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http://mathoverflow.net/questions/112710?sort=newest | ## Is the n-torsion of an extension of an abelian variety by a torus, finite and flat?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for reference or hints how to prove the following result.
Let $G$ be a commutative $S$-group scheme which is the extension of an abelian scheme $A$ by a torus $T$. Then the n-torsion $G[n]$ is a finite flat $S$-group scheme.
Specifically, I have difficulties in showing that $G[n]$ is finite. For a general semi-abelian scheme we know that it is quasi-finite and flat, but not necessarily finite (see e.g. the book Neron Models, Lemma 7.3/2).
Thanks in advance,
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Is $S$ an arbitrary scheme? – Angelo Nov 17 at 20:36
The statement should be true for an arbitrary scheme but I would be happy with an answer for $S= Spec R$ when $R$ is the ring of integers of a finite $\mathbb{Q}_p$-extension. – Tzanko Matev Nov 17 at 20:48
I have seen a generalization of this statement in several papers on 1-motives, however no proof or reference is given there. For example: Deligne [Hodge III, 10.1.10] or M. Raynaud, [1-Motifs et Monodromie Géométrique, 3.1]. This is why I think that it should be true. – Tzanko Matev Nov 17 at 20:56
@Tzanko Matev: In "Néron models", semi-abelian means flat and semi-abelian fibers, the scheme itself is not necessarily an extension of an abelian scheme by a torus. For example if we consider the Néron model of an elliptic curve with multiplicative reduction, it is semi-abelian in the sense of "Néron models", and the $n$-torsion is not finite in general. – Qing Liu Nov 17 at 21:36
@Qing Liu: I am sorry if the question was not stated well. I know that for a general semi-abelian scheme what I ask is false. I am only interested in the case when the scheme is an extension of an abelian scheme by a torus. – Tzanko Matev Nov 18 at 8:50
show 1 more comment
## 1 Answer
It is an exercise with descent theory and the snake lemma for fppf abelian group sheaves to deduce the result for $G[n]$ from the cases of $T[n]$ and $A[n]$.
In more detail, by the snake lemma $G[n]$ is an extension of $A[n]$ by $T[n]$ in the sense of such abelian sheaves. Since $A[n]$ and $T[n]$ are each finite fppf over $S$, the same then holds for $G[n]$. Indeed, rather generally, if $$1 \rightarrow G' \rightarrow G \rightarrow G'' \rightarrow 1$$ is a complex of $S$-group schemes with $G'$ affine fppf over $S$ and the diagram is short exact for the fppf topology (so $G'$ is the scheme-theoretic kernel of $G \rightarrow G''$) then the functor of points of $G$ as a $G''$-scheme is a $G'$-torsor for the fppf topology on $G''$, so the $G''$-scheme $G \rightarrow G''$ becomes isomorphic fppf-locally on $G''$ to $G'$ (over the base) as a scheme. Hence, by fppf descent for properties of morphisms, $G \rightarrow G''$ inherits many "nice" properties that may be satisfied by $G' \rightarrow S$, such as: proper, flat, smooth, etale, finite, etc. In particular, $G$ is fppf over $G''$ and if $G'$ is finite over $S$ then so is $G \rightarrow G''$ (and hence so is $G$ if $G''$ is also finite over $S$).
See Oort's LNM book on commutative group schemes for generalizations with the fpqc topology (around section 18, IIRC).
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@nosr Thanks a lot! I will accept the answer when I manage to verify it. – Tzanko Matev Nov 18 at 8:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9054416418075562, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Sigmoidal_curve | # Logistic function
(Redirected from Sigmoidal curve)
For the recurrence relation, see Logistic map.
Standard logistic sigmoid function
A logistic function or logistic curve is a common sigmoid function, given its name in 1844 or 1845 by Pierre François Verhulst who studied it in relation to population growth. A generalized logistic curve can model the "S-shaped" behaviour (abbreviated S-curve) of growth of some population P. The initial stage of growth is approximately exponential; then, as saturation begins, the growth slows, and at maturity, growth stops.
A simple logistic function may be defined by the formula
$P(t) = \frac{1}{1 + e^{-t}}$
where the variable P might be considered to denote a population, where e is Euler's number and the variable t might be thought of as time.[1] For values of t in the range of real numbers from −∞ to +∞, the S-curve shown is obtained. In practice, due to the nature of the exponential function e−t, it is sufficient to compute t over a small range of real numbers such as [−6, +6].
The logistic function finds applications in a range of fields, including artificial neural networks, biology, biomathematics, demography, economics, chemistry, mathematical psychology, probability, sociology, political science, and statistics. It has an easily calculated derivative:
$\frac{d}{dt}P(t) = P(t)\cdot(1-P(t)).\,$
It also has the property that
$1-P(t) = P(-t).\,$
Thus, the function $t\mapsto P(t) - 1/2$ is odd.
## Logistic differential equation
The logistic function is the solution of the simple first-order non-linear differential equation
$\frac{d}{dt}P(t) = P(t)(1-P(t))$
with boundary condition P(0) = 1/2. This equation is the continuous version of the logistic map.
The qualitative behavior is easily understood in terms of the phase line: the derivative is 0 at P = 0 or 1 and the derivative is positive for P between 0 and 1, and negative for P above 1 or less than 0 (though negative populations do not generally accord with a physical model). This yields an unstable equilibrium at 0, and a stable equilibrium at 1, and thus for any value of P greater than 0 and less than 1, P grows to 1.
One may readily find the (symbolic) solution to be
$P(t)=\frac{e^{t}}{e^{t}+e^{c}}$
Choosing the constant of integration ec = 1 gives the other well-known form of the definition of the logistic curve
$P(t) = \frac{e^t}{e^t + 1} \! = \frac{1}{1 + e^{-t}} \!$
More quantitatively, as can be seen from the analytical solution, the logistic curve shows early exponential growth for negative t, which slows to linear growth of slope 1/4 near t = 0, then approaches y = 1 with an exponentially decaying gap.
The logistic function is the inverse of the natural logit function and so can be used to convert the logarithm of odds into a probability; the conversion from the log-likelihood ratio of two alternatives also takes the form of a logistic curve.
The logistic sigmoid function is related to the hyperbolic tangent, A.p. by
$2 \, P(t) = 1 + \tanh \left( \frac{t}{2} \right).$
## In ecology: modeling population growth
Pierre-François Verhulst (1804–1849)
A typical application of the logistic equation is a common model of population growth, originally due to Pierre-François Verhulst in 1838, where the rate of reproduction is proportional to both the existing population and the amount of available resources, all else being equal. The Verhulst equation was published after Verhulst had read Thomas Malthus' An Essay on the Principle of Population. Verhulst derived his logistic equation to describe the self-limiting growth of a biological population. The equation is also sometimes called the Verhulst-Pearl equation following its rediscovery in 1920. Alfred J. Lotka derived the equation again in 1925, calling it the law of population growth.
Letting P represent population size (N is often used in ecology instead) and t represent time, this model is formalized by the differential equation:
$\frac{dP}{dt}=rP\left(1 - \frac{P}{K}\right)$
where the constant r defines the growth rate and K is the carrying capacity.
In the equation, the early, unimpeded growth rate is modeled by the first term +rP. The value of the rate r represents the proportional increase of the population P in one unit of time. Later, as the population grows, the second term, which multiplied out is −rP2/K, becomes larger than the first as some members of the population P interfere with each other by competing for some critical resource, such as food or living space. This antagonistic effect is called the bottleneck, and is modeled by the value of the parameter K. The competition diminishes the combined growth rate, until the value of P ceases to grow (this is called maturity of the population).
Dividing both sides of the equation by K gives
$\frac{d}{dt}\frac{P}{K}=r\frac{P}{K}\left(1 - \frac{P}{K}\right)$
Now setting $x=P/K$ gives the differential equation
$\frac{dx}{dt} = r x (1-x)$
For $r = 1$ we have the particular case with which we started.
In ecology, species are sometimes referred to as r-strategist or K-strategist depending upon the selective processes that have shaped their life history strategies. The solution to the equation (with $P_0$ being the initial population) is
$P(t) = \frac{K P_0 e^{rt}}{K + P_0 \left( e^{rt} - 1\right)}$
where
$\lim_{t\to\infty} P(t) = K.\,$
Which is to say that K is the limiting value of P: the highest value that the population can reach given infinite time (or come close to reaching in finite time). It is important to stress that the carrying capacity is asymptotically reached independently of the initial value P(0) > 0, also in case that P(0) > K.
### Time-varying carrying capacity
Since the environmental conditions influence the carrying capacity, as a consequence it can be time-varying: K(t) > 0, leading to the following mathematical model:
$\frac{dP}{dt}=rP\left(1 - \frac{P}{K(t)}\right)$
A particularly important case is that of carrying capacity that varies periodically with period T:
$K(t+T) = K(t).\,$
It can be shown that in such a case, independently from the initial value P(0) > 0, P(t) will tend to a unique periodic solution P*(t), whose period is T.
A typical value of T is one year: in such case K(t) reflects periodical variations of weather conditions.
Another interesting generalization is to consider that the carrying capacity K(t) is a function of the population at an earlier time, capturing a delay in the way population modifies its environment. This leads to a logistic delay equation,[2] which has a very rich behavior, with bistability in some parameter range, as well as a monotonic decay to zero, smooth exponential growth, punctuated unlimited growth (i.e., multiple S-shapes), punctuated growth or alternation to a stationary level, oscillatory approach to a stationary level, sustainable oscillations, finite-time singularities as well as finite-time death.
## In statistics and machine learning
### Logistic regression
Logistic functions are used in several roles in statistics. Firstly, they are the cumulative distribution function of the logistic family of distributions. Secondly they are used in logistic regression to model how the probability p of an event may be affected by one or more explanatory variables: an example would be to have the model
$p=P(a + bx)\,$
where x is the explanatory variable and a and b are model parameters to be fitted.
An important application of the logistic function is in the Rasch model, used in item response theory. In particular, the Rasch model forms a basis for maximum likelihood estimation of the locations of objects or persons on a continuum, based on collections of categorical data, for example the abilities of persons on a continuum based on responses that have been categorized as correct and incorrect.
Logistic regression and other log-linear models are also commonly used in machine learning.
### Neural networks
Logistic functions are often used in neural networks to introduce nonlinearity in the model and/or to clamp signals to within a specified range. A popular neural net element computes a linear combination of its input signals, and applies a bounded logistic function to the result; this model can be seen as a "smoothed" variant of the classical threshold neuron.
A common choice for the activation or "squashing" functions, used to clip for large magnitudes to keep the response of the neural network bounded[3] is
$g(h) = \frac{1}{1 + e^{-2 \beta h}} \!$
which is a logistic function. These relationships result in simplified implementations of artificial neural networks with artificial neurons. Practitioners caution that sigmoidal functions which are antisymmetric about the origin (e.g. the hyperbolic tangent) lead to faster convergence when training networks with backpropagation.[4]
A generalisation and extension of the logistic function to multiple inputs is the softmax activation function.
## In medicine: modeling of growth of tumors
See also: Gompertz curve#Growth of tumors
Another application of logistic curve is in medicine, where the logistic differential equation is used to model the growth of tumors. This application can be considered an extension of the above mentioned use in the framework of ecology (see also the Generalized logistic curve, allowing for more parameters). Denoting with X(t) the size of the tumor at time t, its dynamics are governed by:
$X^{\prime}=r\left(1 - \frac{X}{K}\right)X$
which is of the type:
$X^{\prime}=F\left(X\right)X, F^{\prime}(X) \le 0$
where F(X) is the proliferation rate of the tumor.
If a chemotherapy is started with a log-kill effect, the equation may be revised to be
$X^{\prime}=r\left(1 - \frac{X}{K}\right)X - c(t)X,$
where c(t) is the therapy-induced death rate. In the idealized case of very long therapy, c(t) can be modeled as a periodic function (of period T) or (in case of continuous infusion therapy) as a constant function, and one has that
$\frac{1}{T}\int_{0}^{T}{c(t)\, dt} > r \rightarrow \lim_{t \rightarrow +\infty}x(t)=0$
i.e. if the average therapy-induced death rate is greater than the baseline proliferation rate then there is the eradication of the disease. Of course, this is an oversimplified model of both the growth and the therapy (e.g. it does not take into account the phenomenon of clonal resistance).
## In chemistry: reaction models
The concentration of reactants and products in autocatalytic reactions follow the logistic function.
## In physics: Fermi distribution
The logistic function determines the statistical distribution of fermions over the energy states of a system in thermal equilibrium. In particular, it is the distribution of the probabilities that each possible energy level is occupied by a fermion, according to Fermi–Dirac statistics.
## In linguistics: language change
In linguistics, the logistic function can be used to model language change:[5] an innovation that is at first marginal begins to spread more quickly with time, and then more slowly as it becomes more universally adopted.
## In economics: diffusion of innovations
The logistic function can be used to illustrate the progress of the diffusion of an innovation through its life cycle. Historically, when new products are introduced there is an intense amount of research and development which leads to dramatic improvements in quality and reductions in cost. This leads to a period of rapid industry growth. Some of the more famous examples are: railroads, incandescent light bulbs, electrification, the Ford Model T, air travel and computers. Eventually, dramatic improvement and cost reduction opportunities are exhausted, the product or process are in widespread use with few remaining potential new customers, and markets become saturated.
Logistic analysis was used in papers by several researchers at the International Institute of Applied Systems Analysis (IIASA). These papers deal with the diffusion of various innovations, infrastructures and energy source substitutions and the role of work in the economy as well as with the long economic cycle. Long economic cycles were investigated by Robert Ayres (1989).[6] Cesare Marchetti published on long economic cycles and on diffusion of innovations.[7][8] Arnulf Grübler’s book (1990) gives a detailed account of the diffusion of infrastructures including canals, railroads, highways and airlines, showing that their diffusion followed logistic shaped curves.[9]
Carlota Perez used a logistic curve to illustrate the long (Kondratiev) business cycle with the following labels: beginning of a technological era as irruption, the ascent as frenzy, the rapid build out as synergy and the completion as maturity.[10]
## Double logistic function
Double logistic sigmoid curve
The double logistic is a function similar to the logistic function with numerous applications[citation needed]. Its general formula is:
$y = \mathrm{sgn}(x-d) \, \Bigg(1-\exp\bigg(-\bigg(\frac{x-d}{s}\bigg)^2\bigg)\Bigg),$
where d is its centre and s is the steepness factor. Here "sgn" represents the sign function.
It is based on the Gaussian curve and graphically it is similar to two identical logistic sigmoids bonded together at the point x = d.
One of its applications is non-linear normalization of a sample, as it has the property of eliminating outliers.
## Notes
1. Verhulst, Pierre-François (1838). "Notice sur la loi que la population poursuit dans son accroissement" (PDF). Correspondance mathématique et physique 10: 113–121. Retrieved 09/08/2009.
2. V.I. Yukalov, E.P. Yukalova and D. Sornette, Punctuated Evolution due to Delayed Carrying Capacity, Physica D 238, 1752–1767 (2009)
3. Gershenfeld 1999, p.150
4. LeCun, Y.; Bottou, L.; Orr, G.; Muller, K. (1998). "Efficient BackProp". In Orr, G.; Muller, K. Neural Networks: Tricks of the trade (Springer). ISBN 3-540-65311-2
5. Bod, Hay, Jennedy (eds.) 2003, pp. 147–156
6.
7.
8.
9. Perez, Carlota (2002). Technological Revolutions and Financial Capital: The Dynamics of Bubbles and Golden Ages. UK: Edward Elgar Publishing Limited. ISBN 1-84376-331-1.
## References
1. Jannedy, Stefanie; Bod, Rens; Hay, Jennifer (2003). Probabilistic Linguistics. Cambridge, Massachusetts: MIT Press. ISBN 0-262-52338-8.
2. Gershenfeld, Neil A. (1999). The Nature of Mathematical Modeling. Cambridge, UK: Cambridge University Press. ISBN 978-0-521-57095-4.
3. Kingsland, Sharon E. (1995). Modeling nature: episodes in the history of population ecology. Chicago: University of Chicago Press. ISBN 0-226-43728-0. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 25, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9093478918075562, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/47056/interpretation-of-e-psi2-as-electron-density/47065 | # Interpretation of $e|\psi|^2$ as electron density
In solid state physics the electron density is often equated to $e|\psi|^2$. However, the Sakurai says (Chapter 2.4, Interpretation of the Wave Function, p. 101) that adopting such a view leads "to some bizarre consequences", and that Born's statistical interpretation of $|\psi|^2$ as a probability density is more satisfactory.
I am aware that a position measurement of an electron leads to (in the Copenhagen interpretation) a collapse of the wave function into a position eigenstate $x$ with the probability given by $|\psi(x)|^2$, and that it is known from some scattering experiments that the electron behaves as a point-like particle.
However, the electron density is experimentally observable, e.g. by X-ray scattering. One could argue that X-ray scattering is done with a large ensemble of atoms, so that what we actually observe is the average over many position measurements.
I am wondering if this the only "bizarre consequence", since this distinction seems to be a minor difference to me. My questions aims at clarifying the difference between the probabilistic interpretation of $|\psi(x)|^2$ and the interpretation as an electron density.
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## 2 Answers
The statement that the right interpretation of $|\psi(x)|^2$ is probabilistic means that the value of this expression may only be "measured" in the same sense as other probabilistic distributions – by a repeated experiment starting with the same initial conditions. Every probability distribution may be approximately reconstructed by "throwing the dice" many times and recording the frequencies.
However, $|\psi(x)|^2$ cannot be measured in one repetition of the situation. There can't exist a gadget that would show $|\psi(x)|^2$ on its display. That's why we say that the wave function (or its squared absolute value) isn't observable: it's not observable in one copy of the situation. This word "observable" may be interpreted colloquially as well as technically. Technically, an "observable" is a Hermitian operator. The squared absolute value of the wave function isn't an operator, so it's not an observable, so it can't be measured.
The charge density is a classical quantity in classical physics and may be an observable in quantum field theory – like for other observables, only the probabilities of different values may be predicted. However, in one-particle quantum mechanics, a particle is either located in the volume $dV$ or it's not. Nothing in between is possible and a measurement may answer whether the question is Yes or No. The probability of Yes is given by $|\psi(x)|^2 dV$.
If you understand the things above and your wording indicates that you do, then you understand why the right interpretation of the wave function is probabilistic. This assertion means nothing else than the fact than the fact that to "measure" $|\psi(x)|^2$ at a given point, one needs to repeat the same situation many times and use the laws of probability and statistics. The function $|\psi(x)|^2$ doesn't correspond to a property of the region of space that may be measured instantly, by one measurement in one repetition of the experiment.
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Thanks for the clarification. I'm still wondering about the "bizarre consequences" mentioned in the Sakurai. Is there any experiment which would give a fundamentally different result if $|\psi|^2$ actually was an electron density instead instead of a probability density? – m4r73n Dec 17 '12 at 13:33
Hi, if it were an electron (and electric charge) density, the results would be completely different. In a double-slit-like interference experiment, each electron would actually create a whole cloud on the photographic plate. Atoms would lose their sharp identity - the nuclei's charge would be neutralized by thousands of "piece of electrons" from various other places. And so on. When we see that the electron may always be seen at particular locations, it's clear that assuming that it's actually spread would lead to a totally different world. – Luboš Motl Dec 20 '12 at 7:04
Obviously, the interpretation of $e |\psi|^2$ as an electron density will hold as soon as we consider an ensemble of electrons, since the probability density then results in a statistical distribution. In the case of just one electron we can still interpret $e |\psi|^2$ as an electron density as long as we think about longer time scales. Please note, that this only works as all observables (that I can think of now) depend linearly on the electron charge.
In any other case the electron still has to be considered point-like and is not smeared out in space, only that its position in the phase space is not determined. This means e.g. that the detection of an electron will always (after the aforementioned collapse of the wave function) result in one single point of detection.
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http://ldtopology.wordpress.com/2009/08/31/forms-of-the-virtual-haken-conjecture/ | # Low Dimensional Topology
## August 31, 2009
### Forms of the Virtual Haken Conjecture
Filed under: 3-manifolds,Hyperbolic geometry,Virtual Haken Conjecture — Nathan Dunfield @ 2:04 pm
Daniel Moskovich suggested that, in light of the recent advances, I summarize the state of the Virtual Haken Conjecture.
So let M be a closed hyperbolic 3-manifold. Then one has the following sequence of increasingly strong conjectures.
Conjecture 1: $\pi_1(M)$ contains a surface subgroup.
Conjecture 2: M has a finite cover N which is Haken, i.e. contains a closed incompressible surface.
Conjecture 3: M has a finite cover N with $b_1(M) = \mathrm{rank}(H^1(M; Z)) > 0$.
Conjecture 4: For each $n > 0$, the manifold M has a finite cover N with $b_1(N) > n$.
Conjecture 5: M has a finite cover N which is large, i.e. $\pi_1(N)$ surjects onto a free group of rank 2.
Here Conjecture 2 is the classical form of the VHC. Conjecture 3 could also be stated as N contains a finite cover which contains a non-separating incompressible surface. In addition one has the question about virtual fibering
Conjecture 6: M has a finite cover N which fibers over the circle.
Notice that Conjecture 6 immediately implies Conjectures 1-3, though it is not currently know to imply Conjectures 4-5. A surface subgroup of $\pi_1(M)$ is either geometrically finite or geometrically infinite; in the former case, it’s a quasi-fuchsian group, and in the latter, Thurston and Bonahon showed it comes from a virtual fiber. Thus, in the language of Conjecture 1, Conjecture 6 says that $\pi_1(M)$ contains a geometrically infinite surface group.
Tying these these questions together is
Conjecture 7: The fundamental group of M is subgroup seperable (a.k.a LERF), i.e. finitely generated subgroups are closed in the profinite topology.
In particular, Conjectures 1 and 7 imply all the rest; for Conjectures 2-5 a nice argument for this was given in [Lubotzky 1996], for Conjecture 6 this is the recent work of Agol and Bergeron-Wise.
The recent announcements are: Kahn-Markovic claim a proof of Conjecture 1, and Wise claims a proof of Conjecture 7 in the special case that M is a Haken manifold containing an embedded quasi-fuchsian surface. Assuming both proofs are correct, this would reduce all of the above questions to:
Conjecture 8: At least one of the quasi-fuchsian surface groups constructed by Kahn-Markvoic has a finite index subgroup which corresponds to an embedded surface in some finite cover of M.
Wise’s claimed result almost gives that the classical VHC (Conjecture 2) implies all the rest, but not quite; it mean that Conjecture 2 implies Conjectures 3 and 6, but not Conjectures 4, 5, and 7 in certain special cases. In particular the cover manifold N given by Conjecture 2 could fiber over the circle with $b_1 = 1$ and then Wise’s result does not apply. Conjecture 4, even for $n = 2$, does suffice to prove them all, since not every class in $H^1(N)$ can come from a fibration when $b_1 > 1$. This suggests it would be particularly interesting to show that Conjecture 6 implies Conjecture 4, but currently this is only known when the fiber has genus 2, by a nice result of Masters.
Edit: Last paragraph clarified.
## 3 Comments »
1. You state that Wise’s claimed result gives that the VHC (conjecture 2) implies conjecture 6. Doesn’t your statement then imply that Wise’s claimed result currently gives conjecture 6 for all Haken manifolds?
Comment by Mayer A. Landau — September 1, 2009 @ 11:45 pm
• Yes, that’s correct.
Comment by — September 2, 2009 @ 7:08 am
2. We live in exciting times…
Markovic will be talking here in Kyoto next week, and I’ll certainly try to understand as much as I can of his argument (already well-outlined in Danny Calegari’s blog)… Wise’s claim seems much more mysterious at the moment.
Comment by dmoskovich — September 6, 2009 @ 2:56 am
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http://mathoverflow.net/questions/37127?sort=votes | ## Can Convergence Radii of Padé Approximants Always Be Made Infinite?
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I've found (as have others), that for some analytic functions, a Padé approximant of it has an infinite convergence radius, whereas its associated Taylor series has a finite convergence radius. $f(x)=\sqrt{1+x^2}$ appears to be one such function. My questions are:
1) Is there any function where the Taylor series has the largest convergence radius of all associated Padé approximants? If so, is the Taylor series radius strictly larger, or only equal to the convergence radius of other Padé approximants (i.e. excluding the Taylor series itself)?
2) If not, is there any function that is analytic everywhere, and yet for which there is no (limit of) Padé approximant(s) that has an infinite convergence radius?
It would be both very cool and very useful if there is always a (limit of) Padé approximant(s) that has an infinite convergence radius for any function that is analytic everywhere, though I haven't the slightest how one checks/analyzes convergence of Padé approximants if the degrees of numerator and denominator both approach infinity. :)
One extra question, if there is always such a Padé approximant:
3) Is there always a numerically stable method of computing this approximant up to a finite order?
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I don't have a good answer to your questions (I'm still trying to read the book by Baker and Graves-Morris), but I wish to direct your attention to the theorem by Montessus de Ballore, which guarantees the uniform convergence of a sequence of Padé approximants of increasing numerator degree and fixed denominator degree, assuming that the function being approximated is meromorphic within a disk, and all the poles of the function within said disk are simple. – J. M. Aug 30 2010 at 13:10
Look at the function more closely - it has a branch cut (in the complex plane). So, over the complex, you won't have good convergence properties. Over the real line, sure, it will behave very well. But that seems more like an accident, no? Just rotate your function by a quarter turn, and now things are ugly again. Pade works best when you have simple poles, not branch cuts. – Jacques Carette Sep 9 2010 at 0:44
Touché about the example I gave. It is quite odd and curious that I've been working lately with a ton of such functions whose Padé approximants appear to have infinite convergence along the real axis, but clearly do not converge infinitely along the imaginary axis. I suppose I shouldn't be too surprised, since they come from recursively applying a real, non-negative perturbation to rational functions. However, I was expecting that, like with the Taylor series of them, the rational perturbation would only be valid up to a finite value of the parameter. Anyway, I'm just rambling now. – Neil Dickson Sep 9 2010 at 7:23
## 1 Answer
Don't be lured into thinking Pade approximates are 'nice'. Here is why:
Notation:
Let $[L/M]_f(z)$ be the Pade approximation $\frac{p_n(z)}{q_m(z)}$ to $f$ where $\text{deg}(p_n)\leq n$ and $\text{deg}(q_m)\leq m$, $q_m(0)=1$.
Parital Answer:
1) The partial theta function $h_q(z)$ with $q=e^{iz}$ and $z/2\pi$ real and irrational is a counterexample by Lubinsky and Saff. It is analytic in $|z|<1$ but there is no subsequence of pade approximants $[L/M]_f(z)$ with $M$ fixed that converges for all $|z|<1$. See Pade Approximants by Baker and Braves Morris, p279.
2) A theorem by Wallin shows that there is an entire function such that the approximant sequence $[L/L]_f(z)$, $L\in\mathbb{N}$ is unbounded in $z\neq0$. (I cannot remember a reference for this.) However, I don't have an answer if you consider pade sequences $[L/M]_f(z)$ where $L$ and $M$ are two independent sequences.
3) For special sequences of Pade approximations (i.e. $[L/L]_f(z)$ -the diagonal sequence) there are numerically stable ways to calculate the Pade approximant. Not sure in general.
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Wynn ε is the standard method if you're interested only in evaluation of the diagonal approximants; getting the coefficients of the rational function require the solution of an appropriate Toeplitz system built from the series coefficients. For following arbitrary paths on the Padé table, Rutishauser's quotient-difference algorithm is used, but it is unstable in inexact arithmetic. – J. M. Sep 8 2010 at 20:40
Thanks for the insights! It seems reasonable that if M is fixed, there would be some function for which the approximant sequence doesn't fully converge. It's very interesting that it also happens for M=L on some function. – Neil Dickson Sep 9 2010 at 7:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9354614019393921, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/99705/change-the-sign-of-volatility/99708 | ## change the sign of volatility
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Assume the time inhomogeneous SDE $dX(t)=\mu(t,X(t))dt+\sigma(t,X(t))dW(t)$ has a solution $X(t)$. If we replace $\sigma$ with its absolute value, does the new SDE $dY(t)=\mu(t,Y(t))dt+|\sigma(t,Y(t))|dW(t)$ have a solution $Y(t)$ which has the same distribution as $X(t)$? Here, the initial condition for both SDEs are assumed to have the same distribution.
My guess is yes. And I have a non rigorous proof coming from the numerical solution of this SDE $X(t_{k+1})=X(t_k)+\mu(t_k,X(t_k))(t_{k+1}-t_k)+\sigma(t_k,X(t_k))(W(t_{k+1})-W(t_{k}))$. For any path of $X(t_k)$ such that $X(t_k)=x$, on the corresponding path of $Y(t_k)$ such that $Y(t_k)=x$, the drift parts are the same, and the volality parts should have the same distribution. But these are all purely heuristic. Can someone offer some ideas?
Thank you!
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## 1 Answer
There is no change into the probability distribution due to the absolute value. The reason can be traced back on the Ito's lemma applied to the derivation of the (Fokker-Planck) equation for the probability distribution: This will depend on $\sigma^2$.
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Thank you very much, Jon! This exactly answers my question. – perfectconan Jun 15 at 14:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9123377203941345, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/3184/philosophical-meaning-of-the-yoneda-lemma/121397 | ## “Philosophical” meaning of the Yoneda Lemma
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The Yoneda Lemma is a simple result of category theory, and its proof is very straightforward.
Yet I feel like I do not truly understand what it is about; I have seen a few comments here mentioning how it has deeper implications into how to think about representable functors.
What are some examples of this? How should one think of the Yoneda Lemma?
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## 10 Answers
One way to look at it is this:
for $C$ a category, one wants to look at presheaves on $C$ as being "generalized objects modeled on $C$" in the sense that these are objects that have a sensible rule for how to map objects of $C$ into them. You can "probe" them by test objects in $C$.
For that interpretation to be consistent, it must be true that some $X$ in $C$ regarded as just an object of $C$ or regarded as a generalized object is the same thing. Otherwise it is inconsistent to say that presheaves on $C$ are generalized objects on $C$.
The Yoneda lemma ensures precisely that this is the case.
I wrote up a more detailed expository version of this story at motvation for sheaves, cohomology and higher stacks.
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More specifically, the category of presheaves on C is the free cocomplete category on C. This fact is probably about as fundamental as Yoneda's lemma. So the Yoneda lemma is a kind of categorification of the fact that the map from a set to the free commutative monoid on that set is an injection. – Reid Barton Oct 29 2009 at 2:08
Good point. By the way, a gentle introduction by John Baez on what "free cocompletion" means is here: ncatlab.org/nlab/show/free+cocompletion – Urs Schreiber Oct 29 2009 at 2:14
So, for example (if I understand what you're getting at), sometimes you want to add "Pro-objects" and "Ind-objects" to your favorite category C which lacks limits over (co-)filtered systems, and you can do this by identifying C with its image under the Yoneda embedding (which implicitly uses Yoneda's Lemma). – John Goodrick Feb 13 2010 at 20:50
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@Hans, the precise statement that you are looking for is simply the Yoneda lemma. Those words are meant to highlight its "philosophical meaning", as requested. Those "objects modeled on C" are precisely presheaves on C. Their "set of probes" by some X in C is the value of the presheaf on X. Plenty of further technical details are linked to at ncatlab.org/nlab/show/… – Urs Schreiber Feb 16 2012 at 10:29
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@HansStricker: The nLab article on "Space and Quantity" (ncatlab.org/nlab/show/space+and+quantity) is my favorite exposition of these ideas. The "Details" section talks a little about the role of the Yoneda lemma. – Vectornaut Feb 10 at 21:45
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In his Algebraic Geometry class a few years back, Ravi Vakil explained Yoneda's lemma like this: You work at a particle accelerator. You want to understand some particle. All you can do are throw other particles at it and see what happens. If you understand how your mystery particle responds to all possible test particles at all possible test energies, then you know everything there is to know about your mystery particle.
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To me, this is a beautiful explanation of particle physics in the familiar terms of Yoneda's lemma! – Thanos D. Papaïoannou Oct 29 2009 at 9:31
It is, indeed, to some extent also good explanation of some string theoretic thinking is suggesting: that whatever space is, you find out by testing it by throwing worldsheets of strings into it. One can make this Yoneda-way of thinking about string theory sigma-models pretty precise in cases. One rarely sees string theorists talk this way, of course. One exception is Eric Sharpe, who promoted that point of view here: golem.ph.utexas.edu/string/archives/000677.html – Urs Schreiber Oct 29 2009 at 9:53
One does not need to get into any fancy string theory to see this idea. Fundamental groups and homotopy groups are already a manifestation of this idea! Indeed I think Manin once suggested that we can view Gromov-Witten invariants as an algebro-geometric analogue of fundamental group... – Kevin Lin Oct 29 2009 at 14:20
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A lower-level manifestation of the same idea: a locally integrable function on an open domain is determined by the knowledge of the values of the integrals against test functions. – Gian Maria Dall'Ara Nov 1 2009 at 15:51
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@Agusti: There are, of course, two Yoneda lemmas. One says you can recover A from hom(A,-) and the other says you can recover it from hom(-,A). The latter is covariant in A, so makes a better statement. But, yes, from the hom(A,-) perspective, I'd say "you can know your particle by looking at its emissions". – Theo Johnson-Freyd Feb 14 2010 at 18:16
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Lazily, I'll just point to some notes on this question: What's the Yoneda Lemma all about?
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Could you explain the last statement on page 8? (Any functor C^op \mapsto Set can be built out of representables Hom(-,A) in very roughly the same way that any number is built as a product of primes). Should I make it a separate question? – Max M Nov 2 2009 at 1:34
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This refers to the "co-Yoneda lemma", which says that every presheaf is a colimit of representable presheaves: ncatlab.org/nlab/show/co-Yoneda+lemma – Urs Schreiber Nov 17 2009 at 17:04
Max M, sorry, I only just saw your question. Urs is right: I was referring to the fact that every presheaf is a colimit, in a canonical way, of representable presheaves. I'd call it that the 'Density formula'. Another reference is Theorem 5.1.16 of these notes: maths.gla.ac.uk/~tl/msci . Don't take the analogy with prime factorization too seriously. – Tom Leinster Nov 18 2009 at 1:26
@Tom: Do you know some applications of Yoneda lemma to combinatorics? – Yannic Apr 19 2012 at 7:16
Not off the top of my head. The only thing I can think of is that Joyal's theory of species is a categorical approach to enumerative combinatorics, and is almost certain to involve the Yoneda lemma in its development. – Tom Leinster Apr 19 2012 at 13:45
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If you have basic experience with abstract algebra, the ideas in the Yoneda lemma should be quite familiar and even intuitive; the apparent difficulty is only in recognizing them in this new presentation.
You can think of "category" as meaning the same thing as "algebraic theory in a multisorted language with only unary functions" (the objects of the category being the sorts of the language, the morphisms being the definable functions, and the equalities between (composites of) morphisms being the laws of the theory). From this perspective, a functor from C to Set is simply a model of the theory corresponding to C, and natural transformations of such functors are homomorphisms of models. The Yoneda lemma then is about free models: specifically, it says that for every sort s, the "term model" of terms with a single variable, of sort s (equivalently, definable functions with domain s) is the free model on a single generator of sort s. [It may be unfamiliar when expressed as "Nat(Hom(s, -), M) ~= M(s), naturally in M", but that is indeed all this categorical expression is saying]
The so-called co-Yoneda lemma mentioned in the other comments also has a nice interpretation from this perspective, amounting to the demonstration that every model can be specified by generators and relations.
(I wouldn't say this is The One Right Way to think about the Yoneda lemma, because it's useful to view it from many different perspectives, but this is certainly One Right Way to think about the Yoneda lemma.)
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Neat! And under this analogy, representable functors (with representing object s) are just models that are freely generated by a single element (of sort s). To me, at least, this clears things up a lot. – John Goodrick Feb 13 2010 at 20:55
A good and frequent use of the Yoneda lemma is internalization: If e.g. I have monoid valued representable contravariant functor Hom(-,A):C-->Set, then the representing object A must be a monoid object in C. This is because the structure morphism Hom(-,A)xHom(-,A)=Hom(-,AxA)-->Hom(-,A) is a natural transformation and thus, by Yoneda comes from a morphism AxA-->A inside C, same for the other structure morphisms and the commuting diagrams.
The same goes through for other algebraic (or limit) structures and also for covariant Hom-functors which, if they are algebra-valued are represented by an coalgebra-object. An excellent example for the latter is the fact that affine algebraic groups are represented by Hopf algebras.
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Barr and Wells (Toposes, Triples, and Theories, 84) talks about arrows as a general kind of elements. In Set, arrows from {*}→A are the usual elements of A, and arrows from bigger sets X→A are the X-elements of A, or elements of A parameterised in X. Of course the latter makes sense in any category, so we can use this language the state the Yoneda lemma as:
The Hom(-,A)-elements of F are just the usual elements of FA.
I find this to be, at least, a useful mnemonic, but also justifies the intuition that an object "is" its collection of probes.
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Nice interpretation! – Martin Brandenburg Feb 18 2011 at 19:05
Another way to think about the Yoneda lemma is in terms of universal things. Consider, for instance, the existence of classifying spaces for bundles. The statement is that for any suitable group G, there is a space BG such that for any nice enough space X, homotopy classes of maps X → BG are in natural bijection with isomorphism classes of G-structured bundles over X. In categorical terms, that means there is a natural isomorphism between the functors
X ↦ {G-structured bundles over X}
and
X ↦ [X,BG]
The Yoneda lemma implies that this natural isomorphism is uniquely determined by a specific G-structured bundle over BG. That is, the existence of a "classifying space" BG with the above property implies the existence of a universal bundle EG → BG such that every bundle over any space X is the pullback of the universal one along a map X → BG, unique up to homotopy.
The search for representing objects, and hence for universal data, lies at the heart of a lot of modern algebraic topology, algebraic geometry, and even category theory.
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Why do you say "G-structured bundle" instead of G-principal bundle? – Urs Schreiber Oct 29 2009 at 7:39
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It could be a vector bundle with transition functions in G. – Mike Shulman Oct 30 2009 at 0:42
If you don't mind thinking of category theory in terms of functional programming there is an interpretation here. Fix a type A and a functor F. If you have a machine that can give you back an object of type FB every time you give it a function of type A->B, can you reverse engineer fully what the machine is doing? Essentially the machine must contain an element of FA and you can recover that FA from how it responds to your functions. This is very similar to Theo's physical perspective.
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You might also want to think about the Yoneda Lemma as a statement about functors.
A locally small category $\mathcal{C}$ is embedded by the $\mathrm{hom}$ functor in the category $\mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})$. This is called the Yoneda embedding. Thus, the $\mathrm{hom}$ functor is fully faithful (this itself is a corollary of the Yoneda lemma), but is not an equivalence of categories, because it isn't essentially surjective. In other words, not every functor $F$ from $\mathcal{C}^{\mathrm{op}}$ to $\mathrm{set}$ is representable- for example, the empty functor which maps each object in $\mathcal{C}$ to the empty set, is never representable. The problem is that the Yoneda embedding does not commute with colimits. But the Yoneda lemma tells you that every functor $F\in \mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})$ becomes representable when extended appropriately. In other words, every functor $F$ from $\mathcal{C}^{\mathrm{op}}$ to $\mathrm{set}$ extends to a functor from $\left(\mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})\right)^{\mathrm{op}}$ to $\mathrm{set}$ (this is a special case of the Yoneda extension) which does commute with colimits, and is representable.
So one philosophical interpretation'' of the Yoneda lemma is the following:
Every functor $F$ from $\mathcal{C}^{\mathrm{op}}$ to $\mathrm{set}$ can be extended to a representable functor from $\left(\mathrm{Hom}(\mathcal{C}^{\mathrm{op}},\mathrm{set})\right)^{\mathrm{op}}$ to $\mathrm{set}$.
One reference for this point of view is these notes by Akhil Mathew.
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Not only is it not essentially surjective in general, it's never essentially surjective, because the functor $C^{op} \to Set$ that is constantly empty cannot be isomorphic to a representable. (I must confess: I don't understand what you are trying to say in your answer.) – Todd Trimble Feb 11 at 20:26
Thank you for your comment. I've used it to clarify and expand this answer- does it make more sense now? – Daniel Moskovich Feb 12 at 8:39
Here is an example on representable functors. Yoneda's lemma gives down-to-earth, morpshim oriented interpretation of representable functors, and vice versa.
I will explain this with an example.
In a category $\mathscr{C}$, the product of $A$ and $B$ is the pair of object $A\times B$ in $\mathscr{C}$ and a fixed natural isomorphism $$\sigma \colon \mathrm{Hom}(-,A\times B)\to \mathrm{Hom}(-,A)\times \mathrm{Hom}(-,B).$$
This definition of products only uses terminology of functors. By applying Yoneda's lemma, we arrive at a morphism oriented definiton of products. Yoneda's lemma says that there is a bijection $$\Psi \colon \mathrm{Hom}\left( \mathrm{Hom}(-,A\times B),\mathrm{Hom}(-,A)\times \mathrm{Hom}(-,B)\right) \to \mathrm{Hom}(A\times B,A)\times \mathrm{Hom}(A\times B,B).$$ In particular, we apply this to $\sigma$ and denote $$\Psi(\sigma)=\sigma(A\times B)(\mathrm{id}_{A\times B})=(\pi^{A}\colon A\times B\to A,\pi^{B}\colon A\times B\to B).$$ Next, by applying the inverse of $\Psi$, we compute $$\sigma(X)=\Psi^{-1}\left( \Psi(\sigma)\right)(X):\mathrm{Hom}(X,A\times B)\to \mathrm{Hom}(X,A)\times \mathrm{Hom}(X,B)$$ $$f\colon X\to A\times B\mapsto (\pi^{A}\circ f,\pi^{B}\circ f).$$ Since $\sigma$ is a natural isomorphism, $\sigma(X)$ is a bijection. This bijectivity is the usual definition of product based on morphisms (universality):
For any pair of morphisms $f^{A}\colon X\to A$ and $f^{B}\colon X\to B$, there exists a unique morphism $f\colon X\to A\times B$ with $\pi^{A}\circ f=f^{A}$ and $\pi^{B}\circ f=f^{B}$.
I think the Philosophy behind Yoneda's lemma is that, it connects the world of functors (and natural transformations) $\mathfrak{Set}^{\mathscr{C}^{\mathrm{op}}}$ and the world of morphisms $\mathscr{C}$.
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http://mathhelpforum.com/algebra/102726-simplifying-help.html | # Thread:
1. ## Simplifying help.
Can someone show me how to simplify this.
$6x\ln(1-xy)-\frac{3x^2y}{1-xy}+\frac{3x^2-2x^3y^2}{(1-xy)^2}$
I have had a few goes but I can't remember how to make them the same denominator. and keep coming up with a mash of rubbish , your help is appreciated.
2. For a start, note that in the term on the RHS you've got a factor of $1-xy$ in the top and the bottom, so you can cancel that out of top and bottom to make it yay simpler. That means the 2nd and 3rd terms *do* have a common denominator. That should help for a start.
3. so
$\frac{3x^2-2x^2y}{1-xy}$??
4. Actually, now I look at it I may be wrong! I don't think that there is such a common factor after all.
So, take your initial equation and multiply top and bottom of the middle term by 1-xy to put it over a common denominator. Then it's a matter of sorting out the stuff on top and seeing whether there's a factor. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9680774807929993, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/53220/list | ## Return to Question
2 fixed formatting
Suppose that over some (algebraically closed) field $K$ of characteristic $p>0$ we have: numerical equivalence of cycles coincides with homological one with respect to ${\mathbb{Q}}_{l}$ and `${\mathbb{Q}}{l'}$-adic {\mathbb{Q}}_{l'}$`-adic etale cohomology for two (distinct) primes $l,l'\neq p$ (i.e. the Standard Conjecture D is fulfilled). Then it was proved in: Smirnov O., Graded associative algebras and Grothendieck standard conjectures, Invent. Math., 128 (1997), 201-206 that the Lefschetz standard conjecture holds also; hence the Kunneth decompositions of the motif of a smooth projective $P$ exists both with respect to $\mathbb{Q}_l$-adic and with respect to `$\mathbb{Q}{l'}$-adic \mathbb{Q}_{l'}$`-adic etale cohomology. Are these two decompositions necessarily isomorphic? I suspect that that the answer is 'Yes' and the proof is easy, but I am not sure.
P.S. I don't understand why writing ${\mathbb{Q}}_{l'}$ in my question leads to catostrophic appearance.
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# Could the Kunneth decomposition of a motif depend on the choice of $l$?
Suppose that over some (algebraically closed) field $K$ of characteristic $p>0$ we have: numerical equivalence of cycles coincides with homological one with respect to ${\mathbb{Q}}_{l}$ and ${\mathbb{Q}}{l'}$-adic etale cohomology for two (distinct) primes $l,l'\neq p$ (i.e. the Standard Conjecture D is fulfilled). Then it was proved in: Smirnov O., Graded associative algebras and Grothendieck standard conjectures, Invent. Math., 128 (1997), 201-206 that the Lefschetz standard conjecture holds also; hence the Kunneth decompositions of the motif of a smooth projective $P$ exists both with respect to $\mathbb{Q}_l$-adic and with respect to $\mathbb{Q}{l'}$-adic etale cohomology. Are these two decompositions necessarily isomorphic? I suspect that that the answer is 'Yes' and the proof is easy, but I am not sure.
P.S. I don't understand why writing ${\mathbb{Q}}_{l'}$ in my question leads to catostrophic appearance. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.92570561170578, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/116767/list | Return to Answer
2 Corrected definition, added link to Yoshinaga's paper.
The answer is probably no.
In his paper Transcendence of Periods: the State of the Art (Pure and Applied Mathematics Quarterly, Volume 2, Number 2, p. 435-463, 2006), Michel Waldschmidt conjectures that no period is a Liouville number (see questions 2 & 3 in the introduction). This expectation is supported by the main advances of transcendental number theory in 20th century, notably Baker's theory of linear forms in logarithms and its extensions to commutative algebraic groups.
Here, the word period has to be understood in the sense of Kontsevich and Zagier: the a complex number whose real and imaginary parts are values of absolutely convergent integral of an algebraic $d$-form rational functions with rational coefficientson a , over domains in $d$-dimensional semi-algebraic set defined \mathbf R\$ given by polynomials polynomial inequalities with rational coefficients. (The preceding definition is quoted from the paper Periods and elementary real numbers by Masahiko Yoshinaga, who was apparently the first to prove that periods belong to the field of elementary complex numbers, those whose real and imaginary parts can be effectively approximated by Cauchy sequences of rationals.)
1
The answer is probably no.
In his paper Transcendence of Periods: the State of the Art (Pure and Applied Mathematics Quarterly, Volume 2, Number 2, p. 435-463, 2006), Michel Waldschmidt conjectures that no period is a Liouville number (see questions 2 & 3 in the introduction). This expectation is supported by the main advances of transcendental number theory in 20th century, notably Baker's theory of linear forms in logarithms and its extensions to commutative algebraic groups.
Here, the word period has to be understood in the sense of Kontsevich and Zagier: the integral of an algebraic $d$-form with rational coefficients on a $d$-dimensional semi-algebraic set defined by polynomials with rational coefficients. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9002984166145325, "perplexity_flag": "head"} |
http://www.wikiwaves.org/Category:Linear_Water-Wave_Theory | # Category:Linear Water-Wave Theory
From WikiWaves
## Introduction
Linear water waves are small amplitude waves for which we can linearise the equations of motion (Linear and Second-Order Wave Theory). It is also standard to consider the problem when waves of a single frequency are incident so that only a single frequency needs to be considered, leading to the Frequency Domain Problem. The linear theory is applicable until the wave steepness becomes sufficiently large that non-linear effects become important.
## Equations in the Frequency Domain
We assume small amplitude so that we can linearise all the equations (see Linear and Second-Order Wave Theory). We also assume that Frequency Domain Problem with frequency $\omega$ and we assume that all variables are proportional to $\exp(-\mathrm{i}\omega t)\,$ The water motion is represented by a velocity potential which is denoted by $\phi\,$ so that
$\Phi(\mathbf{x},t) = \mathrm{Re} \left\{\phi(\mathbf{x},\omega)e^{-\mathrm{i} \omega t}\right\}.$
The coordinate system is the standard Cartesian coordinate system with the $z-$axis pointing vertically up. The water surface is at $z=0$ and the region of interest is $-h<z<0$. There is a body which occupies the region $\Omega$ and we denote the wetted surface of the body by $\partial\Omega$ We denote $\mathbf{r}=(x,y)$ as the horizontal coordinate in two or three dimensions respectively and the Cartesian system we denote by $\mathbf{x}$. We assume that the bottom surface is of constant depth at $z=-h$. Variable Bottom Topography can also easily be included but we do not consider this here.
The equations are the following
$\begin{align} \Delta\phi &=0, &-h<z<0,\,\,\mathbf{x} \in \Omega \\ \partial_z\phi &= 0, &z=-h, \\ \partial_z \phi &= \alpha \phi, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align}$
(note that the last expression can be obtained from combining the expressions:
$\begin{align} \partial_z \phi &= -\mathrm{i} \omega \zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \\ \mathrm{i} \omega \phi &= g\zeta, &z=0,\,\,\mathbf{x} \in \partial \Omega_{\mathrm{F}}, \end{align}$
where $\alpha = \omega^2/g \,$)
$\partial_n\phi = \mathcal{L}\phi, \quad \mathbf{x}\in\partial\Omega_B,$
where $\mathcal{L}$ is a linear operator which relates the normal and potential on the body surface through the physics of the body.
