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http://math.stackexchange.com/questions/260594/order-sequence-of-n-numbers-with-repetition-having-4-consecutive-increasing-nu | Order sequence of n numbers (with repetition) having 4 consecutive increasing numbers!!
Context:
I recently came across a question like “what’s the probability that a random number (integer) generator will generate four successive numbers as consecutive increasing numbers?”. Many people have solved this assuming each sample point consists of only 4 random numbers and counting number of cases four of them are consecutive & increasing and finally concluding the probability as $\frac{(N-4+1)}{N^4}$ assuming the generators produces integers in the range $[1,N]$. However this problem has induced many interesting thoughts in my mind such as what if we try to find the probability for the same event (i.e. 4 consecutive numbers) but in $n (n\ge4)$ picks! Of course in this case 4 consecutive increasing number would be rephrased as at least one occurrence of 4 or more consecutive increasing numbers in successive picks. Yet another interesting observation I made here is that with $n$ changing the probability also changes. I spent ample time to solve this question but couldn't figure out how to count size of the event space. In this regard I’m rephrasing my question below with equivalent Urn & Ball model and any help solving this question would be much appreciated.
Question:
An urn has $N$ balls marked with a integer number from $1$ to $N$. Thus there would be exactly one ball in the urn with a given number from $[1,N]$. If we randomly pick $n$ balls with replacement then we know that there would be $N^n$ possible ways of picking $n$ balls. The question is how many of such ways we’ll have at least one occurrence of $4(=c)$ or more successive picks having consecutive increasing numbers? We can assume $n\ge4$.
Example: For N=9 (range of numbering 1-9) , n=7 (number of pick), c=3(number of picks with consecutive numbers)
Sequences that are included:
i) $1,2,3,1,2,3,1$ (2 occurrences of 3 consecutive numbers)
ii) $1,2,3,3,4,5,6$ (1 occurrence of 3 consecutive numbers and 1 occurrence of 4 consecutive numbers)
iii) $8,9,9,9,4,5,6$ (1 occurrence of 2 consecutive numbers and 1 occurrence of 3 consecutive numbers)
Sequences that are NOT included:
i) $1,2,4,5,7,8$ (3 occurrences of 2 consecutive numbers)
@Hurkyl: I'm afraid that I could not guess whether you're criticizing the post or providing a hint to solution. In fact I didn't get why did you mention "no 4-sequence can begin with a digit in [N-2,N]"! To me if N $\ge$ 4+2=6 then a 4-sequence can start with N-2. Am I missing something! – BuckCherry Dec 17 '12 at 11:45
@BuckCherry: What you seem to be missing is that once you have chosen one of $\{N-2,N-1,N\}$, there is no possible way to continue for at least three steps, each time choosing the number that is one more than the previously chosen number. Therfore once such a number is drawn (and it is not itself one more than the previous number) one goes to a "nothing useful found yet" state. What Hurkyl suggests is that the process of checking whether a desired subsequence exists can be done by an automaton with only very few states. For me it is not convincing. – Marc van Leeuwen Dec 17 '12 at 16:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9609491229057312, "perplexity_flag": "head"} |
http://quantummoxie.wordpress.com/2010/04/17/spacetime-a-more-convincing-argument/ | # Quantum Moxie
thoughts of a selective subjectivist
## Spacetime: a more convincing argument
Following up on my previous post, here’s an even more convincing argument that it is possible to construct a toy universe in which the curvature of spacetime is due to electromagnetism and not gravity and that I was unjustly vilified over at nForum (remember while reading the rest of my post below that it was asserted that I lacked a rudimentary knowledge of general relativity).
Consider again a charged massless universe, that is a universe in which there are electromagnetic fields but no gravitational fields. In such a case the complete stress-energy tensor only includes the electromagnetic portion. Einstein’s field equations are
$G^{\alpha\beta}=8\pi T^{\alpha\beta}$.
In order to prove my point I need to show that $G$ is solely determined by $T$ (which I always thought was the standard interpretation, but after the tongue-lashing I received I’ll prove it just to be on the safe side).
In my toy universe I will assume that Riemannian geometry still exists (since, as a mathematical tool, it is independent of anything physical anyway). The definition of $G$ is
$G \equiv R^{\alpha\beta}-\frac{1}{2}g^{\alpha\beta}R$
where the $R$s may be written in terms of $g$. Thus it boils down to determining whether $g$, which is the metric, can be independently determined (i.e. from conditions not present in Einstein’s field equations).
As a simple case, let’s take the weak field approximation,
$g_{\alpha\beta} = \eta_{\alpha\beta} + h_{\alpha\beta}$
where $|h_{\alpha\beta}|<<1$ and $\eta_{\alpha\beta}$ is the Minkowski metric for flat spacetime. If we further restrict ourselves to the linearized theory, the linearized, weak field Einstein equations are
$\Box\bar{h}^{\mu\nu} = -16\pi T^{\mu\nu}$
where $\Box$ is the d'Alembertian. It is easier to explain the next step by showing what is done in normal GR, i.e. not in my toy universe. In such a case, for example,
$\bigtriangledown^{2}\bar{h}^{00}=-16\pi\rho$
and this is compared to the Newtonian
$\bigtriangledown^{2}\phi=4\pi\rho$
where $\phi$ is a scalar potential identified with Newtonian gravity. Thus, we choose
$\bar{h}^{00} = -4\phi$
in order to force this to match the Newtonian gravity!
Let’s switch back to my toy universe now. The energy density, $\rho$, is given by $T^{00}$ which, in normal GR is the density of the gravitational field. But in my toy universe,
$T^{00} = \frac{1}{2}(E^{2} + B^{2})$
where we have employed units such that $\mu_{0} = \epsilon_{0} = c = 1$. Further, in the electrostatic case, we note that,
$\bigtriangledown^{2}\phi = -\rho$
where $\rho$ is the charge density and where we again are employing units with $\mu_{0} = \epsilon_{0} = c = 1$. Thus in the electrostatic case of my toy universe we may choose
$\bar{h}^{00} = -16\pi\phi$
where $\phi$ is the charge density! The metric and thus the curvature of spacetime in my toy universe has absolutely nothing to do with gravity!
Addendum: If the full stress-energy tensor is employed in general relativity, i.e. with both gravitational and electromagnetic portions, one could presumably do a similar weak field construction in which $\bar{h}$ depends on both gravitational and electromagnetic fields which means I’m even right in this universe: the metric encodes the curvature of spacetime due to field sources and is not necessarily due solely to gravity.
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### 43 Responses to “Spacetime: a more convincing argument”
1. BlacGriffen Says:
April 17, 2010 at 6:06 pm
No good, your math looks fine but your conceptual framework is wrong. All you’ve shown is that you can get a space-time curvature, which is by definition gravity, from an energy density caused by an electromagnetic field. They are still entities which you could consider to be separate.
Put another way, you’ve demonstrated that photons couple to the gravitational field at the classical level even in the absence of anything else in the universe.
If you want to do this right start looking at QED, explicitly write in the U(1) gauge theory metric, and start making it dynamic. The next, though perhaps not necessary, step is to find a way of making the low energy effective Lagrangian linear in the field strength/curvature. If you’re up for a real challenge, though, you can consider a non-abelian gauge theory.
Even then, however, there’s still going to be one factor that separates gauge theories from gravity. And that is: the vector being parallel transported in the case of the gauge theories is purely orthogonal to the space-time manifold on which the theory is defined, whilst for gravity the vectors are tangent. So what in other theories is a clear distinction between Lorentz and group indices gets all mixed up in gravitational theories.
The other route, that I don’t know whether it’s been tried, would be to find a way to remove the dynamics from the metric in relativity. That one’s less of a tall order mathematically, since it would actually simplify the equations describing gravity, but more of an empirical problem to see whether or not you can explain all of the data (especially since you’d be basically throwing out big bang cosmology).
BG
2. quantummoxie Says:
April 17, 2010 at 6:25 pm
a space-time curvature, which is by definition gravity
And this is where I disagreed with the folks on nForum. It is precisely my point that this definition is arbitrary. Read the very beginning of Ch. 8 in Schutz’ book on GR (first edition). The logical premise that space-time curvature is gravity is axiomatic, i.e. it’s not derivable, it’s simply chosen. If you walk through the full derivation of Einstein’s field equations it is quite clear that this definition is a choice. In fact, it seems fairly clear given what I’ve shown above.
(By the way, you’re a mysterious figure, Mr./Ms. BlackGriffen. )
3. BlacGriffen Says:
April 17, 2010 at 6:58 pm
Heh. I’ve had this pseudonym since I was an undergrad, and I use it all over the place. My real name is Sean Lake, and I’m a grad student at UCLA. I don’t have anything published, yet, but I’ve got some stuff up my sleeve that I’m sure you’ll like.
As for gravity being space-time curvature I was under the impression that it is in an iff relationship with the gravitational/inertial mass equivalence principal coupled with the SR classification of inertial mass as a type of energy. Certainly you can imagine a universe in which all particles have a fixed charge to rest mass ratio, including the force carriers, and in that case it would likely not be possible to cleanly separate the effects of one force from another. In fact that is hinted at by the inquiries in the previous post about the use of different objects to probe the field. You can find a similar discussion in Jackson in the section where he discusses the presence/absence of magnetic monopoles. He shows that it is possible to mix the electric and magnetic fields in such a way that the correct question is not, “Why are there no monopoles,” but, “Why is the electric charge/magnetic charge ratio the same for all particles?”
Speaking of magnetic monopoles, I had an interesting experience during one journal club a couple of years back. We were getting into a discussion of topological solitons and part of the lead-up required a discussion of monopoles. I’ve never been a fan of the Dirac string since it throws out the property of the vector potential used to justify B = curl(A) and E = – grad(phi) – diff(A, t) in the first place (continuous/differentiable). So, I introduced another vector potential (can’t recall what letter I gave it at the time, so I’ll use ~A) which corresponded to another gauge symmetry. Specifically, if I recall correctly, the symmetry that produced the correct properties was a chiral U(1) (ie exp( i q_b * gamma_5 * phi(x) ) ). It had all the right properties, and I even tried to gauge the electric/magnetic charge rotations since that would immediately rule out being able to observe such a difference. Needless to say, that effort failed but I hadn’t completed it in time for the presentation. It did have the correct rotation properties between q and q_b, though, so I was happy with it. I wonder if something interesting could be said by combining this treatment with a full QFT treatment including the fact that the global chiral symmetry is “anomalously” broken by triangle diagrams. Well, the chiral symmetry isn’t necessarily broken so much as we have to choose between breaking U(1) chiral and U(1), so we keep the U(1).
After the presentation I was politely informed that all of the monopole dynamics were typically accounted for by using the ordinary vector potential and allowing for Dirac strings. Nobody mentioned whether anyone had bothered to look at it in the light that I had, though.
So I still think that a Dirac string monopole is something that we’ll never observe, but we’ll see.
4. quantummoxie Says:
April 17, 2010 at 7:14 pm
Aha! But by revealing your true identity you have now taken all the mystery out of it. Either way, publications or not, you clearly know what you’re talking about so kudos. If you’re interested in some deeper details, both Joe Fitzsimons and I are exploring a bit of this offline.
As for gravity being space-time curvature I was under the impression that it is in an iff relationship with the gravitational/inertial mass equivalence principal coupled with the SR classification of inertial mass as a type of energy.
Right, but, again that’s an axiomatic assumption about the nature of space-time curvature. Space-time curvature is described by a metric, but a metric is just a mathematical thing. Mathematicians (as is their wont) have completely generalized metrics.
Let me put it this way: my assertion does not “break” the equivalence principle nor does it “break” the notion that inertial mass is a form of energy. That’s still as true as ever.
5. BlacGriffen Says:
April 17, 2010 at 8:09 pm
Thanks for the offer, but I’ve got a pretty full plate as it is.
Right, but, again that’s an axiomatic assumption about the nature of space-time curvature. Space-time curvature is described by a metric, but a metric is just a mathematical thing. Mathematicians (as is their wont) have completely generalized metrics.
Well, that skips a step. The connection is what defines the curvature. Making the metric dynamical is a step that’s always taken but I haven’t seen a convincing argument for why the connection has to be metric compatible and thus why the theory should be defined in terms of a dynamic metric instead of a connection. Maybe I missed that day in GR.
As for the axiomatic assumptionness of it, I think there’s a pretty good case for dividing forces between those that arise due to parallel transport of vectors tangent to the manifold (ie with indices that correspond to directions of travel along the manifold) and those that are orthogonal (ie elements of a vector space that has no a priori connection to the manifold). Now you may argue that I’m artificially dividing a larger manifold into space-time and “other,” and that would be a fair criticism. I don’t actually have a good reason for doing so, but it is standard practice. Consider, for example, the functional integral form of QFT. The n-point functions are all of the basic form:
integral( D[field strength] * (field strength factors) * exp(i integral( d(spacetime) Lagrangian density )))
So in some sense you’re integrating along part of the extended manifold outside of the exponential and part inside. I understand that it works, but am kind of curious if there’s an explanation for why. Or, equivalently, if there’s a way to unify everything onto a single manifold with a single unified treatment for the variables that describe the configuration of, for lack of a better term, reality where all of the parameters are treated on an equal footing.
BG
6. BlacGriffen Says:
April 17, 2010 at 8:10 pm
D’oh! Using HTML like quote tags doesn’t work. I’ll have to try something else next time…
7. quantummoxie Says:
April 17, 2010 at 9:51 pm
For including LaTeX, embed the LaTeX code inbetween \$latex on the left and a single dollar sign on the right.
I need to hit the sack, as they say, so a fuller response will have to wait until tomorrow, but here’s a quick response. You’re absolutely right that what you’re saying is the standard interpretation. My point is that we’ve abstracted almost beyond meaning in some cases (especially in certain aspects of QFT). So, for instance, there are certain things that are taught in QM in a certain way that, if you start asking questions, hinge on whether or not Hilbert spaces are real or not. People either brush this aside or ignore the contradictions. To me, as an empiricist, there are similar problems in interpretations of GR.
As for the connection defining the curvature, that’s an interesting point. Let me sleep on it.
8. Dear Ian,
In 1925, G.Y. Rainich provided the necessary and sufficient conditions for a Lorentz spacetime to be solution of Einstein-Maxwell equations. One account, simplified by the use of spinors, can be found in “Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields (Cambridge Monographs on Mathematical Physics)”. There are three algebraic conditions and one differential. These conditions were rediscovered by Misner and Wheeler in 1957, and are presented e.g. in “Gravitation – An Introduction To Current Research”, 1962, ed. Louis Witten, chapter 9. In addition to rediscovering the condition, they showed that a pair electron-positron can be interpreted, based on cohomology, as the two mouths of a wormhole – the electric field lines don’t need to originate and end in a source – they can be with no beginning and no end (chapter 10, see also “Gauge Fields, Knots and Gravity” by John Baez and Javier Muniain).
Rainich’s algebraic conditions for the energy-momentum tensor to be sourced by the electromagnetic field are
$A1. T^i{}_i=0$
$A2. T^i{}_k T^k{}_j \cong \delta^i{}_j$
$A3. T_{ij}v^iv^j\geq 0$ for any timelike vector $v^i$
Rainich’s differential condition:
$D0. \nabla_i S_j=\nabla_j S_i,$
where
$S_j:=\frac {\epsilon_{ij}{}^{kl}T_{mk}\nabla_l T^{im}} {T_{rs}T^{rs}}$
If these conditions are satisfied, there is an electromagnetic field which can source it, uniquely determined up to a duality rotation – a complex rotation where the complex structure is given by the Hodge * operator acting on 2-forms.
Is it possible for other sources of the gravity field to respect Rainich’s conditions? Take for example a fluid in thermodynamic equilibrium, characterized by the density $\rho$, pressure $p$, and 4-velocity u. It’s energy-momentum tensor is
$T^{ij} = (\rho + \frac {p} {c^2}) u^i u^j + p g^{ij}$
In diagonal form, for the metric signature (-+++), the energy-momentum tensor looks like
$T^i{}_j = diag(-c^2\rho,p,p,p)$
The Rainich algebraic condition A2 is satisfied because of the diagonal form, and the energy condition A3 is satisfied too. In order to be trace-free (condition A1), it has to satisfy
$3p = c^2 \rho.$
The difficult problem seems to be to find solutions which satisfy the differential condition. How can the fields $\rho$, p and u be taken so that both Einstein’s equation, the condition above, and the condition D0 be satisfied?
Cheers,
Cristi
9. Jose Ignacio Says:
April 18, 2010 at 7:58 am
The statement:
“The energy density, $\rho$ , is given by $T^{00}$ which, in normal GR is the density of the gravitational field”
is false, in normal GR $T^{\mu\nu}$ is the energy-momentum tensor of matter (or fields other than gravity). In normal GR the gravitational field has not definite energy-momentum tensor because gravity is replaced by geometry, and this is a valid replacement because the equivalence principle.
10. quantummoxie Says:
April 18, 2010 at 11:12 am
Cristi,
Thanks for the reference. I will need to find it and read it.
Jose,
My words were poorly chosen there. What I mean was that the associated energy density in GR is the energy density of the gravitational field which is exactly what you’re saying.
My point (and the work by Rainich seems to support this interpretation) is that you can interpret the geometry of spacetime differently without breaking the equivalence principle.
Ian
11. Dear Ian,
You’re welcome.
I think that it should be plausible that the same geometry of spacetime can be obtained from different fields, knowing that the matter fields have, in total, more degrees of freedom than the Ricci tensor. Even in the solution of Rainich, the same curvature can be obtained from different electromagnetic fields, obtained from one another by a duality rotation. In the case of the wormhole “charge without charge” solution of Wheeler and Misner, this ambiguity is removed by requiring the source to be electric charge only, not magnetic monopole or combination.
In the case of the fluid, I checked only the algebraic conditions, and not the differential one. The differential condition translates in a PDE on the fluid’s flow satisfying the constraint $3p(t,x) = c^2 \rho(t,x)$, but I did not investigated whether it has solution. Maybe it is an interesting research to see what other matter fields curve the spacetime like the electromagnetic field does.
The main point is that there must be some matter to mimic the energy of an electromagnetic field, because in vacuum the energy-momentum vanishes, and the geometry of a vanishing energy-momentum is different from that of a non-vanishing one. Thinking at an example with a box in vacuum, the geometry outside the box is very different if the box contains a massless charge, than if it contains a neutral massive object. Outside the box, if the source is the mass only, the spacetime is Einstein, being a solution to Einstein’s vacuum equation. If inside the box there is a charge, even a massless one, outside the box cannot be, in fact, vacuum, because of the electromagnetic field’s energy. This is why I interpreted your affirmation “The energy density, $\rho$, is given by $T^{00}$ which, in normal GR is the density of the gravitational field” as referring, in fact, to the energy density of a gravitational source, possibly a material fluid, not the gravitational field itself.
Best regards,
Cristi
12. [...] In the responses to my previous post, Cristi Stoica provided some very useful references that, for now, I am going to classify as “Rainich unification” since I needed something [...]
13. Tim van Beek Says:
April 21, 2010 at 8:48 am
Hi there,
let me start with some (hopefully) simple question about nomenclature: Ian, you write that in your toy universe there are no gravitational fields, but only electro-magnetic fields.
I am used to gravity = curvature of spacetime = spacetime metric = gravitational field.
So “graviational field” has to mean something else, right?
You write: “In order to prove my point, I have to show that G is solely determined by T”.
The field equations are usually seen as the only equations that determine G and T (and if the folks at the nForum said otherwise that has escaped me), so I guess the problem is more about what the interpretation or the content of G and T are?
14. quantummoxie Says:
April 21, 2010 at 11:55 am
Tim,
Yeah, the standard interpretation is that curvature = gravity which one could still insist on, but it would then be an argument over semantics.
1. Yakir Aharonov’s interpretation of a version of the Aharonov-Bohm effect is that the shifting of the electron’s interference pattern in said effect is due to the topology of space (Aharonov & Rohrlich, p.49). If the curvature (topology) of space is just gravity, there must be some gravitational effect going on here. Why hasn’t anyone considered it?
2. Let’s take the standard interpretation that gravity is the curvature of spacetime. Now we know from QED (actually from simple relativity, really – mass = energy) that photons (E&M fields) can produce curvature. Urs says this is simply gravity (to which I say, fine, but it’s a semantic thing). But if this is true, suppose we have a phenomenal concentration of photons in one location such that spacetime is warped enough that it’s a macroscopic effect. You should, in some way, be able to recover Newtonian gravity from this situation if the curvature is simply gravitational. But will we? And does it hold up experimentally (admitted this is hard to test)?
Pure speculation: I have a hunch there’s a greater symmetry (duality?) between electromagnetism and gravity than people are willing to admit. Even Rainich questions this (in fact that is the point of his paper).
15. Ian,
1. The Aharonov-Bohm effect is a topological effect, indeed. Only it is related to the topology of the gauge bundle of the electromagnetic field, rather than that of the spacetime itself.
16. BlackGriffen Says:
April 21, 2010 at 2:36 pm
Are you looking to re-invent Kaluza-Klein theory here? http://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory
17. quantummoxie Says:
April 21, 2010 at 2:57 pm
Cristi: as an empiricist, my first question is, what physical significance does the gauge bundle have? Spacetime has some kind of physical meaning and interpretation. I’m not sure gauge bundles do.
Sean: LOL, God no. My PhD dissertation was on Eddington’s Fundamental Theory which built heavily on Kaluza-Klein theory. I’ve seen enough of that stuff. It’s much closer to Wheeler’s geometrodynamics, but taking a step back and trying to “reconcile” the “language” as it were. I know that sounds vague, but I have to run to a meeting.
18. Tim van Beek Says:
April 21, 2010 at 3:37 pm
I’ll try if I can use some tags here…
Ian wrote:
…the standard interpretation is that curvature = gravity which one could still insist on, but it would then be an argument over semantics.
I guess we agree upon the meaning of curvature as a mathematical object, living on a smooth manifold. But then what is gravity to you?
If the curvature (topology) of space is just gravity, there must be some gravitational effect going on here…
Like Cristi Stoica said:
The mathematical model is R^3 with the z-axis removed and the electrons moving in the x-y-plane (you may add a time dimension as you wish), the point is: the curvature is zero everywhere, ergo (using my nomenclature) no gravity. The effect is explained by the fact that the manifold the electrons move in is not simply connected, i.e. has a nontrivial fundamental group. Same thing as with Dirac’s monopole, that’s a purely topological effect in a flat space, too.
Ian wrote:
…to which I say, fine, but it’s a semantic thing…
Perhaps if you try to rephrase what you mean here in one or two different ways I will stand a better chance to follow you.
Ian wrote:
suppose we have a phenomenal concentration of photons in one location such that spacetime is warped enough that it’s a macroscopic effect.
Theoretically that’s the case – from the viewpoint of an external observer at infinity – when looking at the event horizon of a (Schwartzschild) black hole that sucks in a nearby star, plus the black whole could have been created – theoretically – by the gravitational collapse of a bunch of photons. My point being: I don’t see anything special or new or paradox in this Gedankenexperiment.
You should, in some way, be able to recover Newtonian gravity from this situation…
That’s possible in the weak field approximation only, and – again – if you like to calculate the gravitational field of a dense bunch of photons, go ahead, I don’t get the point where there is something mysterious (that’s not critisism, I’m trying to indicate where I’m stuck in following you).
Ian wrote:
Pure speculation: I have a hunch there’s a greater symmetry (duality?) between electromagnetism and gravity than people are willing to admit.
If we are talking about a unification of gravitation and electromagnetism as classical field theories, then this has of course a long long history (see the review I linked to over at the nForum).
Ian wrote:
Spacetime has some kind of physical meaning and interpretation. I’m not sure gauge bundles do.
The gauge bundle keeps track of the phases of (localized) excitations of (quantum) fields, as a mathematical device it allows to calculate e.g. interference patterns. I suppose you know the cute little book “QED: The Strange Theory of Light and Matter” by Feynman and his clock-analogy of the phase of an electron?
19. quantummoxie Says:
April 21, 2010 at 9:13 pm
Couple of clarifying points:
I agree that there really isn’t anything remarkable here. What I’m arguing for is consistency in thinking and rationalizing definitions. It’s very, very similar to arguments I made in my FQXi essay and similar to a general philosophy espoused by Tom Moore. It’s born out of the desire to explain physics to the undergraduate non-majors that I teach without oversimplifying it.
Regarding the Aharonov-Bohm effect, see my comment below that I hope will clarify things (but you never know since I’m watching hockey at the moment). Ditto that for fiber bundles and semantics. I don’t know what I was thinking in regard to the ball of photons.
Anyway, here’s another attempt at explaining my position:
Spacetime clearly is some kind of physical “surface” that can be modeled mathematically by a manifold. This manifold clearly also includes areas of local curvature near large concentrations of mass/energy. Thus the standard interpretation is that the curvature of spacetime is gravity. I actually have no objection to this statement.
But the standard interpretation goes further and says the curvature of spacetime is and only is gravity, i.e. it defines gravity to be the curvature of spacetime. That’s the part I have a problem with (hence, I agree with David – let’s just dispose of the word entirely).
Here’s why I object. First, the sole motivation for this is historical because that was what Einstein was attempting to figure out at the time. Second, it is my contention that charge ought to do the same thing independent of its association with any mass. Now the usual response to this is that I’m automatically associating gravity with mass forgetting that gravity can be associated with pure energy since mass and energy are equivalent. This is the motivation for saying that energy causes curvature. But, the only self-consistent definition of mass, i.e. that makes sense in both single and multi-particle systems, is as the magnitude of the four-momentum vector (Tom Moore has an excellent discussion of this in his textbook). This implies, if you follow the logic all the way through, that while mass is energy, energy isn’t necessarily always mass. Thus, since the word “mass” was historically associated with “gravity” thanks to Newton, the assumption is that all of energy should be as well.
I find this reasoning to be somewhat circular particularly when you start viewing things from the standpoint of the so-called “interaction picture” envisioned by the ideal that the Standard Model strives for since it would imply that photons and other massless particles could exchange gravitons. But that throws all sorts of wrenches into the particle interpretation which has nice and neat classifications for particles based on their characteristics (mass, charge, spin, etc.). Nevertheless, what we’re arguing about here is a word, not the mathematics or even how to use it. Hence it’s really just semantics.
As for fiber bundles, this relates again to my FQXi essay. You say,
The gauge bundle keeps track of the phases of (localized) excitations of (quantum) fields…
That tells me what the bundle “keeps track of.” It doesn’t tell me what the bundle is physically. This is part of what I tend to think of as the “over-abstractification” of physics. When does it stop being physics and simply become math? One never knows when this stuff could actually make its way into technology. In fact quantum gravity might be more important technologically than people think – gravity is known to degrade entanglement and they’re currently trying to develop satellite-based entanglement systems. Suppose we actually reach the point where this stuff matters technologically. Spacetime makes intuitive sense to most engineers. It’s “substantive.” You can work with it. How do you work with a fiber bundle?
20. BlackGriffen Says:
April 21, 2010 at 10:52 pm
Yeah, it gets tricky when you define mass as $p_\mu g^{\mu\nu}p_\nu$ and you start to let the metric vary, but what else is new? I mean, last I heard there isn’t even a single agreed upon definition of energy in GR, so why should the mass picture be any clearer?
On the subject of mass, there’s a fun example in E&M where the photons behave like massive particles. I’m not talking about the difficult picture of superconductivity, I’m talking about sending signals down a wave guide or coaxial cable. When you do that you get a dispersion relation that looks exactly like the equation for mass/energy relation in SR, modulo units of course.
BG
21. Ian,
you said “as an empiricist, my first question is, what physical significance does the gauge bundle have? Spacetime has some kind of physical meaning and interpretation. I’m not sure gauge bundles do.”
The empirical component of the gauge bundle is given by the Aharonov-Bohm effect itself. It is an effect of the holonomy of a connection (which is the electromagnetic potential) on the U(1)-bundle.
To avoid loading your page too much with my own viewpoint, and to have more control of the equations, I provide more details here: http://www.unitaryflow.com/2010/04/vec-bund-fundam-physics.html
22. Tim van Beek Says:
April 22, 2010 at 7:23 am
I have some minor remarks below, but the main point, I think, is this (don’t know if it is entirely clear, but I hope it let’s us advance a bit):
In GR space and time do not have any “reality” on their own, they are defined as causal relationships with respect to the gravitational field (I know, I should not use this term, but did not come up with something better). You cannot define a universe without any effects related to gravity. If you visualize a manifold as spacetime and then put fields on it, the gravitational field being one of them, then this picture is misleading.
If you put an electromagnetic field on Minkowski space, you are implicitly saying “there is gravitation=curvature and is is zero everywhere and the energy of the electromagnetic field is too weak to couple to it”.
Example (very famous, has a name that I don’t remember): Fill a bucket with water. The water is still. Now let the bucket rotate. Let the water rotate with the bucket, now the surface is curved. How does the water “know” that it is in motion and has to form a curved surface? The answer of GR is: It is in motion with respect to the local gravitational field. And that is there, even when we are in Minkoswki space. But being in Minkowski space enables you to tell when you are not in an inertial system, by comparing your motion to the local gravitational field.
Ian wrote:
But the standard interpretation goes further and says the curvature of spacetime is and only is gravity, i.e. it defines gravity to be the curvature of spacetime. That’s the part I have a problem with (hence, I agree with David – let’s just dispose of the word entirely).
I’ve no problem with disposing with the term gravity, but you seem to see a crucial difference between the curvature of spacetime and gravity, and that is a reason not to dispose with it, isn’t it?
Ian wrote:
Here’s why I object. First, the sole motivation for this is historical because that was what Einstein was attempting to figure out at the time.
That’s not really an argument against anything, unless you would like to stress that some label is a historical accident due to a misprint, or that we should rename general relativity because all students keep asking which side of the civil war he fought for; so as an advisor in rhetoric I’d recommend to drop that point entirely
Ian wrote:
Tom Moore has an excellent discussion of this in his textbook.
I don’t know that book, could you provide a reference?
Ian wrote:
…while mass is energy, energy isn’t necessarily always mass.
The standard interpretation is that mass is one particular form of energy, and that vice versa all forms of energy couple to gravity. I know that there is some more important point here that you try to make, but – again in my role as rhetoric advisor – most physicist will understand this as stressing the perfectly obvious
Ian wrote:
But that throws all sorts of wrenches into the particle interpretation…
I’m no expert in string theory but string theorists see no problem here (as didn’t Feynman when he reinvented the graviton in his lectures about gravitation), could you elaborate on that?
Ian wrote:
Spacetime makes intuitive sense to most engineers. It’s “substantive.” You can work with it. How do you work with a fiber bundle?
Only if we are talking about the concepts of time and space built from everyday intuition. One lession of 20th century physics is that nature is essentially different in certain aspects from our everyday intuition. In GR the question “what happens now on Alpha centauri?” is meaningless, because there is no objective meaning to “now”. Given that, spacetime defines causal relationships of events. Gauge bundles add to that description. Neither of them is really close to what our brain learns from everyday input.
23. quantummoxie Says:
April 22, 2010 at 8:49 am
Let me just lead off by saying, you all haven’t said anything I’m not completely aware of. I know I’m bucking a commonly held view here and everyone is free to disagree with me. But my should be respected. It should be obvious by now that it is not born out of ignorance but out of a desire to reinterpret certain cherished ideas (which is why, I suppose, it gets attacked).
If you visualize a manifold as spacetime and then put fields on it, the gravitational field being one of them, then this picture is misleading.
I agree. The manifold should be the field. To quote Rainich,
…when the space is given all the gravitational features are determined; on the contrary, it seemed that the electromagnetic tensor is superposed on the space, that it is something external with respect to the space, that after space is given[,] the electromagnetic tensor can be given in different ways.
This was historically unsettling to many folks who tried to find ways around it: Weyl, Eddington, Rainich, Kaluza, Klein, et. al. But this philosophical approach to unification has waned, largely because most people think it was a failure – which it was, but, personally, I think it was a failure due to a lack of knowledge rather than a poor base model/idea. So much advancement has occurred since those theories were first considered that I think revisiting the base idea might be worthwhile (ok, so maybe I lied about Kaluza-Klein theory, though my approach is entirely different since I haven’t gone as far as postulating additional dimensions).
I’ve no problem with disposing with the term gravity, but you seem to see a crucial difference between the curvature of spacetime and gravity, and that is a reason not to dispose with it, isn’t it?
I guess I’d agree to keep it if it were agreed that gravity didn’t define spacetim curvature (curvature clearly defines gravity, but I think it should be open to interpreting spacetime curvature as being a more general notion).
That’s not really an argument against anything,
Logic compels me to disagree. Prior to 1905, time was treated as being absolute. Why? Because, historically, that’s what Newton did and no one thought to question it until relativity appeared on the scene.
most physicist will understand this as stressing the perfectly obvious
I’m a bit confused about which point you’re referring to here. Because I know a number of physicists who are sympathetic to my view.
One lession of 20th century physics is that nature is essentially different in certain aspects from our everyday intuition. In GR the question “what happens now on Alpha centauri?” is meaningless, because there is no objective meaning to “now”.
I both agree and disagree. Yes, relying on our intuition is not the safest course of action as 20th century physics has taught us. But there is still room for rational, well-formulated experiment/observation and, like it or not, science will never be completely objective (good luck to Harry on trying to fix that). So completely ignoring our everyday experience is also not a wise idea. There needs to be a balance. Where that balance point is, is what we’re debating. So, suppose you were Newton but you stumbled onto essentially the same observations that Eddington did in 1919. It seems to me that the warping of at least space (not time if you’re Newton) is something even Newton would have arrived at eventually given that kind of observational evidence. I can’t say the same about gauge bundles.
24. quantummoxie Says:
April 22, 2010 at 8:53 am
Oh, and, while it is an introductory textbook, Moore’s book espouses an entirely unique philosophy (in my opinion) and is thus worth reading solely for that (I use it to teach intro classes):
Six Ideas That Shaped Physics, by Thomas A. Moore
Comes in six thin volumes: Units C, N, R, E, Q, and T. While I don’t agree with everything he says, I agree with most of it. His motivating principle for writing it (and it was a 10-year project) was to explain things correctly the first time so that it wasn’t necessary to go back and say, “well, we lied to you in intro physics.” He is a frequent collaborator of Dan Schroeder’s. Dan wrote a phenomenal text called Introduction to Thermal Physics which is based on similar guiding principles. Dan also co-authored a QFT text with Peskin.
25. Tim van Beek Says:
April 22, 2010 at 10:46 am
Thanx for the reference to Moore, I will have a look at it.
…curvature clearly defines gravity, but I think it should be open to interpreting spacetime curvature as being a more general notion…
My problem is that I understand “define” in a mathematical sense, and that means both notions are equal. And I think the same happend over at the nForum. Maybe an example of what you have in mind for the notion of gravity would help (sorry, but I still did not get that).
Ian wrote:
I’m a bit confused about which point you’re referring to here
To the statement “matter is always energy, but energy is not always matter”.
…even Newton would have arrived at eventually given that kind of observational evidence. I can’t say the same about gauge bundles.
Ok, one should not completly dismiss everyday intuition, gauge bundles are far removed from that, farther than the notion of spacetime, and maybe one should try to remedy that. I think we can agree on that (but I suspect that some folks over at the nForum would dismiss this kind of reasoning as utterly irrelevant, but then not everybody has to teach non-major undergraduates .
I think I understand your viewpoint a little bit better now (modulo my first question above), did I succeed in explaining what troubled some people about your statements over at the nForum?
26. Tim van Beek Says:
April 22, 2010 at 12:55 pm
I wrote:
Maybe an example of what you have in mind for the notion of gravity would help…
A telltaling mistake, I thought you meant gravity should be open to a more general interpretation. About spacetime curvature I think: Well, it defines locally what an inertial system is and if you are not in one then you will experience forces that we know from everyday live as gravity. Being in an inertial system means free falling means movement along a geodesic means gravitational force is zero etc.etc.
I have a hard time imagining that curvature could have some additional role that is not connected to this set of ideas and rephrase my former question to refer to curvature now. Is your idea that curvature should in addition model electromagnetic forces?
(I hope that my questions provide an indication what kind of introduction you could formulate for your ideas, to avoid misunderstandings, if you talk to people who think like me
27. BlackGriffen Says:
April 22, 2010 at 2:45 pm
Gravity and spacetime curvature are in principal separate concepts for one very basic reason: we can easily imagine a universe in which the spacetime manifold were rigid and thus did not curve in the presence of the stress energy tensor. The force we identify as gravity would then be an ordinary force that has the macroscopic appearance of the manifold flexing due to stress-energy, but that is independent of the manifold microscopically. Please correct me if I’m wrong because this is what I understand the string theory interpretation of gravity to be.
Toward that end we also still perform tests to look for a difference between inertial and gravitational mass. I don’t recall to what level they’ve been verified to be the same, but like tests for a photon mass just about everyone expects the results to be negative.
28. Aha! Excellent points Sean! And what about all those nutty 5-D gravity folks who were involved with Gravity Probe-B? I’m pretty sure they see gravity and spacetime curvature as separate.
I think I understand your viewpoint a little bit better now (modulo my first question above),
LOL, I love the modulo reference. Total math geek, you are. :p
but I suspect that some folks over at the nForum would dismiss this kind of reasoning as utterly irrelevant, but then not everybody has to teach non-major undergraduates
But that’s not the only thing that motivates me. Any description of the universe ought to be both consistent and reasonable (there are better words to use here but I can’t remember them – there’s a whole stack of writing about this, in fact, by people who work in foundations). I work in foundations most of the time and so I ask questions that are unpopular, subtle, difficult, and supposedly resolved because there are always overlooked nuggets of useful ideas floating around out there just waiting to be discovered (or re-discovered in many instances).
My problem is that I understand “define” in a mathematical sense
Bad choice of word on my part, though that is the word that is generally used. I would have used some mathematics-like analogy to sets and subsets or something, but, knowing mathematicians I would have been picked apart on semantics when it was really a bigger picture concept I was after.
did I succeed in explaining what troubled some people about your statements over at the nForum?
I think what you did is drive home a point I already knew (I had a similar bad experience at MathOverflow). I just couldn’t find a way to properly formulate my point (I have that problem sometimes in discussiony situations).
29. Tim van Beek Says:
April 23, 2010 at 2:53 am
Ian wrote:
I think what you did is drive home a point I already knew (I had a similar bad experience at MathOverflow). I just couldn’t find a way to properly formulate my point…
That’s what I meant by “cultural clash”. If you visit a different continent you know there will be some problems due to misunderstandings. The problem here is that we encounter a cultural difference when we do not expect to see one, and one that is not easy to spot, because we all use “the same language” (by which I mean both English, math etc.).
So what I try to do here is explain what I am thinking and why I am mislead by (the specific formulation of) certain statements you make, I do not try to make any specific points about some interpretation of GR
Total math geek, you are.
I’ll take that as a compliment
But seriously, that’s the way of talking people over at mathOverflow and the nForum are used to, and if you would like to be understood you have to adapt to a certain degree.
BlackGriffen wrote:
The force we identify as gravity would then be an ordinary force that has the macroscopic appearance of the manifold flexing due to stress-energy, but that is independent of the manifold microscopically. Please correct me if I’m wrong because this is what I understand the string theory interpretation of gravity to be.
I can’t speak for the whole string community and maybe there are some who would agree with you, the interpretation that I know of is this: In string theory you fix a spacetime as a background, the objects of string theory generate gravitons. The situation of having a (classical) spacetime with gravitons on it is a quantum perturbation of a classical background.
String theory does not change anything in the interpretation of classical GR, it just tries to incorporate quantum gravity effects as a perturbation of the classical situation. One critisism of string theory therefore is that one cannot know if perturbation theory is applicable. Example: If you have a gravitational collapse in a classical situation, can this be modelled by increasing the number of gravitions when starting from a flat spacetime?
Both sides (stringy and antistringy) agree that this is an open problem, they disagree about the importance of it.
• BlackGriffen Says:
April 23, 2010 at 3:09 am
Tim wrote:
[quote]I can’t speak for the whole string community and maybe there are some who would agree with you,[/QUOTE]
I can’t speak for any of them. I just don’t see how the collective action of wiggling strings alters the geometry of the spacetime manifold, or the target space. Unless space itself is somehow constructed out of these strings, of course, but I never delved deep enough to figure out what string theorists interpret these strings to be. If you start trying to reduce space to some kind of web of 1-dimensional objects (or 1+1 dimensional, depending on how you look at it), then it begins to sound vaguely loop quantum gravity-esque to my essentially outsider’s perspective. In such a picture, perhaps, the strings could be the edge of the simplexes that the LQG people use to construct the normal 4-dimensional spacetime.
in all seriousness, what are the strings? What is it that is supposed to be vibrating?
Thanks,
BG
• Tim van Beek Says:
April 23, 2010 at 9:30 pm
BlackGriffen wrote:
…in all seriousness, what are the strings? What is it that is supposed to be vibrating?
That is the input to the theory, there is no further explanation of what a “string” is. Looking for one will be in vain.
30. quantummoxie Says:
April 23, 2010 at 7:08 pm
I actually had thought that the Rainich conditions led to geometrodynamics which in turn gave birth to (among other things) loop quantum gravity. Did someone say that in this discussion thread? I can’t remember and I’m too tired to check.
Sean’s points about strings are one of the main reasons I’ve always had issues with string theory. I was enamored with them for awhile in the ’90s and nearly went into it at one point, but the empiricist in me started to ask questions and I didn’t find the canned answers to be all that satisfying.
Well, anyway, I still think everything seemed to be going fine until Urs went postal on me. And Urs is supposedly a physicist so he should have understood what I was talking about.
Look, I don’t preclude that I am wrong. I just think it was a point worth discussing rather than not only dismissing it but getting personal about it (though that, I think, was born out of something unrelated that didn’t occur to me until today and I’ll leave it at that).
31. Tim van Beek Says:
April 23, 2010 at 9:43 pm
Well, anyway, I still think everything seemed to be going fine until Urs went postal on me.
No. The discussion if “Lorentzian spacetime” is the correct term for manifolds that are candidates for spacetimes set the stage. That was the first step into the wrong direction.
And Urs is supposedly a physicist so he should have understood what I was talking about.
My whole reason for being here is to point out that one cannot conclude that “x is a physicist, therefore he should understand that…”
That is not true. You have to anticipate that there will be misunderstandings. Its unavoidable. Its nooone’s fault. It happens. You have to understand this or you will never overcome them. I was not successful to explain that, but I hope someone else will be.
A dieux.
32. quantummoxie Says:
April 24, 2010 at 8:46 am
What I meant was that the tone of the conversation changed when Urs got fed up. Yes, you’re absolutely right, I should anticipate that there will be misunderstandings. I wrote the above comment after a long and tiring week. My point was that I’m still mystified as to why a discussion about a scientific point has to turn personal. Why not simply say, “OK, majority rules, it will be X” rather than adding “you’re a crappy physicist, go away.” I would have simply shrugged my shoulders and shut up.
• phorgyphynance Says:
April 27, 2010 at 2:42 am
Where did anyone ever say you were a crappy physicist? Harry doesn’t count
33. Steve Dufourny Says:
April 26, 2010 at 6:54 am
Hi all, Cristi, Ian,
Very beautiful blog and of course discussions.
Best Regards
Steve
• quantummoxie Says:
April 26, 2010 at 7:48 pm
Hi Steve! Thanks for the comment!
• Steve Dufourny Says:
June 1, 2010 at 4:53 am
You are welcome, with pleasure.
34. phorgyphynance Says:
April 27, 2010 at 2:57 am
I pretty much just relived that spacetime discussion, i.e. I just reread it.
I still think any disagreement boiled down to semantics. I created a page on geometrodynamics that I hope we can fill with some content (after vetting away from the nLab, i.e. the nLab is for polished material).
35. quantummoxie Says:
April 27, 2010 at 6:23 am
We could do it here if you want. I could give you posting access and we could start a page on it.
Incidentally, there is an interesting article on gravity over at FQXi. It’s about loop quantum gravity, but I seem to recall that geometrodynamics is a historical antecedent of LQG. I don’t know enough about LQG to say for certain, though.
36. Hi Steve, glad to meet you here!
Hi Ian,
You are right, geometrodynamics is related to LQG. More precisely, as part of the geometrodynamics program, Arnowitt, Deser and Misner provided a Hamiltonian for General Relativity. That is, an explicit time evolution in General Relativity. To canonically quantize a “classical” theory, you need a Hamiltonian formulation. This Hamiltonian was used in Wheeler’s and deWitt’s Quantum Gravity (quantum geometrodynamics).
The ADM Hamiltonian had some issues, e.g. its constraints were not closed under Poisson brackets, and the equations had a complicated dependence in the basic variables. Abhay Ashtekar provided new variables for the Hamiltonian. Consequently, the equations become polynomial, and the Poisson brackets closed. Moreover, his formulation, being more like Yang-Mills’s theory, allowed a direct quantization in terms of Wilson loops (Rovelli and Smolin).
In Ashtekar’s formulation, the gauge group is SO(3), or SU(2), depending on the formulation and of whether you include fermions or not. One of the variables is a connection, and the other is analogous to the “electric” field conjugate to that connection.
Wilson loops are in fact the traces of the holonomy of a gauge connection, and are gauge invariants. They work fine with all gauge theories.
• Steve Dufourny Says:
June 1, 2010 at 4:54 am
Thanks dear Christi.
It’s a beautiful and relevant blog.
Best Regards to all.
