cases.
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http://physics.stackexchange.com/questions/30157/at-what-g-is-terminal-velocity-not-terminal/30162
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# At what g is terminal velocity not terminal?
How weak would gravity need to be in order for a human to reliably survive the terminal velocity of falling through air?
(Context: watching scifi on a space station with a variety of artificial gravities, it occurred to me that medium-strength gravity would have some advantages; also I note that insects already seem to have this luxury on earth, with their small masses and high air resistances...)
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I imagine the answer is low enough to make walking impossible, but I don't know for sure. – Nathaniel Jun 15 '12 at 17:59
Related question that I thought .. what is the biggest size animal that can survive a terminal velocity drops. Ants can do it, humans can't. Where is the transition? – ja72 Jun 16 '12 at 1:58
## 2 Answers
References for the terminal falling velocity of a skydiver give numbers in the range of $54 m/s$ to $76 m/s$. I expect that the real range is even larger, since it's strongly affected by the orientation and body position of the skydiver.
For normal atmospheric drag on this scale, we have the fairly good approximation that force is proportional to velocity squared, $F_{air} \propto v^2$. For normal terminal falling conditions, the upward air force is exactly equal to gravity, and the air force is a function of velocity exclusively because we're assuming the same air composition for now.
You ask about a fall that is "reliably survivable", I should note that this would still NOT be a comfortable fall. In low gravity environments, you could expect some gnarly bounces to top it all off. I don't have a good reference for this, but I would anecdotally put it in the neighborhood of a $40 ft$ fall, which would equate to about $15 m/s$.
I will refer to the air force during a normal terminal velocity fall on Earth as $F_{air}$ and introduce $F_{air}'$ for the force on the new planet. I'll do some simple equivalencies to get an answer for the new gravity needed.
$$m g = F_{air} = (\text{const}) v^2$$
$$\frac{F_{air}}{F_{air}'} = \frac{v^2}{v'^2} = \frac{g}{g'}$$
$$\frac{v}{v'} = \frac{60 m/s}{15 m/s}$$
$$\frac{g}{g'} = \left(\frac{60}{15}\right)^2 = 16$$
So my answer is quite simply that gravity would have to be 1/16th as strong as it is on Earth, or $0.6 m/s$. Are there any bodies in the solar system quite like this? Wikipedia is helpful here. Several bodies come close, like Pluto, Eris, or Triton, but none of them have much of an atmosphere. It is fun to think about, but I doubt that an atmosphere of such a high density with such a low gravity will be found in our local celestial neighborhood.
Walking would be difficult if such a planet existed, but not impossible. The moon is 1/3rd the gravity of Earth, so this hypothetical planet would be roughly 5 times less gravity than the moon. It would be very bouncy, but still very different from zero-gravity.
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Your figure of $0.6m/s^2$ doesn't match the moons you've selected; off by a factor of 10. Pluto would be a better example. – Mark Beadles Jun 15 '12 at 20:13
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@MarkBeadles This was apparently a careless mistake where I was reading from the column with the wrong units. Edited to hopefully be correct now. – AlanSE Jun 15 '12 at 20:35
– Nathaniel Jun 15 '12 at 21:28
@Nathaniel You're entirely right. I tried to avoid saying it would be easy. For low gravity environments less than 1/10th Earth gravity, it might feel more like zero gravity than a gravity environment, except for the fact that objects still collect on the surface. You might not be able to do much resembling normal walking – AlanSE Jun 17 '12 at 15:24
Also, note that the air pressure is going to be dependent on the size of $g$ (pressure is determined by the weight of overhead air), so a smaller $g$ will imply less pressure, thus, less air resistance, so this is probably an overestimate. – Jerry Schirmer Sep 25 '12 at 17:45
The smallest body that we know has an atmosphere is Titan, which has about 1/7 of Earth's surface gravity ($1.4\ m/s^2$) but an atmospheric pressure of 1.45 Earth's ($146.7\ kPa$). Using $pV=nRT$ where $T=95K$ and a mean molar mass of Titan's atmosphere being $28.6\ g/mol$ allows us to calculate an atmospheric density of $5.87 kg/m^3$, greater than Earth's.
So since there is less gravity but more atmosphere, I decided to look at the specific question of whether a terminal velocity fall on Titan is survivable. The answer was enlightening.
Using the formula for terminal velocity $$V_t = \sqrt{\frac{2mg}{\rho AC_d}}$$
where (using reasonable estimates for the human coefficients): $$m_{human} = 75\ kg\\ g_{Titan}=1.4\ m/s^2\\ C_{d\ human}= 1.0\\ \rho_{Titan}= 5.87\ kg/m^3 \\ A_{human}= 0.75\ m^2$$
we get the interestingly low figure of
$$6.9\ m/s$$
Such a fall would be survivable.
Of course, the extreme cold and unbreathable atmosphere would not be, but at least the fall wouldn't kill you.
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http://mathoverflow.net/questions/17819?sort=newest
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## What’s the origin of the naming convention for the standard basis of sl_2?
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$\mathfrak{sl}_2(\mathbb{C})$ is usually given a basis $H, X, Y$ satisfying $[H, X] = 2X, [H, Y] = -2Y, [X, Y] = H$. What is the origin of the use of the letter $H$? (It certainly doesn't stand for "Cartan.") My guess, based on similarities between these commutator relations and ones I have seen mentioned when people talk about physics, is that $H$ stands for "Hamiltonian." Is this true? Even if it's not, is there a connection?
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For me, it is usually named $E$, $F$, $H$... – Mariano Suárez-Alvarez Mar 11 2010 at 4:19
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X, Y, H is the convention used by the Wikipedia article and by Fulton and Harris. I hope the answer isn't just that H is the next letter after G... – Qiaochu Yuan Mar 11 2010 at 5:05
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The connection to Hamiltonians seems unlikely to me. As the maximal torus of G is a subgroup, calling it H and the corresponding Lie algebra element h is quite natural. I hope someone can pinpoint the etymology though. – Q.Q.J. Mar 11 2010 at 5:25
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I had always imagined that H was called H because Cartan subalgebras are always called mathfrak{h}. Of course I now realise that it could have been the other way around! Which convention came first I wonder? – Kevin Buzzard Mar 11 2010 at 11:43
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Well, at least we can be sure this h wasn't due to Dirichlet. (anyone who doesn't catch the joke should look at mathoverflow.net/questions/17062/…) – KConrad Mar 11 2010 at 14:54
show 5 more comments
## 2 Answers
I always thought that $H$ did stand for Cartan—at least, for ‘Henri’. However, I seem to recall that I had this discussion with Brian Conrad, and that he said it was actually Élie, not Henri, after whom the subgroups were named.
For what it's worth, the $(X, Y, H)$ convention (in preference to $(E, F, H)$) is the one to which I'm accustomed; Carter uses it, for example, in his discussion of Jacobson–Morosov triples. Embarrassingly, I don't know where the Jacobson–Morosov theorem was proven; but that's where I'd look for the history of the name.
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Following Lie and Killing, Elie Cartan (father of Henri) laid foundations for Lie groups and "infinitesimal" groups (later Lie algebras): Paris thesis 1894, etc. Killing studied "Cartan" subalgebras first but didn't invent the "Killing" form. Even in Bourbaki, notation varies: `$(H, X_+, X_1), (X,Y,H), (x,h,y)$`. Jacobson-Morozov comes from separate work (Duke J. 1938?, Doklady note 1942): see Jacobson (1962 book) pp. 98-100, where `$(x,h,y)$` is used. Upper case is common for Lie algebra elements viewed as vector fields or as matrices. Lots of obscure history, but does it matter? – Jim Humphreys Mar 31 2010 at 13:31
Hmmm, my math got distorted in the previous comment. Also, I should have typed `$(H,X_+, X_-)$`. This kind of notation has been fairly popular with physicists but also occurs in Bourbaki. – Jim Humphreys Mar 31 2010 at 16:29
Jim, thanks for the history, and especially for the attribution! I'm not sure what to make of the "does it matter?" question—I guess it matters to Qiaochu, and certainly I find it interesting. – L Spice Mar 31 2010 at 17:05
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Both terminology and notation in Lie theory have varied over time, but as far as I know the letter H comes up naturally (in various fonts) as the next letter after G in the early development of Lie groups. Lie algebras came later, being viewed initially as "infinitesimal groups" and having labels like those of the groups but in lower case Fraktur. Many traditional names are not quite right: "Cartan subalgebras" and such arose in work of Killing, while the "Killing form" seems due to Cartan (as explained by Armand Borel, who accidentally introduced the terminology). It would take a lot of work to track the history of all these things. In his book, Thomas Hawkins is more concerned about the history of ideas. Anyway, both (h,e,f) and (h,x,y) are widely used for the standard basis of a 3-dimensional simple Lie algebra, but I don't know where either of these first occurred. Certainly h belongs to a Cartan subalgebra.
My own unanswered question along these lines is the source of the now standard lower case Greek letter rho for the half-sum of positive roots (or sum of fundamental weights). There was some competition from delta, but other kinds of symbols were also used by Weyl, Harish-Chandra, ....
ADDED: Looking back as far as Weyl's four-part paper in Mathematische Zeitschrift (1925-1926), but not as far back as E. Cartan's Paris thesis, I can see clearly in part III the prominence of the infinitesimal "subgroup" `$\mathfrak{h}$` in the structure theory of infinitesimal groups which he laid out there following Lie, Engel, Cartan. (Part IV treats his character formula using integration over what we would call the compact real form of the semisimple Lie group in the background. But part III covers essentially the Lie algebra structure.) The development starts with a solvable subgroup `$\mathfrak{h}$` and its "roots" in a Fitting decomposition of a general Lie algebra, followed by Cartan's criterion for semisimplicity and then the more familiar root space decomposition. Roots start out more generally as "roots" of the characteristic polynomial of a "regular" element. Jacobson follows a similar line in his 1962 book, reflecting his early visit at IAS and the lecture notes he wrote there inspired by Weyl.
In Weyl you also see an equation of the type `$[h e_\alpha] = \alpha \cdot e_\alpha$`, though his notation is quite variable in different places and sometimes favors lower case, sometimes upper case for similar objects. Early in the papers you see an infinitesimal group `$\mathfrak{a}$` with subgroup `$\mathfrak{b}$`. All of which confirms my original view that H is just the letter of the alphabet following G, as often encountered in modern group theory. (No mystery.)