The simplest case is for a fixed body where the operator is $L=0$ but more complicated conditions are possible.
The equation is subject to some radiation conditions at infinity. We assume the following. $\phi^{\mathrm{I}}\,$ is a plane wave travelling in the $x$ direction,
$\phi^{\mathrm{I}}(x,z)=A \phi_0(z) e^{\mathrm{i} k x} \,$
where $A$ is the wave amplitude (in potential) $\mathrm{i} k$ is the positive imaginary solution of the Dispersion Relation for a Free Surface (note we are assuming that the time dependence is of the form $\exp(-\mathrm{i}\omega t)$) and
$\phi_0(z) =\frac{\cosh k(z+h)}{\cosh k h}$
In two-dimensions the Sommerfeld Radiation Condition is
$\left( \frac{\partial}{\partial|x|} - \mathrm{i} k \right) (\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|x|\rightarrow\infty\mathrm{.}$
where $\phi^{\mathrm{{I}}}$ is the incident potential.
In three-dimensions the Sommerfeld Radiation Condition is
$\sqrt{|\mathbf{r}|}\left( \frac{\partial}{\partial|\mathbf{r}|} - \mathrm{i} k \right) (\phi-\phi^{\mathrm{{I}}})=0,\;\mathrm{{as\;}}|\mathbf{r}|\rightarrow\infty\mathrm{.}$
where $\phi^{\mathrm{{I}}}$ is the incident potential.
## Subcategories
This category has the following 10 subcategories, out of 10 total.
## Pages in category "Linear Water-Wave Theory"
The following 23 pages are in this category, out of 23 total. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9293099045753479, "perplexity_flag": "head"} |
http://soffer801.wordpress.com/category/logic/ | #### Andy Soffer
UCLA Mathematics Ph.D Student
## A nonstandard proof
May 21, 2012
We’re going to use a nonstandard model of the real numbers to make this work. The construction is a bit wacky: Take a non-principal ultrapower of $\mathbb R$. Call this the hyperreals. Here’s another way to put it:
A hypernumber a sequence of regular numbers $(a_1,a_2,\dots)$. To make decisions about hypernumbers, we put it to a vote:
Is it true that $(3,0,0,\dots)=(0,0,0,\dots)$? Well, the first position votes no, because $0\ne 3$. The rest of them all vote yes, however, so the vote is infinity to one. They’re equal. Wackier things such as $(0,1,0,1,\dots)$ we wont get into. If you’re interested, look here. (The basic idea is that the ultrafilter decides whether the evens or the odds win in the vote. The ultrafilter always decides which sets are more important. The fact that its non-principal implies that it prefers cofinite sets to finite ones).
It’s easy to see that the standard real numbers are embedded as hyperreal numbers. If you have a real number $x$, then $(x,x,\dots)$ is the standard real number embedded in the hyperreals as a hypernumber.
Nonstandard models are fun because now we really do have “infinitesimal numbers.” The hypernumber $(1,\frac12,\frac13,\dots)$ is smaller than every standard positive number $(x,x,x,\dots)$, but its still bigger than $(0,0,0,\dots)$. We can also have infinite numbers $(1,2,3,\dots)$, and even larger infinite numbers $(2,3,4,\dots)$.
Lots of things you can do in the hyperreals are essentially the same in the standard real numbers. Some people call this the “transfer principle.” I never liked that name. All it says is that, in model theoretic terms, the standard real numbers and the hyperreals are “first-order equivalent.” That is, they have the same first order theories. This means that anything you can write down using quantifiers $\forall$ and $\exists$, where you only quantify over elements (not subsets of elements) is true in one model if and only if it is true in the other.
Anyway, enough logic. Let $P$ be the set of hyperprimes. These are the hypernatural numbers which aren’t hyperdivisible by any hypernatural numbers other than hyper-one and themselves. That sentence was fun.
Let $N=(1!,2!,3!,\dots)$. Note that $N$ is hyperdivisible by every standard natural number. This means that $N!+1=(1!+1,2!+1,3!+1,\dots)$ cannot be divisible by any standard prime. But there does have to be some hyperprime dividing any hypernumber. This is because every standard natural number is divisible by a standard prime. Since divisibility is a first-order property, there must be a hyperprime hyperdividing any hypernatural number. This tells us that $P$, the set of hyperprimes, contains some nonstandard element.
Now I just want to argue why a set which contains nonstandard elements must have its standard counterpart be infinite. The idea is that, if it were finite, I could define the set by saying $n$ is in $P$ if and only if $n=p_1$ or $n=p_2$ or … or $n=p_k$. I can’t do this if $P$ is infinite, because the sentences I write down must be finite. Then, when I transfer this set to the hyperreal numbers, if it were finite, it would satisfy this property in the hyperreals too. Thus, in the hyperreals, it would only contain the same standard elements.
The fact that $P$ contains a nonstandard hyperprime tells us that there must be infinitely many primes. I left out many details, but hopefully the sketch gives you a reasonable idea of what is going on.
At some level, this is no different from Euclid’s proof, but it just seems so much more badass. I mean, we used the axiom of choice (to construct the non-principal ultrafilter) in a number theory proof. That is decidedly awesome.
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## Why Programming is Hard
March 18, 2012
Take your favorite programming language and write a program in it. Your compute stores the source code as a string of zeros and ones. We can interpret that code as a number in binary. So to each program, we have a number. We may have many numbers whose programs that do the same thing, but any program I can write down has a number associated with it. Let $P_e$ be the program which is encoded by $e$. This might not do anything. It might have lots of spelling mistakes and syntactical errors, but every program is $P_e$, for some number $e$.
Programs take in inputs and spit out outputs. Or at least, sometimes they do. Sometimes programs run forever. Suppose I am handed a bunch of programs and told to run them on some inputs, but I’m told I only have an hour to do this. Obviously I can’t run the ones that go one forever, so how do I know which ones those are? How do I tell the difference between a program that runs for an obscenely long time, and one that will never end?
The short of it is that I can’t. There is no algorithmic way to determine which programs halt and which programs run forever.
Theorem: There is no program $H$ which takes input $e$ (an encoding of a program) and $x$ (some input to the program encoded by $e$), and spits out $1$ if $P_e(x)$ halts and $0$ otherwise.
Proof: Suppose I had such a program. I’m going to make a slightly different program. What it will do is take in some input $e$, and determine whether or not $P_e(e)$ halts. If $P_e(e)$ never halts, it will return $17$. Otherwise, it will enter an infinite loop and never halt. This program sometimes stops and sometimes goes without stopping, so we’ll call it the “Yielding Program.” Here’s some pseudocode for that program:
To save space, I’m going to write $Y$ for the “yielding program,” and I’m going to let $p$ be the number encoding it. That is, $Y=P_p$. Ready for the magic? Does $Y(p)$ halt? That is, does $Y(p)$ return $17$? If it did, that would mean that we were in the first case in the if-statement, so $H(p,p)=0$. But $H(p,p)=0$ means that $P_p(p)$ (a.k.a. $Y(p)$) doesn’t halt. That’s a contradiction.
What about the other possibility $Y(p)$ goes on forever. This means that $H(p,p)$ is not $0$, so it must be $1$. But this too is a problem, for then $P_p(p)$ (a.k.a. $Y(p)$) does halt. That’s another contradiction.
Since we used the existence of a program $H$ to construct $Y$, which lead to a contradiction, it must be that $H$ cannot not really exist.
$\square$
If you didn’t like this proof, then perhaps this one in a Dr. Seuss style will suit you better: Scooping the Loop Snooper
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## Yoneda’s Lemma (part 1)
February 13, 2012
So I think we’re ready, at least for the statement of Yoneda’s lemma. It says that for any locally small category $\mathcal C$, if $A$ is an object in $C$, and $F:\mathcal C\to\textsc{Set}$ a functor, then
$\mbox{Nat}(h^A,F)\cong F(A)$
Moreover, the isomorphism is natural in both $A$ and $F$.
Wow that looks complicated. Let’s parse some of the notation that I haven’t even explained yet. So certainly we know what the righthand-side means. It’s $F$ applied to $A$. That’s just a set. As for the lefthand-side, $\mbox{Nat}$ and $h^A$ I haven’t explained.
So $h^A$ is a functor we mentioned briefly, but I used a different notation. It’s the representable functor $\hom(A,-)$. I write it as $h^A$ here to avoid over using parentheses and therefore complicating this business well beyond it’s current level of complication. As a reminder, $\hom(A,-)$ is a functor from $\mathcal C$ to $\textsc{Set}$ (the same as $F$). It takes objects $B$ in $\mathcal C$ to the set of homomorphisms $\hom(A,B)$. Remember we’re in a locally small category, so $\hom(A,B)$ really is a set. It takes morphisms $f:B\to C$ to a map $h^A(f):\hom(A,B)\to\hom(A,C)$ by sending
$h^A(f):\phi\mapsto f\circ\phi$
We checked all the necessary details in an earlier post to make sure this really was a functor.
So the last thing is that $\mbox{Nat}$. It’s the collection of all natural transformations between the functors $h^A$ and $F$. So Yoneda’s lemma claims that there is a one to one correspondence between the the natural transformations from $h^A$ to $F$ and the set $F(A)$.
In particular, it claims that $\mbox{Nat}(h^A,F)$ is a set. This is because $\textsc{Set}$ is a locally small category, and so each natural transformation (defined as a collection of morphisms, each of which was a set) is a set. But there are only so many collections of morphisms, not even all of which are natural transformations. The collection is small enough to be a set. If you don’t care about this set theory business. Then disregard the paragraph you probably just read angrily.
It’s worth mentioning now, that Yoneda’s lemma is a generalization of some nice theorems. We can (and will) use it to derive Cayley’s theorem (every group embeds into a symmetric group). We can (and will) use it to derive the important fact that $\hom_R(R,M)\cong M$ in the category of $R$-modules. I bet in that one you can already start to see the resemblance.
We’ll prove Yoneda’s lemma over the next few posts.
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## Coproducts in the category of Sets
January 31, 2012
As with products. Not every category has coproducts. However, $\textsc{Set}$ does. I claim that given two sets $A$ and $B$, their coproduct $A\coprod B$ is the disjoint union of $A$ and $B$. Since coproducts are unique, all we need to do is show that the disjoint union satisfies the properties of a coproduct. So we can let $A\coprod B$ denote the disjoint union, and then verify that it really is the coproduct.
#### Proof:
We have the natural injections $i_A:A\to A\coprod B$ and $i_B:B\to A\coprod B$. Suppose we have a set $C$ and $f_A:A\to C$ and $f_B:B\to C$. Define $g:A\coprod B\to C$ by $g(x)=f_A(x)$ if $x\in\mbox{im}\ i_A$, and $g(x)=f_B(x)$ if $x\in\mbox{im}\ i_B$. Notice that every $x\in A\coprod B$ is in the image of precisely one of the maps $i_A$ and $i_B$, so this map $g$ is well defined. Now we just need to check that the diagram
commutes. I’ll leave that as an exercise.
$\square$
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## More Pedantry
January 17, 2012
Similar to last time, I want to talk about something mildly pedantic, but again, very important. There’s something in-between large categories and small categories. We don’t call it a medium category (I suppose my letters to the AMS and MAA haven’t been that persuasive). The collection of homomorphisms between $X$ and $Y$ we denoted as $\hom(X,Y)$. This may be a set or a proper class. If it is a set, we’ll call it the “hom-set.” Easy enough. What if for every $X,Y\in \mathcal C$, $\hom(X,Y)$ is a hom-set? This is what we call a locally small category. Make sense? We may have way too many objects to have it actually be a small category, but there may be few enough maps between them to be locally small. All small categories are locally small. The obvious example of a locally small category which is not small is the category of sets with set maps as the morphisms. There is no “set of all sets,” but given any two sets, the collection of maps between them is a set. Check your set theory axioms.
I’d also like to point out in slightly further detail the mistake I made last time (and have since edited). The definitions of large versus small categories are accurate. Most examples that we know of are large categories. But many of these examples are at least locally small. It seems that smallness is a very restrictive property. Too restrictive to say anything about the categories we care about. On the other hand largeness is just too wide open to say anything interesting. Being locally small is in some sense a balance, and it turns out to have some interesting properties.
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## Pedantry
January 16, 2012
We’ve mentioned the category $\textsc{Set}$ (the category of sets with functions as morphisms). Consider another category $\textsc{Two}$ which has two objects $A$ and $B$ and a morphism between them (along with the required identity morphisms)
First of all, the small-caps notation and naming convention is not standard, there is no standard, but I think it is as good as any other, so I’ll use it.
Secondably [sic], there’s a fundamental difference between these two categories. The difference is in the objects. Not the objects themselves, but $Ob(\textsc{Set})$ and $Ob(\textsc{Two})$. The first one is not a set. After all, we can’t talk about the set of all sets. In some sense, the collection of all sets is just to big to be a set. We can however talk about the set $\{A, B\}$. That is $Ob(\textsc{Fin})$ is a set. (One caveat here is that $A$ and $B$ must be sets themselves, but we can choose them to be, and it’s not the point of this post, nor will it be relevant later).
This is why, in the definition of categories, I specifically mentioned that we had a collection of objects, not a set of objects. But if it’s not a set, what is it? It’s called a class, and it pushes us closer to axiomatic set theory than I want to go. It does however give rise to the following definitions that the pedantic reader will care to think about:
#### Definition:
A category $\mathcal C$ is small if $Ob(\mathcal C)$ is an honest-to-goodness set. Otherwise, we say that $\mathcal C$ is large (in the situation that $Ob(\mathcal C)$ is a class and not a set).
I must say that really this stuff is important despite how I’ve presented it. But if you trust me not to lie to you (probably a bad move), you can just read and trust that I’m not breaking any mathematical laws.
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## Guest post: Isaac Solomon
January 2, 2012
Read Isaac’s blog here.
From a philosophical point of view, the Axiom of Choice (AC) is often a tough sell. On the one hand, there’s something wrong with vector spaces without bases, or fields without algebraic closures (consequences of $\neg$ AC). On the other hand, non-measurable sets.
The most common example of a non-measurable set, the so-called “Vitali set,” is a tranversal of the quotient space $\mathbb{R}/\mathbb{Q}$. While $\mathbb{R}/\mathbb{Q}$ is certainly a repulsive creature, there’s no reason to believe, a priori, that all other examples of non-measurable sets are as abstruse. To see that that is indeed the case, we look to the opening line of Solovay’s 1970 paper, “A model of set theory in which every set of reals is Lebesgue measurable”:
We show that the existence of a non-Lebesgue measurable set cannot be proved in Zermelo-Frankel set theory (ZF) if the axiom of choice is disallowed.
Ouch. Unfortunately, it gets worse. There is no formula $\varphi(x)$ for which ZFC proves, “there is a unique non-measurable subset of reals satisfying $\varphi(x)$.” In other words, while non-measurable sets are definable in certain models of ZFC, the theory does not have the capacity to prove that a particular formula “talks about” a non-measurable set. Thus, it appears that AC furnishes a very large collection of sets that even it cannot get a handle on.
In a recent conversation, Andy and I were wondering if there was some formal way of saying that a proof is constructive. We suggested that a proof be called constructive if every existential statement in its deduction had a (first-order) definable witness. So, for example, the proof of the compactness of a metric space would be constructive if, given an arbitrary sequence in that space, one could explicitly outline an algorithm for producing a convergent subsequence.
Thinking more about this, I suspect our definition may be flawed, in that it shifts responsibility for definability to (the theory of the) model at hand, as opposed to the theory we used to generate it. By that I mean that in certain models of ZFC, in which non-measurable sets are not definable, the proof of their existence is non-constructive, but in other models of ZFC, in which non-measurable sets are definable (without parameters), the same proof would be constructive. To me, that seems to miss the whole point of why we pretend to care about constructability in the first place.
Conversely, if you want to switch your perspective to the theory, things become very arbitrary. Given a model $M$, and a deduction in the theory of that model, do we want to look at provably definable objects in all of $\mbox{Th}(M)$, or in some sub-theory? In that light, we’re stuck with philosophy again, wondering which theory is most natural for considering a certain deduction.
For the philosophically-minded, the remedy probably lies in intuitionistic, or constructive, logic. Whereas in classical logic, all statements are assumed to be true or false, in constructive logic, a statement is only deemed true or false if there is a constructive proof witnessing that assertion. Put differently, the law of the excluded middle, and double negation elimination, are not axioms of constructive logic. Thus, a better definition of constructability would be that a deduction is constructible if and only if it is valid in intuitionistic logic.
Luckily, because we are mathematicians and not philosophers, we don’t have to worry ourselves with these insipid considerations. The workaround is never to adopt a more conservative approach to mathematics, but to rephrase paradoxes as counterexamples and leave reality for the physicists.
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## Philosophy and choice
December 29, 2011
Seriously, category theory is coming soon. I promise. In the meantime, a friend (who posts on this awesome blog) and I had a conversation about the axiom of choice from a philosophical standpoint. It started when I made the bold claim “whatever, the axiom of choice is obviously true.” Originally, this was meant half as a provocative joke, but really, I think it makes sense. My friend, who I’ll call Steve Neen for the sake of this post, declared, as many mathemematicians do, that he preferred constructive mathematics. After all, how can you say something exists without being able to construct it. This is why mathematicians prefer constructive proofs to non-constructive ones.
I had always sort of accepted this as a reasonable point of view (which I still think it is), but the following thought dawned on me: At their core, each of the axioms of ZFC simply say “there exists a something such that something else.” We may have notation for sets constructed from a specific axiom, but symbology shouldn’t have any bearing on whether or not an axiom is constructive. There is simply nothing intrinsically constructive or non-constructive about any of the axioms.
Of course, Steve then pointed out the difference may not be subjective. Look at what computers can do. They can represent unions, pairs, power sets (though not efficiently), etc. all in an effective manner. But look at the axiom of choice. Computers just aren’t built to do this constructively.
Fair point, Steve. One that had me thinking all of last night. One answer is that what computers can do is somewhat arbitrary, but that feels like a cop out. So here’s my two part answer, that I’m reasonably satisfied with.
• Look at the Axiom of Infinity. It essentially says there is an infinite set. More precisely,
$\exists x[\varnothing\in x\wedge\forall y(y\in x\to y\cup\{y\}\in x)]$
Here’s an example of a set a computer can’t represent explicitly. One could argue that an implicit representation is possible, because in finite time one could decide whether or not a given set was an element of this one, but there is no bound on how long it would take to a computer to make this decision. Fair enough. But the power set of an infinite set (i.e., an uncountable set) cannot have such a decision procedure.
This is all to say, that if you discount the axiom of choice for it’s non-constructibility, you should also discount the axiom of infinity.
• Jokingly, and to prove his point, Steve asked me to pick one element from each coset of $\mathbb R/\mathbb Q$. First of all, it’s important to note that one’s ability to do so is reliant on the axiom of choice. This is one of the reasons I think it should be true. Morally, I can pick out an element from each coset. I told Steve, all I need to do is produce a choice function on the coset. I claimed (again, jokingly) that I had such a function. In fact, I had uncountably many. If he named a coset, I’d consult my function and tell him the answer.
Steve did point out, and is correct in doing so, that just because I claim it exists, and can verify as much as he asks doesn’t mean that I have such a function. But from a computational point of view, it sort of does. Any function that’s consistent with a choice function can be extended. If you accept the axiom of choice, then you can extend it to a choice function. If you don’t, then it can just be extended further, so long as the extension is countable. Regardless, all computations are going to be finite, so from a computational point of view, we never really need the axiom of choice, just as we don’t need the axiom of inifinty. We only need finite choice, which is provable from ZF, even without choice.
So that’s a recap of the discussion we had. I’m not entirely convinced anymore that the axiom of choice is “no different” from the other axioms, but it’s an interesting philosophical question I invite you to think about.
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## Global choice and algebraic closures
December 28, 2011
It was pointed out to me today that I glazed over a set-theoretic point in my proof that every field has an algebraic closure. We appealed to Zorn’s lemma, which says:
Given a partially ordered set $\mathcal P$, if every chain $\mathcal C\subseteq\mathcal P$ has an upper bound, then $\mathcal P$ has a maximal element.
An assumption we made here that seems innocuous is that $\mathcal P$ is a set. If it is a proper class (class and not a set), we may run into problems. So what if the collection of all field extensions of a given field is a proper class? Then we can’t use Zorn’s Lemma.
I think, though haven’t attempted to prove it, that this collection of field extensions is really a set, meaning we’re safe in applying Zorn’s lemma. But it’s an interesting diversion to see what happens if we try to prove Zorn’s lemma for classes.
It’s well known that Zorn’s lemma is equivalent to the axiom of choice, so it would be nice to find an “axiom of choice for classes.” Such a thing exists, and it’s called the “axiom of global choice.” It says the same thing about classes that the axiom of choice says about sets. So if you accept the axiom of global choice, we just need to prove a “global Zorn’s lemma.”
Unfortunately, the standard proof of Zorn’s lemma goes something like this: Assume it’s false. Then use the axiom of choice to construct (via transfinite induction) larger and larger elements. These form a well-ordering. We can keep going and we’ll eventually get to a well-ordering that has too many elements to be contained in the set $\mathcal P$.
The problem is that if we replace the axiom of choice with the axiom of global choice, we’ll be unable to make this well-ordering bigger than our “poclass” $\mathcal P$.
If you want more information on this sort of set theory stuff, you should read Zach Norwood’s posts (this is a great blog to which you should definitely subscribe):
Basic facts about AC, part I
Basic facts about AC, part II
On an unrelated note, coming soon will be a series of posts on some generalized abstract nonsense.
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Filed under Field Theory, Logic, Set Theory Tagged with axiom of choice, classes, global choice
## Every field has an algebraic closure
October 25, 2011
Yeah. That’s right. Every field. Yesterday we showed how we could add a root of a polynomial to a field and that would make it bigger. So we can just keep adding roots over and over again until we get them all, right? Well, it’s not that simple. See, it’s conceivable that when I add a new root, now I have a lot of new polynomials whose roots I need to add. And if I add all of them, I could get even more polynomials. Will it ever end?
Luckily, we have a handy tool. The careful reader will have noticed that “Every field has an algebraic closure” kind of looks like “Every vector space has a basis.” The choice of titles was not a coincidence. The proofs both rely on Zorn’s lemma. Once again, this is here for the sake of completeness. If you’re comfortable taking this on faith, I won’t hold it against you. Without further ado…
#### Proof:
Take all the fields that contain a given field $k$, and order them with $E\ge F$ if $E$ is a field extension of $F$. Some things might not be comparable. That’s fine. Then neither will be bigger than the other (this is called a partially ordered set, or poset for short). Last time we showed we can always add on a root to a polynomial (or at least waved our hands in that direction). So if we can produce a maximal field, that means that it must already have all the roots. This is exactly what an algebraically closed field is.
Zorn’s lemma gives us a maximal element under certain conditions, so we’ll just have to satisfy those conditions. The conditions are that “every chain has an upper bound.” What’s a chain? It’s what we call a poset that is totally ordered (I voted for toset, but no one else seemed to like that). If you draw a picture, you’ll see why. Everything lies in one line.
So suppose we have a bunch of fields in a chain. That means that if I pick any two fields, one must contain the other. Let’s call them $F_\alpha$, for $\alpha$ in some indexing set $\mathcal A$. It’s important to note that if $\mathcal A$ is finite, this not hard. One of the $F_\alpha$s is a maximum. But if it’s infinite, or really infinite, things aren’t so clear. In any event, like many Zorn’s lemma arguments, unions come to save the day.
$\displaystyle\bigcup_{\alpha\in\mathcal A}F_\alpha$ is an upper bound for every $F_\alpha$.
Well, duh. The big union contains everything. That was the point. We do however need to check that it’s a field. (Left as an exercise to the reader.) The trick is to take notice that we’re only allowed finite sums/products/etc., and we already know how to do those, since they always come from one of the $F_\alpha$s.
Okay, so any chain has an upper bound. We invoke Zorn’s lemma and we have a maximal element. Like I said, maximal means that we can’t extend it any more (so it must have the root of every polynomial).
$\square$
One can also show that algebraic closures are unique, though I don’t care to do so. We can actually talk about the algebraic closure of a field $k$, instead of just talking about an algebraic closure. We’ll write $\overline k$ for the algebraic closure of $k$. For example, $\mathbb C=\overline{\mathbb R}$
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http://mathoverflow.net/questions/44275/hopf-algebra-and-group-structure-correspondence-for-algebraic-varieties/44284 | ## Hopf algebra and group structure correspondence for algebraic varieties
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Let $V$ be a real algebraic variety and let ${\cal O}(V)$ denote its algebra of regular functions. If we put a group structure on $V$ (not necessarily an algebraic group structure) it will induce a Hopf algebra structure on ${\cal O}(V)$ in the usual manner. My question is, is there a bijective correspondence between the possible group structures on $V$ and the possible Hopf algebra structures on ${\cal O}(V)$?
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What definition of "real algebraic variety" are you using here? If it is an integral separated scheme of finite type over $\mathbb{R}$, then how are you defining a group structure? – S. Carnahan♦ Oct 31 2010 at 9:06
Just the plain ordinary zero set of an ideal of polynomials definition of an algebraic variety. – Abtan Massini Oct 31 2010 at 18:23
## 1 Answer
If $V$ is an affine algebraic variety over any field $k$, then there is a bijection between the algebraic group structures on $V$ and the Hopf algebra structures on $O(V)$. The reason is that the category of affine varieties over $k$ is the just the contravariant category (the category with arrows reversed) of algebras over $k$. So a group law $m:V \times V \to V$ becomes a coproduct $m:O(V) \to O(V) \otimes O(V)$, and this correspondence is a bijection. (Because also, the tensor product of algebras corresponds to the Cartesian product of varieties.)
If you are interested in group structures on $V$ that are not necessarily algebraic group structures, then there isn't necessarily a good relation to $O(V)$ either.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8919865489006042, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/30402?sort=oldest | ## Parabolic envelope of fireworks
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The envelope of parabolic trajectories from a common launch point is itself a parabola. In the U.S. this weekend many will have a chance to observe this fact direcly, as the 4th of July is traditionally celebrated with fireworks.
If the launch point is the origin, and the trajectory starts off at angle $\theta$ and velocity $v$, then under unit gravity it follows the parabola $$y = x \tan \theta - [x^2 /(2 v^2)] (1 + \tan^2 \theta)$$ and the envelope of all such trajectories is another parabola: $$y = v^2 /2 - x^2 / (2v^2)$$
These equations are not difficult to derive. I have two questions. First, is there a way to see that the envelope of parabolic trajectories is itself a parabola, without computing these equations? Is there a purely geometric argument? Perhaps there is a way to nest cones and obtain the above picture through conic sections, but I couldn't see it.
Second, of course the trajectories are actually pieces of ellipses, not parabolas, if we follow the true inverse-square law of gravity. Is the envelope of these elliptical trajectories also an ellipse? (I didn't try to work out the equations.) Perhaps the same geometric viewpoint (if it exists) could apply, e.g., by slightly tilting the sections.
I ask these questions in a weekend recreational spirit.
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Somewhat off-topic: there is a neat online game where one can get some practice with parabolic trajectories :) Or, are they in fact elliptical? onlinegames.com/basketball – Andrey Rekalo Jul 4 2010 at 15:27
## 3 Answers
• E. Torricelli, who was the last Galileo's secretary, suggested a purely geometrical method to find the envelope in his De motu Proiectorum. He also coined the term `parabola of safety'. Apparently it was the first example of computation of an envelope. The method is briefly described in this note.
• Another approach is to launch identical missiles with the same velocity at all possible angles simultaneously. At time $t$, their positions describe a circle $$x^2+\left(y-\frac{t^2}{2}\right)^2=(vt)^2.$$ The latter equation has a unique solution in $t$ provided $(x,y)$ belongs to the parabola $$y=\frac{v^2}{2}-\frac{x^2}{2v^2}.$$
• In the case of missiles moving in a Kepler field (with the attractive potential $\sim -1/r$), the envelope of elliptic trajectories is indeed an ellipse. A web search gave the nice short article which contains several elementary geometric proofs of this and related results.
Edit. A free version of J.-M. Richard article can be found here.
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The first links seems to be broken and the last isn't open-access, unfortunately; the second point is quite cool though. – Daniel Litt Jul 3 2010 at 19:37
Daniel, thank you for the comment. I've added an arxiv link to the second article. As for the first one, the link is jstor.org/stable/3620177?seq=1 – Andrey Rekalo Jul 3 2010 at 20:12
Anyway, it is probably a MO bug, as both links work in my web browser. – Andrey Rekalo Jul 3 2010 at 20:14
@Andrey: Thanks so much!! It is cool that this question has such deep historical roots. I cannot access the JSTOR papers (must be a JSTOR permissions issue?), but I can access the arXiv paper. The freely falling expanding sphere is a beautiful viewpoint! And that the envelope of the elliptical trajectories is also an ellipse is quite satisfying. I am still hopeful there is a nested/tangent cones viewpoint, perhaps extractable from Richard's article. Will ponder this as I watch the fireworks... :-) – Joseph O'Rourke Jul 3 2010 at 23:11
@Joseph: Thank you for the comment. Unfortunately, I have not found a free version of the JSTOR paper. – Andrey Rekalo Jul 4 2010 at 12:18
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It is easy to see that all these parabolas have the same directrix. Height of a directirix correspond to energy of the body. So you have the family of parabolas with the common point $P$ and the directrix $l$. It is easy to prove, (using just definition of parabola as a locus of points...) that all of the touched the parabola with the focus at $P$ and the directrix $l_1$, which parallel $l$ (actually $l$ is midline of $P$ and $l_1$).
The same holds for sun-earth set. If Earth decides to fly in other direction (but with the same speed) its path will be always touch the fixed ellipse with foci in Sun and this position of Earth.
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I found another article to supplement those to which Andrey linked: Eugene I Butikov, "Comment on 'The envelope of projectile trajectories'," Eur. J. Phys. 24 L5-L9, 2003. He also explains the expanding-circles viewpoint that is Andrey's second bullet. He imagines first that there is no gravity, in which case the particles are on the surface of an expanding sphere whose radius $r$ equals $v t$. "With gravity, this uniformly expanding sphere is falling freely as a whole with the acceleration of free fall." He then finds the envelope of these falling, expanding circles.
Later in the note he considers water drops spun off a spinning, wet bicycle wheel. To continue the 4th-of-July theme, these could be sparks from a spinning sparkler wheel. He proves that again, the envelope of the drops/sparks is a parabola.
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This is interesting. So it's not only mathematicians who love to go back and and rederive classical results. Some physicists seem to have this `bug' too. – Andrey Rekalo Jul 4 2010 at 14:59 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9331494569778442, "perplexity_flag": "middle"} |
http://mathhelpforum.com/pre-calculus/2742-3-intercepts-between-f-x-inverse.html | # Thread:
1. ## 3 intercepts between f(x) and inverse??
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points.
My question is, simply, how do you figure out how many intersections there will be between a function and its inverse? Even my teacher was perplexed.
2. Originally Posted by freswood
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points.
My question is, simply, how do you figure out how many intersections there will be between a function and its inverse? Even my teacher was perplexed.
You have, $f(x)=\sqrt{x}$ since this is a bijective function it have an inverse, $y=x^2 \mbox{ for } x\geq 0$. To find intersection points you need to consider,
$\sqrt{x}=x^2$ for $x\geq 0$
Square both sides,
$x=x^4$
Thus,
$x^4-x=0$
Thus,
$x(x^3-1)=0$
Thus,
$x=0$ and $x^3-1=0$
All the real solutions are,
$x=0,1$
I do not see your problem?
3. In the example they gave, it was:
f(x) = 4 - 2[root](2x+6)
We've always been taught that a function and its inverse will intersect on the line y=x, but never any mention of anything else.
4. Originally Posted by freswood
We've always been taught that a function and its inverse will intersect on the line y=x, but never any mention of anything else.
What is that supposed to mean?!?!
A function and its inverse are line relfections in line $y=x$ that is probably what you mean. Not that they intersect at all point on $y=x$
5. Originally Posted by ThePerfectHacker
What is that supposed to mean?!?!
A function and its inverse are line relfections in line $y=x$ that is probably what you mean. Not that they intersect at all point on $y=x$
In order for a graph to intersect its inverse we have to have the following situation: Let y = f(x) be the original graph. The inverse function is represented by x = f(y), or y = g(x). The only way these will intersect is if there is a point (x,y) in the original graph and a point (y,x) in the inverse graph such that (x,y)=(y,x), ie. x = y. Thus all such intersection points will be on the line y = x.
(BTW: The graph of a square root function is essentially the graph of part of a parabola either opening to the right or left. So I am going to refer to your square root function as a parabola. In the following paragraph, then, the "parabola" mentioned can either be your square root function, or its inverse.)
Now, we are looking for the set of intersection points of a parabola, its inverse, and the line y = x. A parabola can intersect with a line in AT MOST 2 points. These are the same two points that the inverse function will cross the line y = x at. So your solution set of the crossing points of your parabola and its inverse is at most 2 points, not 3.
I must therefore conclude that something was either solved or graphed incorrectly for you to have 3 points.
-Dan
6. Originally Posted by freswood
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points.
My question is, simply, how do you figure out how many intersections there will be between a function and its inverse? Even my teacher was perplexed.
Hello,
in addition to all posts I only want to point out, that you have to look thoroughly at the domain and range of a function and its inverse.
$f(x)=4-\sqrt{2x+6}\ \wedge \ f(x) \leq 4$ you'll get the inverse:
$g(x)=\frac{1}{8} x^2-x-1\ \wedge \ x\leq4$
If you take g without the restriction you'll get indeed 3 intercepting points.
Greetings
EB
7. Originally Posted by freswood
Something I find really odd.... A question provides us with a square root function, and asks us to sketch it and its inverse. Strangely enough, f(x) and its inverse intersect not only along y=x but at two other points....
Hello,
it's me again.
I've attached a diagram, which shows that your teacher is right.
Greetings
EB
Attached Thumbnails
8. Originally Posted by earboth
Hello,
it's me again.
I've attached a diagram, which shows that your teacher is right.
Greetings
EB
Thanks for clarifying it. So how would I know (without using a calculator) whether a graph intersects its inverse more than once? Our end of year exam has a part with no calc.
9. Originally Posted by freswood
Thanks for clarifying it. So how would I know (without using a calculator) whether a graph intersects its inverse more than once? Our end of year exam has a part with no calc.
I don't know if this is of any help, but the number of points of intersection
of a function $f$ and its inverse $f^{-1}$ is equal to the number of roots of
$f(f(x))=x$ where $x$ is in the domain of both $f$ and $f^{-1}$.
This is of course equivalent to saying it is equal to the number of roots of
$f(x)=f^{-1}(x)$ in the intersection of the two domains, so its not saying
anything new really.
RonL
10. Originally Posted by earboth
Hello,
it's me again.
I've attached a diagram, which shows that your teacher is right.
Greetings
EB
My apologies. I was obviously only considering symmetric points!
-Dan | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 23, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9639151096343994, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2011/10/04/inner-products-on-differential-forms/?like=1&_wpnonce=9d779cb1c0 | # The Unapologetic Mathematician
## Inner Products on Differential Forms
Now that we’ve defined inner products on $1$-forms we can define them on $k$ forms for all other $k$. In fact, our construction will not depend on the fact that they come from a metric at all.
In fact, we’ve basically seen this already when we were just dealing with vector spaces and we introduced inner products on tensor spaces. Pretty much everything goes just as it did then, so going back and reviewing those constructions will pay dividends now.
Anyway, the upshot: we know that we can write any $k$-form as a sum of $k$-fold wedges, so the bilinearity of the inner product means we just need to figure out how to calculate the inner product of such $k$-fold wedges. And this works out like
$\displaystyle\begin{aligned}\langle \alpha_1\wedge\dots\wedge \alpha_k,\beta_1\wedge\dots\wedge \beta_k\rangle&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\hat{\pi}\in S_k}\mathrm{sgn}(\pi\hat{\pi})\langle \alpha_{\pi(1)}\otimes\dots\otimes \alpha_{\pi(k)},\beta_{\hat{\pi}(1)}\otimes\dots\otimes \beta_{\hat{\pi}(k)}\rangle\\&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\hat{\pi}\in S_k}\mathrm{sgn}(\pi\hat{\pi})\langle \alpha_{\pi(1)},\beta_{\hat{\pi}(1)}\rangle\dots\langle \alpha_{\pi(k)},\beta_{\hat{\pi}(k)}\rangle\\&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\hat{\pi}\in S_k}\mathrm{sgn}(\pi^{-1}\hat{\pi})\langle \alpha_1,\beta_{\pi^{-1}(\hat{\pi}(1))}\rangle\dots\langle \alpha_{k},\beta_{\pi^{-1}(\hat{\pi}(k))}\rangle\\&=\frac{1}{k!}\frac{1}{k!}\sum\limits_{\pi\in S_k}\sum\limits_{\sigma\in S_k}\mathrm{sgn}(\sigma)\langle \alpha_1,\beta_{\sigma(1)}\rangle\dots\langle \alpha_k,\beta_{\sigma(k)}\rangle\\&=\frac{1}{k!}\sum\limits_{\sigma\in S_k}\mathrm{sgn}(\sigma)\langle \alpha_1,\beta_{\sigma(1)}\rangle\dots\langle \alpha_k,\beta_{\sigma(k)}\rangle\\&=\frac{1}{k!}\det\left(\langle\alpha_i,\beta_j\rangle\right)\end{aligned}$
Now let’s say we have an orthonormal basis of $1$-forms $\{\eta^i\}$ — a collection of $1$-forms such that $\langle\eta^i,\eta^j\rangle$ is the constant function with value $1$ if $i=j$ and $0$ otherwise. Taking them in order gives us an $n$-fold wedge $\eta^1\wedge\dots\wedge\eta^n$. We can calculate its inner product with itself as follows:
$\displaystyle\begin{aligned}\langle\eta^1\wedge\dots\wedge\eta^n,\eta^1\wedge\dots\wedge\eta^n\rangle&=\frac{1}{n!}\det\left(\langle\eta^i,\eta^j\rangle\right)\\&=\frac{1}{n!}\det\left(\delta^{ij}\right)=\frac{1}{n!}\end{aligned}$
We’ve seen this before when talking about the volume of a parallelepiped, but it still feels like this should have volume $1$. For this reason, many authors will rescale the inner products on $k$-forms to compensate. That is, they’ll define the inner product on $\Omega^k(U)$ to be the determinant above, rather than $\frac{1}{k!}$ times the determinant like we wrote. We’ll stick with this version, but remember that not everyone does it this way.
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Posted by John Armstrong | Differential Geometry, Geometry
## 2 Comments »
1. [...] say that is an orientable Riemannian manifold. We know that this lets us define a (non-degenerate) inner product on differential forms, and of course we have a wedge product of differential forms. We have almost everything we need to [...]
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2. [...] these inner products are those induced on differential forms from the metric. This doesn’t quite hold, but we can show that it does hold “up to [...]
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## Descriptions in physics
Let suppose we have a frame of reference, in which a particle is decribed by $$\vec{x}(t)$$...I don't give, on purpose, the context, if it is a classical framework, or a relativistic one...(Like somebody new, that does not even know what those word mean)...The question is : does this trajectory, which is, after some discussion with the non-knower, as seen (or oberseved or measured ??..not measured, because this would mean a radar like method...at least I suppose) by an observer at the origin, hence at x=0. So the question is : does this take into account the delay imposed by the speed of light limit, after another discussion with the same person who has to give an answer to the question ?
Do you think it takes the delay into account ?
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Quote by kleinwolf Let suppose we have a frame of reference, in which a particle is decribed by $$\vec{x}(t)$$...I don't give, on purpose, the context, if it is a classical framework, or a relativistic one...(Like somebody new, that does not even know what those word mean)...The question is : does this trajectory, which is, after some discussion with the non-knower, as seen (or oberseved or measured ??..not measured, because this would mean a radar like method...at least I suppose) by an observer at the origin, hence at x=0. So the question is : does this take into account the delay imposed by the speed of light limit, after another discussion with the same person who has to give an answer to the question ? Do you think it takes the delay into account ?
No, normally not. The standard usage of x(t) is to take it to be the *instantaneous* value of x at t. In non-relativistic physics, that's unambiguous, in relativistic physics, the t is the time parameter of the "obvious" foliation of spacetime (usually observer-related when observers are moving). But it does normally NOT take into account the delay for x to be observed. You can do that, of course, but that means t is not parametrising a spacelike foliation anymore, but is parametrising light cones.
As in non-relativistic physics, the lightcone is flat, it coincides with the spacelike foliation, so both notions are identical. But in relativistic physics, clearly the cone is not flat.
So do you think that the following transformation $$x_o(t_o)=x(t_o-x_o(t_o))$$ could eventually be a transformation of a "time foliation" (in which I don't really understand that global time valid on the whole space)...to a sort of "observed trajectory" (the observer reamains at O)...or is there another formula, because i get trapped in kind of circular reasoning...: the third time argument should be t ot$$t_o$$ ?
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## Descriptions in physics
Quote by kleinwolf So do you think that the following transformation $$x_o(t_o)=x(t_o-x_o(t_o))$$ could eventually be a transformation of a "time foliation" (in which I don't really understand that global time valid on the whole space)...to a sort of "observed trajectory"
Well, in string theory, one often works with the transformation
x+ = x + t
x- = x - t
y = y
z = z
I'm forgetting right now the name of this coordinate system... it has a specific name...
Thanks, I suppose it is linked to the old-fashioned : retarded or advanced terms...which correspond two the two x+, x- (with c=1 of course) (depending on your consider observer->observed, or observed->observer transformation i suppose : if you observe x at t, it was at t-x (it is not there at 't'), whereas if it is at x in t, it will be observed at t+x...but do you know for the formula above, because it is not only an event coordinate transformation, it involves the whole trajectory (hence a dependence between x and t, or other way expressed : x+ and x-)...? (it seems to be an implicit notion..)
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http://mathoverflow.net/questions/54252/ | ## Are there smooth bodies of constant width?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The standard Reuleaux triangle is not smooth, but the three points of tangential discontinuity can be smoothed as in the figure below (left), from the Wikipedia article. However, it is unclear (to me) from this diagram whether the curve is $C^2$ or $C^\infty$.
Meissner’s tetrahedron is a 3D body of constant width, but it is not smooth, as is evident in the right figure below.
My question is:
Are there $C^\infty$ constant-width bodies in $\mathbb{R}^d$ (other than the spheres)?
The image of Meissner’s tetrahedron above is taken from the impressive work of Thomas Lachand–Robert and Edouard Oudet, "Bodies of constant width in arbitrary dimension" (Math. Nachr. 280, No. 7, 740-750 (2007); pre-publication PDF here).
I suspect the answer to my question is known, in which case a reference would suffice. Thanks!
Addendum. Thanks to the knowledgeable (and rapid!) answers by Gerry, Anton, and Andrey, my question is completely answered—I am grateful!!
-
I trust you mean, other than the sphere. – Gerry Myerson Feb 3 2011 at 22:40
@Gerry: Yes, that's what I meant, thanks; updated the question. – Joseph O'Rourke Feb 3 2011 at 22:48
You ask whether the curve on the diagram is smooth. It isn't: it's made from six circle arcs. – Zsbán Ambrus Feb 4 2011 at 10:25
Thanks, Zsbán. So I guess just $C^1$? – Joseph O'Rourke Feb 4 2011 at 10:45
## 5 Answers
Fillmore showed that there are sets of constant width in $\mathbb R^d$ with analytic boundaries which have a trivial symmetry group (so these are very different from spheres; see "Symmetries of surfaces of constant width", J. Differential Geom., Vol 3, (1969), pp. 103-110).
Moreover, the set of bodies of constant width with analytic boundaries is dense in the space of all convex bodies of constant width in $\mathbb R^d$ with respect to the Hausdorff metric (see e.g. "Smooth approximation of convex bodies" by Schneider).
-
1
In the second paragraph, a "of constant width" is missing; it is implicit, but it disturbed me for a second. – Benoît Kloeckner Feb 11 2011 at 12:11
@Benoît Kloeckner: Edited, thanks! – Andrey Rekalo Feb 11 2011 at 12:16
2
For an ignoramus like me, these results are quite impressive and highly counter-intuitive. – Olivier Dec 11 2011 at 14:19
I took a look at the Fillmore paper, and just before his Corollary to Theorem 2 -- which reads "Corollary. There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex." -- he writes "If we imitate the construction of a Reuleux triangle . . .. Thus:" This seems to imply that he is assuming that [the intersection of four balls in 3-space, centered at the vertices of a regular tetrahedron and each with radius = the side-length of the tetrahedron] is a body of constant width. But this is known to be false. – Daniel Asimov Jan 11 at 4:50
### 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.
Jay P Fillmore, Symmetries of surfaces of constant width, J Differential Geometry 5 (1969) 103-110, says: the curve $$x_1=h\cos\theta-{dh\over d\theta}\sin\theta,\qquad x_2=h\sin\theta+{dh\over d\theta}\cos\theta$$ where $h=a+b\cos3\theta$, $0\lt8b\lt a$, is analytic and of constant width. If we rotate this curve in Euclidean space ${\bf E}^n$ about an $(n-2)$-dimensional axis perpendicular to the line $\theta=0$, we obtain an analytic surface, not a sphere, of constant width in ${\bf E}^n$.
The paper may be available at http://www.intlpress.com/JDG/archive/1969/3-1&2-103.pdf
-
Take any odd $C^\infty$-function $f$ on the sphere. Consider convex set $$R_\epsilon=\{\,x\in\mathbb R^n\mid\langle x,u\rangle\le 1+\epsilon{\cdot}f(u)\ \ \text{for any}\ \ u\in\mathbb{S}^{n-1}\,\}.$$ Clearly for all sufficiently small $\epsilon>0$, $R_\epsilon$ is a smooth body of constant width.
-
Take any odd smooth function h on the unit (d-1)-sphere and take a constant r>0 large enough to ensure that h+r is the support function of a convex body K
(the condition for h+r to be the support function of a smooth convex body whose boundary has positive Gaussian curvature is that the eigenvalues of Hess(h)+(h+r).Id be positive).
This convex body K is of constant width 2r.
Moreover, any smooth convex body with constant width 2r whose boundary has positive Gaussian curvature can be constructed in this way :
If S is a closed convex hypersurface of constant width 2r, then S is the sum of a sphere of radius r with a "projective hedgehog" H whose support function h is the odd part of the support function of S (and which can be regarded as the locus of the middles of S's diameters)." ;
See for instance:
Y. Martinez-Maure, Arch. Math., Vol. 67, 156-163 (1996), page 157.
-
Michael Kallay characterized the set of all planar sets with a given width functions: See M. Kallay, Reconstruction of a plane convex body from the curvature of its boundary. Israel J. Math. 17 (1974), 149–161. and M. Kallay, The extreme bodies in the set of plane convex bodies with a given width function. Israel J. Math. 22 (1975), no. 3-4, 203–207.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9106763601303101, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2011/06/27/what-does-the-bracket-measure-part-2/?like=1&source=post_flair&_wpnonce=a832ba55c5 | # The Unapologetic Mathematician
## What Does the Bracket Measure? (part 2)
Today we’ll prove the assertions we made last time: if $X$ and $Y$ are vector fields with flows $\Phi$ and $\Psi$, respectively in some neighborhood $U$ of some point $p\in M$, we define the curve
$\displaystyle c(t)=\left[\Phi_{-t}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)$
We assert that for any smooth $f\in\mathcal{O}U$ we have
$\displaystyle\begin{aligned}(f\circ c)'(0)&=0\\\frac{1}{2}(f\circ c)''(0)&=[X,Y]_p(f)\end{aligned}$
To show the first, we define the three “rectangles”
$\displaystyle\begin{aligned}V_1(s,t)&=\left[\Psi_s\circ\Phi_t\right](p)\\V_2(s,t)&=\left[\Phi_{-s}\circ\Psi_t\circ\Phi_t\right](p)\\V_3(s,t)&=\left[\Psi_{-s}\circ\Phi_{-t}\circ\Psi_t\circ\Phi_t\right](p)\end{aligned}$
Notice that $c(t)=V_3(t,t)$, $V_3(0,t)=V_2(t,t)$, and $V_2(0,t)=V_1(t,t)$. The chain rule lets us then calculate:
$\displaystyle\begin{aligned}(f\circ c)'(0)=&\frac{d}{dt}f(c(t))\Big\vert_{t=0}\\=&\frac{d}{dt}f(V_3(t,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_3(s,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{d}{dt}f(V_3(0,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{d}{dt}f(V_2(t,t))\Big\vert_{t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_2(t,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_2(t,t))\Big\vert_{s=t=0}\\=&\frac{\partial}{\partial s}f(V_3(s,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial s}f(V_2(t,t))\Big\vert_{s=t=0}\\&+\frac{\partial}{\partial s}f(V_1(t,t))\Big\vert_{s=t=0}+\frac{\partial}{\partial t}f(V_1(t,t))\Big\vert_{s=t=0}\\=&-Y_pf-X_pf+Y_pf+X_pf=0\end{aligned}$
as asserted. As for the other assertion, we start by observing
$\displaystyle(f\circ c)''(0)=\frac{\partial^2(f\circ V_3)}{\partial s^2}(0,0)+2\frac{\partial^2(f\circ V_3)}{\partial s\partial t}(0,0)+\frac{\partial^2(f\circ V_3)}{\partial t^2}(0,0)$
Using the fact that $\frac{\partial(f\circ V_3)}{\partial s}=-(Yf)\circ V_3$ we can turn the first term on the right into
$\displaystyle\frac{\partial^2(f\circ V_3)}{\partial s^2}(0,0)=\frac{\partial(-(Yf)\circ V_3)}{\partial s}(0,0)=Y_pYf$
Now, similar tedious calculations that make the big one above look like idle doodling give us two more identities:
$\displaystyle\begin{aligned}\frac{\partial^2(f\circ V_3)}{\partial s\partial t}(0,0)&=-Y_pYf\\\frac{\partial^2(f\circ V_3)}{\partial t^2}(0,0)&=Y_pYf+2[X,Y]_pf\end{aligned}$
and we conclude that
$\displaystyle(f\circ c)''(0)=2[X,Y]_pf$
as asserted.