37. quantummoxie Says:
April 29, 2010 at 8:24 pm
Thanks for the overview, Cristi. I’m actually reading some of Rovelli’s stuff right now, though it’s not on LQG…
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http://unapologetic.wordpress.com/2012/03/07/minkowski-space/?like=1&source=post_flair&_wpnonce=9a71b83933 | # The Unapologetic Mathematician
## Minkowski Space
Before we push ahead with the Faraday field in hand, we need to properly define the Hodge star in our four-dimensional space, and we need a pseudo-Riemannian metric to do this. Before we were just using the standard $\mathbb{R}^3$, but now that we’re lumping in time we need to choose a four-dimensional metric.
And just to screw with you, it will have a different signature. If we have vectors $v_1=(x_1,y_1,z_1,t_1)$ and $v_2=(x_2,y_2,z_2,t_2)$ — with time here measured in the same units as space by using the speed of light as a conversion factor — then we calculate the metric as:
$\displaystyle g(v_1,v_2)=x_1x_2+y_1y_2+z_1z_2-t_1t_2$
In particular, if we stick the vector $v=(x,y,z,t)$ into the metric twice, like we do to calculate a squared-length when working with an inner product, we find:
$\displaystyle g(v,v)=x^2+y^2+z^2-t^2$
This looks like the Pythagorean theorem in two or three dimensions, but when we get to the time dimension we subtract $t^2$ instead of adding them! Four-dimensional real space equipped with a metric of this form is called “Minkowski space”. More specifically, it’s called 4-dimensional Minkowski space, or “(3+1)-dimensional” Minkowski space — three spatial dimensions and one temporal dimension. Higher-dimensional versions with $n-1$ “spatial” dimensions (with plusses in the metric) and one “temporal” dimension (with minuses) are also called Minkowski space. And, perversely enough, some physicists write it all backwards with one plus and $n-1$ minuses; this version is useful if you think of displacements in time as more fundamental — and thus more useful to call “positive” — than displacements in space.
What implications does this have on the coordinate expression of the Hodge star? It’s pretty much the same, except for the determinant part. You can think about it yourself, but the upshot is that we pick up an extra factor of $-1$ when the basic form going into the star involves $dt$.
So the rule is that for a basic form $\alpha$, the dual form $*\alpha$ consists of those component $1$-forms not involved in $\alpha$, ordered such that $\alpha\wedge(*\alpha)=\pm dx\wedge dy\wedge dz\wedge dt$, with a negative sign if and only if $dt$ is involved in $\alpha$. Let’s write it all out for easy reference:
$\displaystyle\begin{aligned}*1&=dx\wedge dy\wedge dz\wedge dt\\ *dx&=dy\wedge dz\wedge dt\\ *dy&=dz\wedge dx\wedge dt\\ *dz&=dx\wedge dy\wedge dt\\ *dt&=dx\wedge dy\wedge dz\\ *(dx\wedge dy)&=dz\wedge dt\\ *(dz\wedge dx)&=dy\wedge dt\\ *(dy\wedge dz)&=dx\wedge dt\\ *(dx\wedge dt)&=-dy\wedge dz\\ *(dy\wedge dt)&=-dz\wedge dx\\ *(dz\wedge dt)&=-dx\wedge dy\\ *(dx\wedge dy\wedge dz)&=dt\\ *(dx\wedge dy\wedge dt)&=dz\\ *(dz\wedge dx\wedge dt)&=dy\\ *(dy\wedge dz\wedge dt)&=dx\\ *(dx\wedge dy\wedge dz\wedge dt)&=-1\end{aligned}$
Note that the square of the Hodge star has the opposite sign from the Riemannian case; when $k$ is odd the double Hodge dual of a $k$-form is the original form back again, but when $k$ is even the double dual is the negative of the original form.
## 2 Comments »
1. Hi John,
I just wanted to let you know that I’ve been loving this series. Connecting what you talked about before to something slightly more familiar has been really helpful in my understanding.
That’s all I wanted to say, but I wanted to say it before I forgot to.
Comment by | March 7, 2012 | Reply
2. [...] First off, I’m going to use a coordinate system where the speed of light is 1. That is, if my unit of time is seconds, my unit of distance is light-seconds. Mostly this helps keep annoying constants out of the way of the equations; physicists do this basically all the time. The other thing is that I’m going to work in four-dimensional spacetime, meaning we’ve got four coordinates: , , , and . We calculate dot products by writing . Yes, that minus sign is weird, but that’s just how spacetime works. [...]
Pingback by | July 17, 2012 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 22, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9294022917747498, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/14165/how-can-area-be-a-vector | # How can area be a vector?
My professor told me recently that Area is a vector. A Google search gave me the following definition for a vector:
Noun: A quantity having direction as well as magnitude, esp. as determining the position of one point in space relative to another.
My question is - what is the direction of area? I can relate to the fact that velocity is a vector. The velocity of a moving motorbike for example, has a definite direction as well as a definite magnitude assuming that the bike is moving in a straight line & not accelerating.
My friend gave me this explanation for the direction of Area vector. Consider a rectangular plane in space. He argued that the orientation of the plane in space can only be described by considering area as a vector & not a scalar.
I still wasn't convinced. Suppose the plane was placed such that its faces were perpendicular to the directions, North & South for example. Now the orientation of the plane is the same irrespective whether the so called vector points to north or to the south. Further what is the direction of a sphere's area?
Does considering area as a vector have any real significance? Please explain.
Thanks in advance.
-
2
Since this question is really mathematical in nature, would it be appropriate for migration to the math site? I think that most questions which deserve the "mathematics" tag (not to be confused with "mathematical-physics") are probably better off on math.SE. – David Zaslavsky♦ Aug 31 '11 at 8:26
3
@David Honestly I can not think of a better example of clear overlap between physics and math. While I don't doubt that mathematics wouldn't have a problem vectorizing an area, it seems like the entire point is so that it can be used in some physical sense. It also depends, if you're talking about differential surfaces for integration (like I think you are), then yes I'd agree it's a math topic. But what about using the area vector for a current loop in calculating the magnetic field? That's almost certainly physics material. – AlanSE Aug 31 '11 at 13:15
## 4 Answers
This might be more of a math question. This is a peculiar thing about three-dimensional space. Note that in three dimensions, an area such as a plane is a two dimensional subspace. On a sheet of paper you only need two numbers to unambiguously denote a point.
Now imagine standing on the sheet of paper, the direction your head points to will always be a way to know how this plane is oriented in space. This is called the "normal" vector to this plane, it is at a right angle to the plane.
If you now choos the convention to have the length of this vector ("the norm") equal to the area of this surface, you get a complete description of the two dimensional plane, its orientation in three dimensional space (the vector part) and how big this plane is (the length of this vector).
Mathematically, you can express this by the "cross product" $$\vec c=\vec a\times\vec b$$ whose magnitude is defined as $|c| = |a||b|sin\theta$ which is equal to the area of the parallelogram those to vectors (which really define a plane) span. To steal this picture from wikipedia's article on the cross product:
As I said in the beginning this is a very special thing for three dimensions, in higher dimensions, it doesn't work as neatly for various reasons. If you want to learn more about this topic a keyword would be "exterior algebra"
Update:
As for the physical significance of this concept, prominent examples are vector fields flowing through surfaces. Take a circular wire. This circle can be oriented in various ways in 3D. If you have an external magnetic field, you might know that this can induce an electric current, proportional to the rate of change of the amount flowing through the circle (think of this as how much the arrows perforate the area). If the magnetic field vectors are parallel to the circle (and thus orthogonal to its normal vector) they do not "perforate" the area at all, so the flow through this area is zero. On the other hand, if the field vectors are orthogonal to the plane (i.e. parallel to the normal), the maximally "perforate" this area and the flow is maximal.
if you change the orientation of between those two states you can get electrical current.
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+1 for mention of magnetic fields. Not all surface vectors used in physics are differential. – AlanSE Aug 31 '11 at 13:16
Thanks. Just a few clarifications. You asked me to imagine a person standing on a paper & consider the direction of his head as representing the normal vector. But suppose this person was standing on the exact opposite face, then won't the orientation of the paper still remain the same? But now the direction of the vector is in the opposite direction. Please clarify. – Green Noob Sep 1 '11 at 6:26
Secondly, you said that this concept doesn't work so well in higher dimensions. So does that mean that my question about the direction of a sphere's area is invalid? If so then is area a scalar in this particular case since considering it as a vector cannot specify its orientation in space? – Green Noob Sep 1 '11 at 6:32
@Green Noob: The orientation is in fact ambiguous. This is why the cross product deals with this in a defined (but arbitrary way), the so called right hand rule. Take you thumb to be the first factor $a$, you index two be $b$ then $c$ will point in the direction of your middle finger. As you see $a\times b = - b\times a$. The spheres area is still two dimensional. So locally you can still apply this concept, as locally the sphere looks like euclidian space. Imagine standing on Earth, you head points upwards away from the surface. This is called the "surface element". – luksen Sep 1 '11 at 8:04
What you said makes sense though I'm not fully satisfied. Thanks a lot :) – Green Noob Sep 2 '11 at 7:36
show 1 more comment
The main regime of use is when an area is infinitesimally small, like one would use in an integral. In that case, we can easily see that it is flat, and the shape doesn't really matter. In which case, we can encode the information as a vector, with the magnitude representing the (scalar) area; the choice (as you noticed) of pointing out of any given side is exactly that --- a choice --- but one that can be made consistently. We can extend this to non-infinitesimal planes, but it doesn't work that well for curved surfaces.
To be precise, what you really want is a co-vector. This is an abstract gadget which takes a vector and spits out a scalar. For a plane, you want this to represent the "amount" of the vector that goes through the plane --- so it should be linear in the vector (doubling the vector doubles the output) and it should take into account the angle at which the vector hits it (gives a factor of $\cos$). Now, we can ask the question of how to represent this abstract co-vector, and it turns out that a vector is a good idea! Specifically, we can represent the action as taking the dot product, which naturally encodes the linearity and the cosine. Now, in general, this happens to have the same number of dimensions as a proper vector, but this only encodes an area (a 2D surface) in 3D --- in 2D you would get a line, in 4D a volume (yes! A 4-vector intersects a volume at one point!).
If you want to learn more about this sort of thing, you want to investigate differential geometry, where all it is necessary to be clear about this sort of thing and not mix up vectors and co-vectors (called forms in that field). A good readable reference is Gauge Fields, Knots and Gravity which starts from a basic overview of the mathematics and develops it for physical use.
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+1 I second the book suggestion. great book. – luksen Aug 31 '11 at 8:11
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In the context of field theories, such as with electromagnetism, the concept of "the amount of a vector (field) that goes through a plane segment" is given the name flux. So you may think of area as being characterized by a function which maps vectors (or a vector field) to the flux of that vector (field) through the area. – Niel de Beaudrap Aug 31 '11 at 10:27
Think of Force is Pressure times Area ($F = P\cdot A$). You know pressure is a scalar (there is no direction associated with it), and a force is a vector (it acts along an axis). So what does that mean for pressure.
Take a small area and see it's contribution to the total force due to pressure
$${\rm d}F = P(x,y,z)\,{\rm d}A$$
The direction of the force is normal to the area, and its magnitude is proportional to the size of the area. This is why an infinitesimal area ${\rm d} A$ can be a vector. It is convenient to think of (vector)=(scalar)*(vector).
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There is an especially picturesque example of the Law of Pythagors in three dimensions applied to the areas of a simplex. (Where by "simplex" I believe I mean a section of space bounded by three orthogonal planes and one arbitrary plane.) The sum of the squares (of the areas) of the three small faces are equal to the square of the area of the oblique face. It is easily explained by the pressure/flow type arguments put forward in the other answers posted here, plus the obvious physical condition that an undisturbed fluid is in equilibrium with itself.
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http://math.stackexchange.com/questions/179775/is-this-a-legitimate-way-to-show-that-ze-z2-2-infty-infty-0-pr | Is this a legitimate way to show that $[-ze^{-z^2/2}]_{-\infty}^{\infty}=0$? (Proving a statement about a function of a normal random variable)
The problem is, let Z be a standard normal variable and $n\geq1$ be an integer. Show that $E[Z^{n+1}]=nE[Z^{n-1}]$. Here's what I've got so far, miraculously:
$E[Z^{n+1}]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^{n+1}e^{-z^{2}/2}dz$
Doing integration by parts with $u=z^n$ and $dv=ze^{-z^{2}/2}$ gives
$$[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty+n\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-z^{2}/2}z^{n-1}dz=[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty +nE[Z^{n-1}]$$
Now my Calc II teacher actually told me never to do this $[f(x)]_{-\infty}^\infty$ but my probability textbook does it so I'm just gonna throw caution to the wind. I'm thinking I turn $[\frac{1}{\sqrt{2\pi}}(-z^{n}e^{-z^{2}/2})]_{-\infty}^\infty$ into $$\frac{1}{2\pi}(\lim_{z\rightarrow\infty}-z^{n}e^{-z^{2}/2}-\lim_{z\rightarrow -\infty}-z^{n}e^{-z^{2}/2})$$Wolfram Alpha tells me that each of those limits is 0 but I don't quite see how—as far as I can see both of those limits are in indeterminate forms. What I was thinking is maybe since $n$ is an integer greater than 1 that I can argue that if I write $lim_{z\rightarrow\infty}-z^{n}e^{-z^{2}/2}$ as $lim_{z\rightarrow\infty}-z^n/e^{z^2 /2}$ then repeated applications of l'Hopital's rule will eventually put a 0 in the numerator while you'll still have some exponential function in the denominator, and so that'll be zero and hence the limit is zero. But that feels cumbersome and I'm not sure what I can do beyond just asserting that, unless I want, like, prove that by induction on n or something. Am I missing some obvious better way to evaluate this limit?
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Exponential function $f(x) = a^x$ of the base $a$ greater than 1 overwhelms any polynomial as $x \to \infty$. There are many ways to prove this fact, including the method using power series of $e^x$. That is, for $x > 0$, we have $$e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \geq \sum_{k=0}^{n} \frac{x^k}{k!}.$$ – sos440 Aug 7 '12 at 5:28
1 Answer
The calculation using integration by parts is the important part. Now to the main question.
Your idea of using L'Hospital's Rule is good. Think of the limit of $|z|^n e^{-z^2/2}$ as $|z|\to \infty$. (This is just to collapse the two limit calculations into one.)
In order to make the repeated differentiation easier, one might as well let $u=z^2/2$. So $|z^n|=2^{n/2}u^{n/2}$. We are trying to find $$\lim_{u\to \infty}\frac{2^{n/2}u^{n/2}}{e^u}.$$ It is enough to show that for any positive integer $k$, $$\lim_{u\to\infty}\frac{u^k}{e^u}=0.$$ This is a routine application of L'Hospital's Rule.
Much more informally (but enough for probability), $e^{z^2/2}$ grows enormously faster in the long run than any power of $z$.
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Thanks, this is great! I had to go through carefully on paper to make sense of the u substitution but once I got it I got it. And yeah, the integration by parts was a wild stab in the dark—I mean there wasn't anything else to do with that expression but do integration by parts but I still didn't expect anything useful to come out of it and then, lo and behold, there's the result I want staring me in the face. Thanks for your help! – crf Aug 7 '12 at 16:33
– André Nicolas Aug 7 '12 at 16:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 5, "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": 0.9665717482566833, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/96584?sort=oldest | ## Ternary “Lie structure”
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One of the motivation of the theory of Lie Algebras is that every associative algebra $A$ is a LA when the bracket is defined by $[a,b]=ab-ba$ : this is skew-symmetric and satisfies the Jacobi identity $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$. Conversely, every abstract LA can be embedded into an associative algebra (its envelopping algebra). And for some good reason, one is really interested in sub-LAs rather than sub-algebras. A similar attitude, with different motivation lead to the notion of Jordan algebras.
If $A$ is an associative algebra, one may consider instead the ternary bracket $$[a,b,c]_3=abc+bca+cab-acb-cba-bac.$$ Does $[.,.,.]_3$ satisfy non-trivial identities, besides skew-symmetry? Is there any theory of abstract objects, vector spaces endowed which a ternary skew-symmetric product satisfying these identities?
More generally, we may consider a $d$-bracket, which bears the name of standard non-commutative polynomial in $d$ non-commuting variables. For $d=2$, it is nothing but the standard bracket. When $d=2p$, the $d$-bracket does satisfy non-trivial identies, for instance $$\sum_{i\in\frak A_7}[[a_{i_1},a_{i_2},a_{i_3},a_{i_4}],a_{i_5},a_{i_6},a_{i_7}]=0,\qquad\forall a_1,\ldots,a_7\in A.$$ I don't know if something non-trivial exists when $d$ is odd.
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Have you checked if these are $L_\infty$ structures? – Jim Conant May 10 2012 at 16:50
(Ginzburg and Berger have studied a family of algebras, called antisymmetrizer algebras. The thirds one is the quotient of the free algebra on three generators $a$, $b$, $c$ modulo what you write as $[abc]_3$; the algebra is encodes then properties of triads of elements which "ternary-Lie-commute". The algebra is $3$-Koszul in the sense of Berger, so quite special: maybe you can extract information from that fact) – Mariano Suárez-Alvarez May 10 2012 at 18:21
If you remove $bca-acb$ (that is, the only two terms with $c$ in the middle) then you get a Lie triple system jstor.org/discover/10.2307/… en.wikipedia.org/wiki/… – Fernando Muro May 10 2012 at 19:40
You might be interested in my answer to a previous MO question concerning n-Lie algebras: mathoverflow.net/questions/49437/… – José Figueroa-O'Farrill May 12 2012 at 9:54
## 4 Answers
"Identities for the ternary commutator" by Bremner classifies all such identities up to degree 7. A recent exposition can also be found in "Ternutator Identities" by C. Devchand, D. Fairlie, J. Nuyts, G. Weingart. Similar identities for n-ary commutators are proven in "Multi-operator brackets acting thrice" by T. Curtright, X. Jin, L. Mezincescu. I don't know if there is an accepted definition for what a Lie n-ary algebra should be for $n\geq 3$ (not to be confused with Lie n-algebras from nlab). For $n=3$ the most standard object one encounters are Lie triple systems.
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Ternutators is a painful neologism... – Mariano Suárez-Alvarez May 10 2012 at 18:17
Thanks! This is definitely the answer I was looking for. I'm impressed by the complexity of such a short paper. Page 618 is impressive! The referee must have had hard time... – Denis Serre May 10 2012 at 20:10
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"Ternutation" = the act of neezing. Compare "Sternutation". – Noam D. Elkies May 11 2012 at 2:06
Which gives éternuer in French. – Chandan Singh Dalawat May 11 2012 at 9:29
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
The proper setting is the theory of operads, which allows to deal with any number of generators at a single stroke.
The Lie operad is the sub-operad of the associative operad generated by 12-21.
There exists some n-ary Lie operads, but I do not remember whether they are exactly what you ask.
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One way to define a Lie structure on a vector space $V$, is as a map $\wedge^2V\to V$ such that its natural extension to $d\colon\wedge^k V\to \wedge ^{k-1}V$ satisfies $d^2=0$. This exactly gives the Jacobi identity. Similarly one can define a Lie infinity structure to be any map $d\colon \wedge V\to \wedge V$ with square $0$. In your case $[\cdot,\cdot,\cdot]_3$ gives a map $\wedge^k V\to\wedge^{k-2}V$ and the question becomes whether it squares to $0$, which I haven't had time to work out. But if it does, then you have a L-infinity structure.
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You might want to look at the paper "On Lie k-Algebras" by P. Hanlon by M. Wachs (http://www.sciencedirect.com/science/article/pii/S0001870885710389). They consider algebras satisfying the generalized Jacobi identity you specify. I wanted to leave this as a comment but I don't have enough reputation.
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http://physics.stackexchange.com/questions/4937/what-is-the-need-for-the-higgs-mechanism-and-electroweak-unification/4951 | # What is the need for the Higgs mechanism and electroweak unification?
The Higgs mechanism allows massless fields to acquire mass through their coupling to a scalar field. But if the masses cannot be predicted because the couplings have to be fixed, what really is the utility of the Higgs mechanism? Instead of saying "Here are a priori couplings; the Higgs mechanism generates mass.", I could just as well say "Here are a priori masses. Period.".
I understand that the Higgs mechanism is crucial to electroweak unification, but I have the same question there. Why does electromagnetism and the weak force have to be unified? Even if the couplings of the photon, Z and W bosons became related on unification, this is still at the cost of introducing new parameters - so it's not really clear to me that something has been explained or tidied up.
Do either the Higgs mechanism or electroweak unification tell us something new? Do either make any predictions that don't come at the cost of extra parameters? (I'm not really challenging anything here; I'm sure the answer to both questions is 'yes' - I just want to fill the gaps in my understanding as I study the Standard Model)
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## 4 Answers
Dbrane, aside from "beauty", the electroweak unification is actually needed for a finite theory of weak interactions. The need for all the fields found in the electroweak theory may be explained step by step, requiring the "tree unitarity".
This is explained e.g. in this book by Jiří Hořejší:
http://www.amazon.com/dp/9810218575/
The sketch of the algorithm is as follows:
Beta-decay changes the neutron to a proton, electron, and an antineutrino; or a down-quark to an up-quark, electron, and an anti-neutrino. This requires a direct four-fermion interaction, originally sketched by Fermi in the 1930s, and improved - including the right vector indices and gamma matrices - by Gell-Mann and Feynman in the 1960s.
However, this 4-fermion interaction is immediately in trouble. It's non-renormalizable. You may see the problem by noticing that the tree-level probability instantly exceeds 100% when the energies of the four interacting fermions go above hundreds of GeV or so.
The only way to fix it is to regulate the theory at higher energies, and the only consistent way to regulate a contact interaction is to explain it as an exchange of another particle. The only right particle that can be exchanged to match basic experimental tests is a vector boson. Well, they could also exchange a massive scalar but that's not what Nature chose for the weak interactions.
So there has to be a massive gauge boson, the W boson.
One finds out inconsistency in other processes, and has to include the Z-bosons as well. One also has to add the partner quarks and leptons - to complete the doublets - otherwise there are problems with other processes (probabilities of interactions, calculated at the tree level, exceed 100 percent). It goes on and on.
At the end, one studies the scattering of two longitudinally polarized W-bosons at high energies, and again, it surpasses 100 percent. The only way to subtract the unwanted term is to add new diagrams where the W-bosons exchange a Higgs boson. That's how one completes the Standard Model, including the Higgs sector. Of course, the final result is physically equivalent to one that assumes the "beautiful" electroweak gauge symmetry to start with.
It's a matter of taste which approach is more fundamental and more logical. But it's certainly true that the form of the Standard Model isn't justified just by aesthetic criteria; it can be justified by the need for it to be consistent, too.
By the way, 3 generations of quarks are needed for CP-violation - if this were needed. There's not much other explanation why there are 3 generations. However, the form of the generations is tightly constrained, too - by anomalies. For example, a Standard Model with quarks and no leptons, or vice versa, would also be inconsistent (it would suffer from gauge anomalies).
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Nice outline, Lubosh! I like you mentioned some difficulties. – Vladimir Kalitvianski Feb 10 '11 at 17:31
Now that is the kind of answer I was looking for :) I'm all for beauty and elegance too, but in the end I would only let an inconsistency like non-unitarity or anomalies make me add things to my theory. Thanks for the great answer again Lubos - I'll see if I can find a copy of the book you referred to. – dbrane Feb 10 '11 at 17:31
Thanks, @dbrane, and good luck in finding it for free (there are 3 copies left at amazon). The author was my QFT instructor in Prague. @Vladimir, right, there are problems, but the goal - and a successful one - of the procedure is to solve the difficulties. And indeed, they're solved at the end. – Luboš Motl Feb 10 '11 at 17:52
Great, I just discovered that there are physics books that you can't find in any of the Cambridge libraries. :/ – dbrane Feb 10 '11 at 18:05
@Vladimir, why do you name Lubos as Lubosh? He doesn't name himself that way and gives the impression of you lacking a respectful, fair attitude towards people. – John McVirgo Feb 10 '11 at 18:50
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Electroweak unification means that there is a symmetry between electromagnetic and weak interactions -- you can use them interchangeably. In reality this is not the case -- $W$ and $Z$ bosons have mass, while photon haven't.
Higgs mechanism provides a way for a spontaneous symmetry breaking between those interactions: Lagrangian of standard model is electroweak symmetric, while the vacuum is not due to non-zero vacuum expectation value for the Higgs field.
The same applies for fermions -- you cannot introduce mass terms for quarks and leptons in the Lagrangian, while preserving the electroweak symmetry. But it is possible to introduce electroweak symmetric Yukawa terms, coupling Higgs field to the fermions.
Edit:
I don't think that Higgs mechanism can "tell us something new". It is just a simplest way to ensure spontaneous symmetry breaking. While electroweak unification means that those interactions are gauge interactions and establishes the gauge symmetry itself. The classification of fermions into three generations is also done from the "electroweak point of view".
Of course you can argue that this systematization or classification is not "something new". But from such a point of view one can criticise almost every theoretical construction that attempts to predict results of future experiments.
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Thanks, but you haven't answered my question, you've just made it a bit more explicit. Introducing mass terms breaks electroweak symmetry. So instead, we impose electroweak symmetry on a massless theory and use the Higgs mechanism to generate back the masses. My question is, why not assume there never was an electroweak symmetry to start with, and all along work with a massive theory without any symmetry breaking? That's one less scalar field and one less particle to look for. – dbrane Feb 10 '11 at 16:14
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Of course, it's great that this leads to the prediction of a new particle. But I can't, at the moment, see why Nature would want to burden itself with an additional scalar field and associated particle when it could do everything as an SSB-less theory with the same number of free parameters. – dbrane Feb 10 '11 at 16:16
Very good reasoning! I support it with two my hands! – Vladimir Kalitvianski Feb 10 '11 at 16:17
It is true that recognizing that the data have an SU2xU1 symmetry one has a number of parameters and one may say that problem number one has been reduced to problem number two which has the same number of unknowns.
Let me give the often quoted example of the epicycles in the geocentric system and the ellipses in the heliocentric one. The number of parameters are the same, and if you go to a planetarium program and go to the geocentric frame, the epicycles will appear in all their glory. Nevertheless there is nobody now who would ask, "what is the use of ordering the data in the heliocentric system".
Asking "why should electromagnetism be unified with weak theory" is a bit like asking "why have a heliocentric system".The answer is that in both cases, the data fall into an ordered form effortlessly. And then we were led to higher symmetries (SU3xSU2xU1) and more inclusive theories.
I should have added that the clarity introduced by symmetries/order once manifested lead to calculable theories which can describe the data and make predictions for new observations. The Higgs is one such prediction coming out from the theories describing the quark model symmetries. If it is not found, a rethink will be in order, (though there are papers which claim that a Higgs mechanism can be a composite manifestation, not necessarily one particle).
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"effortlessly"? How much fog they supply to hide failures of this approach! – Vladimir Kalitvianski Feb 10 '11 at 16:36
@Anna: No, it's not the same because requiring unification introduces a new physical particle. It's not at all like looking at things from a more natural frame of reference (as in heliocen. vs. geocen.) because there the two descriptions are equally valid in that they don't differ in predictions - but differ in how they fit in naturally with known theories of celestial mechanics. – dbrane Feb 10 '11 at 16:36
@dbrane Well, the heliocentric system introduces a force, that was not present in the geocentric view of the world, no? – anna v Feb 10 '11 at 16:39
But the geocentric view involves forces too of course - only they are much more complicated to calculate because of the "unnatural" choice of coordinates. And again, in the end, the predictions are the same. – dbrane Feb 10 '11 at 16:42
To Anna: I disagree with your analogy with celestial mechanics. And you speak of introducing a "new physical particle" as a human intellect achievement whereas it is a fix of worthless (massless) theory. I can remind you bare particles, virtual particles, ghosts - all invented to stay in this failed approach. I would love to learn how physical electron looks like. Nobody can describe it yet. – Vladimir Kalitvianski Feb 10 '11 at 16:42
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In a phenomenological approach one introduces masses without problem.
It is in the "local gauge invariance" approaches where one needs a fix because the gauge approach does not give masses, unfortunately. Technically it is is implemented as "coupling" to Higgs in some way.
I never considered the "local gauge invariance principle" as reasonable or physical. It is one of many blind ways to "construct" interacting theories; it does not guarantee anything physical and does not save us from infinities in calculations. It's been done by analogy with QED which itself has problems. This way needs fixes.
I would say the Higgs is a price for choosing a "mathematical" rather than physical approach to constructing theories. It is a wrong guess, in my opinion. There are more physical ideas on this subject but they remain unknown and thus unexploited.
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what is this "phenomenological approach" you refer to? – user346 Feb 10 '11 at 16:23
Do you remember the Schroedinger equation with $m_e$ or Dirac equation with $m_e$? Something like this, if you like. – Vladimir Kalitvianski Feb 10 '11 at 16:33 | {"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": 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": 0.9409183859825134, "perplexity_flag": "middle"} |
http://mathhelpforum.com/math-topics/40306-rectangular-piece-paper-folded.html | Thread:
1. A Rectangular Piece of Paper is folded
The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.
10 X 6
B________________________ A
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| / |
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| ________/______________ |
C Y D
I realize that both AB and CD are 16 across. I've tried turning parts of this rectangel into a triangle. But that didn't yield results, as I couldn't determine the new figure's length. The closest I got to a lead was My thought of drawing a line from Y to A. That would give me a triangle with a base of 6, but no angles, so no way to determine XY.
2. I just added the diagram. I couldn't make it on this form.
Attached Thumbnails
3. Since you have a rectangle, the length is half the width. So $BC$ and $AD$ are 8.
And you can make this picture: link
Now apply the Pythagorean Theorem a bunch of times.
$(XB)^2 + (BC)^2 = (XC)^2$
$8^2 + 10^2 = (XC)^2$
$164 = (XC)^2$
$XC = 2\sqrt{41}$
Then apply it again: $(XA)^2 + (AD)^2 = (DX)^2$
$6^2 + 8^2 = (DX)^2$
$36 + 64 = (DX)^2$
$80 = (DX)^2$
$DX = 4\sqrt{5}$
Now let's fill that in our pic: link
Since $XZ$ is an angle bisector, then $ZC = DZ$
So both are 8.
Apply Pythagorean Theorem to get $XZ$
$(ZC)^2 + (XZ)^2 = (2\sqrt{41})^2$
$64 + (XZ)^2 = 164$
$(XY)^2 = 100$
$XY = 10$
And without more info, I don't see how you can find $XY$
4. Thank you
Thank you, very much. That was really helpful and I wouldn't have figured out the problem by myself before tomorrow. So I really appreciate your assistance.
5. Originally Posted by Jonboy
Since you have a rectangle, the length is half the width. So $BC$ and $AD$ are 8.
And you can make this picture: link
Now apply the Pythagorean Theorem a bunch of times.
$(XB)^2 + (BC)^2 = (XC)^2$
$8^2 + 10^2 = (XC)^2$
$164 = (XC)^2$
$XC = 2\sqrt{41}$
Then apply it again: $(XA)^2 + (AD)^2 = (DX)^2$
$6^2 + 8^2 = (DX)^2$
$36 + 64 = (DX)^2$
$80 = (DX)^2$
$DX = 4\sqrt{5}$
Now let's fill that in our pic: link
Since $XZ$ is an angle bisector, then $ZC = DZ$
So both are 8.
Apply Pythagorean Theorem to get $XZ$
$(ZC)^2 + (XZ)^2 = (2\sqrt{41})^2$
$64 + (XZ)^2 = 164$
$(XY)^2 = 100$
$XY = 10$
And without more info, I don't see how you can find $XY$
I'm just wondering, how did you deduce that the length was half the width?
6. Honestly, I assumed too much.
The only way I could've got the width was knowing the perimeter, which we don't know.
That's because:
$P = 2L + 2W$
$P = 2(16) + 2W$
$2W = P - 32$
$W = \frac{P - 32}{2} = \frac{1}{2}P - 16$
But, look on the bright side, it further proves you can't solve this problem.
7. Originally Posted by D. Martin
The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.
10 X 6
B________________________ A
| / |
| / |
| / |
| ________/______________ |
C Y D
I realize that both AB and CD are 16 across. I've tried turning parts of this rectangel into a triangle. But that didn't yield results, as I couldn't determine the new figure's length. The closest I got to a lead was My thought of drawing a line from Y to A. That would give me a triangle with a base of 6, but no angles, so no way to determine XY.
I don't have time to get to it right now, but I'll try (no promises) to get back to it later.
Note that whereas we don't know the width of the rectangle, we do know that points B and D lie the same distance away from the fold along the red line (the diagonal) because when the paper is folded these points overlap. My diagram stinks, but it gives you the idea behind what I'm talking about anyway.
Edit: Oh! I almost forgot. BD and XY have to be perpendicular as well. Do you see why that is?
-Dan
Attached Thumbnails
8. Originally Posted by D. Martin
The Problem: A rectangular piece of paper, ABCD, is folded so that point B lies on point D. It is then unfolded to show the crease XY. If AX = 6 and XB = 10, find the length of the crease XY.
10 X 6
B________________________ A
| / |
| / |
| / |
| ________/______________ |
C Y D
I've sketched the rectangle.
BX = XD = 10 and XA = 6. Use Pythagorean theorem in the triangle XAD. Therefore AD = 8.
Use Pythagorean theorem with the grey triangle to calculate the length of the crease. I've got $c = \sqrt{16+64} = 4 \cdot \sqrt{5}$
Attached Thumbnails
9. I'd forgotten how much I love this stuff. | {"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": 43, "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": 0.9460605978965759, "perplexity_flag": "middle"} |
http://drorbn.net/index.php?title=08-401/The_Fundamental_Theorem | # 08-401/The Fundamental Theorem
### From Drorbn
1 Jan 9 About, Notes, HW1
2 Jan 16 HW2, Notes
3 Jan 23 HW3, Photo, Notes
4 Jan 30 HW4, Notes
5 Feb 6 HW5, Notes
6 Feb 13 On TT, Notes
R Feb 20 Reading week
7 Feb 27 Term Test (and solution)
8 Mar 5 HW6, Notes
9 Mar 12 HW7, Notes
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12 Apr 2 FT, HW10, Notes
13 Apr 9 Notes
S Apr 14-25 Study Period:
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Add your name / see who's in!
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The statement appearing here, which is a weak version of the full fundamental theorem of Galois theory, is taken from Gallian's book and is meant to match our discussion in class. The proof is taken from Hungerford's book, except modified to fit our notations and conventions and simplified as per our weakened requirements.
Here and everywhere below our base field F will be a field of characteristic 0.
## Statement
Theorem. Let E be a splitting field over F. Then there is a bijective correspondence between the set {K:E / K / F} of intermediate field extensions K lying between F and E and the set $\{H:H<\operatorname{Gal}(E/F)\}$ of subgroups H of the Galois group $\operatorname{Gal}(E/F)$ of the original extension E / F:
$\{K:E/K/F\}\quad\leftrightarrow\quad\{H:H<\operatorname{Gal}(E/F)\}$.
The bijection is given by mapping every intermediate extension K to the subgroup $\operatorname{Gal}(E/K)$ of elements in $\operatorname{Gal}(E/F)$ that preserve K,
$\Phi:\quad K\mapsto\operatorname{Gal}(E/K):=\{\phi:E\to E:\phi|_K=I\}$,
and reversely, by mapping every subgroup H of $\operatorname{Gal}(E/F)$ to its fixed field EH:
$\Psi:\quad H\mapsto E_H:=\{x\in E:\forall h\in H,\ hx=x\}$.
This correspondence has the following further properties:
1. It is inclusion-reversing: if $H_1\subset H_2$ then $E_{H_1}\supset E_{H_2}$ and if $K_1\subset K_2$ then $\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_2)$.
2. It is degree/index respecting: $[E:K]=|\operatorname{Gal}(E/K)|$ and $[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]$.
3. Splitting fields correspond to normal subgroups: If K in E / K / F is the splitting field of a polynomial in F[x] then $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ and $\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.
The Fundamental Theorem of Galois Theory, all in one.
## Lemmas
The four lemmas below belong to earlier chapters but we skipped them in class (the last one was also skipped by Gallian).
### Zeros of Irreducible Polynomials
Lemma 1. An irreducible polynomial over a field of characteristic 0 has no multiple roots.
Proof. See the proof of Theorem 20.6 on page 362 of Gallian's book. $\Box$
### Uniqueness of Splitting Fields
Lemma 2. Let $\phi:F_1\to F_2$ be an isomorphism of fields, let $f_1\in F_1[x]$ be a polynomial and let f2 = φ(f1), and let E1 and E2 be splitting fields for f1 and f2 over F1 and F2, respectively. Then there is an isomorphism $\bar\phi:E_1\to E_2$ (generally not unique) that extends φ.
Proof. See the proof of Theorem 20.4 on page 360 of Gallian's book. $\Box$
### The Primitive Element Theorem
The celebrated "Primitive Element Theorem" is just a lemma for us:
Lemma 3. Let a and b be algebraic elements of some extension E of F. Then there exists a single element c of E so that F(a,b) = F(c). (And so by induction, every finite extension of E is "simple", meaning, is generated by a single element, called "a primitive element" for that extension).
Proof. See the proof of Theorem 21.6 on page 375 of Gallian's book. $\Box$
### Splitting Fields are Good at Splitting
Lemma 4. (Compare with Hungerford's Theorem 10.15 on page 355). If E is a splitting field of some polynomial f over F and some irreducible polynomial $p\in F[x]$ has a root v in E, then p splits in E.
Proof. Let L be a splitting field of p over E. We need to show that if w is a root of p in L, then $w\in E$ (so all the roots of p are in E and hence p splits in E). Consider the two extensions
E = E(v) / F(v) and E(w) / F(w).
The "smaller fields" F(v) and F(w) in these two extensions are isomorphic as they both arise by adding a root of the same irreducible polynomial (p) to the base field F. The "larger fields" E = E(v) and E(w) in these two extensions are both the splitting fields of the same polynomial (f) over the respective "small fields", as E / F is a splitting extension for f and we can use the sub-lemma below. Thus by the uniqueness of splitting extensions (lemma 2), the isomorphism between F(v) and F(w) extends to an isomorphism between E = E(v) and E(w), and in particular these two fields are isomorphic and so [E:F] = [E(v):F] = [E(w):F]. Since all the degrees involved are finite it follows from the last equality and from [E(w):F] = [E(w):E][E:F] that [E(w):E] = 1 and therefore E(w) = E. Therefore $w\in E$. $\Box$
Sub-lemma. If E / F is a splitting extension of some polynomial $f\in F[x]$ and z is an element of some larger extension L of E, then E(z) / F(z) is also a splitting extension of f.
Proof. Let $u_1,\ldots,u_n$ be all the roots of f in E. Then they remain roots of f in E(z), and since f completely splits already in E, these are all the roots of f in E(z). So
$E(z)=F(u_1,\ldots,u_n)(z)=F(z)(u_1,\ldots,u_n)$,
and E(z) is obtained by adding all the roots of f to F(z). $\Box$
## Proof of The Fundamental Theorem
### The Bijection
Proof of $\Psi\circ\Phi=I$. More precisely, we need to show that if K is an intermediate field between E and F, then $E_{\operatorname{Gal}(E/K)}=K$. The inclusion $E_{\operatorname{Gal}(E/K)}\supset K$ is easy, so we turn to prove the other inclusion. Let $v\in E-K$ be an element of E which is not in K. We need to show that there is some automorphism $\phi\in\operatorname{Gal}(E/K)$ for which $\phi(v)\neq v$; if such a φ exists it follows that $v\not\in E_{\operatorname{Gal}(E/K)}$ and this implies the other inclusion. So let p be the minimal polynomial of v over K. It is not of degree 1; if it was, we'd have that $v\in K$ contradicting the choice of v. By lemma 4 and using the fact that E is a splitting extension, we know that p splits in E, so E contains all the roots of p. Over a field of characteristic 0 irreducible polynomials cannot have multiple roots (lemma 1) and hence p must have at least one other root; call it w. Since v and w have the same minimal polynomial over K, we know that K(v) and K(w) are isomorphic; furthermore, there is an isomorphism $\phi_0:K(v)\to K(w)$ so that φ0 | K = I yet φ0(v) = w. But E is a splitting field of some polynomial f over F and hence also over K(v) and over K(w). By the uniqueness of splitting fields (lemma 2), the isomorphism φ0 can be extended to an isomorphism $\phi:E\to E$; i.e., to an automorphism of E. but then φ | K = φ0 | K = I so $\phi\in\operatorname{Gal}(E/K)$, yet $\phi(v)=w\neq v$, as required. $\Box$
Proof of $\Phi\circ\Psi=I$. More precisely we need to show that if $H<\operatorname{Gal}(E/F)$ is a subgroup of the Galois group of E over F, then $H=\operatorname{Gal}(E/E_H)$. The inclusion $H<\operatorname{Gal}(E/E_H)$ is easy. Note that H is finite since we've proven previously that Galois groups of finite extensions are finite and hence $\operatorname{Gal}(E/F)$ is finite. We will prove the following sequence of inequalities:
$|H|\leq|\operatorname{Gal}(E/E_H)|\leq [E:E_H]\leq |H|$
This sequence and the finiteness of | H | imply that these quantities are all equal and since $H<\operatorname{Gal}(E/E_H)$ it follows that $H=\operatorname{Gal}(E/E_H)$ as required.
The first inequality above follows immediately from the inclusion $H<\operatorname{Gal}(E/E_H)$.
By the Primitive Element Theorem (Lemma 3) we know that there is some element $u\in E$ so that E = EH(u). Let p be the minimal polynomial of u over EH. Distinct elements of $\operatorname{Gal}(E/E_H)$ map u to distinct roots of p, but p has exactly degp roots. Hence $|\operatorname{Gal}(E/E_H)|\leq\deg p=[E:E_H]$, proving the second inequality above.
Let $\sigma_1,\ldots,\sigma_n$ be an enumeration of all the elements of H, let ui: = σiu (with u as above), and let f be the polynomial
$f=\prod_{i=1}^n(x-u_i)$.
Clearly, $f\in E[x]$. Furthermore, if $\tau\in H$, then left multiplication by τ permutes the σi's (this is always true in groups), and hence the sequence $(\tau u_i=\tau\sigma u_i)_{i=1}^n$ is a permutation of the sequence $(u_i)_{i=1}^n$, hence
$\tau f=\prod_{i=1}^n(x-\tau u_i)=\prod_{i=1}^n(x-u_i)=f$,
and hence $f\in E_H[x]$. Clearly f(u) = 0, so p | f, so $[E:E_H]=\deg p\leq \deg f=n=|H|$, proving the third inequality above. $\Box$
### The Properties
Property 1. If $H_1\subset H_2$ then $E_{H_1}\supset E_{H_2}$ and if $K_1\subset K_2$ then $\operatorname{Gal}(E/K_1)>\operatorname{Gal}(E/K_1)$.
Proof of Property 1. Easy. $\Box$
Property 2. $[E:K]=|\operatorname{Gal}(E/K)|$ and $[K:F]=[\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)]$.
Proof of Property 2. If K = EH, then $|\operatorname{Gal}(E/K)|=|\operatorname{Gal}(E/E_H)|=[E:E_H]=[E:K]$ as was shown within the proof of $\Phi\circ\Psi=I$. But every K is EH for some H, so $|\operatorname{Gal}(E/K)|=[E:K]$ for every K between E and F. The second equality follows from the first and from the multiplicativity of the degree/order/index in towers of extensions and in towers of groups:
$[K:F] = \frac{[E:F]}{[E:K]} = \frac{|\operatorname{Gal}(E/F)|}{|\operatorname{Gal}(E/K)|} = [\operatorname{Gal}(E/F):\operatorname{Gal}(E/K)].\quad\Box$
Property 3. If K in E / K / F is the splitting field of a polynomial in F[x] then $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ and $\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.
Proof of Property 3. We will define a surjective (onto) group homomorphism $\rho:\operatorname{Gal}(E/F)\to\operatorname{Gal}(K/F)$ whose kernel is $\operatorname{Gal}(E/K)$. This shows that $\operatorname{Gal}(E/K)$ is normal in $\operatorname{Gal}(E/F)$ (kernels of homomorphisms are always normal) and then by the first isomorphism theorem for groups, we'll have that $\operatorname{Gal}(K/F)\cong\operatorname{Gal}(E/F)/\operatorname{Gal}(E/K)$.
Let σ be in $\operatorname{Gal}(E/F)$ and let u be an element of K. Let p be the minimal polynomial of u in F[x]. Since K is a splitting field, lemma 4 implies that p splits in K[x], and hence all the other roots of p are also in K. As σ(u) is a root of p, it follows that $\sigma(u)\in K$ and hence $\sigma(K)\subset K$. But since σ is an isomorphism, [σ(K):F] = [K:F] and hence σ(K) = K. Hence the restriction σ | K of σ to K is an automorphism of K, so we can define ρ(σ) = σ | K.
Clearly, ρ is a group homomorphism. The kernel of ρ is those automorphisms of E whose restriction to K is the identity. That is, it is $\operatorname{Gal}(E/K)$. Finally, as E / F is a splitting extension, so is E / K. So every automorphism of K extends to an automorphism of E by the uniqueness statement for splitting extensions (lemma 2). But this means that ρ is onto. $\Box$ | {"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": 90, "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": 0.9125238656997681, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23505 | ## Examples of common false beliefs in mathematics. [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.
Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are
(i) a bounded entire function is constant; (ii) sin(z) is a bounded function; (iii) sin(z) is defined and analytic everywhere on C; (iv) sin(z) is not a constant function.