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Pretty sure Knapp uses $\delta$ for the half-sum. The real showdown is between $\Pi$ and $\Delta$ or $\Delta$ and $R$ for the simple roots and all roots. – Q.Q.J. Mar 13 2010 at 8:18
Probably I picked up the use of `$\delta$` from Jacobson's book on Lie algebras, but conventions have varied for a long time. The names of root systems or simple roots also vary from one source to another. After 1968 Bourbaki standardized usage somewhat, but I think made a mistake in switching to roman letters `$R$` and `$B$` (the former common in English for rings and the latter ubiquitous for Borel subgroups). I've pretty much followed Borel and Tits, using `$\Phi$` and `$\Delta$` (the latter often being replaced by a numbered list of simple roots). – Jim Humphreys Mar 22 2010 at 17:09
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http://mathoverflow.net/questions/40337?sort=newest
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## Ingenuity in mathematics
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[This is just the kind of vague community-wiki question that I would almost certainly turn my nose up at if it were asked by someone else, so I apologise in advance, but these sorts of questions do come up on MO with some regularity now so I thought I'd try my luck]
I have just been asked by the Royal Society of Arts to come along to a lunchtime seminar on "ingenuity". As you can probably guess from the location, this is not a mathematical event. In the email to me with the invitation, it says they're inviting me "...as I suppose that some mathematical proofs exhibit ingenuity in their methods." :-)
The email actually defines ingenuity for me: it says it's "ideas that solve a problem in an unusually neat, clever, or surprising way.". My instinct now would usually be to collect a bunch of cute low-level mathematical results with snappy neat clever and/or surprising proofs, e.g. by scouring my memory for such things, over the next few weeks, and then to casually drop some of them into the conversation.
My instinct now, however, is to ask here first, and go back to the old method if this one fails.
## Question: What are some mathematical results with surprising and/or unusually neat proofs?
Now let's see whether this question (a) bombs, (b) gets closed, (c) gets filled with rubbish, (d) gets filled with mostly rubbish but a couple of gems, which I can use to amuse, amaze and impress my lunchtime arty companions and get all the credit myself.
This is Community Wiki of course, and I won't be offended if the general consensus is that these adjectives apply to the vast majority of results and the question gets closed. I'm not so sure they do though---sometimes the proof is "grind it out". Although I don't think I'll be telling the Royal Society of Arts people this, I always felt that Mazur's descent to prove his finiteness results for modular curves was pretty surprising (in that he had enough data to pull the descent off). But I'm sure there are some really neat low-level answers to this.
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I'm in a mellow mood and shan't vote to close, but I think that as penance for asking the question you should do a report on the seminar for us. Slightly more seriously, I think it'll be quite an amazing result that a) has a surprising proof, b) is explainable to non-mathematicians, and c) the fact that the proof is surprising is explainable to non-mathematicians. – Andrew Stacey Sep 28 2010 at 16:56
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+1 for 'I suppose that some mathematical proofs exhibit ingenuity'. Two cultures, anyone? – HW Sep 28 2010 at 17:31
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Check out mathoverflow.net/questions/8846/… – Martin Brandenburg Sep 28 2010 at 17:36
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Kevin: doesn't do it for me. Why would I be interested in the sum of the first 1000 numbers? Sounds like the sort of thing my kids would do before they knew any better. – Andrew Stacey Sep 28 2010 at 19:33
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I think, by the way, that "I need some examples for a talk" is a completely legitimate excuse for a big-list community wiki question. – gowers Sep 28 2010 at 20:03
show 9 more comments
## 33 Answers
Mathematicians are used to seeing these sorts of problems and are generally more intrigued by their solutions, but I came across this recently:
Colour every point in the plane either black or white.
Fix any positive distance $d$.
Can I find two identically coloured points which are a distance $d$ apart?
Answer: Of course. Look at the vertices of an equilateral triangle of side length $d$.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Let me mention that the question is closely related to a more recent one http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking which I think also is about ingenuity in a sense. So indeed I was not sure where this answer better fits but I think the proof I would like to mention belongs here more.
This is Galvin's proof of the Dinitz conjecture: The Dinitz conjecture asserted that if you give me an n by n array and in each square you put a set of size n then you can chose one element from each set so that all the chosen elements in a row or in a column are distinct.
If all the sets are the same (say 1,2,...,n) then you just want a Latin square. You can simply chose at position (i,j) i+j modulo n,
Galvin's proof derive Dinitz conjecture from the Gale-Shapley marriage theorem (Actually he used a theorem of Maffray, related to the stable marriage theorem.). It is short, elementary and extremely surprising. ( I will try to find a link.)
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a. Euler's computation of Zeta(2), (at first with only very weak, handwaving, or no convergence arguments), as redacted in Polya or here. Obviously amazingly ingenious, and interesting for artists, connecting numbers and circles. Requires unerstanding that a polynomial is equal to the product of its first degree factors, possibly it is too well known, however. It also allows one to then revisit the convergence argument and show what mathematicians actually worry about, after the "flash of ingenuity"
b. A lovely agument I heard a long time ago on sci.math relating to Sagan's book "Contact" in which a message is encoded in the bits of Pi. Someone asked whether a deity could arrange for Pi to be a different real number. Opinions went back and forth relating to spacetime, etc. Then someone asked, in light of any of the familiar series expansions, "If Pi were different, which natural number would be missing or duplicated, and how might that be?" Leads to a discussion of the Peano axioms. Everyone goes home wondering. :-)
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I like very much the proof of fundamental theorem of algebra (using the winding number around the origin), but it will probably take too long to explain...
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Consider a random walk on $\mathbb{Z}/n\mathbb{Z}$ starting at 0. At each step either add 1 or subtract 1, with probability 1/2 each. Let $0\neq i\in\mathbb{Z}/n\mathbb{Z}$. What is the probability $P_n(i)$ that $i$ is the last point to remain unvisited?
Solution. Consider the first time the walk reaches a point adjacent to $i$ (either $i-1$ or $i+1$). The probability that $i$ is the last point to remain unvisited is then the probability that the walk visits the other point adjacent to $i$ before visiting $i$. This probability is independent of $i$. Hence $P_n(i)=1/(n-1)$. (I don't know the origin of this classic problem.)
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One of my students created the following proof that the medians of a triangle are concurrent while waiting to talk to me in office hours.
Choose any two medians.
1) the two medians do meet somewhere inside the triangle.
2) If we join the centers of the three sides, obtaining a "central triangle",
then the two given medians are also medians of that central triangle,
hence their meeting point is also inside that triangle.
3) We are done.
I.e. iterating the procedure shows that any two medians meet at the unique point common to all central triangles.
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You need to show that the triangles doesn't have more than one common point: You could make this "proof" for any three lines from the vertices to the opposite sides. The difference is, that for the median, the area of the central triangles goes to 0, but it doesn't in general. – Sune Jakobsen Dec 15 2010 at 18:38
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I don't know if this meets your requirement for being elementary, but the Eilenberg Swindle is very easy, very clever, and widely applicable.
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This obviously doesn't meet the requirement for being elementary. And widely applicable? Only if you are a mathematician; this talk is for artists!!! – Sheikraisinrollbank Nov 30 2010 at 15:50
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I think the 3 children with dirty faces puzzle is a good one for this type of talk (also referred to as the "blue-eyed islanders puzzle" on Tao's blog http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/):
I came home yesterday to find that my 3 kids had been playing rugby and all had dirty foreheads. I told them that whoever could tell me whether their own forehead was dirty would get a shiny new bike. After looking at each other for a moment, they seemed uncertain. I asked again, "No ones knows?" Still no reply. Finally I shrugged and said, "Well, if you can't tell me then no one gets a bike," at which point all three simultaneously shouted: "My forehead is dirty!".
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Alex R's answer reminds me of another sort of clever "strategy" problem. The way I've heard it phrased is as follows (with apologies to the vegetarians in the audience):
You are given 1000 cups filled with water, exactly one of which (unknown to you) is laced with a lethal poison that is guaranteed to kill within 24 hours if ingested. You are given ten rats with which to determine the poisoned cup. What is the minimum amount of time needed to be certain?
One naive approach would be to break up the cups into groups of 100 to be fed to each rat (i.e. rat 1 gets cups 1 through 100, rat 2 gets 101 through 200 etc). After 24 hours one rat will be dead and you have narrowed the poisoned cup to one in 100. Repeat two more times: you have your poisoned cup identified within 72 hours.
There is in fact a much better solution, which identifies the cup in the minimal 24 hours. Since $1000~<~2^{10} = 1024$, assign each cup a number expressed in binary. Then give the water in that cup to those rats for which there is a '1' in the binary representation (so, for example, cup 17 = 10001 is fed to rats 1 and 5). After 24 hours, you can simply read off the number of the poisoned cup by the numbers of the dead rats.
You can impress your audience with the remarkable (some might say brutal) efficiency of this procedure: to identify one in a million cups you would need only 20 rats (ignoring, of course, any deaths caused by overhydration...)
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Anyone who likes this problem might also enjoy thinking about the generalization where two or more cups are poisoned, which seems much harder to me: math.stackexchange.com/questions/639/… – Qiaochu Yuan Nov 30 2010 at 21:54
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A recent reference is "Street fighting mathematics" by Sanjoy Mahajan.
AT http://www.amazon.com/Street-Fighting-Mathematics-Educated-Guessing-Opportunistic/dp/026251429X
I have browse and read some part. It look as if it almost manages to render NavierStokes equation edible for an hard die finitist. It uses mainly dimensionality but is full of ingenuity.
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One of my favorites is von Neumann's randomness extractor:
Suppose you have a biased coin that comes up heads with probability $p$ and tails with $q=1-p$. How do you construct an unbiased coin from the biased one?
The answer is simple but really nice. Simply toss the bad coin twice, and discard $HH$ and $TT$. The other two events occur with equal probabilities, $pq$ and $qp$.
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I like evaluating the integral $$\int_{-\infty}^{\infty} e^{-x^2} dx.$$
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The RSA algorithm publicly described in 1978 by Ron Rivest, Adi Shamir, and Leonard Adleman at MIT is a classic example of "Ingenuinity in Mathematics". Its the famous Public Key Cryptography scheme widely used everywhere and is solidly founded on Mathematics.
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How about the fact that in hex the first player has a winning strategy?