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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
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http://mathhelpforum.com/advanced-algebra/60673-cofactor-expansion.html | # Thread:
1. ## Cofactor Expansion
Let A be an n x n matrix.
a) Show that f(t) = det(tIn - A) is a polynomial in t of degree n.
b) What is the coefficient of t^n in f(t)?
c) What is the constant term in f(t)?
I don't understand how I would show it in a simpler way, than the way
that I am currently doing. I just need feedback about the way I'm
solving the problem!
I know that tIn would be an identity matrix multiplied by t. If I
subtract A from tIn than I would get a matrix like this:
|t - a11 ... -a1n|
|-a21 ...... -a2n|
|-an1 ... t - ann|
For the first part, I understand how the determinant is a polynomial
in the form of t^n.
The first part of the determinant that would be calculated would be
the diagonal, (a11)(a22)(a33)..(ann). That would give me the equation
(t-a11)(t-a22)...(t-a33).
The resulting equation would end in
t^n +- .... +- a11a22a33..ann. My first question is, is there an
easier way to prove this, or is this good enough.
For part b, I'm not sure if I was supposed to come up with a different
coefficient, besides 1. I did some examples with n = 3,4,5 and I
always got that the coefficient was 1.
For part c I got that the constant from part a (+- a11a22..ann) would
be added to the opposite diagonal (+-a1n a2n-1...an1).
I'm not sure if I did that right! Thanks so much for your help!
2. By proving this by induction on $n$ when you expand along on of the cofactors. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9553664922714233, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/147627-proof-convergence-recursive-sequence-print.html | # Proof of Convergence for Recursive Sequence
Printable View
• June 3rd 2010, 07:18 AM
moemoe
Proof of Convergence for Recursive Sequence
I am trying to figure out how to prove that the following sequence converges:
$a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$
with $a_{1}=\sqrt{2}$
Thoughts anyone?
• June 3rd 2010, 07:53 AM
moemoe
Quote:
Originally Posted by General
Since the sequence converges, then $\lim_{n\to\infty} a_n = L$ , where L is a finite number ..
Recall that, for the convergent sequence $a_n$:
$\lim_{n\to\infty} a_n=\lim_{n\to\infty} a_{n+1}$ ..
Since : $a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$
By taking the limit of both sides, we will conclude the following equation :
$L=\sqrt{2+\sqrt{L}}$ , solve it for L ..
Thanks for the quick replies. I believe that in a formal proof
$L=\sqrt{2+\sqrt{L}}$
can be used to solve for the limit L if we know that the series converges. However, I need to first prove that the series converges given
$a_{1}=\sqrt{2}$
• June 3rd 2010, 07:54 AM
parkhid
I think we should say this :
$|a_{n+1} - L|<\epsilon$
and $-\epsilon < a_{n+1} - L < \epsilon$
it means :
$0 < a_{n+1} - L < 2\epsilon$
and $a_{n+1} < 2\epsilon+L$
also if $2\epsilon<<L$ we can unspot $2\epsilon$ .
and $a_{n+1} < L$
this is the proof. but to complete it
(Cool)
we should proof that $2\epsilon<<L$ . or
$\epsilon<<\frac{L}{2}$
We Know L is 1 as here :
$\lim_{n\rightarrow\infty}\sqrt{\frac{2}{a_{n}} +1} = 1$
so the $\epsilon<<0.5$
• June 3rd 2010, 07:57 AM
Plato
Quote:
Originally Posted by moemoe
However, I need to first prove that the series converges given
Show that the sequence is increasing and bounded.
Use induction to do that.
• June 3rd 2010, 08:00 AM
Defunkt
Quote:
Originally Posted by parkhid
I think we should say this :
$|a_{n+1} - L|<\epsilon$
...
You can't assume convergence if you're trying to prove it...
@moemoe: The usual way to prove that this type of recursive sequences converge is to prove by induction that the sequence is monotone and bounded.
In your case, plug in the values for the first few terms to see whether the sequence is increasing or decreasing, and then prove it by induction. After you do that, prove that it is bounded (this can be done by induction as well).
EDIT: Plato ahead of me :<
• June 3rd 2010, 03:53 PM
moemoe
So assuming
$<br /> a_{n} \geq a_{n-1} \geq 0<br />$
we have
$<br /> a_{n+1} = \sqrt{2+\sqrt{a_{n}}} \geq \sqrt{2+\sqrt{a_{n-1}}}= a_{n}<br />$
Does anyone have a slick way of showing the sequence is bounded?
• June 3rd 2010, 03:59 PM
moemoe
Maybe define a new sequence
$<br /> b_{n+1}=\sqrt{2+\sqrt{b_{n}}}<br />$ with $<br /> b_{1}=2<br />$
where $<br /> a_{n} \leq b_{n}<br />$ for each $n$ and $<br /> b_{n}<br />$ is decreasing by the same induction argument above?
Is there another simpler way?
• June 3rd 2010, 04:08 PM
Defunkt
There is a simpler way.
First, validate that $a_1, a_2 < 2$.
Now, assume $a_n < 2$ and use the induction hypothesis to prove that $a_{n+1} < 2$.
This will prove that the sequence is bounded by 2 (it doesn't mean that 2 is the limit, though..)
• June 3rd 2010, 05:47 PM
galactus
Perhaps we can try this.
By the induction step it can be shown that $a_{n}<2$
$a_{n+1}=\sqrt{2+\sqrt{a_{n}}}$
$(a_{n+1})^{2}=2+\sqrt{a_{n}}$
$(a_{n+1})^{2}-1=1+\sqrt{a_{n}}$
$(a_{n+1}+1)(a_{n+1}-1)=1+\sqrt{a_{n}}$
Since for all n, $a_{n}<2$
$a_{n+1}-1<1$
$a_{n+1}+1>a_{n}+1$
$a_{n+1}>a_{n}$
and it is monotone increasing.
• June 4th 2010, 01:59 AM
chisigma
The sequence is defined recursively as...
$\Delta_{n} = a_{n+1} - a_{n} = \sqrt{2 + \sqrt{a_{n}}} - a_{n} = f(a_{n})$ (1)
The function $f(x)= \sqrt{2 + \sqrt{x}} - x$ is represented here...
http://digilander.libero.it/luposabatini/MHF63.bmp
It has a single zero at $x_{0} = 1.83117720721...$ and because $|f(x)|$ is less that the line crossing the x axis in $x=x_{0}$ with unity negative slope, any $a_{0} \ge 0$ will produce a sequence converging at $x_{0}$ without oscillations...
Kind regards
$\chi$ $\sigma$
All times are GMT -8. The time now is 02:12 PM. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 49, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9427016973495483, "perplexity_flag": "middle"} |
http://mathhelpforum.com/geometry/108700-area-quadrilateral-given-diagonals-1-side.html | # Thread:
1. ## Area of Quadrilateral given diagonals and 1 side.
Can some one help me out please...
In a quadrilateral ABCD, the diagonals AC and BD intersect at O. Let OA = 2, OB = 2, OC = 3, OD = 4 and AB = 3. The area of the quadrilateral is ?
Im sorry if this is too simple....
2. Originally Posted by itsmaximhere
Can some one help me out please...
In a quadrilateral ABCD, the diagonals AC and BD intersect at O. Let OA = 2, OB = 2, OC = 3, OD = 4 and AB = 3. The area of the quadrilateral is ?
Im sorry if this is too simple....
The area (K) of a quadrilateral can be found by
$K = \dfrac{1}{2} pq \sin \theta$
the diagonals are p & q
In this case, from the data give:
p = 2+3 = 5 & q = 4+2 = 6
You have given the three sides of a triangle: 2,2,3
Use the cosine law to compute $\angle AOB = \theta$
Probably easier (since $\triangle AOB$ is isosceles)
$\theta = 2 \, sin^{-1} \left ( \dfrac{1.5}{2} \right)$
Compute $\theta$ and $\sin \theta$, then plug them into the equation above. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8881721496582031, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/3904/how-can-i-count-the-number-of-colored-combinations-in-a-set-of-regions/3907 | # How can I count the number of colored combinations in a set of regions?
Let me first start out by saying as you might have guessed, this is a homework problem. Therefore I am not looking for an answer to the question. I am looking for help in how to analyze it (My textbook is horrid).
I have a grid of 9 squares surrounded by a circle. I can use three colors to color each region with no region touching any other region with the same color (corners don't count). **`
I am asked to find the number of possible different colorings.
I drew a set of nodes with lines connecting neighboring regions, but I still have no idea how to figure out the number of combinations.
I also made a guess of three (which I know is wrong) since you can color the regions once with the three colors then map each color to a different one two times after the first coloring.
No blue is touching any other blue regions, no white is touching any other white regions, no red is touching any other red regions.
-
1
Do the corners of the grid touch the circle? – Weltschmerz Sep 3 '10 at 0:06
No they do not. – Joshua Enfield Sep 3 '10 at 0:12
## 3 Answers
I'm not sure about this, maybe I don't understand the problem, but here it goes. If you colour the region outside the grid A, the 8 external cells of the grid must be coloured alternately with B and C, and the one in the center, with A. Thus you have two possible colourings. Same thing if you colour the region outside B, or C. Thus the total number of colourings would be 6.
Update Indeed, there are more. If the external cells are coloured B and C, then the one in the centre may be A, and either B or C, so there are 12 different colourings now.
-
This did help. So I can color it 3 different colors for each "pattern." So for the one I have in the original post, I get 3 combinations, then if I put blue in the middle like you said (colour in circle) then I get another 3 for 6. I know there is still more, but this gives me a way of approach, which is what I was looking for thanks! I just need to look for all the different color patterns then multiply by 3. – Joshua Enfield Sep 3 '10 at 0:25
While it is valuable to attack such problems "by hand," here are some generalities. The general version of this problem is called graph coloring. (The graph being colored here is the graph whose vertices are the regions of the diagram where there is an edge between two regions if they are adjacent.) Counting colorings is known to be #P-hard, so it is not all that easy to do in general. For small graphs there is a reasonable way to compute the number of colorings by any fixed number of colors $n$ by using the deletion-contraction recurrence for the chromatic polynomial.
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The problem of counting the number of valid graph colourings is a harder problem than just determining if a colouring exists (the graph colouring problem you refer to). Counting number of valid colourings is #P-Complete. – Aryabhata Sep 3 '10 at 2:18
Expanding on this answer a bit, your first step should be drawing the graph of your above picture. Each region is a node and two nodes are connected by an edge only if they are adjacent in your drawing. Like Qiaochu hinted at, once you've drawn the graph you can use all the tools already developed to solve the problem. – Hooked Sep 3 '10 at 2:48
@Moron: thanks for the correction. – Qiaochu Yuan Sep 3 '10 at 3:17
I've found that the simplest way to attack problems like this is really by straight enumeration. Start by coloring the outer region, then see what colors that forces or eliminates; in this case, it means that none of the eight outer cells of your grid can share that color. Now, pick another unforced cell (e.g., the top-left corner of the grid) and see how many ways that you can color it; in this case we have two separate colors that we can use, and whichever one we use then forces the colors on all of the outer cells of the grid, leaving us with just one un-forced color (the central square of the grid can be one of two colors), etc. While this is sort-of a graph coloring problem, it's really closer to a constraint satisfaction problem in disguise, and the methods used for solving a lot of the popular pen-and-paper puzzles like Sudoku/Kakuro/Picross/etc. are pretty closely related to this.
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http://mathhelpforum.com/calculus/174528-find-number-values-x-satisfying-given-equation-print.html | # Find number of values of x satisfying the given equation.
Printable View
• March 14th 2011, 12:05 AM
findmehere.genius
Find number of values of x satisfying the given equation.
limit 0 to 2[x+14] $\int [\frac{t}{2}].dt$=limit 0 to {x} $\int [t+14] dt$
where {.} is fractional part of x and [.] is integer number less than or equal to x.
actually speaking, I am not able to get anything from the language of question whether is it asking integral values or real number.
But one thing I hope am doing correct is that R.H.S. comes out to be 14 (Happy).
P.S.: i tried my best to solve this question and also use LaTex to type it here. (Nerd)
• March 14th 2011, 01:19 AM
NOX Andrew
Is the following equation the equation you are trying to solve for in terms of $x$?
$\displaystyle \int_0^{2(x+14)} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^x \lfloor t + 14 \rfloor \, dt$
• March 14th 2011, 01:52 AM
findmehere.genius
1 Attachment(s)
Quote:
Originally Posted by NOX Andrew
Is the following equation the equation you are trying to solve for in terms of $x$?
$\displaystyle \int_0^{2(x+14)} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^x \lfloor t + 14 \rfloor \, dt$
No..the upper limit in LHS is enclosed by [.] (greatest integer function)...and that of RHS by { }(fractional part of x)...so it will look like...i am nt so good at latex so...just trying to modify yours
Quote:
$\displaystyle \int_0^{[2x+14]} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^{/{x/}} \lfloor t + 14 \rfloor \, dt$
and if it still doesnt seems clear...so a jpg i am attaching!
• March 14th 2011, 02:35 AM
NOX Andrew
Don't worry about the LaTeX. It took me a while to familiarize myself with the syntax, and I'm still not that good! Here is the equation for anyone else who may be able to help:
$\displaystyle \int_0^{2 \lfloor x+14 \rfloor} \bigg \lfloor \dfrac{t}{2} \bigg \rfloor \, dt = \int_0^{\{x\}} \lfloor t + 14 \rfloor \, dt$
I have to leave for school now, but I will work on the problem during the day. I'll let you know how I fare.
Edit: $x = -14$ is the only solution I found.
• March 14th 2011, 07:01 AM
findmehere.genius
answer given is 14 i.e. 14 solutions of x!
anyway, would please you elaborate, or rather just hint out how did you proceed with to get to that answer at least!
• March 14th 2011, 02:02 PM
NOX Andrew
I made an educated guess. At $x = -14$, $2\lfloor x - 14 \rfloor = 0$ and the left-hand side of the equation is $0$. Furthermore, at $x = -14$, $\{x\} = 0$ and the right-hand side is also $0$. Therefore, $x = -14$ is a solution to the equation.
I don't know how to solve the equation using analytic methods, but I will explore the properties of the greatest integer and fractional part functions. In the meantime, someone else with more knowledge may come along and help.
• March 14th 2011, 03:03 PM
Plato
Quote:
Originally Posted by findmehere.genius
limit 0 to 2[x+14] $\int [\frac{t}{2}].dt$=limit 0 to {x} $\int [t+14] dt$
where {.} is fractional part of x and [.] is integer number less than or equal to x.
Simplify by using some properties of the floor function.
Note that $0\le \left\{ x \right\} = x - \left\lfloor x \right\rfloor <1,\quad \left\lfloor {x + n} \right\rfloor = \left\lfloor x \right\rfloor + n,\;\& \,\left\{ {x + n} \right\} = \left\{ x \right\}$
Notice that $\displaystyle \int_0^{\left\{ x \right\}} {\left\lfloor {t + 14} \right\rfloor dt}= \int_0^{\left\{ x \right\}} {\left\lfloor t \right\rfloor dt} + \int_0^{\left\{ x \right\}} {14dt} = 0 + 14\left\{ x \right\} = 14\left\{ x \right\}$
• March 15th 2011, 03:42 AM
findmehere.genius
Thanks to you both of you for trying this question, even I finally got somewhere through it and the integration part reduced to(Do confirm it) [x+14]^2 =14{x}
then plotting the graph of both the function would give us 7 solutions?!
I would be pleased with just this much if I get to confirm that my integration is correct, rest answer given in book doesn't matter!
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http://mathoverflow.net/questions/69143?sort=votes | ## Is every algebraic space the quotient of a scheme by a finite group?
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In this MO question it is claimed that a catchphrase for "algebraic spaces" could be that they are "the result of looking at the orbit space of the action of a finite group on a scheme".
Hence my question:
Given an algebraic space $X$, is it true that there exists a scheme $S$ and an action of a finite group $G$ on $S$ such that $X=S/G$ ?
If I remember correctly, every algebraic space is the quotient of an affine scheme by an étale equivalence relation, so I tend to think that there could exist such equivalence relations that are not "implemented" by a group action...
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Do you mean a group scheme which is of finite type over $S$? – Martin Brandenburg Jun 29 2011 at 20:10
I don't know :) Probably, I mean that everything is -say- over $\mathbb{C}$ and $G$, as a $\mathbb{C}$-scheme, is a finite discrete collection of closed points. Anyway, I mean any situation that resembles the setting in the question quoted above. – Qfwfq Jun 29 2011 at 22:50
1
If X is a geometric quotient for G acting on S, then won't it be the coarse moduli space for the stack quotient of G acting on S? Not all DM stacks are global quotients (even if you allow groups which aren't finite - there is an example in the paper by Edidin, Hassett, Kresch, and Vistoli - Brauer Groups and Quotient Stacks) so this would give a negative answer to your question. – mdeland Jun 30 2011 at 18:03
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@mdeland: The coarse moduli space of the example you mention (I assume you mean Example 2.2.1) is actually a scheme. – ulrich Jul 1 2011 at 10:43
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@ulrich - you are right, I hadn't looked at the paper again.... there must be other examples, but none that I can think of off the top of my head. – mdeland Jul 1 2011 at 14:46
show 4 more comments
## 3 Answers
As I just remembered, the answer is yes if $X$ is quasiseparated, noetherian and normal: see Laumon and Moret-Bailly, Champs algébriques, (16.6.2). The quesiseparated assumption is needed (see Scott Carnahan's answer). Without the normality condition, I don't have a counterexample but my guess is that there is one.
EDIT after Chris' and Jason's comments: in fact the proof in the case of normal algebraic spaces can be made substantially simpler than in the book (which proves a more general result about noetherian Deligne-Mumford stacks). It goes like this:
Assume $X$ noetherian, integral, normal and, to simplify, separated (I am not sure how much this helps). Cover $X$ by étale maps $X_i\to X$ with each $X_i$ integral and affine. There is a dense open subspace $U$ of $X$ which is a scheme and such that each induced map $U_i:=X_i\times_X U\to U$ is finite. Let $V\to U$ be an étale Galois cover, with Galois group $G$, dominating all the `$U_i$`'s. Now let $\overline{V}\to X$ (resp. $\overline{X_i}\to X$) be the normalization of $X$ in $V$ (resp. in $X_i$); we have dense open immersions $V\subset \overline{V}$ and $X_i\subset \overline{X_i}$. By functoriality, $G$ acts on $\overline{V}$, with quotient $X$.
I claim that $\overline{V}$ is a scheme. Indeed, for each $i$, $\overline{X_i}$ is also the normalization of $X$ in $U_i$. In particular there is an $X$-morphism $f_i:\overline{V}\to\overline{X_i}$ (deduced from $V\to U_i$) which must be finite surjective (everyone is integral, finite and surjective over $X$). Put $V_i:=f_i^{-1}(X_i)$: this is an open subspace of $\overline{V}$ which is finite over $X_i$, hence an affine scheme. So, the union $W$ of the $V_i$'s is an open subspace of $\overline{V}$ which is a scheme and maps surjectively to $X$ (since $V_i\to X_i$ is surjective), hence $\overline{V}$ is covered by `$\{gW\}_{g\in G}$`.
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@Chris: Are you asking Laurent to repeat the arguments from his book? – Jason Starr Feb 7 2012 at 11:15
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One class of counterexamples arises from quotients by actions of infinite discrete groups acting freely, but it only works under some definitions of algebraic space. For example, if you specify an action of $\mathbb{Z}$ on $\mathbb{G}_m$ generated by multiplication by an infinite order element (say, over a characteristic zero field), the sheaf quotient $\mathbb{G}_m/\mathbb{Z}$ is an algebraic space that is not a quotient of a scheme by a finite group. Since it is non-quasi-separated, it does not satisfy Knutson's definition of algebraic space, but it does satisfy Demazure-Gabriel's. This appeared a couple years ago in Chris Schommer-Pries's question.
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I am not too familiar with algebraic spaces, but maybe this is an example of an algebraic space that cannot be a quotient of a scheme by a finite group. The property of such quotients that I use is that if an irreducible algebraic space $X$ of dimension at least two is the quotient of a scheme $S$ by a finite group $G$, then there is no point on $X$ with the property that every closed curve in $X$ contains that point. The reason is that given any point in $S$, it is certainly possible to find a curve in $S$ missing the $G$-orbit of that point, and thus we obtain a curve in $X$ missing any given point.
On the other hand, there are algebraic spaces with a point contained in every closed curve. A classical example is obtained as follows. Let $X'$ denote the blow up of $\mathbb{P}^2$ at 10 points lying on a smooth plane cubic. Suppose that the points on the cubic are chosen so that the 10 divisor classes determined by them and the class of a line in the Picard group of the cubic are independent. It is a fact (proved by Artin, I believe) that a curve with negative definite self-intersection on a smooth surface can be contracted to an algebraic space. In particular, denoting by $C$ the strict transform of the cubic in $X'$, the curve $C$ can be contracted on $X'$ to an algebraic space $X$. It is also a consequence of the construction that the curve $C$ intersects every curve in $X'$, and therefore every curve in $X$ contains the point to which $C$ is contracted.
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"given any point in $S$, it is certainly possible to find a curve in $S$ missing the $G$-orbit of that point": why is that? – Laurent Moret-Bailly Jan 27 2012 at 10:46
@Laurent: I was thinking $S$ of finite type over a field and a finite set of closed points. If $S$ is a surface, resolve it and reduce to the quasi-projective case (which should be clear). If $S$ is not a surface, restrict to a divisor and proceed by induction. Does this not work? – MP Jan 27 2012 at 15:37
@MP: "If S is a surface, resolve it": but then what? If $p:S'\to S$ is a resolution, and $C$ is a curve in $S'$ avoiding some finite set $F$, $p(C)$ need not avoid $p(F)$. – Laurent Moret-Bailly Jan 27 2012 at 20:14
@Laurent: although I cannot find a counterexample, I cannot also find a proof of the fact that on a surface you can always find a closed curve missing any finite subset. Thus, for the moment, the above is not an argument. – MP Jan 30 2012 at 23:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 82, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9361690878868103, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/239?sort=votes | ## Has anyone tabulated 2-knots? Would anyone like to try?
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I'd love to have a list of 'small' 2-knots, for some sense of small. It's not clear what one should filter by, but there are two obvious candidates
• Write a movie presentation, and count the frames.
• Project the 2-knot to R^3, and count the triple points.
Does anyone know if this has been attempted? Such a list could be quite useful.
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## 2 Answers
There is a table in Yoshikowa's paper based upon the representation as graphs with certain markings of vertices. The markings indicate which directions are the A and B smoothings. The table is quite small.
Kamada's book on Surface braids has a nice table of 2-knots that don't have triple points. There are tons of these. So your candidate for small is sort of strange. Braid index might be better.
Quandle cocycles are the only known ways to get lower bounds for triple points, and there are not nearly enough 3-cocycles that have been calculated.
Last spring Dennis Roseman told me a neat way to generate "random" 2-knots. I don't think he has written anything down about it though.
So if you look at movies, you have to be careful about what you define as an event. In the CRS point of view each of Reid. I,II,III, birth, death, saddle, switchback, and psi (pitchfork) moves is an event, but exchanges of distant critical points are also events. Any such event leaves its trace on the chart.
There are other notions of simplicity. For example, there is a notion of thickness. Project into 3-space and subsequently onto the plane (that gives the chart). Take a spear perpendicular to the plane and see how many times it generically passes through the surface. Find the maximum of these, and then minimize over all projections.
Probably the real reason for making a census would be to discover new invariants. So it is a good problem in that sense.
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A problem with the Yoshikawa technique is that having one of those decorated knotted 4-valent graphs is not enough to specify an embedding $S^2 \to S^4$ -- when you resolve the 4-valent graph either way you get a trivial link. One also has to know the spanning discs for those two trivial links. Change your spanning discs and (in general) you change the knot. – Ryan Budney Aug 8 2010 at 23:55
As far as I know, the assumption is not that one adds a generic spanning disk to the trivial link $L_t$. The assumption is that one adds a trivial disk system to $L_t$, for the definition of a trivial disk system see section 8.5 of "Braids and Knot Theory in Dimension Four" by Kamada. Any two trivial disk systems bounding a given trivial link $L_t$ are ambient isotopic, Proposition 8.6 of Kamada. Thus, the 2-knot type is independent of the trivial disk system which closes the trivial link $L_t$, Kamada Proposition 9.11. – Kelly Davis Aug 3 2011 at 10:30
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Another complexity filter could be the number of 4-dimensional simplices in a triangulation of the complement.
2-knots complements have simple ideal triangulations. The proof is very similar to the proof for 1-knots, viewed through the lens of Morse theory on manifolds with boundary. ie: the Wirtinger presentation is viewed as the fundamental group computation coming from the cell decomposition of a Morse height function (in the stratified sense) on the knot complement.
It'd be nice if there was a SnapPea-like algorithm for triangulating 2-knot complements. As far as I know those kinds of details haven't been worked out by anyone. But it should be reasonably doable.
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http://mathhelpforum.com/calculus/111125-how-use-ftc-finding-limit.html | # Thread:
1. ## How to use FTC in finding this limit?
$\lim_{x\,\to\,\infty}e^{-e^{x}}\int_0^x e^{e^{t}}dt$
Thanks in advance.
2. Originally Posted by xcluded
$\lim_{x\,\to\,\infty}e^{-e^{x}}\int_0^x e^{e^{t}}dt$
Thanks in advance.
Express the limit as $\lim_{x\to\infty} \frac{\int_0^x e^{e^t}dt}{e^{e^x}}$. Both the numerator and denominator approach infinity, so by L'Hopital's rule and the fundamental theorem of calculus, the limit equals $\lim_{x\to\infty} \frac{e^{e^x}}{e^{e^x}e^x} = \lim_{x\to\infty} e^{-x} = 0$.
3. Edited.
Thanks ! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8697606325149536, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/64839/involution-on-hyperelliptic-curves-and-their-jacobians/64845 | ## Involution on Hyperelliptic curves and their Jacobians
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Let $X$ be a hyperelliptic curve and let $i:X\to X$ denote the hyperelliptic involution. Once we fix a point $x_0\in X$ we get the Abel-Jacobi map $AJ:X\to J$ where $J$ denotes the Jacobian variety. Now the Jacobian is also equipped with an involution, namely $x\mapsto x^{-1}$. Is it possible to choose the base point $x_0$ in such a way that the restriction of the involution on the Jacobian is the involution on $X$.
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## 2 Answers
Yes, pick $x_0$ to be a Weierstrass point, i.e. a fixed point of the hyperelliptic involution.
Let $\sigma$ denote the hyperelliptic involution on $X$. Under the Abel-Jacobi map we have $x \mapsto [x-x_0]$ and $\sigma(x) \mapsto [\sigma(x) - x_0]$. Now $[x + \sigma(x) - 2x_0]$ is the divisor of a function, since it is the pullback under the hyperelliptic map of a degree zero divisor on $\mathbf P^1$. Hence $AJ(x)$ and $AJ(\sigma(x))$ are inverses.
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Another way to think of this is the following (at least over $\mathbb{C}$).
Consider the diagram $$\begin{array}{ccccc} X & \xrightarrow{AJ \times AJ \circ \sigma} & J \times J & & \\ \downarrow & & \downarrow & & \\ \mathbb{P}^1 & \to & Sym^2J & \xrightarrow{+} & J \end{array}.$$ Note that the composition along the bottom row must be constant, as there are no non-trivial maps from $\mathbb{P}^1$ to any Abelian variety.
What this tells you is that $f(x) = AJ(x) + AJ\big(\sigma(x)\big)$ is constant. So if you translate it (i.e. choose a different basepoint): $$\tilde{AJ}(x) = AJ(x) - \frac{f(x)}{2}$$ then you find that the corresponding $\tilde{f}(x) = 0$ for all $x$. That is, the two involutions agree with each other as desired.
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There seems to be a small argument missing here, since not every translation on $\mathrm{Pic}^0$ comes from a different choice of basepoint. – Dan Petersen May 13 2011 at 6:55
If $x_0$ is chosen to be a Weierstrass point then the image of $x_0$ (and hence all of $\mathbb{P}^1$) in $J$ is 0, so there is no need to translate. – Jim Bryan May 13 2011 at 14:40 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9144161343574524, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/3889/is-a-normal-ordered-product-of-free-fields-at-a-point-a-wightman-field | # Is a normal-ordered product of free fields at a point a Wightman field?
$:\!\!\hat\phi(x)^2\!\!:$, for example, constructed from the real Klein-Gordon quantum field.
For a Wightman field, the Wightman function $\left<0\right|\hat\phi(x)\hat\phi(y)\left|0\right>$ is a distribution, which is certainly the case for the real Klein-Gordon quantum field --- call it $C(x-y)$ in this case. In contrast, the expected value $\left<0\right|:\!\!\hat\phi(x)^2\!\!:\,:\!\!\hat\phi(y)^2\!\!:\left|0\right>$ is a product of distributions, in fact for the real Klein-Gordon quantum field it's $2C(x-y)^2$, which seems to make this normal-ordered product of quantum fields not a Wightman field. $C(x-y)^2$ is well-enough behaved off the light-cone, but on the light-cone it has a $[\delta((x-y)^2)]^2$ component.
If $:\!\!\hat\phi(x)^2\!\!:$ is not a Wightman field, then is it nonetheless in the Borchers' equivalence class of the free field? If so, why so? A (mathematically clear) citation would be nice!
Finally, if $:\!\!\hat\phi(x)^2\!\!:$ is not in the Borchers' equivalence class of the free field, because it isn't a distribution, is it nonetheless empirically equivalent to the free field at the level of S-matrix observables, as is proved to be the case for Borchers' equivalence classes (Haag, Local Quantum Physics, $\S$ II.5.5), even though it is manifestly not empirically equivalent to the free field at the level of Wightman function observables?
My reading of the Wightman fields literature of the late 1950s and 1960s is far from complete, which may be why I haven't so far found clear answers for these questions.
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## 2 Answers
If I am not wrong the product is well defined, because the 2 point funtion is boundary value of an analytic function in some tube (see Streater/Wightman) with certain bounds. More genral there is 1-1 correspondence between translation invariant tempered distributions which satisfy the spectrum condition and analytic functions in this mentioned tube and an easy way to define the product of the distribution is the product of the analytic functions which lies in the same class and therefore gives a well defined distribution on the boundary.
Really easy is the massless case in let's say 4D.
$W(x-y) \sim \frac1{(x-y+i\epsilon(x_0-y_0))^2}$
$W^2(x-y) \sim \frac1{(x-y+i\epsilon(x_0-y_0))^4}$ where the two-point function is just the boundary value of the analytic function $1/z^2$.
In 1 dimension, that means a chiral field on a light ray, it is even more drastic. There the Wick (=normal ordered) square of the free fermion is a free boson.
edit
The answer to the question "Is a normal-ordered product of free fields at a point a Wightman field?" is: YES!
For convinience I give you a reference, which is "General Principles of Quantum Field Theory" by Bogolubov, Logunov, Oksak, Todorov (1990) p. 344
Ex. 8.16 Prove that the Wick Monomials satisfy the Wightman Axioms
edit Tim: I can not comment on your yet. I did not say you can multiply general distribution which are boundary values by multiplying the function, just for a certain class but maybe I was sloppy because I thought this was standard. The $\delta$ is not in this class, because it does not fullfill a "spectrum contion", i.e. the fourier transform is supported in some convex cone with non-empty dual cone. The fourier transform of $\delta(x)$ is constant so does not fall into this class. But distributions which have this property on their fourier transformation you can multiply: see the standard textbook Reed Simon Volume 2 - page 92 Ex 4.
I hope my references clear the confusion.
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I think you can't take the product of limits like that. $W(x-y)$ is the limit as $\epsilon\rightarrow 0$, which in other terms is a delta function. In your expression for $W(x-y)^2$, there's a question how we should take the distinct limits of the two different $\epsilon$'s. There are mathematical constructions that make products of generalized functions possible, up to a point, for example an approach due to Colombeau, but that puts us outside Wightman's axiomatization. The limitations on what is and is not OK when we take products of generalized functions are non-trivial. – Peter Morgan Jan 27 '11 at 0:36
Normally this is studied in microlocal analysis, but for this kind of functions one don't need this big machinery. I'll search the ref – Marcel Jan 27 '11 at 10:07
Voilà: Reed Simon Volume 2 - page 92 Ex 4. – Marcel Jan 27 '11 at 10:10
And no the limit for $\epsilon \rightarrow 0$ is not a delta function. The Commutator function is like a delta function on the light cone (Huygens principle). – Marcel Jan 27 '11 at 10:17
Thanks, Marcel, I'll look at your references as soon as I can. – Peter Morgan Jan 27 '11 at 15:31
show 3 more comments
the real free bosonic quantum field with mass m > 0 is a Wightman field, because one can define it as an operator valued distribution that satisfies the Klein-Gordon equation in the sense that $\phi(\Box f + m^2 f) = 0$ for all test functions f. The product $\phi(x) \phi(x)$ is then simply not defined for well known reasons (you don't have to look at the two point functions, already the field itself is not well defined). I assume that you know all this and your question was if physicists know a trick around this problem: As far as I know there is no such trick within the Wightman framework.
In order to make sense of this "field" I think we would need to leave the Wightman axioms and allow more severe singularities of two point functions on the diagonal, which are usually described via operator product expansions.
Addendum: When we have a representation of a distribution as a boundary value of an analytic function, it is not true that we can consistently define e.g. the square of this distribution as the boundary value of the square of the analytic function (if this would work, we would have found a solution to a problem where our math friends have already proven that there is no solution, namely to define $\delta^2(x)$). The reason for this is that we are actually talking about an equivalence class of analytic functions instead of one unique analytic function (equivalence means equal up to an analytic function that has an analytic extension to the boundary). Multiplication is not well defined on the equivalence classes. For more details see the buzzword "hyperfunction".
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Although I somewhat agree with your sentiment, it's not clear whether "Leaving the Wightman axioms" is on the table, insofar as we would then have to provide an alternative, motivated axiomatization (at a comparable level of abstraction -- not, in other words, the Haag-Kastler axiomatization). If we "allow more severe singularities", that would put us into a different order of mathematics, but it seems that people who have worked within the Wightman axioms do take normal-ordered products of quantum fields not to be too singular to work with within the Wightman axioms. – Peter Morgan Jan 27 '11 at 0:26
Your Addendum seems to me very sensible. I can't fit my response, with much LaTeX, into 600 characters, however. (1) Ordinarily, we smear $\hat\phi(x)$ to give $\hat\phi_f=\int\hat\phi(x)f(x)d^4x$. Suppose we similarly construct $\hat\Phi_f=\int:\!\!\hat\phi(x)^2\!\!:f(x)d^4x$. If two test functions $f(x),g(x)$ have space-like or time-like separated supports, assuming(!) we take $\int\delta(x^2)^2\cdot ZERO d^4x=0$, we can work out $\int f(x)C(x-y)^2g(y) d^4x$ fairly happily. If $f(x),g(x)$ have light-like separated supports, not so much. We'll see if it will let me do a 2nd comment... – Peter Morgan Jan 27 '11 at 1:00
(2) If $f(x),g(x)$ have light-like separated supports, the Fourier transform of $C(x-y)^2$ is a convolution, which, with the test functions, we can integrate fairly happily, IF there isn't a resonance between the frequencies in the Fourier space representations of the two test functions. In these terms, the resonance is what makes $:\!\!\hat\phi(x)^2\!\!:$ not a distribution, because a distribution is not supposed to care what test functions we give it. In a Physics point of view, a resonance is sort-of-OK, but it's also what one expects to see in an interacting theory, not in a free field. – Peter Morgan Jan 27 '11 at 1:11
Hi there, this software is unfit to host ongoing discussions, so I'll be brief: Making sense of $\phi(x) \phi(x)$ as a Wightman field would, that's my educated guess, result in the very first construction of an interacting Wightman theory. I doubt that it can be done. Products of distributions can be defined, however, under certain circumstances. Note that in the two point function for a Wightman field the product is actually a tensor product (you feed every factor one test function separately). – Tim van Beek Jan 27 '11 at 9:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9303968548774719, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/50092/computing-the-fundamental-group-of-a-cell-complex | # Computing the fundamental group of a cell complex
In Hatcher on page 84 there is the following proposition: For a connected graph $X$ with maximal tree $T$, $\pi_1 (X)$ is a free group with basis the classes $[f_\alpha]$ corresponding to the edges $e_\alpha$ of $X - T$.
I tried to apply this to the torus $T^2$ with the two edges $e_a, e_b$ and the maximal tree $T = \{ x_0\}$ where $x_0$ is the point connecting the two edges.
The problem is that then I get the free group $F(a,b)$ instead of $\mathbb{Z} \oplus \mathbb{Z}$.
Where am I making the mistake? Many thanks for your help!
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3
Hmm. How is the torus a graph (= 1-dim CW-complex)? In fact, you're not that far off, the relation $[a,b] = aba^{-1}b^{-1}$ from the two-cell gives you the desired $\mathbb{Z}^2$. – t.b. Jul 7 '11 at 13:02
well, I thought I could just use the $1$-skeleton to get the fundamental group of the whole complex. Need to work out how to put things together correctly I think... – Matt N. Jul 7 '11 at 13:46
## 1 Answer
In this proposition a graph is a regular, 1-dimensional CW-complex.
In fact you computed the fundamental group of the 1-Skeleton of the Torus, which is a bouquet of $2$ $S^1$ and has therefore $\mathbb Z * \mathbb Z$ as fundamental group.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9171671271324158, "perplexity_flag": "head"} |
http://stats.stackexchange.com/questions/4023/chi-square-test-for-equality-of-distributions-how-many-zeroes-does-it-tolerate | # Chi-square test for equality of distributions: how many zeroes does it tolerate?
I am comparing two groups of mutants each of which can have only one out of 21 different phenotypes. I would like to see whether distribution of these outcomes is similar between two groups. I found an online test that calculates the "Chi-square test for equality of distributions" and gives me some plausible results. However, I have quite a few zeroes in this Table, so can I use chi-square in this case at all?
Here is the table with two groups and counts of particular phenotypes:
2 1
2 3
1 6
1 4
13 77
7 27
0 1
0 4
0 2
2 7
2 3
1 5
1 9
2 6
0 3
3 0
1 3
0 3
1 0
1 2
0 1
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The Table did not come out right. Every odd number is a count from group 1 and every even number is the respective count from group 2 – Membran Oct 27 '10 at 14:42
I've reformatted your question. Is the table now correct? – csgillespie Oct 27 '10 at 15:04
yes, many thanks! – Membran Oct 27 '10 at 15:50
## 2 Answers
Perfectly feasible these days to do Fisher's 'exact' test on such a table. I just got p = 0.087 using Stata (tabi 2 1 \ 2 3 \ .... , exact. Execution took 0.19 seconds).
EDIT after chl's comment below (tried adding as a comment but can't format):
It works in R 2.12.0 for me, though i had to increase the 'workspace' option over its default value of 200000:
> fisher.test(x)
Error in fisher.test(x) : FEXACT error 7.
LDSTP is too small for this problem.
Try increasing the size of the workspace.
> system.time(result<-fisher.test(x, workspace = 400000))
user system elapsed
0.11 0.00 0.11
> result\$p.value
[1] 0.0866764
(The execution time is slightly quicker than in Stata, but that's of dubious relevance given the time taken to work out the meaning of the error message, which uses 'workspace' to mean something different from R's usual meaning despite the fact that fisher.test is part of R's core 'stats' package.)
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1
Interesting, Fisher's test crashed on R. – chl♦ Oct 27 '10 at 18:51
Cannot upvote more, sorry. It seems I hadn't increase the wksp enough :) – chl♦ Oct 27 '10 at 20:59
Isn't it that Fisher's "exact" test actually addresses slightly different question: "...it is used to examine the significance of the association (contingency) between the two kinds of classification" (wiki page). In my case I sought to confirm (or refute) the hypothesis that distributions of phenotypes between 2 groups are similar (equal). When I found that online test (see the first post) named "Chi-square test for equality of distributions" I thought it was precisely for my problem... – Membran Oct 27 '10 at 22:38
Also, if you think that mentioned version of Fisher's test is fine for comparing two distributions, can it also be used for checking uniformity of distribution (i.e. to say that phenotypes within one group were distributed non-uniformly between a finite number of possible phenotypes)? One can do this even in Excel using CHITEST function, but what if I have a distribution similar to the ones above, with lots of phenotypes observed less than 5 times? – Membran Oct 27 '10 at 22:42
@Membran # 1: It is a slightly different question as Fisher's exact test conditions on both sets of marginal totals. This seems something of an academic statistical nicety to me though, and I'm a statistician in academia. (BTW could you clarify to which wiki you refer?) @Membran #2: I would not call the conditional exact test "Fisher's exact test" in the case of a one-way table, but such a test should be possible.and I would have thought more straightforward for one-way tables, but I can't currently find software to assist and I don't have time to perform the calculation without. – onestop Oct 28 '10 at 4:07
The usual guidelines are that the expected counts should be greater than 5, but it can be somewhat relaxed as discussed in the following article:
Campbell , I, Chi-squared and Fisher–Irwin tests of two-by-two tables with small sample recommendations, Statistics in Medicine (2007) 26(19): 3661–3675.
See also Ian Campbell's homepage.
Note that in R, there's always the possibility to compute $p$-value by a Monte Carlo approach (chisq.test(..., sim=TRUE)), instead of relying on the asymptotic distribution.
In you case, it appears that about 80% of the expected counts are below 5, and 40% are below 1. Would it make sense to aggregate some of the observed phenotypes?
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Thank you for suggestions. Logically, it is not quite possible to merge phenotypes as each of them is a unique combination of three recorded parameters. Since each of these parameters can go "up", "down" or stay "unchanged" as a result of a mutation, so there can be 3^3=27 distinct phenotypes. In the example above I removed those phenotypes for which both groups scored "0", so there were only 21 of them. I do see the prevalence of certain phenotypes but i would like to have some statistical proof that distributions of such phenotypes in various groups of mutants is similar (or not). Thank you! – Membran Oct 27 '10 at 16:03
1
@Membran Aggregation doesn't have to be meaningful: you're free to combine bins any way you please. A subtle problem, though, is that post-facto aggregation casts the p-values in doubt; the aggregation ought to be independent of the data. – whuber♦ Oct 27 '10 at 21:16 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9403173923492432, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/6210/what-are-some-different-cryptography-methods | What are some different cryptography methods?
Some of the most effective cryptography methods and algorithms are based of factoring large prime numbers (e.g. RSA). I'm curious whether there are some other cryptography methods. Somethings that is very mathematical or physical based. Of course, I know about quantum cryptography, but I'm interested in other things also. For example, braid groups which have connection to knot theory can have application in cryptography. What I am asking is for a similar thing. I mean, is there any branch or equations in mathematics or physics that can lead to a cryptography methods.
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Elliptic curves and chaos theory are among the answers IMHO. – Mok-Kong Shen Feb 3 at 16:20
1
– AntonioFa Feb 3 at 17:08
1 Answer
All mathematical groups can be used to perform an ElGamal encryption, so that is a first kind of math. That's where elliptic curves are useful: they have a group structure. If you find a group, you can build a cryptosystem out of it. However, as @poncho pointed out, different groups have different properties with regards to security. For instance, elliptic curves are better than modular groups because they do not have a specific structure which can be leveraged to solve the discrete logarithm problem (see http://en.wikipedia.org/wiki/Pohlig%E2%80%93Hellman_algorithm for an example of algorithm working only in modular groups).
You also have cryptosystems based on lattice related problems. In particular, lattices have been used to create the first fully homomorphic cryptosystem (see the references in http://en.wikipedia.org/wiki/Homomorphic_encryption).
A system which turned out to be very bad was built from the knap sack problem (see the Naccache–Stern knapsack cryptosystem on wikipedia).
For symmetric crypto, the theory of boolean functions can explain how the functions inside Feistel Networks or substitution permutation network are chosen. This yields connexions with coding theory and of course with finite fields of size $2^n$.
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3
Yes, you can use a arbitrary group in an ElGamal structure; however, that doesn't mean that such an encryption is even slightly secure. – poncho Feb 6 at 1:58
Very true, I'm updating my answer to mention this. – picarresursix Feb 6 at 17:08
Pedantically, the cryptographic hardness comes from the representation of group elements, not the structure of the group. For example, all cyclic groups have the same structure (that of addition mod $n$). The natural representation of $(\mathbb{Z}_{p-1}, +)$ and the natural representation of $(\mathbb{Z}_p^*, \times)$ have very different cryptographic properties though they are isomorphic groups. – Mikero Mar 9 at 3:33 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9288725256919861, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?p=4030737 | Physics Forums
## vector help needed
Ok, so I'm trying to find the vectors v1 and v2 that point from the positions to a sighting. It says the form must be in <a,±1,b> form, and calculations to 3 significant digits. Position 1: (3500, 450, 800) and the sighting is 30° south of west, and 0.83° above the horizontal. Position 2: (200, 1650, 600) and the sighting is due south, 4.70° above the horizontal.
I honestly have no idea where to start, but I was thinking maybe it would be something along the lines of the magnitude of position 1 which is 3,618.356, and do that multiplied by cos30°, and then sin30°, then I'm not sure what to do with the 0.83* above the horizontal part. But, I'm not sure.
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Hmmmm, it's pretty hard to tell what you're asking for. If you're asking for vectors that point in the direction of the sightings, then one could choose any magnitude for the vector and position 1 and 2 would be irrelevant because vectors are absolute position independent (it wouldn't matter which point you use!) You may at some point want the unit vectors in the sighting directions so I can walk you through that for now. To form a unit vector pointing 30° south of west, and 0.83° above the horizon (and assuming positive $\hat{x}$ is east, positive $\hat{y}$ is north, and positive $\hat{z}$ is up..) you can first project a unit vector onto the xy-plane and then project THAT vector into the 'x' and 'y' directions. So that would be something like <(1 * cos(0.83°)) * -cos(30°), (1 * cos(0.83°)) * -sin(30°), sin(0.83°)> or <-cos(0.83°)cos(30°), -cos(0.83°)sin(30°), sin(0.83°)>. You can verify that the magnitude is 1. I have a feeling that that may be useful for you depending on what your problem is.
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Quote by whiskeygirl Ok, so I'm trying to find the vectors v1 and v2 that point from the positions to a sighting. It says the form must be in <a,±1,b> form, and calculations to 3 significant digits. Position 1: (3500, 450, 800) and the sighting is 30° south of west, and 0.83° above the horizontal. Position 2: (200, 1650, 600) and the sighting is due south, 4.70° above the horizontal. I honestly have no idea where to start, but I was thinking maybe it would be something along the lines of the magnitude of position 1 which is 3,618.356, and do that multiplied by cos30°, and then sin30°, then I'm not sure what to do with the 0.83* above the horizontal part. But, I'm not sure. Help please!
The magnitude of position 1 or 2 has nothing to do with it. Those positions just give locations. The problem appears to require you to write the equations of the lines through the given locations using direction vectors calculated from the given data. Then find where they intersect. They might not exactly intersect due to round off errors in calculations.
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## vector help needed
Quote by whiskeygirl Ok, so I'm trying to find the vectors v1 and v2 that point from the positions to a sighting. It says the form must be in <a,±1,b> form, and calculations to 3 significant digits. Position 1: (3500, 450, 800) and the sighting is 30° south of west, and 0.83° above the horizontal. Position 2: (200, 1650, 600) and the sighting is due south, 4.70° above the horizontal. I honestly have no idea where to start, but I was thinking maybe it would be something along the lines of the magnitude of position 1 which is 3,618.356, and do that multiplied by cos30°, and then sin30°, then I'm not sure what to do with the 0.83* above the horizontal part. But, I'm not sure. Help please!
If a = (3500,450,600) and b = (200,1650,600) and if p = first sight-line vector (defined by the first angles you gave) and if q = second sight-line vector, you get two lines in 3-space. These are L1: x1(s) =a + s*p and L2: x2(t) = b + t*q, where s and t are scalars. As s and t range over the real line, x1 and x2 trace out two lines, which are the estimated sightlines that pass through the object you want to locate. The intersection of these lines gives the position of the object; this will happen if the three simultaneous equations a1+s*p1 = b1 + t*q1, a2+s*p2 = b2+t*q2 and a3+s*p3 = b3+t*q3 have a consistent solution (s,t).
However, because of experimental (measurement) errors and roundoff, the two estimated lines L1 and L2 may not intersect in practice. In that case it makes sense to find the two points (one on L1 and the other on L2) that are as close as possible to one another, that is, to minimize the distance ||x1(s) - x2(t)||. That gives an unconstrained optimization problem in s and t. If the resulting two points are very close together (using some definition of 'very close') then it makes some kind of sense to use a point close to them (for example, one of them, or their average, or something similar) as your estimate of the object's location.