Obviously, it is (ii) that is false. I think probably many people visualize the extension of sin(z) to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.
A second example is the statement that an open dense subset U of R must be the whole of R. The "proof" of this statement is that every point x is arbitrarily close to a point u in U, so when you put a small neighbourhood about u it must contain x.
Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.
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I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan May 6 2010 at 0:55
12
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – To be cont'd May 22 2010 at 9:04
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wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. – S. Sra Sep 20 2010 at 12:39
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It's a thought -- I might consider it. – gowers Oct 4 2010 at 20:13
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Meta created meta.mathoverflow.net/discussion/1165/… – quid Oct 8 2011 at 14:27
show 8 more comments
## 169 Answers
"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This is more of a false philosophy than a clear mistake, but nevertheless it is very common:
A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $\le 2^{\aleph_0}$, etc.
This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.
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False statement: If $A$ and $B$ are subsets of $\mathbb{R}^d$, then their Hausdorff dimension $\dim_H$ satisfies
$$\dim_H(A \times B) = \dim_H(A) + \dim_H(B).$$
EDIT: To answer Benoit's question, I do not know about a simple counterexample for $d = 1$, but here is the usual one (taken from Falconer's "The Geometry of Fractal Sets"):
Let $(m_i)$ be a sequence of rapidly increasing integers (say $m_{i+1} > m_i^i$). Let $A \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_j + 1 \leq r \leq m_{j+1}$ and $j$ is odd. Let $B \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_{j} + 1 \leq r \leq m_{j+1}$ and $j$ is even. Then $\dim_H(A) = \dim_B(A) = 0$. To see this, you can cover $A$, for example, by $10^k$ covers of length $10^{- m_{2j}}$, where $k = (m_1 - m_0) + (m_3 - m_2) + \dots + (m_{2j - 1} - m_{2j - 2})$.
Furthermore, if $\mathcal{H}^1$ denotes the Hausdorff $1$-dimensional (metric) outer measure of $E$, then the result follows by showing $\mathcal{H}^1(A \times B) > 0$. This is accomplished by considering $u \in [0,1]$ and writing $u = x + y$, where $x \in A$ and $y \in B$. Let $proj$ denote orthogonal projection from the plane to $L$, the line $y = x$. Then $proj(x,y)$ is the point of $L$ with distance $2^{-1/2}(x+y)$ from the origin. Thus, $proj( A \times B)$ is a subinterval of $L$ of length $2^{-1/2}$. Finally, it follows:
$$\mathcal{H}^1(A \times B) \geq \mathcal{H}^1(proj(A \times B)) = 2^{-1/2} > 0.$$
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Actually, the situation is worse than I say: there exist sets $A, B \subset \mathbb{R}$ with $dim_H(A \times B )= 1$, and yet $\dim_h(A) = \dim_H(B) = 0$. – DJC Apr 5 2011 at 6:22
show 2 more comments
A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.
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Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere.
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An incredibly common false belief is:
For a (say smooth, projective) algebraic variety $X$ the $K_X$-negative part of the cone $NE(X)$ is locally polyhedral.
A right statement of the theorem of the cone is
$\overline{NE(X)} = \overline{NE(X)}_{K_X \geq 0} + \sum_{i} \mathbb{R}[C_i]$ for a denumerable set ${ C_i }$ of rational curves, which accumulate at most on the hyperplane $K_X = 0$.
At a first glance this seems to imply that $\overline{NE(X)}_{K_X < 0}$ is locally poyhedral, but this is not true. It depends on the shape of the intersection `$\overline{NE(X)} \cap \{ K_X = 0 \}$`.
For instance if this latter intersection is round, and there is only one curve $C_i$, the half-cone $\overline{NE(X)}_{K_X < 0}$ is actually a circular cone! Definitely not polyhedral in any sense. I believe this behaviour can happen even with varieties birational to abelian varieties.
The strange thing about this false belief is that it is held true by many competent mathematicians (and indeed I don't believe that many undergraduates meet the theorem of the cone!).
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Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... – Victor Protsak May 5 2010 at 6:57
5
Yes, but among those, almost all believe that the wrong version is true. – Andrea Ferretti May 5 2010 at 10:13
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And about 50% of the large community who cannot understand the point will believe that the right version is true! Rather high percentage... – Wadim Zudilin May 5 2010 at 11:41
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Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. This leads us to the mistaken:
“Let $(g_n)$ be a sequence in $H$. If, for every $f\in H$, $\langle f,g_n\rangle\to0$, then $g_n\to 0$.”
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@Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! – Peter LeFanu Lumsdaine Dec 2 2010 at 0:43
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Duality reverses inclusions of vector spaces.
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That's funny, because I don't imagine this kind of idea would occur to someone who has just learned the definition of a dual space. That would be a strangely sophisticated mistake to make. – Thierry Zell Apr 7 2011 at 0:21
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A random $k$-coloring of the vertices of a graph $G$ is more likely to be proper than a random $(k-1)$-coloring of the same graph.
(A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.)
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2
For some graphs $G$ and integers $k$, the opposite. The easiest example is the complete bipartite graph $K_{n,n}$ with $k=3$. The probability a $2$-coloring is proper is about $(1/4)^n$ while the same for a $3$-coloring is about $(2/9)^n$, where I've ignored minor terms like constants. The actual probabilities cross at $n=10$, so as an explicit example, a random $2$-coloring of $K_{10,10}$ is more likely to be proper than a random $3$-coloring. – aorq May 10 2011 at 0:37
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This seems like a good example of a counterintuitive statement, but to call it a common false belief would mean that there are lots of people who think it's true. The question would probably never have occurred to me it I hadn't seen it here. The false belief that Euclid's proof of the infinitude of primes, on the other hand, actually gets asserted in print by mathematicians---in some cases good ones. – Michael Hardy May 10 2011 at 15:36
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An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$
$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .
Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.
In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.
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This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. – Qiaochu Yuan Feb 24 2011 at 21:08
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I have heard the following a few times :
"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."
$f(z)=e^z$ of course is a counterexample.
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No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. – Tom Goodwillie May 4 2011 at 0:16
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When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.
I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.
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You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. – Qiaochu Yuan May 4 2010 at 22:14
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In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. – Leandro May 4 2010 at 22:34
Many people believe that Cantor proved the uncountability of the real line using a diagonal argument. This paper does not that provide that proof; Cantor's stated purpose was to prove the existence of `uncountable infinities' without using the theory of irrational numbers.
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More to the point, I think, is that the paper proves that the power set of any set has greater cardinality than the set itself. This is the first proof that there is no greatest cardinality. (The uncountability of the real line easily follows, even if Cantor does not mention it because he has bigger fish to fry.) – John Stillwell May 31 2010 at 5:12
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Just to fill in some history here: if I remember right, Cantor first proved the uncountability of the reals by other arguments, then later (as you reference) found the diagonal argument, as a proof of the more general statement about power sets. – Peter LeFanu Lumsdaine Sep 27 2010 at 3:01
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The fundamental group of the Klein bottle is $D_\infty$, the infinite dihedral group (which is $\mathbb Z \rtimes \mathbb Z_2$).
I believed this for some time, and I seem to recall some others having the same confusion.
The group that has been mistaken for $D_\infty$ is in fact $\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation $x^2y^2=1$. The former abelianizes to $\mathbb Z_2\oplus \mathbb Z_2$, the latter to $\mathbb Z\oplus \mathbb Z_2$.
A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian.
I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations $x\mapsto ax+b$, where $a>0, b\in \mathbb R$ are homeomorphic to $\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular $\mathbb R^2$ cross any finite non-abelian group is non-abelian.
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True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).
False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.
There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.
But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.
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What I mean is: People use these formulations as a protective charm against a danger they don't see but intuitively feel is there, although closer inspection shows that it is pure superstition. – darij grinberg Mar 15 2011 at 17:26
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I saw many students using the "fact" that for a subset `$S$` of a group one has `$SS^{-1}=\{e\}$`
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This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! – Thierry Zell Apr 14 2011 at 15:50
If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.
False for $\alpha>1$, because this set contains only constant functions.
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In group theory, if `$G_1 \cong G_2$` and `$H_1 \cong H_2$`, then
`$G_1 / H_1 \cong G_2 / H_2$`.
For example, `$\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$`. The point is that the inclusion of `$H_j$` into `$G_j$` is needed in order to define the quocient.
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False belief: A function being continuous in some open interval implies that it is also differentiable in that interval:
Counterexample:
The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:
$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$
Where $a \in (0, 1)$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though!
Another example that is maybe not a false belief so much as something that is very hard to believe at first is the Monty Hall problem. I remember spending most of a day in catatonic despair when I first learned of it...
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Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. – Thierry Zell Aug 31 2010 at 2:34
This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the `$Y^*$` (the dual of $Y$) is isomorphic to a subspace of `$X^*$` (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris
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I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^*$ is a subobject of $X^*$”, where the correct dual is “$X^*$ is a quotient of $Y^*$”. – Peter LeFanu Lumsdaine Dec 1 2010 at 15:19
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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.
Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.
Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.
The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.
This statement happens to be true.
Here is a reference:
http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Hodgegaillard/
The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.
Here is the (trivial) algebra background.
Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.
We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.
Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.
Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.
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Here's another howler some people commit: If m, n are integers such that m divides n^2 then m divides n.
It's true sometimes, for example if m is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers p, q, r with p squarefree such that $m = p q^2$ and $n = p q r$
The usual counterexample is that 8 divides 4^2 but not 4 ;-)
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An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P – Peter LeFanu Lumsdaine Feb 23 2011 at 9:40
If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.
True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.
False over other fields. For example, over $\mathbb{C}$, `$\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$` and `$\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$` are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.
False for other nondegenerate symmetric bilinear forms: `$\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$` and `$\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$` are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to `$\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$` and `$\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$`.
You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form `$\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$`.
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You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. – Mark Meckes Jan 28 2011 at 16:46
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Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) – Mark Meckes Jan 28 2011 at 21:24
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"The set A = {a, b} has two elements..."
It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...
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I'm not sure there is a false belief here, as much as awkward writing. Depending on context, I might very well write "The set $\{a,b \}$ (where $a$ and $b$ might be equal)..." if this issue mattered. – David Speyer May 6 2010 at 11:16
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There are many situations where one needs to speak of a set of two numbers that may or may not be equal. E.g.: "Let x<sub>1</sub>, x<sub>2</sub> ∈ ℝ. Then among all the open intervals containing the set {x<sub>1</sub>, x<sub>2</sub>}, none of them is contained in all the others." If one is addressing mathematicians, there is no need to specify that x<sub>1</sub> might be equal to x<sub>2</sub>. – Daniel Asimov Jun 17 2010 at 23:34
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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.
Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.
Belief 2. It's simpler and more efficient to use the following fact.
Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.
(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references
http://en.wikipedia.org/wiki/Matrix_exponential#Alternative
http://www.iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients/
Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.
Here is another attempt to describe the situation.
Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.
This can be summarized as follows:
There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.
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"$\emptyset$ is a basis for {0}" is an immediate consequence of the definitions. There is no false belief in this point. – Johannes Hahn May 12 2010 at 15:14
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Dear Johannes, please reread my post. – Pierre-Yves Gaillard May 12 2010 at 15:18
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Your opinions are normative statements: "one *should*" and "it is *better*". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. – Robin Chapman May 15 2010 at 14:07
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I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. – Qiaochu Yuan May 17 2010 at 6:19
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These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.
Another mistake is if one i asked to solve this equation, $\displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.
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False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"
The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.
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I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. – Tom Goodwillie Oct 5 2010 at 0:35
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True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. – Peter Arndt Oct 5 2010 at 10:52
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If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.
More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:
1. Any expression inside a radical evaluates to a real number
2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$
Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.
$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$
$f(x) = 4x^3 - 3x + \frac{1}{2}$.
So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.
Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine
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This has a name: en.wikipedia.org/wiki/Casus_irreducibilis – Qiaochu Yuan Apr 6 2011 at 21:21
For a bounded subset of a metric space the diameter is two times the radius!
Let $S\subset X$ be bounded. The definitions are:
`$\mathrm{diameter}(S):=\sup\{d(x,y)\,|\,x,y\in S\}$`
`$\mathrm{radius}(S):=\inf\{r>0\,|\,\exists x\in X:\,S\subset B(x,r)\}$`
where `$B(x,r)$` denotes the open ball of radius `$r$` around `$x$`.
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Hםw do you define the radius of an arbitrary bounded subset? – Mark Schwarzmann Apr 11 2011 at 15:34
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Disproved nicely by Reuleaux triangles. – darij grinberg Apr 12 2011 at 8:10
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Disproved nicely by a two-point metric space. – Tom Goodwillie Apr 17 2011 at 1:36
Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)
For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.
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http://mathoverflow.net/revisions/102129/list | ## Return to Answer
2 Adeded two notes
I agree with the answer given by Peter Dalakov, and with the comments made so far. On the other hand, there is a solution as close as possible to the requirement.
We consider a default Riemannian metric $g_0$ on the vector bundle $E$. This is the special metric mentioned by Johannes Nordström in his comment. Any other metric can be obtained from this one, by applying a section from the bundle $GL(E)$ (having as fiber at $p\in M$ the general linear group of $E_p$). Any section of $GL(E)$, when applied to $g_0$, will give another Riemannian metric. Also, any section of $GL(E)$ can be written as $\exp(s)$, where $s$ is a section of the vector bundle $SL(E)$.
Hence, we can take as the desired vector bundle the bundle $SL(E)$. Any section $s$ of it will provide a transformation $\exp(s)$ which, when applied to the specially chosen metric $g_0$, gives another metric. Any metric can be obtained this way.
Note 1:
This works similarly for the Hermitian case.
Note 2:
If for the particular problem is necessary to work with generic linear combinations of the metrics themselves, then the solution proposed here will not be of use. Instead, one needs geometric methods which apply to metrics having variable signature, hence also being degenerate. Such methods were developed in arXiv:1105.0201, arXiv:1105.3404, and arXiv:1111.0646. One can define for example covariant derivatives for special tensor fields and differential forms, and also define the Riemann tensor $R_{abcd}$, although these constructions are apparently forbidden by $g_{ab}$ not being always invertible.
1
I agree with the answer given by Peter Dalakov, and with the comments made so far. On the other hand, there is a solution as close as possible to the requirement.
We consider a default Riemannian metric $g_0$ on the vector bundle $E$. This is the special metric mentioned by Johannes Nordström in his comment. Any other metric can be obtained from this one, by applying a section from the bundle $GL(E)$ (having as fiber at $p\in M$ the general linear group of $E_p$). Any section of $GL(E)$, when applied to $g_0$, will give another Riemannian metric. Also, any section of $GL(E)$ can be written as $\exp(s)$, where $s$ is a section of the vector bundle $SL(E)$.
Hence, we can take as the desired vector bundle the bundle $SL(E)$. Any section $s$ of it will provide a transformation $\exp(s)$ which, when applied to the specially chosen metric $g_0$, gives another metric. Any metric can be obtained this way. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9325823783874512, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/19261/drawing-graphs-by-numbers-a-minimality-question/19324 | ## Drawing (graphs) by numbers: a minimality question
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Every simple graph $G$ can be represented ("drawn") by numbers in the following way:
1. Assign to each vertex $v_i$ a number $n_i$ such that all $n_i$, $n_j$ are coprime whenever $i\neq j$. Let $V$ be the set of numbers thus assigned.
2. Assign to each maximal clique $C_j$ a unique prime number $p_j$ which is coprime to every number in $V$.
3. Assign to each vertex $v_i$ the product $N_i$ of its number $n_i$ and the prime numbers $p_k$ of the maximal cliques it belongs to.
Then $v_i$, $v_j$ are adjacent iff $N_i$ and $N_j$ are not coprime,
i.e. there is a (maximal) clique they both belong to. Edit: It's enough to assign $n_i = 1$ when $v_i$ is not isolated and does not share all of its cliques with another vertex.
Being free in assigning the numbers $n_i$ and $p_j$ lets arise a lot of possibilites, but also the following question:
QUESTION
Can the numbers be assigned systematically such that the greatest $N_i$ is minimal (among all that do the job) — and if so: how?
It is obvious that the $n_i$ in the first step have to be primes for the greatest $N_i$ to be minimal. I have taken the more general approach for other - partly answered - questions like "Can the numbers be assigned such that the set $\lbrace N_i \rbrace_{i=1,..,n}$ fulfills such-and-such conditions?"
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Since you're only using multiplicative properties of integers, and coprimeness, you're really just assigning sets of integers to the vertices and cliques, and coprimeness translates into disjointness. As far as the minimality, you're assigning weights of $\log p_i$ to the integer $i$, where $p_i$ is the $i$-th prime. – Victor Miller Mar 25 2010 at 14:30
## 2 Answers
I'm not sure this qualifies as an answer, but I hope these remarks are useful to you; they re-present your problem in a format which is likely to be answerable by experts in discrete optimization.
The abstract: I suspect the problem of computing the smallest maximum Nj is intractible, and suggest approaches to obtaining upper and lower bounds for Nj . I also make brief remarks about the case of low clique number.
### Reformulation
Let K be the set of maximal cliques Cj , and consider the bipartite graph graph H with vertex-set V(G) ∪ K, and adjacency defined by $$v~C_j \;\in\; E(H) \;\;\;\iff\;\;\;\; v \in C_j \;.$$ Your weighting scheme then amounts to a weighting of the vertices of H by (co-)prime integers. Instead of considering products of such (co-)prime integers, we may consider the sum of their logarithms. So:
• weight the vertices of H with real numbers ω(v) = ln(p) for distinct primes p;
• define the "weight" Ω(v) of a neighborhood of a vertex v as the sum of ω(x) for x ranging over v and its neighbors. For vertices v ∈ V(G), its neighbors are the maximal cliques Cj to which it belongs in G; for vertices C ∈ K, its neighbors are all of the vertices in G which C contains.
We are interesting in minimizing $$\large \Omega(G) \equiv \max_{v \in V(G)} \Omega(v)$$ for v ∈ V(G) subject to the above definitions/constraints. The minimum Nj which you describe above is then eΩ(G).
Now, the weights ω(v) for v ∈ V(H) form a vector of logarithms of primes. There's no reason to take any coefficient to be larger than ln(ph), where h = |V(H)| and ph is the hth prime. So we may as well fix the column vector p = ( ln 2, ln 3, ... , ln ph )T, and describe the weight function ω in terms of permutations of the coefficients of this vector. So really, we would like to obtain $$\Omega^\ast(G) \;\;=\; \large \min_{\Pi \in \mathfrak S_h} \;\max_{v \in V(G)} \big(\mathbf{e}_v^\top A(H) \: \Pi \;\mathbf{p} \big)$$ where A(H) is the adjacency matrix of H, and $\mathfrak S_h$ is the group of permutation matrices on ℝh.
### Remarks on the reformulation
Evaluating Ω*(G) is likely to be difficult, as in computationally intractible. (Disclaimer: I am not an expert on such problems, and I have not given this instance a lot of thought; but some similar problems are NP complete.) A better question is whether you can get "nice" upper or lower bounds for Ω*(G).
• You can obtain a lower bound for Ω*(G) by taking a convex relaxation. For instance, instead of optimizing over $\mathfrak S_h$, oprimize over the convex closure of that set, which is the set of doubly-stochatic matrices over ℝh. You can then exploit the fact that the maximum is the uniform norm of the [restriction to ℝV(G) of the] vector ω(H) = A(H) Π p ; as such, it is a convex function (as the uniform norm satisfies the triangle inequality). It should be possible to optimize this function efficiently using steepest descent techniques.
• Obviously, you're more interested in upper bounds for Ω*(G). The function f(x) = ln(px) grows asymptotically like ln(x) + ln(ln x); therefore, the contribution of a large log-prime weight to some sum Ω(v) is not much different than the contribution of a slightly larger log-prime weight. Optimizing the location of the larger primes among themselves is then unlikely to be useful; in practise it is more useful to optimize the location of the smaller primes.
The weights of the clique-vertices Cj contribute to many different neighborhood weights Ω(v). This suggests that a reasonable approach is to allocate the smallest log-prime weights to cliques according (roughly) to the number of vertices they contain. Obviously this will fail if there is a very large clique which "interacts" with very few other cliques (i.e. shares vertices in common with few other cliques), and there exists elsewhere a large congregation of cliques which each share something like half of their elements with other cliques (i.e. for a large subset S of V(G), each vertex in S belongs to approximately half of a large collection of cliques). It may be worthwhile to investigate the graph of incidence of maximal cliques.
A final remark: in the case of a bipartite graph, the maximal cliques are all edges, in which case the graph H is just a subdivision of G. In this case, attributing weights to the vertices v ∈ V(G) does not aid in the representation of the graph. For graphs with low clique number, it may be worthwhile to investigate a similar scheme where only the edges or maximal cliques are given weights, or more generally where almost every vertex is given weight 1.
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Thanks! I didn't think of "discrete optimization" but that's what my question seems to be about. (You are providing so many details, it's hard for me to follow.) – Hans Stricker Mar 25 2010 at 18:01
Sorry (?). For whatever reason, I often find it difficult to give impressionist depictions of mathematical ideas. – Niel de Beaudrap Mar 26 2010 at 9:27
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As an alternative to my earlier computational answer for particular graphs G, here is a worst-case description of the asymptotic growth of the minimum size of the integers Nj
Let h be the sum of |V(G)| and the number of maximal cliques in G (bounded above by |E(G)|, which is saturated for bipartite graphs, whose edges are the maximal cliques). Let c be the maximum number of maximal cliques to which a vertex v ∈ V(G) may belong. Because ph ≤ h (ln(h) + ln ln(h)), we can bound $$\large \Big(\min_{\small\text{weightings}} \; \max_{v \in V(G)}\; N_v\Big) \;\;\in\;\; \mathrm O\Big(h^c \log^c(h)\Big).$$ This bound is asymptotically saturated by placing the largest prime weights on the vertex in the largest number of maximal cliques, and those maximal cliques of which it is a part, which is obviously a bad thing to do. But e.g. in Cayley graphs G, every vertex belongs to the same number of maximal cliques, this asymptotic growth cannot be avoided, as there will exist vertices v for which Nv will consist exclusively of a product of primes pt, for t bounded below by a constant fraction of h.
One can construct bipartite Cayley graphs in which the degree of each vertex is a constant fraction of n = |V(G)|. We then have h = α n for some 0 < α < 1; and h = |V(G)| + |E(G)| ∈ O(n2), so that $$\large \Big(\min_{\small\text{weightings}} \; \max_{v \in V(G)}\; N_v\Big) \;\;\in\;\; \mathrm O\Big(n^{2\alpha n} \log^{\alpha n}(n)\Big)$$ for such graphs. Thus there exist graphs for which the coefficients Nj grow much more quickly than e.g. the factorial function.
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http://mathoverflow.net/questions/102461/concerning-the-classification-of-transversally-integral-affine-structures-on-symp | ## Concerning the classification of transversally integral affine structures on symplectic foliations $F$
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Recently, in http://arxiv.org/pdf/1207.3655.pdf, the authors have determined that an element $c$ in $H^2(P, P_v)$ is the Chern class of a twisted isotropic realization $p$: $(M, \sigma)$\rightarrow$(P, \pi, \phi)$ if and only if $D_\Lambda(c)$= $e_\phi$, increasing progress towards determining the constructions of twisted isotropic realizations. This problem is divided into 2 steps:
1. Classify, up to isomorphism, all transversly integral affine structures $F$
2. For a fixed transversal integral affine structure \Lambda\rightarrow$P$, determine which elements of $H^2(P, P_\Lambda)$ isomorphic to $H^1(P; C^\infty($v^*$F/A$)\$
How completely classified are the transversally affine structures on a sympletic foliation? Any references will be greatly appreciated.
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http://mathoverflow.net/revisions/55006/list | 2 edited tags
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# n-th roots of Pythagorean numbers
Let $F$ be the field ${\mathbb Q}(i)\subset \mathbb C$ and let $T\subset F$ be the set of all elements of complex absolute value 1. Let $n$ be a natural number $\ge 2$ and let $\mu_n(T)\subset\mathbb C$ be the set of all $n$-th roots of elements of $T$. Finally, let $E=F(\mu_n(T))$.
Question: Is the field extension $E/F$ finite or infinite? | {"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": 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": 0.7617565393447876, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/43362/the-consistency-of-zfc-ch-gives-the-ability-to-travel-to-a-universe-which-model | ## The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \neg CH? [closed]
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Why does forcing seem to be so vacuously true?
It seems like you are just reversing the subset containment of the model of ZFC + CH to be the other way in the poset. So, why is this valid? Why are you allowed to just put the empty set at the top of the partial order*?
*I have just started learning set theory so I have only seen one example of forcing and it was the forcing of the set of reals to be equal to the second uncountable cardinal.
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Con(ZFC + CH) -> Con(ZFC + not CH) – Martin Brandenburg Oct 24 2010 at 8:17
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I do not understand your first question. Forcing is a technique, not a result. What do you mean by "vacuously true"? Also, I do not understand your "why are you allowed". Care to elaborate? – Andres Caicedo Oct 24 2010 at 14:34
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Erin, welcome to MathOverflow! We look forward to more questions from you, as you learn forcing better... – Joel David Hamkins Oct 24 2010 at 19:54
Thanks Martin, I realize this is what I should have written. The title was for intrique :) – Erin K Carmody Oct 25 2010 at 22:44
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The edit seems to be based on a misunderstanding. You can get universes where CH is false by adjoining a lot of Cohen reals to any universe, whether or not this ground universe satisfies CH. (Of course, if your ground model violates CH, then you don't need to force over it at all.) – Andreas Blass Mar 10 2012 at 17:21
## 2 Answers
The empty set $\emptyset$ here is the forcing condition that has the least amount of information about the generic object being constructed.
Since it has the least information, you might think it should be at the bottom of the partial order. There are two replies to this:
(1) Mathematically, we are free to define our partial order however we want, even if it confuses the reader.
(2) In this case, there is no intent to confuse the reader. Having the least information, the condition $\emptyset$ also leaves the greatest amount of possibilities still open. So it is really a 50-50 decision which way one should define it. (Others may know the history of this particular convention.)
A similar convention question is whether the Turing degree of the computable sets, $\mathbf 0$, is the smallest or the largest Turing degree. Conventionally one speaks of the degrees of unsolvability and so $\mathbf 0$ is the smallest, but if the convention were degrees of solvability it would be the greatest. Since none of the other degrees are very solvable at all, the current convention is perhaps the best in this case.
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Thanks for your response. What do you mean it isa 50-50 decision which way to define it? – Erin K Carmody Oct 25 2010 at 22:41
I mean you have to define $\emptyset$ to be the smallest, or the largest, and it doesn't matter which you choose as long as you stick to your decision. – Bjørn Kjos-Hanssen Oct 25 2010 at 23:38
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Bjorn already mentioned one important point: We can define a partial order in any way we want. It should also be mentioned that Cohen defined the order in the other direction: He considered the empty condition the smallest and defined everything else accordingly (i.e., just the other way around). Shelah for instance still uses Cohen's original version of forcing.
Considering $\emptyset$ the largest condition in this particular forcing context agrees with the usual order on Boolean algebras in the following way:
When forcing with Boolean algebras, the elements of the Boolean algebra correspond to truth values of sentences. If a sentence implies another, the truth value of the first sentence is less or equal to the truth value of the second sentence. The empty condition in Cohen forcing is the truth value of all sentences that are true in every forcing extension by Cohen forcing. This is the largest possible truth value in this context. (This is not to be confused with the fact that $\emptyset$ is the smallest element of the Boolean algebra $\mathcal P(\omega)$. The Boolean algebra corresponding to Cohen forcing is very different from $\mathcal P(\omega)$.) This interpretation of the direction of the order agrees with Bjorn's explanation saying that $\emptyset$ leaves open the most possibilities.
Anyhow. The whole discussion above in somewhat unnecessary since your problem is actually at a different level: The partial order that we are talking about is just a technical tool in order to construct a forcing extension. It has little to do with the containment relation on the subsets of $\mathbb N$. In fact, the containment relation among subsets is the same in the ground model (the one that satisfies CH) and the generic extension. I.e., if $a,b\subseteq\mathbb N$ are both in the ground model, then $a\subseteq b$ holds in the extension iff it holds in the ground model. In the extension we just have many more subsets of $\mathbb N$.
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Thanks for your response. I see the distinction now between the empty set and the empty condition as you defined it. This relieved some of my philosophical concerns :) – Erin K Carmody Oct 25 2010 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": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9358017444610596, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2007/12/01/miscellany/?like=1&_wpnonce=401286028f | # The Unapologetic Mathematician
## Miscellany
Well, yesterday was given over to exam-writing, so today I’ll pick up a few scraps I mentioned in passing on Thursday.
First of all, the rational numbers are countable. To be explicit, in case I haven’t been before, this means that there is an injective function from the set of rational numbers to the set of natural numbers. Really, I’ll just handle the positive rationals, but it’s straightforward how to include the negatives as well. To every positive rational number we can get a pair of natural numbers — the numerator and the denominator. Then we can send the pair $(m,n)$ to the number $\frac{(m+n)(m+n+1)}{2}+n$, which is a bijection between the set of all pairs of natural numbers and all natural numbers. Clearly they contain the natural numbers, so the set of rational numbers is countably infinite.
Now, equivalence relations. Given any relation $R\subseteq X\times X$ on a set $X$ we can build it up into an equivalence relation. First throw in the diagonal $\{(x,x)|x\in X\}$ to make it reflexive. Then throw in all the points $(y,x)$ for $xRy$ to make it symmetric. For transitivity, we can similarly start throwing elements into the relation until this condition is satisfied.
But that’s all sort of ugly. Here’s a more elegant way of doing it: given any relation $R\subseteq X\times X$, consider all the relations $S$ on $X$ that contain $R$ — $R\subseteq S\subseteq X\times X$. Some of these $S$ will be equivalence relations. In particular, the whole product $X\times X$ is an equivalence relation, so there is at least one such. It’s simple to verify that the intersection of any family of equivalence relations on $X$ is again an equivalence relation, so we can take the intersection of all equivalence relations on $X$ containing $R$ to get the smallest such relation. Notice, by the way, how this is similar to generating a topology from a subbase, or to taking the closure of a subset in a topological space.
Finally, absolute values. In any totally ordered group we have the positive “cone” $G^+$ of all elements $g\in G$ with $g\geq1$. and the negative “cone” $G^-$ of all elements $g\in G$ with $1\geq g$. In the latter case, we can multiply both sides by the inverse of $g$ to get $g^{-1}\geq1$ in the positive cone. Notice that the identity $1$ is in both cones, but the reflection described leaves it fixed. So for every element in $g\in G$ we get a well-defined element $\left|g\right|\in G^+$ called its absolute value. Of course, we often assume that $G$ is abelian and write this all additively instead of multiplicatively.
This function has a number of nice properties. First of all, $\left|g\right|$ is always in $G^+$. Secondly, $\left|g\right|$ is the identity in $G$ if and only if $g$ itself is the identity. Thirdly, $\left|g\right|=\left|g^{-1}\right|$. And finally, if $G$ is abelian we have the “triangle inequality” $\left|a+b\right|\leq\left|a\right|+\left|b\right|$.
Okay, does that catch us up?
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http://physics.stackexchange.com/questions/tagged/research-level+spin-model | # Tagged Questions
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From what I gather, a Boltzmann machine (BM) is essentially a spin glass with no applied field evolving under Glauber dynamics (if this is at all mistaken, I don't think it will be off enough to ...
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### Random bond Ising model and computational efficiency
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Measurements of consecutive sites in a many body qudit system (e.q. a spin chain) can be interpreted as generating a probabilistic sequence of numbers $X_1 X_2 X_3 \ldots$, where \$X_i\in ...
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### Do any entanglement measures for mixed states exist that use only single site correlation functions?
For a pure state $\rho_{AB}$, the entropy of entanglement of subsystem $A$ is \begin{equation} S( \rho_A) = -tr (\rho_A \log \rho_A) \end{equation} where $\rho_A$ is the reduced density matrix of A. ...
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### How does one geometrically quantize the Bloch equations?
I've just now rated David Bar Moshe's post (below) as an "answer", for which appreciation and thanks are given. Nonetheless there's more to be said, and in hopes of stimulating further posts, I've ...
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### Fluctuations of an interface with hammock potential
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### Phase Transition in the Ising Model with Non-Uniform Magnetic Field
Consider the Ferromagnetic Ising Model ($J>0$) on the lattice $\mathbb{Z}^2$ with the Hamiltonian with boundary condition $\omega\in\{-1,1\}$ formally given by ...
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### Mermin-Wagner theorem in the presence of hard-core interactions
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http://mathforum.org/mathimages/index.php?title=Markus-Lyapunov_Fractals&oldid=17023 | # Markus-Lyapunov Fractals
### From Math Images
Revision as of 10:07, 26 May 2011 by Dpatton1 (Talk | contribs)
Markus-Lyapunov Fractal
A representation of the regions of chaos and stability over the space of two population growth rates.
Markus-Lyapunov Fractal
Fields: Dynamic Systems and Fractals
Created By: BernardH, using Mathematica 5
# Basic Description
The Markus-Lyapunov fractal is much more than a pretty picture; it is a map. The curving bodies and sweeping arms of the image are in fact a color-coded plot that shows us how a population changes as its rate of growth moves between two values. All the rich variations of color in the fractal come from the different levels of stability and chaos possible in such change.
Before anything else, in order to generate a Markus-Lyapunov fractal, we must be able to represent animal population change mathematically. This entails more than simply writing an equation for constant growth or constant reduction – a population cannot grow infinitely, but rather is constrained by the amounts of food, space, etc., available to it. To model the pattern of growth and reduction that occurs as populations approach and retreat from their maximum sizes, mathematicians have developed the logistic formulaThis is an iterative formula that determines each new iteration based on the difference between 1 and the value of the previous iteration. In this way, the values cannot become larger than 1., which models population growth fairly accurately by including a factor that diminishes as population size grows, just as food and space would diminish.
The logistic formula is driven by the initial population size and by the potential rate of change of that population. Mathematically, the more powerful of these values is the rate of change; it will determine whether the size of the population settles to a specific value, oscillates between two or more values, or becomes chaotic. To help determine which of these outcomes would occur, a mathematician named Aleksandr Lyapunov developed a method for comparing changes in growth and time in order to calculate what has been dubbed the Lyapunov exponentThe Lyapunov exponent represents the overall rate of change of a system over many iterations, expressed logarithmically.. This is a handy little indicator, and here's why:
• If it is zero, the population change is neutral; at some point in time, it reaches a fixed point and remains there.
• If it is less than zero, the population will become stableStability is different from a fixed point; A system that oscillates between two values is stable, and a system that oscillates between sixteen values is still stable.. The lower the number, the faster and more thoroughly the population will stabilize.
• If it is positive, the population will become chaotic.
Another example of a Markus-Lyapunov fractal, this one with chaos in black and stability in gold.
What does all this have to do with the fantastical shapes of the Markus-Lyapunov fractal? Well, a scientist named Mario Markus wanted a way to visualize the potential represented by the Lyapunov exponent as a population moved between two different rates of growth. So he created a graphical space with one rate of growth measured along the x-axis and the other along the y. Thus for any point, (x,y), there is one specific Lyapunov exponent that predicts how a population with those rates of change will behave. Markus then assigned a color to every possible Lyapunov exponent – one color for positive numbers and another for negative numbers and zero. This second color he placed on a gradient, so that lower negative numbers are lighter and those closer to zero are darker, with zero itself being black. Some Markus-Lyapunov fractals also display superstable$\lambda=-\infty$, as the lowest possible Lyapunov exponent, indicates the fastest possible approach to stability. points in a third color or black. By this code, Markus could color every point on his graph space based on its Lyapunov exponent.
Consider the main image on this page. The blue "background" shows all the points where the combination of the rates of change on the x and y axes will result in chaotic population growth. The "floating" yellow shapes show where the population will move toward stability. The lighter the yellow, the more stable the population.
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
### The Logistic Forumula
This comes from the field of Verhulst Dynamics. Basic, unrestricted growt [...]
[Click to hide A More Mathematical Explanation]
### The Logistic Forumula
This comes from the field of Verhulst Dynamics. Basic, unrestricted growth can be represented by
$x_{(n+1)}=Rx_n$
Where x(n+1) is the population size at time n+1. But this, as discussed above, is not a realistic model of population growth in the ecological world. To account for the changing rate of change, R, of an actual population, Verhulst constructed
$R=\mathbf{r}(1-x_n)$
Where r is a parameter for the potential rate of change of the population. In this way, the overall rate of change, R, is higher when xn is lower and lower when xn is higher. Re-inserting this in our initial representation of growth, we have:
$x_{(n+1)}=(1-x_n)\mathbf{r}x_n$
This is the logistic formula.
A bifurcation diagram for the logistic formula. The populations sizes resulting from 120 iterations of the formula based each r value are plotted above that r value (after an initial, unrecorded period of 5000 iterations to level the systems). Note the self-similarity shown in the enlarged section.[1]
#### Bifurcation
The logistic equation is interesting partly for its properties of bifurcation. Bifurcation occurs when a system "branches" (hence the name) into multiple values. In the logistic formula, this means that, as the r value grows, xn goes from a single value to oscillating among two or more values; the population volume ceases to be constant and begins fluctuating between multiple volumes.
The diagram to the left shows how the logistic formula bifurcates as the value of r changes. Population sizes, xn, (y-axis) are plotted against the r values (x-axis) that generate them. The most stable state therefore appears as a single horizontal line. When this line appears to "branch" into two, we are observing bifurcation; the r value has changed so that the population is now oscillating between two volumes. As the branching continues, so does bifurcation: Three lines show oscilation among three volumes, four lines show oscillation among four volumes, and so forth. The grey areas show where the system bifurcates to the extent that it essentially "oscillates" among all possible xn values. That is, it becomes chaotic. As r values increase, we see wider and wider "bands" of chaos where ranges of r values yield only chaotic systems. Above the r value of 3, these "bands" become continuous, and all r values yield chaos.
In the enlarged portion, we can see that the diagram of logistic bifurcation is self-similar. This is the fractal property that carries through into Markus-Lyapunov fractals.
### The Lyapunov Exponent
The discrete form of the Lyapunov exponent is
$\lambda=\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N \log_2 \frac{dx_{(n+1)}}{dx_n}$
In other words, the Lyapunov exponent $\lambda$ represents the limit of the meanThe presence of $\frac{1}{N}\sum_{n=1}^N ...$ shows that this is a mean. of the exponentialThe presence of $\log_2$ makes the relationship between $\lambda$ and the original equation exponential. rates of change$\frac{dx_{(n+1)}}{dx_n}$ defines the rate of change. that occur in each transition, xn $\rightarrow$ x(n+1), as the number of transitions approaches infinity.
What does this have to do with stability? The key is the log2 component, which renders numbers under 1 negative and those over 1 positive. This is what yields the properties of Lyapunov exponents laid out in the "Basic Explanation" – those mean overall rates of change that make the system finiteA geometric series or sequence is finite if is multiplied by a factor, r < 1, that makes it converge to a discrete value or set of values. must be less than 1, giving us a negative Lyapunov exponent, while those rates of change that expand the system to the point of chaos must be greater than one, giving us a positive exponent. When the mean overall rate of change is zero, the logarithmic component no longer exists, showing exactly what happens in a superstable system; the rate of change ceases to exist.
In other words, the Lyapunov exponent is a method for examining the rate of change of a system considered over infinite iterations, then taking that rate of change and making it easily identifiable as a value that induces either chaos or stability.
#### In the Logistic Formula
Basic differentiation shows us that, for the logistic formula (3):
$\frac{dx_{(n+1)}}{dx_n}=\mathbf{r}-2\mathbf{r}x_0^2$
Using this and a sufficiently large N number of iterations, we can approximate the Lyapunov exponent for the logistic formula to be:
$\lambda \approx \frac{1}{N}\sum_{n=1}^N \log_2 \mathbf{r}-2\mathbf{r}x_0^2$
Here we can see much more clearly a property that we have been assuming -- the variable that has the greatest impact on the stability of the logistic equation is r, not xn. This is clear here because, as we differentiate the logistic formula in order to examine its overall rate of change, xn is reduced to the constant value x0 (the starting volume of the population). It is therefore no longer a variable when we look at the formula on the level of its Lyapunov exponent, and so changing the value of xn does not change whether the logistic function yields chaos or stability.
### Forcing the Rates of Change
Mathematically, the important part of Markus's contribution to understanding this type of system was not his method for generating fractals, but his use of periodic rate-of-change forcing. We have been discussing the great impact of the r value in determining the output of the logistic formula, but this value can have still greater impact if we do not choose to keep it constant. Anyone who has studied biology, as Markus has, knows that the rates of change of a populations size do not simply fluctuate with changing supplies of food and space, but also often alternate between two or more specific potential rates of change depending on such things as weather and mating seasons.
A Markus-Lyapunov fractal with rate-of-change pattern ab
In terms of the logistic formula, this means we choose a set of rates of change, r1, r2, r3,..., rp, where p is the period over which the rates of change loop. When we force the rates of change to follow such a loop, we have a new, modular logistic equation (3):
$x_{(n+1)}=\mathbf{r}_{n \text{mod} p}x_n(1-x_n)$
It is in these forced alterations in rates of change that the fascinating shapes of the Markus-Lyapunov fractal come out. Each of the fractals is formed from some pattern of two rates of change, a and b. So a pattern aba would mean each point on the fractal is colored based on the Lyapunov exponent of the logistic formula 7, where r1 = a, r2 = b, and r3 = a. That is, the r values would cycle a,b,a,a,b,a,a,b,a....
Because the axes used to map these fractals are measurements of changes in a and b, the pattern a would simply yield a set of vertical bars, just as the pattern b would yield horizontal bars. However, once the patterns start to become mixed, more interesting results come out. The image to the right shows an ab pattern. Note that it is much simpler than other images shown on this page; the main image, for instance, is a bbbbbbaaaaaa pattern.
# Why It's Interesting
An enlargement of a section of "Zircon Zity," showing self-similarity.
### Fractal Properties
The movements from light to dark and the dramatic curves of the boundaries between stability and chaos here create an astonishing 3D effect. But the image is striking not only for its beauty but also for its self-similarity. Self-similarity is that trait that makes fractals what they are – zooming in on the image reveals smaller and smaller parts that resemble the whole. Consider the image to the right, enlarged from a section of the main image above. Here we see several shapes that repeat in smaller and smaller iterations. Perhaps ironically, this type of pattern is a common property of chaos.
For more images of the fractal properties of chaotic systems, see the Henon Attractor, the Harter-Heighway Dragon Curve, and Julia Sets.
One artist superimposed and edited several real Markus-Lyapunov fractals to create this piece of art.
### Artistic Extensions
After Markus saw the incredible beauty and intriguing three-dimensionality of the images generated by his plotting system, he immediately sent the images to a gallery in the hopes that it would display his images in an exhibition.[2] It's easy to see why he did so, and in fact, pictures based on these fractals have become a large part of what is called "fractalist" art. As with all domains of fractalist art, there is a great deal of debate in the art community over whether these images are truly "art" given their intrinsic reliance on a purely scientific, algorithmically-generated chart. One could say that such a process is devoid of creativity, but it is equally valid to say that the identification and presentation of the beauty in the science is an art in itself – a concept that is critical in modern art. Either way, there has been an undeniable artistic fascination with Markus-Lyapunov fractals; if the image seems familiar, you have likely seen it on posters, t-shirts, or any other canvas for graphic design.
# Teaching Materials
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# References
1. ↑ Peitgen, H., & Richter, P. (1986). The Beauty of Fractals: Images of Complex Dynamic Systems. Berlin: Springer-Verlag.
2. ↑ Dewdney, A. K. (1991). Leaping into Lyapunov Space. Scientific American, (130-132).
Other Sources Consulted
Elert, G. (2007). The Chaos Hypertextbook. http://hypertextbook.com/chaos/
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page. | {"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": 14, "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": 0.9304956197738647, "perplexity_flag": "middle"} |
http://mathhelpforum.com/discrete-math/125160-equivalence-classes-8x8-chessboard.html | # Thread:
1. ## Equivalence classes on an 8x8 chessboard
Hi there, I'm struggling a bit with a uni question I've been given. I think I understand the basics but possibly not!! Any help greatly appreciated!
Take a standard 8×8 chessboard, and label the squares
(i, j) 1 ≤ i ≤ 8, 1 ≤ j ≤ 8, with (1, 1) being the bottom lefthand square,
and (8, 8) the top righthand square. Define a relation on these squares by
(i, j) ∼ (i′, j′) iff i + j = i′ + j′.
(a) Show that ∼ defines an equivalence relation on the chessboard.
(b) How many equivalence classes are there, and what are their sizes?
(c) Choose equivalence class representatives for the classes.
Here are my thoughts...
a) would I assume ~ if i+j not equal to i'+j' and show this is not the case?
b) I'm not entirely sure but I would assume that as the canonical projection map is surjective and that there are 64 ways to write i+j that there are 64 equivalence classes of size 2, then again I'm well prepared to believe I've missed the point!
c) As a set of class representatives is a subset of X which contains exactly one element from each equivalence class could the class representative just be either the i's or the j's
Luckily this is over the computer so if people tell me I'm completly wrong you won't see me blushing!!
Thanks for the help, Laura
2. Very often, when approaching a new problem, is it useful to consider several examples. What are all the squares for which the sum of coordinates is 4? 5? This will help you find out almost everything about equivalence classes.
To show that ∼ is an equivalence relation (as well as to show anything in mathematics) one should start with definitions. Once you know the definitions of an equivalence relation and related notions, you need to know what it means to prove the statements involved. Those statements have the form "For all x, ...", so you need to know what it means to prove them.