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Among the fairly recent results, I like the proof of the joints conjecture (the final clean version that can be explained in a few sentences to anyone familiar with polynomials, derivatives, and elementary linear algebra). This may be a bit hard to the general audience though. Among all combinatorial geometry theorems, Bang's solution of Tarski's plank problem is unbeatable in my humble opinion. When presenting it, just do it for the unit ball in $\mathbb R^n$ assuming that all strips pass through the origin. It is then a one-liner (If the strips are `$S_j=\{x:|(x,v_j)|\le |v_j|^2\}$` and $\sum_j |v_j|<1$, the point $y=\sum_j \pm v_j$ with the largest possible norm lies in the ball but in none of the strips). Once you have this idea in its pure form, the rest can be figured out in finite time. But how does one come up with ideas like that? (this is not a rhetoric question, by the way).
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Here is a proof of Pythagoras's theorem that I like:
• You want to prove that the sum of the squares on each of the non-hypotenuse sides equals the square on the hypotenuse.
• You generalize, and instead prove that for any shape, if you scale it by $a$, and then by $b$, the sum of the resulting areas is the area of the shape scaled by $c$. (We began with the case of the unit square.)
• By thinking about how areas scale, it suffices to check for one particular shape.
• We check it by taking the shape to be the original triangle (to be pedantic: scaled so that its hypotenuse has length one). This case is clear: just drop a perpendicular from the vertex opposite the hypotenuse to the hypotenuse, and see note that the triangle with hypotenuse length $c$ is the sum of two similar triangle of hypotenuse lengths $a$ and $b$.
Obviously you are not going to drop this into casual conversation: it requires a focused effort at explanation. But I think it illustrates something true, and fairly general, about how mathematicians argue.
For example, the squares that we are adding get transmuted from areas of specific shapes, to scaling factors for areas of quite general shapes. This lets us go from checking something tricky (or trying to make tricky geometric constructions with the squares on the three sides, to see how they are related) to checking something that is immediately obvious. So we also see two kinds of ingenuity: the ingenious reinterpretation of the meaning of the squares (ingenious on a conceptual level, which is certainly a very common form of mathematical ingenuity), and the ingenuity of drawing a single line on the original triangle to break it into two triangles similar to itself (which is ingenious on a more visceral level --- a single stroke of ingenuity).
One other advantage: the other participants quite likely have heard of Pythagoras's theorem, and may even remember it, but are unlikely to know a proof. (Or do painters have some training in Euclidean geometry?) Knowing the result, they may be better positioned to appreciate the ingenuity of the proof.
Also: this argument appeared somewhere else recently on MO, I think, but I forget where. (Added: here, in a reponse of Dick Palais to Timothy Gowers's question about making something easier by generalization.)
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There are irrational numbers $x$ and $y$ such that $x^y$ is rational. For consider $\sqrt{2}^\sqrt{2}$. This number is either rational or irrational. If it's rational then we're done. Otherwise, it's irrational, and $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = 2$, so we're still done.
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This is my favorite example of a non-constructive proof. – Derrick Stolee Nov 30 2010 at 7:40
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@Dinesh: but then you first have to prove that $e$ and $ln 2$ are irrational, which is slightly harder than the proof of irrationality of $\sqrt{2}$ – Jan Jitse Venselaar Nov 30 2010 at 15:58
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Well, let's choose $x=\sqrt{2}$, $y=2\log_2 3$. It is easy to prove that both $x$, $y$ are irrational. – Fedor Petrov Nov 30 2010 at 22:35
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While I like this result, I'm not sure if this is the sort of thing that will really "wow" a room full of artists, for whom (I imagine) notions of rationality and irrationality aren't familiar enough for this to be anything more than a remote curiosity. – Nick Salter Dec 1 2010 at 20:47
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i taught a class for gifted 8-10 year olds this summer and one of them showed me this proof! – roy smith Aug 23 2011 at 2:53
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The game of chomp: http://en.wikipedia.org/wiki/Chomp where player 1 has a winning strategy, but no one really knows how to find it.
It's a neat example of a nonconstructive proof of existence (and has a short proof).
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I know how to find it. Wanna play chomp? – Sheikraisinrollbank Dec 3 2010 at 12:04
"At any party, there are at least two people with the same number of friends there".
This typically won't be obvious to someone who isn't a mathematician. However, it's quite easy for many people to see this by picturing everyone at the party wearing a t-shirt displaying how many of their friends are present. The number of possible t-shirts is one less than the number of attendees, so there must be a double somewhere.
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You need a little more to complete the proof, since, for $n$ people at the party, there are, a priori, $n$ possible t-shirts, bearing the numbers 0 through $n-1$. The extra step you need is that 0 and $n-1$ can't both occur (though either one can). – Andreas Blass Sep 29 2010 at 13:03
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First, I'd suggest Erathostenes method of computing the radius of the Earth, unless it is really well-known to them.
Otherwise, here's a nice one. You go for a bike ride. At the end, each wheel has made a closed simple curve, and, suppose, you noticed that these two curves never met. What's the area encosed between them? The distance between the centers of the wheels is $r$.
Answer : $\pi r^2.$ Explanation: the line passing for the two contact points of the wheels on the ground, is tangent to the back wheel's curve, and during the trip, it made a complete $2\pi$ rotation. To convince your audience: imagine a disk (a pizza) cut into several very thin slices (with vertex in the center as usual). Make the slices slide on each other, so that a hole appears in the middle of the pizza. The curve bounding the hole is the one traced by the back wheel; the exterior boundary is the curve traced by the other wheel.
```` <
/___~
() \- ()
````
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done. A pizza bike for you! – Pietro Majer Sep 29 2010 at 10:30
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Linear algebra proof of Binet's formula for Fibonacci numbers.
Fibonacci numbers satisfy
$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} = \begin{pmatrix} F_{n+1} & F_{n} \\ F_{n} & F_{n-1} \end{pmatrix}$
Diagonalize the matrix on left.
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That's a really cool one,Fubini. John Kennedy showed to me originally and it shows the power of linear algebra manifesting in a really unexpected place. – Andrew L Sep 29 2010 at 0:33
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In fact, I'd go with the identity $F_{n+1}F_{n-1}-F_n^2=(-1)^n$ instead of the Binet formula. – Victor Protsak Sep 29 2010 at 5:55
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This is indeed a cool fact. However, I am certain any non-math person would be lost at the term matrix, let alone diagonalize, especially when said non-math people merely suppose some mathematical proofs exhibit ingenuity :) – Peter Luthy Sep 29 2010 at 8:26
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@AndrewL: It's never surprising when the power of linear algebra manifests itself. – Sheikraisinrollbank Nov 30 2010 at 15:29
Sylvester-Gallai Theorem:
n points in the 2D plane, not all collinear. There is a line which passes through exactly two points.
Proof: Dualize. Points<->Lines. In dual, we need to show no more than 3 lines intersect. Consider intersection points and lines as planar graph. Planar graphs have nodes with degree no more than 5. Done.
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The problem:
A rectangle R can be tiled with smaller rectangles such that
• The sides of the smaller rectanges are parallel the sides of R.
• At least one side of each of the smaller rectangle is integral.
Show that at least one side of R is integral.
The proof:
Consider $\iint_{R} e^{2 \pi i (x+y)} dxdy$
This is zero, by adding up along each small rectangle. The result follows.
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I think an analogous proof that can be explained to artists is to tile the rectangle w/ a black and white checkerboard parallel to the rectangles with squares of width 1/2. You can then show that a rectangle having equal black and white areas (of overlapped checkerboard squares) is necessary and sufficient for determining that it has an integral side. Each rectangle with integral sides then has equal black and white squares, which combine to show that the large rectangle has the same property and thus an integral side itself. – Randy Qian Sep 28 2010 at 19:45
The impossibility of tiling a chessboard with two opposite corners removed using dominos is quite good for this purpose I think, especially if you start by giving a boring case-analysis proof for a 4-by-4 board.
The bridges of Königsberg is also pretty good. Marcus du Sautoy spoke about it last night in his series A Brief History of Mathematics on Radio 4 (though he overdid it when he claimed that the solution had "revolutionized the internet").
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+1 for the chess board. It's my canonical example for such occasions. – Alex Bartel Sep 29 2010 at 0:25
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The chessboard problem can be nicely continued by proving that the chessboard can be tiled with dominos if two differently coloured squares are removed. Check out the solution of puzzle B at gurmeetsingh.wordpress.com/2008/09/12/… – Tamas Hausel Sep 29 2010 at 16:24
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How about the classic 20 prisoner problem. A warden tells 20 prisoners that they can live if they survive the following game: tomorrow the warden will line up the prisoners single file (at random), facing forward and place either a red or blue hat on each prisoner. Each prisoner can only see the color of hats infront of him. The warden starts from the back of the line and asks each prisoner in turn what color hat they think they are wearing. If a prisoner answers correctly they live, otherwise they die. No communication is allowed between prisoners once the game starts and they can only shout "red" or "blue" when it's their turn. The prisoners have the evening to discuss a strategy.What is the best strategy for the prisoners? How many prisoners can be guaranteed to live.
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Apart from being not for the general audience, I think that it's also wrong. It relies on a kind of constructive version of the axiom of choice, which of course doesn't exist, otherwise it wouldn't have to be an axiom. The AC is an existence statement and not a procedure for making a choice. In particular the sentence from the wikipedia page "Using the axiom of choice, they select and memorize a representative sequence from each equivalence class" doesn't make any sense to me. – Alex Bartel Sep 29 2010 at 4:55
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The Monty Hall Problem is one possibility. I usually pose the problem as, "Should you always switch or always stay or does it even matter?" The argument that switching is the best strategy is itself not especially neat from a mathematical point-of-view, but since the result completely defies human intuition, I think the proof seems quite ingenious, especially to people outside math. It certainly solves a problem in a surprising or unusual way in that the answer is shocking. If you decide to do it, I would recommend bringing in 3 cards (two spades and a heart, say) as props to aid in your explanation and discussion.
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Much easier explanation: Suppose there are a hundred doors... – Harry Gindi Nov 30 2010 at 17:37
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The problem with explaining this to a general audience is not that they won't believe your explanation of why it's 1/3 to 2/3, but rather that they will generate one or more explanations of why it's even probability, believe those too, and get confused. – Elizabeth S. Q. Goodman Dec 1 2010 at 7:24
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On the walk home I remembered the following. You play a game with a friend. The friend deals from a pack of cards, turning each card face up one by one. After each card, you have the opportinity to say "STOP" (and after the 51st card has been turned over you have to say "STOP"). The next card is then turned over, and you win if it's red. Can you devise a strategy which enables you to win on average with probability over 50 percent?