RGV
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Quote by Ray Vickson However, because of experimental (measurement) errors and roundoff, the two estimated lines L1 and L2 may not intersect in practice. In that case it makes sense to find the two points (one on L1 and the other on L2) that are as close as possible to one another, that is, to minimize the distance ||x1(s) - x2(t)||. That gives an unconstrained optimization problem in s and t. If the resulting two points are very close together (using some definition of 'very close') then it makes some kind of sense to use a point close to them (for example, one of them, or their average, or something similar) as your estimate of the object's location. RGV
I ran it through Maple and the two lines do in fact not intersect. So the system is inconsistent. I simply solved the x and y equations and used those values of s and t and not worry about the fact that the whole system is inconsistent. The z values came out very "close" as you might expect.
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Quote by LCKurtz I ran it through Maple and the two lines do in fact not intersect. So the system is inconsistent. I simply solved the x and y equations and use those values of s and t and not worry about the fact that the whole system is inconsistent. The z values came out very "close" as you might expect.
I used Maple to solve the minimum distance problem and as you found, the lines do come close.
RGV
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http://stats.stackexchange.com/questions/33666/chi2-distance-and-multivariate-gaussian-distribution | # $\chi^2$ distance and multivariate gaussian distribution
I want to know how to write the $\chi^{2}$ distance between two multivariate Gaussian distributions $f$ and $g$ in terms of their parameters only. The parameters of $f$ is the vector $\mu_{1}$ and a covariance matrix $\Sigma_{1}$. The parameters of $g$ is the vector $\mu_{2}$ and a covariance matrix $\Sigma_{2}$.
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I would think that distance would be a measure of the separation of the mean vectors. But that would require a common scale. If the covariance matrices were equal I would think the Mahalanobis distance might be what the OP is referring to. – Michael Chernick Aug 4 '12 at 14:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9447941780090332, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/38172/linear-rising-potential-from-a-gribov-propagator | # Linear rising potential from a Gribov propagator
It is common wisdom that a gluon propagator (Gribov-)like $$G(p^2)=\frac{a+bp^2}{cp^4+dp^2+e}$$ should give rise to a linear rising potential. So far, I have not seen a proof of this and I would like to get a mathematical derivation, or a reference, displaying how a potential emerges from it.
Thanks.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9188928604125977, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/67977-continuous-function-metric-spaces.html | # Thread:
1. ## Continuous function on metric spaces.
Let X, Y be metric spaces and f: X->Y be a continous function on the indicated metric spaces. Prove the followings.
(a) For each subset A of X, $f(\overline {A}) \subset \overline{f(A)}.$
(b) For each subset B of Y, $f^{-1}(int B) \subset int f^{-1}(B).$
----------------------------------------------------------------------
In my text, "int A" denotes the interior of A, which is the set of all interior points of A. A point x in A is an interior point of A, or A is neighborhood of x, provided that there is an open set O which contains x and is contained in A.
----------------------------------------------------------------------
2. Originally Posted by aliceinwonderland
Let X, Y be metric spaces and f: X->Y be a continous function on the indicated metric spaces. Prove the followings.
(a) For each subset A of X, $f(\overline {A}) \subseteq \overline{f(A)}.$
We need a fact: $a\in \bar A$ if and only if $B_X(a,\epsilon) \cap A \not = \emptyset$ for any $\epsilon > 0$.
Here, $B_X(a,\epsilon) = \{x\in X | d_X(x,a) < \epsilon \}$ i.e. the ball in $X$ at $a$ with radius $\epsilon$.
Let $y\in f(\bar A)$ this means $y = f(a)$ for some $a\in \bar A$. Let $\epsilon > 0$.
There is $\delta > 0$ such that if $x\in B_X(a,\delta) \implies d_Y(f(x),f(a)) < \epsilon$.
However, $B_X(a,\delta) \cap A \not = \emptyset$ and so pick some $x\in B_X(a,\delta) \text{ and }x\in A$.
Therefore, $f(x) \in f(A)$ and $f(x) \in B_Y(f(a),\epsilon)$.
Thus, $f(A) \cap B_Y(y,\epsilon) \not = \emptyset$ for any $\epsilon>0$.
This means by above fact that $y\in \overline{f(A)}$.
Thus, $f(\bar A)\subseteq \overline{f(A)}$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8854220509529114, "perplexity_flag": "head"} |
http://scicomp.stackexchange.com/questions/5325/clustering-algorithm-for-a-congruence-relation | # Clustering Algorithm for a congruence relation?
Say we are given a congruence relation$~\sim$ in a dataset with $n$ elements. I am looking for an algorithm for optimally sorting the $n$ elements into $m$ clusters according to given congruence relations. For instance if the data contains ${a,b,c,d,e,f,g,h}$, and: $$a\sim b,\ d\sim b,\ e \sim h,\ f \sim c$$ The data should be sorted into the following clusters: $$\{a,b,d\},\ \{c,f\},\ \{e,h\},\ \{g\}$$ As said I'm looking for an efficient algorithm to solve this, I am led to believe this can be done in $O(n)$, but I can't seem to work out the details.
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## 2 Answers
I've always heard this referred to as a "Union Find". It's described here, as well as the optimizations you can do to beat the naive implementation: http://www.algorithmist.com/index.php/Union_Find
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2
– nbubis Feb 19 at 21:26
Write your relation as a sparse graph and use a "connected components" function, like this.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9362465739250183, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/32318/is-there-a-simple-explanation-for-schwingers-relation-g-2alpha-pi-for-the-g-fa | # Is there a simple explanation for Schwinger's relation g=2+alpha/pi for the g-factor of the electron?
Schwinger has on his grave (it seems) the relation between the g-factor of the electron and the fine structure constant:
$g = 2 \ + \ \alpha / \pi \ + \ ...$
Did Schwinger or somebody else ever give a simple explanation for the second term of the right hand side? The 2 appears from Dirac's equation. The second term is due to the emission and absorption of a photon. Is there a simple way to see that this process leads to the expression alpha/pi?
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This is a loop integral which is done elegantly in Weinberg, Schwinger's way. This is as simple as it gets. – Ron Maimon Jul 18 '12 at 18:08
Also see chapter 6 of Peskin. – DJBunk Jul 18 '12 at 18:34 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9312825202941895, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/77172/calculating-the-fourier-transform-of-frac-sinhkx-sinhx | # Calculating the Fourier transform of $\frac{\sinh(kx)}{\sinh(x)}$
I'm trying to compute $$\int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx$$ i.e. the Fourier transform of $x\mapsto \frac{\sinh(kx)}{\sinh(x)}$, where $0<k<1$ is fixed.
But I'm having trouble with it.
Motivation: I'm trying to derive an expression for the solution of the Dirichlet problem $\Delta u = 0$ on the strip $[0,1]\times \mathbb R$ with values on the boundary $f_0, f_1$ (assuming necessary niceness conditions for all functions involved).
For this I took the Fourier transform of the solution $u$ and got a formula for $\hat u$ in terms of $\hat{ f_0}, \hat{ f_1}$ and now I want to transform it back. Doing this led (more or less) to the integral expression: $u(x,y) = \int \frac{\sinh(kx)}{\sinh(k)}e^{-iky} \hat f(k) \ dk$.
Now I'm trying to apply the product formula $\int f \hat g = \int \hat f g$ to get everything in terms of $f$. This is why I'm interested in computing the above integral.
My attempt: (which led nowhere, so you may actually ignore everything below)
I think it should be possible using residues. For this I thought of the path having the following components:
$$[-R,R], \ [R,R+i\pi], \ [R+i\pi, \delta + i\pi],$$ $$\ \text{semicircle from $\delta + i\pi$ to $-\delta + i\pi$ below $i\pi$},$$ $$[-\delta + i\pi, -R + i\pi], \ [-R+i\pi, -R]$$
with the intention of letting $\delta \to 0$ eventually.
The integrals over the vertical components will vanish for $R\to\infty$, so the integral over the path then becomes
\begin{align} 0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx + \left(\int_{\infty}^{\delta} + \int_{-\delta}^{-\infty}\right) \frac{\sinh(k(x+i\pi))}{\sinh(x+i\pi)}e^{-i\omega (x+i\pi)} \ dx \\ & \qquad + \int_{\text{semicircle}} \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \end{align}
Using $\sinh(a+ib) = \sinh(a)\cos(b) + i \cosh(a)\sin(b)$ for real $a,b$, we get
\begin{align} 0 &= \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\ &\qquad + \left(\int_{\delta}^{\infty} + \int_{-\infty}^{-\delta}\right) \frac{\sinh(kx)\cos(k\pi) + i \cosh(kx)\sin(k\pi)}{\sinh(x)}e^{-i\omega x}e^{\omega \pi} \ dx \\ & \qquad + \int_{\text{semicircle}} \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \end{align}
The integral over the semicircle should go to $$(-\pi i) \ \mathrm{Res}_{x = \pi i}\left(\frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x}\right) = -\pi \sin(k\pi)e^{\omega \pi}$$ as $\delta \to 0$. Therefore
\begin{align} \pi \sin(k\pi) e^{\omega \pi} &= \int_{-\infty}^\infty \frac{\sinh(kx)(1+\cos(k\pi)e^{\omega \pi}) + i \cosh(kx) \sin(x\pi)e^{\omega \pi}} {\sinh(x)} e^{-i\omega x} \ dx \\ &= (1+\cos(k\pi)e^{\omega \pi}) \int_{-\infty}^\infty \frac{\sinh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \\ & \qquad + i \sin(k\pi)e^{\omega \pi}\int_{-\infty}^\infty \frac{\cosh(kx)}{\sinh(x)}e^{-i\omega x} \ dx \end{align}
I don't see whether this has brought me any closer to my goal?
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I thought $(e^{kx} -e^{-kx})/(e^x - e^{-x})$ might help but nothing yet. – user13838 Oct 30 '11 at 13:55
What happens if you now start over and try to compute the Fourier transform of $\cosh(kx)/\sinh(x)$ using the same contour? Maybe you will get a second relation between the two transforms, which together with the relation you already have will let you solve for both of them. (I haven't tried it, though.) – Hans Lundmark Oct 30 '11 at 15:10
– user13838 Oct 30 '11 at 15:25
Out of curiosity: is this homework or something you bumped into in an application? – J. M. Oct 30 '11 at 15:58
1
@J.M. I now have added some motivation above. – Sam Oct 30 '11 at 16:26
show 3 more comments
## 1 Answer
The result is doable by method of residues. We complete the integration path by the arc crossing from $+\infty$ to $-\infty$ over the upper-half complex plane. Then $$\begin{eqnarray} \mathcal{F}(\omega, \kappa) &=& \int_{-\infty}^\infty \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \mathrm{d} x = 2 \pi i \sum_{n=1}^\infty \operatorname{Res}_{x = i \pi n} \frac{\sinh(\kappa x)}{\sinh(x)} \mathrm{e}^{i \omega x} \\ &=& \sum_{n=1}^\infty 2 \pi (-1)^{n-1} \mathrm{e}^{-\omega \pi n} \sin(\pi \kappa n) = \frac{2 \pi e^{\pi \omega } \sin (\pi \kappa )}{2 e^{\pi \omega } \cos (\pi \kappa )+e^{2 \pi \omega }+1} \\ &=& \frac{\pi \sin (\pi \kappa )}{\cos (\pi \kappa )+\cosh\left( \pi \omega \right)} \end{eqnarray}$$
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Very nice, sir. =) This contour crossed my mind at first, but I didn't investigate it further, since I just thought "infinite sum" and went on to look for a different one. Thanks a bunch! – Sam Oct 30 '11 at 17:17
I just managed to derive the expression I wanted. =) Thanks to your help. Great! – Sam Oct 30 '11 at 17:51
1
I wonder whether somebody could also post the estimate necessary to show that the integral along the half-circle vanishes? thx, I've been trying for some hours, but I don't get a very clean estimate. – Karl Jan 16 at 23:09 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8196489214897156, "perplexity_flag": "middle"} |
http://mathhelpforum.com/number-theory/144362-isosceles-triangle-integer-sides.html | # Thread:
1. ## Isosceles Triangle With Integer Sides
Hello. I thought up the following problem this morning:
Is it possible to construct an isosceles triangle with integer side length and area?
Mathematica was unable to find any solutions, but it did not say that there were none. I tried to work with the associated Diophantine equation, but I couldn't see immediately how to show that there weren't any solutions.
I'd appreciate any thoughts on the matter.
2. Originally Posted by roninpro
Hello. I thought up the following problem this morning:
Is it possible to construct an isosceles triangle with integer side length and area?
Mathematica was unable to find any solutions, but it did not say that there were none. I tried to work with the associated Diophantine equation, but I couldn't see immediately how to show that there weren't any solutions.
I'd appreciate any thoughts on the matter.
Maybe I don't understand the problem correctly, but it seems easy to come up with examples using Pythagorean triples.
Take a 3-4-5 right triangle. Put two of them together to make an isosceles triangle and you're done.
3. Originally Posted by undefined
Maybe I don't understand the problem correctly, but it seems easy to come up with examples using Pythagorean triples.
Take a 3-4-5 right triangle. Put two of them together to make an isosceles triangle and you're done.
Problem is: he required the area of the triangle to be an integer (an integral multiple of the unit area) as well.
4. Originally Posted by Failure
Problem is: he required the area of the triangle to be an integer (an integral multiple of the unit area) as well.
Say we place the triangles together so that the sides of length four are touching.
Then the area is $\left(\frac{1}{2}\right)(6)(4) = 12 \in \mathbb{Z}$.
(Fixed typo.)
5. Originally Posted by undefined
Say we place the triangles together so that the sides of length four are touching.
Then the area is $\left(\frac{1}{2}\right)(6)(4) = 12 \in \mathbb{Z}$.
(Fixed typo.)
Right, how foolish of me: I just didn't really do the math but thought that you had not even considered the question of the area in your reply.
6. I guess I should have tried to construct an example explicitly.
Now I'm wondering why Mathematica failed to find a solution. Maybe I typed in the formula incorrectly.
Thanks.
7. Originally Posted by roninpro
I guess I should have tried to construct an example explicitly.
Now I'm wondering why Mathematica failed to find a solution. Maybe I typed in the formula incorrectly.
Thanks.
Regarding Mathematica.
Heron's formula:
$A=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}{16}}$
Let $a=b$. Then
$A=\sqrt{\frac{(2a+c)(2a-c)(c)(c)}{16}}$
$=\sqrt{\frac{((2a)^2-c^2)(c^2)}{16}}$
$16A^2=(4a^2-c^2)(c^2)$
$16A^2-(4a^2-c^2)(c^2)=0$
Screenshot from Mathematica:
Edit: Of course it's also possible to avoid Heron's formula by dividing the isosceles triangle into two right triangles to begin with.
Label the two congruent sides $a$ and the other side $b$. Treat $b$ as the base and draw an altitude from the base to the opposite vertex.
$\left(\frac{b}{2}\right)^2+h^2=a^2$
and
$A=\left(\frac{1}{2}\right)bh$
$h = \frac{2A}{b}$
Substitute
$\left(\frac{b}{2}\right)^2+\left(\frac{2A}{b}\righ t)^2=a^2$
etc.
8. I actually used that formula, originally. I now see that I screwed up the syntax.
Thanks for the computation. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9435318112373352, "perplexity_flag": "head"} |
http://www.conservapedia.com/Compton_Scattering | # Compton Scattering
### From Conservapedia
Compton Scattering is the collision process between a X-ray or a gamma ray and a bound atomic electron where only part of the energy of the electromagnetic ray is transferred to the electron.
The effect was at first observed by Arthur Holly Compton in 1923 at Washington University in St. Louis and explained in his article "A Quantum Theory of the Scattering of X-ray by Light Elements"[1]. Compton was rewarded the 1927 Nobel Prize in Physics for this discovery.
Arthur H. Compton treated the x-ray photons as particles and applied conservation of energy and conservation of momentum to the collision of a photon with a stationary electron.[2]. He used the Planck relationship and the relativistic energy expression to derive the standard Compton formula:
$\Delta \lambda = \frac{h}{m_e c} (1-\cos \theta)$
Here, Δλ denotes the difference between the wavelengths of the incoming and the scattered ray, while θ is the angle of scattering.
## The Compton Experiment
A collimated beam of high-energy photons, e.g. emitted from a radioactive 137Cs source or an X-ray emitter hits a target, e.g. a rod of graphite. A scintillation counter is used to measure the number and the energy of the deflected photons at various angles.
## Classical Expectation
The number of photons should vary with the angle of the deflection, while the energy (i.e., the wavelength) of the photons should remain unchanged.
## Observation
While indeed many of the scattered photons have an unchanged wavelength, there are some which lost energy, i.e., their wavelength lengthened. This lengthening depends on the angle of the deflection only.
## Explanation
The lengthening of the wave-length happens when a photon interacts with a free electron in the material. The effect can be calculated along the following lines.
1) Conservation of energy: We assume that before the collision, the electron is nearly at rest, so its kinetic energy is zero: Ee = 0. The photon has an energy of Eγ = hf, where h is Planck's constant, and f is its initial frequency. After the interaction, the photon's frequency changed to f', so its energy is now E'γ = hf'. The kinetic energy of the electron after the collision is $E'_e = \sqrt{p^2_e c^2 + m^2_e c^4} - m^2_e c^2$. (Here, me is the mass of the electron.) We get the equation;
$h f = h f' +\sqrt{p^2_e c^2 + m^2_e c^4} - m_e c^2$
(here, $\vec{p_e}$ is the momentum of the electron after the event)
2) Conversation of momentum: As we assume that the electron is nearly at rest at first, its initial momentum is zero. The momentum of the photon is at first $\vec{p_p}$ and then $\vec{p'_p}$. The momentum of a photon can be calculated via p = E / c = hf / c.
Looking at the picture, we see that
$p^2_e = (\vec{p_p}-\vec{p'_p})\cdot (\vec{p_p}-\vec{p'_p}) =$ p2 + p'2 − 2pp'cosθ, thus
I: (pec)2 = (hf)2 + (hf')2 − 2h2ff'cosθ
Squaring $h (f-f') + m_e c^2 = \sqrt{p^2_e c^2 + m^2_e c^4}$ and rearranging leads to:
II: (pec)2 = (hf)2 + (hf')2 − 2h2ff' + 2mec2(hf − hf')
We equate our I and II and get:
− 2h2ff'cosθ = − 2h2ff' + 2mec2(hf − hf')
This we rearrange to:
$\frac{1}{h f'} - \frac{1}{h f'} = \frac{1}{m_e c^2} (1 - \cos \theta)$
As for a photon fλ = c, we can bring this into Compton's form:
$\lambda' - \lambda = \Delta \lambda = \frac{h}{m_e c} (1 - \cos \theta)$
This describes exactly the effect observed by Compton! Compton himself used a similar derivation, also including the relativistic energy expression.
## Probability of Compton Scattering
The probability for Compton scattering is approximately proportional to the atomic number Z, and for energies greater than 500 keV approximately proportional to $\frac{1}{E^\gamma}$,[3] the energy of the gamma ray photon.
## Reference
1. ↑ Arthur H. Compton: A Quantum Theory of the Scattering of X-ray by Light Elements, The Physical Review, Vol. 21, No. 5, May, 1923
2. ↑ Compton Scattering Equation, Hyperphysics, C. R. Nave, Georgia State University
3. ↑ Glossary of Nuclear Science Terms. Retrieved on January 10, 2013. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 11, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8892294764518738, "perplexity_flag": "middle"} |
http://www.mashpedia.com/Prime_factor | # Prime factor
Language
For the Star Trek: Voyager episode, see Prime Factors (Star Trek: Voyager).
In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The process of finding these numbers is called integer factorization, or prime factorization. A prime factor can be visualized by understanding Euclid's geometric position. He saw a whole number as a line segment, which has a smallest line segment greater than 1 that can divide equally into it.
For a prime factor p of n, the multiplicity of p is the largest exponent a for which pa divides n. The prime factorization of a positive integer is a list of the integer's prime factors, together with their multiplicity. The fundamental theorem of arithmetic says that every positive integer has a unique prime factorization.
To shorten prime factorization, numbers are often expressed in powers, so
$288 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^5 \times 3^2.$
For a positive integer n, the number of prime factors of n and the sum of the prime factors of n (not counting multiplicity) are examples of arithmetic functions of n that are additive but not completely additive.
Determining the prime factors of a number is an example of a problem frequently used to ensure cryptographic security in encryption systems; this problem is believed to require super-polynomial time in the number of digits — it is relatively easy to construct a problem that would take longer than the known age of the Universe to solve on current computers using current algorithms.
Two positive integers are coprime if and only if they have no prime factors in common. The integer 1 is coprime to every positive integer, including itself. This is because it has no prime factors; it is the empty product. It also follows from defining a and b as coprime if gcd(a,b)=1, so that gcd(1,b)=1 for any b>=1. Euclid's algorithm can be used to determine whether two integers are coprime without knowing their prime factors; the algorithm runs in a time that is polynomial in the number of digits involved.
The function ${\omega(n)}$ represents the number of distinct prime factors of n, while ${\Omega(n)}$ represents the total number of prime factors of n. If $n = \prod_{i=1}^{\omega(n)} p_i^{\alpha_i}$, then $\Omega(n) = \sum_{i=1}^{\omega(n)} \alpha_i$.
For example, $24=2^3 \times 3^1$, so: $\omega(24)=2$ and $\Omega(24)=3+1=4$.
ω(n) for n = 1, 2, 3, ... is 0, 1, 1, 1, 1, 2, 1, 1, 1, ... (sequence in OEIS)
Ω(n) for n = 1, 2, 3, ... is 0, 1, 1, 2, 1, 2, 1, 3, 2, ... (sequence in OEIS)
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http://mathoverflow.net/questions/26350/linear-algebra-inequality/26353 | ## Linear algebra inequality
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I'm wondering (hoping) if an inequality is true. Please can anyone help me?
Let $V$ be a complex vector space $dim_{\mathbb{C}}(V)=n$ with a hermitian scalar product $h$. Let $v,a, b \in V$.
Is it true that
$(h(v,v)h(a,a)-{|h(v,a)|}^{2})(h(v,v)h(b,b)-{|h(v,b)|}^{2})\geq |(h(v,v)h(a,b)-h(a,v)\overline{h(b,v)}|^{2}$?
With the overline meaning complex conjugate.
Thank you in advance.
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## 2 Answers
Yes. The case where $v=0$ is trivial so suppose $v\ne0$. Consider the projection map from $V$ to the hyperplane orthogonal to $v$ and let $a'$ and $b'$ be the images of $a$ and $b$ under this projection. Then your inequality reduces to $$h(a',a')h(b',b')\ge\vert h(a',b') \vert^2,$$the Cauchy-Schwarz inequality.
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Thank you very much Robin! – Italo May 29 2010 at 14:21
### 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.
Cauchy-Schwarz in the orthogonal complement to v?
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Ah, Robinned again. – Charles Matthews May 29 2010 at 13:19
Charles, you had a plenty of time before Robin came: the question was asked long time ago. :) +1 to both (not to offend somebody). – Wadim Zudilin May 29 2010 at 13:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9180211424827576, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/1453/great-unsolved-physics-problems?answertab=oldest | # Great unsolved physics problems [closed]
We all know that some theoretical ideas lack experimental evidence while in other cases there's a lack of a suitable theory for known phenomena and established facts and concepts.
But what problem in physics, according to you, deserves a mention? And why you think solving that particular problem is of utmost importance and/or how far-reaching its effects/repercussions would be.
One unsolved problem per post.
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Some qualified person (i.e. not me) should write an answer about the AdS/CFT conjecture! – Greg P Dec 28 '10 at 19:14
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@arivero: perhaps. I just wanted to point out that in my opinion greatest unsolved problems are hiding in the nature not in the mathematical foundations of our theoretical models. But this was never intended as a criticism, just an amusing fact to note :) – Marek Jan 20 '11 at 14:34
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## closed as not constructive by David Zaslavsky♦May 15 '12 at 21:22
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## 33 Answers
One fundamental problem in mathematical physics that has to my knowledge not been resolved, but I might be mistaken since I haven't been working in this particular field for already 5 years and 5 years is a long time, is the following:
In macroscopic bodies, there are several laws existing describing transport phenomena like heat conduction. They are described by precise mathematical laws which are well established experimentally. A well-known one is Fourier's law:
$$c\frac{\partial}{\partial t}T(\vec{r},t)=\nabla \cdot \left(\kappa \nabla T(\vec{r},t)\right) \; ,$$
where $T(\vec{r},t)$ is the local temperature, $c=c(T)$ the specific heat and $\kappa=\kappa(T)$ the heat conductivity. There is however no rigorous mathematical derivation of this equation for any classical or quantum model with a Hamiltonian microscopic evolution. This problem is related to the question wether deterministic microscopic systems can fully explain the behaviour of macroscopic matter. (Remember this question, well basically, we have no rigorous proof that it's all just probability applied on huge amounts of microscopic particles. At least not for out of equilibrium processes, which transport phenomena are.)
Now, I don't expect anyone but mathematical physicists invested in this particular field to loose any sleep over it. We have a good heuristic understanding of how things work and there's nobody seriously doubting that the explanation of macroscopic phenomena lies in understanding the underlying microscopic phenomena (except maybe some crackpots and fringe physicists). Still, a mathematical derivation could bring us a deeper understanding of why it works.
The Green-Kubo relation gives $\beta V \int_0^\infty d\tau \left\langle J(0)J(\tau) \right\rangle$ for the transport coefficients. The time autocorrelation function can be computed using the closed time path formalism.
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Interesting. I had no idea that Fourier's law had no derivation from microscopic principles! – Noldorin Nov 30 '10 at 11:25
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Q: Are there magnetic monopoles?
Why: If there are, Maxwell's equations become more symmetric and, more important, it immediately explains why charge is quantized. If there aren't, then the interesting question might be: Why?
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– Robin Maben Nov 30 '10 at 4:45
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@user16307 Well, "explain" is relative. But we already know that angular momentum is quantized, and if magnetic monopoles exist, you can find a system whose EM field has an angular momentum that is proportional to the electric charge, so that then has to be quantized as well. – Lagerbaer Mar 4 at 16:00
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I think one important problem (at least from the point of mathematical physics) is to establish the Standard Model as a mathematically complete and consistent quantum field theory. This is related to: http://www.claymath.org/millennium/Yang-Mills_Theory/.
The point is that the Standard Model of particle physics deals with fundamental structures of matter and is one of the most successful models in physics in terms of accuracy of predictions. However it seems to consist of a bunch of unjustified rules, people argue using undefined objects etc. Up to now there is no conceptual clear, mathematical and logical consistent description of the Standard Model available. So I think it would give us a quite deeper understanding how nature works if we had such a consistent formualtion of the Standard Model.
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I don't think this is an important problem because I don't think it is true that the Standard Model is mathematically complete and consistent. The Clay prize refers only to QCD, which is asymptotically free. Other parts of the Standard Model (the U(1) gauge part, the Higgs sector) are not asymptotically free and thus need a UV completion. This is one of the reasons why people think the Standard Model is only an effective field theory, valid at low energies rather than a theory that makes mathematical sense at all distance scales. – pho Jan 20 '11 at 0:49
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I think any theory that needs renormalizations and IR problem resolving is physically and mathematically inconsistent by definition. A consistent theory calculates energies, scattering cross sections, etc., from the fundamental constants, not "renormalizes" them. – Vladimir Kalitvianski Jan 20 '11 at 11:29
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Vladimir, you'll have to admit this is a minority opinion. – pho Jan 20 '11 at 13:05
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A general existence and uniqueness theorem for the Navier-Stokes equations. The lack of one is probably the largest open problem in classical mechanics, considering that the study of fluid dynamics is based on the Navier-Stokes equations and generalizations thereof. If there were an equation of state/initial data set where the solution was either non-existent or non-unique, even if such a thing were not experimentally realizable (for instance, the "good" initial data formed a dense set in the space of possible initial data), it would still fundamentally change how we look at the evolution of classica systems.
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– user346 Jan 20 '11 at 6:00
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I would have thought the question was only of mathematical interest. Real fluids are not continuous, but made of particles. This makes the Navier-Stokes equations an approximation. – Hugh Allen Feb 7 '11 at 6:15
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Is supersymmetry (SUSY) a symmetry of our world?
This has to do with many outstanding problems in modern physics, like hierarchy problem, nature of dark matter, cosmological constant problem and others. If true, it explains some of them in one stroke and for others it provides clues as to what to do next. But note that it would also bring many other questions (e.g. precisely what SUSY model is realized in nature and how exactly is spontaneous breaking of SUSY realized). It also points a way to quantum gravity (via supergravity).
In short, this symmetry is very appealing theoretically and many people are quite sure it is indeed correct despite not yet being observed (so, in this regard it is quite similar to Higgs boson). Now, not only is the resolution of this problem (in either way) very interesting but LHC should actually be able to give the answer in next few years and this makes it all the more exciting.
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I would like to know how to compute the charge of an electron from first principles. This would likely have major implications that would depend on the form of the solution.
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Do mean the value of e? At what energy scale? – pho Jan 20 '11 at 2:37
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I mean I want to derive the fine structure constant from first principles. – Matt Mar 2 '11 at 7:07
Since no one mentioned the confinement:
"no analytic proof exists that quantum chromodynamics should be confining"
Briefly: it turns out that one cannot isolate color-charged particles. Intuitively one can understand it -- gluons, being the carriers of strong interaction, carry the color-charge themselves. So they are "screening" the color-charge of a carrier. And when one tries to, say, separate two quarks, the "gluon tube" appears between them. Which leads to production of new quarks, and those new quarks combine with the quarks being separated, producing colorless states in the end.
Up to now there is no analytic calculation, that supports this picture.
I think that the solution of that would be a great achievement.
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A completely general method for overcoming the fermion sign problem in quantum monte carlo sampling.
As it is, any special circumstances in which we can guarantee the weights of a fermion path integral to be nonnegative are highly prized, but in general, in spatial dimensions greater than d = 1 we just can't accurately do QMC simulations for fermions.
Consequences? Well, in addition to making my life (exponentially!) easier, it would also prove P=NP.
(see, e.g.: Phys. Rev. Lett. 94, 170201)
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There is not (as far as I know) a satisfactory microscopic description of high-Tc superconductors.
Classical (s-wave) superconductors are described by BCS theory, which qualitatively assembles charge-carriers into phonon-mediated Cooper pairs, and successfully predicts many macroscopic phenomena like the Meissner effect. This theory has been known for over 50 years. The behavior of high-Tc superconductors seems to be a different animal, with substantially more complexity.
If we were to have a good grasp of the microscopic theory, this would be a major insight into the behavior of certain highly-correlated electronic systems, and one could perhaps build on such an understanding to analyze more complicated systems. On a more applications-oriented level, it is conceivable that we would have an easier time cooking up better-performing or more industrially economical superconductors.
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Is physical world ultimately describable by a renormalizable theory?
Most physicists tend to assume that this has to be so. The truth is that this assumption is just a convenience so we can add a cutoff that will keep the low-energy models safe from interacting with features of the high-energy underlying models.
There is nothing that will force this; in fact, is not out of the park to believe that most of the (27? 28? can't remember) "free" parameters in the standard model might be predicted by a non-perturbative nonrenormalizable high-energy model
More importantly though, this very assumption is directly related to the fact that current physical theories have remained largely unfalsifiable in practical terms. So this assumption (may) eventually turn out to be a self-induced dead-end
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Non-perturbatively coupling a charge and its quantized electromagnetic field.
It is a better initial approximation for QED. It may prevent the theory from UV and IR infinities at one stroke and provide correct physical description even in the first Born approximation. The perturbative series will become quite different - with numerically small terms. If successful, this formulation of QED may serve as a model to other phenomena (interactions) in particle physics.
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That would be helpful, right @Vladimir :) Don't let the anonymous cowards get to you! – user346 Jan 20 '11 at 5:56
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-1 This is neither a problem nor is it important. The QED has been worked out for some 70 years now. I honestly don't understand why you are trying to fix problems that aren't there :) – Marek Jan 20 '11 at 9:36
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Greg P asked, so here it is. I think a proof of the AdS/CFT conjecture relating $N=4$ SYM with gauge group $SU(N_c)$ to IIB string theory on $AdS_5 \times S^5$ with $N_c$ units of five-form flux would be extremely important, even if the proof was only a proof by physics standards. I say this not because I think the conjecture may be false, indeed there is overwhelming evidence that the conjecture is true. Rather I think it would be important because many of the most important applications of gauge/gravity duality require going beyond AdS/CFT to theories without supersymmetry or conformal symmetry (such as gravity duals of QCD) or to theories without Lorentz invariance (various condensed matter applications) or to theories on the gravity side which are not asymptotically AdS. For example there are proposals for a de Sitter correspondence and Verlinde has proposed "emergent gravity" bases on holographic ideas which as far as I can tell do not involve the input of data from the boundary of spacetime as one needs in AdS. A proof of the original conjecture might indicate more clearly whether these extensions of the idea are correct, or whether at some point things go wrong when you try to generalize.
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Does it make sense to talk about quantum gravity? Is it possible that gravity is a classical field or a condensate which appears largely classical, but where the underlying quantum physics is entirely different. Thirring fermions ~ sine Gordon solitons, where the SG solitons are a Euclideanize form of the hyperbolic dynamics on $AdS_2$. The $AdS_2~\sim~ CFT_1$ tells us the isometry of the spacetime is equivalent to the group for conformal quantum mechanics on the boundary. Might it then be the actual quantization is not with the spacetime, but with an underlying fermionic physics? If so can this be generalized to $AdS_n$, for $n~>~2$?
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Find a consistent and complete theory of quantum gravity combining quantum mechanics with general relativity.
Clearly, our universe is described by both quantum mechanics and general relativity. So, a more complete theory of the universe would have to incorporate quantum gravity somehow. However, combining the two has resisted decades of effort so far, and consistency and completeness turn out to be extremely stringent criteria in this case.
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@Marek: It may be too general, but as we are not really sure what form the final theory of quantum gravity will look like, making it more specific might run the risk of excluding the final correct theory. – QGR Jan 20 '11 at 9:49
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Why is the cosmological constant $10^{-120}$ times smaller than the Planck scale? Why is the electroweak scale so many orders of magnitude smaller than the Planck scale? Why is the QCD scale comparable to the electroweak scale?
The cosmological constant problem is especially acute because of zero-point energy corrections coming from quantum field theory. In a nonsupersymmetric theory, with a mismatch between the number of bosonic and fermionic fields, this would require an incredible fine-tuning. Even in a nonsupersymmetric theory where the number of bosonic and fermionic fields match, unless the masses also match, we still require an enormous fine-tuning. Even if we have supersymmetry, it has to be broken below the TeV scale.
Dynamical mechanisms to solve this problem typically run into problems with the renormalization group.
Do the anthropic principle and the multiverse answer these questions?
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Why was the initial state of the universe so special? Why was the initial entropy so low? See recent book by Sean Carroll.
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Is information preserved in evaporating black holes in quantum gravity?
I know there are already too many answers here relating to quantum gravity, but I think this question is important.
According to quantum field theory in curved spacetime, information crossing the event horizon will end up at the singularity where it will be destroyed. If we produce a pair of entangled particles outside the black hole, and throw one of them inside, it might appear as if we have converted a pure state into a mixed state, which is extremely problematic.
A lot of weird and undesirable things will happen if unitary time evolution is violated in quantum mechanics, which is why it's likely a complete theory of quantum gravity will lead to unitary time evolution. So, the question is now where is the information hidden?
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What is the nature and parameters of the Higgs sector?
I.e.
Does the Higgs particle exist?
If so, what is its mass?
Are there Higgs multiplets e.g. as predicted by supersymmetry?
This is the last remaining question mark over the standard model and its resolution may take us to the next step beyond the standard model.
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I can't believe this answer (well, question) doesn't have more up-votes. How many billions of dollars were just spent building the LHC in an effort to answer it?? (Not to mention man-hours of the thousands of physicists contributing to it.) Needless to say, it gets my vote. – qftme May 10 '11 at 10:30
What makes up dark matter.
The existence of dark matter is well established cosmologically especially through recent gravitational lens observations of galaxies that collided. But we have no idea what it consists of. Ordinary baryonic matter is virtually rules out and galactic structure formation models suggest a weakly interacting massive particle is the best fit. So is that right and if so how do such particles fit into particle physics beyond the standard model?
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Shucks - i just repeat the question that M. Veltman asks about a dozen times in Facts and Mysteries - WHY ARE THERE THREE GENERATIONS OF FERMIONS ? Halzen & Martin ask the same, as do a dozen others I could find.
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What mechanism is responsible for the imbalance between matter and anti-matter in the universe?
Did the universe start with such an imbalance already in place or is it due to some CP violating mechanism at high energy that was significant in the early universe? If so what interactions are involved?
Currently known CP violating mechanisms do not appear to be strong enough to account for the observed amount of matter in the universe, so the explanation is liekly to be someting new beyond the known standard model.
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+1 Baryon asymmetry is definitely one of the most perplexing aspects of cosmology. – Noldorin Jan 22 '11 at 23:47
What is the explanation of sonoluminescence? Under the right circumstances, sound waves can cause a bubble in a liquid to emit light. But the mechanism by which this happens is not understood.
Some of the theories proposed to explain it are impressively exotic and far-fetched. Who knows what could come out of it?
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Why is the expansion of the universe accelerating?
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Are there extra spatial dimensions?
Since many problems become exceptionally easier to solve with an assumption of extra spatial dimensions, I think this is one of the greatest questions of all time.
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What is the explanation for all the seemingly arbitrary dimensionless parameters?
This is the biggest mystery of science today I think
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– Gordon Feb 3 '11 at 23:49
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Given that the fine structure constant is a running constant that scales with energy, it's worth noting that it has this value only at zero energy. – qftme May 10 '11 at 10:25
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The Fundamental Properties of Space and Time at an Event.
Some specifics are:
1. Is Time a logical derivable construct from some non-time based fundamental theory?
2. Is Time discrete or continuous?
3. Is Space(-Time) discrete - and if so at the Planck level?
4. Directionality (and perhaps dimensionality) of Time
5. Dimensionality of Space (the String Theory issue)
6. Substructure (and any physics) below any Planck scale "minimum distance"
These kinds of issues are of course at the root of many conceptual and calculational issues in the theories around today.
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What is mass/Inertia? What is gravity?
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What is the exact form of Exchange-correlation (xc) functional in Density Functional Theory (DFT)?
The entire community of Condensed Matter Physics and Quantum Chesmitry want to know that answer. (:
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What the heck happens when we do a measurement in quantum theory? That is why should the system jump to an Eigen state?
Why should nice unitary evolution suddenly do something weird just because someone did a measurement? What exactly is a measurement anyway?
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The atom is invariant thru time?
NO - The atom is invariant thru time: This answer is naturally the expected one. It is 'common sense'. The physics building is rooted upon this answer, but this question has not being asked, nor answered, nor discussed. And Physics continues as usual.
YES - The atom is NOT invariant thru time: It is NOT the expected answer. If it happens to be the valid answer then Physics continues as usual, after a revolution.
How sure we are that the atom is 'invariant'?
Because our local lab is our reference and we can not measure directly any variability it follows that we are not sure. This fact justifies our natural answer, but we can not know the rightness of this answer until we explore. One possible way to explore it is to look back the distant universe in search of clues.
How to decide that the space expansion model is better than a shrinking or 'evanescent matter' model ? Without further reasoning one can say that one model appears to be the dual of the other, like an image in a mirror.
If we can not assert that the atom is invariable then the consequences can be dramatic:
• No space expansion
• No Big Bang
• No Inflation Era
• No Dark Energy
• No Dark Matter
• A better understanding of the nature of TIME
• A model of the Universe with only one parameter: The Hubble Constant
• A better understanding of the past and the future history of the Earth and of Life
• A different and interesting Large Scale Structure evolution of the Universe is attainable.
We must be conscient that a long future awaits the Humanity and the Physics must evolve. Under a closer scrutiny several fundamental notions, our present believes, can peril.
I tried to persuade our community, whith the above lines, that there exists merit in this quest and it is of the utmost importance, and surely it is an unsettled question.
This subject of a possible 'atom evanescence' was already studied in the arxiv paper linked bellow, and it was never criticized nor discussed. In there is presented a model that conforms to the Universe evolution and the experimental barionic data, with only one parameter - the Hubble Constant. The equations of the model can be written in the back of an envelope, somewhat ironically.
As any model it was constructed upon one hypothesis, as above described, and applying it upon data. The paper includes a comparison between the SM and the new model.
A relativistic time variation of matter/space fits both local and cosmic data (arxiv astro-ph 0208365)
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## protected by David Zaslavsky♦Jun 4 '11 at 20:45
This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9357807040214539, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/6394/lecture-notes-on-representations-of-finite-groups/6395 | ## Lecture notes on representations of finite groups
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Next term I am supposed to teach a course on representation of finite groups. This is a third year course for undegrads. I was thinking to use the book of Grodon James and Martin Liebeck "Representations and characters of groups", but also looking for other references.
The question is: could you advise some other books (or lecture notes)? Maybe you had a nice experience of teaching or listening to a course with a similar title? It would be really nice if this book (notes) has also exercises.
ADDED. I would like to thank everybody who answered the question, very helpful answers!!! The answer of John Mangual below contains a "universal" reference. For the moment my favourites are Serre (very clear and short introduction of main ideas), some bits from notes of Teleman and Martin, and Etingof for beautiful exposition. My last problem is to have enough of exercises, in particular to write down a good exam. So I would like to ask if there are some additional references for exercises (with or without solutions)?
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Since you are planning a course in representation theory could you share if you found a reference/notes which shows a lot of actual examples of computing roots and weights and Dynkin diagrams? <Apart from Fulton and Harris's book> – Anirbit Dec 27 2009 at 19:28
Anirbit, this is not planned in my course :)), this is too advanced. Have you seen below the reference to the page of Khovanov that John Mangual gave? This seem to contain everything you can imagine! – Dmitri Dec 27 2009 at 21:27
I'm wikifying this question. See meta.mathoverflow.net/discussion/6 – Anton Geraschenko♦ Dec 30 2009 at 20:08
## 11 Answers
Some material from the undergrad rep theory course in Cambridge: Example sheets, A recent set of notes (by Martin), and a less recent (but very nice) set of notes (by Teleman).
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Finally, I decided to use the notes of Teleman for the course :) Hope it will go fine. – Dmitri Jan 13 2010 at 21:45
Since Dmitri just edited, I figured now might be a good time to comment that the first link above is broken. Perhaps those example sheets were taken down? – David White Jun 28 2011 at 17:24
1
While I'm at it, I agree with Dmitri's choice, and gave this answer +1. I think representation theory is one of the hardest subjects an undergrad can learn. I recall how painful it was for me to try to learn as a 3rd year undergrad out of Isaac's Character Theory of Finite Groups. That book (and Serre's below) are great for grad students but I think Teleman's notes are some of the best I've seen at the undergrad level. I also retagged above to reflect that it's a request for a textbook-recommendation – David White Jun 28 2011 at 17:28
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The middle third of Serre's "Linear Representations of Finite Groups" is excellent. It's in 3 totally seperate sections, the first third is ok but very elementary and the last third is tough going. But the middle is "just right."
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10
+1. I am now imagining Noah as Goldilocks. – S. Carnahan♦ Nov 22 2009 at 1:22
I enjoyed Pavel Etingof's lecture notes for his representation theory class, which can be found here: http://www-math.mit.edu/~etingof/replect.pdf (there is a link to it on his website) They move fast, but without skipping too much and still providing insightful proofs of results.
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Thanks a lot! This looks cool indeed! May be the level is a bit advanced, but it could be that I'll need only first 50 pages, even less – Dmitri Nov 21 2009 at 20:08
Since I can't figure out a better place to ask this: Can anyone help me with two exercises from the above notes? In Problem 1.26 (c) I would like to know whether the field $k$ is assumed to be algebraically closed (no hints, please; just an answer to this question). In Problem 1.34 (d), am I right in assuming that the grading on $P_Q$ is given by $\mathrm{deg}p_i=0$ and $\mathrm{deg}a_h=1$ ? I really enjoy this text, by the way - it's concise and straight to the point (and not analysis-biased as Fulton-Harris). – darij grinberg Dec 22 2009 at 10:57
1
@darij, it's better to post a new question -- many more people will see it. – Ilya Nikokoshev Dec 31 2009 at 18:20
Mikhail Khovanov lists a bunch of materials for his course Representations of Finite Groups. Of course, he would be more interested in Hopf Algebras, their Representations, Applications, and Categorifications...
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This answer seem to contain everything!!! Amazing :) Huge thanks! – Dmitri Dec 27 2009 at 19:11
Artin's Algebra has a good chapter on representations of finite groups. The exercises are nice.
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Thanks, I'll have a look! – Dmitri Nov 21 2009 at 20:09
The first section of Representation Theory by Fulton and Harris is a great introduction to representations of finite groups (about a quarter of the book, if I remember correctly). There are lots of examples and exercises. The rest of the book is devoted to Lie theory.
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This is nice! I have the book :)) – Dmitri Nov 21 2009 at 20:28
1
I second Fulton and Harris: I taught such a course several times initially with James and Liebeck but with a lot of changes after I read F&H. – Fran Burstall Nov 21 2009 at 21:22
I was surprised to see Dummit and Foote's "Abstract algebra" book has a decent amount of basics on representations of finite groups. I believe this was our 2nd year algebra text, when I was an undergraduate. If your university uses this book, it might be cost effective for your students.
When I took a course on representations we used Serre's book. It's quite nice though it sounds like you want something that's a little more rich in examples and exercises.
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Thanks for this too! – Dmitri Nov 21 2009 at 20:10
Alperin and Bell's book is where I got my introduction to the representation theory of finite groups. I didn't understand much at the time, but it helped a lot with getting my bearings as I delved deeper into representation theory in general.
If you're interested (as I am) specifically in representations of finite groups, there's a book by Digne and Michel; but this is probably far too much for undergraduates.
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I like M. Isaacs, "Representation theory of finite groups" since it has lots of exercises.
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Do you mean "Character theory of finite groups"? If so, it should be pointed out that the book is much more focused on character theory than on representation theory per se, both in terms of the language/terminology used and the kinds of question it discusses – Yemon Choi Jul 1 2011 at 22:30
Here is one more reference (kindly communicated to me by one of my colleagues), the course of Iain Gordon. You can find the photos of all blackboards (I still have to go through this course, but it seem a bit similar in spirit to the courses of Teleman and Martin).
http://www.maths.ed.ac.uk/~igordon/4rt/rt2008.htm
ADDED.
I just read the article "Representation Theory" of Ian Grojnowski in the book "Princeton companion of mathematics". I find it really vivid, inspiring and amazingly well written. The first 6 pages are on finite groups, then it proceeds to compact Lie groups and non-compact Lie groups, and smoothly finishes with Langlands correspondence :). The beginning will serve perfectly for the introductory lecture of my course :).
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I think it's a really good introduction, but unfortunately one of the photos is broken. If you are reading Iain... ;) – Grétar Amazeen Dec 27 2009 at 18:54
At the risk of tooting my own horn, I have a new book that will be published by Springer which is a course on representation theory of groups intended for undergrads and beginning grads. It assume only linear algebra, group theory and basic ring theory. It assumes no module theory. Included are applications to combinatorics and probability. The link is
http://www.springer.com/mathematics/algebra/book/978-1-4614-0775-1?detailsPage=authorsAndEditors
and it should be out by the end of the year.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9489215016365051, "perplexity_flag": "middle"} |
http://cms.math.ca/10.4153/CJM-2011-094-6 | Canadian Mathematical Society
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# A Stochastic Difference Equation with Stationary Noise on Groups
Read article
[PDF: 238KB]
http://dx.doi.org/10.4153/CJM-2011-094-6
Canad. J. Math. 64(2012), 1075-1089
Published:2011-12-23
Printed: Oct 2012
• Chandiraraj Robinson Edward Raja,
Stat-Math Unit, Indian statistical institure, 8th Mile Mysore Road, Karnataka 56059, INDIA
Features coming soon:
Citations (via CrossRef) Tools: Search Google Scholar:
Format: LaTeX MathJax PDF
## Abstract
We consider the stochastic difference equation $$\eta _k = \xi _k \phi (\eta _{k-1}), \quad k \in \mathbb Z$$ on a locally compact group $G$ where $\phi$ is an automorphism of $G$, $\xi _k$ are given $G$-valued random variables and $\eta _k$ are unknown $G$-valued random variables. This equation was considered by Tsirelson and Yor on one-dimensional torus. We consider the case when $\xi _k$ have a common law $\mu$ and prove that if $G$ is a distal group and $\phi$ is a distal automorphism of $G$ and if the equation has a solution, then extremal solutions of the equation are in one-one correspondence with points on the coset space $K\backslash G$ for some compact subgroup $K$ of $G$ such that $\mu$ is supported on $Kz= z\phi (K)$ for any $z$ in the support of $\mu$. We also provide a necessary and sufficient condition for the existence of solutions to the equation.