3. Hello, leshields!
With terms like "canonical projection map" and "surjective",
. . you're overthinking the problem.
Take a standard 8×8 chessboard, and label the squares: . $(i, j)\quad 1 \leq i \leq 8,\;\;1 \leq j \leq 8$
with (1, 1) being the bottom lefthand square, and (8, 8) the top righthand square.
Define a relation on these squares by: . $(i,j) \sim (i',j') \;\;\Longleftrightarrow\;\;i+j \,=\,i' + j'$
Did you make a "sketch"?
. . $\begin{array}{cccccc}<br /> 81 & 82 & 83 & 84 & \hdots & 88 \\<br /> \vdots & \vdots & \vdots & \vdots & & \vdots \\<br /> 31 & 32 & 33 & 34 & \hdots & 38 \\<br /> 21 & 22 & 23 & 24 & \hdots & 28 \\<br /> 11 & 12 & 13 & 15& \hdots & 18\end{array}$
Two squares are related by $\sim$ if the sums of their coordinates are equal.
For example: . $(1,4) \sim (2,3) \sim (3,2) \sim (4,1)$
(a) Show that $\sim$ defines an equivalence relation on the chessboard.
Reflexive
Is $(a,b) \sim (a,b)$ ?
$a+b \:=\:a+b\quad\hdots$ Yes, "they" have the same sum.
Symmetric
If $(a,b) \sim (c,d)$, is $(c,d) \sim (a,b)$ ?
Since $(a,b) \sim(c,d)$, we have: . $a+b \,=\,c+d$
Then we have: $c+d \,=\,a+b$, which gives us: . $(c,d) \sim (a,b)$
Yes, the relation is symmetric.
Transitive
If $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$, is $(a,b) \sim (e,f)$ ?
Since $(a,b) \sim (c,d)$, we have: . $a+b \,=\,c+d$
Since $(c,d)\sim(e,f)$, we have: . $c+d\,=\,e+f$
Then we have: . $a+b\,=\,e+f$ which gives us: . $(a,b)\sim(e,f)$
Yes, the relation is transitive.
Therefore, $\sim$ is an equivalence relation.
(b) How many equivalence classes are there, and what are their sizes?
There are 15 equivalance classes . . .
. . $\begin{array}{cc}<br /> \text{Sum} & \text{Size} \\ \hline<br /> 2 & 1 \\ 3 & 2 \\ 4 & 3 \\ 5 & 4 \\ 6 & 5 \\ 7 & 6 \\ 8 & 7 \\ 9 & 8 \end{array}$
. . . $\begin{array}{cc}<br /> 10 & \quad\; 7 \\ 11 & \quad\;6 \\ 12 & \quad\;5 \\ 13 & \quad\;4 \\ 14 & \quad\;3 \\ 15 & \quad\;2 \\ 16 & \quad\;1\end{array}$
(c) Choose equivalence class representatives for the classes.
I don't know what this means . . .
4. Originally Posted by Soroban
I don't know what this means . . .
You said there were 15 classes. you need to explicitly state one member of each class to represent that class. for example, partitioning the integers according to parity is an equivalence relation. i can choose 0 to be the class representative of the even integers while 1 can be that of the odd integers. | {"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": 26, "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": 0.9320781826972961, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/208890/the-real-line-with-its-additive-group-is-a-topological-group | # The real line with its additive group is a topological group?
Maybe it's a stupid question, I'm starting to study topological groups, I'm struggling to prove that the real line is a topological group with its additive group structure and Euclidean topology, precisely the part which we have to prove that if $g_1$ and $g_2$ $\in$ R, then the multiplication map $G\times G \to G$ defined by $m(g_1,g_2)=g_1 + g_2$ is continuous. Anyone can help me please.
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7
If you want to prove that the additive group of real numbers is a topological group, the relevant map is the addition map $(g_1,g_2) \mapsto g_1 + g_2$. The real numbers under multiplication are not even a group, because 0 is not invertible. – Logan Maingi Oct 7 '12 at 20:23
1
@LoganMaingi yes, I know that, when I said multiplication I meant the additive operation. – user42912 Oct 7 '12 at 20:32
2
@LoganMaingi Actually you need to show that both $p(x,y)=x+y$ and $n(x)=-x$ are continuous. A clever little observation is that it's enough to show that $m(x,y) = x-y$ is continuous because $n(x) = m(0,x)$ and $p(x,y) = m(x,n(y))$ so if $m$ is continuous, then $n$ and $p$ are composites of continuous maps. – kahen Oct 7 '12 at 20:32
## 4 Answers
Addition is continuous because of the familiar 'addition law for limits':
$$\lim x_n=x \text{ and }\lim y_n=y\ \Rightarrow \lim (x_n+y_n)=x+y.$$
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In order to show that addition is continuous you're going to have to use the definition of continuity. I'm not sure which definition of continuity your book/class/notes is using, but once you write down the details of the definition it should be relatively clear. If not, you can edit your question to explain where you got stuck.
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Let $g_1,g_2 \in \mathbb{R}$ and $\epsilon>0$.
Then for each $(r_1,r_2)\in (g_1-\frac{\epsilon}{2},g_1+\frac{\epsilon}{2}) \times (g_2-\frac{\epsilon}{2},g_2+\frac{\epsilon}{2})$ we have that $$m(r_1,r_2)=r_1+r_2 \in (m(g_1,g_2)-\epsilon,m(g_1,g_2)+\epsilon) \ .$$
So if the definition for continuous functions you have is:
$f:A\longrightarrow B$ is continuous at $a \in A$ if for every open neighborhood $U$ of $f(a) \in B$ exist an open neighborhood $V$ of $a \in A$ such that $$\forall x \in V \; \Rightarrow\; f(x) \in U \ .$$
then $m$ is continuous.
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Let $\mathbb R$ be the real line with its additive group structure and euclidean topology. We want to prove that the real line is indeed a topological group. Let $i:\mathbb R \to \mathbb R$, defined by $i(x)=-x$ and $m:\mathbb R \times \mathbb R \to \mathbb R$ defined by $m(x,y)= x+y$. we have to prove the following: (a) $i$ is continuous (b) $m$ is continuous.
(a) By a real analysis argument we know that $i$ is continuous, because $i$ is a polynomial function with real coefficients.
(b) We know that the projection $\Pi_1:X\times Y \to X$, $\Pi_1(x,y)=x$, where $X$ and $Y$ are topological spaces, is always continuous because for any open subset $U$ of X, we have $\Pi^{-1}(U)=U\times Y$ a open subset of $X \times Y$.
With the same argument the projection $\Pi_2:X\times Y \to Y$, $\Pi_2(x,y)=y$, where $X$ and $Y$ are topological spaces, is also continuous.
So by a real analysis argument which claims that the sum of continuous functions is continuous and as we know $m =\Pi_1 +\Pi_2$, then m is continuous.
- | {"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": 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": 0.963748574256897, "perplexity_flag": "head"} |
http://mathhelpforum.com/advanced-algebra/206840-f-g-wy.html | # Thread:
1. ## f|g wy?
Let $f(x),g(x)$ be polynomials with rational coefficients, and $f(x)$ is irreducible. Suppose further that there exists a complex number $\alpha$ such that $f(\alpha)=g(\alpha)=0$. Prove that $f(x)|g(x).$
I do not know how to prove. and I've thought it for many times....Would you help me out?
2. ## Re: f|g wy?
consider the map from Q[x] to Q(α) given by: p(x)-->p(α). this is a ring homomorphism.
furthermore, since α is algebraic over Q (it satisfies f(x) in Q[x]), Q[a] = Q(α) (that is, Q[α] is actually a FIELD).
now the image of this homomorphism contains Q (we can take as the pre-image of q in Q the constant polynomial q) and α (which has pre-image x), and is therefore surjective.
by the fundamental isomorphism theorem (for rings), Q(α) = Q[x]/I, for some ideal I of Q[x] (namely, the kernel of our homomorphism).
now Q[x] is a principal ideal domain, so I is generated by some polynomial, h(x) in Q[x].
since f(x) is in the kernel (since f(α) = 0), we have f(x) = h(x)k(x), for some polynomial k(x) in Q[x].
but f(x) is irreducible, so k(x) must be a unit (that is, an element of Q*). but then f(x) = qh(x), so h(x) = (1/q)f(x).
this means that (h(x)) = (f(x)).
now g(α) is also in the kernel, which is generated by f(x), so g(x) = f(x)s(x), for some polynomial s(x) in Q[x], that is: f|g.
3. ## Re: f|g wy?
Thank you very much. I need to spend some time writting your words in the language of only linear algebra...
4. ## Re: f|g wy?
divisibility is not a linear algebra concept, as it deals with multiplication, and a vector space need not have one. the set of polynomials over a field does indeed form a vector space, but it is also a RING, this is important.
5. ## Re: f|g wy?
divisibility is indeed a linear algebra concept. And this problem is a problem from linear alegebra in a entrance examination for graduate.
6. ## Re: f|g wy?
shall i ask you how? what does it mean for one vector to divide another?
i'm not sure what you MEAN by "the language of linear algebra". if you explain it to me, perhaps i can put what i said above in different terms. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.942827045917511, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/73279/integrate-frac1-sqrt6x-x2-5?answertab=votes | # Integrate $\frac{1}{\sqrt{6x-x^2-5}}$
Question: Integrate $\frac{1}{\sqrt{6x-x^2-5}}$
My Working:
$$\int{\frac{1}{\sqrt{6x-x^2-5}}} = \int{\frac{1}{\sqrt{-(x^2-6x+5)}}} = \int{\frac{1}{\sqrt{-((x-3)^2-3^2+5)}}} = \int{\frac{1}{\sqrt{-((x-3)^2-4)}}}$$
Is it right so far? Because I think I should be using the formula
$$\int{\frac{1}{\sqrt{a^2-(x+b)^2}}} = \sin^{-1} \left( \frac{x+b}{a} \right)+c$$
$-4$ can't fit into $a^2$ can it? I haven't covered complex numbers yet, so I must have made a mistake?
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1
There's two minus signs in front of the 4. You have $a^2=4$ and $b=-3$ by my reckoning. – anon Oct 17 '11 at 7:16
$6x-x^2-5=2^2-(x-3)^2$ – pedja Oct 17 '11 at 7:20
## 2 Answers
Your derivation is correct, but you aren't looking at the last line correctly. Notice that $$-((x-3)^2-4)=4-(x-3)^2,$$ so your parameters are $a=2$ and $b=-3$ to plug into the formula. We rule out $a=-2$ because the square root function is always nonnegative, and so following holds only for $a$ positive: $$\frac{d}{dx}\sin^{-1}\left(\frac{x+b}{a}\right)=\frac{1}{a\sqrt{1-(x+b)^2/a^2}}=\frac{1}{\sqrt{a^2-(x+b)^2}}.$$
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You are doing just fine. You completed the square, which is the key step, and ended up with $$\int \frac{dx}{\sqrt{4-(x-3)^2}}.$$
Now you have a formula that you want to use. It may be better not to try to remember too many formulas. We will instead find a substitution that gives us a very familiar integral.
The above integral looks kind of like $$\int \frac{dw}{\sqrt{1-w^2}}.$$ We look for a substitution that brings out the resemblance. Suppose that we let $x-3=2u$. Then the bottom will look like $\sqrt{4-4u^2}$, which is good, because we can then "take the $4$'s out."
So let's do it. Let $x-3=2u$, or if you prefer, let $(x-3)/2=u$. Then $dx=2\,du$ and our integral becomes $$\int\frac{2u}{\sqrt{4-4u^2}}.$$
But $\sqrt{4-4u^2}=2\sqrt{1-u^2}$. So our integral simplifies to $$\int \frac{du}{\sqrt{1-u^2}}=\arcsin u +C.$$ Finally, replace $u$ by $(x-3)/2$.
Another way to get to the right substitution is to note that $$\sqrt{4-(x-3)^2}=2\sqrt{1-((x-3)/2)^2},$$ which makes letting $u=(x-3)/2$ very natural.
- | {"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": 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": 0.3, "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": 0.94255530834198, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/72518/list | ## Return to Answer
1 [made Community Wiki]
As a counterpoint to gowers's devil's advocacy, I'd mention that some formulas for $\pi$ have been discovered experimentally, and in some cases we still don't know how to prove them. For example, in the paper "About a New Kind of Ramanujan-type Series" by Jesús Guillera (Experimental Mathematics 12 (2003), 507–510), the following conjecture by Gourevitch is stated: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$ As far as I know, this is still unproved. In principle this kind of formula should be WZ-able, but it seems to be just out of reach of current computers. And probably there ultimately does exist some "motivic explanation" as Torsten Ekedahl said, but since we don't currently know of one, I think that one answer to Qiaochu's question is, "experimental observation." | {"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": 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": 0.9695830345153809, "perplexity_flag": "middle"} |
http://mathhelpforum.com/pre-calculus/138014-find-equation-parabola.html | # Thread:
1. ## Find the equation for the parabola...
Find the equation for the parabola that has its focus on the positive x-axis, 4 units away from the directrix.
2. Hello, uselessjack!
Is there more information?
There are four forms of the parabola, each with its own equation.
I'll derive one of them . . .
Find the equation for the parabola that has its focus on the positive x-axis,
4 units away from the directrix.
This is a "vertical" parabola, opening upward.
The focus is at $F(f,0)$
The directrix is: $y = -4$
Code:
``` |
|
* | *
| F
- - -*- + - - - o - - - - -* - -
* | (f,0) *
* : *
| * : *
| o V
| (f,-2)
| :
| :
- - - + - - - + - - - - -
-4|
|```
The equation has the form: . $(x-h)^2 \:=\:4p(y-k)$
The vertex is at: . $V(f,-2)$
The value of $p$ is: . $p \,=\,2$
The equation is: . $(x-f)^2 \:=\:4(2)\left(y-[-2]\right) \quad\Rightarrow\quad (x-f)^2 \:=\:8(y+2)$
3. Thank you for your help and time. I am left with confusion still. I thought there would be a specific answer. I am given five choices and still can't figure out which one is correct.
4y^2 = 4x
y^2 = 4x
y^2 = 8x
x^2 = 16y
4y^2 = x
#### Search Tags
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Copyright © 2005-2013 Math Help Forum. All rights reserved. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9372711777687073, "perplexity_flag": "middle"} |
http://anhngq.wordpress.com/2012/07/21/the-yamabe-problem-the-work-by-thierry-aubin/ | # Ngô Quốc Anh
## July 21, 2012
### The Yamabe problem: The work by Thierry Aubin
Filed under: PDEs, Riemannian geometry — Tags: Yamabe problem — Ngô Quốc Anh @ 15:05
Following the previous note about the work of Trudinger, today we talk about the work of Aubin regarding to the Yamabe problem, that is the following simple PDE
$\displaystyle -\Delta \varphi + R\varphi = C_0 \varphi^\frac{n+2}{n-2}.$
In his elegant paper entitlde “Équations différentielles non linéaires et problème de Yamabe concernant la courbure scalaire” published in J. Math. Pures Appl. in 1976, Aubin proved the existence for almost all manifolds for $n\geqslant 6$.
By using the notations used in the note about the work of Yamabe, they are
$\displaystyle {F_q}(u) = \frac{{\displaystyle\int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}|\nabla u{|^2} + R{u^2}} \right)d{v_g}} }}{{{{\left( {\displaystyle\int_M {|u{|^q}d{v_g}} } \right)}^{\frac{2}{q}}}}}$
where $q \leqslant \frac{2n}{n-2}$ and
$\displaystyle {\mu _q} = \mathop {\inf }\limits_{u \in {H^1}(M)} {F_q}(u),$
Aubin proved that
Theorem. If $\mu_\frac{2n}{n-2}$ satisfies
$\displaystyle\mu_\frac{2n}{n-2}<n(n-1)\omega_n^\frac{2}{n},$
then the Yamabe problem is solvable where $\omega_n$ is the volume of the unit sphere in $\mathbb R^n$.
In fact, he proved a stronger result saying that in any case, there holds
$\displaystyle\mu_\frac{2n}{n-2} \leqslant n(n-1)\omega_n^\frac{2}{n},$
and the equality occurs if and only if $M$ is conformally equivalent to the sphere with standard metric. Having this result, to solve the Yamabe problem, we have only to exhibit a test function $\psi$ such that $F_\frac{2n}{n-2}(\psi)<n(n-1)\omega_n^\frac{2}{n}$.
To conclude the paper, Aubin proved the following
Theorem. If $(M, g)$ ($n \geqslant 6$) is a compact nonlocally conformally flat Riemannian manifold, then
$\displaystyle\mu_\frac{2n}{n-2}<n(n-1)\omega_n^\frac{2}{n}.$
Therefore, the cases that $n=3, 4, 5$ and that $M$ is locally conformally flat are still open.
See also:
• The Yamabe problem: A Story
• The Yamabe problem: The work by Hidehiko Yamabe
• The Yamabe problem: The work by Neil Sidney Trudinger
• The Yamabe problem: The work by Thierry Aubin
• The Yamabe problem: The work by Richard Melvin Schoen | {"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": 18, "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": 0.8478214144706726, "perplexity_flag": "middle"} |
http://en.wikipedia.org/wiki/Davies%e2%80%93Bouldin_index | Davies–Bouldin index
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The Davies–Bouldin index (DBI) (introduced by David L. Davies and Donald W. Bouldin in 1979) is a metric for evaluating clustering algorithms.[1] This is an internal evaluation scheme, where the validation of how well the clustering has been done is made using quantities and features inherent to the dataset. This has a drawback that a good value reported by this method does not imply the best information retrieval.
Preliminaries
Let Ci be a cluster of vectors. Let Xj be an n dimensional feature vector assigned to cluster Ci.
$S_i = \sqrt[q]{\frac{1}{T_i} \displaystyle\sum_{j=1}^{T_i} {|X_j-A_i|^q} }$
Here $A_i$ is the centroid of Ci and Ti is the size of the cluster i. Si is a measure of scatter within the cluster. Usually the value of q is 2, which makes this a Euclidean distance function between the centroid of the cluster, and the individual feature vectors. Many other distance metrics can be used, in the case of manifolds and higher dimensional data, where the euclidean distance may not be the best measure for determining the clusters. It is important to note that this distance metric has to match with the metric used in the clustering scheme itself for meaningful results.
$M_{i,j} = \left|\left|A_i-A_j\right|\right|_p = \sqrt[p]{\displaystyle\sum_{k=1}^{n}\left|a_{k,i}-a_{k,j}\right|^p }$
$M_{i,j}$ is a measure of separation between cluster $C_i$ and cluster $C_j$.
$a_{k,i}$ is the kth element of $A_i$, and there are n such elements in A for it is an n dimensional centroid.
Here k indexes the features of the data, and this is essentially the Euclidean distance between the centers of clusters i and j when p equals 2.
Definition
Let Ri,j be a measure of how good the clustering scheme is. This measure, by definition has to account for Mi,j the separation between the ith and the jth cluster, which ideally has to be as large as possible, and Si, the within cluster scatter for cluster i, which has to be as low as possible. Hence the Davies Bouldin Index is defined as the ratio of Si and Mi,j such that these properties are conserved:
1. $R_{i,j} \geqslant 0$.
2. $R_{i,j} = R_{j,i}$.
3. if $S_j \geqslant S_k$ and $M_{i,j} = M_{i,k}$ then $R_{i,j} > R_{i,k}$.
4. and if $S_j = S_k$ and $M_{i,j} \leqslant M_{i,k}$ then $R_{i,j} > R_{i,k}$.
$R_{i,j} = \frac{S_i + S_j}{M_{i,j}}$
This is the symmetry condition. Due to such a formulation, the lower the value, the better the separation of the clusters and the 'tightness' inside the clusters.
$D_i \equiv \max_{j : i \neq j} R_{i,j}$
If N is the number of clusters:
${DB} \equiv \frac{1}{N}\displaystyle\sum_{i=1}^N D_i$
DB is called the Davies Bouldin Index. This is dependent both on the data as well as the algorithm. Di chooses the worst case scenario, and this value is equal to Ri,j for the most similar cluster to cluster i. There could be many variations to this formulation, like choosing the average of the cluster similarity, weighted average and so on.
Explanation
These conditions constrain the index so defined to be symmetric and non-negative. Due to the way it is defined, as a function of the ratio of the within cluster scatter, to the between cluster separation, a lower value will mean that the clustering is better. It happens to be the average similarity between each cluster and its most similar one, averaged over all the clusters, where the similarity is defined as Si above. This affirms the idea that no cluster has to be similar to another, and hence the best clustering scheme essentially minimizes the Davies Bouldin Index. This index thus defined is an average over all the i clusters, and hence a good measure of deciding how many clusters actually exists in the data is to plot it against the number of clusters it is calculated over. The number i for which this value is the lowest is a good measure of the number of clusters the data could be ideally classified into. This has applications in deciding the value of k in the kmeans algorithm, where the value of k is not known apriori. The SOM toolbox contains a MATLAB implementation.[2] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "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": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9333992004394531, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/tagged/lie-algebras+pde | # Tagged Questions
0answers
107 views
### First-order derivatives in differential forms calculus
Let $d$ denote the Cartan differential, and let $\delta$ denote the codifferential. The underlying domain is not important for what follows. The canonical generalization of the Laplace-operator ... | {"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": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8584214448928833, "perplexity_flag": "middle"} |
http://mathhelpforum.com/statistics/8092-marbles.html | # Thread:
1. ## Marbles
I really suck with probability questions can someone help me out?
Each of two boxes contains 20 marbles, and each marble is either black or white. The total number of black marbles is different from the total number of white marbles. One marble is drawn at random from each box. The probability that both marbles are white is .21. What is the probability that both are black?
2. Originally Posted by MathMage89
I really suck with probability questions can someone help me out?
Each of two boxes contains 20 marbles, and each marble is either black or white. The total number of black marbles is different from the total number of white marbles. One marble is drawn at random from each box. The probability that both marbles are white is .21. What is the probability that both are black?
Let,
$x$ be the amount of white and $y$ be the amount of black. We are told that $x\not y$. (And of course $x\geq 2$).
The probability is .21 meaning the ratio of favorable outcomes to possible outcomes is this fraction. The number of favorable outcomes is $_xC_2$. The number of possible outcomes is, $_{20}C_2$
Thus,
$\frac{_xC_2}{_{20}C_2}=\frac{\frac{x(x-1)}{2}}{190}=.21$
More simply,
$\frac{x(x-1)}{380}=.21$
Thus,
$x(x-1)=79.8$
Which is impossible for $x$ is an integer.
3. Hello, MathMage89!
Is there a typo in the problem?
Each of two boxes contains 20 marbles, and each marble is either black or white.
The total number of black marbles is different from the total number of white marbles. ?
One marble is drawn at random from each box.
The probability that both marbles are white is 0.21 ?
What is the probability that both are black?
The contents of the two boxes are:
. . $\boxed{\begin{array}{cc}W_1 &\text{white} \\ 20-W_1 & \text{black}\end{array}}\;\; <br /> \boxed{\begin{array}{cc}W_2 & \text{white} \\ 20-W_2 & \text{black}\end{array}}$
The probability that both marbles are white is $0.21$
Hence, we have: . $\frac{W_1}{20}\cdot\frac{W_2}{20} \:=\:0.21\quad\Rightarrow\quad W_1\cdot W_2\:=\:84<br />$
$W_1,\,W_2$ are positive integers $\leq 20\quad\Rightarrow\quad\{W_1,\,W_2\} \:=\:\{6,\,14\}$
But this makes the contents:
. . $\boxed{\begin{array}{cc}6 &\text{white} \\ 14 & \text{black}\end{array}}\;\; <br /> \boxed{\begin{array}{cc}14 & \text{white} \\ 6 & \text{black}\end{array}}$
And there are 20 blacks and 20 whites . . . ??
edit: How did I overlook $7$ and $12$? . . . *blush*
4. Hey Soroban
There isn't a typo in the problem, but if you use the factors 12 and 7 rather than 6 and 14 you will be lead to the current answer, I believe. Thanks for helping me start this up! | {"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": 18, "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": 0.9374820590019226, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/250799/in-linear-algebra-why-does-the-dimension-of-a-vector-space-bbb-r-n-equal/250819 | # In Linear Algebra, why does the dimension of a vector space $\Bbb R ^ n$ equal $n$?
In my attempts to learn linear algebra from Khan Academy I've come across several concepts that I can't completely connect.
Every vector space has a base. This base consists of the minimal elements that can span the entire vector space.
The number of elements in this vector space is defined to be the vector space's dimension.
When trying to conclude the power of a vector space (as in how many vectors there are in the space), I know that $|V| = |F| ^ n$
I also know that this small n = dimension of V. What I don't understand is why does the dimension equals that $n$?
Thanks!
-
3
You are talking about two different things. The dimension of a vector space V is the number of elements of its basis (or rather of any basis), not the number of elements in V. – Fant Dec 4 '12 at 17:08
1
The subject is a bit confused. The dimension is not "$\mathbb R^n$. You are confusing terminology there – Thomas Andrews Dec 4 '12 at 17:12
I know, what I said is that dimension equals N, the power of R. Maybe the question isn't clear enough. I'll edit it – vondip Dec 4 '12 at 17:15
You are right, I just edited my question. thanks! – vondip Dec 4 '12 at 17:18
1
what do you mean ? – vondip Dec 4 '12 at 17:35
show 4 more comments
## 1 Answer
A basis for a vector space is a linearly independent set that spans the space. For the Euclidean space $\mathbb R^n$, a basis is $(1,0,...,0),(0,1,0,...,0),...,(0,...,0,1)$. Since there are $n$ elements in this set, the dimension of $\mathbb R^n$ is $n$.
Bases for a vector space are not unique. It can be shown that any two bases have the same number of elements, so that this dimension is well-defined.
-
while this makes sense, if I were to say that I have a vector space which follows a certain rule such that it only has 2 vectors in its base, why would that mean that the entire space is only F^2? Why wouldn't there be more? – vondip Dec 4 '12 at 17:48
Is there a particular name for this basis in your answer? Something like 'canonical vector space basis'? – 000 Dec 4 '12 at 17:54 | {"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": 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": 0.9559744000434875, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/141377/ranking-probability-problem?answertab=votes | # Ranking probability problem
$A, B, C$ are independently sampled from an uniform distribution in $[0, 1]$.
We know $P(A > B) = 0.7, P(B > C) = 0.6$, what is $P(A > C)$?
Is this a well defined problem? Does it have a sensible answer?
EDIT: Suppose we have two careless observers. An observer observes $A > B$ and there are 70% probability that she is right. Another observer observes $B > C$ and there are 60% probability that she is right. So what is the probability of $A > C$ in the underlying event?
-
2
Wait, if they are all sampled from the same uniform distribution on $[0,1]$, how can we have $P(A > B) \neq 0.5$? – TMM May 5 '12 at 13:37
@TMM I edited the question. Is it well defined now? – lqhl May 5 '12 at 14:05
There is a potentially interesting Bayesian problem here, struggling to get out. – André Nicolas May 5 '12 at 14:29
## 3 Answers
I wrote following MATLAB code. Simulation results show the probability is around 0.602. I hope someone could confirm this with an analytic answer.
````N = 1000000;
A = rand(N, 1);
B = rand(N, 1);
C = rand(N, 1);
p1 = 0.7;
p2 = 0.6;
c1 = rand(N, 1);
c2 = rand(N, 1);
ob1 = ((A > B) & (c1 < p1)) | ((A < B) & (c1 > p1));
ob2 = ((B > C) & (c2 < p2)) | ((B < C) & (c2 > p2));
ob = ob1 & ob2;
pos = ob & (A > C);
sum(pos) / sum(ob)
````
=======================update==============================
I enumerate all the 6 possibilities of relative order of $A, B, C$. They all appear with probability 1/6.
The following lists shows with how much probability each case passes the two observers
• $A>B>C$, $0.7\times 0.6$
• $A>C>B$, $0.7\times 0.4$
• $B>A>C$, $0.3\times 0.6$
• $B>C>A$, $0.3\times 0.6$
• $C>A>B$, $0.7\times 0.4$
• $C>B>A$, $0.3\times 0.4$
Among them, $A>B>C$, $A>C>B$, $B>A>C$ are the valid cases. So
$\frac {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6} {0.7\times 0.6+0.7\times 0.4+0.3\times 0.6+0.3\times 0.6+0.7\times 0.4+0.3\times 0.4} = 0.6027$
-
What does "$A > C > B, 0.7 \times 0.4$" mean? Certainly it cannot mean $P(A > C > B) = 0.7 \cdot 0.4$. – TMM May 5 '12 at 15:35
@TMM It means the probability that $A>C>B$ passes the two observers. Since $A>C$, it passes the first observer probability with $70\%$ (she makes a correct observation) probability. Since $C>B$, it passes the second observer with $40\%$ probability (she makes a mistake). And the two events are independent. Hope this solves your problem :) – chtlp May 5 '12 at 15:41
Nope, it doesn't. The two observations are fixed, while the values of $A,B,C$ are not. So what does "the probability that [it] passes the two observers" mean? – TMM May 5 '12 at 15:45
@TMM Imagine we repeat sampling $\langle A, B, C\rangle$ many times, some of them fit the description $P(A>B)=0.7$, $P(B>C)>0.6$ (pass the observers''). And we want to know in these events, how many of them have $A>C$. – chtlp May 5 '12 at 16:16
+1: Despite the downvoting, the simulated answer and the maths is absolutely correct, under the assumption that the values of A,B and C are independent of the observed probabilities. Nice job. – Ronald May 5 '12 at 23:27
Let $P(A > B) = 0.7$ and $P(B > C) = 0.6$, and suppose that these events are independent. Then
• $P(A > B > C) = P(A > B) \cdot P(B > C) = 0.42$ in which case $\color{blue}{A > C}$.
• $P(B > A,C) = P(B > A) \cdot P(B > C) = 0.18$ in which case $\color{blue}{A > C}$ or $\color{red}{C > A}$.
• $P(A,C > B) = P(A > B) \cdot P(C > B) = 0.28$ in which case $\color{blue}{A > C}$ or $\color{red}{C > A}$.
• $P(C > B > A) = P(B > A) \cdot P(C > B) = 0.12$ in which case $\color{red}{C > A}$.
If we further assume that in those two events where we do not know which of $A,C$ is larger, both events occur with the same probability, then $\color{blue}{P(A > C) = 0.42 + \frac{1}{2}(0.18 + 0.28) = 0.65}$ and $\color{red}{P(C > A) = 0.35}$.
-
I think $P(B>A,C)=P(B>A)P(B>C)$ may be wrong, because the two events $B>A$ and $B>C$ are not independent. Can you explain @chtlp's simulation? – lqhl May 5 '12 at 14:44
I assumed the two events are independent, so by definition they are independent. This could be if e.g. $B$ is "fixed" and the distributions of $A,C$ depend on the value of $B$. (Example: If $A,C = B + U(-1,1)$, with $U(a,b)$ random variables uniformly distributed on $[a,b]$ the statement would be true.) – TMM May 5 '12 at 15:30
This seems plausible, but I do not believe that A>C and C>A are equally likely in the middle cases, because of the conditional information we are given - the probabilities in the middle cases are not symmetrical. – Ronald May 5 '12 at 15:42
@Ronald: From the description of the question, I assumed that the observations are fixed, and with probability $0.7$ the first observation is correct, and with probability $0.6$ the second ovservation is correct. With only these conditionals I do not see why $A > C$ or $C > A$ would be more likely in the two middle cases. (Of course "both events occur with the same probability" above is just an assumption, but without assumptions we cannot get an answer.) – TMM May 5 '12 at 17:16
If we observe that P(A>B) = 0.7 then we somehow get the impression that A is larger than B. If we observe that P(B>C) = 0.6 then we similarly observe that B is larger than C, but not so strongly. I believe this makes a difference. If the observations are assumed to be fixed, as you say, then I think the simulation results seem to be more correct. I will check it independently. – Ronald May 5 '12 at 23:11
Ah. It depends strongly on the method for making those probabilistic observations.
For example: If we observe that A=0.7, then we should note P(A>B)=0.7.
If we observe that C=0.4, then we should note P(B>C)=0.6.
(This is perhaps the most obvious, natural way of accessing those probabilities. An observation of B would affect both probabilities)
And, if those were our observations, then it's absolutely guaranteed that A>C. P(A>C) = 1.
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You are assuming $A$ is fixed in $P(A > B) = 0.7$, but it could also be that $B$ is fixed, e.g. $B = 0$ and $A = B + U(-0.3, 0.7)$ and $C = B + U(-0.4, 0.6)$ with $U(a,b)$ a uniformly distributed random variable on $[a,b]$. In that case $P(A > C) > 0.5$ but $P(A > C) \neq 1$. – TMM May 5 '12 at 17:24
As I said, it depends on the method of making the probabilistic observations. What I said is consistent with the observations, and I would argue is the most natural way for those observations to occur, but there are other possible cases. – Ronald May 5 '12 at 23:02 | {"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": 76, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9465669989585876, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/34676/simple-set-exercise-seems-not-so-simple/34678 | # Simple set exercise seems not so simple
Exercise about sets from Birkhoff's "Modern Applied Algebra".
Prove that for operation $\ \Delta$ , defined as
$\ R \Delta S = (R \cap S^c) \cup (R^c \cap S)$
following is true:
$\ R \Delta ( S \Delta T ) = ( R \Delta S ) \Delta T$
($\ S^c$ is complement of $\ S$)
It's meant to be very simple, being placed in the first excercise block of the book. When I started to expand both sides of equations in order to prove that they're equal, I got this monster just for the left side:
$\ R \Delta ( S \Delta T ) = \Bigl( R \cap \bigl( (S \cap T^c) \cup (S^c \cap T) \bigr)^c \Bigr) \cup \Bigl(R^c \cap \bigl( (S \cap T^c) \cup (S^c \cap T) \bigr) \Bigr)$
For the right:
$\ ( R \Delta S ) \Delta T = \Bigl(\bigl( (R \cap S^c) \cup (R^c \cap S) \bigr) \cap T^c \Bigr) \cup \Bigl( \bigl( (R \cap S^c) \cup (R^c \cap S) \bigr)^c \cap T \Bigr)$
I've tried to simplify this expression, tried to somehow rearrange it, but no luck. Am I going the wrong way? Or what should I do with what I have?
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I usually prefer to mark the complement as $\cdot^c$ as $'$ and $\bar\cdot$ have way too many uses. Also, as joriki mentioned the $+$ is the symmetric difference which many mark by $\Delta$. – Asaf Karagila Apr 23 '11 at 9:47
## 3 Answers
This is the symmetric difference. It includes all elements that are in exactly one of the two sets. In binary terms, it's the XOR operation. Independent of the order of the operations on the three sets, the result will contain exactly the elements that are in an odd number of the sets.
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– furikuretsu Apr 23 '11 at 9:07
Reading your answer, I thought those four sentences were exactly the four right points to mention. +1. – Did Apr 23 '11 at 9:12
Nice answer! – Matt N. Apr 23 '11 at 9:15
joriki's answer it the best way to understand this conceptually. Still, it's perfectly possible to continue formally the way you started: simply repeatedly apply De Morgan's laws:
$(A \cup B)' = A' \cap B'$ and $(A \cap B)' = A' \cup B'$.
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Thank you. Now I realise that I've just gave up with formal derivation too easily. – furikuretsu Apr 23 '11 at 9:09
You're doing the right thing. I just did it just to check.
As a hint: you can transform either side into $$(R \cap S^c \cap T^c) \cup (R\cap S \cap T) \cup (R^c \cap S \cap T^c) \cup (R^c \cap S^c \cap T)$$
-
1
To connect the answers: This is the union of the set of all elements contained in all three sets with the three sets of elements contained in exactly one set. – joriki Apr 23 '11 at 9:17 | {"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": 14, "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": 0.9593749046325684, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/118661/good-book-on-representation-theory-of-gln/118677 | ## Good book on representation theory of GL(n)
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I am interested in a recommendation for a good book which discuses representation theory of GL(n)(say over field of complex numbers). I know only a basic representation theory. The question I am interested in are how looks decomposition of $GL(n)$ module $V\otimes W$, where $V$,$W$ irreps. I am interested in book or chapter in book which will not require too much preliminary.
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Are you interested in infinite-dimensional representations or only finite-dimensional representations? – BR Jan 11 at 20:23
Made community wiki, I am interested only in finite dim reps. – Klim Efremenko Jan 12 at 23:45
## 3 Answers
As previous questions about books on representation theory and Lie theory indicate, there are a lot of them out there aimed at different parts of the subject. (So maybe community-wiki is indicated?) It's good to be clear at the outset that the problem of finite dimensional tensor product decomposition over `$\mathbb{C}$` is essentially the same for general linear groups and for their Lie algebras. However, the group approach is sometimes more flexible.
The "classical" Schur-Weyl approach is exposed in many books, including classical texts by Boerner and others aimed at physicists. One mathematical source I'm fond of is the straightforward chapter in the Springer GTM 225 book Symmetry, representations, and invariants by Goodman and Wallach. Also, the symmetric group background needed is well covered in the first part of the earlier Springer GTM 129 Representation Theory by Fulton and Harris; but it may be alittle harder to extract from their book a focused account of Schur-Weyl duality.
The more "modern" thinking of Littelmann and others (which is combinatorial and aims at avoiding formulas with many cancellations) may not yet be as readily accessible in book form.
There are also some Russian alternatives which others will surely want to advocate for.
Once you get into the decomposition of arbitrary tensor products, rather than just tensor powers of the natural representation, life does get more complicated and you can't expect miracles from the textbook sources.
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I would second the recommendation for Fulton and Harris. It covers the basics of the representation theory of $GL(n)$ in a friendly and accessible way. – Chuck Hague Jan 14 at 17:38
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
From the top of my head, I think those should answer the question nicely (isn't it a duplicate?) :
• Goldfeld's "Automorphic forms and $L$-functions for the group $GL(n,\mathbb R)$"
• Bump's "Automorphic forms and representations"
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It's easier to mention the keyword "Littlewood-Richardson coefficients" that gives the answer to your question than to come up with the best possible source explaining it. If you are only interested in the answer, I would suggest to read books on combinatorics, such as "Symmetric group" by Sagan, rather than trying to learn all the background from the representation theory justifying it.
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http://psychology.wikia.com/wiki/Mathematical_function | # Function (mathematics)
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In mathematics, a function relates each of its inputs to exactly one output. A standard notation for the output of the function f with the input x is f(x). The set of all inputs which a function accepts is called the domain of the function. The set of all outputs is called the range.
For example, the expression f(x) = x2 describes a function, f, that relates each input, x, with one output, x2. Thus, an input of 3 is related to an output of 9. Once a function, f, has been defined we can write, for example, f(4) = 16.
It is a usual practice in mathematics to introduce functions with temporary names like f; in the next paragraph we might define f(x) = 2x+1, and then f(4) = 9. When a name for the function is not needed, often the form y=x2 is used.
If we use a function often, we may give it a permanent name as, for example
$\mathrm{Square}(x)\, =\, x^2$.
The essential property of a function is that for each input there must be one unique output. Thus, for example,
$\mathrm{Root}(x) = \pm \sqrt x$
does not define a function, because it may have two outputs. The square roots of 9 are 3 and − 3, for example. To make the square root a function, we must specify which square root to choose. The definition
$\mathrm{Posroot}(x) = \sqrt x$,
for any non-negative input chooses the non-negative square root as an output.
A function need not involve numbers. An example of a function that does not use numbers is the function that assigns to each nation its capital. In this case $Capital(France)=Paris$.
A more precise, but still informal definition follows. Let A and B be sets. A function from A to B is determined by any association of a unique element of B with each element of A. The set A is called the domain of the function; the set B is called the codomain.
In some contexts, such as lambda calculus, the function notion may be taken as primitive rather than defined in terms of set theory.
In most mathematical fields, the terms map, mapping, and transformation are usually synonymous with function. However, in some contexts they may be defined with a more specialized meaning. For example, in topology a map is sometimes defined to be a continuous function.
## Mathematical definition of a function
A precise definition is required for the purposes of mathematics.
A function is a binary relation, f, with the property that for an element x there is no more than one element y such that x is related to y. This uniquely determined element y is denoted f(x).
Because two definitions of binary relation are in use, there are actually two definitions of function, in effect.
### First definition
The simplest definition of binary relation is "A binary relation is a set of ordered pairs". Under this definition, the binary relation denoted by "less than" contains the ordered pair (2, 5) because 2 is less than 5.
A function is then a set of ordered pairs with the property that if (a,b) and (a,c) are in the set, then b must equal c. Thus the squaring function contains the pair (3, 9). The square root relation is not a function because it contains both the pair (9, 3) and the pair (9, −3), and 3 is not equal to −3.
The domain of a function is the set of elements x occurring as first coordinate in a pair of the relation. If x is not in the domain of f, then f(x) is not defined.
The range of a function is the set of elements y occurring as second coordinate in a pair of the relation.
### Second definition
Some authors require that the definition of a binary relation specify not only the ordered pairs but also the domain and codomain. These authors define a binary relation as an ordered triple $(X, Y, G)$, where X and Y are sets (called the domain and codomain of the relation) and G is a subset of the cartesian product of X and Y (G is called the graph of the relation). A function is then a binary relation with the additional property that each element of X occurs exactly once as the first coordinate of an element of G. Under this second definition, a function has a uniquely determined codomain; this is not the case under the first definition.
## History of the concept
As a mathematical term, "function" was coined by Gottfried Leibniz in 1694, to describe a quantity related to a curve, such as a curve's slope at a specific point of a curve. The functions Leibniz considered are today called differentiable functions, and they are the type of function most frequently encountered by nonmathematicians. For this type of function, one can talk about limits and derivatives; both are measurements of the output or the change in the output as it depends on the input or the change in the input. Such functions are the basis of calculus.
The word function was later used by Leonhard Euler during the mid-18th century to describe an expression or formula involving various arguments, e.g. f(x) = sin(x) + x3.
During the 19th century, mathematicians started to formalize all the different branches of mathematics. Weierstrass advocated building calculus on arithmetic rather than on geometry, which favoured Euler's definition over Leibniz's (see arithmetization of analysis).
At first, the idea of a function was rather limited. Joseph Fourier, for example, claimed that every function had a Fourier series, something no mathematician would claim today. By broadening the definition of functions, mathematicians were able to study "strange" mathematical objects such as continuous functions that are nowhere differentiable. These functions were first thought to be only theoretical curiosities, and they were collectively called "monsters" as late as the turn of the 20th century. However, powerful techniques from functional analysis have shown that these functions are in some sense "more common" than differentiable functions. Such functions have since been applied to the modeling of physical phenomena such as Brownian motion.
Towards the end of the 19th century, mathematicians started to formalize all of mathematics using set theory, and they sought to define every mathematical object as a set. Dirichlet and Lobachevsky independently and almost simultaneously gave the modern "formal" definition of function.
In this definition, a function is a special case of a relation, in particular a function is a relation in which every first element has a unique second element.
The notion of function as a rule for computing, rather than a special kind of relation, has been formalized in mathematical logic and theoretical computer science by means of several systems, including the lambda calculus, the theory of recursive functions and the Turing machine.
## Functions in other fields
Functions are used in every quantitative science, to model relationships between all kinds of physical quantities — especially when one quantity is completely determined by another quantity. Thus, for example, one may use a function to describe how the temperature of water affects its density.
Functions are also used in computer science to model data structures and the effects of algorithms. However, the word is also used in computing in the very different sense of procedure or sub-routine; see function (computer science).
## The vocabulary of functions
An input to a function is called argument of the function. For each argument x, the corresponding unique y in the codomain is called the function value at x, or the image of x under f. The image of x can be written as f(x) or as y. Written mathematics sometimes omits the parentheses around the argument, thus: sin x, but calculators and computers require parentheses around the argument. In some branches of mathematics, such as automata theory, the notation x f is used instead of f(x). The notation f$\;:\;$x${}\mapsto{}$y is sometimes used to mean f(x) = y.
The graph of a function f is the set of all ordered pairs (x, f(x)), for all x in the domain X. If X and Y are subsets of R, the real numbers, then this definition coincides with the familiar sense of "graph" as a picture or plot of the function, with the ordered pairs being the Cartesian coordinates of points.
The concept of the image can be extended from the image of a point to the image of a set. If A is any subset of the domain, then f(A) is the subset of the range consisting of all images of elements of A. We say the f(A) is the image of A under f.
This extension is consistent as long as no subset of the domain is also an element of the domain. A few authors write f[A] instead of f(A), to emphasize the distinction between the two concepts; a few others write f` x instead of f(x), and f``A instead of f[A].
Notice that the range of f is the image f(X) of its domain, and that the range of f is a subset of its codomain.
The preimage (or inverse image) of a subset B of the codomain Y under a function f is the subset of the domain X defined by
f −1(B) = {x in X | f(x) is in B}
So, for example, the preimage of {4, 9} under the squaring function is the set {-3,-2,+2,+3}.
In general, the preimage of a singleton set (a set with exactly one element) may contain any number of elements. For example, if f(x) = 7, then the preimage of {5} is the empty set but the preimage of {7} is the entire domain. Thus the preimage of an element in the codomain is a subset of the domain. The usual convention about the preimage of an element is that f −1(b) means f −1({b}), i. e.
f −1(b) = {x in X | f(x) = b}
Three important kinds of function are the injections (or one-to-one functions), which have the property that if f(a) = f(b) then a must equal b; the surjections (or onto functions), which have the property that for every y in the codomain there is an x in the domain such that f(x) = y; and the bijections, which are both one-to-one and onto. This nomenclature was introduced by the Bourbaki group.