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To complete the triviality of the situation, after your audience proposes a bunch of complex strategies, propose an alternative game whereby after "STOP" it's the last card in the deck that gets turned over. Everyone sees that this game is strategy-indifferent, and then you deliver the coup de grace... I've used this several times in math circles. Gets them every time. – Alon Amit Nov 30 2010 at 5:19
Christian Blatter gave a wonderful proof of Pick's Theorem using thermodynamics (okay, it's really not that fancy, but it involves a thought experiment about heat distribution). It originally appeared in Math. Mag. Here is a free link: http://www.math.ethz.ch/~blatter/Pick.pdf.
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There is the old classic of a fly flying between two trains.
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I thought it was Gordon Baker's dog ... – gowers Sep 28 2010 at 19:23
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There is a great story about von Neumann being presented with this problem, which he answered instantaneously. "Aha! So you knew how to do it in a clever way. Most people are trying to sum the series, you know". - "What do you mean? That's what I did!". – Victor Protsak Sep 29 2010 at 5:59
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You run a knockout tournament. Some good players get seeded to later rounds, and some players get byes if there are an odd number of players left. How many matches are necessary? (OK, it's well known and completely trivial, but you did imply that you would be talking to artists....)
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http://topologicalmusings.wordpress.com/tag/identity/
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Todd and Vishal’s blog
Topological Musings
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A note on a simple polynomial-logarithmic algebraic identity
November 10, 2008 in Elementary Math Problem Solving, Problem Corner, Some theorems | Tags: gauss, identity, logarithmic, lucas, polynomial | by Vishal | 12 comments
The following “polynomial-logarithmic” algebraic identity that one encounters on many occasions turns out to have a rather useful set of applications!
POLYNOMIAL-LOGARITHMIC IDENTITY: If $P(z)$ is a polynomial of degree $n \ge 1$ with roots $a_1, a_2, \ldots, a_n$, then $\displaystyle \frac{P'(z)}{P(z)} = \frac1{z-a_1} + \frac1{z-a_2} + \ldots + \frac1{z-a_n}$.
PROOF: This one is left as a simple exercise. (Hint: Logarithms!)
A nice application of the above identity is found in one of the exercises from the chapter titled Analysis (p120) in Proofs from the Book by Aigner, Ziegler and Hofmann.
EXERCISE: Let $p(x)$ be a non-constant polynomial with only real zeros. Show that $p'(x)^2 \ge p(x) p''(x)$ for all $x \in \mathbb{R}$.
SOLUTION: If $x = a_i$ is a zero of $p(x)$, then the right hand side of the above inequality equals zero, and we are done. So, suppose $x$ is not a root of $p(x)$. Then, differentiating the above identity w.r.t. $x$, we obtain $\displaystyle \frac{p''(x)p(x) - p'(x)^2}{p(x)^2} = - \sum_{k=1}^n \frac1{(x - a_k)^2} < 0$, and we are done.
It turns out that the above identity can also used to prove the well-known Gauss-Lucas theorem.
GAUSS-LUCAS: If $P$ is a non-constant polynomial, then the zeros of $P'$ lie in the convex hull of the roots of $P$.
PROOF: See this.
HISTORY: The well-known Russian author V.V. Prasolov in his book Polynomials offers a brief and interesting historical background of the theorem, in which he points out that Gauss’ original proof (in 1836) of a variant of the theorem was motivated by physical concepts, and it was only in 1874 that F. Lucas, a French Engineer, formulated and proved the above theorem. (Note that the Gauss-Lucas theorem can also be thought of as some sort of a generalization (at least, in spirit!) of Rolle’s theorem.)
Even though I knew the aforesaid identity before, it was once again brought to my attention through a nice (and elementary) article, titled On an Algebraic Identity by Roberto Bosch Cabrera, available at Mathematical Reflections. In particular, Cabrera offers a simple solution, based on an application of the given identity, to the following problem (posed in the 2006 4th issue of Mathematical Reflections), the solution to which had either escaped regular problem solvers or required knowledge of some tedious (albeit elementary) technique.
PROBLEM: Evaluate the sum $\displaystyle \sum_{k=0}^{n-1} \frac1{1 + 8\sin^2 (k\pi /n)}$. (proposed by Dorin Andrica and Mihai Piticari.)
SOLUTION: (Read Cabrera’s article.)
There is yet another problem which has a nice solution based again on our beloved identity!
PROBLEM: (Putnam A3/2005) Let $p(z)$ be a polynomial of degree $n$, all of whose zeros have absolute value 1 in the complex plane. Put $g(z) = p(z)/z^{n/2}$. Show that all zeros of $g'(z) = 0$ have absolute value 1.
SOLUTION: (Again, read Cabrera’s article.)
Sophie Germain Identity
February 20, 2008 in Problem Corner | Tags: identity, Sophie Germain | by Vishal | 7 comments
Those who love (elementary) problem-solving eventually come across the Sophie Germain identity. It has lots of applications in elementary number theory, algebra and so on. The identity states
$x^4 + 4y^4 = (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$.
Indeed, note that
$x^4 + 4y^4$
$= (x^2)^2 + (2y^2)^2 + 2\cdot x^2 \cdot 2y^2 - 2\cdot x^2 \cdot 2y^2$
$= (x^2 + 2y^2)^2 - (2xy)^2$
$= (x^2 - 2xy + 2y^2)(x^2 + 2xy + 2y^2)$
More can read about Marie-Sophie Germain, a brilliant mathematician, here and here.
Let us use the above identity in solving a couple of problems. Here is the first one.
Problem 1: Evaluate $\displaystyle \sum_{k=1}^{n}\frac{4k}{4k^4 + 1}$.
Solution: A first glance tells us that the sum should somehow “telescope.” But the denominator looks somewhat nasty! And, it is here that the above identity comes to our rescue. Using the Sophie Germain identity, we first note that
$1 + 4k^4 = (1 -2k + 2k^2)(1 + 2k + 2k^2).$
We thus have
$\displaystyle \sum_{k=1}^{n} \frac{4k}{4k^4 + 1}$
$\displaystyle = \sum_{k=1}^{n} \left(\frac{(1 + 2k + 2k^2)-(1 -2k + 2k^2)}{(1 -2k + 2k^2)(1 + 2k + 2k^2)}\right)$
$\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 + 2k + 2k^2}\right)$
$\displaystyle = \sum_{k=1}^{n} \left(\frac1{1 -2k + 2k^2} - \frac1{1 - 2(k+1) + 2(k+1)^2} \right)$
$\displaystyle = 1 - \frac1{1 + 2n + 2n^2}$.
Here is another one.
Problem 2: Show that $n^4 + 4^n$ is a prime iff $n=1$, where $n \in \mathbb{N}$.
Solution: Note that if $n$ is even, then the expression is clearly composite. If $n$ is odd, say, $2k+1$ for some $k \in \mathbb{N}$, then we have
$n^4 + 4^n$
$= n^4 + 4^{2k+1}$
$= n^4 + 4\cdot (2^k)^4$
$= (n^2 - n\cdot 2^{k+1} + 2^{2k+1})(n^2 + n\cdot 2^{k+1} + 2^{2k+1})$
And, if $n > 1$, then both the factors above are greater than $1$, and hence the expression is composite. Moreover, if $n = 1$, we have $1^4 + 4^n = 5$, which is prime. And, we are done.
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http://www.physicsforums.com/showthread.php?t=144171
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Physics Forums
## Electric field arising from a uniform charge distribution of infinite extent
Consider a uniform charge distribution occupying all of (flat) spacetime,
$$\rho(t,x,y,z) = \text{constant} \;\;\;\;\; ,\; (t,x,y,z) \in R^{1,3}$$
Because this charge distribution is translationally invariant, it seems reasonable to expect that the electric field arising from the charge distribution is zero,
$$E(t,x,y,z) = 0 \;\;\;\;\;\;\;\;\; ,\; (t,x,y,z) \in R^{1,3}$$
But then the electric field does not appear to satisfy Poisson's equation,
$$\nabla\cdot E = \rho/\epsilon_0$$
Presumably this problem has a simple, well-known solution, but I have not encountered it before. Can anyone provide a reference or some insight?
PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus
Recognitions: Science Advisor The solution is simple, E = 0 everywhere as symmetry demands. The crux of the paradox is the boundary conditions of the problem. It is implicitly assumed in Poisson's equation that the charge density tends to zero as one approaches infinity. Your scenario of course, does not have the charge density tending to zero, hence Poisson's equation is actually rendered invalid in this case. Poissons equation is rendered invalid not just for this equation, but for an infinite line charge, infinite sheet charge, in fact, any charge distribution where the boundary condition of charge density tends to zero as we approach infinity. Why does Poisson's equation possess this condition? Because it is derived from how we define work and potential, which requires the condition that potential and thus charge density is zero at infinity. Claude.
Claude, thankyou for casting some light on the matter! You mentioned that Poisson's equation is not applicable to charge distributions of infinite extent. Are Maxwell's equations defined for such systems? Also, do you know of a book which discusses this?
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## Electric field arising from a uniform charge distribution of infinite extent
Maxwell's equations in integral form are valid for any distribution of (electric) charge, it's just that you have to take into account the boundary conditions on th charge & current which occur when you pass from the (always) correct integral form to the differential form. The point i'm trying to make is that you don't have to take the differential equations for granted.
Daniel.
Quote by dextercioby Maxwell's equations in integral form are valid for any distribution of (electric) charge, it's just that you have to take into account the boundary conditions on th charge & current which occur when you pass from the (always) correct integral form to the differential form. Daniel.
Daniel, thanks for the clarification; I can see how, for an infinite charge distribution, Gauss's Law applies but is not equivalent to Poisson's equation.
I find it surprising that the equations in integral form are more fundamental than the equations in differential form. Contrast this with particle physics, where the need to keep all interactions local (no action-at-a-distance) leads to the equations of motion being expressed in differential form.
I wonder whether it would be possible to obtain a differential formulation by dividing the charge distribution into two patches and introducing a potential for each patch? Poisson's equation could be applied to a single patch because each patch has a boundary. This is somewhat reminiscent of the need to use two patches, each with its own vector potential, to describe the field arising from a magnetic monopole. The two vector potentials are related by a gauge transformation; could something similar apply to the two scalar potentials? Perhaps there is a topological charge hidden in this problem???