Keywords: dissipating, distal automorphisms, probability measures, pointwise distal groups, shifted convolution powers
MSC Classifications: 60B15 - Probability measures on groups or semigroups, Fourier transforms, factorization 60G20 - Generalized stochastic processes
© Canadian Mathematical Society, 2013
© Canadian Mathematical Society, 2013 : http://www.cms.math.ca/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.822963535785675, "perplexity_flag": "middle"} |
http://mathhelpforum.com/algebra/108615-find-least-value-n-arithmetic-sequence.html | Thread:
1. Find the least value of n in an arithmetic sequence
Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$
How do I do this?
2. Originally Posted by Viral
Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$
How do I do this?
Furst thing you do is simplify it down by extracting the constants outside:
$\sum_{r=1}^n (4r-3) = 4 \sum_{r=1}^n r - \sum_{r=1}^n 3 = 4 \sum_{r=1}^n r - 3n$.
Then you know (or ought to know, or be able to find out, or work out) what $\sum_{r=1}^n r$ is.
Then you have an inequality in n to solve. Good luck.
3. Hmm, I'm not really sure what you did there =\ . We've never covered that.
4. Originally Posted by Viral
Find the least value of n for which $\sum_{r=1}^n (4r-3) > 2000$
How do I do this?
$\sum_{r=1}^n (4r-3) = 4 \left( \sum_{r=1}^n r\right) - 3n$ and you should know the formula to substitute for $\sum_{r=1}^n r$. Then solve for smallest positive integer value of n such that it's greater than 2000.
If you need more help please show all your working and say where you get stuck.
5. Originally Posted by Viral
Hmm, I'm not really sure what you did there =\ . We've never covered that.
Sigma sums preserve multiplication by a constant and preserve addition. You can prove this using the definition of sigma sums.
6. Is $\sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))$?
7. Originally Posted by Viral
Is $\sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-3))$?
No, but $\sum_{r=1}^n(4r-3) = 4(\sum_{r=1}^n(r-(3/4))$
8. Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.
9. Originally Posted by Viral
Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.
Fair comment. Can you document on this thread everything you've been taught? (joke)
10. To be honest, I would but I don't know how I'd explain it. We've learned how to calculate all terms in a sequence, how to find a, d, n and l. And this is our first look at the summation notation. If there are other methods, and we've been taught it, it should hit me when I see it.
11. Originally Posted by Viral
Hmm, I kind of understand that. Are there other ways to solve this as our teacher specifically said we should be able to solve all questions with what we have been taught.
The first term in the sum is 4 - 3 = 1. Therefore a = 1.
The common difference between each term is 4. Therefore d = 4.
There are n terms.
Substitute all that into your formula for the sum of an arithmetic series and then do with the result what I said to do in my first reply.
12. $S_n = \frac{n}{2}[2a + (n-1)d]$
Just to check, is this the formula for calculating the sum?
13. Originally Posted by Viral
$S_n = \frac{n}{2}[2a + (n-1)d]$
Just to check, is this the formula for calculating the sum?
It's just as easy for you to go to a textbook or use Google as it is for any of us ....
14. It's a help forum after all >> . My textbook has been misplaced, hence why I'm here. I'm also here because google brought me here.
15. Originally Posted by Viral
It's a help forum after all >> . My textbook has been misplaced, hence why I'm here. I'm also here because google brought me here.
Yes, help as in help you to help yourself. Really, why should we be the ones that check a formula using Google when you can easily do it yourself? That's your job, not our job. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9700508117675781, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/280613/bound-on-the-growth-of-a-singular-measure | # Bound on the growth of a singular measure
Working through some measure theory, I came upon the Lebesgue-like decomposition for monotonic functions. In that context, I've cooked up a singular measure $\nu$ on $[0,1]$ about which I know only that it is regular Borel, continuous, and satisfies $\nu K = 0$ for some $K \subset [0,1]$ of full Lebesgue measure.
I would like to conclude that $K$ contains a subset $K'$ of full Lebesgue measure for which each element $x$ satisfies $$\nu\bigl((x-1/n,x+1/n)\bigr) = \mathcal O(1/n)~~.$$ This would be sufficient to demonstrate that the function mapping $x \mapsto \nu\bigl([0,x]\bigr)$ is singular, in the sense of having an a.e. defined and vanishing derivative (of course, it is continuous as well by continuity of $\nu$).
I'd appreciate any comments on how to show the above bound, and whether that's a good strategy for approaching the problem of defining a singular function from a singular measure. My other thought was that perhaps there is a standard way to 'remove' density from the complement of $K$ (using continuity) sufficiently to form an open set containing $K$ with zero $\nu$-measure, from which the result follows (and the singular set looks much more like the complement of a Cantor set).
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You do not need that condition. Your measure is already singular continuous, so your mapping must give you a singular continuous function. – Thomas Jan 17 at 11:12
@Thomas Sure, but the condition I gave is equivalent to the mapping having a vanishing derivative at $x$, which for singularity must be true for all $x$ in a set of full measure; it's not clear to me how without some work in that direction, the result could come obviously. – Eugene Shvarts Jan 17 at 18:56
Ah, I see, so you want to show that if a measure is singular, then the cumulative distribution function is singular, right? – Thomas Jan 17 at 20:49
@Thomas Spot-on. After doing some more reading, it looks like going through bounded-variation function methods will be easier, but I'm still curious if this is a good approach (since, all else equal, it's still true). – Eugene Shvarts Jan 17 at 20:53
## 1 Answer
Let us call $$D^+ \nu (x) := \limsup_{\varepsilon \to 0} {\frac{\nu(]x-\varepsilon,x+\varepsilon[)}{2\varepsilon}},\, D^- \nu (x) := \liminf_{\varepsilon \to 0} {\frac{\nu(]x-\varepsilon,x+\varepsilon[)}{2\varepsilon}}.$$ You can show with the help of the Vitali covering theorem, that for a measurable set $A \subseteq [0,1]$ we have $$D^+ \nu(x) \geq q \quad \forall\, x \in A \qquad \Longrightarrow \qquad \nu(A) \geq q \lambda(A).$$ Now consider $A_n := \{x \in K \: | \: D^+ \nu(x) \geq 1/n\}$. Then we get $\nu(A_n) \geq \lambda(A_n)/n$, but the left-hand side is $0$, so $\lambda(A_n) = 0$ and by taking the limit $\lambda(\{x \in K \: | \: D^+ \nu(x) > 0\}) = 0$, which means $\lambda(\{x \in K \: | \: D \nu(x) = 0\}) = \lambda(K) - 0 = 1$. So the derivative exists almost everywhere and is zero where it exists.
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Thanks Thomas! It's interesting to see things done the measure-theoretic way. – Eugene Shvarts Jan 17 at 22:43 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9558460712432861, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/158352/how-to-prove-the-following-inequality | # How to prove the following inequality?
Show that there exist constant $C= C(n,p)$ depending only on $n$ dimension and $p \in \mathbb{N}$ such that \begin{equation} C|a-b|^{p} \le \langle |a|^{p-2} a - |b|^{p-2}b, a-b \rangle \end{equation} for any $a,b \in \mathbb{R^{n}}$. Thank you.
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## 1 Answer
Vector inequalities of this type are constantly used in the study of the $p$-Laplace equation. Chapter 10 of Notes on the p-Laplace equation has enough of them to answer this question, see (I) and (VII) therein. However, I don't like Lindqvist's approach via integrals: I'd much rather see a clean algebraic derivation.
A few comments follow.
1. The dimension $n$ is irrelevant. Taking the span of $a$ and $b$, you can assume $n=2$.
2. The stated inequality is false for $p=1$.
3. There is no need for $p$ to be an integer. However, you do need $p\ge 2$ to have an inequality as stated.
4. The way in which the inequality is stated is suboptimal. Here is a better one: $$c|a-b|^2(|a|+|b|)^{p-2}\le \langle |a|^{p-2}a-|b|^{p-2}b,a-b \rangle \le C|a-b|^2(|a|+|b|)^{p-2}$$ which is true for $1<p<\infty$ (the constants depend on $p$).
I wish I had a clean proof of 4 up my sleeve, but I don't right now. I do have a proof of the following fact: the inner product is comparable to the product of norms. I actually inserted the following computations, elementary as they are, in a paper upon referee's request.
Let $S_{\alpha}(x)=|x|^{\alpha-1}x$, $0<\alpha<\infty$. (So, $\alpha=p-1$ in the above notation.)
Claim. The mapping $S_{\alpha}\colon\mathbb R^n\to\mathbb R^n$ satisfies the inequality $$\langle S_{\alpha}(x)-S_{\alpha}(y),x-y\rangle\ge \delta|S_{\alpha}(x)-S_{\alpha}(y)||x-y|$$ with $$\delta=\begin{cases}\alpha/(2-\alpha), \quad &0<\alpha\le 1;\\ 1/(2\alpha-1),\quad &1\le\alpha<\infty.\end{cases}$$
Proof. Since $S_{1/\alpha}=S_{\alpha}^{-1}$, it suffices to consider the case $0<\alpha\le 1$. The homogeneity of $S_{\alpha}$ reduces our task to proving the inequality $$\langle{x-S_{\alpha}(y),x-y}\rangle\ge \frac{\alpha}{2-\alpha}|x-S_{\alpha}(y)||x-y|$$ under the assumptions $|x|=1$, $|y|\le 1$. We estimate both sides as follows. $$\langle x-S_{\alpha}(y),x-y\rangle =|{x-y}|^2+\langle (1-|y|^{\alpha-1})y,x-y \rangle \ge |x-y|^2+(1-|y|^{\alpha-1})|y||x-y| \\ =|x-y|(|x-y|+|y|-|y|^{\alpha})$$ and $$|x-S_{\alpha}(y)|\le |x-y|+|y-|y|^{\alpha-1}y|=|x-y|+|y|^{\alpha}-|y|.$$ The desired inequality will follow once we prove that $$|x-y|+|y|-|y|^{\alpha}\ge\frac{\alpha}{2-\alpha}(|x-y|+|y|^{\alpha}-|y|).$$ The latter is equivalent to $$(1-\alpha)|x-y|\ge |{y}|^{\alpha}-|y|.$$ Note that $|y|^{\alpha}\le 1+\alpha(|y|-1)$ because the function $t\mapsto t^{\alpha}$ is concave for $t\ge 0$. Hence $$|y|^{\alpha}-|{y}|\le (1-\alpha)(1-|y|)\le (1-\alpha)|x-y|,$$ which is what we wanted. QED
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Thank you very much. – user29999 Jun 15 '12 at 21:21 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9539306163787842, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2009/07/13/positive-definite-transformations/?like=1&source=post_flair&_wpnonce=18b5244a97 | # The Unapologetic Mathematician
## Positive-Definite Transformations
We’ll stick with the background vector space $V$ with inner product $\langle\underline{\hphantom{X}},\underline{\hphantom{X}}\rangle$. If we want another inner product to actually work with, we need to pick out a bilinear form (or sesquilinear, over $\mathbb{C}$. So this means we need a transformation to stick between bras and kets.
Now, for our new bilinear form to be an inner product it must be symmetric (or conjugate-symmetric). This is satisfied by picking our transformation to be symmetric (or hermitian). But we also need our form to be “positive-definite”. That is, we need
$\displaystyle\langle v\rvert B\lvert v\rangle\geq0$
for all vectors $v\in V$, and for equality to obtain only when $v=0$.
So let’s look at this condition on its own, first over $\mathbb{R}$. If $A$ is antisymmetric, then we see $\langle v\rvert A\lvert v\rangle=-\langle v\rvert A\lvert v\rangle$ by taking the adjoint, and thus $A$ must be zero. But an arbitrary transformation $B$ can be split into a symmetric part $S=\frac{1}{2}\left(B+B^*\right)$ and an antisymmetric part $A=\frac{1}{2}\left(B-B^*\right)$. It’s easy to check that $S+A=B$. So the antisymmetric part of $B$ must be trivial, and the concept of being “positive-definite” only makes real sense for symmetric transformations.
What happens over $\mathbb{C}$? Now we want to interpret the positivity condition as saying that $\langle v\rvert B\lvert v\rangle$ is first and foremost a real number. Then, taking adjoints we see that $\langle v\rvert B^*\lvert v\rangle=\langle v\rvert B\lvert v\rangle$. Thus the transformation $B-B^*$ must always give zero when we feed it two copies of the same vector.
But now we have the polarization identities to work with! The real and imaginary parts of $\langle v\rvert B-B^*\lvert w\rangle$ are completely determined in terms of expressions like $\langle u\rvert B-B^*\lvert u\rangle$. But since these are always zero, so is the rest of the form. And thus we conclude that $B=B^*$. That is, positive-definiteness only really makes sense for Hermitian transformations.
Actually, this all sort of makes sense. Self-adjoint transformations (symmetric or Hermitian) are analogous to the real numbers sitting inside the complex numbers. Within these, positive-definite matrices are sort of like the positive real numbers. It doesn’t make sense to talk about “positive” complex numbers, and it doesn’t make sense to talk about “positive-definite” transformations in general.
Now, there are three variations that I should also mention. The most obvious one is for a transformation to be “negative-definite”. In this case, we have $\langle v\rvert H\lvert v\rangle\leq0$, with equality only for $v=0$. We can also have transformations which are “positive-semidefinite” and “negative-semidefinite”. These are just the same as the definite versions, except we don’t require that equality only obtain for $v=0$.
### Like this:
Posted by John Armstrong | Algebra, Linear Algebra
## 9 Comments »
1. [...] Forms I The notion of a positive semidefinite form opens up the possibility that, in a sense, a vector may be “orthogonal to itself”. That [...]
Pingback by | July 15, 2009 | Reply
2. [...] this last inequality holds because the bra-ket pairing is an inner product, and is thus positive-definite. Indeed, a positive-definite (or negative-definite) form must be nondegenerate. Thus it is [...]
Pingback by | July 17, 2009 | Reply
3. [...] off, note that no matter what we use, the transformation in the middle is self-adjoint and positive-definite, and so the new form is symmetric and positive-definite, and thus defines another inner product. [...]
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4. [...] if is not only self-adjoint, but positive-definite? We would like the determinant to actually be a positive real [...]
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5. [...] going to want to save and to denote transformations). We also have its adjoint . Then is positive-semidefinite (and thus self-adjoint and normal), and so the spectral theorem applies. There must be a unitary [...]
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6. [...] that unitary transformations are like unit complex numbers, while positive-semidefinite transformations are like nonnegative real numbers. And so this “polar decomposition” is like the polar form of a complex number, where we [...]
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7. [...] Here’s a neat thing we can do with the spectral theorems: we can take square roots of positive-semidefinite transformations. And this makes sense, since positive-semidefinite transformations are analogous to nonnegative [...]
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8. [...] if all of these eigenvalues are positive at a critical point , then the Hessian is positive-definite. That is, given any direction we have . On the other hand, if all of the eigenvalues are negative, [...]
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9. [...] already seen that the composition of a linear transformation and its adjoint is self-adjoint and positive-definite. In terms of complex matrices, this tells us that the product of a matrix and its conjugate [...]
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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. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 25, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9122832417488098, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/106144-more-complex-integration-please-help.html | # Thread:
1. ## More Complex Integration Please Help
I'm working through problems on my book and I have one problem, It seems like for the more complex functions they just memorize the table, which to me is pointless if you do not know how to get to the shortcut without using some integration techinque so I have a few problems that I would like a "hint" on what would be the first thing to do like integrations by part or substitution or the best way to expand or break down the integral so I can figure it out on my own
$\int{\frac{1}{3+y^2}}$
I seen one fellow expand the denom. out and have a square root come out I have no clue where that one came from.
$\int{\frac{1}{cos^3x}}$
No idea what to do here, I am not content with memorizing the shortcut or whatever it maybe be called in that table.
Those are all for now
2. Originally Posted by The Power
I'm working through problems on my book and I have one problem, It seems like for the more complex functions they just memorize the table, which to me is pointless if you do not know how to get to the shortcut without using some integration techinque so I have a few problems that I would like a "hint" on what would be the first thing to do like integrations by part or substitution or the best way to expand or break down the integral so I can figure it out on my own
$\int{\frac{1}{3+y^2}}$
I seen one fellow expand the denom. out and have a square root come out I have no clue where that one came from.
$\int{\frac{1}{cos^3x}}$
No idea what to do here, I am not content with memorizing the shortcut or whatever it maybe be called in that table.
Those are all for now
For 1
$\int {\frac{{dy}}{{3 + {y^2}}}} .$
$\frac{1}<br /> {{3 + {y^2}}} = \frac{1}<br /> {3} \cdot \frac{1}<br /> {{1 + \frac{{{y^2}}}<br /> {3}}} = \frac{1}<br /> {3} \cdot \frac{1}<br /> {{1 + {{\left( {\frac{y}<br /> {{\sqrt 3 }}} \right)}^2}}}.$
So $\int {\frac{{dy}}{{3 + {y^2}}}} = \frac{1}{3}\int {\frac{{dy}}<br /> {{1 + {{\left( {\frac{y}{{\sqrt 3 }}} \right)}^2}}}} = \left\{ \begin{gathered} \frac{y}{{\sqrt 3 }} = u, \hfill \\dy = \sqrt 3 du \hfill \\ <br /> \end{gathered} \right\} = \frac{{\sqrt 3 }}{3}\int {\frac{{du}}{{1 + {u^2}}}} =$
$= \frac{{\sqrt 3 }}{3}\arctan u + C = \frac{{\sqrt 3 }}{3}\arctan \frac{y}{{\sqrt 3 }} + C.$
3. I seen you try to explain this in a earlier thread but I do not see where the sqrt comes from
My problem is the reference for every problem is done by looking at the table of integrals my annoyance is how they get to these integrals in the table, or for test time sake's is it better just to accept this is how the table of integrals is and make reference and plug in for corresponding values?
I guess I just want to learn the steps leading to the final integrals in the table for the "non elementary functions" | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9548497796058655, "perplexity_flag": "head"} |
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http://stats.stackexchange.com/questions/12021/an-inconsistency-between-the-concept-of-subindependence-and-the-chi-square-tes | # An inconsistency between the concept of “subindependence” and the chi-square test for independence?
Two random variables are defined as subindependent if their covariance is zero--in other words, if they are uncorrelated. The latter link notes that "not all uncorrelated variables are independent. For example, if $X$ is a continuous random variable uniformly distributed on $[−1, 1]$ and $Y = X^2$, then $X$ and $Y$ are uncorrelated even though $X$ determines $Y$ and a particular value of $Y$ can be produced by only one or two values of $X$." So subindependence, as you can guess from the name, is a weak form of independence.
Soon after reading this, I was looking at Pearson's chi-square test for independence. The wikipedia page says that "for the test of independence, a chi-square probability of less than or equal to 0.05 (or the chi-square statistic being at or larger than the 0.05 critical point) is commonly interpreted by applied workers as justification for rejecting the null hypothesis that the row variable is independent of the column variable. The alternative hypothesis corresponds to the variables having an association or relationship where the structure of this relationship is not specified." A previous CV answer (here) also indicates that this is what the chi-square test is looking for.
Now, how can you test a null hypothesis of independence by looking for a correlation but not define the lack of a correlation as indicative of independence? Granted, my skepticism is based on my reading of the Wikipedia pages for these concepts, which could easily be flawed. But it seems to me like Pearson's chi-square method must be testing for the lack of *sub*independence. Is there something wrong with my conclusions? Or, is this already common knowledge?
-
1
Subindependence is a stronger property than being uncorrelated. – Henry Jun 17 '11 at 6:43
@Henry's point is good but should be made more strongly: the characterization of "subindependent" in this question is incorrect. – whuber♦ Jun 17 '11 at 14:24
2
Your scepticism is based on reading too much into statistical testing. An hypothesis test can falsify the null but never affirm it. Demonstrating that two variables are not uncorrelated a fortiori demonstrates they are not independent. That's all that is going on. – whuber♦ Jun 17 '11 at 14:28
## 1 Answer
Pearson's chi-squared test looks at how closely sample observations match a theoretical distribution. It works with discrete or categorised data.
Taking your example and a sample of 1000, you might get something like
```` -1<=X<0 0<=X<=1
0 <=Y< 0.5 378 329
0.5<=Y<=1 142 151
````
which has a Chi-squared statistic of 1.8803 and 1 degree of freedom: this does not show a significant relationship.
but if you categorise the $X$ data more finely you might get something like
```` -1<=X<-0.5 -0.5<=X<0 0<=X<0.5 0.5<=X<=1
0 <=Y< 0.5 102 276 232 97
0.5<=Y<= 1 142 0 0 151
````
and this time Chi-squared statistic is 428.3342 with 3 degrees of freedom, and is highly significant.
So in some circumstances the Pearson's chi-squared test can spot non-independent data even when it is uncorrelated. In this case, potting $Y$ against $X$ would suggest the relationship more quickly.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9356582164764404, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/24243/electric-dipole-transitions-expectation-value-of-position?answertab=votes | # Electric dipole transitions/expectation value of position
Part of a homework question asks to show that for $\ell=0$ in both $\Psi_i$ and $\Psi_f$, we have $$\int \Psi_i^\ast \vec{r} \Psi_f \; d\tau = 0$$ for the position vector $\vec{r}$. (This is for the electron in hydrogen and the integral is over all space.) The physical interpretation of this is that since the expectation value is zero, such a transition is forbidden. I am having trouble showing the above integral is zero. Since we are asked to show this in general, and not for a special case, it seems the only thing to do is use orthogonality of the $\Psi$'s. Is this correct? Can someone nudge me in the right direction?
-
Can someone fix the box after the $r$? I used \vec{} but as I usually do, but it apparently did not render here. – unit3000-21 Apr 22 '12 at 23:10
1
Looks fine to me... – David Zaslavsky♦ Apr 23 '12 at 3:32
## 1 Answer
Unless I'm missing something this is straightforward. If $\ell=0$ the wavefunction is spherically symmetric, so $\Psi_i^\ast \vec{r} \Psi_f$ is antisymmetric and automatically integrates to zero.
-
For a (electric-dipole) "forbidden" transition $i$ and $f$ have in general differnt $l$'-s, what matters is that $|l_i -l_|f$ must be exactly 1 for the integral to be non-zero. You have shown the case of $l_i=l_f=0$ only. – Slaviks Apr 23 '12 at 8:15
@Slaviks: that's because the original question asked for a proof when $\ell_i = \ell_f = 0$! – John Rennie Apr 23 '12 at 9:15
You're not missing anything, it really is that simple! However, I couldn't flesh out the details until late last night. I ended up converting to Cartesian coordinates, in which case $\Psi^\ast \Psi$ is even in each of $x,y$ and $z$ since $r \mapsto \sqrt{x^2+y^2+z^2}$. Thus, $\Psi^\ast x \Psi$ is odd in $x$, and similarly for $y$ and $z$. So the integral of each is zero, which means the integral of the original thing is zero. – unit3000-21 Apr 23 '12 at 19:59
@JohnRennie You are totally right, I've not been careful with reading the question. – Slaviks Apr 24 '12 at 10:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 22, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9439647197723389, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/67342-can-anyone-help-my-level-revision-i-have-few-practise-quests-but-no-ans-print.html | # can anyone help with my AS level revision i have a few practise quests. but no ans.
Printable View
• January 8th 2009, 11:52 AM
adeshSB
can anyone help with my AS level revision i have a few practise quests. but no ans.
i attmepted this one and got as far as possible but not sure where to go next
find the solutions to x^2 -6x+2 leave answer as a surd i attempted it and ended up with 3+/- (sq. root of 7)
and then it asks to then get the exact solution to the inequality x^2-6x+2 </= 0
by substitution of (3+/- sq.root of 7) into x
i got to a point where -5</= 0
am i going right and does this mean X >/=5
Thanks for help
• January 8th 2009, 01:37 PM
craig
You were right with your roots,
$x^2-6x+2$ roots are, , however the equation in factorised form is, $(x-3-\sqrt7)(x-3+\sqrt7)$.
I presume that the inequatities are therefore;
$x\leq3+\sqrt7$ and $x\leq3-\sqrt7$.
It has been a while since I have done inequalities so I might be wrong?
• January 8th 2009, 02:27 PM
HallsofIvy
Quote:
Originally Posted by adeshSB
i attmepted this one and got as far as possible but not sure where to go next
find the solutions to x^2 -6x+2 leave answer as a surd i attempted it and ended up with 3+/- (sq. root of 7)
and then it asks to then get the exact solution to the inequality x^2-6x+2 </= 0
by substitution of (3+/- sq.root of 7) into x
i got to a point where -5</= 0
am i going right and does this mean X >/=5
Thanks for help
Unfortunately, craig was not completely right. The factorization craig gives makes your inequality $(x-3-\sqrt7)(x-3+\sqrt7)\le 0$
In order that a product of two numbers be negative, the numbers must have opposite sign. So either $x-3-\sqrt{7}\le 0$ and $x-3+\sqrt{7}\ge 0$ or [tex]x-3-\sqrt{7}\ge 0[\math] and $x-3+\sqrt{7}\le 0$. In the first case $x\le 3+ \sqrt{7}$ and $x\ge 3- \sqrt{7}$: that is $3-\sqrt{7}\le x\le 3+ \sqrt{7}$. In the second case, $x\ge 3+ \sqrt{7}$ and $x\le 3- \sqrt{7}$ which is impossible. All x such that $3- \sqrt{7}\le x\le 3+ \sqrt{7}$.
Another way to do this is to recognize that since $x^2- 6x+ 2$ is a continuous function so can change form negative to positive only where it is equal to 0. It is sufficient to check one value in each of the three intervals $3- \sqrt{7}$ and $3+\sqrt{7}$ break the real line into. $0< 3- \sqrt{7}$ and $0^2- 6(0)+ 2= 2> 0$ so no $x< 3- \sqrt{7}$ satisfies $x^2- 6x+ 2< 0$. $3-\sqrt{7}\le 3\le 3+\sqrt{7}$ and $(3)^2- 6(3)+ 2= -7< 0$ so every number between $3-\sqrt{7}$ and $3+ \sqrt{7}$ satisfies $x^2- 6x+ 2< 0$. Finally, $6> 3+ \sqrt{7}$ and $(6)^2- 6(6)+ 2= 2> 0$ so no number larger than $3+ \sqrt{7}$ satisfies the inequality.
Finally, we can recognise that the graph of $y= x^2- 6x+ 2$ is a parabola opening upward. y will be less than 0 only for the portion of the parabola between the two roots $3- \sqrt{7}$ and $3+ \sqrt{7}$.
• January 8th 2009, 02:32 PM
craig
Sorry you are quite right :) had a feeling there was something missing when I posted (Worried)
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http://mathoverflow.net/questions/74777/what-conditions-on-a-probability-distribution-defined-by-long-time-averaging-do-i/75045 | ## What conditions on a probability distribution defined by long-time averaging do I need to satisfy a central limit theorem?
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For integer $n$, $1 \le n \le N$, consider the random variables
$X_n = \cos[t \omega_n]$
For any fixed $N$, we can take the mean
$Y_N = \frac{1}{N} \sum_{n=1}^N X_n$
and define a (cumulative) distribution by averaging over long times:
$P(Y_N \le y) = \lim_{T \to \infty} \frac{1}{2 T} \lambda(\{t \in [-T,T] \; \vert \; Y_N \le y \} )$,
where $\lambda$ is the Lebesgue measure. If the $\omega_n$ are linearly independent over the rationals for all $n$, then the random variables $X_n$ are independent and identically distributed over long times. Since they are i.i.d., we can apply the central limit theorem to show that $p(Y_N = y)$, the probability density of $Y_N$, approaches a normal distribution with zero mean and variance $\sigma_0^2/N$, where $\sigma_0^2=1/2$ is the variance of each $X_n$.
Now, suppose that the $\omega_n$ are not all linearly independent, but are incommensurate (i.e. pairwise linearly independent). This would mean that the $X_n$ are not independent. In particular, suppose we consider only $N=2^M$ for integer $M$ and take
$w_n^{(M)} = \sum_{k=0}^{M-1} (-1)^{r_k} h_k$
where
$r_k = (n/2^k) \mod 2$
are the digits of $n$ in base 2. Though the $h_k$ may be linearly independent over the rationals, there are $2^M$ frequencies $w_n^{(M)}$ for each fixed $M$. Since $k$ only ranges from $0$ to $M-1$, these frequencies are not linearly independent.
So for fixed $M$, not only are the $2^M$ random variables $X_n^{(M)} = \cos[t \omega_n^{(M)}]$ not independent, their joint probability density is actually only has support on a $M$-dimensional subspace. This is not a stationary sequence (I think). I'm not sure if it is ergodic.
Now, we could express the $Y$ as a mean of just $M$ independent random variables by combining the $X_n^{(M)}$. This would seem to guarantee a variance only as small as $\sigma_0^2/M$ rather than $\sigma_0^2/2^M$.
However, numerical work confirmed my intuition that the the $Y$ actually approaches a normal distribution with the small variance $\sigma_0^2/2^M$. This suggests there is a CLT I could apply to get this result analytically. But when I read the standard texts, I can't find much in the way of CLT's for non-stationary processes. I'm trying to read about ergodicity, but I can't even tell if this sequence fits the descriptions.
Is this sequence ergodic? Does it satisfy a CLT?
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## 1 Answer
In your example, algebraic manipulation gives $$2^{-M} \sum_{n=0}^{2^M-1} X_n^{(M)} = \prod_{j=0}^{M-1} \cos t h_j.$$ If the $h_j$'s are linearly independent over $\Bbb Q$, then, as you point out, the random variables $\cos t h_j$ approach independence as $t$ is chosen over larger and larger intervals. Therefore, in the limit, $$Y = Z_1 ... Z_M, \qquad Z_i = \cos W_i, \qquad W_1,...,W_M {\rm\ \ i.i.d.\ uniform\ on\ } [0,2\pi).$$ By symmetry, ${\bf E}[Y]=0$, and since ${\bf E}[Z_i^2]=1/2$ for each $i$, we have ${\bf Var}[Y]=2^{-M}$. This may explain the result of your numerical experiments. However, $Y$ does not approach normal after rescaling: if $Y$ is rescaled to unit variance by setting $Y'_M:=2^{M/2} Y$, then $\log |Y'_M|=\log(\sqrt{2} |Z_1|)+...+\log(\sqrt{2} |Z_M|)$, so, since ${\bf E}[\log(\sqrt{2} |Z_i|)]<0$ and ${\bf Var}[\log(\sqrt{2} |Z_i|)]<\infty$, we can apply the CLT to $\log |Y'_M|$ to show that, as $M\to\infty$, $\log |Y'_M|$ will converge weakly to normal after rescaling, but $Y'_M$ converges weakly to 0.
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Thanks very much for this compact and clear explanation. You are right about $Y$ not approaching a normal distribution. I discovered an an error in my code this morning, and the true distribution it approaches has much heavier tails than a Gaussian. Could you please elaborate on how you know that Var[$Y$] = $2^{-M}$? Is it correct to express Var[$Y$] in terms of the moments of the r.v. ln($Y$) (which is normally distributed), yielding a taylor series? Are there convergence issues I should worry about as a physicist? – Jess Riedel Sep 12 2011 at 17:19
1
To compute the variance of $Y$, observe that $Y^2$ is the product of $M$ independent random variables, each distributed as $\cos^2 W$, where $W$ is uniform. Since $\cos^2$ has average value $\frac{1}{2}$, and the r.v.s are independent, ${\bf E}[Y^2]$ is the product of $M$ copies of $\frac{1}{2}$, which is $2^{-M}$. – David Moews Sep 13 2011 at 1:07
1
Setting $Y_M:=Y$, there is no way to pick scaling constants $a_M$ such that $a_M Y_M$ converges to something nontrivial. $|Y_M|^{1/\sqrt{M}}$ will converge if rescaled appropriately. – David Moews Sep 13 2011 at 1:58
Ha, very embarrassing on my part, David. Good thing this is saved on the internet for perpetuity. Thanks so much for the help. – Jess Riedel Sep 14 2011 at 14:14 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 78, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9377729296684265, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/26898?sort=newest | ## Normal Ordering with Vertex Operators in Conformal Field Theory
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The "definition" of the normal ordering in CFT looks a bit vague to me.
I found the definition in terms of exponentiated functional derivative pretty opaque.
Also in this context it might help if someone can give a reference or if there is a short explanation to understand how the Operator Product Expansion is derived using products of normal ordered operators.
I don't see the conceptual framework in which these ideas fit together. Some of the books I looked at gave a very disparate view as a collection of some complicated formulas.
Let me give a precise example of the kind of calculation that I am stuck with,
Refer to these lecture notes
I can understand equation 4.26 of this but not the next 4 equations that seem to follow from it leading to 4.28.
It would be helpful if someone can decrypt the calculation.
In light of the kinds of references that came in as responses, I think it would help if I make the problematic calculation a little more explicit.
This has to do with what are called "Vertex Operators" in CFT given as $:e^{ikX(z)}:$ where $::$ is the notation for normal ordering and $k$ is some scalar and $X$ is a conformally invariant free Bosonic field. Then I would like to understand the derivation of this equality,
(all expressions are understood to be valid under the Feynman Path Integral)
$:\partial X(z)\partial X(z)::e^{ikX(w)}: = -\frac{k^2\alpha ^2}{4}\frac{:e^{ikX(w)}:}{(z-w)^2}-ik\alpha\frac{:\partial X(z) e^{ikX(w)}:}{(z-w)}$
where we have $X(z)X(w) = -\frac{\alpha}{2}ln \vert z - w \vert$
and what would be the similar simplification of
$:e^{ikX(z)}::e^{ikX(w)}: = ?$
Some more elaboration on what about normal ordering I am concerned about.
The problem is that I can't these books give an honest definition of what it means to "normal order" operators in CFT. Like there is a very clean definition in rest of QFT whose relation to time-ordering is given by the Wick's Theorem. Here in CFT one is supposed to understand that while normal ordering a string of operators inserted at different points on the space-time one is subtracting away from the product every possible way in which one or more pairs of insertion points can coincide and produce a singularity
Like if A,B,C,D are 4 different Bosonic operators say inserted at 4 different space-time points. Then one would define normal ordering as,
$:ABCD: = ABCD - (AB):CD: - (AC):BD: - (AD):BC:-(BC):AD:-(BD):AC:$ $$-(CD):AB:-(AB)(CD)-(AC)(BD)-(AD)(BC)$$
where () denotes the correlation function of the operators.
Now the point is whether one is supposed to take the above kind of equations as being just well-motivated definition or is there is anything more fundamental from which it is derivable?
There is definitely an issue about defining difference of two divergent expressions here.
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2
What are "some books" that you've tried? One axiomatic approach derives from associativity and commutativity of Frenkel, Lepowski, Meurman. A related alternative is in Kac's "Vertex algebras for beginners". – Victor Protsak Jun 3 2010 at 9:58
To emphasize and reiterate the comments by José below: one has to distinguish between (a) the Operator Product Expansion (OPE), which is one of the features of the mathematical formulation of CFT and (b) Wick's theorem, which explains how to rewrite certain products of free boson or fermion fields in the normal form. The $::$ notation used in (a) and (b) is the same, for historical reasons and because (a) is more general. But the definition $:AB:$ in (a) is direct. Provided OPE is local, i.e. commutativity and associativity that I mentioned, you can do calculations like the one you wanted. – Victor Protsak Jun 10 2010 at 1:01
## 2 Answers
The field $L=:\partial X\partial X:$ is Virasoro and your first formula says that a vertex operator is a primary field (highest weight vector) with respect to the Virasoro action, with appropriate weight. This is a basic computation with vertex operators done in nearly any textbook on conformal field theory (Kaku, Di Francesco, etc). You may also want to look at one of the first mathematical treatments of vertex operators, very clearly written:
Frenkel, I. B.; Kac, V. G. Basic representations of affine Lie algebras and dual resonance models. Invent. Math. 62 (1980/81), no. 1, 23--66
By the way, the proper way to think about OPE for bosonic fields is
$$\partial X(z) \partial X(w) = \frac{1}{(z-w)^2}.$$
The "raw" $X(z)$ occurs only as a part of the exponential expression defining the vertex operator, never by itself.
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The best reference I know for this sort of calculations is the PhD thesis of Kris Thielemans. Chapter 2 is all about the calculations of operator product expansions in two-dimensional conformal field theory. In particular in Section 2.6.1 you will find the kind of calculation you are interested in explained in detail. The formalism is clean and not very difficult at all.
Added
Here's a sketch of the calculation of the operator product expansion between two vertex operators $V_k(z)$ and $V_\ell(w)$ where $$V_k(z) = :e^{i k X(z)}:,$$ where $$X(z) = x - i p \log z + i \sum_{n\neq 0} \tfrac1n a_n z^{-n},$$ with canonical commutation relations (CCRs) $[p,x]=1$ and $[a_n,a_m] = n \delta_{n+m,0}$.
The normal ordering prescription consists of writing the creation operators to the left of the annihilation operators: $$V_k(z) = e^{\sum_{n>0} \frac{k}{n} a_{-n} z^n} e^{ikx} z^{kp} e^{-\sum_{n>0} \frac{k}{n} a_n z^{-n}}.$$
To compute the operator product expansion $V_k(z) V_\ell(w)$ one simply has to take the formal product of the above expressions for the vertex operators and commute the operators through using the CCRs in order to bring the creation operators to the left. The basic calculations are simple Weyl identities of the form $$e^{-\frac{k}{n}a_n z^{-n}} e^{\frac{\ell}{n} a_{-n} w^n} = e^{\frac{\ell}{n} a_{-n} w^n} e^{-\frac{k}{n}a_n z^{-n}} e^{\frac{k\ell}{n} (\frac{w}{z})^n}$$ and a similar one for $x$ and $p$.
The resulting power series converges provided that the fields are radially ordered so that $|z|>|w|$, but the result can be analytically continued with either a pole at $z=w$ or a branch cut singularity depending on the values of $k$ and $\ell$.
The final step is to expand the $z$-dependent fields around $z=w$.
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Thanks for this reference. I had a look through it but it doesn't seem to address the specific calculation I am looking for. I have now more explicitly written out the equation whose derivation I am trying to understand. – Anirbit Jun 4 2010 at 13:59
I've added a sketch of the calculation. – José Figueroa-O'Farrill Jun 5 2010 at 1:52
Thanks for this detailed reply. But this language of doing CFT is starkly different from the one seen in the books by Polchinski or the book by Francesco et. al. A clear definition of how products of normal ordered operators is defined is not easily found from these books. They mention that it is "analogous" to Wick's Theorem as in QFT but the analogy is far from obvious. Can you give a reference which explains this technology better? – Anirbit Jun 8 2010 at 5:54
Starkly different?! I don't think so. It's at most a notational difference. The above formulae are strictly speaking true inside radially ordered correlation functions, but some people (e.g., me) never write them explicitly. In addition to Thieleman's PhD thesis (linked above), I have found Paul Ginsparg's "Applied Conformal field theory" and the "Lecture notes in string theory" by Dieter Lüst and Stefan Theisen to be good references. – José Figueroa-O'Farrill Jun 8 2010 at 10:33
2
The normal-ordered product between two fields $A,B$ say, is unambiguously defined as follows. You compute the operator product expansion $A(z)B(w)$. This is a Laurent series in $z-w$ whose coefficients are fields at $w$. You discard the polar piece and take the limit $z\to w$. The resulting field is by definition the normal ordered product $:AB:(w)$. The normal order product of any number of fields can be done by iterating the above. It's best not to think in term of Wick contractions, with which it agrees only for free fields. – José Figueroa-O'Farrill Jun 9 2010 at 23:03
show 1 more comment | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9383249282836914, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/281274/why-is-the-petersson-inner-product-positive-definite?answertab=votes | Why is the Petersson inner product positive definite?
The Petersson inner product is defined on the space $\mathcal{S}_k(\Gamma)$ of weight $k$ cusp forms of level $\Gamma$, and takes values in $\mathbb{C}$.
First of all, I wonder: what does it mean for a complex-valued inner product to be positive definite?
Then, can anyone show me why the Petersson inner product is positive definite?
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1
Positive definite means that $\langle x,x \rangle \ge 0$ for all $x$ and equality holds iff $x = 0$. That Petersson inner product is positive definite should be self evident. – Sanchez Jan 18 at 8:37
2
I cleaned up the spelling and grammar in your post. If you're expecting people to take the trouble to answer your question, you should take a little more trouble over asking it. – David Loeffler Jan 18 at 11:33
@DavidLoeffler, sorry, I will pay attention to it. – hxhxhx88 Jan 18 at 16:05
@Sanchez, maybe I will have once more try, thanks. – hxhxhx88 Jan 18 at 16:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9037631750106812, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/55055/list | ## Return to Answer
2 added 319 characters in body
I'm too lazy to type-up the proof myself, so I'll send you to a reference.
Chang, S.-Y. A., Wang, L. and Yang, P. C. (1999), "Regularity of harmonic maps". CPAM has the proof in Section 3. Once you get $C^{1,\gamma}$ you immediately get RHS is in $C^\gamma$ and the rest follow by standard elliptic regularity.
Note that the structure of the equation (RHS being of the form $d(u\cdot du)$) is only used for Wente's lemma. For the upgrade of regularity one uses a Caccioppoli type inequality.
(BTW, the Chang-Wang-Yang result bypasses the Hardy space estimates. For that the result can be found in the original paper of Helein, though I'd guess the material is also in his book if you don't read French.)
1
I'm too lazy to type-up the proof myself, so I'll send you to a reference.
Chang, S.-Y. A., Wang, L. and Yang, P. C. (1999), "Regularity of harmonic maps". CPAM has the proof in Section 3. Once you get $C^{1,\gamma}$ you immediately get RHS is in $C^\gamma$ and the rest follow by standard elliptic regularity.
Note that the structure of the equation (RHS being of the form $d(u\cdot du)$) is only used for Wente's lemma. For the upgrade of regularity one uses a Caccioppoli type inequality. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9259200096130371, "perplexity_flag": "middle"} |
http://mathhelpforum.com/discrete-math/187854-determine-relation-transitive-print.html | # Determine is relation is transitive
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• September 12th 2011, 01:57 PM
tangibleLime
Determine is relation is transitive
Problem:
$nRm$ in $\mathbb{Z}$ if $nm > 0$. Determine if this is an equivalence relation.
------------
I know that an equivalence relationship requires it to be reflexive, symmetric and transitive. I know how to do reflexive and symmetric, but I'm having some trouble with transitive. I think I may have gotten it, or maybe I'm close.
Here's my attempt:
First introduced $p$ as a third variable for the transitive test. I said that if $nRp$ and $pRm$ were true, then $np>0$ and $mp>0$.
$(nRp \land pRm) \to (np>0 \land pm>0)$
Since $np>0$ as proven in the first step by assuming $nRp$, and likewise for $mp$, n, p and m are all not zero.
$(np>0) \to (n \neq 0 \land p \neq 0)$
$(mp>0) \to (m \neq 0 \land p \neq 0)$
And therefore $nRp$ and $pRm$ are both true. Referring to the definition of a transitive relation that says $(xRy \land yRz) \to (xRz)$, I say that this IS a transitive relation because both of the required relations in the implication of the definition (replacing x with n, y with p and z with m) are true, and therefore xRz is true, or in my case, nRm, which it what I was out to prove.
But now that I look at it more closely... what if p was negative? Then this wouldn't be transitive. I think I'm assuming something I'm not supposed to assume.
I was looking at an example in my book, and they are proving that it's not transitive if it's simply nm >= 0. They said that if aRb and bRc, then ab>=0 and bc>=0. Thus acb^2 >= 0. What allowed them to combine those two expressions to get that b^2? If I know that, I think I can just say that (in my problem) p^2 must be > 0 because all numbers squared are positive except for zero, but I've already proved that it's not zero.
A push in the right direction would be perfect. Please don't just spit out the answer :)
Thanks.
• September 12th 2011, 02:40 PM
Plato
Re: Determine is relation is transitive
Quote:
Originally Posted by tangibleLime
Problem:
$nRm$ in $\mathbb{Z}$ if $nm > 0$. Determine if this is an equivalence relation.
Is this true $0R0~?$
If no then why do any more?
• September 12th 2011, 02:49 PM
TheChaz
Re: Determine is relation is transitive
Quote:
Originally Posted by Plato
Is this true $0R0~?$
If no then why do any more?
It might be a useful exercise...
• September 12th 2011, 03:14 PM
Plato
Re: Determine is relation is transitive
Quote:
Originally Posted by TheChaz
It might be a useful exercise...
But that is not what was posted.
I think that it is important to deal only what was actually posted.
Otherwise, this forum could dissolve into wild guessing.
• September 12th 2011, 03:40 PM
TheChaz
Re: Determine is relation is transitive
Quote:
Originally Posted by Plato
But that is not what was posted.
I think that it is important to deal only what was actually posted.
Otherwise, this forum could dissolve into wild guessing.
...or into speculation about the future of the forum :)
I think the exercise would be useful.
• September 12th 2011, 04:29 PM
tangibleLime
Re: Determine is relation is transitive
Oooh, I think I was assuming that we were supposed to assume that nm > 0 was true. Now I see that it's not necessarily true... and if it's false, then something like 0R0 can arise... correct?
• September 12th 2011, 04:58 PM
TheChaz
Re: Determine is relation is transitive
There is a time to make such an assumption.
Usually, we prove that - FOR ALL elements "x" in the set - xRx.
This is not true, since 0R0 <=> (0*0 > 0)
Since reflexivity fails, there is no need to continue.
However, suppose that we are dealing with the integers (or reals) except (set difference) 0.
Then it is indeed true that for all nonzero integers "x", xRx since x*x > 0
For symmetry, we need to prove that
(xRy) => (yRx)
And HERE we DO assume the hypothesis (that x is related to y) and show that the conclusion (yRx) is true
Assume xRy <=> x*y > 0
Well, since integers commute in multiplication, this is equivalent to (y*x > 0) <=> yRx as desired.
For transitive, we would assume that xRy AND that yRz
xy > 0 AND yz > 0
We multiply these together to get
xy*yz > 0
x(y^2)z > 0
Since y^2 is positive, so is xz. QED
• September 13th 2011, 02:53 AM
emakarov
Re: Determine is relation is transitive
Quote:
Originally Posted by tangibleLime
Oooh, I think I was assuming that we were supposed to assume that nm > 0 was true.
Without saying what n and m are, this statement is meaningless. It is true for some n, m and false for others. Usually when people say that some statement P(n, m) is true without specifying n and m, they mean "For all n and m it is the case that P(n, m) is true." But if you mean this, then the relation R is total, i.e., nRm holds for all n and m. Then it is automatically reflexive, symmetric and transitive (e.g., transitivity holds just because the conclusion is true regardless of the premise).
Moral: always define the objects you use (like n and m above).
• September 13th 2011, 02:59 AM
Plato
Re: Determine is relation is transitive
Quote:
Originally Posted by emakarov
Without saying what n and m are, this statement is meaningless.
Moral: always define the objects you use (like n and m above)
Actually it was defined in the OP.
Quote:
Originally Posted by tangibleLime
Problem:
$nRm$ in $\color{blue}\mathbb{Z}$ if $nm > 0$. Determine if this is an equivalence relation.
• September 13th 2011, 03:10 AM
emakarov
Re: Determine is relation is transitive
I am still not sure which n and m in Z the OP had in mind when he/she said "I was assuming that we were supposed to assume that nm > 0." Either "nm > 0" is not a proposition and cannot be assumed, or it stands for "for all n and m, nm > 0," i.e., "for all n and m, nRm." In the latter case, as I said, the problem is trivial since a total relation is an equivalence relation.
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The photon tag has no wiki summary.
learn more… | top users | synonyms
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http://physics.stackexchange.com/questions/39526/geodesic-equation-from-energy-momentum-conservation | # Geodesic Equation from energy-momentum conservation
I've been reading the excelent review from Eric Poisson found here. While studying it I stumbled in a proof that I can't make... I can't find a way to go from Eq.(19.3) to the one before Eq.(19.4) (which is unnumbered).
I've been able to do some progress (which I present below), but can't get the right answer... Please somebody help me. It's getting really frustating...
Thank you
Given the energy-momentum tensor$$T^{\alpha\beta}\left(x\right)=m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\left(x,z\right)g^{\beta}{}_{\nu}\left(x,z\right)\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)d\lambda,}$$
one can take it's divergence\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda+m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\beta}\left[\delta_{4}\left(x,z\right)\right]d\lambda.}} \end{alignedat}
But using Eq.13.3 of the reference one finds that the divergence of the energy-momentum tensor is also given by\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]g^{\beta}{}_{\nu}\delta_{4}\left(x,z\right)d\lambda-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}} \end{alignedat}
Which means that$$m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\beta}\left[\delta_{4}\left(x,z\right)\right]d\lambda=-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}}$$
so it must be zero (please correct me if I'm wrong).
Then, using Eqs.(5.14) and (13.3), the divergence of the energy-momentum tensor is simply\begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\frac{D}{d\lambda}\left[\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda}+\\ & \,\,\,+m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}\dot{z}^{\mu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\beta}\left[g^{\beta}{}_{\nu}\dot{z}^{\nu}\right]\delta_{4}\left(x,z\right)d\lambda.} \end{alignedat}
If what I've done is correct then comparing with the reference's result the last term must be zero. Can anyone think why?
I thought that, since the covariant derivative is taken in the point $x$ then $g^{\beta}{}_{\nu}\nabla_{\beta}\dot{z}^{\nu}$ is zero but then, what forbids $g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\nu}\nabla_{\beta}\dot{z}^{\mu}$ to be equally zero?
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## 1 Answer
I will abuse notation a little, but I hope you don't find it terrible. After all, I only abuse notation: not children!