When the first definition of function given above is used, since the codomain is not defined, the "surjection" must be accompanied with a statement about the set the function maps onto. For example, we might say f maps onto the set of all real numbers.
The function composition of two or more functions uses the output of one function as the input of another. For example, f(x) = sin(x2) is the composition of the sine function and the squaring function. The functions f: X → Y and g: Y → Z can be composed by first applying f to an argument x to obtain y = f(x) and then applying g to y to obtain z = g(y).
The composite function formed in this way is denoted g o f: X → Z defined by (g o f)(x) = g(f(x)) for all x in X. Note that the function on the right acts first, the function on the left acts second.
Informally, the inverse of a function f is one that "undoes" the effect of f, by taking each function value f(x) to its argument x. The squaring function is the inverse of the non-negative square root function. Formally, since every function f is a relation, its inverse f−1 is just the inverse relation. That is, if f has domain X, codomain Y, and graph G, the inverse has domain Y, codomain X, and graph
G−1 = { (y, x) : (x, y) ∈ G }
For example, if the graph of f is G = {(1,5), (2,4), (3,5)}, then the graph f−1 is G−1 = {(5,1), (4,2), (5,3)}.
The relation f−1 is a function if and only if for each y in the codomain there is exactly one argument x such that f(x) = y; in other words, the inverse of a function f is a function if and only if f is a bijection. In that case, f−1(f(x)) = x for every x in X, and f(f−1(y)) = y for any y in Y. Sometimes a function can be modified, often by replacing the domain with a subset of the domain, and making corresponding changes in the codomain and graph, so that the modified function has an inverse that is a function.
For example, the inverse of y = sin(x), f(x) = arcsin (x), defined by y = arcsin (x) if and only if x = sin(y), is not a function, because its graph contains both the ordered pair (0, 0) and the ordered pair (0, 2π). But if we change the domain of y = sin(x) to -π/2 $\leq$ x $\leq$ π/2 and change the codomain to — 1 $\leq$ y $\leq$ 1, then the resulting function does have an inverse, denoted with a capital letter A, f(x) = Arcsin (x).
This does not work with every function, however, and inverses are sometimes difficult or impossible to find.
## Specifying a function
If the domain X is finite, a function f may be defined by simply tabulating all the arguments x and their corresponding function values f(x).
More commonly, a function is defined by a formula, or more generally an algorithm — that is, a recipe that tells how to compute the value of f(x) given any x in the domain. More generally, a function can be defined by any mathematical condition relating the argument to the corresponding value. There are many other ways of defining functions. Examples include recursion, algebraic or analytic closure, limits, analytic continuation, infinite series, and as solutions to integral and differential equation.
There is a technical sense in which most mathematical functions cannot be defined at all, in any effective way, explicit or implicit. A fundamental result of computability theory says that there are functions that can be precisely defined which cannot be computed.
## Functions with multiple inputs and outputs
### Functions of two (or more) variables
The concept of function can be extended to an object that takes a combination of two (or more) argument values to a single result. This intuitive concept is formalized by a function whose domain is the Cartesian product of two or more sets.
For example, consider the multiplication function that associates two integers to their product: f(x, y) = x·y. This function can be defined formally as having domain Z×Z , the set of all integer pairs; codomain Z; and, for graph, the set of all pairs ((x,y), x·y). Note that the first component of any such pair is itself a pair (of integers), while the second component is a single integer.
The function value of the pair (x,y) is f((x,y)). However, it is customary to drop one set of parentheses and consider f(x,y) a function of two variables, x and y.
### Functions with output in a product set
The concept can still further be extended by considering a function that also produces output that is expressed as several variables. For example consider the function mirror(x, y) = (y, x) with domain R×R and codomain R×R as well. The pair (y, x) is a single value in the codomain seen as a set.
### Binary operations
The familiar binary operations of arithmetic, addition and multiplication, can be viewed as functions from Z×Z to Z. This view is generalized in abstract algebra, where n-ary functions are used to model the operations of arbitrary algebraic structures. For example, an abstract group is defined as a set X and a function f from X×X to X that satisfies certain properties.
Traditionally, addition and multiplication are written in the infix notation: x+y and x×y instead of +(x, y) and ×(x, y).
## Set of all functions
The set of all functions from a set X to a set Y is denoted by X → Y, by [X → Y], or by YX. The latter notation is justified by the fact that |YX| = |Y||X|. See the article on cardinal numbers for more details.
It is traditional to write f: X → Y to mean f ∈ [X → Y]; that is, "f is a function from X to Y". This statement is sometimes read "f is a Y-valued function of an X-valued variable". One often writes informally "Let f: X → Y" to mean "Let f be a function from X to Y".
## Is a function more than its graph?
Most mathematicians define a binary relation (and hence a function) as an ordered triple (X, Y, G), where X and Y are the domain and codomain sets, and G is the graph of f. However, some mathematicians define a relation as being simply the set of pairs G, without explicitly giving the domain and co-domain.
There are advantages and disadvantages to each definition, but either of them is satisfactory for most uses of functions in mathematics. The explicit domain and codomain are important mostly in formal contexts, such as category theory.
## Partial functions and multi-functions
The condition for a binary relation f from X to Y to be a function can be split into two conditions:
1. f is total, or entire: for each x in X, there exists some y in Y such that x is related to y.
2. f is single-valued: for each x in X, there is at most one y in Y such that x is related to y.
| | | |
|----------------------------|----------------------------|-------------------------------------------|
| File:NotMap1.png | | File:Mathmap2.png |
| Total butnot single-valued | Single-valued butnot total | Total andsingle-valued(a proper function) |
In some contexts, a relation that satisfies condition (1), but not necessarily (2), may be called a multivalued function; and a relation that satisfies condition (2), but not necessarily (1), may be called a partial function.
### Other properties
There are many other special classes of functions that are important to particular branches of mathematics, or particular applications. Here is a partial list:
## Restrictions and extensions
Informally, a restriction of a function f is the result of trimming its graph to a smaller domain.
More precisely, if f is a function from a X to Y, and S is any subset of X, the restriction of f to S is the function f|S from S to Y such that f|S(s) = f(s) for all s in S.
The restriction f|S can also be expressed as the composition f$\circ$ incS,X, where incS,X is the inclusion function of S in X.
If g is any restriction of f, we say that f is an extension of g.
## Pointwise operations
If f: X → R and g: X → R are functions with common domain X and codomain is a ring R, then one can define the sum function f + g: X → R and the product function f × g: X → R as follows:
(f + g)(x) = f(x) + g(x)
(f × g)(x) = f(x) × g(x)
for all x in X.
This turns the set of all such functions into a ring. The binary operations in that ring have as domain ordered pairs of functions, and as codomain functions. This is an example of climbing up in abstraction, to functions of more complex types.
By taking some other algebraic structure A in the place of R, we can turn the set of all functions from X to A into an algebraic structure of the same type in an analogous way.
## Computable and non-computable functions
The number of computable functions from integers to integers is countable, because the number of possible algorithms is. The number of all functions from integers to integers is higher: the same as the cardinality of the real numbers. This argument shows that there are functions from integers to integers that are not computable. For examples of noncomputable functions, see the articles on the halting problem and Rice's theorem.
## Lambda calculus
The lambda calculus provides a powerful and flexible syntax for combining functions of several variables. In particular, composition becomes a special case of variable substitution; and n-ary functions can be reduced to functions with fewer arguments by a process called currying.
## Functions in category theory
The notion of function can be generalized to the notion of morphism in the context of category theory. A category is a collection of objects and morphisms. The collection of objects is completely arbitrary. The morphisms are the relationships between the objects. Each morphism is an ordered triple (X, Y, f), where X is an object called the domain, Y is an object called the codomain, and f connects X to Y. There are a few restrictions on the morphisms which guarantee that they are analogous to functions (for these, see the article on categories). However, because category theory is designed to work with objects that may not be sets or may not behave like sets do, morphisms are not the same as functions.
In a concrete category, each morphism is associated with an underlying function.
## References
• Lawrence S. Husch (2001). Visual Calculus. University of Tennessee.
• João Pedro da Ponte (1992). The history of the concept of function and some educational implications. The Mathematics Educator 3(2), 3-8. available online in Microsoft Word and HTML formats.
• Anton, Howard (1980). Calculus with analytical geometry.. New York:John Wiley and Sons. ISBN 0-471-03248-4.
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• Content is available under CC-BY-SA. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9101846218109131, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/266450/do-we-have-maximal-abelian-algebras-maas/282098 | # Do we have Maximal Abelian Algebras (MAAs)?
Let $\mathcal{H}$ be a Hilbert space and $B(\mathcal{H})$ the algebra of bounded linear operators on $\mathcal{H}$. A MASA $\mathcal{M}$ is a subalgebra of $B(\mathcal{H})$ that is abelian and self-adjoint, and is maximal with respect to these conditions.
MASAs are quite standard materials in operator theory, but I wonder whether we have studied MAAs, that is, we drop the condition that the algebra has to be closed under involution.
Just like for MASAs, MAA exists if we assume Zorn's lemma. But MASAs can be constructed by hand (for instance, $\mathcal{H}=\mathcal{L}^2(X,\mu)$, a MASA is the multiplication algebra). I am not sure whether we have such kind of explicit examples of MAA.
SO, do we have a nice example of MAA (on $\mathcal{L}^2$ maybe)? Do we know something about MAAs?
Thanks!
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2
This might be appropriate for mathoverflow, if you don't receive answers here. – Potato Jan 2 at 6:09
@Potato Is there a convenient way to migrate questions to MO? – Hui Yu Jan 14 at 16:32
I think you could just repost it with a link to this question and a note saying you are reposting it because you received no answers. – Potato Jan 14 at 17:54
## 1 Answer
The usual proof that $L^\infty$ is a MASA in actuality proves that it's an MAA in your sense; that is, to prove that a given operator $T$ is a multiplication operator all you need to know is that it commutes with all the other multiplication operators (and oftentimes all you need to know is that it commutes with one particular multiplication operator; see e.g. Arveson's "A short course on spectral theory", section 4.1).
On the other hand, there are MAA's which are not MASAS. This is easiest to see in the finite-dimensional case. Let $S$ be the $n\times n$ matrix with $1$'s along the diagonal just below the main diagonal, and $0$'s elsewhere (so $S$ is just the unilateral shift). It's not hard to prove that any matrix which commutes with $S$ must be lower-triangular and constant along all the diagonals, and these are exactly the matrices given by polynomials in $S$; hence they form an MAA. But obviously this algebra is not self-adjoint. A similar argument also works in the infinite-dimensional case.
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http://mathhelpforum.com/advanced-algebra/92322-group-theory-question.html | # Thread:
1. ## group theory question
prove that if G is a finite group that the number of elements in G is even
So
e≠x belong to G (e is the neutral element)
that provide x^2 = e (in other words: x=x^-1).
2. Hello,
Let $A=\left\{x\ :\ x\in G\ \text{and}\ x\neq x^{-1}\right\}$ and $B=\left\{x\ :\ x\in G\ \text{and}\ x=x^{-1}\right\}$. We have $A\cap B=\emptyset$ and $G=A\cup B$ so $|G|=|A|+|B|$. Can you show that $|B|$ is even and that there exists an element $x\neq e$ lying in $B$?
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http://mathoverflow.net/questions/104742/exponentiable-objects-in-a-category-valued-in-a-larger-containing-category | ## Exponentiable objects in a category, valued in a larger, containing category
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Recall that when dealing with topological spaces one usually likes dealing with a subcategory of $Top$ which is convenient, one facet of which is that it is cartesian closed. However to get to a similar point with smooth manifolds one needs to consider things like diffeological spaces. Not that there is anything wrong with that. But we have a partial solution if we are just looking for exponentiable objects, and willing to consider infinite-dimensional smooth manifolds (usually Frechet manifolds).
More formally, an object $A$ of category $C$ with binary products is exponentiable if the functor $-\times A\colon C\to C$ has a right adjoint. The classification of which topological spaces are exponentiable is well known, and cartesian closed categories are defined by the fact that every object is exponentiable.
But in the category of (Hausdorff, finite-dimensional) smooth manifolds the only exponentiable objects are the compact manifolds of dimension at most zero. But we can still sensibly talk about smooth mapping spaces between a general compact manifold and an arbitrary manifold, where the mapping space is an object in the category of Frechet manifolds $Frech$, in which the category $Diff$ of finite-dimensional smooth manifolds sits as a full subcategory.
There are clear analogies with, say, finite CW-complexes, where the 'internal' hom is a topological space of a rather more infinite nature. Similarly, we can consider the mapping presheaf $X \mapsto C(X\times A,Y)$ on C.
What I would like to know is if there is a name for this sort of phenomenon, that we have a category $C$, and full embedding $C\hookrightarrow D$ and $D$-valued mapping objects for certain objects of $C$: these are objects of $C$ which are exponentiable as objects of $D$. It seems to fall into some gap between cartesian closedness and enrichment, but I don't have a way of making that precise.
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I don't think there's a name for this, but this property can probably be rephrased as the existence of some Kan extensions along the inclusion. – Fernando Muro Aug 15 at 15:37
## 1 Answer
Fernando is right that it has something to do with Kan extensions. However, it is not about Kan extensions along an inclusion, but more about extension of an inclusion.
I thought about similar issues a few years ago (and recently --- yesterday) , but in a slightly different context --- after the excellent answer by Todd Trimble to my question http://mathoverflow.net/questions/59291/completion-of-a-category, I wondered if there was a general 2-categorical setting that could explain such constructions (I was mainly interested in carrying to a 2-categorical setting the highly related concept of Day convolution).
Now I try to slowly reproduce some of these ideas.
I shall introduce the concept of a Yoneda triangle (perhaps I should call it the right Yoneda triangle, because there are obvious dual concepts).
Let $\mathbb{W}$ be a 2-category. A Yoneda triangle in $\mathbb{W}$ consists of 1-morphisms $y \colon A \rightarrow \overline{A}$, $f \colon A \rightarrow B$, $g \colon B \rightarrow \overline{A}$ together with a 2-morphism $\eta \colon y \rightarrow g \circ f$ which exhibits $g$ as a pointwise left Kan extension of $y$ along $f$ and exhibits $f$ as an absolute left Kan lifting of $y$ along $g$. (BTW: these data are exactly what led Mark Weber to strengthen the definition of a Yoneda structure introduced by Street and Walters).
The idea of a Yoneda triangle is that, we have a morphism $y \colon A \rightarrow \overline{A}$ which plays the role of a "defective identity" and for a given morphism $f \colon A \rightarrow B$ we try to characterise its right adjoint up to the "defective identity" $y$.
Example: [Yoneda triangles in $\mathbf{Cat}$] If we take $\mathbb{W}$ to be the 2-category $\mathbf{Cat}$ of locally small categories, functors and natural transformations, then the condition that $G$ is a pointwise left Kan extension of $Y$ along $F$ reduces to: $$G(-) = \int^{A\in\mathbb{A}} \hom(F(A), -) \times Y(A)$$ (where the coend has to be interpreted as the colimit of $Y$ weighted by $\hom(F(=), -)$ in case the category is not tensored over $\mathbf{Set}$). And the condition that $F$ is an absolute left Kan lifting of $Y$ along $G$ reduces to: $$\hom(Y(-), G(=)) \approx \hom(F(-), =)$$
Particularly, if $Y$ is dense, than $G$ is canonically a pointwise Kan extension --- from density we have: $$G(-) \approx \int^{A\in\mathbb{A}} \hom(Y(A), G(-)) \times Y(A)$$ and using the formula for an absolute lifting: $$G(-) \approx \int^{A\in\mathbb{A}} \hom(F(A), -) \times Y(A)$$
Example: [Adjunction as Yoneda triangle] It is folklore that an adjunction $f \dashv g$ in a 2-category $\mathbb{W}$ may be equally characterised in the following way: $f$ is an absolute left lifting of the identity along $g$. In such a case $g$ is automatically a pointwise left extension of the identity along $f$ and $\mathit{id}, f, g$ together with the unit of the adjunction form a Yoneda triangle.
Example: [Yoneda triangle as a relative adjunction] There is an old concept of so called "relative adjunction", which is defined in the same way as the Yoneda triangle, but without the requirement that $g$ is a left Kan extension. Note however, that in such a case $g$ need not be uniquely determined by $f$.
Let me move to the more specific example that you asked about.
Example: [Yoneda triangle along Yoneda embedding] Let $F \colon \mathbb{A} \rightarrow \mathbb{B}$ be a functor between locally small categories (or more generally, a locally small functor). There is also an inclusion $y_\mathbb{A} \colon \mathbb{A} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$. One may easily verify that these data may be always extended to the Yoneda triangle with $G(-) = \hom(F(=), -) \colon \mathbb{B} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$ --- which reassembles the fact that every functor always has a "distributional" right adjoint. The same is true for internal categories and for categories enriched in a complete and cocomplete symmetric monoidal closed category, and generally (almost by definition) for any 2-category equipped with a Yoneda structure.
The essence of the above example is that because the Yoneda functor $y_\mathbb{B} \colon \mathbb{B}\rightarrow \mathbf{Set}^{\mathbb{B}^{op}}$ is a full and faithful embedding, functors $F\colon\mathbb{A} \rightarrow \mathbb{B}$ may be thought as of distributors $$y_\mathbb{B} \circ F = \hom(=, F(-))$$ Every distributor arisen in this way has a right adjoint distributor $\hom(F(=), -)$ in the bicategory of distributors. The distributor $\hom(F(=), -)$ has actually the type $\mathbb{B} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$, which is the only think that may prevent $F$ of having the ordinary (functorial) right adjoint $G \colon \mathbb{B} \rightarrow \mathbb{A}$ --- just recall, that we say that $F$ has a right adjoint, if there exists $G$ such that: $$y_\mathbb{A} \circ G \approx \hom(F(=), -)$$ which means: $$\hom(=, G(-)) \approx \hom(F(=), -)$$
Unfortunately, as a non-mathematician I will not help you with your other examples involving highly mathematical and completely non-understandable terms like a topological space or a manifold, so perhaps you have to calculate the other examples yourself :-)
However, I will give you another example that actually led me to the above considerations. One may similarly define the concept of a Yoneda bi-triangle and a Yoneda monoidal bi-triangle.
Example: [2-powers from Yoneda triangle] The motivating example is to start with a 2-functor $J \colon \mathbb{W} \rightarrow \mathbb{D}$ equipping a 2-category $\mathbb{W}$ with proarrows, and an extension $Y \colon \mathbb{W} \rightarrow \overline{\mathbb{W}}$ embedding "small objects" into "locally small" (or large) objects in $\overline{\mathbb{W}}$. Then to extend these data to the Yoneda triangle, we have to find a functor $P \colon \mathbb{D} \rightarrow \overline{\mathbb{W}}$ representing a proarrow $A \nrightarrow B$ as a morphism $A \rightarrow P(B)$ in $\overline{\mathbb{W}}$, and a natural transformation $\eta \colon Y \rightarrow P\circ J$ playing the role of a familly of Yoneda morphisms $\eta_A \colon A \rightarrow P(A)$.
The archetypical situation is when we take $\mathbb{W} = \mathbf{cat}$, $\overline{\mathbb{W}} = \mathbf{Cat}$, $\mathbb{D} = \mathbf{Dist}$, where $\mathbf{cat}$ is the 2-category of small categories, $\mathbf{Cat}$ is the 2-category of locally small categories, and $\mathbf{Dist}$ is the bicategory of distributors between small categories. Then $J \colon \mathbf{cat} \rightarrow \mathbf{Dist}$, $Y \colon \mathbf{cat} \rightarrow \mathbf{Cat}$ are the usual embeddings, $P \colon \mathbf{Dist} \rightarrow \mathbf{Cat}$ is the covarinat 2-power pseudofunctor $\mathbf{Set}^{(-)^{op}}$ defined on distributors via left Kan extensions, and $\eta_\mathbb{A} \colon \mathbb{A} \rightarrow \mathbf{Set}^{\mathbb{A}^{op}}$ is the Yoneda embedding of a small category $\mathbb{A}$.
We know that there are isomorphisms of categories: $$\hom_{\mathbf{Dist}}(\mathbb{A}, \mathbb{B}) \approx \hom_{\mathbf{Cat}}(\mathbb{A}, \mathbf{Set}^{\mathbb{B}^{op}})$$ where $\mathbb{A}$ and $\mathbb{B}$ are small. Therefore, to show that $P$ is a (bi)pointwise left Kan extension it suffices to show that $Y$ is 2-dense. However, $Y$ is obviously 2-dense, because the the terminal category is a 2-dense subcategory of $\mathbf{Cat}$ and $Y$ is fully faithful.
The point is that in most situations $\mathbb{D}$ is a monoidal (bi)category, where the monoidal structure is inherited from the closed structure on $\mathbb{W}$. Moreover, functors and the natural transformation constituting the Yoneda triangle are (lax)monoidal. This means that monoids in $\mathbb{W}$ are mapped to the (pro)monoids in $\mathbb{D}$ which are mapped to monoids in $\overline{\mathbb{W}}$. If I am not mistaken this observation leads to an abstract characterisation of the concept of the Day convolution (and in a similar manner one may try to define a Dedekind-MacNeille completion of an object).
In our archetypical situation, a category $\mathbb{A} \times \mathbb{B}$ is mapped by $P$ to $\mathbf{Set}^{\mathbb{A}^{op} \times \mathbb{B}^{op}}$ and the missing morphisms making the unit of the triangle lax monoidal: $$\mathbf{Set}^{\mathbb{A}^{op}} \times \mathbf{Set}^{\mathbb{B}^{op}} \rightarrow \mathbf{Set}^{\mathbb{A}^{op} \times \mathbb{B}^{op}}$$ is given by the convolution of the distributional identity $\mathbb{A} \times \mathbb{B} \nrightarrow \mathbb{A} \times \mathbb{B}$: $$\langle F, G \rangle \mapsto \int^{A \in \mathbb{A}, B \in \mathbb{B}} F(A) \times G(B) \times \hom(-, A) \times \hom(=, B) = F(-) \times G(=)$$ Now, a promonoidal category $M \colon \mathbb{A} \times \mathbb{A} \nrightarrow \mathbb{A}$ is mapped by $P$ to: $$H \mapsto \int^{\langle A, B \rangle \in \mathbb{A}\times \mathbb{A}} H(A, B) \times M(-, A, B)$$ and by composing it with the above map: $$\langle F, G \rangle \mapsto \int^{\langle A, B \rangle \in \mathbb{A}\times \mathbb{A}} F(A) \times G(B) \times M(-, A, B)$$ we obtain the well-known formula for convolution.
One may also go in the other direction --- starting from the composition $P \circ J$ satisfying monoidal-like laws and try to find a right or left resolution in the category of (right/left) modules over monoid on $P \circ J$. If I am not mistaken, the left resolution (the Eilenberg-Moore object) of $P \circ J$ in our archetypical situation consists of the category of cocomplete categories and cocontinous functors and the right resolution (the Kleisli object) consists of the bicategory of distributors (i.e. the category of free cocomplete categories and cocontinous functors).
(BTW: this in some sense relates the concept of a proarrow equipment with the concept of a Yoneda structure.)
(BTW: perhaps the concept of a 2-topos should be defined as a Yoneda monoidal bi-triangle induced by the embedding of a 2-category of small objects into a category of bigger objects relatively to a category of "relations" in $\mathbb{W}$, which, for some purposes may be defined as the 2-category of discrete fibred spans, and for another purposes may be defined as the 2-category of codiscrete cofibred cospans).
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This will take some digesting! Thanks for thinking about this. – David Roberts Apr 18 at 5:33
+1. I only gave this a cursory read and can't comment on anything past where the word "topos" appears, but the beginning makes a lot of sense to me and the fact that distributors came up in the middle convinces me that this is the right thing. I'm not sure, but I suspect there's a simpler way to do this, relying on distributors and Kan extensions without mention of Yoneda triangles. BTW, I think Michal should identify at least in part as a mathematician. I certainly count this as math, and good math at that! – David White Apr 18 at 14:16
1
@David Roberts, let me know when you find any interesting examples! @David White, I have expanded the example of a 2-topos (now it is called 2-power) --- do hope it is more readable now. – Michal R. Przybylek Apr 19 at 22:00 | {"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": 108, "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": 0.9405092000961304, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/19627/why-is-a-gaussian-fixed-point-called-gaussian | # Why is a gaussian fixed point called gaussian?
I know what a gaussian fixed point is, and I did read the wikipedia entry, but it wasn't helpful. It says because the probability distribution is gaussian, but what probability distribution?
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## 1 Answer
The probability distribution of any observable that is a linear function of fields is Gaussian i.e. $$A\exp(-(x-x_0)^2/2\Delta x^2)$$ in the ground state of a Gaussian fixed point. Also, the path integral of a Gaussian fixed point (free field theory) is the integral of the exponential of a bilinear function of the fields which is also called Gaussian.
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http://math.stackexchange.com/questions/132514/prove-that-the-integral-of-an-odd-step-function-on-1-1-is-0 | # Prove that the integral of an odd step function on [-1,1] is 0
Let $\varphi\colon[-1,1]\to \mathbb R$ be an odd step function.Prove that: $\int_{-1}^1\! \varphi(t)\, dt = 0$
Thanks!
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## 1 Answer
Let $f$ be an odd integrable function, that is, $f(x)=-f(-x)$ for all $x\in\mathbb{R}$. Then:
$\int_{-R}^R f(x) dx=\int_{-R}^0f(x)dx+\int_0^Rf(x)dx=-\int_0^Rf(x)dx+\int_0^Rf(x)dx=0$
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1
Note that this doesn't even use the assumption that $\varphi$ is a step function -- the conclusion holds for every odd integrable function. – Henning Makholm Apr 16 '12 at 14:16 | {"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": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.8417587280273438, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/161192/lifting-additive-characters | Lifting additive characters
Let $K$ a finite extension of $\mathbb{Q}_p$ ($p$ prime different from 2) and let $G_K$ the absolute Galois group of $K$.
Let $\bar{u} : G_K \longrightarrow \mathbb{F}_p$ a continuous additive character. Is it always possible to lift $\bar{u}$ to an additive character $u : G_K \longrightarrow \mathbb{Z}_p$ ?
I know the answer is yes when $K$ does not contains the $p^{th}$-root of unity. What happens in the other case ?
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1 Answer
By local class field theory, such a character corresponds to a character $K^{\times} \to \mathbb F_p$. Now $K^{\times} \cong \mathbb Z \times \mathcal O_K^{\times}.$ Certainly a character $\mathbb Z \to \mathbb F_p$ can be lifted to a character $\mathbb Z \to \mathbb Z_p$. So the question is whether $\mathcal O_K^{\times} \to \mathbb F_p$ can be lifted. (This corresponds to the restriction of your Galois character to inertia.)
Now $\mathbb O_K^{\times} \cong \mu \times \Gamma$, where $\Gamma$ is isomorphic to a product of copies of $\mathbb Z_p$, and $\mu$ is the subgroup of roots of unity in $K$. Again, a character $\Gamma \to \mathbb F_p$ can always be lifted, so the question is whether a character $\mu \to \mathbb F_p$ can be lifted to a character $\mu \to \mathbb Z_p$.
Since $\mu$ is finite, this is possible if and only if $\mu \to \mathbb F_p$ is trivial. This will be automatic if and only if $\mu$ contains no elements of order $p$, i.e. if and only if $K$ contains no $p$-power roots of unity.
So, if $K$ contains $p$-power roots of unity, then you have to check whether or not your given character $K^{\times} \to \mathbb F_p$ is trivial on these roots of unity. It lifts to a character $K^{\times} \to \mathbb Z_p$ if and only if it is trivial on them.
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http://mathhelpforum.com/algebra/207682-simplyfying-equation-low-level-math.html | # Thread:
1. ## Simplyfying an equation (low level math)
Hey guys, math newbie here. I've run into a bit of a hazzle, trying to simplify this equation:
I honestly have no clue where to begin. My first intuition, was to simplify the brackets one at the time: (6x-5)(2x+7) and then multiply that result with the next bracket and so on, until
finally multiplying it all with 2 and leaving the denominator intact, but I realize that I'm completely off.
If someone could show me the first step or point me in the right direction, I'd be very happy.
Thank you!
2. ## Re: Simplyfying an equation (low level math)
$\frac{(6x-5)(2x+7) - (3x^2-5x) \cdot 2}{(2x+7)^2}$
$\frac{12x^2 + 32x - 35 - 6x^2 + 10x}{(2x+7)^2}$
$\frac{6x^2 + 42x - 35}{(2x+7)^2}$
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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.946398913860321, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/100233?sort=votes | ## Question on Hilbert Manifolds
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I have a very basic question on Hilbert manifolds.
Consider the Hilbert space $$\mathcal{H}:= L^2(S^1)$$ with $S^1$ the unit circle. On $\mathcal{H}$ let us introduce the equivalence relation $$f\sim g : \Leftrightarrow f(\cdot ) = g(\cdot + \alpha)\quad \mbox{for some }\alpha \in S^1.$$ Now define the factor space $$\overline{\mathcal{H}}:= \mathcal{H}/\sim.$$ What is the structure of $\overline{\mathcal{H}}$? Is it a Hilbert manifold? If so, how to construct the smooth structure? I am particularly interested in computing a (Riemannian) distance between two elements of $\overline{\mathcal{H}}$.
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Sorry, I has an answer that was wrong, using Fourier transform. If it can be corrected, I will repost. – BS Jun 21 at 13:40
I would be equally interested in the answer for $L^2(G)/G$ if $G$ is a (compact) topological group. – Spice the Bird Jun 21 at 23:58
1
See Andrew Stacy's answer to the question mathoverflow.net/questions/10200/… – Spice the Bird Jun 22 at 0:10
2
Not a Hilbert manifold. As the answer(s) have pointed out, you get problems at fixed points of the circle action (indeed, at any point where the circle action has a non-trivial stabiliser). There's another problem which is that the circle does not act continuously on the Hilbert space - how much of a problem this is will depend on how you want to fix the first problem (for example, you could go for a stratified space). – Andrew Stacey Jun 25 at 9:28
## 1 Answer
It probably is NOT a smooth manifold. I think finding a chart around the point corresponding to constants, namely, the fixed points of the action of the group of rotation, is problematic.More precisely, at a fixed point, there is not a well-defined tangent space.
- | {"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": 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": 0.9117489457130432, "perplexity_flag": "head"} |
http://unapologetic.wordpress.com/2007/07/30/transformations-of-adjoints/?like=1&source=post_flair&_wpnonce=e94a57fbb5 | # The Unapologetic Mathematician
## Transformations of Adjoints
And now we go back to adjoints. Like every other structure out there, we want to come up with some analogue of a homomorphism between two adjunctions. Let’s consider the adjunctions $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$ and $F'\dashv G':\mathcal{C}'\rightarrow\mathcal{D}'$, and try to find a good notion of a transformation from the first to the second.
We’ll proceed by considering an adjunction to consist of the pair of categories $(\mathcal{C},\mathcal{D})$ with the functors giving extra structure. Down in the land of groups and rings and such, we’d consider sets with extra structure and functions that preserved that structure. So here, naturally, we want to consider functors which preserve this extra structure. That is, a map of adjunctions consists of a pair of functors $K:\mathcal{C}\rightarrow\mathcal{C}'$ and $L:\mathcal{D}\rightarrow\mathcal{D}'$. These must preserve the structure in that $K\circ G=G'\circ L$ and $L\circ F=F'\circ K$.
But hold up a second, we’ve forgotten something else that goes into an adjunction: the isomorphism $\phi:\hom_\mathcal{D}(F(C),D)\rightarrow\hom_\mathcal{C}(C,G(D))$. Here’s a diagram showing how the map of adjunctions should play nicely with them:
Equivalently we can specify an adjunction by its unit and counit. In this case the compatibility in question is a pair of equations of natural transformations: $1_K\circ\eta=\eta'\circ1_K$ and $1_L\circ\epsilon=\epsilon'\circ1_L$.
What if we’re looking at two different adjunctions between the same pair of categories? Well then we may as well try to use the appropriate identity functors for $K$ and $L$. But then it’s sort of silly to insist that $G=1_\mathcal{C}\circ G=G'\circ\mathcal{D}=G'$ on the nose, and similarly for $F'$. Instead, as we do so often, let’s weaken this equality to just a natural transformation.
We’ll say that a pair of natural transformations $\sigma:F\rightarrow F'$ and $\tau:G'\rightarrow G$ are “conjugate” if $\hom_\mathcal{C}(1_C,\tau_D)\circ\phi'_{C,D}=\phi_{C,D}\circ\hom_\mathcal{D}(\sigma_C,1_D)$. This is equivalent, in terms of the unit and counit, to any one of the following four equalities:
• $\tau=(1_G\circ\epsilon')\cdot(1_G\circ\sigma\circ1_{G'})\cdot(\eta\circ1_{G'})$
• $\sigma=(\epsilon\circ1_{F'})\cdot(1_F\circ\tau\circ1_{F'})\cdot(1_F\circ\eta')$
• $\epsilon\cdot(1_F\circ\tau)=\epsilon'\cdot(\sigma\circ1_{G'})$
• $(1_G\circ\sigma)\cdot\eta=(\tau\circ1_{F'})\cdot\eta'$
Now it’s easily verified that given a pair of categories $(\mathcal{C},\mathcal{D})$ we can build a category whose objects are adjunctions $F\dashv G:\mathcal{C}\rightarrow\mathcal{D}$ and whose morphisms are conjugate pairs of natural transformations, which we write out in full as $(\sigma,\tau):(F,G,\phi)\rightarrow(F',G',\phi'):\mathcal{C}\rightarrow\mathcal{D}$. We compose conjugate pairs in this category in the obvious way, which we write $(\sigma',\tau')\cdot(\sigma,\tau)$.
On the other hand, if we have a pair $(\sigma,\tau):(F,G,\phi)\rightarrow(F',G',\phi'):\mathcal{C}\rightarrow\mathcal{D}$ and another $(\bar{\sigma},\bar{\tau}):(\bar{F},\bar{G},\bar{\phi})\rightarrow(\bar{F}',\bar{G}',\bar{\phi}'):\mathcal{D}\rightarrow\mathcal{E}$, then we can form the composite $(\bar{\sigma}\circ\sigma,\tau\circ\bar{\tau}):(\bar{F}\circ F,G\circ\bar{G},\bar{\phi}\cdot\phi)\rightarrow(\bar{F}'\circ F',G'\circ\bar{G}',\bar{\phi}'\cdot\phi'):\mathcal{C}\rightarrow\mathcal{E}$, which we’ll write as $(\bar{\sigma},\bar{\tau})\circ(\sigma,\tau)$. Notice the similarity of this situation with the two different compositions of natural transformations between functors.
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Posted by John Armstrong | Category theory
## 3 Comments »
1. I’m sorry to leave such an unrelated comment, but how do you pingback to previous posts. It doesn’t seem to work for me.
Comment by | July 31, 2007 | Reply
2. It’s sort of sporadic. If I link to five other posts I’ve made, maybe one gets a pingback. I really don’t understand it very much, but it’s automatic to the extent that it does work.
Comment by | July 31, 2007 | Reply
3. [...] with parameters Now that we know how to transform adjoints, we can talk about whole families of adjoints parametrized by some other category. That is, for [...]
Pingback by | July 31, 2007 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"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": 29, "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": 0.9159541726112366, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/179882-arithmetic-progression.html | # Thread:
1. ## Arithmetic Progression
The sum of the first three terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121, find the number of terms in this sequence.
$S_3=\frac{3}{2}[2a+(3-1)d]=9$ -----(1)
$S_n-S_{n-3}=121-\frac{n-3}{2}[2a+(n-3-1)d]=57$
$121-\frac{n-3}{2}[2a+(n-4)d]=57$
$\frac{n-3}{2}[2a+(n-4)d]=64$ -----(2)
$S_n=\frac{n}{2}[2a+(n-1)d]=121$ -------(3)
2. Originally Posted by Punch
The sum of the first three terms and that of the last three terms of an arithmetic progression are 9 and 57 respectively. It is known that the sum of the whole sequence is 121, find the number of terms in this sequence.
$S_3=\frac{3}{2}[2a+(3-1)d]=9$ -----(1)
$S_n-S_{n-3}=121-\frac{n-3}{2}[2a+(n-3-1)d]=57$
$121-\frac{n-3}{2}[2a+(n-4)d]=57$
$\frac{n-3}{2}[2a+(n-4)d]=64$ -----(2)
$S_n=\frac{n}{2}[2a+(n-1)d]=121$ -------(3)
3. If we take a as the first term of the sequence and d as the common difference, the first three terms are a, a+ d, and a+ 2d so their sum is 3a+ 3d= 9.
I am also making use of the fact that an arithemetic sequence has the useful property that the average of all terms is the same same average of the first and last terms, $(a_1+ a_n)/2$ so that their sum is $(a_1+ a_n)(n/2)$.
If there are n terms in the sequence, the last three terms are a+ (n- 3)d, a+ (n- 2)d, and a+ (n-1)d so their sum is 3a+ 3nd- 6d= 57. Subtracting the previous equation, 3nd- 9d= 3d(n- 3)= 48. Now, we can argue that since n is an integer, so is n-2 and so 3d must be an integer divisor of 48. $48= 3(2^4)$, so 3d must be one of 1, 3, 6, 12, 24, or 48.]
If 3d= 1 then d= 1/3 and n-3= 48 so n= 51. In that case, a= (9- 3(1/3))/3= 8/3 so that the last three terms are 8/3+ (48)(1/3)= 56/3, 8/3+ 49(1/3)= 57/3, and 8/3+ 50(1/3)= 58/3 which add to 57, of course. The sum of all 51 terms is (1/2)(8/3+ 58/3)(51)= (11)(51)= 561, not 121.
If 3d= 3 then d= 1 and n-3= 16 so n= 19. In that case, a= (9- 3)/2= 2. The last term is 2+ 18(1)= 20. The sum of all 19 terms is (1/2)(2+ 20)(19)= 11(19)= 209, not 121.
If 3d= 6, then d= 2 and n-3= 8 so n= 11. In that case, a= (9- 6)/3= 1 so the last term is 21. The sum of all 11 terms is (1/2)(1+ 21)(11)= 11(11)= 121!
If 3d is larger than 6, a= (9- 3d)/3 is not positive.
4. Originally Posted by abhishekkgp | {"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": 0.9354934692382812, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2011/01/25/straightening-a-polytabloid/?like=1&_wpnonce=401286028f | # The Unapologetic Mathematician
## “Straightening” a Polytabloid
Let’s look at one example of “straightening” out a polytabloid to show it’s in the span of the standard polytabloids, using the Garnir elements.
$\displaystyle t=\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}$
Now, it’s slightly abusive to the notation, but we’ll just write a tableau $t$ and know that we actually mean the polytabloid $e_t$ in our linear combinations. Using this, we’ve seen that we can write
$\displaystyle\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}=\begin{array}{ccc}1&2&3\\4&5&\\6&&\end{array}-\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}-\begin{array}{ccc}1&2&3\\4&6&\\5&&\end{array}+\begin{array}{ccc}1&4&3\\2&6&\\5&&\end{array}-\begin{array}{ccc}1&5&3\\2&6&\\4&&\end{array}$
Now, by the way we’ve selected our Garnir elements, we know that none of these can have any column descents. And we also know that they can’t have a row descent in the same place $e_t$ did. Indeed, the only three that have a row descent all have it between the second and third entries of the first row. So now let’s look at
$\displaystyle u=\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}$
We can write down another table, just like before:
$\displaystyle\begin{array}{cccc}A'&B'&\pi&\pi u\\\hline\\\{4,5\}&\{3\}&e&\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}\\\{3,5\}&\{4\}&(3\,4)&\begin{array}{ccc}1&3&4\\2&5&\\6&&\end{array}\\\{3,4\}&\{5\}&(3\,5\,4)&\begin{array}{ccc}1&3&5\\2&4&\\6&&\end{array}\end{array}$
which lets us write
$\displaystyle\begin{array}{ccc}1&4&3\\2&5&\\6&&\end{array}=\begin{array}{ccc}1&3&4\\2&5&\\6&&\end{array}-\begin{array}{ccc}1&3&5\\2&4&\\6&&\end{array}$
Similarly we can write
$\displaystyle\begin{array}{ccc}1&4&3\\2&6&\\5&&\end{array}=\begin{array}{ccc}1&3&4\\2&6&\\5&&\end{array}-\begin{array}{ccc}1&3&6\\2&4&\\5&&\end{array}$
and
$\displaystyle\begin{array}{ccc}1&5&3\\2&6&\\4&&\end{array}=\begin{array}{ccc}1&3&5\\2&6&\\4&&\end{array}-\begin{array}{ccc}1&3&6\\2&5&\\4&&\end{array}$
Putting these all together, we conclude that
$\displaystyle\begin{aligned}\begin{array}{ccc}1&2&3\\5&4&\\6&&\end{array}=&\begin{array}{ccc}1&2&3\\4&5&\\6&&\end{array}-\begin{array}{ccc}1&3&4\\2&5&\\6&&\end{array}+\begin{array}{ccc}1&3&5\\2&4&\\6&&\end{array}-\begin{array}{ccc}1&2&3\\4&6&\\5&&\end{array}\\+&\begin{array}{ccc}1&3&4\\2&6&\\5&&\end{array}-\begin{array}{ccc}1&3&6\\2&4&\\5&&\end{array}-\begin{array}{ccc}1&3&5\\2&6&\\4&&\end{array}+\begin{array}{ccc}1&3&6\\2&5&\\4&&\end{array}\end{aligned}$
All of these tabloids are standard, and so we see that our original — nonstandard — $e_t$ is in the span of the standard polytabloids.
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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": 12, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.8951968550682068, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/198373/choosing-an-abstract-algebra-text/198416 | # Choosing an abstract algebra text
I was wondering if some one here would please recommend an abstract algebra text book. In particular, I seeking a text from among one of these:
1)Abstract Algebra by Dummit and Foote
2)Algebra by Michael Artin
3)Topics in Algebra by Herstein.
I was wondering which text is best for a high school student and has great exposition and good problems. My goal is study some algebraic number theory some time later in the near future as I liked elementary number theory.
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Have you been exposed to proof-based mathematics before? – Alex Becker Sep 18 '12 at 4:54
1
Dummit and Foote is a great text, with many examples and diagrams, but the sheer amount of information can be a little overwhelming to someone just getting started. I would probably check out Herstein first for a friendlier introduction. – Tarnation Sep 18 '12 at 4:54
1
Of those three I’d be inclined to go with Herstein. It’s a little old-fashioned, but that’s not really a problem for a first text. The exposition isn’t flashy, but it’s very solid, and there are lots of problems, including some very challenging ones. (Incidentally, when I used it some years ago in a university undergraduate algebra course, my best student by a significant margin was a high school student.) – Brian M. Scott Sep 18 '12 at 5:15
1
My favorite Algebra book is Algebra by Thomas Hungerford. It's a pretty big encyclopedia of algebraic concepts. – emka Sep 18 '12 at 5:33
2
Patience, Squid, it has only been an hour. Fermat had to wait 350 years for an answer! – Gerry Myerson Sep 18 '12 at 5:54
show 10 more comments
## 4 Answers
To repeat what I said in the comments, my first choice among the three books that you mention is Herstein’s. It’s a little old-fashioned, but that’s not really a problem for a first text. The exposition isn’t flashy, but it’s very solid, and there are lots of good problems, including some very challenging ones.
Having said that, though, I strongly recommend that you spend some time looking at the early parts of all three before you decide, in order to see how well each of the expository styles works for you. This includes the way the author puts words together, the amount of detail in proofs, the number and placement of concrete examples $-$ even the typographical layout, if that affects the book’s readability for you.
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+1 for typographical layout. This seems to be greatly overlooked, but I think the visual imprint and accessibility make a big difference in absorption and retention. – Andrew Feb 12 at 17:00
I would suggest less on the choice of the textbook and more on solving problems. You do not learn algebra by reading 700 pages of definitions and other's proofs. Pick up any book and work through 70% of the problems you consider not simply a routine manipulation of algebra (for example, verifying the real numbers is a field). Then you will know:
1) Some seemingly difficult problems can be solved by proving simpler lemmas first.
2) Some problems are not clear how to find the best solution, and may need help from others or check on other reference books.
3) Some problems has deep association with other fields, and may provide motivation for your future study.
In the end, no matter which book you use, the goal is if you encounter a mathematical phenomenon you will immediately know what kind of structure may associated with it. For example, if someone is talking about finite groups acting on a vector space or a set, you will be thinking how the representation can be decomposed like or how the group action's stabilizer is. If you can successfully build up a "personal dictionary" that translates mathematical phenomenon on the one hand and abstract mathematical structure on the other hand, then you achieved your goal in learning algebra. In the end you are going to work on problems not in the textbook, and building up a mathematical structure yourself can be very fulfilling if you find its association with other fields of mathematics.
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Is that an implicit nod to Herstein? – Squid Sep 18 '12 at 6:16
no. I have never read Herstein and did not know Michael or Dummit and Foote very well, though I worked through some exercises in both books. – user32240 Sep 18 '12 at 6:18
My favorite Algebra book is Algebra by Thomas Hungerford. It's a pretty big encyclopedia of algebraic concepts. – EMKA
I would suggest this text if it's your first abstract class. If this is graduate level, I would suggest Advanced Modern Algebra - Joseph J. Rotman.
I've read both, Hungerford starts off with rings then works to fields with most of groups at the end. Honestly, I think he does a better job at conveying rings than Rotman or D&F.