Quote by Claude Bile The solution is simple, E = 0 everywhere as symmetry demands. The crux of the paradox is the boundary conditions of the problem. It is implicitly assumed in Poisson's equation that the charge density tends to zero as one approaches infinity. Your scenario of course, does not have the charge density tending to zero, hence Poisson's equation is actually rendered invalid in this case. Poissons equation is rendered invalid not just for this equation, but for an infinite line charge, infinite sheet charge, in fact, any charge distribution where the boundary condition of charge density tends to zero as we approach infinity. Why does Poisson's equation possess this condition? Because it is derived from how we define work and potential, which requires the condition that potential and thus charge density is zero at infinity. Claude.
Actually, you can come up with a large number of electric fields which satisfy the equation in general, so the problem is worse than just "E = 0". You can check to see that
[tex]
\vec{E} = V_0 (x \hat{x} + y \hat{y})
[/tex]
satisfies Gauss' law quite nicely, so it turns out that there are an infinite number of solutions for the given charge distribution.
Quote by Claude Bile The solution is simple, E = 0 everywhere as symmetry demands. The crux of the paradox is the boundary conditions of the problem. It is implicitly assumed in Poisson's equation that the charge density tends to zero as one approaches infinity. Your scenario of course, does not have the charge density tending to zero, hence Poisson's equation is actually rendered invalid in this case. Poissons equation is rendered invalid not just for this equation, but for an infinite line charge, infinite sheet charge, in fact, any charge distribution where the boundary condition of charge density tends to zero as we approach infinity. Why does Poisson's equation possess this condition? Because it is derived from how we define work and potential, which requires the condition that potential and thus charge density is zero at infinity. Claude.
How does Poisson's equation assume the potential vanishes at infinity? Mathematically, you can get there from the integral form of Gauss' Law by applying the divergence theorem and recognizing that any curl-less vector field can be represented as the gradient of a scalar field. All of this is done locally without reference to boundary conditions. So, where does the problem come from?
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Quote by Parlyne How does Poisson's equation assume the potential vanishes at infinity?
The definition of V demands that the potential be zero at infinity. Once you include V in your calculations you therefore are implicity assuming some boundary conditions. Gauss Law and Poisson's equation therefore does make reference to boundary conditions, it is just buried in how quantities are defined.
Claude.
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Quote by su2 Claude, thankyou for casting some light on the matter! You mentioned that Poisson's equation is not applicable to charge distributions of infinite extent. Are Maxwell's equations defined for such systems? Also, do you know of a book which discusses this?
I don't know of any books, my readings on this subject has been limited to journal articles. This journal article may contain some of the discussion you are after.
"Generalization of the electrostatic potential function for an infinite charge distribution" Palma et al. American Journal of Physics, 71 (8) 813-815.
Quote by Claude Bile The definition of V demands that the potential be zero at infinity. Once you include V in your calculations you therefore are implicity assuming some boundary conditions. Gauss Law and Poisson's equation therefore does make reference to boundary conditions, it is just buried in how quantities are defined. Claude.
The definition of V in no way assumes that. V is simply defined to be that scalar field such that $$\vec{E} = -\nabla V$$. This makes no assumptions about behavior at infinity. All it assumes is that $$\nabla \times \vec{E} = 0$$. For that matter, adding an arbitrary constant or scalar function of time to V will not change $$\vec{E}$$; so, if you give me a V that vanishes at infinity, I can give you an equally valid one that doesn't.
It's certainly true that in solving Poisson's equation (or Gauss' Law, or what have you) that you need to impose boundary conditions to get a unique solution. However, no boundary conditions are actually assumed in the definition of V (or $$\vec{E}$$ for that matter).
Recognitions: Science Advisor At some point you need to set a reference point for V. If we use differential laws, we MUST use the reference that V approaches zero at infinity, because this reference is set as we move from the integral form to the differential form (via the definition of an absolute potential). If we stick with the integral form, we can use our own reference point. EDIT: I should include that the definition for absolute potential is the work required to move a +1 C charge from infinity to a given point. The value for absolute potential at infinity must therefore approach zero. Claude.
Quote by Claude Bile At some point you need to set a reference point for V. If we use differential laws, we MUST use the reference that V approaches zero at infinity, because this reference is set as we move from the integral form to the differential form (via the definition of an absolute potential). If we stick with the integral form, we can use our own reference point.
This is simply not correct. To solve the differential equation for the potential, we need to set boundary conditions. However, the differential equation is valid independent of the boundary conditions. Furthermore, we have near complete freedom to set our boundary conditions. If it were necessary that the potential be 0 at infinity, we wouldn't have full gauge invariance.
EDIT: I should include that the definition for absolute potential is the work required to move a +1 C charge from infinity to a given point. The value for absolute potential at infinity must therefore approach zero. Claude.
This definition is implicit in the integral equation for the potential; but it is not quite fundamental. I could choose that energy is not 0 at infinity without changing any physical consequences.
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Quote by Parlyne This is simply not correct. To solve the differential equation for the potential, we need to set boundary conditions. However, the differential equation is valid independent of the boundary conditions. Furthermore, we have near complete freedom to set our boundary conditions.
The problem isn't with the boundary conditions of Poisson's equation etc. it's a problem with the boundary conditions of the quantity V. All this hoo-ha can be avoided by choosing the reference point for V at the origin for example, and everything works out fine. Does this not indicate that it is the boundary conditions of V that results in the apparent paradox present in Poisson's equation for infinite charge distributions?
Quote by Parlyne If it were necessary that the potential be 0 at infinity, we wouldn't have full gauge invariance.
Yes, the definition of an absolute potential in this fashion does introduce a constraint, the very constraint in fact, that results in the paradox.
Quote by Parlyne This definition is implicit in the integral equation for the potential; but it is not quite fundamental. I could choose that energy is not 0 at infinity without changing any physical consequences.
Absolutely, but while physical consequences don't change, mathematical ones do. As I have pointed out, it is our choice of our reference for V that results in the apparent paradox, if we choose a different reference, the paradox is avoided.
Claude.
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http://math.stackexchange.com/questions/73071/a-constrained-topological-sort
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A constrained topological sort?
Suppose that one has a directed, acyclic graph G, and each vertex $v$ contains a (positive) value $a_v$. Additionally, let $r$ be a constant. For my purposes, $r>1$, but this might not matter. Let $n$ be the number of vertices in G and let $[n]:=\{1,2,\ldots, n\}$.
A topological sort of $G$ is a bijection $G\to[n]$ such that if there is a path from $v$ to $w$, then $\tau(v)<\tau(w)$. Alternately, if we view $G$ as defining a partial ordering, a topological sort is a total ordering extending the partial ordering.
I would like to find $\displaystyle\min_{\tau} \sum_v a_v r^{\tau (v)}$, where we are taking the minimum over all topological sorts of G. It may help to generalize the problem and to finding $\displaystyle\min_{\tau} \sum_v a_v p({\tau (v)})$, where $p:[n]\to \mathbb R$ is a penalty/weight function (perhaps assumed to take positive values and be monotonic).
There are two extreme cases. If $G$ has no edges, we would sort things in ascending or descending order depending on if $r<1$ or $r>1$. If $G$ is already a linear order, there is nothing to be done. Already, the problem seems nontrivial given two disjoint linear orderings, where the problem reduces to the optimal riffle shuffle.
So is there a good algorithm for solving this? I know of some heuristics which help in certain cases, and I can use a bubble-sort type algorithm to get "local" minima, but unless there is a way to recast the problem, I don't see a good way to solve it.
Added later: I want to extend my comment and explain why I view dynamic programming to be insufficient. At best, this will clarify what I'm looking for. At worst, this will reveal a gap in my understanding which someone can clarify.
For there to be a dynamic programming solution, there need to be sub-problems which can be built upon to get a larger solution. For example, when searching for a path through a graph with edge lengths, if a minimal length path passes through a particular vertex, then the path from from the start to that vertex must be of minimal length. If we keep track of all the vertices that can be reached in time less than $t$, then we can ignore all paths through those close vertices which do not begin with a minimal path, and so we need to remember at most one path to any vertex, and at every stage we only need to find the shortest path to an unvisited vertex which is an extension of a known minimal path. This gives $O(n^2)$ storage costs and $O(nm)$ time costs where $n$ is the number of vertices and $m$ is the number of edges.
The obvious sub-problem to use for the problem at hand is that, if we know an initial/terminal segment for an optimal solution, the restriction to the subgraph containing just those elements will yield the same initial/terminal segment. It does not appear that we can say anything stronger. The algorithm this yields is as follows:
1. Select all vertices with no predecessors, and put each of these singletons into a list of admissible initial segments.
2. (Definition) For an admissible initial segment of length $n$, we say that an extension of length $n+1$ is admissible if it satisfies both the topological constraints of the directed graph and, if no topologically allowed insertion of the new element has a lower total value. From the collection of all admissible extensions of length $n+1$.
3. Given the collection of all admissible segments of length $n$, form the collection of all admissible extensions of length $n+1$.
4. From the collection generated in (3), if any two admissible extensions use exactly the same collection of vertices, remove the segment with a higher associated cost.
5. Loop through (3) and (4) until you have found the minimal initial segment containing every vertex of the graph.
If the penalty function is monotonic increasing (e.g., if $r>1$), we can improve run time somewhat by adding a heuristic (total cost of adding everything else at the minimum possible distance, ignoring that only one item can be in any particular spot), but even with this improvement, we have the following fundamental problem:
The algorithm doesn't require checking every initial segment, but it does require examining every collection of vertices which could form an allowable initial segment. In the worst case scenario, this is exponential in the number of vertices (though is significantly reduced when there are severe topological constraints). Additionally, in the worst case scenario, the space requirements are on the order of $\binom{n}{n/2}$ where there are $n$ vertices.
The dynamic programming algorithm is still quite an improvement over more naive algorithms, but I would like to find something that runs in polynomial time, or else show that such an algorithm cannot exist.