Trick 1: Parametrize by Proper Length. We will pick for our affine parameter $\lambda=s$ the proper length. Then the stress energy tensor becomes $$\tag{1}T^{\alpha\beta}(x)=m\int_{\gamma}u^{\alpha} u^{\beta}\frac{\delta^{(4)}\bigl(x,z(s)\bigr)}{\sqrt{|g|}}\,\mathrm{d}s$$ where $u^{\alpha}=\mathrm{d}x^{\alpha}/\mathrm{d}s$ and $g=\det{g_{\mu\nu}}$.
Trick 2: Covariant Derivative Trick. We can write $$\nabla_{\mu}f^{\mu}=\frac{1}{\sqrt{|g|}}\partial_{\mu}(\sqrt{|g|}f^{\mu})$$ for arbitrary $f^{\mu}$.
Exercise: Using Poisson's notation (13.2), we have $$\delta(x,x') = \frac{\delta^{(4)}(x-x')}{\sqrt{|g|}} = \frac{\delta^{(4)}(x-x')}{\sqrt{|g'|}}$$ and thus using our Covariant Derivative trick, find $$\nabla_{\mu}\delta(x,x')=???$$
This will tell you that $$\int u^{\alpha}u^{\beta}\nabla_{\alpha}\delta(x,x')\,\mathrm{d}s = \mbox{boundary terms}$$ and thus we can ignore it.
Remark 1. You are in error writing \begin{alignedat}{1}\nabla_{\beta}T^{\alpha\beta} & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\delta_{4}\left(x,z\right)\right]d\lambda}=\\ & =m{\displaystyle \int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta_{4}\left(x,z\right)d\lambda-m{\displaystyle \int_{\gamma}\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\nabla_{\nu}\left[\delta_{4}\left(x,z\right)\right]d\lambda,}} \end{alignedat} This should have been a simple application of the product rule. That is, the minus sign should be a plus sign.
Remark 2. Why should we expect the right hand side of $\nabla_{\beta}T^{\alpha\beta}=0$? Well, because using Einstein's field equation it's $\nabla_{\beta}G^{\alpha\beta}$ and this is identically zero.
This is why we set $$\tag{2}m\int_{\gamma}\nabla_{\beta}\left[\frac{g^{\alpha}{}_{\mu}g^{\beta}{}_{\nu}\dot{z}^{\mu}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\nu}}}\right]\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$ ...which is precisely the geodesic equation for a point particle as discussed in Poisson's article section 3.
Edit We can rewrite (2) since ${g^{\alpha}}_{\beta}={\delta^{\alpha}}_{\beta}$ is the Kronecker delta. So $$\tag{3}m\int_{\gamma}\nabla_{\nu}\left[\frac{\dot{z}^{\alpha}\dot{z}^{\nu}}{\sqrt{-g_{\alpha\beta}\dot{z}^{\alpha}\dot{z}^{\beta}}}\right]\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$ But if we pick the arclength as the parameter, this becomes simply $$\tag{4}m\int_{\gamma}\nabla_{\nu}(u^{\alpha}u^{\nu})\delta\bigl(x,z\bigr)\,\mathrm{d}\lambda=0.$$ Great, but really is $$\tag{5}\nabla_{\nu}(u^{\alpha}u^{\nu})=0?$$ Lets recall for a geodesic using arclength parametrization we have $$u_{\mu}u^{\mu}=1\implies u_{\mu}\nabla_{\nu}u^{\mu}=0.$$ Thus (5), when contracted by a non-negative vector (say $u_{\alpha}$) becomes \begin{alignedat}{1}u_{\alpha}\nabla_{\nu}(u^{\alpha}u^{\nu}) &= u_{\alpha}\underbrace{u^{\nu}\nabla_{\nu}u^{\alpha}}_{=0} + \underbrace{u_{\alpha}u^{\alpha}}_{=1}\nabla_{\nu}u^{\nu}\\ &=\nabla_{\nu}u^{\nu}\end{alignedat} But this is a continuity-type equation (and if you use trick 2, it really resembles electromagnetism's continuity equation!).
Now we can go back, and by inspection we find $$\nabla_{\nu}(u^{\alpha}u^{\nu})=\left(\begin{array}{c}\mbox{Geodesic}\\ \mbox{Equation}\end{array}\right)+\left(\begin{array}{c}\mbox{Continuity}\\ \mbox{Equation}\end{array}\right).$$ This is, of course, $\nabla_{\mu}T^{\mu\nu}$. Why should we expect it to be zero?
Well, if the Einstein field equations hold, then $$G^{\mu\nu}-\kappa T^{\mu\nu}=0$$ and moreover $$\nabla_{\mu}(G^{\mu\nu}-\kappa T^{\mu\nu})=0.$$ However, $\nabla_{\mu}G^{\mu\nu}=0$ identically thanks to geometry.
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Oh I'm sorry I wrote my equations wrong. I copy pasted them and forgot to edit. I wasn't understanding what you meant by product rule since I've done it in the first divergence. It was because the second time I took the divergence I wrote the equation wrong. I believe that what you did is more rigorous but what I did at the very least isn't wrong. I find the same equation you did in the end but my problem is to figure how come that is the geodesic equation... Thank you for your time – PML Oct 11 '12 at 8:25
@PML: I've updated my post with reasoning I was taught as far as "conservation law for point particle gives the geodesic equation". It answers your question, using a few tricks and slightly different notation; the short answer is: continuity type equation + geodesic equation = that last equation you've got. – Alex Nelson Oct 11 '12 at 13:44
Thank you for your time and explaination. To tie things up: after your very nice presentation you basically say that the expression before Eq.(19.4), which is unnumbered, in Poisson review misses the continuity equation, right? – PML Oct 11 '12 at 14:16
Right, which is why it appears as "bonus parts" which --- unless explained as a continuity equation --- seems problematical. Extra Point: when you include the electromagnetic field coupled to the point particle, you get the Lorentz force for free from the conservation of the stress-energy tensor. This miracle should be worked out by the reader! – Alex Nelson Oct 11 '12 at 16:04
This is indeed excinting and really appears out of nowhere! Was that result known?I'm going to try to work out the details of what you're saying. – PML Oct 11 '12 at 16:18
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http://math.stackexchange.com/questions/212031/what-is-lattice-with-arbitrary-suprema-called?answertab=oldest | # What is lattice with arbitrary suprema called?
Topology is a system of sets, which is closed under arbitrary unions and finite intersections.
Is there a name for lattice, which has analogous properties, i.e., every subset has supremum. (And also every finite subset has infimum, but this is already in the definition of lattice.
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## 2 Answers
A lattice with suprema for all subsets is called a complete lattice, or perhaps a cocomplete lattice if you are category-theoretically inclined. Here's a fun fact: any lattice that has suprema for all subsets also has infima for all subsets. The infimum of a subset is, of course, the supremum of all the lower bounds for that subset.
Unfortunately, a homomorphism of lattices need not preserve infinite suprema/infima, and here is where the subtlety is: for example, when one says "(co)complete join semilattice", one is thinking of homomorphisms that preserves all suprema, even though a (co)complete join semilattice is automatically a complete lattice. A continuous map of topological spaces induces a homomorphism on the lattice of open sets that preserves finite meets and infinite joins, which is precisely the definition of a homomorphism of frames. (A frame is a complete lattice that satisfies the infinite distributive law.)
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Normally one defines a complete lattice to be one that has suprema and infima for all subsets. – Brian M. Scott Oct 13 '12 at 7:45
You fun fact is a relatively easy observation, I should have noticed that. But I'm glad I haven't - since otherwise I would not learn from your answer about relation of this question to frames. – Martin Sleziak Oct 13 '12 at 7:51
1
– Martin Sleziak Oct 16 '12 at 11:31
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$\newcommand{\int}{\operatorname{int}}$If $\langle X,\tau\rangle$ is a topological space, $\langle\tau,\subseteq\rangle$ is a complete distributive lattice: for any $\mathscr{U}\subseteq\tau$, $\bigvee\mathscr{U}=\bigcup\mathscr{U}$, and $\bigwedge\mathscr{U}=\int\bigcap\mathscr{U}$, and for any $U,V,W\in\tau$ we have $$U\land(V\lor W)=U\cap(V\cup W)=(U\cap V)\cup(U\cap W)=(U\land V)\lor(U\land W)$$ and $$U\lor(V\land W)=U\cup(V\cap W)=(U\cup V)\cap(U\cup W)=(U\lor V)\land(U\lor W)\;.$$ What distinguishes the two is infinite distributivity: for any $V\in\tau$
$$V\land\bigvee\mathscr{U}=V\cap\bigcup\mathscr{U}=\bigcup_{U\in\mathscr{U}}(V\cap U)=\bigvee_{U\in\mathscr{U}}(V\land U)\;,$$
but it’s not true in general that
$$V\lor\bigwedge\mathscr{U}=V\cup\int\bigcap\mathscr{U}\overset{?}=\int\bigcap_{U\in\mathscr{U}}(V\cup U)=\bigwedge_{U\in\mathscr{U}}(V\lor U)\;.$$
What you really want to talk about, I suspect, are frames; see also pointless topology (a name that I occasionally find regrettably apt!).
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9407632350921631, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/1321/how-does-the-wider-cryptographic-community-view-non-abelian-group-based-cryptogr?answertab=votes | # How does the wider cryptographic community view non-abelian group based cryptography?
Is there perhaps some neural expository article on crypto systems based on non-abelian groups?
I've gleaned that Anshel–Anshel–Goldfeld key exchange is the most well-known cryptographic algorithm based upon non-abelian groups. Do people consider it respectably secure? Why? Why not?
Any opinions about the role of combinatorial group theory here?
I'm curious because, as I understand it, non-abelian finite groups aren't considered particularly useful in cryptography, due to the Coincident Group Orders Theorem that follows from the Classification of the Finite Simple Groups, but obviously infinite non-abelian groups are another matter.
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– Paŭlo Ebermann♦ Nov 26 '11 at 1:32
– Jeff Burdges Nov 26 '11 at 2:54
– Jeff Burdges Nov 26 '11 at 3:05
1
Did you try to google with following terms: "non-abelian group" site:eprint.iacr.org ? – jug Nov 28 '11 at 19:11
1
– jug Nov 28 '11 at 19:14
show 3 more comments
## 2 Answers
In my experience, I never have found that cryptographers base their opinion of a cryptosystem on the properties of the underlying group. If its a braid group, abelian group, or finite field: that does not really matter. What matters is, as @Thomas notes, how hard do we think the problem is in a particular setting?
Cryptography in braid groups usually has security reductions to problems related to the simultaneous conjugacy search problem. This problem has definitely not been studied as closely as problems related to integer factorization or discrete logarithms; and it is likely less studied than finding multiplicands on elliptic curves or the shortest vector problem on ideal lattices.
The community warms up to an idea after "adequate" attention has been given to the underlying hard problem. What is "adequate" is a mix of time and attention. It also helps to have a company pushing and standardizing it (Certicome with ECC, NTRU with lattices). It can also help gather attention if there is some "killer app" other than efficiency (ECC was helped by pairings, lattices will likely be helped by fully homomorphic encryption).
Braid group cryptography has none of these. Its main selling point is efficiency, which is a tough sell. It used to be embedded systems and mobile devices, then smartcards, and now RFIDs and sensor networks. As technology gets better, small computational devices become more capable of implementing standard cryptography.
Further, protocols like AAG have had a number of attacks against them requiring further refinement of how parameters are chosen. This isn't necessarily devastating: it could be viewed as akin to moving to safe primes or away from certain curves, or it could be a sign of deeper problems.
To answer your questions directly... I am not sure what you mean by a neural exploratory article. I don't think many cryptographers consider AAG secure or broken; the jury is still out (and no sign of them coming back any time soon). I don't think the group theory it is based on has any role in people's opinion (other than how the group theory dictates the hardness of conjugacy search problems).
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Almost all cryptographic algorithms which use groups actually work in subgroups generated by a conventional element; even if the group as a whole is non-abelian, the subgroup is cyclic, thus abelian. The Anshel-Anshel-Goldfeld protocol tries to use non-commutativity itself, and relies on "how much non-abelian" the group is.
All asymmetric cryptographic algorithms rely at some point on a presumably hard problem. We do not know whether hard problems really exist, neither asymptotically (that's the whole P=NP problem) nor in practice (when given a specific computational budget, e.g. $2^{80}$ elementary operations). Our only tool is accumulation of work: we take many cryptographers and mathematicians, we let them loose on the problem for a few decades, and we see if they find some way to solve the problem.
To be honest, I had never heard of the Anshel-Anshel-Goldfeld protocol before this morning. It is described in an article which was published in Mathematical Research Letters, which is not usually associated with cryptologic research. There is a bit of research on doing cryptography with braid groups (the AAG protocol is just an instance of that); see this page for a lot of relevant pointers. It is still a recent and widely unexplored area, so, regardless of its inherent merits, the AAG protocol cannot be deemed secure yet. Also, it is only a framework, explained generically over an unspecified non-abelian group; it cannot be declared secure in abstracto.
Infinite non-abelian groups, or anything infinite for that matter, have a slight practical issue: they do not fit well in actual computers which have only a finite amount of RAM. Even for infinite sets where each element can be encoded in a finite (but unbounded) amount of RAM, such as plain integers, variable size easily leads to side-channel leakage, and that's a bad thing.
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Interesting thank! I'd naively assume any implementation would use only a bounded fragment of the infinite object for computational reasons, which avoids that side channel attack. You wouldn't be creating a finite group by imposing this bound, but maybe the CFSG still worms it's way into the picture somehow. – Jeff Burdges Nov 25 '11 at 15:46
It's worth pointing out that we have no proof that factoring, for example, is NP. All that matters is the problem is "hard." The problem's complexity class is a complementary discussion. – Simon Johnson Nov 29 '11 at 7:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9483715891838074, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/149585/is-this-a-good-way-to-explicate-skolems-paradox/149588 | # Is this a good way to explicate Skolem's Paradox?
Skolems Paradox shows an ostensible conflict between Cantor's Thoerem (CT) and the downward Löwenheim–Skolem Theorem (ST).
CT: for any set $A$, the powerset of $A$, $P(A)$, has a strictly greater cardinality than $A$. Cardinality is in terms of bijections: two sets have the same cardinality iff there exists a bijection between them. A set $A$ is countable iff there exists a objection between $A$ and the set of naturals $\omega$. Since (by CT) no function surjects $\omega$ onto its powerset, we learn that $P(\omega)$ is uncountable. Thus CT generally tells us some sets are uncountable.
ST: If a countable first-order theory has an infinite model, it has a countable model. The standard axiomatization of set theory, ZFC, is such a theory.
Assume ZFC has a model (which must be infinite).
By ST, ZFC has a countable model 𝔐.
By CT, we can deduce the existence of uncountable sets from ZFC.
Therefore, there must be some set A ∈ 𝔐 such that 𝔐 satisfies $A$ is uncountable. That is, there is no bijection $f \in$ 𝔐 between $A$ and $\omega$.
However, since 𝔐 has a countable domain, there are only countably many elements available to be members of A.
Thus A appears both countable and uncountable.
1. How did first-order logic come to be the dominant formal logic? (and comments)
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## 2 Answers
Indeed $\mathfrak M$ sees only a small fraction of the universe, in particular he does not see the bijection between $\omega$ and itself, or between $\omega$ and some of the members of $\mathfrak M$.
However it is perfectly fine to have a model which is countable and has elements which are uncountable.
If you start by taking the real numbers and using it to generate a countable model, you will end up with a countable model of ZFC (of course it would not know of its own countability) in which there is a really uncountable set.
Now recall that $\mathfrak M\models A\text{ is countable}$ if and only if there is $f\in\mathfrak M$ which is a bijection between $A$ and $\omega$. If $\mathfrak M$ does not know about such $f$, $\mathfrak M\models A\text{ is uncountable}$. This is the heart of internal-external points of view.
If, on the other hand, we require $\mathfrak M$ to be transitive then every element of $\mathfrak M$ has to be countable as well (since it is a subset of $\mathfrak M$ by transitivity).
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I changed the OP a bit, is it OK? I fear "That is, A in 𝔐 is (at most) countable" might be wrong UNLESS I stipulated that 𝔐 was transitive. – pichael Jun 4 '12 at 18:45
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@pichael: It is fine, however as long as you do not assume that $\frak M$ is transitive there is no guarantee that every element of $\frak M$ is countable. – Asaf Karagila Jun 4 '12 at 18:50
So basically you have been saying that my explanation assumes that $M$ is transitive. What would I have to add/subtract from my explanation to include non-transitive models of ZFC? – pichael Jun 4 '12 at 23:23
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@pichael: If you look at the Skolem-Lowenheim theorem you will see that it does not only guarantee us a countable model, but also it allows us to point at a certain element and ensure that this element would be in the model. Now suppose we have a model in which there is a truly uncountable set $A$; we can consider the countable model in which $A$ is a member. Obviously this model cannot be transitive, as that would imply $A$ is also a subset of that model, which is a contradiction. So you have $A$ which both the universe and the countable model agree is uncountable... – Asaf Karagila Jun 5 '12 at 4:46
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No, the set $A$ is the same set. The cardinality of the set changes. It may be the case that the set has some specific role (being $\omega$, being $\omega_1$, being $\cal P(\omega)$, etc.) which may also change between models. However as sets these are the same, especially when you take submodels of "good" models (i.e. transitive and well-founded models, which may be very large and not just countable). – Asaf Karagila Jun 5 '12 at 4:55
show 8 more comments
A set is countably infinite if it is in bijection to $\mathbb{N}$. If you have a countable model, a set can be uncountably from the "point of view" of the model simply because the model doesn't contain the bijection. So the model doesn't see the bijection, and a set is seen as uncountable.
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Can I give a kind of example for your comment: Let M be a model that satisfies "A is countable." As you say, this just means there is a bijection in M consisting of N and A. Let's throw away all such bijections between A and N and call this new model M'. Thus M sees A as countable, M' sees A as uncountable. – pichael May 25 '12 at 10:49
@pichael: Of course you have to be careful when you say such things. If $A$ is a set which is provably countable (for example, $\omega$ itself) then $M'$ would not be a model of ZFC. – Asaf Karagila May 25 '12 at 10:50
@AsafKaragila: I must be missing something crucial here, as that is not immediately obvious to me. What is M' not satisfying? Pairing? – pichael May 25 '12 at 10:59
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@pichael: You will note that I refrain from using $\mathbb N$. This is because $\mathbb N$ is a mathematical idea which is interpreted often as a model of PA which in ZF is interpreted canonically as $\omega$ along with ordinal operations. The same goes for $\mathbb Z$ or $\mathbb Q$ and so on. All of these can be interpreted as any countably infinite set. $\omega$, on the other hand, is a very concrete set in ZFC - it is the collection of finite ordinals. And yes, everything definable from $\omega$ also has to stay if you want $M'$ to be a submodel. – Asaf Karagila May 25 '12 at 11:08
1
@pichael: It means sets which can be defined from $\omega$. For example the set of finite non-zero ordinals; the ordinal $\omega+1$; the set of ordinals which are decomposable in multiplication; etc. – Asaf Karagila May 25 '12 at 11:23
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http://unapologetic.wordpress.com/2008/04/25/convergence-tests-for-infinite-series/?like=1&_wpnonce=81a70c627a | # The Unapologetic Mathematician
## Convergence Tests for Infinite Series
Now that we’ve seen infinite series as improper integrals, we can immediately import our convergence tests and apply them in this special case.
Take two sequences $a_k$ and $b_k$ with $b_k\geq a_k\geq0$ for all $k$ beyond some point $N$. Now if the series $\sum_{k=0}^\infty a_k$ diverges then the series $\sum_{k=0}^\infty b_k$ does too, and if the series $\sum_{k=0}^\infty b_k$ converges to $b$ then the series of $\sum_{k=0}^\infty a_k$ converges to $a\leq b$.
[UPDATE]: I overstated things a bit here. If the series of $b_k$ converge, then so does that of $a_k$, but the inequality only holds for the tail beyond $N$. That is:
$\displaystyle\sum\limits_{k=N}^\infty a_k\leq\sum\limits_{k=N}^\infty b_k$
but the terms of the sequence $a_k$ before $N$ may, of course, be so large as to swamp the series of $b_k$.
If we have two nonnegative sequences $a_k$ and $b_k$ so that $\lim\limits_{k\rightarrow\infty}\frac{a_k}{b_k}=c\neq0$ then the series $\sum_{k=0}^\infty a_k$ and $\sum_{k=0}^\infty b_k$ either both converge or both diverge.
We read in Cauchy’s condition as follows: the series $\sum_{k=0}^\infty a_k$ converges if and only if for every $\epsilon>0$ there is an $N$ so that for all $n\geq m\geq N$ the sum $\left|\sum_{k=m}^n a_k\right|<\epsilon$.
We also can import the notion of absolute convergence. We say that a series $\sum_{k=0}^\infty a_k$ is absolutely convergent if the series $\sum_{k=0}^\infty|a_k|$ is convergent (which implies that the original series converges). We say that a series is conditionally convergent if it converges, but the series of its absolute values diverges.
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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. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 30, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9238577485084534, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/61155/normal-domains-with-algebraically-closed-quotient-field | normal domains with algebraically closed quotient field
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am looking for an integral domain $A$ with the following properties:
1. $A$ is not integrally closed
2. $A$ has a quotient field $K$ that is algebraically closed and that has characteristic 0
3. There is an integral element $x\in K$ (over $A$) such that $A[x]$ is integrally closed.
Can someone help to tell me if the above is even possible?
Edit: Lubin easily gave me an example. Now I want to consider the case when I replace the condition 2. by:
2'. $A$ has a quotient field $K$ that is real closed.
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I don't understand what "of $A$" means in condition 3. – Gerry Myerson Apr 10 2011 at 0:07
an integral element of $A$ that is a member of $K$ – Jose Capco Apr 10 2011 at 11:21
@Jose, but if $x$ is an element ${\it of\ }A$ then $A[x]$ is just $A$. Maybe you mean $x$ is integral ${\it over\ }A$? – Gerry Myerson Apr 10 2011 at 22:25
yes integral over $A$. Thanks for the correction :) ... english is just too difficult :) – Jose Capco Apr 11 2011 at 4:25
Remark: assume $A$ fullfills all the requirements and let $m$ be a maximal ideal of $A$. There exists a maximal ideal $n$ of $B:=A[x]$ lying over $m$. Then $B/n$ is algebraically closed. On the other hand $B/n = A/m [x+n]$, hence by Schreier's theorem either $A/m = B/n$ or $A/m$ is real closed. – Hagen Apr 12 2011 at 7:28
show 1 more comment
1 Answer
Try this: Let $B_0$ be the ring of real algebraic integers, and let $B=B_0[1/2]$, so the ring of real algebraic numbers integral except possibly at $2$. But $B[i]$ is equal to the ring of algebraic numbers integral except possibly at $2$, and this is integrally closed. And so we take $A=B[3i]$, not integrally closed, and of course the fraction field is algebraically closed.
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This looks really nice! – Todd Trimble Apr 9 2011 at 22:47
Wow, yes indeed. Can we do the same thing if the field were real closed? – Jose Capco Apr 10 2011 at 11:09
1
I'll bet a nickel that that's not possible. But no more than that! – Lubin Apr 10 2011 at 16:36 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 35, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9290644526481628, "perplexity_flag": "head"} |
http://quant.stackexchange.com/questions/tagged/probability+black-scholes | # Tagged Questions
1answer
194 views
### What are $d_1$ and $d_2$ for Laplace?
What are the formulae for d1 & d2 using a Laplace distribution?
1answer
310 views
### Simulating the joint dynamics of a stock and an option
I want to know the joint dynamics of a stock and it's option for a finite number of moments between now and $T$ the expiration date of the option for a number of possible paths. Let $r_{\mathrm{s}}$ ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9007338881492615, "perplexity_flag": "middle"} |
http://mathhelpforum.com/number-theory/202025-can-anyone-help-me-one.html | # Thread:
1. ## Can anyone help me with this one?
How many multiples of 9 are there in the open interval (π^5, π^6)?
2. ## Re: Can anyone help me with this one?
Hello, musngiburger!
Are you waiting for someone to provide a formula?
It is just requires a little Thinking.
How many multiples of 9 are there in the open interval $(\pi^5,\,\pi^6)\,?$
Use your calculator: . $\begin{Bmatrix}\pi^5 &=& 306.019... \\ \pi^6 &=& 961.389... \end{Bmatrix}$
The interval contains the integers:. $\{307, 308, 309\hdots 961\}$ . . . 655 integers.
One-ninth of them are multiple of 9: . $\left\lfloor \frac{655}{9}\right\rfloor \;=\;72$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8760600090026855, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/227005/the-integrals-from-1-to-infty-for-dfrac1x-and-dfrac1x2?answertab=active | # The integrals from $1$ to $\infty$ for $\dfrac{1}{x}$ and $\dfrac{1}{x^2}$
I have two integrals:
$$A=\int\limits_1^\infty \dfrac{1}{x}dx\,, B=\int\limits_1^\infty \dfrac{1}{x^2}dx$$
Calculus says that A is an improper integral as it diverges, but the B converges and is $1$, because $\dfrac{1}{x^2}$ is faster near $y=0$ than $\dfrac{1}{x}$.
I don't understand the reason behind this. So I looked for another way to put down my problem. Multiple sources define that:
$$\frac{1}{\infty} = \frac{1}{\infty^2}$$
What is the reason that the integral of $\dfrac{1}{x}$ is divergent and $\dfrac{1}{x^2}$ is convergent? In the end they both reach $\dfrac{1}{\infty}$ (or $\dfrac{1}{\infty^2}$ which is $\dfrac{1}{\infty}$).
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They're both improper (Riemann) integrals. $A$ is a divergent improper Riemann integral and $B$ is a convergent one. The Riemann integral is only defined for (certain) bounded functions on bounded intervals. They're called improper because they're limits (in a suitable sense) as the (in this case) upper integration boundary "goes to infinity". – kahen Nov 1 '12 at 20:14
What happens at $\infty$ doesn't matter. It's what happens "near" $\infty$ that matters: i.e. how the function behaves for large (finite) real values. – Hurkyl Nov 1 '12 at 20:20
## 4 Answers
The answer has nothing to do with '$\infty$ calculus'. Calculate the finite integral first, and take limits.
$\int_1^x \frac{1}{t} dt = \ln x$, $\int_1^x \frac{1}{t^2} dt = 1-\frac{1}{x}$.
$A = \lim_{x \to \infty} \ln x = \infty$, $B =\lim_{x \to \infty} 1-\frac{1}{x} = 1$.
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Thanks! This is exactly what I was looking for. – Jarvix Nov 1 '12 at 20:19
\begin{eqnarray} \int\limits_1^\infty \dfrac{1}{x}dx & = & \lim_{ \varepsilon \rightarrow \infty } \int\limits_1^\varepsilon \dfrac{1}{x}dx & = & \lim_{ \varepsilon \rightarrow \infty } \log \varepsilon = \infty \end{eqnarray} While \begin{eqnarray} \int\limits_1^\infty \dfrac{1}{x^2}dx & =& \lim_{ \varepsilon \rightarrow \infty } \Bigr(-\dfrac{1}{\varepsilon } +1 \Bigl) = 1 \end{eqnarray}
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You likely mean $\epsilon \to \infty$, not $\epsilon \to 0$. – gt6989b Nov 1 '12 at 20:30
There is something called Cauchy integral test which tells us that we can compare an infinite sum with a corresponding integral and vice versa under certain circumstances (see the link). Because $$\sum\limits_{n = 1}^\infty {\frac{1}{x}}$$ diverges so does the integral and since $$\sum\limits_{n = 1}^\infty {\frac{1}{{{x^2}}}}$$ converges so does the integral. By the way, infinity is not a number. Saying $\infty^2$ is the same as saying $\text{blue}^2$ -- it does not mean anything, so you cannot look at these integral as fractions with $\infty$ as a denominator.
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(assuming the function $f$ is non-negative, decreasing, and continuous) – Hurkyl Nov 1 '12 at 20:19
@Hurkyl Yes, hence the link. I did not want to write it all out. – glebovg Nov 1 '12 at 20:21
– Jarvix Nov 1 '12 at 20:21
@glebovg: It's worth mentioning something conditional in your statement, then. e.g. "... which tells us when we can compare ..." or "... which tells us that we can compare ... if certain properties are met ..." or something. – Hurkyl Nov 1 '12 at 20:22
@Hurkyl Thanks. – glebovg Nov 1 '12 at 20:24
show 1 more comment
Improper integrals, such as $A$ and $B$ are defined as limits: $$A= \int_1^{\infty} \frac{1}{x}dx= \lim_{R \to \infty} \int_1^R \frac{1}{x}dx$$ and $$B= \int_1^{\infty} \frac{1}{x^2}dx= \lim_{R \to \infty} \int_1^R \frac{1}{x^2}dx$$ If you carry out the details you'll get $$\lim_{R \to \infty} \ln R$$ which goes to $\infty$ and $$\lim_{R \to \infty} \frac{1}{R}$$ which converges.
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http://mathhelpforum.com/calculus/24573-challenging-dif-eq.html | Thread:
1. Challenging Dif EQ
Given the following Dif EQ's (linear):
$\textbf{X}'=\left(\begin{array}{rr}-1 & c \\ 1 & -1 \end{array}\right)\textbf{X}$
Determine the values of $c$, where $c$ is a constant, for which every solution $\textbf{X}$ has the property where $\textbf{X}(t)\rightarrow \textbf{0}$ as $t\rightarrow \infty$
2. Is there a determinant property that will solve this?
3. Originally Posted by TKHunny
Is there a determinant property that will solve this?
I was thinking that maybe I have to do the Wronskian on this, but I'm not entirely sure.
4. I plugged this in Maple, and found the eigenvalues and corresponding eigenvectors in terms of c.
The two eigenvalues are:
$-1+\sqrt{c}$
$-1-\sqrt{c}$
The corresponding eigenvectors, respectively, are:
[sqrt(c),1]
[-sqrt(c),1]
The multiplicities of both are 1. Not sure where to go with this...
When c = 0, we have repeated eigenvalues ...
When c is negative, we have complex numbers... I can't imagine complex numbers yielding X(t) -> 0 with t -> infinity ....
Ahh, very challenging.
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http://math.stackexchange.com/questions/9844/how-to-prove-for-any-m-in-z-sum-n-mmn-1-hn-sum-n-0n-1-hn | # How to prove: for any $m \in Z$, $\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$
Suppose $h$ is a function on $Z$ that is periodic with period N, that is $h(n+N)=h(n)$ for all $n$. How to prove: for any $m \in Z$, $\sum_{n=m}^{m+N-1} h(n) =\sum_{n=0}^{N-1} h(n)$. In other words, any sum over an interval of length $N$ yields the same result. This is problem from the area of linear algebra and is a homework
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Why can't you tag it as `linear-algebra` by yourself? – J. M. Nov 11 '10 at 10:46
It might be a problem from linear algebra class but it has nothing to do with linear algebra. Not sure what would be a right tag though. – Marek Nov 11 '10 at 11:09
Maybe you should consider accepting some of the answers to your questions? – AD. Nov 11 '10 at 11:29
@J. M.: Yes he/she could! – AD. Nov 11 '10 at 11:31
## 3 Answers
Let $m \in \{0 \dots, N-1\}$. Then we have $$\sum_{n=m}^{m+N-1} h(n) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n + N) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n) = \sum_{n=0}^{N-1} h(n)$$ Now for $m \in \mathbb{Z}$ write $m = kN + r$ with $k \in \mathbb{Z}$ and $r \in \{0, \dots, N-1\}$. We clearly have $$\sum_{n=kN+r}^{k(N+1) + r -1} h(n) = \sum_{n=r}^{N+r-1} h(n+kN) = \sum_{n=r}^{N+r-1} h(n)$$ Now the last expression equals $\sum_{n=0}^{N-1} h(n)$ by the first part of the argument.
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How do you get $\sum_{n=m}^{m+N-1} h(n) = \sum_{n=m}^{N-1} h(n) + \sum_{n=0}^{m-1} h(n)$? And why you it is must to make argument after the first argument? I mean $\sum_{n=kN+r}^{k(N-1)+r-1} h(n)...$. – alvoutila Nov 11 '10 at 11:47
You just split the sum into two parts: those less than N and those greater or equal to N. The first part is already ok. The second part is in the for h(N+k), so you can use the periodicity assumption to bring it back to h(k). – Marek Nov 11 '10 at 13:23
Regarding my argument... you certainly don't have to do it like that, you are free to come with proof of your own. But this seemed like a simple way to mee: first inspect the case where the summation interval is close to what you want to prove (i.e. [0, N-1]) and then use periodicity to transform any other interval (e.g. [2N +4, 3N + 3], with N > 4) into the one consider in the first part (in this case [4, N + 3]). – Marek Nov 11 '10 at 13:27
HINT:
Expand $$\sum_{n=m}^{m+N-1}h(n)$$ What do you know about $h$?
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The important point here is that any set of $N$ consecutive integers is a complete residue system modulo $N$. See complete residue system. We consider the index modulo $N$ since $h(n)$ has period $N.$ i.e. the numbers $\lbrace m,m+1,m+2,\ldots,m+N-1 \rbrace$ modulo $N$ are $\lbrace 0,1,2,\ldots,N-1 \rbrace$ in some order.
Your identity follows immediately from this fact.
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http://solution-nine.com/The_Mean_Value_Theorem | # The Mean Value Theorem Research Materials
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Rihanna - Take A Bow
Music video by Rihanna performing Take A Bow. YouTube view counts pre-VEVO: 66288884. (C) 2008 The Island Def Jam Music Group.
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Music video by Rihanna performing Rehab. YouTube view counts pre-VEVO: 19591123. (C) 2007 The Island Def Jam Music Group.
Key & Peele: Substitute Teacher
A substitute teacher from the inner city refuses to be messed with while taking attendance.
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Music video by Rihanna performing Unfaithful. (C) 2006 The Island Def Jam Music Group #VEVOCertified on Feb. 15, 2012. http://vevo.com/certified http://youtu...
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Due to so many request I decided to upload this epic track as well. Enjoy it bros!
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"Just One Last Time" feat. Taped Rai. Available to download on iTunes including remixes of : Tiësto, HARD ROCK SOFA & Deniz Koyu http://smarturl.it/DGJustOne...
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YOLO is available on iTunes now! http://smarturl.it/lonelyIslandYolo New album coming soon... Check out the awesome band the music in YOLO is sampled from Th...
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MACKLEMORE & RYAN LEWIS - CAN'T HOLD US FEAT. RAY DALTON (OFFICIAL MUSIC VIDEO)
Macklemore & Ryan Lewis present the official music video for Can't Hold Us feat. Ray Dalton. Can't Hold Us on iTunes: https://itunes.apple.com/us/album/cant-...
Draw My Life- Jenna Marbles
This video accidentally turned out kind of sad, ME SO SOWWY IT NOT POSED TO BE SAD WHO WANTS HUGS AND COOKIES? Also, FYI for anyone attempting this, it takes...
For the theorem in harmonic function theory, see Harmonic function#Mean value property.
Not to be confused with the Intermediate value theorem.
Calculus
• Fundamental theorem
• Limits of functions
• Continuity
• Mean value theorem
• Rolle's theorem
Definitions
Concepts
Rules and identities
Integral calculus
Definitions
Integration by
Formalisms
Definitions
Specialized calculi
For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.
In calculus, the mean value theorem states, roughly: given a planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints.
The theorem is used to prove global statements about a function on an interval starting from local hypotheses about derivatives at points of the interval.
More precisely, if a function f is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that
$f'(c) = \frac{f(b) - f(a)}{b-a} \, .$[1]
This theorem can be understood intuitively by applying it to motion: If a car travels one hundred miles in one hour, then its average speed during that time was 100 miles per hour. To get at that average speed, the car either has to go at a constant 100 miles per hour during that whole time, or, if it goes slower at one moment, it has to go faster at another moment as well (and vice versa), in order to still end up with an average of 100 miles per hour. Therefore, the Mean Value Theorem tells us that at some point during the journey, the car must have been traveling at exactly 100 miles per hour; that is, it was traveling at its average speed.
A special case of this theorem was first described by Parameshvara (1370–1460) from the Kerala school of astronomy and mathematics in his commentaries on Govindasvāmi and Bhaskara II.[2] The mean value theorem in its modern form was later stated by Augustin Louis Cauchy (1789–1857). It is one of the most important results in differential calculus, as well as one of the most important theorems in mathematical analysis, and is essential in proving the fundamental theorem of calculus. The mean value theorem follows from the more specific statement of Rolle's theorem, and can be used to prove the more general statement of Taylor's theorem (with Lagrange form of the remainder term).
## Formal statement
Calculus
• Fundamental theorem
• Limits of functions
• Continuity
• Mean value theorem
• Rolle's theorem
Definitions
Concepts
Rules and identities
Integral calculus
Definitions
Integration by
Formalisms
Definitions
Specialized calculi
Let f : [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a < b. Then there exists some c in (a, b) such that
$f ' (c) = \frac{f(b) - f(a)}{b - a}.$
The mean value theorem is a generalization of Rolle's theorem, which assumes f(a) = f(b), so that the right-hand side above is zero.
The mean value theorem is still valid in a slightly more general setting. One only needs to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit
$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
exists as a finite number or equals +∞ or −∞. If finite, that limit equals f′(x). An example where this version of the theorem applies is given by the real-valued cube root function mapping x to x1/3, whose derivative tends to infinity at the origin.
Note that the theorem, as stated, is false if a differentiable function is complex-valued instead of real-valued. For example, define f(x) = eix for all real x. Then
f(2π) − f(0) = 0 = 0(2π − 0)
while |f′(x)| = 1.
## Proof
The expression (f(b) − f(a)) / (b − a) gives the slope of the line joining the points (a, f(a)) and (b, f(b)), which is a chord of the graph of f, while f′(x) gives the slope of the tangent to the curve at the point (x, f(x)). Thus the Mean value theorem says that given any chord of a smooth curve, we can find a point lying between the end-points of the curve such that the tangent at that point is parallel to the chord. The following proof illustrates this idea.
Define g(x) = f(x) − rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is true for g. We now want to choose r so that g satisfies the conditions of Rolle's theorem. Namely
$\begin{align}g(a)=g(b)&\iff f(a)-ra=f(b)-rb\\ &\iff r(b-a)=f(b)-f(a) \\&\iff r=\frac{f(b)-f(a)}{b-a}\cdot\end{align}$
By Rolle's theorem, since g is continuous and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,
$f '(c)=g '(c)+r=0+r=\frac{f(b)-f(a)}{b-a}$
as required.
## A simple application
Assume that f is a continuous, real-valued function, defined on an arbitrary interval I of the real line. If the derivative of f at every interior point of the interval I exists and is zero, then f is constant.
Proof: Assume the derivative of f at every interior point of the interval I exists and is zero. Let (a, b) be an arbitrary open interval in I. By the mean value theorem, there exists a point c in (a,b) such that
$0 = f'(c) = \frac{f(b)-f(a)}{b-a}.$
This implies that f(a) = f(b). Thus, f is constant on the interior of I and thus is constant on I by continuity. (See below for a multivariable version of this result.)
Remarks:
• Only continuity of f, not differentiability, is needed at the endpoints of the interval I. No hypothesis of continuity needs to be stated if I is an open interval, since the existence of a derivative at a point implies the continuity at this point. (See the section continuity and differentiability of the article derivative.)
• The differentiability of f can be relaxed to one-sided differentiability, a proof given in the article on semi-differentiability.
## Cauchy's mean value theorem
Cauchy's mean value theorem, also known as the extended mean value theorem, is a generalization of the mean value theorem. It states: If functions f and g are both continuous on the closed interval [a,b], and differentiable on the open interval (a, b), then there exists some c ∈ (a,b), such that
Geometrical meaning of Cauchy's theorem
$(f(b)-f(a))g\,'(c)=(g(b)-g(a))f\,'(c).\,$
Of course, if g(a) ≠ g(b) and if g′(c) ≠ 0, this is equivalent to:
$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot$
Geometrically, this means that there is some tangent to the graph of the curve
$\begin{array}{ccc}[a,b]&\longrightarrow&\mathbb{R}^2\\t&\mapsto&\bigl(f(t),g(t)\bigr),\end{array}$
which is parallel to the line defined by the points (f(a),g(a)) and (f(b),g(b)). However Cauchy's theorem does not claim the existence of such a tangent in all cases where (f(a),g(a)) and (f(b),g(b)) are distinct points, since it might be satisfied only for some value c with f′(c) = g′(c) = 0, in other words a value for which the mentioned curve is stationary; in such points no tangent to the curve is likely to be defined at all. An example of this situation is the curve given by
$t\mapsto(t^3,1-t^2),$
which on the interval [−1,1] goes from the point (−1,0) to (1,0), yet never has a horizontal tangent; however it has a stationary point (in fact a cusp) at t = 0.
Cauchy's mean value theorem can be used to prove l'Hôpital's rule. The mean value theorem is the special case of Cauchy's mean value theorem when g(t) = t.
### Proof of Cauchy's mean value theorem
The proof of Cauchy's mean value theorem is based on the same idea as the proof of the mean value theorem.
Define h(x) = f(x) − rg(x), where r is a constant. Since f and g are continuous on [a, b] and differentiable on (a, b), the same is true for h. We now want to choose r so that h satisfies the conditions of Rolle's theorem. Namely
$\begin{align}h(a)=h(b)&\iff f(a)-rg(a)=f(b)-rg(b)\\ &\iff r(g(b)-g(a))=f(b)-f(a) \\&\iff r=\frac{f(b)-f(a)}{g(b)-g(a)}\cdot\end{align}$
By Rolle's theorem, since h is continuous and h(a) = h(b), there is some c in (a, b) for which h′(c) = 0, and it follows from the equality h(x) = f(x) − rg(x) that,
$h'(c)=0=f'(c)-r(g'(c)) \Rightarrow\frac{f'(c)}{g'(c)}= r=\frac{f(b)-f(a)}{g(b)-g(a)}$
as required.
## Generalization for determinants
Assume that $f$, $g$, and $h$ are differentiable function on $(a,b)$ that are continuous on $[a,b]$. Define
$D(x)=\left|\begin{array}{ccc}f(x) & g(x)& h(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|$
There exists $c\in(a,b)$ such that $D'(c)=0$.
Notice that
$D'(x)=\left|\begin{array}{ccc}f'(x) & g'(x)& h'(x)\\ f(a) & g(a) & h(a)\\ f(b) & g(b)& h(b)\end{array}\right|$
and if we place $h(x)=1$, we get Cauchy's mean value theorem. If we place $h(x)=1$ and $g(x)=x$ we get Lagrange's mean value theorem.
The proof of the generalization is quite simple: each of $D(a)$ and $D(b)$ are determinants with two identical rows, hence $D(a)=D(b)=0$. The Rolle's theorem implies that there exists $c\in (a,b)$ such that $D'(c)=0$.
## Mean value theorem in several variables
The mean value theorem in one variable generalizes to several variables by applying the theorem in one variable via parametrization. Let G be an open subset of Rn, and let f : G → R be a differentiable function. Fix points x, y ∈ G such that the interval x y lies in G, and define g(t) = f((1 − t)x + ty). Since g is a differentiable function in one variable, the mean value theorem gives:
$g(1) - g(0) = g'(c) \!$
for some c between 0 and 1. But since g(1) = f(y) and g(0) = f(x), computing g′(c) explicitly we have:
$f(y) - f(x) = \nabla f ((1- c)x + cy) \cdot (y - x)$
where ∇ denotes a gradient and · a dot product. Note that this is an exact analog of the theorem in one variable (in the case n = 1 this is the theorem in one variable). By the Schwarz inequality, the equation gives the estimate:
$|f(y) - f(x)| \le |\nabla f ((1- c)x + cy)| \, |y - x|.$
In particular, when the partial derivatives of f are bounded, f is Lipschitz continuous (and therefore uniformly continuous). Note that f is not assumed to be continuously differentiable nor continuous on the closure of G. However, in the above, we used the chain rule so the existence of ∇f would not be sufficient.
As an application of the above, we prove that f is constant if G is connected and every partial derivative of f is 0. Pick some point x0 ∈ G, and let g(x) = f(x) − f(x0). We want to show g(x) = 0 for every x ∈ G. For that, let E = {x ∈ G : g(x) = 0} . Then E is closed and nonempty. It is open too: for every x ∈ E,
$|g(y)| = |g(y) - g(x)| \le (0) |y - x| = 0$
for every y in some neighborhood of x. (Here, it is crucial that x and y are sufficiently close to each other.) Since G is connected, we conclude E = G.
Remark that all arguments in the above are made in a coordinate-free manner; hence, they actually generalize to the case when G is a subset of a Banach space.
## Mean value theorem for vector-valued functions
There is no exact analog of the mean value theorem for vector-valued functions. Jean Dieudonné in his classic treatise Foundations of Modern Analysis discards the mean value theorem and replaces it by mean inequality as the proof is not constructive and by no way one can find the mean value. In applications one only needs mean inequality. Serge Lang in Analysis I uses the mean value theorem, in integral form, as an instant reflex but this requires use the continuity of the derivative. If one uses Henstock-Kurzweil integral one can have the mean value theorem in integral form without the additional assumption that derivative should be continuous as every derivative is Henstock-Kurzweil integrable. The problem is roughly speaking the following: If f : U → Rm is a differentiable function (where U ⊂ Rn is open) and if x + th, x, h ∈ Rn, t ∈ [0, 1] is the line segment in question (lying inside U), then one can apply the above parametrization procedure to each of the component functions fi (i = 1, ..., m) of f (in the above notation set y = x + h). In doing so one finds points x + tih on the line segment satisfying
$f_i(x+h) - f_i(x) = \nabla f_i (x + t_ih) \cdot h.\,$
But generally there will not be a single point x + t*h on the line segment satisfying
$f_i(x+h) - f_i(x) = \nabla f_i (x + t^* h) \cdot h.\,$
for all i simultaneously. (As a counterexample one could take f : [0, 2π] → R2 defined via the component functions f1(x) = cos(x), f2(x) = sin(x). Then f(2π) − f(0) = 0 ∈ R2, but $\,f_1'(x)=-\sin (x)$ and $\,f_2'(x)=\cos (x)$ are never simultaneously zero as x ranges over [0, 2π].)
However a certain type of generalization of the mean value theorem to vector-valued functions is obtained as follows: Let f be a continuously differentiable real-valued function defined on an open interval I, and let x as well as x + h be points of I. The mean value theorem in one variable tells us that there exists some t* between 0 and 1 such that
$f(x+h)-f(x) = f'(x+t^*h)\cdot h. \,$
On the other hand we have, by the fundamental theorem of calculus followed by a change of variables,
$f(x+h)-f(x) = \int_x^{x+h} f'(u)du = \left(\int_0^1 f'(x+th)\,dt\right)\cdot h.$
Thus, the value f′(x + t*h) at the particular point t* has been replaced by the mean value
$\int_0^1 f'(x+th)\,dt.$
This last version can be generalized to vector valued functions:
Let U ⊂ Rn be open, f : U → Rm continuously differentiable, and x ∈ U, h ∈ Rn vectors such that the whole line segment x + th, 0 ≤ t ≤ 1 remains in U. Then we have:
$\text{(*)} \qquad f(x+h)-f(x) = \left(\int_0^1 Df(x+th)\,dt\right)\cdot h,$
where the integral of a matrix is to be understood componentwise. (Df denotes the Jacobian matrix of f.)
From this one can further deduce that if ||Df(x + th)|| is bounded for t between 0 and 1 by some constant M, then
$\text{(**)} \qquad \|f(x+h)-f(x)\| \leq M\|h\|.$
Proof of (*). Write fi (i = 1, ..., m) for the real valued components of f. Define the functions gi: [0, 1] → R by gi(t) := fi(x + th).
Then we have
$f_i(x+h)-f_i(x)\, =\, g_i(1)-g_i(0) =\int_0^1 g_i'(t)dt = \int_0^1 \left(\sum_{j=1}^n \frac{\partial f_i}{\partial x_j} (x+th)h_j\right)\,dt =\sum_{j=1}^n \left(\int_0^1 \frac{\partial f_i}{\partial x_j}(x+th)\,dt\right)h_j.$
The claim follows since Df is the matrix consisting of the components $\frac{\partial f_i}{\partial x_j}$, q.e.d.
Proof of (**). From (*) it follows that
$\|f(x+h)-f(x)\|=\left\|\int_0^1 (Df(x+th)\cdot h)\,dt\right\| \leq \int_0^1 \|Df(x+th)\| \cdot \|h\|\, dt \leq M\| h\|.$
Here we have used the following
Lemma. Let v : [a, b] → Rm be a continuous function defined on the interval [a, b] ⊂ R. Then we have
$\text{(***)}\qquad \left\|\int_a^b v(t)\,dt\right\|\leq \int_a^b \|v(t)\|\,dt.$
Proof of (***). Let u in Rm denote the value of the integral
$u:=\int_a^b v(t)\,dt.$
Now
$\|u\|^2 = \langle u,u \rangle = \left\langle \int_a^b v(t) dt,u \right\rangle = \int_a^b \langle v(t),u \rangle \,dt \leq \int_a^b \| v(t) \|\cdot \|u \|\,dt = \|u\| \int_a^b \|v(t)\|\,dt,$
thus $\| u\| \leq \int_a^b \|v(t)\|\,dt$ as desired. (Note the use of the Cauchy–Schwarz inequality.) This shows (***) and thereby finishes the proof of (**).