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Rotman's book is not as good as Hungerford, though the later delved hundreds of pages dealing with ring without an identity. You may compare the exercises and see the difference. – user32240 Sep 18 '12 at 6:08
Tom’s Algebra is a graduate text that starts with semigroups and groups and is totally unsuitable. You’re thinking of his undergraduate text, Abstract Algebra: An Introduction, which does indeed start with rings. Unfortunately. But he had in mind an audience that included a lot of prospective high school math teachers, and especially for them he saw a pædagogical advantage in starting with rings. I saw his point, but I still don’t like teaching it that way. – Brian M. Scott Sep 18 '12 at 7:51
Of the three texts that you have mentioned, Dummit and Foote is the only one I have looked at and it is indeed an extremely good entry level book.
I would also recommend A First Course in Abstract Algebra by John B. Fraleigh. This was the first abstract algebra book I encountered and I found it to be an invaluable reference with many useful examples.
As others have said above, the important thing about learning abstract algebra (or any kind of maths for that matter) is doing exercises, rather than just simply reading page after page of theory. This is the way that you learn what is $\textit{actually}$ happening!
- | {"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": 0.9530079960823059, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/89148/inequality-for-the-first-fourier-level-of-a-boolean-function/106670 | Inequality for the first Fourier level of a Boolean function
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
In the study of Boolean functions, the hypercontractive inequality enables one to bound from above the norm of $Tf$ by some norm of $f$, where $T$ is the noise operator depending on the noise parameter. This can be written in terms of the Fourier transform of $f$ as (in a special case of $q=2$):
$\left( \sum\limits_{S \subseteq [n]} (p - 1)^{|S|} \widehat{f}(S)^2 \right)^{1/2} \leq \left( \frac{1}{2^n} \sum\limits_{x \in (0,1)^n} |f(x)|^p \right)^{1/p}$
However, this involves all Fourier levels of $f$. In the application I have in mind, I'm interested only in bounding the norm of the first level of $f$, i.e. restricting the sum on the left to $|S| = 1$. Is it possible to give any inequality of this kind, probably with a different right hand side (but still involving some information about the norm of $f$) and some additional assumptions on $f$? If it's impossible for some trivial reasons, let me know anyway.
Here I'd be mostly interested in matrix-valued Boolean functions (see for example http://arxiv.org/abs/0705.3806), although any answer would be appreciated.
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1 Answer
Here is some general discussion of the level-1 weight of Boolean functions --
http://analysisofbooleanfunctions.org/?p=885
-- including the theorems that:
1. If f is boolean-valued and all of its degree-1 coefficients are small then the weight at level 1 is not more than $2/\pi$;
2. If f is boolean-valued and has very small variance $\alpha$ then the weight at level 1 is at most $2 \alpha \log_2(1/\alpha)$. ["Chang's Lemma", or "Talagrand's Lemma"]
For functions that are not boolean-valued, I don't have a lot to say; the main thing I can suggest is taking $p$ in the hypercontractive inequality as you stated it very close to $1$; if it is, say, $1+\epsilon$ then the LHS will have $\widehat{f}(\emptyset)^2$ (which usually you have information about), plus $\epsilon$ times the weight at level 1, plus at most $\epsilon^2$ times the $2$-norm (neglectable if $\epsilon$ is small enough). So this may allow you to "isolate" the level-1 weight after subtracting $\widehat{f}(\emptyset)^2$ and dividing by $\epsilon$.
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http://mathhelpforum.com/algebra/96248-inequality-check.html | # Thread:
1. ## Inequality check.
Hey folks, I got a problem here and I think that I have a solution, but the wording of the question makes me think I'm short on the problem.
Alice has gone to the hardware store to purchase two different kinds of cord she uses in kite construction. The first type of cord has a mass of $1.50g$ per metre, and the second type has a mass of $2.00g$ per metre. she intends to purchase no more than $6.0g$ in total. If $x$ represents the length of the first type of cord and $y$ represents the length of the second type of cord, shade the region in the coordinate plane that represents the lengths of each type of cord she could purchase. State the inequalities that define this region.
So I started with the inequality I found and manipulated it into a staight line inequality from there. We know that $y$ must be less than or equal to because she wants no more than $6$, right?
So here's what I did.
$1.5x+2y \leq 6$
$2y \leq 6 - 1.5x$
$y \leq -\frac{3}{4}x+3$
So I want to just graph this line and shade underneath it, but is there another inequality I'm missing? The question says "State the inequalities that define this region."
Thanks!
2. You have : $x\geq 0 , y\geq 0$ because they must be positive.
But i'm not sure about the inequality because the condition stated is about mass, whereas x and y are length....i don't know how to do this myself...sorry^^
3. My thinking is that they want $0\leq y \leq -\frac{3}{4}x +3;$ i.e., she needs to buy something?
Should probably get a second opinion, though, because the wording of word problems always trips me up much more than the math.
4. Originally Posted by AlephZero
My thinking is that they want $0\leq y \leq -\frac{3}{4}x +3;$ i.e., she needs to buy something?
Should probably get a second opinion, though, because the wording of word problems always trips me up much more than the math.
You have it right. If you simply graph the linear equation and shade underneath the line, the problem doesn't make sense as it applies to reality. Shading the region bounded by y=0, x=0 and the line y=(-3/4)x+3 will give you the region that corresponds to the problem. However, from the question it looks like she has to buy SOMETHING of each of the chord types so your axis lines should be dashed indicating they are not included in the solution.
5. I think it's not dashed. It's possible if x = 0 because it will give a certain value of y
6. Originally Posted by songoku
I think it's not dashed. It's possible if x = 0 because it will give a certain value of y
Certainly. However that would suggest she only uses one type of chord. Which is valid, but doesn't seem to fit the parameters of the question.
7. Originally Posted by ANDS!
Certainly. However that would suggest she only uses one type of chord. Which is valid, but doesn't seem to fit the parameters of the question.
Craigston's first post is correct.
And yes, it does state that she intends to buy two different cords, but let's be realistic; if you can include a length of cord $x=0.0000000000000001$, you can include 0. Your teacher will not mark you down for this.
Remember, when thinking about what a feasable domain of a function could be, look at all of the possibilities. In other words, it is "feasable" that the lady gets to the store and decides she wants to blow her wad on one cord. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9733982086181641, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/166957/proving-that-an-expression-divides-a-number/166958 | Proving that an expression divides a number
How do you prove that
$$n(n+1)(n+2)$$
is divisible by 6 by using the method of mathematical induction?
According to my book $$\begin{aligned} (n+1)(n+2)(n+3) &= n(n+1)(n+2)+3(n+1)(n+2)\\ &= 6k + 3*2k'\\ &= 6(k+k')\\ &=6k'' \end{aligned}$$
But I wonder, where does that k come from anyway?
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3 Answers
The $k$ comes from the induction hypothesis: you are supposed to prove that $(n+1)(n+2)(n+3)$ is divisible by 6 assuming that $n(n+1)(n+2)$ is divisible by 6.
That's from where you get an integer $k$ for which $$n(n+1)(n+2) = 6k.$$
The $k'$ comes from the observation that $(n+1)(n+2)$ is the product of two subsequent numbers. One of which has to be even, so the product is even.
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Ah, now the k makes sense. The k' still confuses me though. Why to care if a number is even or not? If any number multiplied by 6 instantly becomes divisible by 6 anyway, I think. – Omega Jul 5 '12 at 10:41
The second term in the sum is $3(n+1)(n+2)$. To show that it's divisible by 6 they need to show that $(n+1)(n+2)$ is even and then they substitute $(n+1)(n+2) = 2k'$. – Joni Jul 5 '12 at 10:46
+1 for actually answering the question as asked. Yes, there are easier ways to prove the result without using induction (just observe than one of $n+1$, $n+2$ and $n+3$ must be divisible by 3 and at least one must be even), but that's not what the OP asked about. – Ilmari Karonen Jul 5 '12 at 11:01
It is a bit curious though to use without explanation that $(n+1)(n+2)$ is divisible by $2$ (implicitly checking the possible classes of $n$ modulo $2$), and to not use the very similar argument (checking the possible classes of $n$ modulo $6$) to conclude that $n(n+1)(n+2)$ is divisible by $6$ – Marc van Leeuwen Jul 5 '12 at 11:09
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@MarcvanLeeuwen: mostly a tribute to how textbooks aren't very good at coming up with good examples of induction, I suspect :) – Ben Millwood Jul 5 '12 at 11:45
Since one number out of every $2$ consecutive integers is divisible by $2$ and one out of every $3$ consecutive integers is divisible by $3$, therefore, in the product of $3$ consecutive integers $n(n+1)(n+2)$, atleast one number is divisible by $2$ and one divisible by $3$ and $\gcd(2,3)=1$ $\implies n(n+1)(n+2)$ is divisible by $2\times3=6$. You can also do it this way: Since every integer is one of the forms: $6k,6k+1,6k+2,6k+3,6k+4$ or $6k+5$, therefore, possibilities the product of three consecutive numbers is : $$6k(6k+1)(6k+2)=6(k(6k+1)(6k+2))$$ $$(6k+1)(6k+2)(6k+3)=(6k+1)2(3k+1)3(2k+1)=6((6k+1)(3k+1)(2k+1))$$ $$(6k+2)(6k+3)(6k+4)=2(3k+1)3(2k+1)(6k+4)=6((3k+1)(2k+1)(6k+4))$$ $$(6k+3)(6k+4)(6k+5)=3(2k+1)2(3k+2)(6k+5)=6((2k+1)(3k+2)(6k+5)).$$ Thus product of every possible combination of $3$ consecutive integers is coming out to be divisible by $6$.
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I tend to think it unwise to denote multiplication as you did in your first paragraph $2.3=6$. To those of us for whom $.$ is a decimal separator, it looks like a very strange statement indeed... perhaps `\times` is better? – Ben Millwood Jul 5 '12 at 11:47
@Ben Millwood: Thanks for the suggestion. better now?? – Avatar Jul 5 '12 at 11:49
Yep, I'm happy :) – Ben Millwood Jul 5 '12 at 11:49
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:):):) me too.. – Avatar Jul 5 '12 at 11:50
f(n)=n(n+1)(n+2)
Let f(n) is divisible by 6 be true for n=m.
f(m+1) = (m+1)(m+2)(m+3)=m(m+1)(m+2) + 3(m+1)(m+2) =f(m) + 3(m+1)(m+2). Now, product of two consecutive number is always even. So, the proposition is true for n = m+1 , if it is true for n=m.
By k means there exists an integer k such that n(n+1)(n+2)/6.
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http://mathoverflow.net/questions/81772?sort=oldest | ## How are these number-theoretical constants actually distributed?
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I'm very curious about this and would be really grateful for any help or comments in this direction. If we consider any of the following number-theoretical constants:
1)The various singular series arising from any given system $\Psi: \mathbb{Z}^{d}\rightarrow \mathbb{Z}^t$,$d,t \geq 1$ in the Green-Tao paper "Linear Equations in Primes" (http://arxiv.org/PS_cache/math/pdf/0606/0606088v2.pdf):
$$\prod_{p}\beta_p.$$
2)The Hardy-Littlewood/Bateman-Horn constants arising in the Hardy-Littlewood Conjecture F/ general Bateman-Horn conjecture, say for $f\in \mathbb{Z}[x]$ irreducible and $n_{p,f}$ the number of solutions to $f(n) \equiv 0 \bmod p$ in $\mathbb{Z}/p\mathbb{Z}$, the constant $$C(f) = \prod_{p}\left(\frac{p-n_{p,f}}{p-1}\right),$$ and also the case (also covered by Bateman-Horn) where there is more than one polynomial (thus including the Hardy-Littlewood $k$-tuple conjecture)
3)The analogous constants that would arise naturally from maybe even a combination of (1) and (2) (Is this possible? I think the last slide of http://www.dpmms.cam.ac.uk/~bjg23/papers/icm-handout.pdf already talks about this.), so that one asks for systems $\Psi:\mathbb{Z}^d \rightarrow \mathbb{Z}^t$, $d,t\geq 1$, $\Psi = (\psi_1,\dots,\psi_t)$, where this time the $\psi_i$ need not be linear(!) (But here I'm only asking about the constants, not the actual conjecture.)
The question is, is there an easy way to see whether these constants are dense on any part of the real line or not? If I give you a fixed positive real number, and say you can pick any system from (3) that you want, or even something more general than (3), how would you go about systematically picking the system so that its constant is $\epsilon$-close to the given real number? Thank you very much.
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## 2 Answers
These are Euler products which are convergent (though only conditionally, usually), and such Euler products quite often have a limiting distribution when taken in "reasonable" families. Once something like this is proved, one can conclude that the values are dense in the support of the limiting measure. These limit measures are most of the time characterized by the fact that their moments (or their Fourier transform) is itself an Euler product, the factors of which correspond to the limiting distribution of the $p$-component of the original family of constants. One can then hope to compute the support of the corresponding measure.
The classical example of doing this is for the values of the Riemann zeta-function, which goes back to Bohr and Jessen. It is treated in Chapter XI of Titchmarsh's book (and in that case, one can even deal with divergent Euler products). Another one, probably more directly related, is the study of the distribution of $L(1,\chi)$ for Dirichlet character $\chi$'s. There's been a lot of woork here, in particular recently of Granville and Soundararajan. A good survey is www.math.uiuc.edu/~lamzouri/DistribL2.pdf
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Thank you very much! This is really interesting. – Timothy Foo Nov 24 2011 at 9:23
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This paper by Kowalski also looks highly relevant: http://arxiv.org/abs/0805.4682?context=math
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http://mathoverflow.net/questions/31577?sort=votes | Decision problem restricted to inputs that satisfy some necessary condition.
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Consider the following decision problem:
Problem 1
INPUT: A graph G.
OUTPUT: YES if G is 3-colorable, NO if not.
This is a well-known NP-complete problem. Now suppose that we have a necessary (but not sufficient) condition for a graph to be 3-colorable, called NC. Consider the following problem:
Problem 2
INPUT: A graph G that satisfies NC.
OUTPUT: YES if G is 3-colorable, NO if not.
Now suppose that it is not known whether NC can be determined in polynomial time. Can we say that Problem 2 is in NP? It seems to me that it should be, seeing as there is a succinct certificate for a YES answer.
(However, I've been told by someone who I trust on other matters that this is a "promise problem" and not in NP, which is why I'm posting it here.)
Update
I'm not entirely happy with the answers below (although the subsequent discussions in the comments were useful), so I will attempt to answer the question myself.
Consider:
Problem 3
INPUT: A planar graph G.
OUTPUT: YES if G is 3-colorable, NO if not.
Now, by the usual definitions, an NP problem is one where, from the set of all binary strings, certain ones (those in the "language") must be recognised. Now, it is not easy to come up with a way to represent planar graphs such that every binary string corresponds to a planar graph. So by convention, Problem 3 means the following:
Given an input string, determine if it represents a planar graph that is 3-colorable.
So by the same convention, Problem 2 defines an NP problem: Given an input string, determine if it represents a graph that satisfies NC that is 3-colorable.
Now, the strings that should give "yes" answers are exactly the same as those that should give "yes" answers to Problem 1. So as NP problems, Problems 1 and 2 are exactly the same. (In other words, the languages they define are the same.)
So to make Problem 2 useful, we need to cast it as a Promise problem, where the input is not all binary strings, but is restricted in some way. Posed in this form, it is not an NP problem.
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You cannot say that problem $2$ is in NP, because the condition NC might me "is $3$-colorable"! – Mariano Suárez-Alvarez Jul 12 2010 at 16:48
this is indeed a promise problem. These are common in approximation lower bounds, where you supply a set of inputs that are promised to have either a large or small value for some function, and the problem is to separate them. – Suresh Venkat Jul 12 2010 at 16:50
or, to satisfy your "(but not sufficient)" let NC be "is $3$-colorable or isomorphic to the complete graph in three vertices". – Mariano Suárez-Alvarez Jul 12 2010 at 16:51
Mariano: Yes the NC might be that. In this case Problem 2 is trivial, and so in NP. – Emil Jul 12 2010 at 16:51
Just to clarify, my comment was in respone to Mariano's first comment. – Emil Jul 12 2010 at 16:52
show 5 more comments
2 Answers
Your Problem 2 is indeed a promise problem. By definition, it is not in NP, because NP is a class of decision problems, not promise problems. But if you like, you can say that it's in a class Promise-NP. See Wikipedia or Oded Goldreich's survey On Promise Problems.
For a decision problem, all strings are either YES instances or NO instances. For a promise problem, some strings are allowed to be "invalid": neither YES nor NO.
In general, if you have a nondeterministic algorithm for a promise problem, then when you feed it inputs that don't satisfy the promise, it will either accept or not — so the the sets of strings accepted and not accepted by your predicate will be supersets of the actual YES and NO instances respectively (and have intersection with the "invalid" instances). In this particular problem, because your condition is a necessary one for 3-colorability, the set of accepted strings will be exactly the set of 3-colorable graphs, but its complement will include graphs that don't satisfy the condition. (If you like, you can artificially change the problem to a decision problem with the same set of YES instances, but then your Problem 2 becomes the same as Problem 1 and therefore in NP.)
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Thanks for answering, but you haven't given any specific reason why Problem 2 is not in NP. Would you be able to explain why you think it is not? – Emil Jul 12 2010 at 18:10
Response to edit: Remember that NC is a necessary condition - in other words, all 3-colorable graphs satisfy NC. For Problem 2, the succinct certificate is an explicit 3-coloring, and this certificate also guarantees that NC is satisfied. (I am not sure the survey paper discusses this situation - could you give me a paragraph reference if it does?) – Emil Jul 12 2010 at 19:01
I'm still confused by this myself (and sadly, I don't have a postscript reader handy to check out the reference). Wikipedia explicitly says 'There may be inputs which are neither yes or no. If such an input is given to an algorithm for solving a promise problem, the algorithm is allowed to output anything.' PlanetMath says something similar; it seems like the promise version of a problem can never be more complex than the non-promise version (just ignore the promise!). I'd prefer to say that his problem 2 is surely in NP but not at all guaranteed to be NP-complete. – Steven Stadnicki Jul 12 2010 at 19:30
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Oh sorry, I didn't realise that your condition is one necessary for 3-colorability. Technically, NP is a class of *languages*: sets of strings. What is the set of YES instances for your problem 2? It seems the only reasonable definition would be simply the set of 3-colorable graphs, in which case yes, your problem 2 is in NP, because (as a language) it's the same as Problem 1. – shreevatsa Jul 12 2010 at 20:00
@shreevatsa: could you edit your answer then? – Emil Jul 12 2010 at 20:34
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A promise problem cannot be in NP, just because NP is defined to be a set of languages (or decision problems). It's like asking if the problem "Given n, output 2n" is in P. It's clearly an easy problem, and has a linear time solution, but it cannot be in P as stated because P is a set of decision problems, and the given problem is not a decision problem.
Your problem is in Promise-NP, since it's a promise problem with an efficiently verifiable certificate. See the wikipedia article on promise problems for some more information. Whether NC is a sufficient condition or a necessary condition or a completely arbitrary condition has nothing to do with the problem belonging to Promise-NP. As long as NC is a non-trivial condition which makes this a promise problem, the problem belongs to Promise-NP.
EDIT 1: I thought I should edit this to better answer Emil's question: I just want to know if Problem 2 is in NP. If you think it is not in NP, please could you explain why? It seems to me that "yes" answers do have succinct certificates.
NP is not the set of all things in the universe with succinct certificates! It is the set of all languages that have succinct certificates. Your problem does not define a language. It defines a promise problem. Therefore it cannot be in NP, not because it does not have a short certificate, but because it is not a language at all.
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See comments after shreevatsa's answer. – Emil Jul 12 2010 at 20:33
I answered the comments after Shreevatsa's answer. The fact that NP is defined using decision problems and not promise problems is just a matter of convention. We could equally well have started complexity theory with only promise problems. There is nothing deep here, it's just a matter of convention and definitions. – Rune Jul 12 2010 at 20:57
I'm not saying there is anything deep here. I just want to know if Problem 2 is in NP. If you think it is not in NP, please could you explain why? It seems to me that "yes" answers do have succinct certificates. – Emil Jul 12 2010 at 21:10
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I've added an answer to your question in my original answer. How's that? – Rune Jul 12 2010 at 23:16
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The decision problem associated with a language $L\subseteq$ {0,1}<sup>*</sup> is the following: Given x in L, output yes. Given x not in L (i.e., given x in the complement of L), output NO. If L is the language you claim -- the set of all strings representing graphs that satisfy NC and are 3 colorable, then the complement of L contains all graphs that do not satisfy NC or are not 3-colorable. Thus the output of the algorithm on such graphs will be NO. This is not what you mean by your problem. You want the output to be NO when the graph satisfies NC but is not 3-colorable. – Rune Jul 13 2010 at 3:14
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http://mathoverflow.net/questions/55515/additive-functions-on-a-lattice | ## additive functions on a lattice
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Given a lattice $L$. Can one classify all functions $f:L\rightarrow \mathbb{R}$, that satisfy
$f(a \wedge b)+f(a\vee b) = f(a)+f(b)$.
Some examples are the the set of all finite subsets of a given set $S$. Then every such function is uniquely determined by the element $(f(\{s\}))_{s\in S}\in \prod_S\mathbb{R}$ plus the value on the empty set. Indeed this gives a vector space isomorphism from the set of all such functions to $\mathbb{R}^{|S|+1}$.
For other lattices there are also other additive functions arising naturally. For example if one considers the set of all natural numbers (without $0$) ordered by divisibility and assigns to a natural number (for a fixed prime $p$) the biggest $n$, so that $p^n$ divides this number. The logarithm is another example for such a function.
More general a ultrafilter on the underlying poset can also be viewed as such a function. So there are a lot of interesting examples. There are even more interesting examples like lattices of measurable sets and their measures, Euler-characteristic of subcomplexes and so on. So my question is: Can one classify the set of all those functions, probably in terms of filters / ultrafilters on the underlying poset?
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One example to keep in mind is the case when $L$ is a total order, e.g., $\mathbf{Z}$ with join and meet given by max and min. Then your condition on $f$ is empty, i.e. any function $f\colon \mathbf{Z} \to \mathbf{R}$ is additive. – Guntram Feb 15 2011 at 16:21
One characterization of modular lattices is that a graded lattice is modular if and only if its rank function is additive in your sense. Both of your examples arise in this way: the boolean lattice and $\mathbf N$ under divisibility are both modular, and the additive functions you have are obtained by altering the values of the rank function on the atoms of the lattice. This might give you some more interesting examples. – Dan Petersen Feb 16 2011 at 7:54
## 2 Answers
An additive function with the additional property that $f(a) < f(b)$ whenever $a < b$ (i.e., strict monotonicity) can exist only when the lattice is modular. A partial converse was mentioned in a comment above: If the lattice is modular and has finite height, so that it has an integer-valued rank function, then this rank function is additive. But modular lattices of infinite height need not support any strictly monotone additive function. Any additive function on a non-modular lattice factors through the projection to a modular quotient lattice. I believe all this information is in Birkhoff's classic book "Lattice Theory".
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Update: My answer is incorrect. See comments below.
This answer may be obvious, and it probably isn't what you were looking for, but I realized that for any lattice $L$, there is a subset $M$, in which any $f:M\rightarrow \mathbb{R}$ extends uniquely to an additive $f$ on $L$.
I can't think of any great characterization of this set, except to just pick it using the axiom of choice, i.e. well-order $L$ and build both the set $M$ of generators and $N$ of determined points. For each $x \in L$, add it to $M$ if $M\cup N$ doesn't contain a three elements of $a,b,a\wedge b,a \vee b$ where $x$ is the fourth. Else the value $f(x)$ of any additive $f$ is determined by the values of $f$ on the elements in $M\cup N$, which are in turn determined by the values on $M$. So add $x$ to $N$.
In the linear case, $M$ is just all of $L$. In the case of a finite Boolean algebra with atoms, then $M$ can be set of the atoms plus the bottom element.
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I don't buy it yet. For example if I consider a lattice having a smallest element 0, then 3 noncomparable elements a,b,c and a largest element d. It is a lattice. My well ordering is $a,b,c,0, d$. Then the set $M$ should be $\\{a,b,c,0\\}$. But not any function $M\rightarrow \mathbb{R}$ extends to a additive function. Note that every additive function on that lattice sends $a,b,c$ to the same value. – HenrikRüping Feb 16 2011 at 10:56
But the weaker claim does seem to hold, that every additive function is determined by its values on $M$. – Joel David Hamkins Feb 16 2011 at 11:44
@Hendrik - Yes, I seemed to have neglected the possibility of over-specification. Worst, consider the lattice $0,1,a,b,c,d$ where $a,b,c,d$ are incompatible and $0,1$ are the smallest, largest elements. This has no subset $M$ on which every $f:M \rightarrow R$ extends uniquely to an additive $f$ on the lattice. – Jason Rute Feb 16 2011 at 14:34
Actually, my above comment is incorrect as well. Knowing $a,0$ determines $b,c,d,1$. We must have $a=b=c=d$, and then $0+1=2a$. So maybe there is something salvageable about my argument. – Jason Rute Feb 16 2011 at 14:48
OK, I have it for the finite case at least. Your problem can be reduced to a system of linear equations with one variable for each node $a$ representing the value $f(a)$. Then, if $L$ is finite, so is the system, and we can pick a (possibly empty) set of variables, with corresponding nodes $a_1,\ldots,a_n$, that are free and the rest are determined from them. Hence if $M= \\{ a_1,\ldots,a_n \\}$, then any $f:M\rightarrow \mathbb{R}$ extends uniquely to an additive function $f$ on $L$. I don't know if this property is true of infinite systems of finite linear equations. – Jason Rute Feb 16 2011 at 17:05 | {"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": 64, "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": 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": 0.9254701137542725, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/20272/errata-for-emil-artins-the-gamma-function/20305 | ## Errata for Emil Artin’s ‘The Gamma Function’?
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In the English translation of The Gamma Function by Emil Artin (1964 - Holt, Rinehart and Winston) there appears to be a mistake in the formula given for the gamma function on page 24:
$$\Gamma(x) = \sqrt{2\pi}x^{x-1/2}e^{-x+\mu(x)}$$ $$\mu(x)=\sum_{n=0}^\infty(x+n+\frac{1}{2})\text{log}(1+\frac{1}{x+n})-1=\frac{\theta}{12x},\ \ \ \ \ 0 < \theta < 1$$
and on page 22 where this is derived, it is noted that '$\theta$ is a number independent of $x$ between 0 and 1'.
This sounds incorrect, as $\theta$ does depend on $x$, but since the wording is a little ambiguous it may just be an unclear translation. The original German might have meant that $0< \theta(x) < 1$ for any $x$. That the variable $x$ is suppressed from $\theta$ could be just confusing notation, or someone's misunderstanding (possibly mine.)
The preface does mention that a (different) formula had to be corrected for the English reprint.
I would like to know if there are mistakes in this book, and if so, whether they exist in the German edition. Is there an available list of errata?
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## 1 Answer
It seems clear that `$\theta$` can indeed be chosen to be a number independent of `$x$` as stated, to get Stirling's formulas for the gamma function when `$x$` is large. The wording, at least in English, is not too helpful in this section. But I'm less clear about where in the formula on page 24 there is supposed to be a mistake. Here as in any mathematics book (especially a translation) one has to be wary about misprints or errors. Probably there is no publicly available list of errata for this small monograph published originally in 1931 in German and later republished in 1964 in an English translation by Michael Butler. This English version is included in the 2007 AMS softcover book Exposition by Emil Artin: A Selection edited by Michael Rosen. (There is an older 1965 book The Collected Papers of Emil Artin published by Addison-Wesley and edited by Lang & Tate. This contains Artin's research papers, in the original German or English.) As Zavosh observes, the 1964 preface by Edwin Hewitt reprinted here does indicate one formula corrected in the translation: " ... a small error following formula (59) (this edition) was corrected..." However, the formula seems to be the one actually numbered (5.9). Caveat lector.
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I think $\theta$ actually converges to 1 quickly as $x$ grows large. For just an asymptotic formula for $\Gamma(x)$ (as in the common version of Stirling's formula), there would be no reason to mention $\mu(x)$ or $\theta$ at all, since the $e^{-\mu(x)}$ term converges to 1. It sounds like the author is stating on page 24 an exact formula involving a constant $\theta$, when it's actually not a constant. I suspect it's a mistake of the translator originating from a misunderstanding on page 22. – Zavosh Apr 4 2010 at 16:35
I looked at the AMS 2007 version and it's exactly the same. I will mark your answer as accepted soon, unless someone else comes up with a miraculously clarifying answer (which is unlikely.) Thanks very much. – Zavosh Apr 4 2010 at 16:38
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Since this section of Artin's book is concerned with approximating the gamma function, there may be some tendency to use the equals sign loosely. I guess the point is to find a convenient elementary function giving a good approximation for large `$x$`; the choice might be fine-tuned in various ways. But in the era before computers the shape of an approximating function would have been the most interesting question for many people. – Jim Humphreys Apr 4 2010 at 17:29
That makes sense. Thank you. – Zavosh Apr 4 2010 at 18:10
P.S. At the moment the AMS book mentioned earlier is on sale online: Exposition by Emil Artin: A Selection - Michael Rosen, Brown University, Editor - AMS | LMS, 2006, 346 pp., Softcover, ISBN-10: 0-8218-4172-6, ISBN-13: 978-0-8218-4172-3, List: US$59, All AMS Members: US$47, Sale Price: US\$38, HMATH/30 – Jim Humphreys May 9 2010 at 14:11 | {"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": 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": 0.9433630108833313, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2011/02/17/kostka-numbers/?like=1&source=post_flair&_wpnonce=c1ce40f6e1 | # The Unapologetic Mathematician
## Kostka Numbers
Now we’ve finished our proof that the intertwinors $\bar{\theta}_T$ coming from semistandard tableauxspan the space of all intertwinors from the Specht module $S^\lambda$ to the Young tabloid module $M^\mu$. We also know that they’re linearly independent, and so they form a basis of the space of intertwinors — one for each semistandard generalized tableau.
Since the Specht modules are irreducible, we know that the dimension of this space is the multiplicity of $S^\lambda$ in $M^\mu$. And the dimension, of course, is the number of basis elements, which is the number of semistandard generalized tableaux of shape $\lambda$ and content $\mu$. This number we call the “Kostka number” $K_{\lambda\mu}$. We’ve seen that there is a decomposition
$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}m_{\lambda\mu}S^\lambda$
Now we know that the Kostka numbers give these multiplicities, so we can write
$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}K_{\lambda\mu}S^\lambda$
We saw before that when $\lambda=\mu$, the multiplicity is one. In terms of the Kostka numbers, this tells us that $K_{\mu\mu}=1$. Is this true? Well, the only way to fit $\mu_1$ entries with value $1$, $\mu_2$ with value $2$, and so on into a semistandard tableau of shape $\mu$ is to put all the $i$ entries on the $i$th row.
In fact, we can extend the direct sum by removing the restriction on $\lambda$:
$\displaystyle M^\mu=\bigoplus\limits_\lambda K_{\lambda\mu}S^\lambda$
This is because when $\lambda\triangleleft\mu$ we have $K_{\lambda\mu}=0$. Indeed, we must eventually have $\lambda_1+\dots+\lambda_i<\mu_1+\dots+\mu_i$, and so we can't fit all the entries with values $1$ through $i$ on the first $i$ rows of $\lambda$. We must at the very least have a repeated entry in some column, if not a descent. There are thus no semistandard generalized tableaux with shape $\lambda$ and content $\mu$ in this case.
## 2 Comments »
1. [...] First let’s mention a few more general results about Kostka numbers. [...]
Pingback by | February 18, 2011 | Reply
2. Your math is far beyond my comprehension, but I love reading this blog daily.
Comment by | February 19, 2011 | Reply
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"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": 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": 0.9026748538017273, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like/112 | ## What do epimorphisms of (commutative) rings look like?
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(Background: In any category, an epimorphism is a morphism $f:X\to Y$ which is "surjective" in the following sense: for any two morphisms $g,h:Y\to Z$, if $g\circ f=h\circ f$, then $g=h$. Roughly, "any two functions on $Y$ that agree on the image of $X$ must agree." Even in categories where you have underlying sets, epimorphisms are not the same as surjections; for example, in the category of Hausdorff topological spaces, $f$ is an epimorphism if its image is dense.)
What do epimorphisms of (say commutative) rings look like? It's easy to verify that for any ideal $I$ in a ring $A$, the quotient map $A\to A/I$ is an epimorphism. It's also not hard to see that if $S\subset A$ is a multiplicative subset, then the localization $A\to S^{-1}A$ is an epimorphism. Here's a proof to whet your appetite.
If $g,h:S^{-1}A\to B$ are two homomorphisms that agree on $A$, then for any element $s^{-1}a\in S^{-1}A$, we have
$$g(s^{-1}a)=g(s)^{-1}g(a)=h(s)^{-1}h(a)=h(s^{-1}a)$$
Also, if $A\to B_i$ is a finite collection of epimorphisms, where the $B_i$ have disjoint support as $A$-modules, then $A\to\prod B_i$ is an epimorphism.
Is every epimorphism of rings some product of combinations of quotients and localizations? To put it another way, suppose $f: A\to B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must $f$ be an isomorphism?
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Upvoting for providing background. Let's all try to do this more! – Scott Morrison♦ Oct 6 2009 at 6:19
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Seconded! (And some more characters to take me over the minimum) – Andrew Stacey Oct 6 2009 at 6:50
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@Andrew: you can vote up a comment (click the little up-arrow that appears to the left of the comment when you mouse-over it). It doesn't generate any reputation, but it highlights good comments. – Anton Geraschenko♦ Oct 6 2009 at 19:02
Are you fishing for an up-vote for your answer! In fact I have up-voted both the question and your answer, did they not register? (My comment above was left before your answer, by the way.) – Andrew Stacey Oct 7 2009 at 6:59
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@Anton: incidentally, I think that in $\mathbf{Top}$, epis do lie over surjective functions (see en.wikipedia.org/wiki/Epimorphism). Another way to see this is to note that the forgetful functor $\mathbf{Top}\to\mathbf{Set}$ has a right adjoint (indiscrete topology), and therefore preserves colimits, and in particular pushouts (see also Ex. 4, p. 72 of Mac Lane) – unknown (google) May 10 2010 at 14:28
show 5 more comments
## 4 Answers
No, not every epimorphism of rings is a composition of localizations and surjections.
An epimorphism of commutative rings is the same thing as a monomorphism of affine schemes. Monomorphisms are not only embeddings, e.g., any localization is an epimorphism and the corresponding morphism of schemes is not a locally closed embedding.
Example: Let C be the nodal affine cubic and let X be its normalization. Pick any point x above the node. Then X\{*x*}->C is a monomorphism (see Proposition below). The corresponding homomorphism of rings is injective but not a localization.
Proposition (EGA IV 17.2.6): Let f:X->Y be a morphism locally of finite type between schemes. TFAE:
(i) f is a monomorphism.
(ii) Every fiber of f is either an isomorphism or empty.
Remark: A flat epimorphism A->B is a localization if A is normal and Q-factorial. This is a result by D. Lazard and P. Samuel. [cf. Lazard "Autour de la platitude" (IV, Prop 4.5)]
Remark: There was a seminar on epimorphisms of rings directed by P. Samuel in 1967-68.
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+1 beautiful example, as always. Thanks for the references. – Anton Geraschenko♦ Oct 7 2009 at 1:32
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If there were a book, "Counterexamples in Algebraic Geometry," much like the existing books "Counterexamples in Analysis" or "Counterexamples in Topology," I think the normalization of a nodal curve would have to be in the top 5. – Jack Huizenga Apr 16 2011 at 18:53
This seminar is the definite answer for all questions about epimorphisms ... – Martin Brandenburg Jul 5 at 7:16
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George Bergman gave me a reference (Isbell's "Epimorphisms and dominions, IV") and a very pretty counterexample. In particular, he says that the characterization of epimorphisms Andrew gave us works for non-commutative rings as well:
Recall that an inclusion A in B is an epimorphism if and only if the "dominion" of A in B is all of B, where this dominion is defined as the subring of elements b of B which behave the same under all pairs of homomorphisms on B that agree on elements of A.
Now the Silver-Mazet-Isbell Zigzag Lemma for rings says that the dominion of A in B consists of those elements of B which can be written XYZ, where X is a row, Y a matrix, and Z a column over B, such that XY and YZ have entries in A. (It is easy to verify that such a product is in the dominion of A -- a generalization of the proof that if Y is in A and has an inverse in B, then this inverse is in the dominion of A.)
Let k be a field. Consider the inclusion of k[x, xy, xy2 - y] into k[x,y]. I claim that this is an epimorphism. Note that it is an inclusion, no non-units become units, and k[x,y] has no idempotents.
Suppose f and g are two morphisms from k[x,y] to some other commutative ring which agree on the given subring. Using f(xy)=g(xy) and f(x)=g(x), we see that f(xy2)=g(xy2):
f(yxy) = f(yx)f(y) = g(yx)f(y) = g(y)g(x)f(y) = g(y)f(x)f(y) = g(y)f(xy) = g(y)g(xy) = g(yxy)
Since f and g agree on xy2-y, they agree on y, so they agree on all of k[x,y].
Finally, to see that the inclusion is not an isomorphism, consider the surjective morphism k[x,y] to k[x,x-1] sending y to x-1. This sends the subring to k[x], which is clearly smaller, so the inclusion of k[x,xy,xy2-y] into k[x,y] must be strict.
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That's great! That's also a counter-example to my conjecture above which settles that question as well. I was sure that there was a simple counter-example like that but couldn't quite see it ... – Andrew Stacey Oct 6 2009 at 18:31
A little searching turned up:
Ring epimorphisms and C(X) by Michael Barr, W.D. Burgess and R. Raphael (article).
They consider this question for rings of the form of continuous functions on a topological space. They quote the following characterisation of epimorphisms in the category of commutative rings:
Proposition: A homomorphism f : A → B is an epimorphism if and only if for all b ∈ B there exist matrices C, D, E of sizes 1 × n, n × n, and n × 1 respectively, where (i) C and E have entries in B, (ii) D has entries in f(A), (iii) the entries of CD and of DE are elements of f(A) and (iv) b = CDE. (Such a triple is called a zig-zag for b.)
This seems a little more complicated than localisation, though I haven't checked the details.
They then go on to prove that
2.12: A subspace Y of a perfectly normal first countable space X induces an epimorphism if and only if it is locally closed.
If I understand all the terminology correctly, then this implies that
C([0,1],ℝ) → C((0,1),ℝ)
is an epimorphism.
There are plenty more references in that article, and it would be nice to have an actual zig-zag for this situation. But in the spirit of open-source mathematics, I thought I'd post this and see if someone (possibly me later on) can fill in the details.
Added Later: The example I gave: C([0,1],ℝ) → C((0,1),ℝ) is a localisation. It is obtained by inverting all functions in C([0,1],ℝ) which are zero only at the end-points. Given a function f ∈ C((0,1),ℝ), there will be a function g ∈ C([0,1],ℝ) which is non-zero apart from at 0 and 1 and which goes to 0 at 0 and 1 faster enough that the product g f also goes to 0 at the end-points. Then g f is (the restriction of something in) C([0,1],ℝ) and g becomes invertible in C((0,1),ℝ). So f = g-1 (g f) is in the specified localisation of C([0,1],ℝ).
Indeed, the Barr et. al. paper comments on the fact that in all the examples they consider (function rings), the zig-zag has length 1. I conjecture that if the zig-zags always have length 1 (for a particular function f: A → B), then B is formed by a localisation on A. A possibly stronger version of this conjecture would be that this is an if-and-only-if. In which case, finding a counter-example to Anton's conjecture would involve finding a case where there was a zig-zag of length 2. I suspect that a universal construction would be the best approach to finding one.
In the spirit of wiki-ness and only doing a little at a time, I'll leave this here.
Added Even Later: (Should I timestamp these? I know that the system does so, but is it useful to embed them in the edit?)
Here's one direction for my conjecture above.
If B = S-1A, then for b ∈ B, we have b = s-1a for some s ∈ S and a ∈ A. Then we put C = s-1, D = s, E = b = s-1 a. Then CD = 1, DE = a, D ∈ f(A), and CDE = b. So in a localisation, zig-zags have length 1.
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Condition (iv) used to say "b = CDE". I fixed it; let me know if I screwed up. – Scott Morrison♦ Oct 6 2009 at 6:20
Yup. There was a mistake, but it was earlier: there should have been a "for each b ∈ B" earlier on. But thanks to your attempted fix, I realised that there was a genuine error so it was still worth doing! – Andrew Stacey Oct 6 2009 at 6:49
@Andrew: no need to timestamp. If somebody wants to see the edit history, they can click the "edited X minutes/hours ago" link. – Anton Geraschenko♦ Oct 6 2009 at 14:27
Here is another perspective on your question. As $\mathbb{Z}$ is the initial object of unital (commutative) rings, one might first of all ask: What do epimorphisms from $\mathbb{Z}$ look like?
I.e. if $A = \mathbb{Z}$ in the original question, what can $B$ be? The answer to this is known. In fact, these rings $B$ and their classification seem to have been (re)invented several times, as "solid rings" by Bousfield and Kan (see this question: http://mathoverflow.net/questions/95160/), as "T-rings" by R. A. Bowshell and P. Schultz (Unital rings whose additive endomorphisms commute, Math. Ann. 228 (1977), 197-214, http://eudml.org/doc/162991;jsessionid=07C5F5F5BBD354C0914511776DA20F5E), and the generalisation to Dedekind domains has been done in W. Dicks and W. Stephenson: Epimorphs and Dominions of Dedekind Domains, J. London Math. Soc. (1984) s2-29(2): 224-228, http://jlms.oxfordjournals.org/content/s2-29/2/224.extract . (Also, by Martin Brandenburg and myself this summer, before we found these papers ...)
So here is a positive answer under a restrictive assumption: If $A \rightarrow B$ is an epimorphism and $A$ is a Dedekind domain, then $B$ will be built up from localisations and quotients of $A$ by suitable finite products and direct limits. To make "suitable" more specific, here follows a more concrete description (the literature above mostly says "take colimits/pullbacks"), setting $A = \mathbb{Z}$ for (mostly notational) simplicity:
Let $P$ be the set of prime numbers and let $n: P \rightarrow \mathbb{N} \cup \lbrace 0, \infty \rbrace$ be any map (a "supernatural number"). Let $P_{fin}(n)$ be the set of primes $p$ with $n(p) < \infty$. Define
$B_n := \lbrace (b_p, b_l) \in \prod_{p \in P_{fin}(n)} \mathbb{Z} / p^{n(p)} \times \mathbb{Z}[P_{fin}(n)^{-1}] :$ $b_p \equiv b_l \text{ mod } p^{n(p)} \text{ for all but finitely many } p \in P_{fin}(n)(b_l) \rbrace$
(index "$l$" for "localisation part") where $\mathbb{Z}[P_{fin}(n)^{-1}]$ is the localisation of $\mathbb{Z}$ at the multiplicative set generated by $P_{fin}(n)$, i.e. the subring of $\mathbb{Q}$ generated by $\lbrace p^{-1}: p \in P_{fin}(n) \rbrace$; further, with $v_p$ being the $p$-adic valuation, $P_{fin}(n)(b_l) := \lbrace p \in P_{fin}: v_p(b_l) \ge 0 \rbrace$ and the condition $b_p \equiv b_l \text{ mod } p^{n(p)}$ makes sense and is to be understood in the subring of $\mathbb{Q}$ where only the $p$'s with $v_p(b_l) < 0$ are inverted. Then $B_n$ is in fact a subring of the direct product, and for $n$ ranging over the supernatural numbers, these are all $B$ with injective epimorphisms $\mathbb{Z} \rightarrow B$. (The non-injective ones are just the quotients. With more complicated notation, one could include this case by counting 0 as a prime.)
Here are two easy-to-see properties:
• $B_n$ is noetherian if and only if $|P_{fin}(n) \setminus P_0(n) | < \infty$ if and only if $B_n$ is the direct product of a quotient and a localisation, namely, $\mathbb{Z}/n \times \mathbb{Z}[P_{fin}(n)^{-1}]$ where by abuse of notation $n$ is the natural number $\prod_{p \in P_{fin}(n)} p^{n(p)}$.
• The non-zero primes of $B_n$ correspond to the ones in $P \setminus P_0(n)$, where $P_0(n)$ is the set of primes $p$ with $n(p) = 0$. In particular, $B_n$ is artinian if and only if its Krull dimension is 0 if and only if $|P \setminus P_0(n)| < \infty$. Otherwise, its Krull dimension is 1.
All this remains true cum grano salis for any Dedekind domain $A$ instead of $\mathbb{Z}$. In particular, as soon as $A$ has infinitely many primes, there are epimorphisms $A \rightarrow B$ where $B$ is non-noetherian. On the other hand, if $A$ has only finitely many primes (which by the way makes it a PID), $B$ will be of the form $A/a \times S^{-1}A$ with $a \in A$ and $S \subseteq A$ multiplicative containing all primes dividing $a$ (and possibly 0). In any case, $B$ will be a colimit of products of localisations and quotients as above, so the answer to the question
suppose $f:A \rightarrow B$ is an epimorphism of rings with no kernel which sends non-units to non-units and such that $B$ has no idempotents. Must f be an isomorphism?
seems to be yes if $A$ is a Dedekind domain: E.g. in the above setting, non-units to non-units implies $P_0(n) = \emptyset$ and $B$ having no idempotents implies $P_{fin}(n) \setminus P_0(n) = \emptyset$.