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Sounds like a dynamic programming algorithm might do the trick. Start by assuming that the last node in the sort is $v$, then build up partial topo-sorts from there, taking at each step the partial sort with the minimal cost function for your best sub-problem solution. – Bean Oct 16 '11 at 19:12
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@Bean A dynamic programming algorithm would work if I knew that the restriction of an optimal sort to a subgraph was still optimal. This statement is not obvious to me. If it were true, this would allow a divide and conquer + merge approach. In particular, dynamic programming does solve the "two disjoint linear orders" problem. However, without being able to divide and conquer, there appear to be too many possible members for a decent sized initial/final segment of the sort for this to be tractable. – Aaron Oct 16 '11 at 19:44
And here is a counterexample that the restriction of an optimal solution is optimal. Consider $r=2$ and $(1\to 16)(4)$. The optimal sort is $1,16,4$. However, if we restrict to just $1$ and $4$, the optimal sort is $4,1$. – Aaron Oct 16 '11 at 21:11
Whether r>1 or r<1 does not matter because the case of r<1 can be handled by considering 1/r and reversing the result. – Tsuyoshi Ito Oct 18 '11 at 12:02
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http://quant.stackexchange.com/questions/792/how-to-perform-risk-factor-calculation/801
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# How to perform risk factor calculation?
I am studying Arbitrage Pricing Theory (APT) and I have a question about calculating factor exposures.
Assume:
\begin{equation} r = \beta_1r_1 + \beta_2r_2 + ... + \beta_kr_k + r_e \end{equation}
Where:
$\beta_i$ is the exposure of the asset to a factor
$r$ is the return attributable to a factor
I believe that beta will be the covariance of the factor with the underlying asset. Is this correct? Also how is the return attributable to a specific factor calculated? Is there a single way this is done or are there a variety of approaches?
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## 4 Answers
I believe that beta will be the covariance of the factor with the underlying asset. Is this correct?
Close, it's the covariance divided by the variance of the factor.
\begin{equation} \beta_{f,a} = \frac{\sigma_{f,a}}{\sigma^2_f} \end{equation}
Also how is the return attributable to a specific factor calculated? Is there a single way this is done or are there a variety of approaches?
That depends on how you derive your factors. As mentioned in this earlier question, I once derived factors with cluster analysis. Thus, each factor was really a collection of highly correlated large-cap stocks. That meant the factor return was simply the cap-weighted average of all constituent stock returns, just like in a stock index.
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Does this imply that the return of a factor is equal to the value at the observed point minus the value at the beginning of the series? – Steve Mar 23 '11 at 1:32
@Steve The return is `(begin-end)/begin`. That's true whether it's an asset or a factor. Again, this is for a selection process that allowed a dollar price in the time series, since the factor is a cap-weighted selection of stocks. – chrisaycock♦ Mar 23 '11 at 2:03
This is only the case for simple linear regression with a single factor (i.e. CAPM), and not for multiple linear regression. If computing a multiple linear regression were that simple, there wouldn't be the vast mountain of literature on the topic (do a google search for 'solving normal equations' ) – shabbychef Mar 23 '11 at 4:40
@shabbychef I reviewed my notes from when I did this. The process then was to select a single "best" factor through cluster analysis, then subtract its cap-weighted returns from the universe before iteratively re-running the cluster analysis and selecting another "best" factor. I found a comment that all selected factors at the end were independent because of the subtractions. Is that logic valid? – chrisaycock♦ Mar 23 '11 at 14:08
if you perform e.g. a en.wikipedia.org/wiki/Gram_schmidt normalization before the regression, you will have an orthogonal design matrix. However, it is difficult to interpret the resultant regression coefficients. – shabbychef Mar 23 '11 at 16:34
I don't have much to add, but wanted to address the "price of risk" question.
APT is kind of "economics"-free and tries to price assets without the utility maximization required in CAPM/ICAPM. Ross's APT observes that groups of assets move together (e.g., tech stocks) and that is the risk you're bearing because the idiosyncratic risk, like the firing of HP's CEO, can be diversified away. Because this risk is easily diversifiable, the market won't pay you to take it. So in your APT model these factors are returns to asset classes, industries, etc.
Although the model looks the same, in Merton's ICAPM, the factors are state variables (e.g., industrial production, inflation). These are purely academic points -- in practice you run a multivariate regression with return on the LHS and whatever factors you think are priced on the RHS. OLS and GMM are common. So you'll estimate $$E ~ \left[ ~r_i~ \right] = \alpha_i + \beta_i^1 f_1 + \beta_i^2 f_2 + \ldots + \beta_i^k f_k$$
Also how is the return attributable to a specific factor calculated?
Now you regress the returns back on the betas. $$E ~ \left[ ~r_i~ \right] = \sum_{j \in K} \lambda_j \beta_i^j$$
Where $\lambda_j$ is the return to factor $j$. Typically the Fama-MacBeth approach is used here. If you've done it correct and found something, $\lambda > 0$ (i.e., the market is paying you to take this risk).
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Note that APT assumes the the alpha term is zero... and that the cross-product of exposures and factor returns + the risk-free rate result in the E[Ri]. So you need to think a little bit about whether to include the intercept or not depending on the application (asset-pricing test would include the intercept, for example). – Quant Guy May 29 '12 at 11:50
I think all of the theories have zero intercepts (i.e., only one risk-free rate)? Empirically you include the intercept to avoid forcing $\alpha_i = 0$ so that you can test if there is a return not correlated with the risk factors. – richardh♦ May 30 '12 at 13:32
If you have a series of observations of the return as a vector, $\mathbf{r}$ with corresponding observations of the factor returns in matrix $Z$, then the least squares estimate of the vector of betas is $$\hat{\beta} = \left(X'X\right)^{-1} X'\mathbf{r},$$ where $X$ is the matrix with $Z$ and a column of all ones (for the intercept term). The last value of $\hat{\beta}$ will be the estimate of the 'idiosyncratic' return. In general, the estimate of the $j$th coefficient, $\hat{\beta_j}$ will not be correlation of the return to the return of the $j$th factor, nor will it be that value adjusted for the volatility of the factor.
If you have only one factor (in which case it is CAPM, not APT), then the computation does simplify. Also, if the sample returns of the different factors are independent vectors (highly unlikely to happen by accident), you will get the simplification.
See wikipedia for more on multiple linear regression.
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You can use PCA as well, not the cluster analysis.
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Orthogonal issue. I would use PCA to pick a set of factors that contained unique information. – Steve Mar 23 '11 at 1:33
You should develop more your answer: explaining why PCA is suitable would make it more useful... – SRKX♦ Nov 12 '12 at 23:02
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http://mathoverflow.net/questions/14262/the-problem-of-infinity/14276
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## The problem of infinity [closed]
### Background and motivation
The following is copied from my blog since someone thought it was the clearest statement I had made regarding a problem I recently posed. On their advice, it is a community wiki. I would encourage, however, I discussion somewhere else (meta.MO perhaps?) on what seems to be a serious disconnect between pure mathematicians and physicists (note: I straddle the line but my PhD is in Math). I should add that nowhere on this site does it say that questions are to be restricted to pure mathematics (as opposed to applied mathematics). Some of the greatest discoveries in the history of mathematics came out of the consideration of physical problems. And I can't emphasis enough how important it is that the communities not drift apart (and I see it happening even within my own sub-discipline). On to the question, then.
Background: One particularly contentious question in the foundations of both mathematics and physics is whether mathematics is discovered or invented. Related to this is whether mathematics is the way the world actually is or if it is simply a way in which we can model the world.
This is a particularly difficult question to answer since it is quite clear that there are physically impossible situations that can be spoken about in mathematics.
Example: taking $n$ copies of a quantum channel should start to approach something that can be approximated by unitaries as n approaches infinity, i.e. in the asymptotic limit, $n$ copies of a channel have a unitary representation (roughly speaking and if the theorem is correct). But, from a physical point of view, this is ridiculous since it is literally impossible for an infinite number of channels (or anything else, for that matter) to physically exist. Yet whole branches of mathematics are devoted to working with infinity (infinite sets, for example).
Clearly some mathematical results have been originally thought to be entirely abstract only to much later find some application in the physical world. But it could be argued that concrete, mathematical analyses of infinite things could never, by their very definition, find an application in the physical world.
### Main questions
What are the different notions of infinity as used by mathematicians? To what extent does the use of infinity inform our understanding of the relationships between mathematics and the real world? Specifically:
• To what extent is mathematics invented and/or discovered? Is there variance that correlates with how infinitary the area of mathematics is? How does "infinity" inform this question in other ways?
• Should the use of infinity within mathematics inform our understanding of the (meta-)physical world? One popular interpretation of the quantum-field-theory formalism is in terms of "many worlds", but there's a lack of consensus on exactly how seriously to take this interpretation. Should the use of infinity be taken as “evidence” for the Many-Worlds Interpretation (MWI), e.g. because MWI includes infinitely many universes, and therefore gives physical meaning to "infinity".
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There are many distinct notions of infinity in mathematics. In your question, you are not distinguishing between them, which makes it hard to make sense of your question, other than than it is based on false assumptions. – Douglas Zare Feb 5 2010 at 13:40
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I certainly think that this is a good question, but it's on the border of too discussion-y for MO. Which is to say: because of the way MathOverflow is implemented, answers change orders depending on the number of votes, so it is almost impossible for there to be a real discussion in the answers (and comments are intentionally limited in length to prohibit technical discussions there). Personally, I wish MO had a "discussion forum" section. – Theo Johnson-Freyd Feb 5 2010 at 16:50
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A remark: your question is titled "the problem of infinity", and yet it seems that you actually want to ask the orthogonal question of to what extent mathematics is invented/discovered. (Aside: the "or" in your post is way oversimplifying the situation, and will lead to less-developed answers. Certainly it can be both: eg modern QFT as done by physicists, cf Kaiser 2005 "Drawing Theories Apart".) I think the "infinity" question is more mathematical, and the answers have responded to it. As this is CW, I may edit the post to separate the questions. Feel free to revert. – Theo Johnson-Freyd Feb 5 2010 at 17:03
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I would like to point out that Neel Krishnaswami took the time to craft an informative answer and you responded with an insult. Please make an effort to use your "adult voice" in the future. – S. Carnahan♦ Feb 5 2010 at 19:35
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I wouldn't call this a "borderline discussion" question. It's at least two pure discussion questions. While discussion/philosophical questions can be very interesting, MO is designed for questions what have more concrete answers, and as a side effect, is bad for such questions (as Theo suggested). Consider reposting them to reddit.com/r/math, which seems to host lots of mathematical discussion. If you want to get into a discussion about whether MO should have discussions, go to meta.mathoverflow.net/discussion/21/… – Anton Geraschenko♦ Feb 6 2010 at 3:20
show 1 more comment
## 3 Answers
I agree with Mike that even the example of $\mathbb{R}^3$ shows that talking about infinite mathematical objects can still make finite physical sense. For example, in $\mathbb{R}^3$ we can show that there are only so many regular polyhedra, and then we invent the microscope and find out that these polyhedra occur all over the natural world - in crystals, in the shape of viruses, and so forth - and no other ones do. That's a pretty good indication that $\mathbb{R}^3$ is a good model of at least some parts of the physical world. The way I think this result should be interpreted is that, even if you believe that there are only a finite number of possible locations in the physical universe (or whatever), the "mesh size" of the universe is small enough that we can take a limit and work in the infinitary setting, which boils down to discarding some negligible error terms.