## Mean value theorems for integration
This section does not cite any references or sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (April 2013)
### First mean value theorem for integration
The first mean value theorem for integration states
If G : [a, b] → R is a continuous function and φ is an integrable function that does not change sign on the interval (a, b), then there exists a number x in (a, b) such that
$\int_a^b G(t)\varphi (t) \, dt=G(x) \int_a^b \varphi (t) \, dt.$
In particular, if φ(t) = 1 for all t in [a, b], then there exists x in (a, b) such that
$\int_a^b G(t) \, dt=\ G(x)(b - a).\,$
More commonly written as:$(1/(b-a))\int_a^b G(t) \, dt=\ G(x).$
The value G(x) is called the mean value of G(t) on [a, b].
### Proof of the first mean value theorem for integration
Without loss of generality assume the one-signed function $\varphi(t)\ge 0$ for all t (the negative case just changes direction of some inequalities). It follows from the extreme value theorem that the continuous function G has a finite infimum m and a finite supremum M on the interval [a, b]. From the monotonicity of the integral and the fact that m ≤ G(t) ≤ M, it follows from the non-negativity of $\varphi(t)$ that
$m I= \int_a^b m\varphi(t)\,dt \le \int^b_aG(t)\varphi(t) \, dt \le \int_a^b M\varphi(t)\,dt = M I,$
where
$I:=\int^b_a\varphi(t) \, dt$
denotes the integral of $\varphi(t)$. Hence, if I = 0, then the claimed equality holds for every x in [a, b]. Therefore, we may assume I > 0 in the following. Dividing through by I we have that
$m \le \frac1I\int^b_aG(t)\varphi(t) \, dt\le M.$
The extreme value theorem tells us more than just that the infimum and supremum of G on [a, b] are finite; it tells us that both are actually attained. Thus we can apply the intermediate value theorem, and conclude that the continuous function G attains every value of the interval [m, M], in particular there exists x in [a, b] such that
$G(x) = \frac1I\int^b_aG(t)\varphi(t) \, dt.$
This completes the proof.
### Second mean value theorem for integration
There are various slightly different theorems called the second mean value theorem for integration. A commonly found version is as follows:
If G : [a, b] → R is a positive monotonically decreasing function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b] such that
$\int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt.$
Here G(a + 0) stands for ${\underset{a_+}{\lim}G}$, the existence of which follows from the conditions. Note that it is essential that the interval (a, b] contains b. A variant not having this requirement is:
If G : [a, b] → R is a monotonic (not necessarily decreasing and positive) function and φ : [a, b] → R is an integrable function, then there exists a number x in (a, b) such that
$\int_a^b G(t)\varphi(t)\,dt = G(a+0) \int_a^x \varphi(t)\,dt + G(b-0) \int_x^b \varphi(t)\,dt.$
This variant was proved by Hiroshi Okamura in 1947.[citation needed]
## A probabilistic analogue of the mean value theorem
Let X and Y be non-negative random variables such that E[X] < E[Y] < ∞ and $X\leq_{st} Y$ (i.e. X is smaller than Y in the usual stochastic order). Then there exists an absolutely continuous non-negative random variable Z having probability density function
$f_Z(x)={\Pr(Y>x)-\Pr(X>x)\over {\rm E}[Y]-{\rm E}[X]}\,, \qquad x\geq 0.$
Let g be a measurable and differentiable function such that E[g(X)], E[g(Y)] < ∞, and let its derivative g′ be measurable and Riemann-integrable on the interval [x, y] for all y ≥ x ≥ 0. Then, E[g′(Z)] is finite and[3]
${\rm E}[g(Y)]-{\rm E}[g(X)]={\rm E}[g'(Z)]\,[{\rm E}(Y)-{\rm E}(X)].$
## Generalization in Complex Analysis
As noted above, the theorem does not hold for differentiable complex-valued functions. Instead, a generalization of the theorem is stated such:[4]
Calculus
• Fundamental theorem
• Limits of functions
• Continuity
• Mean value theorem
• Rolle's theorem
Definitions
Concepts
Rules and identities
Integral calculus
Definitions
Integration by
Formalisms
Definitions
Specialized calculi
Let f : Ω → C be a holomorphic function on the open convex set Ω, and let a and b be distinct points in Ω. Then there exist points u, v on Lab (the line segment from a to b) such that
$\textrm{Re}(f ' (u)) = \textrm{Re} \left(\frac{f(b) - f(a)}{b - a} \right),$
$\textrm{Im}(f ' (v)) = \textrm{Im} \left (\frac{f(b) - f(a)}{b - a} \right).$
Where Re() is the Real part and Im() is the Imaginary part of a complex-valued function.
## Notes
1. Weisstein, Eric. "Mean-Value Theorem". . Wolfram Research. Retrieved 24 March 2011.
2. J. J. O'Connor and E. F. Robertson (2000). Paramesvara, .
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from the entire web. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 67, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8923664689064026, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/7968?sort=newest | Two solid N_3 glued by its boundary
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $N_3$ be the genus three non orientable surface. Do we have an analogous 3d manifold as the solid torus and the solid Klein bottle for $N_3$? I don't see how to extend the ideas related to the 3d lens spaces. Any feedback would be super-welcome
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I don't understand this question. Are you asking whether there's a non-orientable handlebody bounded by N_3? If by N_3 you mean the non-orientable surface of Euler characteristic -2, the answer is yes. (Just take a solid Klein bottle and attach a 1-handle.) Please clarify. – HW Dec 6 2009 at 6:23
2
The title of this question should be: "Does $N_3$ bound?". Then you could mention the application you have in mind (which I guess is gluing together copies of the bounded manifold?). Also, could you explain how lens spaces play a role in this question? – Sam Nead Dec 6 2009 at 12:47
Sam, if N_3 would bound we should try to develop a similar theory of lenspaces where the the building block were solid N_3's with respective boundaries a pair of 2-torus with one point $\mathbb{R}$-blowups but, not now... anyway thanks a lot for your interest – ivane Dec 7 2009 at 0:18
3 Answers
$N_k$ is the connect sum of $k$ copies of the real projective plane, so it has Euler characteristic $2 - k$. For $k$ even $N_k$ bounds a connect sum of $k/2$ copies of the solid Klein bottle, as in Henry's comment. When $k$ is odd the rank of $H_1$ (with $Z/2Z$ coefficients) is odd. Because of "half lives, half dies" the boundary of a three-manifold must have even rank in $H_1$. So $N_k$, for $k$ odd, does not bound.
Half lives, half dies can be found in Hatcher's three-manifold notes as Lemma 3.5, but you'll need to use $Z/2Z$ coefficients. My copy of the universal coefficient theorem is all rusty, so if I've made a mistake, it is here.
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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.
It is a general fact that a closed manifold of odd Euler characteristic cannot bound a compact manifold. This can be deduced pretty easily from the fact that a closed manifold of odd dimension has Euler characteristic zero (a consequence of Poincaré duality) as follows. Suppose N is the boundary of a compact manifold P. Let M be the double of P, the union of two copies of P glued along N. Then the Euler characteristics of M, N, and P are related by:
$\chi(M)=\chi(P)+\chi(P)-\chi(N)$
Thus $\chi(M)$ and $\chi(N)$ are congruent mod 2. If the dimension of N is even, then M is a closed manifold of odd dimension so $\chi(M)=0$, hence $\chi(N)$ is even. And if the dimension of N is odd then $\chi(N)=0$ anyhow.
I should have put this in my book!
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yeah, I completely agree, perhaps in the next update... Prof thanks so much! – ivane Dec 7 2009 at 0:27
Ok, it is perfectly clear that N<sub>3</sub> can not bound an ORIENTABLE 3-manifold... but, what about a NON ORIENTABLE 3d-orbifold? Could the cone $N_3\times I/(N_3\times 1\sim *)$ possibly? Is there a similar half live or half die for 3d-orbifolds? – ivane Dec 7 2009 at 19:28
1
You'll have to settle on a precise definition of orbifold cobordism to make sense of this question but if you allow any kind of orbifold locus you get all compact manifolds as boundaries of orbifolds by considering $M \times [-1,1] / \mathbb Z_2$ where the action of $\mathbb Z_2$ on $[-1,1]$ is reflection about the origin. $\mathbb Z_2$ acting trivially on the $M$ factor. – Ryan Budney Dec 11 2009 at 5:13
Alternatively, the characteristic number $\langle w_2, [N_k] \rangle$ is just the mod 2 reduction of the Euler characteristic (as $w_2$ is the mod 2 reduction of the Euler class) so is -k modulo 2. Thus if k is odd this characteristic number is nontrivial, and $N_k$ cannot bound.
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thanks Oscar, you are a real savant on the subject... – ivane Dec 7 2009 at 0:24 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9360544681549072, "perplexity_flag": "head"} |
http://www.impan.pl/cgi-bin/dict?glance | ## glance
[see also: look, sight]
At first glance Lemma 2 seems to yield four possible outcomes.
Neighbourhoods of points in these spaces appear at first glance to have a nice regular structure, but upon closer scrutiny, one sees that many neighbourhoods contain collections of arcs hopelessly folded up.
Now, for arbitrary $n$, a glance at the derivative shows that ......
Go to the list of words starting with: a b c d e f g h i j k l m n o p q r s t u v w y z | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8922174572944641, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/46645/ways-to-experimentally-control-the-chemical-potential-of-a-solid-state-system/46694 | # Ways to experimentally control the chemical potential of a solid state system
When working in the grand canonical ensemble we write the grand potential as $\Omega = \Omega (T,V,\mu)$. In this case we are taking the chemical potential $\mu$ to be an independent variable. This would imply that this ensemble is best suited to situations in which the chemical potential is an experimentally controlled parameter.
If this is the case, then what are some ways in which the chemical potential can be controlled experimentally?
Specifically I am interested in a solid state system which is being modeled as electrons on a lattice. In theory, by adjusting the chemical potential we could control the filling fraction of electrons for this system, taking it from band insulator to conductor or from Mott insulator to conductor. Is this type of tuning possible with real systems? If so, what are some realistic applications and/or specific limitations of these techniques.
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## 1 Answer
The use of the chemical potential $\mu$ as state variable is useful in situations where composition $N$ is variable and/or cannot be easily controlled. From an experimental point of view the chemical potential is fixed when the system is in contact with energy and particle reservoirs. At equilibrium, the chemical potential of the system equals that of the reservoir $\mu = \mu_\mathrm{res}$. Thus by modifying the parameters of the reservoir you can control the chemical potential of the system.
A typical example is when the system is a layer of molecules adsorbed in a surface. This is an open system and composition is variable --and generally unknown--. By using a gas of the same molecules as reservoir you can fix the value of the chemical potential of the open system. The chemical potential $\mu_\mathrm{res}$ of the gas can be obtained from evaluating the fundamental equation of the gas or, if this is not available, from integrating the Gibbs Duhem relation if the equation of state of the gas is known.
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Thanks @juanrga. I have actually done the problem of a surface with absorption sites in contact with a gas. I recall at the end putting the chemical potential of the gas in terms of its number density. Then by controlling the density of the gas we control the chemical potential of the surface. This is what you said, by modifying parameters of the reservoir you can control the $\mu$. Do you by chance know any of the corresponding parameters for solid state systems? – Todd R Dec 13 '12 at 1:31
@ToddR The chemical potential of electrons in solids is defined in exactly the same way as the chemical potential of a chemical species. There is no 'special' parameters for solids but the usual ones: Temperature $T$, composition $N$, external electric potential $\phi$ (if any)... Notice that, in presence of electric fields, some authors rename the chemical potential to "electrochemical potential", whereas solid-state physicists like to use a radically different terminology: "Fermi energy". – juanrga Dec 16 '12 at 12:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 10, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9141784906387329, "perplexity_flag": "head"} |
http://stats.stackexchange.com/questions/20574/maximum-likelihood-estimation-when-parameters-are-functions-of-another-data-seri | # Maximum likelihood estimation when parameters are functions of another data series
We have two time series: $X_t$ and $R_t$, and a model saying that $R_{t+1} = (\mu(X_t) - \frac{1}{2}\sigma^2(X_t))\Delta T + \sigma(X_t) \sqrt{\Delta T} \epsilon_t$, where $\Delta T$ is given constant and $\epsilon_t$-s are independent normally distributed with zero mean and unit variance. Further we assume that the functions $\mu(x)$ and $\sigma(x)$ are linear for simplicity. I would like to use some standard method (MLE comes to my mind) to estimate parameters of functions $\mu(x)$ and $\sigma(x)$, but I am not sure how to do this.
I would be grateful for detailed answers, because I am not really experienced with statistics.
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Are you familiar with a good math programming language, like R? – jbowman Jan 4 '12 at 19:50
1
This looks like the discretization of an affine diffusion (SDE) or something close. – cardinal Jan 4 '12 at 20:10
@jbowman, sadly I don't know R :( – Grzenio Jan 5 '12 at 9:47
@cardinal, indeed it is – Grzenio Jan 5 '12 at 9:48
## 1 Answer
Let $\theta$ be the parameters involved in $\mu(x)$ and $\sigma(x)$.
Your likelihood function will be $$\mathcal{L}(\theta\,|\,\epsilon_1,\ldots,\epsilon_n) = f(\epsilon_1,\epsilon_2,\ldots,\epsilon_n\;|\;\theta) = \prod_{t=1}^n f(\epsilon_t|\theta)= \prod_{t=1}^{n} \frac{1}{\sqrt{2\pi}\ } \exp\big(-\epsilon_t^2/2\big) \>.$$ You may need to take $t=1$ to $n-1$ (for a large sample it doesn't matter, assuming you have $n$ observations).
Substitute $$\epsilon_t=\dfrac{R_{t+1} - (\mu(X_t) - \frac{1}{2} \sigma^2(X_t))\Delta T}{\sigma(X_t) \sqrt{\Delta T}} \>.$$
This will be in terms of $\theta$, $R_t$, and $X_t$. MLE estimates are the parameters which optimize the likelihood function found above.
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– Grzenio Jan 5 '12 at 16:08
Not in the first expression. Since $\epsilon_t$-s are independent normally distributed with zero mean and unit variance. – vinux Jan 5 '12 at 16:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8873879909515381, "perplexity_flag": "head"} |
http://mathhelpforum.com/discrete-math/104021-some-simple-predicate-logic-questions.html | Thread:
1. Some Simple Predicate Logic Questions
Could someone please check my answers for these questions:
p(x): x^2 - 7x + 10 = 0; x = {5, 2}
q(x): x^2 - 2x - 3; x = {-1, 3}
r(x): x < 0
domains:
i) Z (set of all integers)
ii) Z+ (set of all positive integers)
iii) integers 2 and 5
Are these statements True or false (with counterexamples):
$<br /> \forall{x[p(x)\Rightarrow\neg}{r(x)]}<br />$
i) false x = 1
ii) false x = 1
iii) false
$<br /> \forall{x[q(x)\Rightarrow}{r(x)]}<br />$
i) false x = 3
ii) false x = 3
iii) false x = 2 ?
$<br /> \exists{x[q(x)\Rightarrow}{r(x)]}<br />$
i) true
ii) false
iii) false
$<br /> \exists{x[p(x)\Rightarrow}{r(x)]}<br />$
i) false
ii) false
iii) false
Is it possible to find counterexamples for the existential statements? Also, in the first two problems with the universal quantifier, for these to be true, would p(x) and q(x) have to hold for every value in the domain? For example, would x^2 - 7x + 10 = 0 have to be true for every value of x in the domain, or just the solutions, 5 and 2? I understand quantifiers, but the conditional statement is confusing me a bit. Thanks!
2. Hello aaronrj
Originally Posted by aaronrj
Could someone please check my answers for these questions:
p(x): x^2 - 7x + 10 = 0; x = {5, 2}
q(x): x^2 - 2x - 3; x = {-1, 3}
r(x): x < 0
domains:
i) Z (set of all integers)
ii) Z+ (set of all positive integers)
iii) integers 2 and 5
Are these statements True or false (with counterexamples):
$<br /> \forall{x[p(x)\Rightarrow\neg}{r(x)]}<br />$
i) false x = 1
ii) false x = 1
iii) false
No - you haven't got this at all. $\forall{x[p(x)\Rightarrow\neg}{r(x)]}$ means that if $p(x)$ is true, then $r(x)$ is false. And, in all three domains, $p(x)$ is true when $x = 2$ or $x = 5$. In both of these cases $r(x)$ is false, since $x > 0$.
So the proposition is true in all three domains.
$<br /> \forall{x[q(x)\Rightarrow}{r(x)]}<br />$
i) false x = 3
ii) false x = 3
iii) false x = 2 ?
(i) and (ii) are correct.
But (iii) is a bit sneaky, because in the domain $\{2, 5\},\, q(x)$ is false for all $x$. So for every value of $x$ that makes $q(x)$ true, $r(x)$ is true. In other words, there isn't a counterexample, so it's a true proposition.
$<br /> \exists{x[q(x)\Rightarrow}{r(x)]}<br />$
i) true
ii) false
iii) false
$<br /> \exists{x[p(x)\Rightarrow}{r(x)]}<br />$
i) false
ii) false
iii) false
These are all correct.
Is it possible to find counterexamples for the existential statements?
No. If an existential statement is false, it simply means that no value of $x$ exists that makes the statement true.
Also, in the first two problems with the universal quantifier, for these to be true, would p(x) and q(x) have to hold for every value in the domain?
No. When you form a proposition like $\forall x[a(x) \Rightarrow b(x)]$ you are simply saying if $a(x)$ is true, then $b(x)$ is true. You aren't saying anything at all about $a(x)$ actually being true at all.
Indeed $a(x)$ may never be true for any values of $x$ - see for example part (iii) of the statement $\forall x[q(x) \Rightarrow r(x)]$, where $q(x)$ is false for all values of $x$ in the domain $\{2,5\}$.
Grandad | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 32, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8061545491218567, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/23391?sort=votes | ## What is known about this plethysm?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $S^{\lambda}$ be a Schur functor. Is there a known positive rule to compute the decomposition of $S^{\lambda}(\bigwedge^2 \mathbb{C}^n)$ into $GL_n(\mathbb{C})$ irreps?
In response to Vladimir's request for clarification, the ideal answer would be a finite set whose cardinality is the multiplicity of $S^{\mu}(\mathbb{C}^n)$ in $S^{\lambda}(\bigwedge^2 \mathbb{C}^2)$. As an example, the paper Splitting the square of a Schur function into its symmetric and anti-symmetric parts gives such a rule for $\bigwedge^2(S^{\lambda}(\mathbb{C}^n))$.
Formulas involving evaluations of symmetric group characters, or involving alternating sums over stable rim hooks, are not good because they are not positive.
And, yes, it is easy to relate the answers for $\bigwedge^2 \mathbb{C}^n$ and $\mathrm{Sym}^2(\mathbb{C}^n)$, so feel free to answer with whichever is more convenient.
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7
Is there a reason that the question isn't in the title? (And, unrelated, MathWorld spells "plethysm" with a "y", but it's not a word I've seen before, so I don't know.) Nice question, by the way. – Theo Johnson-Freyd May 4 2010 at 4:51
This isn't related, but is "plethysm" actually a word in English? – Peter Samuelson May 4 2010 at 14:00
Spelling fixed in title. Thanks! – David Speyer May 4 2010 at 15:48
2
It is an English word in so far it has been used for ages now when writing about, well, plethysm in English! – Mariano Suárez-Alvarez May 4 2010 at 16:31
Are you interested in this question for a general $\lambda$? – Victor Protsak May 4 2010 at 22:18
## 5 Answers
From Weyman's book "Cohomology of Vector bundles and Syzygies" Chapter 2 gives the following decompositions: $$\mathrm{Sym}^m \left(\bigwedge^2 E\right)=\bigoplus_{\lambda \in A_m}S^{\lambda}E$$ $$\bigwedge^m \left(\bigwedge^2E\right)=\bigoplus_{\lambda \in B_m}S^{\lambda}E$$ where $A_m$ is the set of all $\lambda$ with $|\lambda|=2m$ such that all parts $\lambda_i$ are even. $B_m$ is the set of all partitions $\lambda$ of $2m$ so that when you write it in hook notation $\lambda=(a_1,\dots,a_r|b_1,\dots,b_r)$ you have $a_i=b_i+1$ for all $i$. Also, maybe this article has some useful references.
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Yes, these are also listed in Macdonald's book. – Vladimir Dotsenko May 5 2010 at 13:13
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If I remember this correctly the cases $\mathrm{Sym}^k(\bigwedge^2 \mathbb{C}^n)$ and $\mathrm{Sym}^k(\mathrm{Sym}^2(\mathbb{C}^n))$ are known; and hence $\bigwedge^k(\bigwedge^2 \mathbb{C}^n)$ and $\bigwedge^k(\mathrm{Sym}^2(\mathbb{C}^n))$. I will look up the references tomorrow (if this is of interest).
Edit The result has now been stated. I learnt this from R.P.Stanley "Enumerative Combinatorics" Vol 2, Appendix 2. Specifically, A2.9 Example (page 449) which refers to (7.202) on page 503. This gives as the original reference (11.9;4) of the 1950 edition of:
Littlewood, Dudley E. "The theory of group characters and matrix representations of groups."
P.S. In the Notes at the end of 7.24 (bottom of page 404 in CUP 1999 edition) it discusses the origin and the etymology of "plethysm". It says:
Plethysm was introduced in
MR0010594 (6,41c) Littlewood, D. E. Invariant theory, tensors and group characters.
Philos. Trans. Roy. Soc. London. Ser. A. 239, (1944). 305--365
The term "plethysm" was suggested to Littlewood by M. L. Clark after the Greek word plethysmos $\pi\lambda\eta\theta\nu\sigma\mu o\zeta$ for "multiplication".
(the Greek is an approximation)
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Yes, this is of interest; please let me know what references you know of. – David Speyer May 4 2010 at 22:06
This is a very special case, of course, because the decompositions are multiplicity free, and due to this fact, they are known very explicitly. I heartily recommend Roger Howe's Schur lectures. – Victor Protsak May 4 2010 at 22:13
You may also use SAGE , (for example, the Sage online notebook )
Example:
The Riemann curvature tensor $R$ lives in the space $Sym^2(\Lambda^2 V)$ (after identifying $V$ with $V^{\vee}$)
Decomposing it in Sage:
$s = SFASchur(QQ)$
(let s be the Schur functor)
$s([2])(s([1,1]))$
(compute plethysm $Sym^2 \Lambda^2$)
s[1, 1, 1, 1] + s[2, 2]
-- i.e., $\Lambda^4 V + S_{[2,2]}$, as it should be
\$ s([3])(s([1,1]))
s[1, 1, 1, 1, 1, 1] + s[2, 2, 1, 1] + s[3, 3]
-- though i understand that the explicit formula is better :)
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What kind of a formula will you find satisfactory? Formulas for the plethysm $s_\lambda\circ h_n$ where coefficients are expressed in terms of $S_n$-characters and generalized Kostka numbers are in Macdonald's book (see pp.138-140), so putting $n=2$ and applying the standard involution will give you some result for $e_2$ as well (which is your question, I presume)...
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If I'm remembering correctly, the formulas in Macdonald contain the values of symmetric group characters evaluated on conjugacy classes, which probably doesn't meet David's criterion ("positive rule"). – GS May 4 2010 at 11:33
@Stephen: I surely seem to have overlooked "positive"... – Vladimir Dotsenko May 4 2010 at 12:21
I've added some clarifications above. – David Speyer May 4 2010 at 16:00
Did you check the book by Procesi?
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http://en.wikibooks.org/wiki/Waves/Plane_Waves | # Waves/Plane Waves
This page may need to be reviewed for quality.
Waves : 2 and 3 Dimension Waves
1 - 2 - 3 - 4 - 5 - 6 - 7
Problems
### Plane Waves
A plane wave in two or three dimensions is like a sine wave in one dimension except that crests and troughs aren't points, but form lines (2-D) or planes (3-D) perpendicular to the direction of wave propagation. Figure 2.5 shows a plane sine wave in two dimensions. The large arrow is a vector called the wave vector, which defines (1) the direction of wave propagation by its orientation perpendicular to the wave fronts, and (2) the wavenumber by its length. We can think of a wave front as a line along the crest of the wave. The equation for the displacement associated with a plane sine wave in three dimensions at some instant in time is
$A(x,y,z) = \sin (\mbox{k} \cdot \mbox{r} ) = \sin (k_x x + k_y y + k_z z) .$ (3.9)
Figure 2.5: Definition sketch for a plane sine wave in two dimensions. The wave fronts are constant phase surfaces separated by one wavelength. The wave vector is normal to the wave fronts and its length is the wavenumber.
Since wave fronts are lines or surfaces of constant phase, the equation defining a wave front is simply $\mbox{k} \cdot \mbox{r} = const$.
In the two dimensional case we simply set $k_z = 0$. Therefore, a wavefront, or line of constant phase $\phi$ in two dimensions is defined by the equation
$\mbox{k} \cdot \mbox{r} = k_x x + k_y y = \phi \quad \mbox{(two dimensions)} .$ (3.10)
This can be easily solved for $y$ to obtain the slope and intercept of the wavefront in two dimensions.
As for one dimensional waves, the time evolution of the wave is obtained by adding a term $- \omega t$ to the phase of the wave. In three dimensions the wave displacement as a function of both space and time is given by
$A(x,y,z,t) = \sin (k_x x + k_y y + k_z z - \omega t ) .$ (3.11)
The frequency depends in general on all three components of the wave vector. The form of this function, $\omega = \omega (k_x , k_y , k_z )$, which as in the one dimensional case is called the dispersion relation, contains information about the physical behavior of the wave.
Some examples of dispersion relations for waves in two dimensions are as follows:
• Light waves in a vacuum in two dimensions obey
$\omega = c ( k_x^2 + k_y^2 )^{1/2} \quad \mbox{(light)} ,$ (3.12)
where $c$ is the speed of light in a vacuum.
• Deep water ocean waves in two dimensions obey
$\omega = g^{1/2} ( k_x^2 + k_y^2 )^{1/4} \quad \mbox{(ocean waves)} ,$ (3.13)
where $g$ is the strength of the Earth's gravitational field as before.
• Certain kinds of atmospheric waves confined to a vertical $x-z$ plane called gravity waves (not to be confused with the gravitational waves of general relativity)3.1 obey
$\omega = \frac{N k_x}{k_z} \quad \mbox{(gravity waves)} ,$ (3.14)
where $N$ is a constant with the dimensions of inverse time called the Brunt-Väisälä frequency.
Figure 2.6: Contour plots of the dispersion relations for three kinds of waves in two dimensions. In the upper panels the curves show lines or contours along which the frequency $\omega$ takes on constant values. For light and ocean waves the frequency depends only on the magnitude of the wave vector, whereas for gravity waves it depends only on the wave vector's direction, as defined by the angle $\theta$ in the upper right panel. These dependences for each wave type are illustrated in the lower panels.
Contour plots of these dispersion relations are plotted in the upper panels of figure 2.6. These plots are to be interpreted like topographic maps, where the lines represent contours of constant elevation. In the case of figure 2.6, constant values of frequency are represented instead. For simplicity, the actual values of frequency are not labeled on the contour plots, but are represented in the graphs in the lower panels. This is possible because frequency depends only on wave vector magnitude $(k_x^2 + k_y^2 )^{1/2}$ for the first two examples, and only on wave vector direction $\theta$ for the third. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9203636050224304, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/31172?sort=newest | ## Uniqueness of local Langlands correspondence for connected reductive groups over real/complex field.
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In Langlands' notes "On the classification of irreducible representations of real algebraic groups", available at the Langlands Digital Archive page here, Langlands gives a construction which is now referred to as "the local Langlands correspondence for real/complex groups".
What Langlands does in practice in this paper is the following. Let $K$ denote either the real numbers or a finite field extension of the real numbers (for example the complex numbers, or a field isomorphic to the complex numbers but with no preferred isomorphism). Let $G$ be a connected reductive group over $K$. Langlands defines two sets $\Pi(G)$ (infinitesimal equivalence classes of irreducible admissible representations of $G(K)$) and $\Phi(G)$ ("admissible" homomorphisms from the Weil group of $K$ into the $L$-group of $G$, modulo inner automorphisms). He then writes down a surjection from $\Pi(G)$ to $\Phi(G)$ with finite fibres, which he constructs in what is arguably a "completely canonical" way (I am not making a precise assertion here). Langlands proves that his surjection, or correspondence as it would now be called, satisfies a whole bunch of natural properties (see for example p44 of "Automorphic $L$-functions" by Borel, available here), although there are other properties that the correspondence has which are not listed there---for example I know a statement explaining the relationship between the Galois representation attached to a $\pi$ and the one attached to its contragredient, which Borel doesn't mention, but which Jeff Adams tells me is true, and there is another statement about how infinitesimal characters work which I've not seen in the literature either.
So this raises the following question: is it possible to write down a list of "natural properties" that one would expect the correspondence to have, and then, crucially, to check that Langlands' correspondence is the unique map with these properties? Uniqueness is the crucial thing---that's my question.
Note that the analogous question for $GL_n$ over a non-arch local field has been solved, the crucial buzz-word being "epsilon factors of pairs". It was hard work proving that at most one local Langlands correspondence had all the properties required of it---these properties are listed on page 2 of Harris-Taylor's book and it's a theorem of Henniart that they suffice. The Harris-Taylor theorem is that at least one map has the required properties, and the conclusion is that exactly one map does. My question is whether there is an analogue of Henniart's theorem for an arbitrary connected reductive group in the real/complex case.
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No. My instinct a year ago would be to ask an expert. My instinct now is to post here first and then ask an expert if nothing is forthcoming. Tony Scholl pointed out to me by email that Henniart's theorem in the p-adic case isn't for any one specific GL_n, as it were, it's a set of conditions that characterise all of the bijections for all GL_n at once. So perhaps asking for properties that characterise the correspondence for one G is not the right idea and one should do all G at once. – Kevin Buzzard Jul 10 2010 at 19:55
Ah, I had assumed in your question that you were also thinking to work with all $G$ at once, for exactly the kind of reason you mention in your comment. (In the GL_n story there's the extra wrinkle that the GL_1 case has to be put in "by hand", via CFT.) Perhaps the "safest" bet is to work throughout with "split" groups over the reals on a first pass through the question, since in principle that permits a lot more use of inductive technique with many torus centralizers to hope to build things up (somehow) from GL_1 and SL_2. – BCnrd Jul 12 2010 at 12:43
## 2 Answers
Dear Kevin,
Here are some things that you know.
(1) Every non-tempered representation is a Langlands quotient of an induction of a non-tempered twist of a tempered rep'n on some Levi, and this description is canonical.
(2) Every tempered rep'n is a summand of the induction of a discrete series on some Levi.
(3) The discrete series for all groups were classified by Harish-Chandra.
Now Langlands's correspondence is (as you wrote) completely canonical: discrete series with fixed inf. char. lie in a single packet, and the parameter is determined from the inf. char. in a precise way.
All the summands of an induction of a discrete series rep'n are also declared to lie in a single packet. So all packet structure comes from steps (1) and (2).
The correspondence is compatible in a standard way with twisting, and with parabolic induction.
So:
If we give ourselves the axioms that discrete series correspond to irred. parameters, that the correspondence is compatible with twisting, that the correspondence is compatible with parabolic induction, and that the correspondence is compatible with formation of inf. chars., then putting it all together, it seems that we can determine step 1, then 2, then 3.
I don't know if this is what you would like, but it seems reasonable to me.
Why no need for epsilon-factor style complications: because there are no supercuspidals, so everything reduces to discrete series, which from the point of view of packets are described by their inf. chars. In the p-adic world this is just false: all the supercuspidals are disc. series, they have nothing analogous (at least in any simple way) to an inf. char., and one has to somehow identify them --- hence epsilon factors to the rescue.
[Added: A colleague pointed out to me that the claim above (and also discussed below in the exchange of comments with Victor Protsak) that the inf. char. serves to determine a discrete series L-packet is not true in general. It is true if the group $G$ is semi-simple, or if the fundamental Cartan subgroups (those which are compact mod their centre) are connected. But in general one also needs a compatible choice of central character to determine the $L$-packet. In Langlands's general description of a discrete series parameter, their are two pieces of data: $\mu$ and $\lambda_0$. The former is giving the inf. char., and the latter the central char.]
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I didn't know all of this Emerton, but it does look good. In fact if you like I'll mention an explicit example of something I'd not really realised until this post. I am pretty sure that it's true that for a general representation, the inf char was too weak to read off the parameter. For example, consider a reducible princ series rep for GL_2(R). This splits up into a f.d. piece and a discrete series piece. These two pieces, if I'm not mistaken, will have the same inf char and rather different associated Weil reps. This caused me some grief at some point when figuring out the details of... – Kevin Buzzard Jul 15 2010 at 10:00
...my almost-finished paper with Toby Gee about auto reps and Galois reps in the global setting. But in fact what you're saying perhaps is that if I restrict to discrete series reps then this issue does not occur and the inf char is a stronger invariant than I had realised---and of course now you phrase it this way I guess it's somehow clear from what Harish-Chandra did. – Kevin Buzzard Jul 15 2010 at 10:02
I don't think that the bit about discrete series being determined by their infinitesimal characters is correct: the Harish-Chandra parameters of a d.s. representation is a $K$-dominant weight, hence there are $W_G/W_K$ discrete series representations of $G$ with the same inf char (this is even mentioned at the en.wikipedia.org/wiki/… Wikipedia article). You can see it already in the case of $SL_2(\mathbb{R}):$ holomorphic and antiholomorphic discrete series reps have the same inf char. – Victor Protsak Jul 15 2010 at 17:28
Dear Victor, The claim is not that discrete series are determined by their inf. char; as you note, that is false. But all these discrete series with the given inf. char. lie in the same $L$-packet, essentially by definition. So the Langlands parameter of a disc. series depends only on the inf. char. – Emerton Jul 15 2010 at 17:37
Dear Emerton, I don't think your post implied it, but that's how I read Kevin's response right below it. I may be a bit rusty with my Langlands parameters, because I thought that a Langlands packet may contain at most one discrete series rep, whereas, as you say, a packet consists of $\textit{all}$ d.s. representations with the same inf character. – Victor Protsak Jul 16 2010 at 4:26
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Here is a slight refinement of Emerton's answer.
Any L-homomorphism $\phi:W_\mathbb R\rightarrow \phantom{}^LG$ defines an infinitesimal character $\lambda$ and a central character $\chi$. If $\phi(W_\mathbb R)$ is not contained in a proper Levi this defines an L-packet: the set of relative discrete series with this infinitesimal and central character (relative = discrete series modulo center). Every relative discrete series L-packet arises this way.
Now compatibility with parabolic induction determines all L-packets as follows. If $\phi(W_\mathbb R)$ is contained in a proper Levi subgroup $\phantom{}^LM$, the preceding construction defines a relative discrete series L-packet for $M$. The L-packet for $G$ is defined to be the irreducible summands of $Ind_{MN}^G(\pi_M)$ as $\pi_M$ runs over the L-packet of $M$ (for the correct choice of $N$). This induction can be done in stages: a tempered step, which is completely reducible, followed by an induction which gives unique irreducible summands. (This incorporates the "twisting" mentioned above).
As Emerton says this is canonical since we don't need L or epsilon factors to define relative discrete series L-packets. I'd like to know if anything like this, however speculative, might hold for general p-adic groups; in particular whether epsilon factors really do come to the rescue outside of GL(n) and perhaps classical groups. See http://mathoverflow.net/questions/24376.
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http://mathoverflow.net/questions/69792?sort=newest | ## Can a metric conformal to a Kahler metric be Kahler?
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Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler?
If $\dim_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler.
If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold?
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## 3 Answers
You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$.
This is a linear algebra fact, valid pointwise : if $\alpha \in T_x^*X$ satisfies $\omega_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim_R X \geq 4$.
The short argument is that, $\omega_x^{n-1}\wedge : T^*_x X\to \bigwedge^{2n-1} T^*_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak).
This said, as in Francesco's answer, you can have non proportional conformal riemannian metrics that are Kähler with respect to different complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional.
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Nice. Are we considering $\omega$ as a symplectic form here (instead of a metric)? – Gunnar Magnusson Jul 8 2011 at 13:57
@Gunnar: definitely yes, $\omega$ is a 2-form, otherwise $d\omega$ wouldn't make much sense. The riemannian metric is $\omega(J.,.)$. – BS Jul 8 2011 at 14:04
Looks like we were answering at the same time. At least I gave a slightly different explanation, so I might as well leave my answer there. – Spiro Karigiannis Jul 8 2011 at 14:10
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The paper by Apostolov, Calderbank, and Gauduchon that Francesco mentions find different Kaehler structures whose associated Riemannian metrics are conformal to each other. But they correspond to different complex structures $J_+$ and $J_-$.
I believe what Gunnar is asking is whether or not one can have $f \omega$ be closed and thus Kaehler with respect to the same complex structure $J$ associated to $\omega$. The answer is no, and this has nothing at all to do with compactness or the hard Lefschetz theorem. On any almost Hermitian manifold $(M, J, \omega, g)$, it is a fact that the wedge product with the Kaehler form $\omega$ on the space of $1$-forms is injective, regardless of the compactness of $M$, the integrability of $J$, or the closedness of $\omega$. This follows, for example, from the identity
$$\ast( \omega \wedge (\ast ( \omega \wedge \alpha) ) = - (m-1) \alpha$$
where $\alpha$ is any $1$-form on $M$, where $\ast$ is the Hodge star operator, and the real dimension of $M$ is $2m$. (One sees that the only requirement is that $m>1$.)
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There are examples in real dimension $4$ of manifolds having two conformally equivalent Kahler metrics, inducing the same conformal structure but with opposite orientation.
See the paper Ambikahler geometry, ambitoric surfaces and Einstein 4-orbifolds by Apostolov, Calderbank and Gauduchon.
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1
Thanks Francesco, that's an interesting article. However it does seem to answer a slightly different question. In the notation of the article, we have a fixed complex structure $J$ on $M$ (such that $X = (M,J)$) and ask for structures $(g_1,J,\omega_1)$ and $(g_2, J, \omega_2)$ such that: $g_2 = f^2 g_1$, and ask if the second structure can be Kahler if the first one is and if $f$ is non-constant. In particular, if this is possible, then both structures would induce the same orientation on $M$. – Gunnar Magnusson Jul 8 2011 at 13:54
Ah ok, I did not notice that your complex structure $J$ was fixed. At any rate, I hope you can find this paper useful :-) – Francesco Polizzi Jul 8 2011 at 13:58 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9229717254638672, "perplexity_flag": "head"} |
http://mathhelpforum.com/geometry/2703-radius-arc.html | # Thread:
1. ## Radius from Arc
If the arc length s and the distance h from the centre point of the associated chord to the arc is known, how do you calculate the radius?
Thank You
If the arc length s and the distance h from the centre point of the associated chord to the arc is known, how do you calculate the radius?
Thank You
I don't think you will find a closed form solution for this problem - though
I am prepared to be proven wrong.
RonL
By the theorem of chord in a circle we have that,
$n^2=r(2r-h)$
But, $n=r\sin(s/2r)$
Thus, we have,
$r^2\sin(s/2r)=r(2r-h)$[/tex]
Thus,
$r\sin(s/2r)=2r-h$
----------------------
If you know calculus
Thus, you need to find the zero's of the function,
$f(x)=x\sin(s/2x)-2x+h$
Use Newton's method
Attached Thumbnails
4. Originally Posted by ThePerfectHacker
By the theorem of chord in a circle we have that,
$n^2=r(2r-h)$
Just as well that I checked my note on this problem
The intersection chord theorem would give in this case:
$<br /> n^2=h(2r-h)<br />$
surly?
RonL
5. Originally Posted by CaptainBlack
Just as well that I checked my note on this problem
The intersection chord theorem would give in this case:
$<br /> n^2=h(2r-h)<br />$
surly?
RonL
One thing that confuses me is how an immortal (me) makes such a mistake
6. Originally Posted by ThePerfectHacker
By the theorem of chord in a circle we have that,
$n^2=r(2r-h)$
But, $n=r\sin(s/2r)$
Thus, we have,
$r^2\sin(s/2r)=r(2r-h)$
...
Hello,
I'm a little bit confused: When you plug in the value of n, shouldn't be there a squared sine value too?:
$r^2\left(\sin(s/2r)\right)^2=r(2r-h)$
Greetings
EB
7. Originally Posted by earboth
Hello,
I'm a little bit confused: When you plug in the value of n, shouldn't be there a squared sine value too?:
$r^2\left(\sin(s/2r)\right)^2=r(2r-h)$
Greetings
EB
There should be, now how did I miss that (its in my notes)
RonL
8. Originally Posted by ThePerfectHacker
One thing that confuses me is how an immortal (me) makes such a mistake | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.936032772064209, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/3925/proving-that-this-sum-sum-limits-0-k-frac2p3-p-choose-k-is-div?answertab=votes | # Proving that this sum $\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p^{2}$
How does one prove that for a prime $p \geq 5$ the sum : $$\sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$$ is divisible by $p^{2}$.
Since each term of $\displaystyle \sum\limits_{0 < k < \frac{2p}{3}} { p \choose k}$ is divisible by $p$, only thing remains is to prove that the sum $$\sum\limits_{ 0 < k < \frac{2p}{3}} \frac{1}{p} { p \choose k}$$ is divisible by $p$.
How to evaluate this sum: $\displaystyle \frac{1}{p} { p \choose k} = \frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdot 3 \cdots k}$
-
2
This is identical to a problem that appeared on the Putnam exam a couple decades ago. – whuber Sep 3 '10 at 14:47
## 2 Answers
Since we are working in the field $\mathbb{F}_p$ we can write
$$\frac{(p-1)(p-2) \cdots (p-k+1)}{1 \cdot 2 \cdots k}$$ as
$$\frac{(-1)(-2) \cdots (-(k-1))}{1 \cdot2 \cdots k}$$
= $$\frac{(-1)^{k-1}}{k}$$
Let $N = [\frac{2p}{3}]$ and $M = [\frac{N}{2}]$
Thus what we need is
$$\sum_{k=1}^{N} \frac{(-1)^{k-1}}{k}$$ $$= \sum_{k=1}^{N} \frac{1}{k} - 2 \sum_{k=1}^{M}\frac{1}{2k}$$ $$=\sum_{k=M+1}^{N}\frac{1}{k}$$
Now $N+M+1 =p$ so we can rewrite as
$$\frac{1}{N} + \frac{1}{M+1} + \frac{1}{N-1} + \frac{1}{M+2} + \cdots =$$
$$\frac{p}{N(M+1)} + \frac{p}{(N-1)(M+2)} + \cdots$$
which is $0$.
There are $N-M$ terms, which is even, so each term gets paired off.
-
Staggering proof! One question: What made you to think of an $N$ and $M$ like that. – anonymous Sep 3 '10 at 7:59
4
2p/3 is in the question you gave, and when you have alternating sums, it's natural to consider it as sum it all, and minus twice the negative terms. – Soarer Sep 3 '10 at 8:32
@Chandru: Leaving the - signs wasn't going anywhere. Trying to get rid of them, you come up with the N and M. So yeah, what Soarer said. +1 to Soarer's comment. – Aryabhata Sep 3 '10 at 13:34
From your last equation one gets $$\frac1p{p\choose k}\equiv(-1)^k\frac 1k\pmod p.$$ So the problem is equivalent to $$\sum_{0 < k < 2p/3}(-1)^k\frac1k\equiv0\pmod{p}.$$
The case $p=1979$ was set as the first problem at the 1979 IMO. The method used for this extends to all primes, and no doubt you can find solutions for the IMO problem out on the interweb.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.960269570350647, "perplexity_flag": "head"} |
http://terrytao.wordpress.com/tag/talagrand-inequality/ | What’s new
Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao
# Tag Archive
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## 254A, Notes 1: Concentration of measure
3 January, 2010 in 254A - random matrices, math.PR | Tags: Azuma's inequality, Chernoff inequality, concentration of measure, McDiarmid's inequality, moment method, Talagrand inequality | by Terence Tao | 77 comments
Suppose we have a large number of scalar random variables ${X_1,\ldots,X_n}$, which each have bounded size on average (e.g. their mean and variance could be ${O(1)}$). What can one then say about their sum ${S_n := X_1+\ldots+X_n}$? If each individual summand ${X_i}$ varies in an interval of size ${O(1)}$, then their sum of course varies in an interval of size ${O(n)}$. However, a remarkable phenomenon, known as concentration of measure, asserts that assuming a sufficient amount of independence between the component variables ${X_1,\ldots,X_n}$, this sum sharply concentrates in a much narrower range, typically in an interval of size ${O(\sqrt{n})}$. This phenomenon is quantified by a variety of large deviation inequalities that give upper bounds (often exponential in nature) on the probability that such a combined random variable deviates significantly from its mean. The same phenomenon applies not only to linear expressions such as ${S_n = X_1+\ldots+X_n}$, but more generally to nonlinear combinations ${F(X_1,\ldots,X_n)}$ of such variables, provided that the nonlinear function ${F}$ is sufficiently regular (in particular, if it is Lipschitz, either separately in each variable, or jointly in all variables).
The basic intuition here is that it is difficult for a large number of independent variables ${X_1,\ldots,X_n}$ to “work together” to simultaneously pull a sum ${X_1+\ldots+X_n}$ or a more general combination ${F(X_1,\ldots,X_n)}$ too far away from its mean. Independence here is the key; concentration of measure results typically fail if the ${X_i}$ are too highly correlated with each other.
There are many applications of the concentration of measure phenomenon, but we will focus on a specific application which is useful in the random matrix theory topics we will be studying, namely on controlling the behaviour of random ${n}$-dimensional vectors with independent components, and in particular on the distance between such random vectors and a given subspace.
Once one has a sufficient amount of independence, the concentration of measure tends to be sub-gaussian in nature; thus the probability that one is at least ${\lambda}$ standard deviations from the mean tends to drop off like ${C \exp(-c\lambda^2)}$ for some ${C,c > 0}$. In particular, one is ${O( \log^{1/2} n )}$ standard deviations from the mean with high probability, and ${O( \log^{1/2+\epsilon} n)}$ standard deviations from the mean with overwhelming probability. Indeed, concentration of measure is our primary tool for ensuring that various events hold with overwhelming probability (other moment methods can give high probability, but have difficulty ensuring overwhelming probability).
This is only a brief introduction to the concentration of measure phenomenon. A systematic study of this topic can be found in this book by Ledoux.
Read the rest of this entry » | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 21, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9341869950294495, "perplexity_flag": "head"} |
http://medlibrary.org/medwiki/Trace_class | # Trace class
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In mathematics, a trace class operator is a compact operator for which a trace may be defined, such that the trace is finite and independent of the choice of basis. Trace class operators are essentially the same as nuclear operators, though many authors reserve the term "trace class operator" for the special case of nuclear operators on Hilbert spaces, and reserve nuclear (=trace class) operators for more general Banach spaces.
## Definition
Mimicking the definition for matrices, a bounded linear operator A over a separable Hilbert space H is said to be in the trace class if for some (and hence all) orthonormal bases {ek}k of H the sum of positive terms
$\|A\|_{1}= {\rm Tr}|A|:=\sum_{k} \langle (A^*A)^{1/2} \, e_k, e_k \rangle$
is finite. In this case, the sum
${\rm Tr} A:=\sum_{k} \langle A e_k, e_k \rangle$
is absolutely convergent and is independent of the choice of the orthonormal basis. This value is called the trace of A. When H is finite-dimensional, every operator is trace class and this definition of trace of A coincides with the definition of the trace of a matrix.
By extension, if A is a non-negative self-adjoint operator, we can also define the trace of A as an extended real number by the possibly divergent sum
$\sum_{k} \langle A e_k, e_k \rangle.$
## Properties
1 If A is a non-negative self-adjoint, A is trace class if and only if Tr(A) < ∞. Therefore a self adjoint operator A is trace class if and only if its positive part A+ and negative part A− are both trace class. (The positive and negative parts of a self adjoint operator are obtained via the continuous functional calculus.)
2 The trace is a linear functional over the space of trace class operators, i.e. $\operatorname{Tr}(aA+bB)=a\,\operatorname{Tr}(A)+b\,\operatorname{Tr}(B).$ The bilinear map $\langle A, B \rangle = \operatorname{Tr}(A^* B)$ is an inner product on the trace class; the corresponding norm is called the Hilbert-Schmidt norm. The completion of the trace class operators in the Hilbert-Schmidt norm are called the Hilbert-Schmidt operators.
3 If $A$ is bounded and $B$ is trace class, $AB$ and $BA$ are also trace class and $\|AB\|_1=\operatorname{Tr}(|AB|)\le \|A\|\|B\|_1,\qquad \|BA\|_1=\operatorname{Tr}(|BA|)\le \|A\|\|B\|_1$ besides, under the same hypothesis, $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$
4 If $A$ is trace class, then one can define the Fredholm determinant of $1+A$ ${\rm det} (I+A):=\prod_{n\ge 1}[1+\lambda_n(A)]$ for $\{\lambda_n(A)\}_n$ the elements of the spectrum of $A$; the trace class condition on $A$ guarantees that the infinite product is finite: indeed ${\rm det} (I+A)\le e^{\|A\|_1};$ it also guarantees that ${\rm det} (I+A)\neq 0$ if and only if $(I+A)$ is invertible.