Further remarks:
Remark 1 (cf. David Rydh's first remark): Flat epimorphisms (from any unital ring) are localisations for a certain Gabriel topology and have a kind of a calculus of fractions. For a precise statement, see Quelques observations sur les épimorphismes plats (à gauche) d'anneaux by N. Popescu and T. Spircu, Journ. Alg. vol. 16, no. 1, pp. 40-59, 1970, http://dx.doi.org/10.1016/0021-8693(70)90039-6, or Bo Stenström's book Rings of Quotients, theorem 2.1 in chapter XI.
Remark 2: Further information might be in the papers of H. H. Storrer, e.g. http://retro.seals.ch/digbib/view?rid=comahe-002:1973:48::11
Remark 3: I have not checked all the details in the generalisation to Dedekind domains, so beware (at least, Martin and I had reached the same result for PIDs). Also, I do not know if there is a generalisation beyond Dedekind domains; I guess Krull domains might be attackable, but I have not seriously tried.
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There are further nice descriptions/definitions of $B_n$. Let $Q=P_{fin}(n)$. Then: 1) $B_n$ is the colimit of the (noetherian) $R$-algebras $\prod_{p \in E} R/p^{n(p)} \times (P \setminus Q \cup E)^{-1}R$, where $E$ runs through the finite subsets of $Q$. 2) $B_n$ is the tensor product over $(P \setminus Q)^{-1} R$ of the algebras $R/p^{n(p)} \times p^{-1} R$, where $p \in Q$. 3) $B_n=(P \setminus Q)^{-1} R[(x_p)_{p \in Q}]/(x_p(1−p x_p),p^{n(p)}(1−p x_p))_{p \in Q}$. – Martin Brandenburg Jan 31 at 15:03 | {"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": 101, "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": 0.9220624566078186, "perplexity_flag": "middle"} |
http://mathhelpforum.com/advanced-algebra/1128-need-some-help-part.html | Thread:
1. Need some help with part A
2. Let P=(x, y) be a point on the graph of y = x^2 - 8.
a) Express the distance of d from P to the point (0, -1) as a function of x.
b) What is d if x = 0?
c) What is d if x = -1?
d) Use a graphing utility to graph d = d (x).
e) For what values of x is d smallest?
Answers
a) d(x) = √x^4-13x^2+49
b) d(0)=7
c) d(-1)= √37 ... is similar to ... 6.08
d) Graph d(x) = √x^4-13x^2+49
e) D is the smallest when x ... is similar to ... -2.55 or x ... is similar to ... 2.55
I can do all of them but part “A” can someone help me figure out how to find out part A?
and how do you do a "... is similar to ..." sign on here?
2. Originally Posted by Robthebear
2. Let P=(x, y) be a point on the graph of y = x^2 - 8.
a) Express the distance of d from P to the point (0, -1) as a function of x.
[...]
I can do all of them but part “A” can someone help me figure out how to find out part A?
and how do you do a "... is similar to ..." sign on here?
$\sim$ or $\approx$ or simply ~.
As to part a: The distance between points with coordinates (a,b) and (c,d) is
$\sqrt{(a-c)^2 + (b-d)^2}$.
Now in this case, $a = x, b = y = x^2-8, \, c = 0,\, d = -1$.
Insert these expressions in the distance formula and you are done.
3. Thank you for the help.
Robert | {"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": 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": 0.8809381723403931, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/11261/is-the-ultraproduct-concept-fundamentally-category-theoretic/26285 | ## Is the ultraproduct concept fundamentally category-theoretic?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Once again, I would like to take advantage of the large number of knowledgable category theorists on this site for a question I have about category-theoretic aspects of a fundamental logic concept.
My question is whether the ultraproduct construction is fundamentally a category-theoretic concept.
The ultraproduct/ultrapower construction of Łos is used pervasively in logic, particularly in model theory and also in set theory, where nearly all of the larger large cardinal axioms can be formulated in terms of the existence of certain kinds of ultrapowers of the universe.
My question is, is the ultraproduct fundamentally a category-theoretic construction, in the sense that it is characterized by some natural category-theoretic universal property? How about the special case of ultrapowers?
I would be very interested, if there were a natural universal characterization in terms of the usual Hom sets for these first order structures, namely, first order elementary embeddings and/or homomorphisms. (Needless to say, I would be much less interested in a characterization that amounted merely to a translation of the Łos construction or of Łos's theorem into category-theoretic language.)
Background. Suppose we have a collection of structures Mi for i in J, all of the same first order type (e.g. groups, partial orders, graphs, fields, whatever), and U is an ultrafilter on the index set J. This means that U is a nonempty collection of nonempty subsets of J, containing every set or its complement, and closed under intersection and superset. The ultraproduct ΠMi /U consists of equivalence classes [f]U, where f is a function with domain J, with f(i) in Mi, and f ∼Ug iff {i in J | f(i)=g(i)} in U. One imposes structure on the ultraproduct by saying that a relation holds in the product, if it holds on a set in U, and similarly for functions. Łos's theorem then states that the ultraproduct satisfies a first order formula φ([f]u) if and only if {i in J | Mi satisfies φ(f(i))} is in U. That is, truth in the ultraproduct amounts to truth on a U-large set of coordinates. The special case when all Mi are the same model M, we arrive at the ultrapower MJ /U. In this case, there is a natural map from M into MJ /U, defined by x maps to [cx]U, where cx is the constant function with value x. It is easy to see that this map is an elementary embedding from M into the ultrapower.
This question is a more focused instance of a probably-too-general question I asked here, and I may have several more in the future.
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1
I googled "ultraproduct universal property" and got this: dialinf.wordpress.com/2009/01/21/… . Apparently, the answer to your specific question about UMPs is "no." – Qiaochu Yuan Jan 9 2010 at 22:18
Thanks for the link! Barr expresses the opinion there that ultraproducts are not defined by any universal mapping property. But I'm not really sure how one would prove such a thing. And will the category theorists really give up so easily? – Joel David Hamkins Jan 9 2010 at 23:31
4
Since I know there are several ways to do this, I really want a Category Theorist to answer and sort things out for us. Here is a summary of what I know, I will post details later if necessary. Ultraproducts are particular kinds of directed colimits, and it is often useful to describe them as such. Also, the ultraproduct $\prod_{i \in I} X_i/\mathcal{U}$ can be viewed as a stalk of a particular sheaf on $\beta I$. Anyway, I would really like to know more ways of thinking about ultraproducts in a categorical setting. I second this great question! – François G. Dorais♦ Jan 9 2010 at 23:35
## 5 Answers
Andrej's answer was very helpful to me, but there is yet another (perhaps not completely unrelated) category theoretic view of ultraproducts that I am aware of. I am still hopeful that more category theorists will eventually step in and sort things out...
If $X$ is a discrete space then a sheaf $F:O(X)^{\mathrm{op}}\to Set$ must be such that $F(A) \cong \prod_{i \in A} F_i$ for some family of sets $(F_i)_{i \in X}$. This sheaf can be moved to a sheaf `$F':O(\beta X)^{\mathrm{op}}\to Set$`. Viewing $\beta X$ as the space of ultrafilters on $X$, the stalk of $F'$ at a point `$\mathcal{U} \in \beta X$` is precisely the ultraproduct `$\prod_{i \in X} F_i/\mathcal{U}$`. (There is one subtle difference which occurs when some of the components $F_i$ are empty, in which case this ultraproduct can still be nonempty when $F(A)$ is nonempty for some $A \in \mathcal{U}$.)
From a more global point of view, the embedding $X \to \beta X$ induces a geometric morphism $Sh(X) \to Sh(\beta X)$. Similarly, a point of $\beta X$ can be identified with geometric morphism $\mathcal{U}:Set \to Sh(\beta X)$. The corresponding ultraproduct map is simply the composition $$Sh(X) \to Sh(\beta X) \xrightarrow{\mathcal{U}^*} Set,$$ where the last component is the inverse image part of $\mathcal{U}$. The corresponding ultrapower functor is the composite `$$Set \xrightarrow{\Delta} Sh(X) \to Sh(\beta X) \xrightarrow{\mathcal{U}^*} Set,$$` where $\Delta$ is the diagonal functor.
Of course, there is nothing very special about discrete spaces in the above construction. The same construction exists for any completely regular Hausdorff space $X$ or, more generally, for a completely regular locale. (This makes sense even when the space/locale $X$ is not completely regular, but the map $X \to \beta X$ is not necessarily an embedding.) Of course, Łoś's Theorem takes a different form for this more general construction, the correct form of the theorem for a space/locale X can be found via the Kripke-Joyal semantics, for example.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This paper comes to mind:
Ultrasheaves and double negation. S. Awodey and J. Eliasson, Notre Dame Journal of Formal Logic 45(4), pp. 235--245 (2004). Available at http://www.andrew.cmu.edu/user/awodey/preprints/udn.pdf
Perhaps this is not quite what you are asking for, because the paper takes utrafilters as given, but it certainly gives a useful category-theoretic perspective.
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3
Perhaps I should have mentioned the "obvious" fact that the ultraproduct construction has a topos-theoretic analogue, namely, the filter-quotient construction (which happens to be an ultrafilter-quotient). You can read about it in e.g., MacLane and Moerdijk's "Sheaves in geometry and logic". – Andrej Bauer Jan 10 2010 at 9:52
Let $\Phi$ a filter on the set $I$ and let $X_i \in C\ i\in I$, and for $U \subset I$ let $X_U:=\prod_{i\in U} X_i$ . We have a natural functor $T : P(I)^{op} \to C$ (where $P(I)$ the order of ubset onf $I$) as: $T(U):=X_U$ and for $T(V\subset U) : X_U \to X_V$ the canonical proiection induced by $V \subset U$. Being $I$ the initial object of $P(I)$ and letting $X:=X_I$ follow a "lifting" of $T$ to a functor $T': P(I)^{op} \to X \downarrow C$ as $T'(U)= (T(U\subset I), X_U)$ and $T'(V\subset U)=T(V\subset U)$. For any $U \in \Phi$ let $\alpha_U, \beta_U : K_U \to X$ the kernel pair of $T'(U)$ we obtain a diagram of all these $\alpha_U, \beta_U$ morphisms and the natural $K_{V \subset U}: K_U\to K_V$ induced by $T(V \subset U)$. The colimit of this diagram is the ultrapower of $\prod_{i\in I} X_i$ respecto to $\Phi$. This colimit is the colimits in $X\downarrow C$ of the cokers of the kernel pairs.
But things are more simple too: observe that choise a retraction $r: I\to U$ os the inclusion $U\subset I$ we have a section $T(r): T(U)\to X$ on $T(U\subset I)$ give by $\pi_i\circ T(r) = \pi_{r(i)}$ then $T(U\subset I)$ being a retraction is a regular epimorphism (a coker of some pair), then is (well knowed fact) the coker of its Ker-pair $\alpha_U, \beta_U$. Then follow that the ultrapower is the colimit in $X\downarrow C$ of the diagram of $T(U\subset I)\ U\subset I$ as objects and $T(V\subset U)$ as morphisms.
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The other answers so far have generally taken ultrafilters as a given, or used the Stone-Cech compactification (which has a universal property in Top). I'd like to point out that the set of ultrafilters on a set $I$ has a categorical interpretation in Sets. In particular, consider the diagram in Sets consisting of finite partitions $X_i\subset 2^I$ of $I$, with an arrow $X_i\to X_j$ if $X_i$ is a refinement of $X_j$; the arrow sends a subset of $I$ in $X_i$ to the unique element of $X_j$ containing it. Then the set of ultrafilters on $I$ is the inverse limit of this diagram.
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For a short and painless description of how ultraproducts are colimits see page 6 of this article (it is equivalent, but I found it more readable than the descriptions given in the answers):
H. Mariano, F. Miraglia: Profinite structures are retracts of ultraproducts of finite structures
The article contains a nice application of this categorical description of ultraproducts...
- | {"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": 68, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9238045811653137, "perplexity_flag": "head"} |
http://www.impan.pl/cgi-bin/dict?matrix | ## matrix
Define $A$ to be the matrix with 1 in the $(i,j)$ entry and 0 elsewhere.
Take $A$ to be the matrix with all entries zero except for $i-j$ at $(i,j)$.
a real $n\times n$ matrix
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": 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": 0.7352080941200256, "perplexity_flag": "head"} |
http://mathhelpforum.com/pre-calculus/18551-need-help-something-fuction.html | # Thread:
1. ## Need help with something in a fuction
(Note: When I say delta, I mean the little triangle thing usually meaning change, I can't find the HTML code for it.)
Alright, I have a the function f(x) = x^2+3x-1
I then need to evaluate for f(x+deltax). I know that then should look something like x^2+3x-1 + deltax^2+3x-1.
Please verify that is essentially what should happen, and then help me understand what exactly I am supposed to do with the delta in front of the "x" value.
2. Originally Posted by Lithiux
Alright, I have a the function f(x) = x^2+3x-1
I then need to evaluate for f(x+deltax). I know that then should look something like x^2+3x-1 + deltax^2+3x-1.
$f(x+\Delta x)=(x+\Delta x)^2+3(x+\Delta x)-1$
Can you take it from there?
3. Sorry, I r stoopid.
I'd gotten that far in my head as far as setting it up, but I'm not sure what to do with the delta. I'm not sure what is supposed to change.
4. Originally Posted by Lithiux
Sorry, I r stoopid.
I'd gotten that far in my head as far as setting it up, but I'm not sure what to do with the delta. I'm not sure what is supposed to change.
You are not stupid. Stop that! You simply don't have a lot of experience.
Try thinking of $\Delta x$ as a single variable, not two symbols.
So for example
$(x + \Delta x)^2$
is the same kind of thing as
$(a + b)^2 = a^2 + 2ab + b^2$
where
$a = x \text{ and }b = \Delta x$
So
$(x + \Delta x)^2 = x^2 + 2x(\Delta x) + ( \Delta x)^2$
-Dan
5. Ooooh, so to evaluate the delta, it would need to be an equation of some sort, and as such I just treat it as if it were an x or y or something like that. Cool, thanks.
6. Originally Posted by Lithiux
Ooooh, so to evaluate the delta, it would need to be an equation of some sort, and as such I just treat it as if it were an x or y or something like that. Cool, thanks.
A $\Delta$ usually (not always) indicates the difference in some value. For example
$\Delta x = x_{final} - x_{initial}$
in Physics. But the point here is that $\Delta x$ is just a number.
-Dan
7. Yeah, that was my problem. I was thinking of it in a Physics frame-of-mind. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9672508239746094, "perplexity_flag": "head"} |
http://stats.stackexchange.com/questions/47719/fitting-lognormal-distribution-in-winbugs | # Fitting Lognormal Distribution in WinBugs
Let's suppose that some data points follow $Lognormal(\mu,\sigma^2)$ and both parameters are unknown . My goal is to obtain the posterior distribution by assigning conjugate prior distributions on both $\mu$ and $\sigma^2$,. How this can be done in WinBugs?
-
2
Why not just transform your data so that it is normal, i.e. take the log of your data? – guy Jan 15 at 0:44
2
1) If the priors really are conjugate, there's no need to use WinBUGS, as the posterior will have a nice closed form. 2) If you take the log of your data, it will be Normally distributed, and you can use the conjugate priors for the Normal. – jbowman Jan 15 at 0:44
@guy, jbowman. Yes, this can be solved by transforming the data to normal, but i really need to learn this without transformation (i.e., what is the conjugate prior for lognormal's parameters?) – act00 Jan 15 at 1:13
1
@act00 the conjugate prior for the lognormal distribution is the normal distribution (or normal-inverse-gamma if you want to put a prior on $(\mu, \sigma^2)$). Why? Because you can transform the data so that it is normal, and then use the conjugate prior for the normal. However, if you want to use the lognormal directly it is given by dlnorm(mu, tau) where tau is $1 / \sigma^2$ in the usual parameterization; just stick a normal-gamma prior on those two. – guy Jan 15 at 2:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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": 0.8652651906013489, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/149039/a-question-about-sorting/149041 | # A question about sorting
I've always been thought that the fastest way to sort an array of numbers has complexity $O(n \log (n))$. However, radix sort has complexity $O(kn)$ where $k$ is the number of bits. There are even questions on the internet where it is asked to prove that a sorting algorithm cannot be faster than $n \log (n)$.
Wanted to have a clarification on this. Does radix sort have any limitations? If not, is the lower bound on sorting linear in the number of elements?
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2
Maybe I misunderstand, but isn't the number of bits needed to store $n$ about $\log n$? – froggie May 24 '12 at 1:13
3
nlog(n) is the bound on comparison sorts, where the only thing you are given is a relation R(x,y) which compares x and y. If you have additional information about elements (which is the case with radix sort) you can sort faster. – Mark May 24 '12 at 1:16
I'm not sure if this helps but there are (randomised) algorithms that are certainly faster than $\mathcal{O}(nlog(n))$. For example: M. Thorup. Randomized Sorting in O(nloglogn) Time and Linear Space Using Addition, Shift, and Bit-wise Boolean Operations. Journal of Algorithms, Volume 42, Number 2, February 2002, pp. 205-230(26) [reference from wikipedia]. Also I remember in one of our classes the professor was talking about "bucket sort" which has O(n) as "expected" time! – Keivan May 24 '12 at 1:18
@froggie: If you're counting such things, don't forget that a comparison would also takes $O(\log n)$ time rather than $O(1)$ time in the worst case! – Hurkyl May 24 '12 at 1:35
@froggie Almost, take the example of $256_{(10)}$, the number of bits required for it is $9$, contrary to the output of $log_2(256) = 8$. That's because in an $n$-bit binary value, the highest power is $2^{n-1}$ which would be, in the case of 8 bits, $128$. The highest value expressible, therefore, is $2^n - 1$ (for all bits to be $1$, evaluating to $255$). Therefore, the actual number of bits required for $n$ is $log_2(n)+1$ (trunc the fractional part). Therefore, you actually can only express $n$ with that many bits, even though a lot more falls into that range ($n_{max}=2^{log_2(n)+1}-1$). – Domagoj Pandža May 24 '12 at 1:51
## 1 Answer
The number of bits $k$ cannot be considered constant in general. In fact, if all the $n$ numbers are distinct then $k = \mathcal{\Omega}(\log n)$. Hence, there is no difference between radix sort and other fast sorting algorithms.
More generally, any generic deterministic sorting algorithm cannot better $\mathcal{O}(n \log n)$ complexity. If you have $n$ numbers, then the number of comparisons you need to make is at least $\log_2(n!)$.
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Shouldn't that last term be $\log_2(n!)$? – Gerry Myerson May 24 '12 at 1:23
@GerryMyerson Yes. Edited it. Thanks. – user17762 May 24 '12 at 1:24
If all the $n$ numbers are distinct, isn't $k = \Omega(\log n)$ rather than $O(\log n)$? – Rahul Narain May 24 '12 at 1:26
@RahulNarain Yes. it should be $\Omega(\log n)$ and not $\mathcal{O}(\log n)$. – user17762 May 24 '12 at 1:27
@Marvis: When we're counting the $\log n$ cost for Radix sort, we really should also be counting the $\log n$ cost to read/write numbers for a comparison-based sort, as well as the $\log n$ cost to do a compare. Despite your statement, radix sort doesn't do any comparisons -- instead, it can be viewed as breaking comparison apart into its component operations down into its component operations, and working with those in a more efficient way than a comparison-based sort could. – Hurkyl May 24 '12 at 1:50
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Where can I find a list of important Taylor-Slavnov identities in Spontaneously broken gauge theories? I am looking for not just the generating functional form, but rather a list of explicit ones ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8959439992904663, "perplexity_flag": "middle"} |
http://en.wikipedia.org/wiki/Thermodynamic_systems | # Thermodynamic system
(Redirected from Thermodynamic systems)
A thermodynamic system is a precisely specified macroscopic region of the universe, defined by boundaries or walls of particular natures, together with the physical surroundings of that region, which determine processes that are allowed to affect the interior of the region, studied using the principles of thermodynamics.
All space in the universe outside the thermodynamic system is known as the surroundings, the environment, or a reservoir. A system is separated from its surroundings by a boundary, which may be notional or real but, by convention, delimits a finite volume. Exchanges of work, heat, or matter and energy between the system and the surroundings may take place across this boundary. A thermodynamic system is classified by the nature of the exchanges that are allowed to occur across its boundary.
A thermodynamic system has a characteristic set of thermodynamic parameters, or state variables. They are experimentally measurable macroscopic properties, such as volume, pressure, temperature, electric field, and others. The set of values of the thermodynamic parameters necessary to uniquely define a state of a system is called the thermodynamic state of a system. The state variables of a system are normally related by one or more functional relationships, the equations of state.
An isolated system is in thermodynamic equilibrium when its state does not change with time. Systems that are not isolated can be in thermodynamic equilibrium, or in a state constant or precisely cyclically changing in time - a steady state - that is far from equilibrium, or they can be in a changing state that is not in thermodynamic equilibrium. Classical thermodynamics considers only states of thermodynamic equilibrium or states constant or precisely cycling in time.
Originally, in 1824, Sadi Carnot described a thermodynamic system as the working substance of a heat engine under study.
Thermodynamics
The classical Carnot heat engine
Branches
Systems
State
Processes
Cycles
Specific heat capacity $c=$
$T$ $\partial S$
$N$ $\partial T$
Compressibility $\beta=-$
$1$ $\partial V$
$V$ $\partial p$
Thermal expansion $\alpha=$
$1$ $\partial V$
$V$ $\partial T$
• Internal energy
$U(S,V)$
• Enthalpy
$H(S,p)=U+pV$
$A(T,V)=U-TS$
$G(T,p)=H-TS$
History / Culture
Philosophy
History
Theories
Key publications
Timelines
Art
Education
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## Overview
Thermodynamics describes the physics of matter using the concept of the thermodynamic system, a region of the universe that is under study. All quantities, such as pressure or mechanical work, in an equation refer to the system unless labeled otherwise. As thermodynamics is fundamentally concerned with the flow and balance of energy and matter, systems are distinguished depending on the kinds of interaction they undergo and the types of energy they exchange with the surrounding environment.
Interactions of thermodynamic systems
Type of system Mass flow Work Heat
Open Y Y Y
Closed N Y Y
Thermally isolated N Y N
Mechanically isolated N N Y
Isolated N N N
Isolated systems are completely isolated from their environment. They do not exchange heat, work or matter with their environment. The only truly isolated system there could be is the universe, but even that is up for debate if the Big Bang is considered. Closed systems are able to exchange energy (heat and work) but not matter with their environment. A greenhouse is an example of a closed system exchanging heat but not work with its environment. Whether a system exchanges heat, work or both is usually thought of as a property of its boundary. Open systems may exchange any form of energy as well as matter with their environment. A boundary allowing matter exchange is called permeable. The ocean would be an example of an open system.
In practice, a system can never be absolutely isolated from its environment, because there is always at least some slight coupling, such as gravitational attraction. In analyzing a system in steady-state, the energy into the system is equal to the energy leaving the system [1].
An example system is the system of hot liquid water and solid table salt in a sealed, insulated test tube held in a vacuum (the surroundings). The test tube constantly loses heat in the form of black-body radiation, but the heat loss progresses very slowly. If there is another process going on in the test tube, for example the dissolution of the salt crystals, it probably occurs so quickly that any heat lost to the test tube during that time can be neglected. Thermodynamics in general does not measure time, but it does sometimes accept limitations on the time frame of a process.
## History
The first to develop the concept of a thermodynamic system was the French physicist Sadi Carnot whose 1824 Reflections on the Motive Power of Fire studied what he called the working substance, e.g., typically a body of water vapor, in steam engines, in regards to the system's ability to do work when heat is applied to it. The working substance could be put in contact with either a heat reservoir (a boiler), a cold reservoir (a stream of cold water), or a piston (to which the working body could do work by pushing on it). In 1850, the German physicist Rudolf Clausius generalized this picture to include the concept of the surroundings, and began referring to the system as a "working body." In his 1850 manuscript On the Motive Power of Fire, Clausius wrote:
“ "With every change of volume (to the working body) a certain amount work must be done by the gas or upon it, since by its expansion it overcomes an external pressure, and since its compression can be brought about only by an exertion of external pressure. To this excess of work done by the gas or upon it there must correspond, by our principle, a proportional excess of heat consumed or produced, and the gas cannot give up to the "surrounding medium" the same amount of heat as it receives." ”
The article Carnot heat engine shows the original piston-and-cylinder diagram used by Carnot in discussing his ideal engine; below, we see the Carnot engine as is typically modeled in current use:
Carnot engine diagram (modern) - where heat flows from a high temperature TH furnace through the fluid of the "working body" (working substance) and into the cold sink TC, thus forcing the working substance to do mechanical work W on the surroundings, via cycles of contractions and expansions.
In the diagram shown, the "working body" (system), a term introduced by Clausius in 1850, can be any fluid or vapor body through which heat Q can be introduced or transmitted through to produce work. In 1824, Sadi Carnot, in his famous paper Reflections on the Motive Power of Fire, had postulated that the fluid body could be any substance capable of expansion, such as vapor of water, vapor of alcohol, vapor of mercury, a permanent gas, or air, etc. Though, in these early years, engines came in a number of configurations, typically QH was supplied by a boiler, wherein water boiled over a furnace; QC was typically a stream of cold flowing water in the form of a condenser located on a separate part of the engine. The output work W was the movement of the piston as it turned a crank-arm, which typically turned a pulley to lift water out of flooded salt mines. Carnot defined work as "weight lifted through a height."
## Boundary
A system boundary is a real or imaginary two-dimensional closed surface that encloses or demarcates the volume or region that a thermodynamic system occupies, across which quantities such as heat, mass, or work can flow.[1] In short, a thermodynamic boundary is a geometrical division between a system and its surroundings. Topologically, it is usually considered nearly or piecewise smoothly homeomorphic with a two-sphere, because a system is usually considered simply connected.
Boundaries can also be fixed (e.g. a constant volume reactor) or moveable (e.g. a piston). For example, in a reciprocating engine, a fixed boundary means the piston is locked at its position; as such, a constant volume process occurs. In that same engine, a moveable boundary allows the piston to move in and out. Boundaries may be real or imaginary. For closed systems, boundaries are real while for open system boundaries are often imaginary. For theoretical purposes, a boundary may be declared adiabatic, isothermal, diathermal, insulating, permeable, or semipermeable—but actual physical materials that provide such idealized properties are not always readily available.
Anything that passes across the boundary that effects a change in the internal energy must be accounted for in the energy balance equation. The volume can be the region surrounding a single atom resonating energy, such as Max Planck defined in 1900; it can be a body of steam or air in a steam engine, such as Sadi Carnot defined in 1824; it can be the body of a tropical cyclone, such as Kerry Emanuel theorized in 1986 in the field of atmospheric thermodynamics; it could also be just one nuclide (i.e. a system of quarks) as hypothesized in quantum thermodynamics.
## Surroundings
See also: Environment (systems)
The system is the part of the universe being studied, while the surroundings is the remainder of the universe that lies outside the boundaries of the system. It is also known as the environment, and the reservoir. Depending on the type of system, it may interact with the system by exchanging mass, energy (including heat and work), momentum, electric charge, or other conserved properties. The environment is ignored in analysis of the system, except in regards to these interactions.
## Open system
During steady, continuous operation, an energy balance applied to an open system equates shaft work performed by the system to heat added plus net enthalpy added.
In open systems, matter may flow in and out of the system boundaries. The first law of thermodynamics for open systems states: the increase in the internal energy of a system is equal to the amount of energy added to the system by matter flowing in and by heating, minus the amount lost by matter flowing out and in the form of work done by the system. The first law for open systems is given by:
$\mathrm{d}U=\mathrm{d}U_{in}+\delta Q-\mathrm{d}U_{out}-\delta W\,$
where Uin is the average internal energy entering the system and Uout is the average internal energy leaving the system.
The region of space enclosed by open system boundaries is usually called a control volume, and it may or may not correspond to physical walls. If we choose the shape of the control volume such that all flow in or out occurs perpendicular to its surface, then the flow of matter into the system performs work as if it were a piston of fluid pushing mass into the system, and the system performs work on the flow of matter out as if it were driving a piston of fluid. There are then, two types of work performed: flow work described above, which is performed on the fluid (this is also often called PV work) and shaft work, which may be performed on some mechanical device. These two types of work are expressed in the equation:
$\delta W=\mathrm{d}(P_{out}V_{out})-\mathrm{d}(P_{in}V_{in})+\delta W_{shaft}\,$
Substitution into the equation above for the control volume cv yields:
$\mathrm{d}U_{cv}=\mathrm{d}U_{in}+\mathrm{d}(P_{in}V_{in}) - \mathrm{d}U_{out}-\mathrm{d}(P_{out}V_{out})+\delta Q-\delta W_{shaft}\,$
The definition of enthalpy, H, permits us to use this thermodynamic potential to account for both internal energy and PV work in fluids for open systems:
$\mathrm{d}U_{cv}=\mathrm{d}H_{in}-\mathrm{d}H_{out}+\delta Q-\delta W_{shaft}\,$
During steady-state operation of a device (see turbine, pump, and engine), any system property within the control volume is independent of time. Therefore, the internal energy of the system enclosed by the control volume remains constant, which implies that dUcv in the expression above may be set equal to zero. This yields a useful expression for the power generation or requirement for these devices in the absence of chemical reactions:
$\frac{\delta W_{shaft}}{\mathrm{d}t}=\frac{\mathrm{d}H_{in}}{\mathrm{d}t}- \frac{\mathrm{d}H_{out}}{\mathrm{d}t}+\frac{\delta Q}{\mathrm{d}t} \,$
This expression is described by the diagram above.
## Closed system
Main article: Closed system#In thermodynamics
In a closed system, no mass may be transferred in or out of the system boundaries. The system always contains the same amount of matter, but heat and work can be exchanged across the boundary of the system. Whether a system can exchange heat, work, or both is dependent on the property of its boundary.
• Adiabatic boundary – not allowing any heat exchange: A thermally isolated system
• Rigid boundary – not allowing exchange of work: A mechanically isolated system
One example is fluid being compressed by a piston in a cylinder. Another example of a closed system is a bomb calorimeter, a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Electrical energy travels across the boundary to produce a spark between the electrodes and initiates combustion. Heat transfer occurs across the boundary after combustion but no mass transfer takes place either way.
Beginning with the first law of thermodynamics for an open system, this is expressed as:
$\Delta U=Q-W+m_{i}(h+\frac{1}{2}v^2+gz)_{i}-m_{e}(h+\frac{1}{2}v^2+gz)_{e}$
where U is internal energy, Q is the heat added to the system, W is the work done by the system, and since no mass is transferred in or out of the system, both expressions involving mass flow are zero and the first law of thermodynamics for a closed system is derived. The first law of thermodynamics for a closed system states that the increase of internal energy of the system equals the amount of heat added to the system minus the work done by the system. For infinitesimal changes the first law for closed systems is stated by:
$\mathrm{d}U={\delta Q}-{\delta W}.$
If the work is due to a volume expansion by dV at a pressure P than:
${\delta W}=P\mathrm{d}V.$
For a homogeneous system, in which only reversible processes can take place, the second law of thermodynamics reads:
${\delta Q}=T\mathrm{d}S$
where T is the absolute temperature and S is the entropy of the system. With these relations the fundamental thermodynamic relationship, used to compute changes in internal energy, is expressed as:
${\delta U}=T\mathrm{d}S-P\mathrm{d}V.$
For a simple system, with only one type of particle (atom or molecule), a closed system amounts to a constant number of particles. However, for systems undergoing a chemical reaction, there may be all sorts of molecules being generated and destroyed by the reaction process. In this case, the fact that the system is closed is expressed by stating that the total number of each elemental atom is conserved, no matter what kind of molecule it may be a part of. Mathematically:
$\sum_{j=1}^m a_{ij}N_j=b_i^0$
where Nj is the number of j-type molecules, aij is the number of atoms of element i in molecule j and bi0 is the total number of atoms of element i in the system, which remains constant, since the system is closed. There is one such equation for each element in the system.
## Isolated system
Main article: Isolated system
An isolated system is more restrictive than a closed system as it does not interact with its surroundings in any way. Mass and energy remains constant within the system, and no energy or mass transfer takes place across the boundary. As time passes in an isolated system, internal differences in the system tend to even out and pressures and temperatures tend to equalize, as do density differences. A system in which all equalizing processes have gone practically to completion is in a state of thermodynamic equilibrium.
Truly isolated physical systems do not exist in reality (except perhaps for the universe as a whole), because, for example, there is always gravity between a system with mass and masses elsewhere.[2][3][4][5][6] However, real systems may behave nearly as an isolated system for finite (possibly very long) times. The concept of an isolated system can serve as a useful model approximating many real-world situations. It is an acceptable idealization used in constructing mathematical models of certain natural phenomena.
In the attempt to justify the postulate of entropy increase in the second law of thermodynamics, Boltzmann’s H-theorem used equations, which assumed that a system (for example, a gas) was isolated. That is all the mechanical degrees of freedom could be specified, treating the walls simply as mirror boundary conditions. This inevitably led to Loschmidt's paradox. However, if the stochastic behavior of the molecules in actual walls is considered, along with the randomizing effect of the ambient, background thermal radiation, Boltzmann’s assumption of molecular chaos can be justified.
The second law of thermodynamics for isolated systems states that the entropy of an isolated system not in equilibrium tends to increase over time, approaching maximum value at equilibrium. Overall, in an isolated system, the internal energy is constant and the entropy can never decrease. A closed system's entropy can decrease e.g. when heat is extracted from the system.
It is important to note that isolated systems are not equivalent to closed systems. Closed systems cannot exchange matter with the surroundings, but can exchange energy. Isolated systems can exchange neither matter nor energy with their surroundings, and as such are only theoretical and do not exist in reality (except, possibly, the entire universe).
It is worth noting that 'closed system' is often used in thermodynamics discussions when 'isolated system' would be correct - i.e. there is an assumption that energy does not enter or leave the system.
## Mechanically isolated system
Main article: Mechanically isolated system
A mechanically isolated system can exchange no work energy with its environment, but may exchange heat energy and/or mass with its environment. The internal energy of a mechanically isolated system may therefore change due to the exchange of heat energy and mass. For a simple system, mechanical isolation is equivalent to constant volume and any process which occurs in such a simple system is said to be isochoric.
## Systems in equilibrium
At thermodynamic equilibrium, a system's properties are, by definition, unchanging in time. Systems in equilibrium are much simpler and easier to understand than systems not in equilibrium. Often, when analyzing a thermodynamic process, it can be assumed that each intermediate state in the process is at equilibrium. This also considerably simplifies the analysis.
In isolated systems it is consistently observed that as time goes on internal rearrangements diminish and stable conditions are approached. Pressures and temperatures tend to equalize, and matter arranges itself into one or a few relatively homogeneous phases. A system in which all processes of change have gone practically to completion is considered in a state of thermodynamic equilibrium. The thermodynamic properties of a system in equilibrium are unchanging in time. Equilibrium system states are much easier to describe in a deterministic manner than non-equilibrium states.
In thermodynamic processes, large departures from equilibrium during intermediate steps are associated with increases in entropy and increases in the production of heat rather than useful work. It can be shown that for a process to be reversible, each step in the process must be reversible. For a step in a process to be reversible, the system must be in equilibrium throughout the step. That ideal cannot be accomplished in practice because no step can be taken without perturbing the system from equilibrium, but the ideal can be approached by making changes slowly.
## References
1. Perrot, Pierre (1998). A to Z of Thermodynamics. Oxford University Press. ISBN 0-19-856552-6.
2. I.M.Kolesnikov; V.A.Vinokurov; S.I.Kolesnikov (2001). Thermodynamics of Spontaneous and Non-Spontaneous Processes. Nova science Publishers. p. 136. ISBN 1-56072-904-X.
3. "A System and Its Surroundings". ChemWiki. University of California - Davis. Retrieved May 2012.
4. "Hyperphysics". The Department of Physics and Astronomy of Georgia State University. Retrieved May 2012.
5. Bryan Sanctuary. "Open, Closed and Isolated Systems in Physical Chemistry,". Foundations of Quantum Mechanics and Physical Chemistry. McGill University (Montreal). Retrieved May 2012.
6. Material and Energy Balances for Engineers and Environmentalists. Imperial College Press. p. 7. Retrieved May 2012.
• Abbott, M.M.; van Hess, H.G. (1989). Thermodynamics with Chemical Applications (2nd ed.). McGraw Hill.
• Halliday, David; Resnick, Robert; Walker, Jearl (2008). Fundamentals of Physics (8th ed.). Wiley.
• Moran, Michael J.; Shapiro, Howard N. (2008). Fundamentals of Engineering Thermodynamics (6th ed.). Wiley. | {"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": 30, "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": 0.9420725703239441, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/51419-help-few-problems-don-t-need-answer-all.html | Thread:
1. help with a few problems (don't need to answer all)
i have tried these many times and still can't get them.
1) f(x)= ((5x^3)-(5x^2)+9x-9) / (x-1) when x < 1
f(x)= (6x^2)-1x+a when x>=1
What value is a to make the function continuous at 1?
I tried making them equal to each other and putting x=1 but that didn't work.
2) (e^2x) - (7e^x) + 6 = 0
It says there are 2 solutions but I keep getting -33.1255 and that's not even one of them.
3) y=loga(x) with the subscript being a. It passes through (28,1).
What is the value of a?
I have no idea what to do on this one.
4) log2(x) + log2(x+3) = k
Solve for x in terms of k.
Again, I have no idea what to do. I can solve it if k is an integer but thats about it.
Any help is greatly appreciated.
2. 1. Your approach would normally work, except that the function $\frac{5x^3 - 5x^2 + 9x - 9}{x - 1}$ has a discontinuity at x = 1. But what is the limit of the function? Factor the numerator to yield $\frac{(5x^2 + 9)(x-1)}{x-1}$. Then cancel the factors of $x-1$ and you are left with $5x^2 + 9$. At x = 1, this function is equal to 14, so the limit is 14, and that is the function value you want for the other part of your function.
2. Substitute $u = e^x$, yielding $u^2 - 7u + 6 = 0$. Then solve for u by factoring and identifying the roots: $(u - 6)(u - 1) = 0$, so u = 1 or 6. Hence $e^x = 1$ or $e^x = 6$. From $e^x = 1$ you have the solution x = 0, and from $e^x = 6$ you have the solution x = ln 6.
3. This is an easy one. The equation $y = \log_a x$ can be rewritten as $a^y = x$. With (28, 1) on the graph, we have $a^1 = 28$, or $a = 28$.
4. You can combine the two logarithms to yield $\log_2 (x(x+3)) = k$, which can be written $2^k = x(x + 3)$, or $x^2 + 3x - 2^k = 0$. Now apply the quadratic formula: $x = \frac{-3 \pm \sqrt{9 + 4\cdot 2^k}}{2}$. Recognizing that x must be a positive number in order to evaluate the logarithm, we have $x = \frac{-3 + \sqrt{9 + 2^{k+2}}}{2}$.
3. ooooooo. Those explanations were great. Thanks a lot. | {"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": 20, "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": 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": 0.9504209160804749, "perplexity_flag": "head"} |
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I'm attempting to numerically approximate a simple ODE, I'm using it to describe the motion of a gradient descent search, but it could easily have physical interpretation. In particular, x'(t) = ... | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159994721412659, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/94178-why-does-integration-give-you-area.html | # Thread:
1. ## Why does integration give you area?
Why does integrating a function within a limit of a to b give you the area of the graph from a to b?
2. There are various ways to construct the integral. Typically the integral of a continuous function $f:[a,b] \rightarrow \mathbb{R}$ is constructed as the limit of a sequence of Riemann sums.
3. Originally Posted by chengbin
Why does integrating a function within a limit of a to b give you the area of the graph from a to b?
4. Originally Posted by chengbin
Why does integrating a function within a limit of a to b give you the area of the graph from a to b?
$\int_a^bf(x)dx$ can also be interpreted as $\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\cdot\Delta_ix$ where $f(x_i)$ represents the height of $f$ at some value $x_i$ on the closed interval $[a,b]$, with $x_1=a$ and $x_n=b$. $\Delta_ix=x_i-x_{i-1}$ represents each distinct subinterval on $[a,b]$ (which can be interpreted as width).
Now, with the technical stuff out of the way, you can see that $\lim_{n\to\infty}\sum_{i=1}^nf(x_i)\cdot\Delta_ix$ is just the sum of the area of a bunch of rectangles as the width of these rectangles tends to zero.
$A=w\cdot{h}$, in the riemann sum, you can see that $w=\Delta_ix$, and height is $f(x_i)$.
The sum of the area of a bunch of rectangles is an area.
the $\int$ symbol has stretched out over the years since newton's day but it used to look more like an $S$. stading for sum.
NOTE. I choose a lack of precision here in an attempt to explain the definition of the riemann integral in an intuitive way. The precise definition is
$f$ continuous: if
$|\sum_{i=1}^nf(\xi_i)\cdot\Delta_ix-A|<\epsilon$ whenever $||\Delta||<\delta$.
then we say that $A=\int_a^bf(x)dx$
$\xi_i$ being any number on each $\Delta_ix$, and $||\Delta||$ being the largest subinterval in the partitioned subdivision from $a$ to $b$.
5. Originally Posted by chengbin
Why does integrating a function within a limit of a to b give you the area of the graph from a to b?
Look at how you have defined the definite integral.
CB
6. Originally Posted by CaptainBlack
Look at how you have defined the definite integral.
CB
I'm sorry, I don't quite get that. Do you mind elaborating?
7. Originally Posted by chengbin
I'm sorry, I don't quite get that. Do you mind elaborating?
What definition of an integral are you using?
CB
8. Originally Posted by chengbin
I'm sorry, I don't quite get that. Do you mind elaborating?
Then I guess we have to ask, exactly what definition of the integral you are using?
9. Originally Posted by chengbin
Why does integrating a function within a limit of a to b give you the area of the graph from a to b?
Get The Calculus by Leithold and study chapter 6. Go through that chapter carefully or even just bits and pieces, and I bet you forever understand what's area got to do with definite integrals.
10. Originally Posted by HallsofIvy
Then I guess we have to ask, exactly what definition of the integral you are using?
For example if I integrate this function
$\int_0^2 x^2dx$
Integrating that will give you the area under the function $y=x^2$ from 0 to 2.
Why?
@shawsend
I can't get that. I'm not a university student, just a 14 year old kid that finished the calculus part of Kumon math curriculum.
11. Originally Posted by chengbin
For example if I integrate this function
$\int_0^2 x^2dx$
Integrating that will give you the area under the function $y=x^2$ from 0 to 2.
Why?
Did you not understand my previous post?
12. Originally Posted by chengbin
For example if I integrate this function
$\int_0^2 x^2dx$
Integrating that will give you the area under the function $y=x^2$ from 0 to 2.
Why?
@shawsend
I can't get that. I'm not a university student, just a 14 year old kid that finished the calculus part of Kumon math curriculum.
Ok sorry.
13. Originally Posted by chengbin
I can't get that. I'm not a university student, just a 14 year old kid that finished the calculus part of Kumon math curriculum.
is it illegal for 14 year olds to get calculus books where you're from? or are you saying you don't have the money?
14. hi
what would you do,if you're told to find the colored area in this example ?
Attached Thumbnails
15. I can't get that. I'm not a university student, just a 14 year old kid that finished the calculus part of Kumon math curriculum
I did not know that. Well, then. It is very good that you have come to where you are at such a young age. When you are my age you will have achieved much in the realm of mathematics.
Maybe, if I say to you that, the way in which books are set up tend to favor practical use over conceptual understanding. Math books today are designed to provide engineering students the tools necessary to perform calculations. This means that the authors don' really care if you know why a theorem is what it is, they just want you to be able to go out and apply already proven theorems to real life problems. It will not be until much, much later, if you decide to major in math, that you will enter into the world of whys, and finally escape the drab world of how tos.
If you have trouble with my previous post, and you would still like to know WHY the integral is sometimes interpreted as an area under the curve of the integrand, then you must venture out and resarch it. Continue to ask your questions, because this is a noble enterprise. One day, you will be a man. The road to in depth understanding of math is long and lonely. You will one of the few singularities of peculiararity that will wonder at the language of nature, whilst you marvel at the rabble, thinking to yourself, "Don't they ever wonder WHY?"
Best of luck kid. | {"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": 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": 0.951560378074646, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/20048/what-is-the-inertial-frame-that-explains-the-foucault-pendulum | # What is the inertial frame that explains the Foucault Pendulum?
I know that the Foucault pendulum rotation in relation to Earth is a proof that the object is inertial in relation to the distant stars. But what makes them more important than the Earth? Are they an absolute and universal inertial frame? How can we prove that? Please elaborate.
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Nobody claims that the distant stars are an inertial frame. But the center of mass of a sufficient number of distant stars is expected to be (the total momentum of the universe is 0). – Fabian Jan 26 '12 at 18:06
@Fabian why the center of mass of distant stars are more important than the mass of the Earth? AFAIK The gravitational effects of this small planet over the object are stronger. – Jader Dias Jan 26 '12 at 18:09
1
– Mark Beadles Jan 26 '12 at 19:02
## 2 Answers
Actually the path of the Foucault Pendulum is not "fixed" (even approximately!) to the "fixed" stars. Unless the pendulum is installed at one of the Earth's poles (as someone has done), then the point of suspension is in constant rotation with the Earth itself. $\therefore$ the pendulum is really not in an intertial frame.