Terence Tao's post on hard and soft analysis may also be of interest. I also want to quote somebody who, on MO, once said something like "do we ever talk about the infinite, or do we only talk about symbols that talk about the infinite?"
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Tao's hard-versus-soft article is very good, and I was going to link to it too. The point is that "the limit as n\to \infty" makes very good sense without any supposition of an infinite: if there's a theorem that says "in the limit, it's unitary", then there's a theorem that says "we can approximate unitarity this well with that many channels." – Theo Johnson-Freyd Feb 5 2010 at 16:57
@QY: Thank you for your answer. I don't agree with it, but it is a "hard" answer to a "soft" question which is what I was looking for. – Ian Durham Feb 7 2010 at 21:02
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This question does not have a mathematical answer.
However, different approaches to formalizing the concept of infinity have been considered in philosophy of mathematics for a very long time (since Aristotle, at least), and have resulted in quite a bit of interesting mathematics proper. Most people here are likely familiar with the modern Cantorian concept of infinity (which is what undergirds modern set theory), and so I will describe a somewhat different mathematical conception infinity.
So, one way of interpreting constructive mathematics is by means of "realizability interpretations". Here, we take the view that a proposition is true when it is possible to give evidence for its truth, and then inductively for each proposition we give conditions for evidence:
• $\top$ has the string $()$ for its justifying evidence
• $A \land B$ is justified when we can give a string $(p, q)$, where $p$ is evidence of $A$ and $q$ is evidence of $B$
• $\bot$ has no evidence
• $A \vee B$ is justified when we can give a string $(i, p)$, where $i$ is either 0 or 1. If $i$ is 0, then $p$ is evidence of $A$ and if $i$ is 1 it is evidence of $B$
• $A \implies B$ is justified by a computer program $c$, if $c$ computes evidence for $B$ as an output whenever it is given evidence for $A$ as an input.
• $\forall x. A(x)$ is justified by a computer program $c$ for evidence, if $c$ computes evidence for $A(n)$ as an output whenever it is given a numeral $n$ as an input.
Now, note that in the cases of implication and quantification, we ask for a computer program which is total on its inputs. So, the question arises, how can we tell whether or not a given program is a legitimate realizer for a proposition or not? The Halting Theorem ensures that we cannot accept arbitrary programs and decide after-the-fact whether or not they are evidence. So we must circumscribe what we will accept to some class of total functions.
So we can decide that certain patterns of recursive program definition (for example, primitive recursion) are acceptable forms of looping, which we believe will always lead to halting programs. These patterns correspond to induction principles. Proof-theoretically stronger theories correspond to logics whose realizability interpretation allows more generous recursion schemes as realizers. In this setup, large infinite sets correspond to very strong induction principles. So this gives a way of understanding Cantorian infinities without having to posit the actual existence of sets with huge numbers of elements.
This should illustrate that how we read mathematical statements shapes what ontological commitments we make regarding them, and so you can't answer physical questions only through pure mathematics. That is to say: physicists can't get out of doing experiments. But funny readings of mathematics may help you interpret those experiments!
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No, it doesn't have a purely mathematical answer, but it is a legitimate question about the nature of mathematics. Several people who are regular contributors to this site said it would make a good community wiki question. I cannot believe how myopic the users of this site are. See mathoverflow.net/questions/14162/… – Ian Durham Feb 5 2010 at 14:24
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Don't get angry at someone who attempted to answer your question just because they happened to not have read the discussion about your previous question. – Zev Chonoles Feb 5 2010 at 14:43
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@Ian: I think you misread Neel's answer. It is obvious to me that Neel thinks your question is a legitimate math philosophy question. Neel is pointing out how realizability suggests a different view of the infinite than Cantorian/Platonic infinity, a view which is more compatible with empiricism. – François G. Dorais♦ Feb 5 2010 at 14:44
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@Ian: remember that there are many users here. And reasonable disagreements about what do and don't make good questions. – Theo Johnson-Freyd Feb 5 2010 at 16:52
Apparently my view of 'reasonable disagreement' appears to be very different from many people on this site. What I don't understand is a) if this question has been closed, how people can still answer and comment, and b) if 'reasonable disagreement' is allowed, why questions can be closed in the first place. This question was posted after Qiaochu Yuan posted a link to my blog post saying it was a clear explanation of what I was asking with my previous question. I can't believe people don't realize how contradictory this seems. – Ian Durham Feb 7 2010 at 22:38
show 2 more comments
You say, "There are physically impossible things that can be spoken about in mathematics." When mathematics is used to model a physical system, it is always an approximation. So, the fact that the model is not a perfect match for the system (that is, it describes some "physically impossible things") is not a surprise. There's no need to get fancy to show examples: our space is not plain old Euclidean 3 dimensional space, but that model has been very useful. Or even something as basic as possible: using natural numbers to count stuff. Well, it's physically possible that half of one of the things you were counting could burn up, and it turns out the mathematical model you were using is inadequate to describe the system (because you can't talk about half an object).
The fact that some mathematics is abstract but finds applications seems like a different issue. Some number theory is such an example, but I don't think anyone ever denied that integers were highly useful in modeling real systems (and thus, in some sense I guess, "real"); it's just that the questions people were asking about them seemed to not have practical uses.
So I actually think there is less here than meets the eye. It's not really a big deal that we talk about infinite sets of things even though maybe there is no infinite collection of objects in reality, just like it's not really a big deal that we treat our space like it's continuous even though maybe it's actually discrete.
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It is a bigger deal than one would think when taken in the context of something like the Quine-Putnam indispensability argument which suggests that mathematical objects are as 'real' as the scientific theories they help explain. It gets at the heart of whether mathematics is invented or discovered. It seems that quite a few people on this board seem to be of the opinion that math is invented (and there's nothing wrong with that view, but it is wrong to assume the question is not worth debating). – Ian Durham Feb 7 2010 at 21:00
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http://math.stackexchange.com/questions/166541/ideal-generated-by-differential-forms/167442
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# Ideal generated by differential forms
I have troubles picturing what elements belong to a particular ensemble.
Let $\omega_1$,...,$\omega_r$ be differential 1-forms on a $C^\infty$ n-manifold that are independent at each point. Considering a complete base $\omega_1$,...,$\omega_r$,...,$\omega_{r+1}$,...,$\omega_{n}$, could someone give me some examples and counter-examples of elements in the ideal $\mathscr{I}$ generated by $\omega_1$,...,$\omega_r$ ?
For example, what to do with $\omega^1\wedge\omega^{r+2}$ ?
Thank you,
JD
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... the generated ideal in which ring? – Rasmus Jul 4 '12 at 14:59
Well, this comes from an exercise which does not provides the ring. Maybe the "implied" ring is the space of differential forms with $\wedge$ as the multiplication operator ? Does this make sense ? – vkubicki Jul 4 '12 at 18:42
Since an ideal is a subgroup, $\theta$ has to be a sum of 1-forms $\omega^1$, $\omega^2$,..., $\omega^r$, without wedge products. $\omega^1\wedge\omega^{r+2}$ is not part of the ideal ??? – vkubicki Jul 4 '12 at 20:32
In the exercise they say that if $\theta$ is an element of the ideal, then $\theta$ is a linear combination of exterior products where those forms appear as factors. How come 2 operators be used to generate the ideal ? – vkubicki Jul 4 '12 at 20:46
## 1 Answer
The ideal generated comprises a linear combination of differential forms containing at least one element of $\omega^1,...,\omega^r$ as a factor in exterior products with $\omega^{1},...,\omega^n$. The generation is therefore not made using the additive operation and the base $\omega^1,...,\omega^r$. This is consistent with the rest of the solution of the exercise. Source: the author of the textbook.
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http://mathoverflow.net/questions/24607?sort=votes
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## What is a Module over a Lie algebroid?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $\alpha: \mathfrak g_A \to T_{A/k}$ be a Lie algebroid over a $k$-algebra $A$. Numerous facts about and its universal enveloping algebra comes from the theory of ring differential operators on $A$. A generalization of theory of D-modules has been used to characterize modules over $\mathfrak g_A$ (eg. Sophi Chemla paper on Inverse image functor for Lie algebroids). But I couldn't find is a good reference for the notion itself. What is a $\mathfrak g_A$-module?
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I don't know this stuff, so I will leave only a comment. I think that there is some disagreement over the correct notion of module. In particular, my friend Alfonso Garcia-Saz math.toronto.edu/alfonso has some papers where they look towards a more general notion of module. The test is whether the Lie algebroid is a module over itself; in some older definitions, it is not. – Theo Johnson-Freyd May 14 2010 at 20:09
Yes, the adjoint action will not give a a module structure over the Lie algebroid itself because it is not $A$-linear. Thank you for the paper-- – lemin May 14 2010 at 23:31
## 2 Answers
A $\mathfrak{g}_A$-module $M$ is a $k$-module endowed with structures of $A$-module and $\mathfrak{g}_A$-module satisfying the compatibility equations $(ax)m = a(xm)$ and $x(am) = x(a)m + a(xm)$ for any $a\in A$, $x\in\mathfrak{g}_A$, and $m\in M$. Here $x(a)$ denotes the action of $\mathfrak{g}_A$ in $A$, while the three other actions are denoted by $ax$, $am$, and $xm$.
A $\mathfrak{g}_A$-module is the same that a module over the enveloping algebra $U_A(\mathfrak{g}_A)$.
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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.
Since Leonid already gave the definition, let me give a reference: Beilinson and Bernstein's A Proof of the Jantzen Conjectures.