### Lidskii's theorem
Let $A$ be a trace class operator in a separable Hilbert space $H$, and let $\{\lambda_n(A)\}_{n=1}^N,$ $N\leq \infty$ be the eigenvalues of $A$. Let us assume that $\lambda_n(A)$ are enumerated with algebraic multiplicities taken into account (i.e. if the algebraic multiplicity of $\lambda$ is $k$ then $\lambda$ is repeated $k$ times in the list $\lambda_1(A),\lambda_2(A),\dots$). Lidskii's theorem (named after Victor Borisovich Lidskii) states that
$\sum_{n=1}^N \lambda_n(A)=\operatorname{Tr}(A).$
Note that the series in the left hand side converges absolutely due to Weyl's inequality
$\sum_{n=1}^N |\lambda_n(A)|\leq \sum_{m=1}^M s_m(A)$
between the eigenvalues $\{\lambda_n(A)\}_{n=1}^N$ and the singular values $\{s_m(A)\}_{m=1}^M$ of a compact operator $A$. See e.g. [1]
### Relationship between some classes of operators
One can view certain classes of bounded operators as noncommutative analogue of classical sequence spaces, with trace-class operators as the noncommutative analogue of the sequence space l1(N). Indeed, applying the spectral theorem, every normal trace-class operator on a separable Hilbert space can be realized as a l1 sequence. In the same vein, the bounded operators are noncommutative versions of l∞(N), the compact operators that of c0 (the sequences convergent to 0), Hilbert-Schmidt operators correspond to l2(N), and finite-rank operators the sequences that have only finitely many non-zero terms. To some extent, the relationships between these classes of operators are similar to the relationships between their commutative counterparts.
Recall that every compact operator T on a Hilbert space takes the following canonical form
$\forall h \in H, \; T h = \sum _{i = 1} \alpha_i \langle h, v_i\rangle u_i \quad \mbox{where} \quad \alpha_i \geq 0 \quad \mbox{and} \quad \alpha_i \rightarrow 0$
for some orthonormal bases {ui} and {vi}. Making the above heuristic comments more precise, we have that T is trace class if the series ∑i αi is convergent, T is Hilbert-Schmidt if ∑i αi2 is convergent, and T is finite rank if the sequence {αi} has only finitely many nonzero terms.
The above description allows one to obtain easily some facts that relate these classes of operators. For example, the following inclusions hold and they are all proper when H is infinite dimensional: {finite rank} ⊂ {trace class} ⊂ {Hilbert-Schmidt} ⊂ {compact}.
The trace-class operators are given the trace norm ||T||1 = Tr [ (T*T)½ ] = ∑i αi. The norm corresponding to the Hilbert-Schmidt inner product is ||T||2 = (Tr T*T)½ = (∑iαi2)½. Also, the usual operator norm is ||T|| = supi(αi). By classical inequalities regarding sequences,
$\|T\| \leq \|T\|_2 \leq \|T\|_1 ,$
for appropriate T.
It is also clear that finite-rank operators are dense in both trace-class and Hilbert-Schmidt in their respective norms.
### Trace class as the dual of compact operators
The dual space of c0 is l1(N). Similarly, we have that the dual of compact operators, denoted by K(H)*, is the trace-class operators, denoted by C1. The argument, which we now sketch, is reminiscent of that for the corresponding sequence spaces. Let f ∈ K(H)*, we identify f with the operator Tf defined by
$\langle T_f x, y \rangle = f(S_{x,y}),$
where Sx,y is the rank-one operator given by
$S_{x,y}(h) = \langle h, y \rangle x.$
This identification works because the finite-rank operators are norm-dense in K(H). In the event that Tf is a positive operator, for any orthonormal basis ui, one has
$\sum_i \langle T_f u_i, u_i \rangle = f(I) \leq \|f\|,$
where I is the identity operator
$I = \sum_i \langle \cdot, u_i \rangle u_i .$
But this means Tf is trace-class. An appeal to polar decomposition extend this to the general case where Tf need not be positive.
A limiting argument via finite-rank operators shows that ||Tf ||1 = || f ||. Thus K(H)* is isometrically isomorphic to C1.
### As the predual of bounded operators
Recall that the dual of l1(N) is l∞(N). In the present context, the dual of trace-class operators C1 is the bounded operators B(H). More precisely, the set C1 is a two-sided ideal in B(H). So given any operator T in B(H), we may define a continuous linear functional φT on $C_1$ by φT(A)=Tr(AT). This correspondence between elements φT of the dual space of $C_1$ and bounded linear operators is an isometric isomorphism. It follows that B(H) is the dual space of $C_1$. This can be used to define the weak-* topology on B(H).
## Notes
1. Simon, B. (2005) Trace ideals and their applications, Second Edition, Amer. Math. Soc.
## References
1. Dixmier, J. (1969). Les Algebres d'Operateurs dans l'Espace Hilbertien. Gauthier-Villars.
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http://math.stackexchange.com/questions/21916/finding-an-inverse-function-with-multiple-occurences-of-y | Finding an Inverse function with multiple occurences of $y$
For some reason I cannot figure out how the book is finding the solution to this problem.
find the inverse function of $f(x)=3x\sqrt{x}$.
My steps seem to lead to a dead end:
step 1. switch $f(x)$ with $y$: $y = 3x\sqrt{x}$
step 2. swap $x$ and $y$: $x = 3y\sqrt{y}$
step 3. solve for $y$:
step 3.1: $\displaystyle \frac{x}{3y} = \sqrt{y}$.
step 3.2: $\displaystyle \left(\frac{x}{3y}\right)^2 = \left(\sqrt{y}\right)^2$
step 3.3: $\displaystyle \frac{x^2}{9y^2} = y$ ... uhhh?
The book answer is $\displaystyle y = \left(\frac{x}{3}\right)^{2/3}$.
-
As always, thanks Arturo Magidin for cleaning up the post, and converting the mathematics into LaTeX. – Eric♦ Feb 14 '11 at 0:54
3
+1 for showing your work. – Arturo Magidin Feb 14 '11 at 0:54
5 Answers
Nothing wrong with what you've done so far, you just haven't finished!
From $\displaystyle \frac{x^2}{9y^2} = y$, move all the $y$'s to the same side:
step 3.4: $x^2 = 9y^3$.
Now isolate $y$:
step 3.5: $\displaystyle\frac{x^2}{9} = y^3$.
Now solve for $y$ by taking cubic roots; you might also notice that the left hand side is a perfect square...
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Note that $\frac{x^2}{9}=\left(\frac{x}{3}\right)^2$. Also note that you can multiply both sides of your equation by $y^2$ to get just one $y$ term, and that the cube root is the same as the $\frac{1}{3}$ power.
You could, as lhf indicates, avoid ever breaking up the $y$ terms in the first place. Technically this would be preferred, because when you divided by $y$ you made the assumption that $y\neq 0$, which is not always true.
Beware of squaring when solving equations in general, because it can lead to extraneous solutions. In this case it is a valid step, but for a silly example consider trying to solve the equation $\sqrt{y}=-1$ by squaring both sides.
A final remark. The original function has an implicit domain, $x\geq 0$. Your inverse function has a formula definition that does not appear to require any domain restriction, but it is misleading to say that $f^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ without mentioning that this is only valid when $x\geq 0$. This domain restriction arises from the fact that the range of $f$ is $[0,\infty)$. The function $h(x)=\left(\frac{x}{3}\right)^{2/3}$ on the whole real line is not invertible (it fails the horizontal line test), whereas if $g(x)=-f(x)=-3x\sqrt{x}$, then the inverse function would be $g^{-1}(x)=\left(\frac{x}{3}\right)^{2/3}$ on $(-\infty,0]$. Looking at the graphs of these functions and keeping in mind how inverses are reflected across the line $y=x$ (resulting from switching $x$ and $y$) should make it clearer what is going on.
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$x=3y\sqrt y$ implies $x^2=9y^3$.
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The algorithm you are applying is correct, and will yield the inverse function. In step 3 you should square both sides. Then we get $$x^2=9y^2\cdot y=9y^3$$
Can you solve the problem from here?
Hint: Divide by $9$ and then take cube roots.
Hope that helps.
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Two hints: $\sqrt{x}=x^{\underline{?}}$ and $x^a\cdot x^b=x^{\underline{?}}$
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 44, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9228405356407166, "perplexity_flag": "head"} |
http://psychology.wikia.com/wiki/Likelihood_function?direction=prev&oldid=136973 | # Likelihood function
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Likelihood as a solitary term is a shorthand for likelihood function. In the colloquial language, "likelihood" is one of several informal synonyms for "probability", but throughout this article only the technical definition is used.
In a sense, likelihood works backwards from probability: given $B$, we use the conditional probability $P(A | B)$ to reason about $A$, and, given $A$, we use the likelihood function $P(A | B)$ to reason about $B$. This mode of reasoning is formalized in Bayes' theorem; note the appearance of a likelihood function for $B$ given $A$ in:
$P(B \mid A) = \frac{P(A \mid B)\;P(B)}{P(A)}\!$
since, as functions of $B$, both $P(A | B)$ and $\frac{P(A | B)}{P(A)}$ are likelihood functions for $B$ given $A$.
In statistics, a likelihood function is a conditional probability function considered a function of its second argument with its first argument held fixed, thus:
$b\mapsto P(A \mid B=b), \!$
and also any other function proportional to such a function. That is, the likelihood function for B is the equivalence class of functions
$L(b \mid A) = \alpha \; P(A \mid B=b) \!$
for any constant of proportionality $\alpha > 0$. Thus the numerical value $L(b | A)$ is immaterial; all that matters are ratios of the form
$\frac{L(b_2 | A)}{L(b_1 | A)}, \!$
since these are invariant with respect to the constant of proportionality.
For more about making inferences via likelihood functions, see also the method of maximum likelihood, and likelihood-ratio testing.
## Concentrated likelihood
For a likelihood function of more than one parameter, it is sometimes possible to write some parameters as functions of other parameters, thereby reducing the number of independent parameters. (The function is the parameter value which maximises the likelihood given the value of the other parameters.) This procedure is called concentration of the parameters and results in the concentrated likelihood function.
For example, consider a regression analysis model with normally distributed errors. The most likely value of the error variance is the variance of the residuals. The residuals depend on all other parameters. Hence the variance parameter can be written as a function of the other parameters.
## Historical remarks
Some early thoughts on likelihood were made in a book by Thorvald N. Thiele published in 1889[1]. The first paper where the full idea of the "likelihood" appears was written by R.A. Fisher in 1922[2]: "On the mathematical foundations of theoretical statistics". In that paper, Fisher also uses the term "method of maximum likelihood". Fisher argues against inverse probability as a basis for statistical inferences, and instead proposes inferences based on likelihood functions.
## Likelihood function of a parametrized model
Among many applications, we consider here one of broad theoretical and practical importance. Given a parametrized family of probability density functions
$x\mapsto f(x\mid\theta), \!$
where θ is the parameter (in the case of discrete distributions, the probability density functions are probability "mass" functions) the likelihood function is
$L(\theta \mid x)=f(x\mid\theta),$
where x is the observed outcome of an experiment. In other words, when f(x | θ) is viewed as a function of x with θ fixed, it is a probability density function, and when viewed as a function of θ with x fixed, it is a likelihood function.
Note: This is not the same as the probability that those parameters are the right ones, given the observed sample. Attempting to interpret the likelihood of a hypothesis given observed evidence as the probability of the hypothesis is a common error, with potentially disastrous real-world consequences in medicine, engineering or jurisprudence. See prosecutor's fallacy for an example of this.
## Example
For example, if I toss a coin, with a probability pH of landing heads up ('H'), the probability of getting two heads in two trials ('HH') is pH2. If pH = 0.5, then the probability of seeing two heads is 0.25.
In symbols, we can say the above as
$P(\mbox{HH} \mid p_H = 0.5) = 0.25$
Another way of saying this is to reverse it and say that "the likelihood of pH = 0.5, given the observation 'HH', is 0.25", i.e.,
$L(p_H=0.5 \mid \mbox{HH}) = P(\mbox{HH}\mid p_H=0.5) =0.25$.
But this is not the same as saying that the probability of pH = 0.5, given the observation, is 0.25.
To take an extreme case, on this basis we can say "the likelihood of pH = 1 given the observation 'HH' is 1". But it is clearly not the case that the probability of pH = 1 given the observation is 1: the event 'HH' can occur for any pH > 0 (and often does, in reality, for pH roughly 0.5).
The likelihood function is not a probability density function – for example, the integral of a likelihood function is not in general 1. In this example, the integral of the likelihood density over the interval [0, 1] in pH is 1/3, demonstrating again that the likelihood density function cannot be interpreted as a probability density function for pH. On the other hand, given any particular value of pH, e.g. pH = 0.5, the integral of the probability density function over the domain of the random variables is 1.
## Notes
1. ↑
2. ↑ Ronald A. Fisher. "On the mathematical foundations of theoretical statistics". Philosophical Transactions of the Royal Society, A, 222:309-368 (1922). ("Likelihood" is discussed in section 6.)
## References
• A. W. F. Edwards (1972). Likelihood: An account of the statistical concept of likelihood and its application to scientific inference, Cambridge University Press. Reprinted in 1992, expanded edition, Johns Hopkins University Press.
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See all photos > | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 23, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8679853081703186, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/tagged/locally-compact-groups+algebraic-groups | # Tagged Questions
2answers
144 views
### More on the versions of the Peter-Weyl theorem
The following three statements appear analogous: For a finite group $G$, the group algebra $\mathbb C[G]$ decomposes as $\bigoplus_{V {\rm\ irred}} V^* \otimes V$. (Peter-Weyl) For a compact group ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8149007558822632, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/142309/probability-with-and-without-repeats | # probability with and without repeats
The standard license plates take 3 letters followed by four digits. ex. ABC 1234 Assume that all plates are non-controversial and meet state restrictions (although we know this is not the case0.
a. how many license plates are possible without repetition of letters or numbers?
b. the standard issue license will repeat letters and/or numbers in the same style (3 letters followed by 4 numbers. How many standard issue plates are available.
Thank you in advance! Susan
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## 2 Answers
So for starter, look at how many choices there are for the first letter, 26. Then we have chosen one, so there are 25 choices for the second, and 24 for the third. From there we consider the numbers. There are 10 choices for the first number (because we are including 0) then there are 9 choices for the second because we used one. Then 8 and 7 for the third and fourth. Then multiply these together for $26 \cdot 25 \cdot 24 \cdot 10 \cdot 9 \cdot 8 \cdot 7$
Then for the repeat. Since we can repeat, we have 26 letter choices for each letter slot and 10 number choices for each number slot so we have $26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 \cdot 10$ or $26^310^4$
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So you have an alphabet of $26$ letters, and there are $$\frac{26!}{23!}=26\cdot25\cdot24=25\cdot624=15600$$ ways to choose three letters (accounting for order) with no repetitions. For the digits, you have something very similar: $$\frac{10!}{6!}=10\cdot9\cdot8\cdot7=5040$$ so the answer to part a will be the product of these two independent sub-choices.
Part b is much easier: $26^3\cdot10^4$, since each letter and digit is completely independent of the others.
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http://math.stackexchange.com/questions/242626/expection-operator-defined-on-colors/244257 | # Expection operator defined on colors
I am trying to implement a computer vision algorithm, but I'm having a problem with some notation used in a article. They define an image as a set of RGB colors with an index $\mathbb{z} = (z_1, z_2, ..., z_n)$.
They then use this formula: $\langle \| z_p - z_q \|^2 \rangle$, with $\langle \cdot \rangle$ defined as the expectation operator over the whole image.
My question is, how can a color be passed to an expectation operator? As far as I understand it, an expectation operator needs a random variable. Somehow, I think I'm interpreting this wrong. Can anyone shed some light on the meaning of the $\langle \cdot \rangle$ operator?
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## 2 Answers
It is possible that they mean empirical expectation, ie they don't really know the exact distribution of the pixels, but they define for instance $p_i=\frac{\#(z_j==i)}{N}$, where $N$ is the total number of pixels and $j\in[1,N]$, which yields the relative number of pixels valued $i$ in the entire image. Now you can define the empirical expectation as $\hat{E}[f(x)] = \sum_i p_i f(x)$ .
I do not know if this is the case here, but it occurs sometimes.
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I have found the answer:
It seems like the meaning of $\langle \| z_p - z_q \|^2 \rangle$ means the color variance measured of the whole image. I dont see their logic, but it seems they meant it like that.
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http://physics.stackexchange.com/questions/54750/unclear-how-heat-interacts-with-navier-stokes | # Unclear how heat interacts with Navier Stokes
I am playing around with an Navier stokes solver and I'm having trouble introducing heat.
Am I right in thinking this would be introduced in the ${\bf f}$ term of ${\partial{\bf u}\over\partial t} = -({\bf u}\cdot\nabla){\bf u}+v\nabla^2{\bf u}+{\bf f}$?
I find the lack of mass term disturbing. Please reassure me!
Also, how do I calculate the force vector, given that I have computed the scalar $dQ\over dt$ from Newton's law of cooling? Or is that the wrong approach?
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## 3 Answers
The Navier-Stokes equation to which you refer is more generally the first moment of velocity of the Boltzmann equation. In order to get a proper connection to heating, you need a second-velocity-moment Navier-Stokes equation. The Boltzmann equation keeps track of distributions of particles. This changes the question from "What is the density and flow of a fluid at a point $\mathbf{x}$ at a time $t$?" to "What is the probability of finding a particle between $\mathbf{x}$ and $\mathbf{x}+\mathrm{d}\mathbf{x}$, with a velocity between $\mathbf{v}$ and $\mathbf{v}+\mathrm{d}\mathbf{v}$, at time $t$?" A nice transition between the two formalisms is discussed in these notes (although they include gravity as an external force the way plasma physicists include the Lorentz force... just imagine the equations without those terms for a plain fluid, keeping only the collisonal term).
It's worth noting that each moment depends on a term from the next higher moment ($d\rho/d t$ depends on $\mathbf{u}$, $d\mathbf{u}/d t$ depends on $\overleftrightarrow{P}$, $d E/d t$ (which is the same moment as $\overleftrightarrow{P}$) depends on the heat conduction, which is a 3rd order moment... Any equation that cuts off has to assume some kind of closure method. For example, to close at first order, you might assume that the pressure is isotropic. Or to close at second order, you might assume that the conduction is infinite (compared to the timescales of interest).
To answer your specific question, volumetric heating can result in a change in pressure, but you need an equation of state linking pressure, temperature and density. (Heating steam will have a very different response from heating water.) The modified pressure term can in turn couple to the first velocity moment (the Navier-Stokes equation you have written).
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To allow for heat effects in a fluid, you need to couple the Navier-Stokes equations, (momentum conservation) which BTW contain the continuity equation for mass conservation too, to the energy (or temperature) equation (energy conservation).
Momentum dissipation in the momentum equation
$$\frac{\partial v}{\partial t} + (\vec{v}\cdot\nabla)\vec{v} = -\frac{\nabla p}{\rho} + \frac{1}{\rho}\nabla S$$
is more correctly described by the divergence of the symmetric stress tensor $S$
$$S = \rho \,\nu (\nabla \circ\vec{v} + (\nabla \circ\vec{v})^{T}) + \rho \,\eta\, I (\nabla \cdot \vec{v})$$
The coefficients $\mu$ and $\nu$ denote the dynamic and kinematic viscosity respectively, $\circ$ is the tensor (outer) product, and $I$ is the unity tensor.
In the energy equation, the momentum dissipation leads to a corresponding positive definite dissipation (frictional heating) $\epsilon$
$$\epsilon = \frac{1}{\rho}(S \, \nabla) \cdot \vec{v}$$
The energy equation can, dependent on the system considered, have different sources of diabatic heating apart from the radiative heating (of which Newtonian cooling es a spacial case), such as latent heating due to phase transitions for example.
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In fluid dynamics, especially in modeling, there are different flavors for including heat.
First of all, heat can be a passive tracer which does not influence the flow, this basically means, that you solve the scalar transport-diffusion equation for the temperature (or energy if you which), which reads, in your notation
$${\partial{T}\over\partial t} +({\bf u}\cdot\nabla){T}=\alpha\nabla^2{T}+Q$$
where $\alpha$ is thermal diffusion coefficient and $Q$ are heat sources (which may be dependent on $T$ for radiation).
The density of fluids is temperature-dependent. You should incorporate this in the Navier-Stokes equations. If the density differences are only relevant for the buoyant terms (e.g. gravity), it is enough to include the effect in $\bf f$, by the Boussinesq approximation. This basically means that you linearize the $\rho(T)$ curve and include the force term as $\beta g (T-T_{ref})$ with $\beta$ a thermal expansion coefficient.
The next step in complexity, arises when $\rho$ changes such, that it takes over control for more terms. The Navier-Stokes equation you gave, assumes constant density, which therefor drops out of the equations. This is generally not true, and you have to take into account the variable density in all terms.
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http://mathoverflow.net/questions/29676/is-there-a-name-for-a-family-of-finite-sequences-that-block-all-infinite-sequence/29681 | ## Is there a name for a family of finite sequences that block all infinite sequences?
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Let ${\bf N}^\omega = \bigcup_{m=1}^\infty {\bf N}^m$ denote the space of all finite sequences $(N_1,\ldots,N_m)$ of natural numbers. For want of a better name, let me call a family ${\mathcal T} \subset {\bf N}^\omega$ a blocking set if every infinite sequence $N_1,N_2,N_3,N_4,\ldots$ of natural numbers must necessarily contain a blocking set $(N_1,\ldots,N_m)$ as an initial segment. (For the application I have in mind, one might also require that no element of a blocking set is an initial segment of any other element, but this is not the most essential property of these sets.)
One can think of a blocking set as describing a machine that takes a sequence of natural number inputs, but always halts in finite time; one can also think of a blocking set as defining a subtree of the rooted tree ${\bf N}^\omega$ in which there are no infinite paths. Examples of blocking sets include
1. All sequences $N_1,\ldots,N_m$ of length $m=10$.
2. All sequences $N_1,\ldots,N_m$ in which $m = N_1 + 1$.
3. All sequences $N_1,\ldots,N_m$ in which $m = N_{N_1+1}+1$.
The reason I happened across this concept is that such sets can be used to pseudo-finitise a certain class of infinitary statements. Indeed, given any sequence $P_m(N_1,\ldots,N_m)$ of $m$-ary properties, it is easy to see that the assertion
There exists an infinite sequence $N_1, N_2, \ldots$ of natural numbers such that $P_m(N_1,\ldots,N_m)$ is true for all $m$.
is equivalent to
For every blocking set ${\mathcal T}$, there exists a finite sequence $(N_1,\ldots,N_m)$ in ${\mathcal T}$ such that $P_m(N_1,\ldots,N_m)$ holds.
(Indeed, the former statement trivially implies the latter, and if the former statement fails, then a counterexample to the latter can be constructed by setting the blocking set ${\mathcal T}$ to be those finite sequences $(N_1,\ldots,N_m)$ for which $P_m(N_1,\ldots,N_m)$ fails.)
Anyway, this concept seems like one that must have been studied before, and with a standard name. (I only used "blocking set" because I didn't know the existing name in the literature.) So my question is: what is the correct name for this concept, and are there some references regarding the structure of such families of finite sequences? (For instance, if we replace the natural numbers ${\bf N}$ here by a finite set, then by Konig's lemma, a family is blocking if and only if there are only finitely many finite sequences that don't contain a blocking initial segment; but I was unable to find a similar characterisation in the countable case.)
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## 2 Answers
Intuitionists use the name "bar" for what you called a blocking set. The relevant context is "bar induction," the principle saying that, if (1) a property has been proved for all elements of a bar and (2) it propagates in the sense that, whenever it holds for all the one-term extensions of a finite sequence s then it holds for s itself, then this property holds of the empty sequence. (I'm omitting some technicalities here that distinguish different versions of bar induction.)
There's also a closely related notion in infinite combinatorics, called a "barrier"; this is a collection $B$ of finite subsets of $\mathbf N$ such that no member of $B$ is included in another and every infinite subset of $\mathbf N$ has an initial segment in $B$. This is the subject of a partition theorem due to Nash-Williams: If a barrier is partitioned into two pieces, then there is an infinite $H\subseteq\mathbf N$ such that one of the pieces includes a barrier for $H$ (meaning that every infinite subset of $H$ has an initial segment in that piece).
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If you assume that no element of the family is an initial segment of the other, it is called a barrier in wqo (well quasi order) theory. See mostly Nash-Williams.
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http://physics.stackexchange.com/questions/40895/how-robust-is-kramers-degeneracy-in-real-material | # How robust is Kramers degeneracy in real material?
Kramers theorem rely on odd total number of electrons. In reality, total number of electrons is about 10^23. Can those electrons be so smart to count the total number precisely and decide to form Kramers doublets or not?
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I think Kramers theorem is not really useful for a bulk material, as the concept of degeneracy fails to have a meaning. In a bulk material you will have energy bands, not discrete levels as in individual atoms. The doublets can still exist, but only if the electrons do not interact. – Alexander Oct 16 '12 at 8:18
## 2 Answers
Remember that for crystalline materials, we usually assume an infinite number of particles, and that electrons do not interact. This allows us to Fourier transform and see that each pseudo-momentum $k$ is independent --- essentially to consider a single unit cell. In this context, Kramer's theorem states that if there is an odd number of electrons per unit cell (we ignore proton and neutrons if we don't care about hyperfine structure; otherwise we would), and assuming time reversal invariance, there is (at least) a two-fold degeneracy of all energy levels. Indeed, this may be seen as the basis of topological insulators.
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I think Kramer's theorem is really only useful when you can write down a wavefunction for your system. It then tells you the degeneracy of the ground state of your wavefunction. If you have 1 mole of your material you couldn't write down a wavefunction for it so the degeneracy of the ground state would be meaningless.
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http://physics.stackexchange.com/questions/12129/speed-of-light-observed-speed-while-travelling-at-the-speed-of-light/12130 | # Speed of light, observed speed while travelling at the speed of light
I was watching Discovery channel the other night, they were telling that time slows down when you travel at a higher speed. This means there is a difference between the actual speed you travel at, and the perceived speed.
Does anybody know what the perceived speed is, the speed it seems you're travelling at?
What does this imply? Does it have any meaning, in some theories perhaps, that the perceived speed is lower? Does this mean that you can never actually travel at the speed of light?
(All this is if we don't account for the fact that's impossible to reach the speed of light.)
I'm also not into physics, so try to keep it simple...
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@ Willie; so this is special-relativity? Didn't know what that meant at first, thanks. – Simon Verbeke Jul 10 '11 at 15:23
3
I think you should start reading books about relativity. If you don't read anything, you will never understand anything. I recommend to start with Taylor, Wheeler "Spacetime physics". – Physicsworks Jul 10 '11 at 17:39
If I can find anything at the local library I will ;) Thanks for the suggestion – Simon Verbeke Jul 10 '11 at 20:40
## 2 Answers
I apologise, this answer as it stood originally was sloppy and flat out wrong on key points, and probably still needs attention from a physicist.
When you travel close to the speed of light, you experience less subjective time, according to the Lorentz transform. Moreover, external distances are contracted by the same transform.
The factor by which time slows down for a rapidly moving object is $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. As velocity $v$ increases it approaches equality with the speed of light, $c$, and the factor approaches infinity.
This means that if you had an arbitrarily fast spacecraft (I assume you don't) you could tour the universe in (from your reference frame) a few years, however you may return to find humanity extinct and continents in unexpected locations. You would actually perceive yourself as travelling at arbitrarily high speeds, in terms of the subjective time it takes to travel from point to point. However, in making measurements against objects you move past, you would see that your subjective velocity does not exceed the speed of light as the rest of the universe appears length-contracted. There are other limitations in your ability to enjoy your tour of the universe - if you set your sights on a star that is millions of light-years away it may well rapidly age (again, from your reference frame) and die before you get there.
References:
Lorentz factor at wikipedia
Relativistic Doppler Effect at wikipedia
The 3rd book of Tipler Physics also has a great primer on relativity, albeit with neglect of acceleration, which is critical in generating discrepant clocks between two observers a la the twin paradox.
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Wow.. This is really hard to understand (I mean it in the way of "How the *** is it possible?", not that I don't understand what you mean." – Simon Verbeke Jul 10 '11 at 16:01
– Richard Terrett Jul 10 '11 at 16:30
So, if they wouldn't move, their lifespan isn't long enough to reach us. But their great speed reduces time for their p.o.v., so they CAN reach us? Wow, very interesting – Simon Verbeke Jul 10 '11 at 16:43
@ Simon Verbeke: Yep, that's it. – Richard Terrett Jul 11 '11 at 3:19
@ Simon Verbeke: I was wrong on the 'see the external universe sped up' bit - Carl is correct that two observers moving past each other at high speed will see each others' clocks slowed down. Acceleration breaks the symmetry. – Richard Terrett Jul 11 '11 at 8:12
Your perceived speed is the same as the speed you travel at. That is, if you travel at 0.8c past the earth, people on the earth will measure your speed as 0.8c, and you will measure the speed of the earth going past you also as 0.8c. Thus you will conclude that you are traveling at 0.8c. So there is no difference between the speed you travel at and the speed you perceive that you travel at.
On the other hand, the people on the earth will see you as slowed down (relative to themselves). Similarly, you will see the people on the earth as slowed down. The two of you can't come to agreement on this because you don't have a consistent way of comparing your watches. Whichever one of you does the comparison will conclude that the other is slowed down. This is related to the fact that in relativity there is no way to determine time in an absolute fashion. It's always relative. There's no way to figure out which of you is moving and which is standing still. It is only when you get back to the earth that you might decide that it was you who was moving and it was you who was experiencing a slowing in time.
The fact that when you get back to the earth you determine that you were the one that was slowed down is only due to the fact that you were the one who changed course and returned to the earth. If instead the earth were somehow sped up and caught up with you, everything would be reversed. It would you who grew older than the people on earth.
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http://mathhelpforum.com/calculus/176491-quotient-rule.html | # Thread:
1. ## quotient rule
2. $\displaystyle y= \frac{3}{5r+11}\implies y' = \frac{(5r+11)\times (3)'-(5r+11)'(3)}{(5r+11)^2}$
3. thanks for the answer could you please explain the steps reasoning so i can get it.
4. Yep, I can,
The quotient rule
$\displaystyle y= \frac{u}{v}\implies y' = \frac{vu'-v'u}{v^2}$
Or you can say
$\displaystyle y= \frac{3}{5r+11} = 3(5r+11)^{-1}$
then use the chain rule.
5. thank you very much i now see how you did it thanks a bunch
6. By the way, seeing as this is clearly a Calculus question, why have you posted it in Pre-Calculus?
7. I thought only university questions could go there?
if that's wrong how do i move the thread?
8. To move the thread, report your own post using the Report Post tool at the lower left corner of the OP. Looks like a triangle with an exclamation mark inside it. A mod will move it for you. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9420390725135803, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/37518?sort=oldest | ## Why do statistical randomness tests seem so ad hoc?
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Wikipedia describes Kendall and Smith's 1938 statistical randomness tests like this:
• The frequency test, was very basic: checking to make sure that there were roughly the same number of 0s, 1s, 2s, 3s, etc.
• The serial test, did the same thing but for sequences of two digits at a time (00, 01, 02, etc.), comparing their observed frequencies with their hypothetical predictions were they equally distributed.
• The poker test, tested for certain sequences of five numbers at a time (aaaaa, aaaab, aaabb, etc.) based on hands in the game poker.
• The gap test, looked at the distances between zeroes (00 would be a distance of 0, 030 would be a distance of 1, 02250 would be a distance of 3, etc.).
It is not obvious to me that these four particular tests were chosen with any deep understanding of how best to detect nonrandomness. Rather, it seems each one was probably chosen for simplicity.
Well, it's easy to see why early work in the field would be like that. But fast forward to 1995 and George Marsaglia's Diehard tests seem, on the surface, just as ad hoc:
• Birthday spacings: Choose random points on a large interval. The spacings between the points should be asymptotically Poisson distributed. The name is based on the birthday paradox.
• Overlapping permutations: Analyze sequences of five consecutive random numbers. The 120 possible orderings should occur with statistically equal probability.
(...and so on)
It is not obvious to me that these tests are really independent (i.e. that none is entirely redundant, rejecting only sequences also rejected by at least one of the other tests), or that there aren't obviously better generalizations of them, much less that they were chosen as a set to try to cover any particular space efficiently.
Ideally, a test suite would be designed to reject as many sequences of low Kolmogorov complexity as possible with minimal computation and false alarms. Is a more theoretical approach to this possible? Why hasn't it happened? —Or is there more to the state of the art than meets the eye?
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2
I think the big issue is the meaning of randomness here. Give a precise meaning to it, then you get a good test. The problem with the Kolmogorov complexity, you mention, is that it is a complicated thing to compute. Playing poker is soooo much easier. – Helge Sep 2 2010 at 18:44
4
While an important question, I disagree that the meaning of randomness is the main issue here. Optimizing the computational efficiency of the test, given the sequences it rejects, is not a very subjective question. I imagine at least some thought was given to this in designing the tests, and I'd like to know more about it – Andrew Critch Sep 2 2010 at 19:55
One thing that strikes me about the first set of tests especially, and the second set to a lesser extent, is that most any human-generated set of "random" numbers will miserably fail the first set of tests (unless of course they are trying to design the sequence to pass these randomness tests, in which case the sequence is decently close to random...) – drvitek Sep 3 2010 at 5:05
## 3 Answers
It's not clear that Marsaglia's tests are really good enough. See this Stack Overflow discussion.
Kolmogorov complexity is not the right criterion for statistical randomness tests, since any pseudorandom sequence has low Kolmogorov complexity. What you really want in a random number generator is for the sequence to be computationally pseudorandom; that is, without knowledge of the seed, no polynomial-time test can distinguish the sequence from a truly random sequence.
In fact, there are a number of random number generators which are believed to be computationally pseudorandom. Nobody uses these, however, partly because they are computationally too expensive, and partly because of inertia and partly because the existing pseudorandom generators we have are good enough most of the time. A while ago, for one of the most common methods of pseudorandom number generation (linear congruential), I ran into cases where they unexpectedly produced the wrong answer.
Marsaglia's tests were developed over a number of years, and I believe each was designed to detect certain flaws that many pseudorandom number generators contained at the time. Once good-enough pseudorandom number generators were available, nobody bothered creating a more stringent series of tests.
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5
Ed Coffman and I stumbled across the following example when trying to debug a program (for some related stochastic process). Take $k$ points on a circle, and repeat the following steps indefinitely. (a) Add a random point on the circle. (b) Choose a random number $j$ between $1$ and $k+1$, and delete the $j$'th point (starting from the origin). This process should converge to $k$ random points on the circle, so you can easily compute what the expected square of the distance between a random point and its neighbor should be. With our random number generator, we got the wrong answer. – Peter Shor Sep 3 2010 at 1:53
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(Excessively long comment continued ...) If I remember correctly, we tried several linear congruential random number generators, and they all gave us the wrong answer. (It was close, but statistically significantly wrong in one of the later decimal points; some were better than others.) And again, if I remember correctly, the details of the program are exactly how you would program this in the most obvious way. All of the other types of random number generators we tried worked fine for this problem (but after this experience, I don't trust that they will work for all problems.) – Peter Shor Sep 3 2010 at 1:58
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(Excessively long comment concluded.) My advice for people doing simulations who want to be exceedingly careful would be to try two different types of psuedorandom number generator, and check that they agree. This is computationally cheaper than using a cryptographically good one. The Mersenne Twister was highly recommended in the Stack Overflow discussion I linked to above, and there are lots of linear congruential ones around, which generally have fairly good statistics. If the two different types give the same answer, you should be very safe. And if they don't, you know something's wrong. – Peter Shor Sep 3 2010 at 2:08
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I would say that the tests were designed to detect "certain flaws that many pseudorandom number generators contained at the time", rather than fix them. This would allow testing new pseudorandom number generators to make sure that they don't also contain the same sort of mistakes. I had a similar error initially in some cellular automata simulations, where the size of the lattice and a coincidental state recurrence interval multiplied together equaled the 2^32 cycle length of the RND PRNG function in the GNU C-compiler. Better PRNGs exist:e.g. statmath.wu.ac.at/prng/doc/prng.html – sleepless in beantown Sep 3 2010 at 3:13
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@peter, that's also the same reasoning behind using two hashing algorithms to double check the integrity of a downloaded source package. While there have been a couple of different ways found to successfully find an MD5sum hash collision, finding a way to modify a file which doesn't also change an SHA-1 sum hash has not been discovered. @András Salamon: It's imperative to understand the algorithms underlying PRNGs when using them to avoid a systematic error in using them inappropriately, or in not noticing that there might be too limited an entropy source used to seed them, as with Debian SSL. – sleepless in beantown Sep 4 2010 at 0:25
show 7 more comments
### 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.
Kolmogorov complexity is a universal quantity for infinite strings. But since randomness tests are run on finite strings, particular choice of architecture or encoding will play a large role in how the strings will be ordered. A goal of good randomness test designer is to choose an encoding such that strings coming from known random sources (like random.org) are deemed more complex than strings coming from known non-random sources (like pseudo-random generators). Good tests incorporate knowledge of how typical non-random strings are generated, hence the apparent ad hockery.
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Kolmogorov complexity can be used to inform an entropy that captures the information in the OP's question as well as the latent information in intrasequence correlations. Without accounting for this, you can shuffle up chunks of a sequence so that the doublet, triplet,...,k-let frequencies match and yet the sequences themselves will be very different globally. – Steve Huntsman Sep 2 2010 at 18:59
But how do you define Kolmogorov complexity for finite string... – Yaroslav Bulatov Sep 15 2010 at 18:08
@Yaroslav Isn't Kolmogorov complexity defined in terms of finite objects. According to Li and Vitanyi (arxiv.org/pdf/cs/9901014v1) "the Kolmogorov complexity of a finite object $x$ is the length of the shortest effective binary description of $x$." (Appendix C of the linked manuscript is all about randomness tests and may be relevant to the OP.) – R Hahn Sep 26 2010 at 18:33
Shortest program for which Turing Machine? Here's a cool one rendell-attic.org/gol/utm/index.htm – Yaroslav Bulatov Sep 26 2010 at 19:35
Well, the KC is defined for finite strings. But 1.) it is only defined up to an additive constant (translating between Turing machines) and 2.) it isn't computable...yeah, Conway's Life is super fun. – R Hahn Sep 26 2010 at 20:31
In Chapter 2 of the book Group Theory in the Bedroom, Brian Hayes mentions how difficult it is to extract "true" randomness from the physical world (including constructing tables of random numbers the old-fashioned way, and how Fisher and Yates had to "fix" their tables in post-processing to re-balance the digits, causing their colleagues to comment: "a procedure of this kind may cause others, as it did us, some misgivings").
The point he makes is that physical randomness is hard to achieve, because even if your source is truly random (whatever that means), the measurement process is apt to introduce bias, so that pseudo-random generators can out-perform physical random sources on many statistical test. [I'll add a more precise page reference as soon as I can find it.]
Given this state of affairs, the ideal test suite seems unlikely to exist. I would agree with you that less ad-hoc tests do seem desirable, though.
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That sounds very interesting. What about hardware based random number generation based on chaotic processes, such as when SGI patented and implemented [Lavarand](en.wikipedia.org/wiki/Lavarand), a hardware RNG using an image of a [lava lamp](en.wikipedia.org/wiki/Lava_lamp) as the seed for a PRNG. It's still questionable how chaotic the lava lamp image is; however an improved version [LavaRND](en.wikipedia.org/wiki/LavaRnd) uses noise from a CCD image sensor as an entropy source followed by multiple SHA hash, rotates, and folds. www.lavarnd.org/what/digital-blender.html – sleepless in beantown Sep 3 2010 at 5:00
@sleepless: I need to find a really good quote from that chapter. But he does mention a lot of hardware-based solutions and attendant issues, including the fact that a generator based on radio-active decay may miss disintegration that are too close together. The book is aimed at a wide readership, and the chapter covers as much cryptographic applications as sources of randomness, so there must be better references for this out there. – Thierry Zell Sep 3 2010 at 15:20 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9440647959709167, "perplexity_flag": "middle"} |
http://complexzeta.wordpress.com/2007/09/12/a-mathematicians-take-on-challah-braiding/ | An Idelic Life
Algebraic number theory and anything else I feel like telling the world about
# A Mathematician’s Take on Challah Braiding
Wednesday, September 12, 2007 in food, group theory
For many years, my mother has baked a challah nearly every Friday for Shabbat. Occasionally, however, she asks me to do some portion of the challah making, possibly including the braiding. For reasons we’ll see later, I don’t like her braiding algorithm. This post includes several algorithms, written in a way that someone who knows what a braid group is can understand. (I never had much success following those series of diagrams I sometimes see; I always wished someone would write out the braiding process in terms of generators of the braid group, so that’s what I’m going to do here after I give the preliminary definitions.)
Wikipedia’s page on braid groups has lots of interesting things, so I’ll only write a few essential points here, leaving the reader to explore Wikipedia at eir leisure. I’m finding it a bit tricky to give a good informal definition of braids, so I’ll just assume that my reader knows roughly what a braid is and skip to the formal definition.
The braid group on $n$ strands is the group $B_n=\langle a_1,\ldots,a_{n-1}\mid a_ia_{i+1}a_i=a_{i+1}a_ia_{i+1} \text{ for } 1\le i\le n-2, \ a_ia_j=a_ja_i \text{ for } |i-j|>1\rangle$. In terms of actually playing with braid strands, $a_i$ means interchanging strand $i$ with strand $i+1$ by putting strand $i$ over strand $i+1$. It is pretty simple to see that these generators do indeed induce all possible braids (although I haven’t yet said what a braid is), and that the relations ought to hold. Now, of course, a braid is an element of the braid group.
The braid groups become rather complicated quite quickly. While $B_0=B_1=0$ and $B_2=\mathbb{Z}$, already $B_3$ is nonabelian, and it’s isomorphic to the fundamental group of the complement of a trefoil knot in $\mathbb{R}^3$.
Note also that there is a natural homomorphism $\pi:B_n\to S_n$ that tells us where the strand that started in the $i^\text{th}$ position ends up.
Okay, now it’s time for some challah braiding algorithms. My mother’s usual challah has four strands on the bottom and three on the top. The algorithm for the top braid is pretty natural: $(a_1a_2^{-1})^n$, where $n$ is decided by the length of the dough ropes.
I’m more concerned about the element of $B_4$ used for the bottom braid. She uses $(a_1a_2^{-1}a_3^{-1}a_2)^n$. If $n=1$, we have $\pi(a_1a_2^{-1}a_3^{-1}a_2)=(142)(3)$ (in cycle notation). This is already bad news to me: one step of the algorithm produces a single fixed point! I think one step of the algorithm ought to give an $n$-cycle (here a 4-cycle) or else a pure braid (i.e. a braid in the kernel of $\pi$). But it gets worse: the strand that starts in position 3 has no undercrossings. So when we’re done, it sits on top of every other strand.
It turns out not to be so bad because the three-strand braid sits on top of the four-strand braid, so the central portion of the four-strand braid is not visible in the finished bread. But aesthetically (and mathematically), this feels like a serious flaw to me.
Fortunately, I found an alternate algorithm for four-strand braiding that lacks these flaws: $(a_2a_1a_3^{-1})^n$. If $n=1$, $\pi(a_2a_1a_3^{-1})=(1243)$, which is nice. Furthermore, every strand has both overcrossings and undercrossings. So this is my new preferred braid.
Sometimes, however, it is preferable to braid with six strands. There was an article in the newspaper that explained how to do it, but I was unable to follow it. Fortunately, I found a YouTube video that shows someone doing it (possibly the same way; I can’t tell). I was able to transcribe this method in terms of generators of the braid group. However, I’m not quite sure where it is supposed to end, so my braid may be slightly different from the one shown in the video. The braid is the video is $(e^{-1}d^{-1}c^{-1}b^{-1}a^{-1}(bcdeabd^{-1}c^{-1}b^{-1}a^{-1}e^{-1}d^{-1})^n$, except that it might stop somewhere in the middle of the $(\cdot)^n$. I don’t really want to calculate $\pi$ of this braid (computations like this have never been that easy for me), but I would guess that it is a 6-cycle if it stops at an appropriate moment. (Also, it’s not as complicated as the formula would make it seem, since there’s a lot of stuff like moving the strand on the right all the way over to the left, and it takes a lot of generators to express that, even though it’s not complicated when you’re actually braiding dough.)
## 3 comments
Heheheh. I make Challa Bread. I tried doing the Star of David arrangement last time and it looked like I must’ve failed math completely.
Cool post.
Even though it is a bit old, I just found this site. You can tell, I am sure, that your mathematical discussion of braiding is a completely new language to me. So what I want to know is do these algorithm actually help someone DO the actual braiding? Or is it that this is the type of mental gymnastics that mathematicians like to do?
Well, both. It helps me to think about braids in terms of mathematics even when I’m actually braiding something because I’m already familiar with braid groups and the language of group presentations. Thus it’s easier to learn new braiding schemes by expressing them in this manner. Also, it’s easier to see what’s a nice braid mathematically and aesthetically (which are likely the same in this case) by writing down groups and homomorphisms. (However, I did happen to notice the problem with the (142)(3) braid without writing down any mathematics because something seemed wrong when I was doing it.)
But of course we turn it into mathematics because we can, and that’s the sort of thing that mathematicians like to do.
Braid groups do show up in important ways in other areas of mathematics. For example, if we want to distinguish two knots, one way of doing it is to compute the fundamental group of its complement, which is essentially all the loops we can draw in $\mathbb{R}^3$ that don’t touch the knot up to wiggling them around. For some knots, this group is a braid group.
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http://math.stackexchange.com/questions/tagged/functions+dynamical-systems | # Tagged Questions
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### Continuum limit of cellular automata
Is there any function defined for say the plane, that has interesting nontrivial behaviour similar to Conway's Game Of Life, but where every point's on/off status is decided by something like the ...
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### Cellular automata – like functions
Let's consider one dimensional cellular automaton. It is build upon its rule, i.e. a function $f : S^3 \rightarrow S$, where $S = \{0,1\}$. The case described is the elementary cellular automaton, ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8670939803123474, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/51526/what-is-the-first-appearance-of-the-mv-mclerran-venugopalan-initial-condition/51531 | # What is the first appearance of the MV (McLerran-Venugopalan) initial condition?
First a quick introduction for the unfamiliar: in saturation physics (my research field), a lot of theoretical work centers on the BK (Balitsky-Kovchegov) equation, which is a differential equation which governs the structure of the proton. It basically takes the form
$$\frac{\partial}{\partial Y}N = K\otimes N - N^2$$
$N$ is a function related to the proton's structure, and is what we solve the equation for. It's a function of $\mathbf{r}$, the position in the plane perpendicular to the beam line, and of $Y = -\ln x$, where $x$ is the momentum fraction of the quark or gluon involved in the collision. $K\otimes$ is some integral operator.
Solving an equation of this form starts with an initial condition at some initial $Y = Y_0$. I've seen a number of recent papers (1,2,3,etc.) that use an initial condition of this form:
$$N(\mathbf{r}, Y_0) = 1 - \exp\biggl[-\frac{(r^2 Q_{s0}^2)^\gamma}{4}\ln\biggl(e + \frac{1}{r\Lambda}\biggr)\biggr]$$
Usually the equation is accompanied by the citation of a trio of papers (4,5,6) by Larry McLerran and Raju Venugopalan from 1994, and accordingly the expression is called the MV initial condition. The thing is, I checked those papers from 1994 and I can't seem to find this expression anywhere in them, nor can I find anything that it can obviously be derived from. So I'm wondering, is there a later paper that actually derives or postulates the MV initial condition itself? Perhaps by starting from the results of the McLerran and Venugopalan papers?
Or is it really in one or more of those three 1994 papers, and I'm just missing it?
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I won't be too surprised if nobody can answer this, as saturation physics is really obscure, but I figured there should be the occasional probe of the site's ability to tackle highly specific questions. – David Zaslavsky♦ Jan 18 at 5:11
Can you access the PhysRevD documents? I remember looking up the same thing a few years ago and I know I've looked at the PhysRev PDFs instead of the ones on arXiv. I can't access them from where I am now, though. They might be slightly different. – David M. R. Jan 30 at 3:14
Not from where I am now (home), but when I get on campus, I'll check that. – David Zaslavsky♦ Jan 30 at 3:17
## 1 Answer
I am not familiar with the area but in ref. 4 there are plenty of derivations given (e.g. eqns. (1), (3), (84)). I guess integrating one or some of the eqns. could give you your equation. But I really just browsed through the articles and I am not even sure if their notation matches yours or the ones in the newer articles. Just to have an idea how this may come from.
EDIT: I found another paper in the reference that might clear things a little - Link to the paper. In II. D. you will find the following statement:
Finally we have to specify the initial condition (i.c.) for the evolution or, equivalently, the precise shape of the proton unintegrated gluon distribution (UGD), ...
So what I guess is that it is not that easy to alter the initial conditions to the UGD. I think a more or less complicated parametrization is needed to do so. The paper referenced in that connection is this one: Link to the paper.
I hope this helps a little.
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My reference 1 only has 15 equations. Did you mean a different one? – David Zaslavsky♦ Jan 28 at 23:15
Yes you are correct. I meant ref. 4. Corrected that. I searched through the papers again and found a reference that might help. See edit. – DaPhil Jan 29 at 8:00
Thanks, I will have to look through this. – David Zaslavsky♦ Jan 30 at 3:17 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9457147121429443, "perplexity_flag": "middle"} |
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