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http://luckytoilet.wordpress.com/2011/05/20/simplifying-a-sum-of-consecutive-cubes/
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# Lucky's Notes
Notes on math, coding, and other stuff
## Simplifying a sum of consecutive cubes
If you were asked to find the sum of the first four cubes, you would easily be able to do so:
$1^3 + 2^3 + 3^3 + 4^3 = 100$
But here you may notice something interesting:
$1^3 + 2^3 + 3^3 + 4^3 = (1+2+3+4)^2$
It turns out that this holds for the sum of the first $n$ cubes for any $n$ — an identity. This identity can be written also as:
$\displaystyle \sum_{i=1}^n i^3 = \left( \displaystyle \sum_{i=1}^n i \right)^2$
A second way of expressing the same identity simplifies the right side and expresses it as a combination:
$\displaystyle \sum_{i=1}^n i^3 = \binom{n+1}{2}^2$
### A proof by bijection
This identity can be proven combinatorially by noticing that $\binom{n+1}{2}$ is the number of ways to choose 2 elements with repetition from the set {1..n} — the following is a proof by Benjamin and Orrison.
In this proof, we prove the third form of the identity
$\displaystyle \sum_{i=1}^n i^3 = \binom{n+1}{2}^2$
by constructing one set of size $\displaystyle \sum_{i=1}^n i^3$ and constructing another set of size $\binom{n+1}{2}^2$, then constructing a bijection between the two sets.
Let $S$ be the set of ordered 4-pairs $(a,b,c,d)$ such that $1 \leq a,b,c \leq d \leq n$ — that is, the fourth integer of the pair is greater or equal to each of the other three. If we fix $d$, the number of possibilities for the variables $a,b,c$ is $d^3$. As $d$ ranges from 1 to n, the total number of elements in $S$ is equal to $\displaystyle \sum_{i=1}^n i^3$.
Also, since $\binom{n+1}{2}$ is the number of ways to choose 2 elements with repetition from the set {1..n}, there are $\binom{n+1}{2}$ pairs of integers $(h,i)$ satisfying $1 \leq h \leq i \leq n$. So let $T$ be the set of pairs of such pairs $((h,i),(j,k))$ where $1 \leq h \leq i \leq n$ and $1 \leq j \leq k \leq n$. The number of elements in $T$ is then $\binom{n+1}{2}^2$.
We further partition the sets as follows: let $S_1$ be the subset of $S$ where $a \leq b$ and let $S_2$ be the subset of $S$ where $a > b$. Similarly, let $T_1$ be the subset of $T$ where $i \leq k$ and let $T_2$ be the subset of $T$ where $i > k$. We can construct a bijection between $S$ and $T$ by constructing two bijections:
$S_1 \leftrightarrow T_1$
$S_2 \leftrightarrow T_2$
The following is trivially a bijection from $S_1$ to $T_1$ — the case where $a \leq b$:
$(a,b,c,d) \rightarrow ((a,b),(c,d))$
The equivalent bijection from $S_2$ to $T_2$ — the case where $a > b$:
$(a,b,c,d) \rightarrow ((c,d),(b,a-1))$
We can confirm that the two operations we defined are indeed bijections. This proves the identity.
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### 3 Responses to Simplifying a sum of consecutive cubes
1. JL says:
Would it not be easier to inductively prove that the sum of the first n integers is n(n+1)/2 and that the sum of the first n cubes is n^2(n+1)^2/4, which equals (n(n+1)/2)^2?
• luckytoilet says:
Shhhh
Besides, any bijective proof beats any boring inductive argument xD
2. Anonymous says:
Bijections are awesome
A Western Li
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http://physics.stackexchange.com/questions/3264/thermal-radiation-spectrum-of-a-blackbody/3269
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# Thermal radiation spectrum of a blackbody
Why is it that thermal radiation of a black body usually described by a spectral distribution function rather than an intensity vs frequency curve?
I have a vague explanation for this: Any measured intensity of radiation will contain a spectrum of frequencies which are impossible to separate (even with a very good spectrometer, one can only measure the intensity of an interval). So it is practically impossible to assign a particular intensity to a particular frequency.
Is my reasoning correct?
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um, isn't a "spectral distribution function" the same thing as an "intensity vs frequency curve"? – Jeremy Jan 18 '11 at 20:40
Right, Jeremy. I personally have seen Intensity versus wavelength often. – Georg Jan 18 '11 at 20:50
@Jeremy: In a spectral distribution function, the quantity which is considered is the spectral radiancy $R_T(\nu)$. $R_T(\nu)$ is defined in such a way that $R_T(\nu)d\nu$ gives the intensity of the radiation $dI$ in the interval $(\nu,\nu+d\nu)$ (similarly for wavelength). Integrating the spectral radiancy curve, one would get the total intensity emitted at a particular frequency (which by Stefan's law is proportional to $T^4$). – Bernhard Heijstek Jan 18 '11 at 20:58
## 2 Answers
It's true that you can't talk about intensity without referring to some specific frequency range, but the reason is the definition of intensity itself, not the inherent limits of our measuring devices. Intensity, also called irradiance, is defined as power per unit area. Typically it takes into account radiation at all frequencies (or wavelengths), but you can also talk about the intensity within a particular band like $400\text{ nm}-700\text{ nm}$.
If you wanted to create a graph of intensity versus frequency, you could certainly do so by dividing the frequency axis into bins and plotting one point for each bin representing the total intensity detected within that frequency range. Obviously, the smaller you can make the bins, the more detailed information you get about the spectrum, but the intensity values become correspondingly smaller (and harder to measure accurately) as the bins get smaller. Besides, if everyone did that, different plots from different experiments would be very difficult to compare unless they all decided on a standard bin size.
In order to make data from different experiments comparable, you normalize the intensities by dividing each intensity measurement by the width of the bin. In the limit as the bin width goes to zero, this is just the spectral irradiance.
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Thanks! But I have seen plots of intensity vs frequency (mostly on non-technical sites). So are these incorrect? – Bernhard Heijstek Jan 18 '11 at 21:39
1
@finemann: I couldn't say without knowing how the plots were generated. It could be that they're using the binning procedure I talked about, or it could be that the plots are just mislabeled. – David Zaslavsky♦ Jan 18 '11 at 22:10
Dear Finemann, the "spectral power distribution"
http://en.wikipedia.org/wiki/Spectral_power_distribution
and "intensity vs frequency" contain exactly the same information. They're so easily related that physicists wouldn't even consider them "two different things".
The spectral power distribution is the (infinitesimal) power per unit area and per unit wavelength: $$M_\lambda = \frac{\delta\Phi}{\delta A\delta \lambda}$$ You may simply write $\lambda$ as $c/f$ and use the right differentials of functions, $\delta\lambda=-(c/f^2)\times \delta f$ and you will get the analogous distribution in terms of the frequencies. The graph will be just flipped from the left to right (that's because of the irrelevant minus sign), and on one side, it will be stretched vertically and squeezed horizontally. There is no international ban on using the frequencies, and indeed, it's true that even top physicists usually prefer to parameterize radiation by its frequency rather than the wavelength. You may have seen "popular" sources that used frequencies instead of the wavelength but it is not because they were incorrect or misleading; they did copy conventions that common in research-level physics.
You may also think that the words "intensity" and "power" are different things. But "intensity" is just a popular term for "irradiance" which is just power per unit area, see
http://en.wikipedia.org/wiki/Radiant_intensity
More properly, "radiant intensity" should be counted as power of a localized source per solid angle (note that if the radiation is radial, the power only depends on the solid angle and not the distance at which you absorb the light).
Black body spectral power distribution is continuous: a black body is indeed able to emit at all frequencies. A usual disclaimer for continuous distributions should be added: the power emitted by a continuous power distribution at a particular exact frequency is always exactly zero. You need a whole interval of frequencies (or wavelengths) to get a nonzero answer. The (infinitesimal) power coming from this interval of frequencies will then be proportional to the (infinitesimal) width of the interval of frequencies. That's why we talk about the ratios of differentials etc.
Cheers LM
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http://mathoverflow.net/questions/5268?sort=oldest
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Functorial Whitehead Tower?
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The Whitehead tower of a (pointed) space is a tower of spaces which successively kills the bottom homotopy groups. The first two spaces can be constructed functorially (at least for suitably nice spaces) as the connected component and the universal cover.
Can the remaining spaces be constructed functorially?
For the dual situation the answer is yes. I.e. for the Postnikov tower where we have a tower of spaces where the bottom homotopy groups are intact, but where we have killed off all the higher homotopy groups does have a functorial construction (again for nice spaces). The construction I know passes through simplicial sets. I'm wondering if something similar exists for the Whitehead tower?
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3 Answers
The nth stage of the Whitehead tower of X is the homotopy fiber of the map from X to the nth (or so) stage of its Postnikov tower, so you can use your functorial construction of the Postnikov tower plus a functorial construction of the homotopy fiber (such as the usual one using the path space of the target).
The nth stage of the Whitehead tower of X is also the cofibrant replacement for X in the right Bousfield localization of Top with respect to the object Sn (or so). Since Top is right proper and cellular this localization exists by the result of chapter 5 of Hirschhorn's book on localizations of model categories. You might look there to see how the cofibrant replacement functor is constructed. With some care you should be able to define functorially the maps in the tower as well.
(BTW, the Postnikov tower can similarly be obtained functorially by a left Bousfield localization of Top.)
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I have taken the liberty of copying this comment to the nLab entry here ncatlab.org/nlab/show/… – Urs Schreiber Nov 13 2009 at 9:13
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.
If we have a functorial Postnikov tower of a pointed space X, the tower inherits a basepoint. Then take the tower over the Postnikov tower which is pointwise the path fibration. This is pointwise a fibration. Pull this tower back along the map from the constant tower on X, then this gives a tower over X, which, by the general treatment in Whitehead's 1952 (I think) article, is a Whitehead tower for X.
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1
BTW This works for all spaces, not just locally nice ones. In a piece of shameless advertising I approach this problem from a 2-bundle point of view, and get for spaces that satisfy a 2-dimensional version of the universal covering space condition a functorial topological groupoid I claim is a good substitute for the 2-connected stage. – David Roberts Nov 13 2009 at 6:44
$X \leftarrow B\Omega X \leftarrow B^2 \Omega^2 X \leftarrow \ldots$
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Preamble: If one considers an ideal gas of non interacting charged particles of charge $q$ in a uniform magnetic field $\mathbf{B} = \mathbf{\nabla} \wedge \mathbf{A}$, then the classical partition ...
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