Consider a pendulum at the equator, swinging in a North South plane. It's obvious from symmetry that the plane of this pendulum doesn't rotate with respect to the earth and that, relative to an inertial frame, it rotates once every 24 hours. - UNSW, Austl.
A very good discussion of the forces (real and fictitious) on the pendulum can be found at this UNSW site. The vector that points from the suspension point toward the Earth is in constant acceleration and has a precession period that varies according to latitude.
This animation from the Wikipedia article on the Foucault pendulum may help show how the plane of the pendulum is rotating.
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I thought it rotated in periods of 32 hours (the sidereal day) – Jader Dias Jan 27 '12 at 13:14
The sidereal day is 23.93447 hours. – Mark Beadles Jan 27 '12 at 14:23
1
Maybe he got confused when he read "At the latitude of Paris a full precession cycle takes 32 hours" from the Wikipedia article – Jader Dias Jan 28 '12 at 16:21
It is not true that the Foucault Pendulum is "inertial in relation to the distant stars". The distant stars are moving in various random directions at various random speeds and are certainly not in the same inertial frame as the pendulum. Our galaxy is rotating, so it can't be used as an inertial frame. The visible universe is expanding and probably accelerating, so it's certainly not an inertial frame.
In Foucault's day the movement of the stars wasn't visible because they are so far away, so it was common to assume that the stars were fixed, and therefore represented a fixed framework that you could use as an absolute frame of reference. Actually that's not a bad approximation, but it is only an approximation.
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If the distant stars aren't the inertial frame, why does the Foucault Pendulum rotates? I think it is inertial to something other than the Earth, what is that other thing? – Jader Dias Jan 26 '12 at 18:43
The Foucault's Pendulum doesn't rotate. It carries on swinging in the same plane while the earth rotates beneath it. – John Rennie Jan 26 '12 at 19:06
2
@JohnRennie What you say is true only at the poles. Elsewhere the pendulum must rotate. – Mark Beadles Jan 26 '12 at 19:54
@JohnRennie What determines the "same plane"? What is this plane referece? – Jader Dias Jan 27 '12 at 13:05
1
I suspect this is moving away from what Vam'çá will find helpful. I'm guess he's puzzled why what is apparently "locking" the pendulum in place. For the purposes of this discussion I suggest we assume the pendulum is at one of the poles - if you try the experiment at the equator you'll find there is no apparent rotation of the pendulum anyway. – John Rennie Jan 27 '12 at 15:15
show 3 more comments | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9220728278160095, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/27485/which-are-the-simplest-known-contextual-inequalities | # Which are the simplest known contextual inequalities?
It is well-known that quantum mechanics does not admit a noncontextual ontological model, and there are countless different proofs of it. I'm interested in the simplest proofs that can be cast as an inequality, and bonus points if there's a proof that a simpler one can't be found.
The definition of contextuality that I care about is the one by Spekkens, that is: I don't care about determinism nor require that the proof of impossibility be specifically about measurement contextuality; failure at either measurement or preparation is fine. Spekkens himself provided very simple proofs for two-dimensional Hilbert spaces, but it is not clear to me how to cast his proofs as inequalities.
Also, it's well-known that unlike nonlocality, contextuality admits state-independent proofs. It would be nice to know the simplest one in this category as well.
Of course, "simplicity" is subjective, but I hope not enough to forbid a definite answer. If you want, a criterion could be: First, dimension of the Hilbert space needed to exhibit a contradiction. Second, number of measurements needed. Or maybe the product of these numbers.
My candidates are currently Klyachko's 5-observable inequality, that is violated by 3-dimensional quantum systems, and Yu's 13-observable inequality for 3 dimensions that is violated independently of the quantum state. I have no idea if these are the best, and I find it weird that I couldn't find an inequality violated for qubits.
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## 3 Answers
Spekkens himself and co authors developed a non-contextuality inequality for a 2-dimensional system and 2 observables here: http://arxiv.org/abs/0805.1463. Note that the inequality is derived in the context of a communication protocol with an additional assumption that no information about the parity of the message (a bit string) is transmitted. I sure it's impossible to get a "simpler" proof in terms of fewer dimensions and observables.
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While this is an interesting result, is not the sort of inequality I am looking for, as it uses an additional assumption that's not needed in the general case. – Mateus Araújo Nov 3 '11 at 17:52
In this paper: A. Cabello, S. Severini, and A. Winter, a generalisation of the notion of contextuality is presented in terms of the compatibility structure of a graph, something akin to the graph colourability arguments in Kochen Specker proofs. If you ask questions with binary outcomes, and you say that certain combinations of questions are compatible and exclusive (both answers cannot be true, or $1$) then questions are vertices of a graph, and if they are compatible and exclusive, you assign an edge. Then asking all questions and putting it into an inequality such as the Klyachko inequality, the non-contextual upper bound is for all possible graphs given by the independence number of the graph. The quantum upper bound is given the by the Lovasz theta function. So to find the smallest example of a contextual inequality which a violation by quantum mechanics, one just needs to find the smallest graph (in the number of vertices and edges) where the Lovasz theta function is larger than the independence number. This is the pentagon, or 5-Cycle graph, the one used in the Klyachko inequality.
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In which sense it is a generalization? It seems to me that they just represent the traditional notion of contextuality in terms of graphs. Also, they restrict themselves to the consideration of projective measurements, so they only prove that the Klyachko inequality is the simplest projective one; to claim more would contradict the fact that there can be violations of contextuality for qubits. – Mateus Araújo Nov 3 '11 at 17:23
It seems that their method can also be used to determine the simplest state-independent inequality, but they didn't seem to have done so. – Mateus Araújo Nov 3 '11 at 17:25
A recent paper of Adán Cabello completely characterizes state-independent contextuality, and by doing so proves that Yu and Oh's 13-observable inequality for 3-dimensional systems is in fact the simplest state-independent noncontextual inequality possible.
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http://physics.stackexchange.com/questions/17440/what-is-the-smallest-distance-between-a-decaying-vertex-and-an-interaction-point | # What is the smallest distance between a decaying vertex and an interaction point a detector can measure?
A short lived particle is created at the interaction point and then decaying a distance $d$ away, in some detector. My question is what is the smallest distance $d$ that can be measured experimentally? Of course it will be different from a detector to another.
Any links that explain how this analysis, of discovering a particle, is done experimentally would be appreciated.
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## 1 Answer
See Tao Han's Collider Phenomenology: Basic Knowledge and Techniques, Page 22, and the referenced articles.
As a rough guide, for the ATLAS detector the resolution is of the order $10\mu m$. It's also possible to resolve the vertex displacement in the longtitudinal direction alone, in the case that there's only one charged particle track to extrapolate backwards. In this case the resolution is about an order of magnitude worse.
The full formula for the resultion also depends on the transverse momentum and the angle between the particle track and the beam line.
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You should say what the answer is, so that it is available without digging through the reference. – Ron Maimon Nov 28 '11 at 6:38
That is an excellent reference. But still it does not answer the question what is the smallest distance for a displaced vertex that can be measured. – Revo Nov 28 '11 at 16:57
@Ron Maimon: OK, I've added some details. – felix Nov 28 '11 at 17:02
1
@Revo: I guess the shortest distance should be comparable to the resolution quoted in the article. But if you are really asking for the shortest distance that can be distinguished from zero at a statistical significance, I don't know the answer. – felix Nov 28 '11 at 17:04
@felix Great, thanks alot. – Revo Nov 28 '11 at 17:06 | {"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": 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": 0.9269173741340637, "perplexity_flag": "head"} |
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0answers
30 views
### What is $\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt$?
I want to compute the following integral: $\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt$ with $\alpha, \beta, c$ real constants, and $\alpha>0,\beta=0$. ...
1answer
142 views
### Integrating $\int^{\infty}_0 e^{-x^2}\,dx$ using Feynman's parametrization trick
I stumbled upon this short article on last weekend, it introduces an integral trick that exploits differentiation under the integral sign. On its last page, the author, Mr. Anonymous, left several ...
1answer
44 views
### Improper integral sin(x)/x converges absolutely, conditionaly or diverges?
$$\int_1^{\infty}\frac{\sin x}{x}dx$$ $$u=\frac{1}{x}$$ $$du=-\frac{1}{x^2}dx$$ $$dv=\sin xdx$$ $$v=-\cos x$$ \int_1^{\infty}\frac{\sin x}{x}dx=\frac{1}{x}(-\cos x)-\int_1^{\infty}\frac{\cos ...
2answers
42 views
### Convergence of $\int_0^\infty \sin(t)/t^\gamma \mathrm{d}t$
For what values of $\gamma\geq 0$ does the improper integral $$\int_0^\infty \frac{\sin(t)}{t^\gamma} \mathrm{d}t$$ converge? In order to avoid two "critical points" $0$ and $+\infty$ I've ...
1answer
28 views
### Convergence of $\int_0^1 \sqrt[3]{\ln(1/x)} \mathrm{d}x$
Does $$\int_0^1 \sqrt[3]{\ln\left(\frac{1}{x}\right)} \mathrm{d}x$$ converge? WA says it is equal to $\Gamma(4/3)$, however calculating the antiderivative seems approachless to me and can't compare ...
1answer
69 views
### $\int_0^1\frac{(f(x)-1)^2 -4x^2}{x^{3.5}}\,dx$ exists. Calculate $f(0)$ and $f'(0)$
I've tried somehow using Taylor to try and figure this one out. Unfortunately, I couldn't seem to get a solid answer. Thank you very much for your help! Let f be a continuos function, ...
2answers
94 views
### Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx$
I have a trouble with a integral: Using this Laplace trasform equation: \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] ...
2answers
53 views
### Improper Integral $\int_0^1\frac{dx}{x^p}$
Is this integral convergent only for $p<1$? $$\int_0^1\frac{dx}{x^p}$$
2answers
131 views
### How can I see if this integral is convergent or not $\int_0^\infty \ \frac{1}{1 + x^4\sin x} \,dx$
I think the integral is convergent, but I don't know how to prove it. $\int_0^\infty \ \frac{1}{1 + x^4\sin x} \,dx$
3answers
85 views
### improper integral question - $\int_{1}^{\infty}\!e^{-x}\ln x\,dx$
I ran into this integral question: does this integral converge: $$\int_{1}^{\infty}\!e^{-x}\ln x\,dx$$ ? Thank you very much in advance, Yaron
3answers
52 views
### $\int_0^1 \frac{{f}(x)}{x^p}$ exists and finite $\implies f(0) = 0$
Need some help with this question please. Let $f$ be a continuous function and let the improper inegral $$\int_0^1 \frac{{f}(x)}{x^p}$$ exist and be finite for any $p \geq 1$. I need to prove ...
1answer
69 views
### Improper integral $\int_{0}^{\infty}\frac{x^n}{x^{m+n+1}} \ dx=\frac{n! {(m-1)}!}{(m+n)!}.$
How can I prove that $$\int_{0}^{\infty}\frac{x^n}{x^{m+n+1}} \ dx=\frac{n! {(m-1)}!}{(m+n)!}\quad ?$$ I tried to do induction on $n$ and on $m$, separately, but I could only do the base case ($n=1$ ...
2answers
110 views
### Proving that $\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}$ by induction
I need to prove $$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$ I've seen other demonstrations of this, but they use some identities that I don't understand. ...
2answers
84 views
### How to prove that an integral doesn't exist?
$$\int_{0}^{\infty}\sin^2\left(\pi \left(x + \frac{1}{x} \right) \right) dx$$ Should I use any test for convergence?
4answers
60 views
### How can I evaluate this given improper integral?
How can I evaluate this integral: $$\int _{ 0 }^{3}{ \frac { x }{ (3-x)^{\frac{1}{3}}} dx} \ ?$$
4answers
75 views
### Could you help me with this improper integral
How can I evaluate this improper integral? $$\displaystyle\int_0^{\infty}\frac{1}{x(1+x^2)}\,dx$$
1answer
94 views
### Closed form for $\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$
Please help me to find a closed form for this integral: $$\int_0^{\infty}\frac{\arctan x\ln(1+x^2)}{1+x^2}\sqrt{x}\,dx$$
2answers
32 views
### What's the radius of convergence of the next sum: $\sum_{n=0}^\infty (\int_o^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt)x^n$
What's the radius of convergence of the next sum: $$\sum_{n=0}^\infty \left(\int_0^n\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt\right)x^n$$ I know that $$\int_0^\infty\frac{\sin^2t}{\sqrt[3]{t^7+1}}dt$$ does ...
0answers
28 views
### Is $f$ integrable, in the Darboux sense, on $[0,1]$?
Is $$f(x)= \begin{cases} 0,\quad &x=0,\\ x\sin x,&x>0, \end{cases}$$ integrable, in the Darboux sense, on $[0,1]$? I know the Darboux integral has to do with the upper sum and lower sum ...
1answer
34 views
### integral convergence - does this integration converge
i just ran into this problem and i'm having a hard time solving it: i would like to know if this integral converges or not, and why. i'd prefer the normal convergence tests. ...
2answers
56 views
### improper integral with square roots
Iv'e ran into this improper integral: $$\int_{0}^{1}(\sqrt{x}/\sqrt{1-x^6})dx$$ I've tried to position $t=\sqrt x$ and i didn't get very close. Any help will be greatly appreciated. thank you, ...
1answer
35 views
### A parameterized elliptical integral (Legendre Elliptical Integral)
$$K(a,\theta)=\int_{0}^{\infty}\frac{t^{-a}}{1+2t\cos(\theta)+t^{2}}dt$$ For $$-1<a<1;$$ $$-\pi<\theta<\pi$$ I know this integral to be a known tabulated Legendre elliptic integral, ...
1answer
66 views
### Finding a generalization for $\int_{0}^{\infty}e^{- 3\pi x^{2} }\frac{\sinh(\pi x)}{\sinh(3\pi x)}dx$
$\;\;\;\;$I was reading the introduction of Paul J. Nain's book "Dr. Euler's fabulous formula" where he talks about the sense of beauty in mathematics and quotes the G.N.Watson as saying that a ...
2answers
66 views
### Differential Equation for improper integrals
How do I use the definiton of the improper integral to find the Laplace transform $F(s)$ for the function $f(t)=e^{(t-1)^2}$
1answer
35 views
### is f improperly integrable if g is not
$f,g$ are nonnegative and locally integrable on $[a,b)$ and $L := \lim_{x\to b-}\frac{f(x)}{g(x)}\$ exists as extended real number. If $0 < L \le \infty$ and $g$ is not improperly ...
1answer
53 views
### Improper Integral Question
Express $$\int_0^1x^m(1-x^n)^pdx$$ in terms of gama function and hence evaluate the integral. I used the substitution $x^n=y$ and solving got this integral as equal to the beta-function {1\over ...
1answer
48 views
### Improper Integrals Problem
Find the value of $$\int_0^1(x \ln x)^3dx$$ Taking a substitute $x=e^{-y}$ i get the value as $$-3\over128$$ Does it look good ?
2answers
54 views
### $\frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx$
Prove that: $\frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx=\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx$ This is my proof: $\forall \ t \in \mathbb{R}$ (the improper integral ...
2answers
29 views
### calculate a generalized integral
I want to calculate a generalized integral: $$\int^1_0\frac{dx}{\sqrt{1-x}}$$ I have a theorem : if $f(x)$ is continuous over $[a,b[$ then: $$\int^b_af(x).dx = \lim_{c\to b⁻}\int^c_af(x).dx$$ if ...
2answers
51 views
### Reinterpreting improper integrals that require Cauchy principal value to be defined
This question concerns the Cauchy principal value. Consider the improper integral $$\int_{-∞}^{∞}\frac{1+x}{1+x^2}dx$$ which is divergent, and then its Cauchy principal value \lim_{u \to ∞} ...
4answers
112 views
### Improper Integral:$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$
I want show that this improper integral convergence: $$\int_{0}^{+\infty}\frac{\sin x}{x+\sin x}dx$$ please help me.
1answer
79 views
### Problem evaluating an improper integral $\int_0^{\infty} \frac{(\sin{2x}-2x\cos{2x})^2}{x^6}$ using fourier transform
This is a question from one of the past papers of my university which I am unable to do. I am not being able to do question 2 from below. Let \$f(x)= a^2-x^2 \,\,\,\,\, |x|<a ...
2answers
48 views
### Integral $\int_{0}^{1}{\frac{\exp(-rx)}{\sqrt{1-x}}dx}$ converges for $r\geq{1}$
How do you prove that $\int_{0}^{1}{\frac{\exp(-rx)}{\sqrt{1-x}}dx}$ converges for $r\geq{1}$ ? N.B. : I forgot everything about improper integrals, so please be very explicit :)
2answers
290 views
### Calculate the Fourier transform of $b(x)=1/(x^2+a^2)$
I need help to calculate the Fourier transform of this funcion $$b(x)=\frac{1}{x^{2}+a^{2}}$$ where $$a>0$$ Thanks
1answer
41 views
### Evaluating improper integrals using laplace transform
I want to calculate the following improper integral using Laplace and transforms (and laplace transforms only). $$\int_0^{\infty} x e^{-3x} \sin{x}\, dx$$ I propose the following method. I plan to ...
0answers
36 views
### computing a difficult integral using software
I'd like to compute the following integral. I've tried SAGE but it just runs for 15 minutes then stops.. not sure what that means. If anyone wants to take a crack with mathematica or anything, please ...
2answers
45 views
### Prove the following: Product of Roots
$1^{(1/1)} \cdot 2^{(1/2)} \cdot 3^{(1/3)} \cdot 4^{(1/4)} \cdot 5^{(1/5)}$.... diverges well I don't really know if it does but my gut tells me it does: I can take the log of this product to ...
3answers
91 views
### How can I prove that $\int_1^\infty \left\lvert\frac{\sin x}{x}\right\rvert dx$ diverges?
I know a start could be to try and prove that $\int_1^\infty \frac{\sin^2x}{x} dx$ diverges since $\frac{\sin^2x}{x} \le \left\lvert\frac{\sin x}{x}\right\rvert$ in this interval, but I wouldn't know ...
3answers
75 views
### Cauchy principal value of $\int_{\infty}^{-\infty}e^{-ax^2}\cos(2abx) \,dx$
How do I find out the Cauchy Principal value of $\int_{-\infty}^{\infty}e^{-ax^2}\cos(2abx) \,dx\,\,\,\,\,\,\,\,a,b>0$ using complex integration? The answer is $\sqrt{\frac{\pi}{a}}e^{-ab^2}$, and ...
2answers
94 views
### Calculating $\int_{-\infty}^\infty e^{-ax^2}e^{ibx}dx$
In my syllabus about quantum mechanics, they state that the following integral can be easily calculated: $$\int_{-\infty}^\infty e^{-ax^2}e^{ibx}dx = \sqrt{\frac{\pi}{a}}e^{-b^2/4a}$$ if it is ...
2answers
66 views
### Laplace transforms please help
I really need help to find to Laplace transforms of $f(x)=x+e^{-x}$, and $g(x)=xe^x$. I'm having big troubles on the calculations. Thanks.
1answer
155 views
### How to prove $\frac{\pi^2}{6}\le \int_0^{\infty} \sin(x^{\log x}) \ \mathrm dx$?
I want to prove the inequality $$\frac{\pi^2}{6}\le \int_0^{\infty} \sin(x^{\log x}) \ \mathrm dx$$ There are some obstacles I face: the indefinite integral cannot be expressed in terms of ...
6answers
143 views
### Does the improper integral $\int_0^\infty\sin(x)\sin(x^2)\,\mathrm dx$ converge
Does the following improper integral converge? $$\lim_{B \to \infty}\int_0^B\sin(x)\sin(x^2)\,\mathrm dx$$
4answers
47 views
### Convergence of improper integral
Show that $\displaystyle\int_{0}^{\infty}ln(x)e^{-x}dx$ converges. i used integration by parts but it always diverges. any hints? | {"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": 88, "mathjax_display_tex": 41, "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": 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": 0.8901669979095459, "perplexity_flag": "middle"} |
http://mathoverflow.net/revisions/75703/list | ## Return to Answer
2 links; added 50 characters in body
What numbers are integrally represented by the three-variable inhomogeneous polynomial $$7 x^2 + 3 x y + 9 y^2 + z^3 \; \; ?$$ This is one of two similar problems where it is easy to show that certain integers are not represented owing to their prime factorization. The first inkling that it might be possible to prove if and only if on a characterization of represented numbers is due to Kevin Buzzard, in his answer to my first MO question. The list of similar problems for class number three is at LINKSECONDTRY but in any case works in my comment below
My first MO question, with answer by Kevin Buzzard and evidence about why if-and-only-if is difficult, is http://mathoverflow.net/questions/12486/
Post Undeleted by Will Jagy
Post Deleted by Will Jagy
1 [made Community Wiki]
What numbers are integrally represented by the three-variable inhomogeneous polynomial $$7 x^2 + 3 x y + 9 y^2 + z^3 \; \; ?$$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9563059210777283, "perplexity_flag": "middle"} |
http://nrich.maths.org/8171/note | nrich enriching mathematicsSkip over navigation
### Sport Collection
This is our collection of favourite mathematics and sport materials.
### Olympic Starters
Look at some of the results from the Olympic Games in the past. How do you compare if you try some similar activities?
### Olympic Turns
This task looks at the different turns involved in different Olympic sports as a way of exploring the mathematics of turns and angles.
# Now and Then
## Now and Then
In $1908$ the Olympic Games were held in London, that's just over $100$ years ago. Then, just after World War $2$ they were again in London in $1948$
Here are the results from some track events;
$1908$
$100$ metres $10.8$ secs
$200$ metres $22.6$ secs
$400$ metres $50.0$ secs
$800$ metres $112$ secs
$1500$ metres $240$ secs
$1948$
$100$ metres $10.3$ secs
$200$ metres $21.1$ secs
$400$ metres $46.2$ secs
$800$ metres $109$ secs
$1500$ metres $229$ secs
The 2012 London Olympics were another 64 years later.
How did the results differ?
Could you have predicted the results?
Perhaps more importantly, what's the reason for your answer?
### Why do this problem
This activity gives an opportunity for pupils to examine data and then consider all different kinds of influencing factors in order to predict what will happen another $64$ years on. It can offer a good forum for debate and discussion while sharing views about things that will affect the results. It may help some children to understand that any answer can be deemed 'correct' so long as it can be justified.
### Possible approach
For many classrooms it would be best to give the results of one event to each group of pupils. They will obviously need opportunities to work as a group and come up with an agreed result. When these results are shared towards the end of the session, there could be some further discussions about the results for one year. This could involve looking at the relationships between the timings for different distances.
### Key questions
What do you think is important about deciding the times for that race in $2012$?
Do the whole group agree? If not, why?
### Possible extension
Ask questions such as:
What do you think about these results getting better and better as time goes on?
What about the results in another $100$ years time?
This should give the opportunity for some good creative thoughts to be shared.
Results from other years could also be examined to see whether results steadily improve or whether there are any sudden changes in the patterns. What might be the reasons for this? Children who are beginning to understand what graphs can mean may like to take a look at those in the activity Olympic Records.
### Possible support
Some pupils may need help in making sense of the results. They could look at the distances on the sports field to get a real idea of what the distances are like.
It may also be helpful to see what they can do in periods of $10$ secs, $21$ secs, and $22$ secs so that they understand just how quickly these races are being run.
```Photograph acknowledgement for icon
www.olympic.org.nz/sites/olympic/files/styles/grid-9/public/games/paris-
```
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9576225876808167, "perplexity_flag": "middle"} |
http://citizendia.org/Natural_logarithm | The natural logarithm, formerly known as the hyperbolic logarithm[1], is the logarithm to the base e, where e is an irrational constant approximately equal to 2. In Geometry, a hyperbola ( Greek, "over-thrown" has several equivalent definitions In Mathematics, the logarithm of a number to a given base is the power or Exponent to which the base must be raised in order to produce radix|basis (topologyIn Arithmetic, the base refers to the number b in an expression of the form b n. The Mathematical constant e is the unique Real number such that the function e x has the same value as the slope of the tangent line In Mathematics, an irrational number is any Real number that is not a Rational number — that is it is a number which cannot be expressed as a fraction 718281828459. In simple terms, the natural logarithm of a number x is the power to which e would have to be raised to equal x — for example the natural log of e itself is 1 because e1 = e, while the natural logarithm of 1 would be 0, since e0 = 1. The natural logarithm can be defined for all positive real numbers x as the area under the curve y = 1/t from 1 to x, and can also be defined for non-zero complex numbers as explained below. In Mathematics, the real numbers may be described informally in several different ways The European Space Agency 's INTErnational Gamma-Ray Astrophysics Laboratory ( INTEGRAL) is detecting some of the most energetic radiation that comes from space Complex plane In Mathematics, the complex numbers are an extension of the Real numbers obtained by adjoining an Imaginary unit, denoted
Graph of the natural logarithm function. The function quickly goes to negative infinity as x approaches 0, but grows slowly to positive infinity as x increases.
Part of a series of articles onThe mathematical constant,
Natural logarithm
Applications in: compound interest · Euler's identity & Euler's formula · half-lives & exponential growth/decay
Defining e: proof that e is irrational · representations of e · Lindemann–Weierstrass theorem
People John Napier · Leonhard Euler
Schanuel's conjecture
The natural logarithm function can also be defined as the inverse function of the exponential function, leading to the identities:
$e^{\ln(x)} = x \qquad \mbox{if }x > 0\,\!$
$\ln(e^x) = x.\,\!$
In other words, the logarithm function is a bijection from the set of positive real numbers to the set of all real numbers. The Mathematical constant e is the unique Real number such that the function e x has the same value as the slope of the tangent line Compound interest is the concept of adding accumulated Interest back to the principal so that interest is earned on interest from that moment on In Mathematical analysis, Euler's identity, named after Leonhard Euler, is the equation e^{i \pi} + 1 = 0 \\! where This article is about Euler's formula in Complex analysis. For Euler's formula in algebraic topology and polyhedral combinatorics see Euler characteristic Half-Life (computer-game page here It's already listed in the disambiguation page Exponential growth (including Exponential decay) occurs when the growth rate of a mathematical function is proportional to the function's current value A quantity is said to be subject to exponential decay if it decreases at a rate proportional to its value In Mathematics, the series representation of Euler's number e e = \sum_{n = 0}^{\infty} \frac{1}{n!}\! can be used to prove The Mathematical constant ''e'' can be represented in a variety of ways as a Real number. In Mathematics, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers For other people with the same name see John Napier (disambiguation. In Mathematics, specifically Transcendence theory, Schanuel's conjecture is the following statement Given any n Complex numbers In Mathematics, if &fnof is a function from A to B then an inverse function for &fnof is a function in the opposite direction from B The exponential function is a function in Mathematics. The application of this function to a value x is written as exp( x) In Mathematics, a bijection, or a bijective function is a function f from a set X to a set Y with the property More precisely it is an isomorphism from the group of positive real numbers under multiplication to the group of real numbers under addition. In Abstract algebra, an isomorphism ( Greek: ἴσος isos "equal" and μορφή morphe "shape" is a bijective In Mathematics, a group is a set of elements together with an operation that combines any two of its elements to form a third element Represented as a function:
$\ln : \mathbb{R}^+ \to \mathbb{R}$
Logarithms can be defined to any positive base other than 1, not just e, and are useful for solving equations in which the unknown appears as the exponent of some other quantity. The Mathematical concept of a function expresses dependence between two quantities one of which is given (the independent variable, argument of the function
Notational conventions
Mathematicians, statisticians, and some engineers generally understand either "log(x)" or "ln(x)" to mean loge(x), i. e. , the natural logarithm of x, and write "log10(x)" if the base-10 logarithm of x is intended. The common logarithm is the Logarithm with base 10 It is also known as the decadic logarithm, named after its base
Some engineers, biologists, and some others generally write "ln(x)" (or occasionally "loge(x)") when they mean the natural logarithm of x, and take "log(x)" to mean log10(x) or, in the case of some computer scientists, log2(x) (although this is often written lg(x) instead). The common logarithm is the Logarithm with base 10 It is also known as the decadic logarithm, named after its base Computer science (or computing science) is the study and the Science of the theoretical foundations of Information and Computation and their In Mathematics, the binary logarithm (log2 n) is the Logarithm for Base 2
In most commonly-used programming languages, including C, C++, MATLAB, Fortran, and BASIC, "log" or "LOG" refers to the natural logarithm. A programming language is an Artificial language that can be used to write programs which control the behavior of a machine particularly a Computer. tags please moot on the talk page first! --> In Computing, C is a general-purpose cross-platform block structured C++ (" C Plus Plus " ˌsiːˌplʌsˈplʌs is a general-purpose Programming language. MATLAB is a numerical computing environment and Programming language. Fortran (previously FORTRAN) is a general-purpose, procedural, imperative Programming language that is especially suited to In Computer programming, BASIC (an Acronym for Beginner's All-purpose Symbolic Instruction Code) is a family of High-level programming languages
In hand-held calculators, the natural logarithm is denoted ln, whereas log is the base-10 logarithm. A calculator is device for performing mathematical calculations distinguished from a Computer by having a limited problem solving ability and an interface optimized for interactive
Why it is called “natural”
Initially, it might seem that since our numbering system is base 10, this base would be more “natural” than base e. The decimal ( base ten or occasionally denary) Numeral system has ten as its base. But mathematically, the number 10 is not particularly significant. Its use culturally — as the basis for many societies’ numbering systems — likely arises from humans’ typical number of fingers. [2] Other cultures have based their counting systems on such choices as 5, 20, and 60. [3][4][5]
Loge is a “natural” log because it automatically springs from, and appears so often, in mathematics. For example, consider the problem of differentiating a logarithmic function:
$\frac{d}{dx}\log_b(x) = \frac{\log_b(e)}{x} =\frac{1}{\ln(b)x}$
If the base b equals e, then the derivative is simply 1/x, and at x = 1 this derivative equals 1. In Calculus, a branch of mathematics the derivative is a measurement of how a function changes when the values of its inputs change radix|basis (topologyIn Arithmetic, the base refers to the number b in an expression of the form b n. Another sense in which the base-e logarithm is the most natural is that it can be defined quite easily in terms of a simple integral or Taylor series and this is not true of other logarithms. In Mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its Derivatives
Further senses of this naturalness make no use of calculus. As an example, there are a number of simple series involving the natural logarithm. In fact, Pietro Mengoli and Nicholas Mercator called it logarithmus naturalis a few decades before Newton and Leibniz developed calculus. Pietro Mengoli (1626-1686 was an Italian Mathematician and clergyman from Bologna, where he studied with Bonaventura Cavalieri at the University of Nicholas ( Nikolaus) Mercator (c 1620 Eutin -1687 Versailles) also known by his Germanic name Kauffmann, was a 17th-century mathematician Sir Isaac Newton, FRS (ˈnjuːtən 4 January 1643 31 March 1727) Biography Early years See also Isaac Newton's early life and achievements [6]
Definitions
Ln(x) defined as the area under the curve f(x) = 1/x.
Formally, ln(a) may be defined as the area under the graph of 1/x from 1 to a, that is as the integral,
$\ln(a)=\int_1^a \frac{1}{x}\,dx.$
This defines a logarithm because it satisfies the fundamental property of a logarithm:
$\ln(ab)=\ln(a)+\ln(b) \,\!$
This can be demonstrated by letting $t=\tfrac xa$ as follows:
$\ln (ab) = \int_1^{ab} \frac{1}{x} \; dx = \int_1^a \frac{1}{x} \; dx \; + \int_a^{ab} \frac{1}{x} \; dx =\int_1^{a} \frac{1}{x} \; dx \; + \int_1^{b} \frac{1}{t} \; dt = \ln (a) + \ln (b)$
The number e can then be defined as the unique real number a such that ln(a) = 1. The European Space Agency 's INTErnational Gamma-Ray Astrophysics Laboratory ( INTEGRAL) is detecting some of the most energetic radiation that comes from space The Mathematical constant e is the unique Real number such that the function e x has the same value as the slope of the tangent line
Alternatively, if the exponential function has been defined first using an infinite series, the natural logarithm may be defined as its inverse function, i. The exponential function is a function in Mathematics. The application of this function to a value x is written as exp( x) In Mathematics, a series is often represented as the sum of a Sequence of terms That is a series is represented as a list of numbers with In Mathematics, if &fnof is a function from A to B then an inverse function for &fnof is a function in the opposite direction from B e. , ln(x) is that function such that $e^{\ln(x)} = x\!$. Since the range of the exponential function on real arguments is all positive real numbers and since the exponential function is strictly increasing, this is well-defined for all positive x.
Derivative, Taylor series
The derivative of the natural logarithm is given by
$\frac{d}{dx} \ln(x) = \frac{1}{x}.\,$
The Taylor polynomials for loge(1+x) only provide accurate approximations in the range -1 < x ≤ 1. In Calculus, a branch of mathematics the derivative is a measurement of how a function changes when the values of its inputs change Note that, for x > 1, the Taylor polynomials of higher degree are worse approximations.
This leads to the Taylor series for ln(1 + x) around 0; also known as the Mercator series
$\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad{\rm for}\quad \left|x\right| \leq 1\quad$
${\rm unless}\quad x = -1$
At right is a picture of ln(1 + x) and some of its Taylor polynomials around 0. In Mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its Derivatives In Mathematics, the Mercator series or Newton-Mercator series is the Taylor series for the Natural logarithm. In Calculus, Taylor's theorem gives a sequence of approximations of a Differentiable function around a given point by Polynomials (the Taylor These approximations converge to the function only in the region -1 < x ≤ 1; outside of this region the higher-degree Taylor polynomials are worse approximations for the function.
Substituting x-1 for x, we obtain an alternative form for ln(x) itself, namely
$\ln(x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} (x-1) ^ n$
$\ln(x)= (x - 1) - \frac{(x-1) ^ 2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} \cdots$
${\rm for}\quad \left|x-1\right| \leq 1\quad {\rm unless}\quad x = 0.$[7]
By using the Euler transform on the Mercator series, one obtains the following, which is valid for any x with absolute value greater than 1:
$\ln{x \over {x-1}} = \sum_{n=1}^\infty {1 \over {n x^n}} = {1 \over x}+ {1 \over {2x^2}} + {1 \over {3x^3}} + \cdots$
This series is similar to a BBP-type formula. In Combinatorial Mathematics the binomial transform is a Sequence transformation (ie a transform of a Sequence) that computes its Forward
Also note that $x \over {x-1}$ is its own inverse function, so to yield the natural logarithm of a certain number n, simply put in $n \over {n-1}$ for x.
The natural logarithm in integration
The natural logarithm allows simple integration of functions of the form g(x) = f '(x)/f(x): an antiderivative of g(x) is given by ln(|f(x)|). The European Space Agency 's INTErnational Gamma-Ray Astrophysics Laboratory ( INTEGRAL) is detecting some of the most energetic radiation that comes from space In Calculus, an antiderivative, primitive or indefinite integral of a function f is a function F whose Derivative This is the case because of the chain rule and the following fact:
$\ {d \over dx}\left( \ln \left| x \right| \right) = {1 \over x}.$
In other words,
$\int { 1 \over x} dx = \ln|x| + C$
and
$\int { \frac{f'(x)}{f(x)}\, dx} = \ln |f(x)| + C.$
Here is an example in the case of g(x) = tan(x):
$\int \tan (x) \,dx = \int {\sin (x) \over \cos (x)} \,dx$
$\int \tan (x) \,dx = \int {-{d \over dx} \cos (x) \over {\cos (x)}} \,dx.$
Letting f(x) = cos(x) and f'(x)= - sin(x):
$\int \tan (x) \,dx = -\ln{\left| \cos (x) \right|} + C$
$\int \tan (x) \,dx = \ln{\left| \sec (x) \right|} + C$
where C is an arbitrary constant of integration. In Calculus, the chain rule is a Formula for the Derivative of the composite of two functions. In Calculus, the Indefinite integral of a given function (ie the set of all Antiderivatives of the function is always written with a constant the constant
The natural logarithm can be integrated using integration by parts:
$\int \ln (x) \,dx = x \ln (x) - x + C.$
Numerical value
To calculate the numerical value of the natural logarithm of a number, the Taylor series expansion can be rewritten as:
$\ln(1+x)= x \,\left( \frac{1}{1} - x\,\left(\frac{1}{2} - x \,\left(\frac{1}{3} - x \,\left(\frac{1}{4} - x \,\left(\frac{1}{5}- \ldots \right)\right)\right)\right)\right) \quad{\rm for}\quad \left|x\right|<1.\,\!$
To obtain a better rate of convergence, the following identity can be used. In Calculus, and more generally in Mathematical analysis, integration by parts is a rule that transforms the Integral of products of functions into other
$\ln(x) = \ln\left(\frac{1+y}{1-y}\right)$ $= 2\,y\, \left( \frac{1}{1} + \frac{1}{3} y^{2} + \frac{1}{5} y^{4} + \frac{1}{7} y^{6} + \frac{1}{9} y^{8} + \ldots \right)$
$= 2\,y\, \left( \frac{1}{1} + y^{2} \, \left( \frac{1}{3} + y^{2} \, \left( \frac{1}{5} + y^{2} \, \left( \frac{1}{7} + y^{2} \, \left( \frac{1}{9} + \ldots \right) \right) \right)\right) \right)$
provided that y = (x−1)/(x+1) and x > 0.
For ln(x) where x > 1, the closer the value of x is to 1, the faster the rate of convergence. The identities associated with the logarithm can be leveraged to exploit this:
$\ln(123.456)\!$ $= \ln(1.23456 \times 10^2) \,\!$
$= \ln(1.23456) + \ln(10^2) \,\!$
$= \ln(1.23456) + 2 \times \ln(10) \,\!$
$\approx \ln(1.23456) + 2 \times 2.3025851 \,\!$
Such techniques were used before calculators, by referring to numerical tables and performing manipulations such as those above.
High precision
To compute the natural logarithm with many digits of precision, the Taylor series approach is not efficient since the convergence is slow. An alternative is to use Newton's method to invert the exponential function, whose series converges more quickly. In Numerical analysis, Newton's method (also known as the Newton–Raphson method, named after Isaac Newton and Joseph Raphson) is perhaps the
An alternative for extremely high precision calculation is the formula
$\ln x \approx \frac{\pi}{2 M(1,4/s)} - m \ln 2$
where M denotes the arithmetic-geometric mean and
$s = x \,2^m > 2^{p/2},$
with m chosen so that p bits of precision is attained. In Mathematics, the arithmetic-geometric mean (AGM of two positive Real numbers x and y is defined as follows First compute the Arithmetic In fact, if this method is used, Newton inversion of the natural logarithm may conversely be used to calculate the exponential function efficiently. (The constants ln 2 and π can be pre-computed to the desired precision using any of several known quickly converging series. IMPORTANT NOTICE Please note that Wikipedia is not a database to store the millions of digits of π please refrain from adding those to Wikipedia as it could cause technical problems )
Computational complexity
See main article: Computational complexity of mathematical operations
The computational complexity of computing the natural logarithm (using the arithmetic-geometric mean) is O(M(n) ln n). The following tables list the Computational complexity of various Algorithms for common Mathematical operations Here complexity refers to the Time complexity Computational complexity theory, as a branch of the Theory of computation in Computer science, investigates the problems related to the amounts of resources Here n is the number of digits of precision at which the natural logarithm is to be evaluated and M(n) is the computational complexity of multiplying two n-digit numbers.
Complex logarithms
Main article: Complex logarithm
The exponential function can be extended to a function which gives a complex number as ex for any arbitrary complex number x; simply use the infinite series with x complex. In Complex analysis, the complex logarithm is the extension of the Natural logarithm function ln( x) &ndash originally defined for Real numbers Complex plane In Mathematics, the complex numbers are an extension of the Real numbers obtained by adjoining an Imaginary unit, denoted This exponential function can be inverted to form a complex logarithm that exhibits most of the properties of the ordinary logarithm. There are two difficulties involved: no x has ex = 0; and it turns out that e2πi = 1 = e0. Since the multiplicative property still works for the complex exponential function, ez = ez+2nπi, for all complex z and integers n.
So the logarithm cannot be defined for the whole complex plane, and even then it is multi-valued – any complex logarithm can be changed into an "equivalent" logarithm by adding any integer multiple of 2πi at will. In Mathematics, the complex plane is a geometric representation of the Complex numbers established by the real axis and the orthogonal imaginary axis The complex logarithm can only be single-valued on the cut plane. In Mathematics, the complex plane is a geometric representation of the Complex numbers established by the real axis and the orthogonal imaginary axis For example, ln i = 1/2 πi or 5/2 πi or −3/2 πi, etc. ; and although i4 = 1, 4 log i can be defined as 2πi, or 10πi or −6 πi, and so on.
z = Re(ln(x+iy)) z = |Im(ln(x+iy))| z = |ln(x+iy)| Superposition of the previous 3 graphs | {"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": 37, "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": 0.9109295606613159, "perplexity_flag": "head"} |
http://crypto.stackexchange.com/questions/5607/why-are-protocols-often-proven-secure-under-the-random-oracle-model-instead-of-a/5614 | # Why are protocols often proven secure under the random oracle model instead of a hash assumption?
Is this true that whenever you design a protocol using a hash function, you must prove its security under the random oracle?
I mean, is it possible to devise a protocol $P$ using a function $H$, and then prove a theorem saying that $P$ is secure in a model $M$ given that $H$ is a collision-resistant function? What I most often see is some thing like: given a random oracle $H$, $P$ is secure in the model $M$.
Is this because it's more difficult or even impossible to prove security without the random oracles?
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## 1 Answer
A random oracle is an idealization of a hash function $H$: if hash functions were perfect they would be random oracles. This is why it is always easier to consider a hash function a random oracle when one proves something about a larger scheme. Those are "proofs in the random oracle model". [1]
That being said it is still possible to prove things using different, weaker, assumptions about the hash function or even the compression function in the case of a MD hash function (collision resistance for example). Those are "proofs in the standard model"
Is it possible to go from one to the other? Not always since there are separations: there exists schemes that can be proven secure in the RO model that will become insecure as soon as you instanciate the RO by a real world hash function. [2]
Is it the end of the world? Not really, since we also have strong results that use the notion of Indifferentiability from RO: simply put, if your hash function is indifferentiable from a RO then you can replace your RO by your hash function and the scheme will remain secure. [3]
There are (as always) subtleties and what I've just said is not always true but this is good enough to make us feel more comfortable with our assumptions. [4]
Is this RO-indifferentiability just theoretical crypto stuff? No! I haven't checked for all of them but I know that Skein and Keccak 2 of the SHA-3 finalist come with proofs regarding that very property. I believe all of the finalist do. [5] [6]
If you want you can start by reading the following:
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1
– fgrieu Dec 7 '12 at 11:42
Also: beware that the length-extension property of many hashes, including SHA-2 (but not SHA-3), makes them easily distinguishable from a random oracle. – fgrieu Dec 7 '12 at 11:45
2
– Alexandre Yamajako Dec 7 '12 at 13:34
I was veering more towards the proof technique in the random oracles. It is not by simply replacing the random oracle with a perfect random function we can arrive at the same proof in the standard model. But what seems troubling to me is the fact that security proof in the random oracle allows the simulator to see the oracle queries made by the adversary. Is there any justification for this? Or is this simply that we cannot do without this assumption? – Anh Dec 10 '12 at 2:48 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9346996545791626, "perplexity_flag": "middle"} |
http://mathhelpforum.com/pre-calculus/141621-how-solve-x-equation.html | # Thread:
1. ## How to solve for x in an equation
Solve for x.
x^n = x^(1/n)
How would we solve for x in this case? Would we move the x^(1/n) over to the left side and set it to zero. Then we factor out the left side? Please help me with this. Thanks
2. Why have you encountered this equation? Perhaps it is only a thought / property question. Rephrase in words and think about it.
What number is alterered identically by a power and a root.
The square is the square root.
The cube is the cube root.
There are not too many candidates.
Does it make a difference if x is positive or negative?
Does it make a difference if n is odd or even?
Again, why have you encountered this equation?
3. This is the problem. I am stuck on b). I have concluded that in order to find point A we must set x^n = x^(1/n) equal to each other because that is where they intersect. After finding x we can plug it back in the equation to find y and thus we would have a coordinate. But the problem is I don't know how to find x. Is there a better way to solve this or am I on the right track? Thanks for the help.
4. Originally Posted by florx
This is the problem. I am stuck on b). I have concluded that in order to find point A we must set x^n = x^(1/n) equal to each other because that is where they intersect. After finding x we can plug it back in the equation to find y and thus we would have a coordinate. But the problem is I don't know how to find x. Is there a better way to solve this or am I on the right track? Thanks for the help.
In the given graphs, suppose n =2. So, In the parabola $y = x^2$, the coordinate pairs are $(x, x^2)$. You should be able to see that the following points are on the graph: (1, 1), (−1, 1), (2, 4), (−2, 4), and so on.
The graph of the square root function is related to $y = x^2$. The coördinate pairs are $(x,\sqrt{x})$. For example, (1, 1), (4, 2), (9, 3), and so on.
So for any n, the graph of $y=x^{\frac{1}{n}}$ will not be negative.
5. So how would we explain that point A is (1,1)? The back of the book says that is the answer.
6. Hello, florx!
Your game plan is absolutely correct!
Solve for $x\!:\;\;x^n \:=\: x^{\frac{1}{n}}$
We have: . $x^n - x^{\frac{1}{n}} \;=\;0$
Factor: . $x^{\frac{1}{n}}\left(x^{\frac{n^2-1}{n}} - 1\right) \;=\;0$
And we have two equations to solve:
. . $x^{\frac{1}{n}} \:=\:0 \quad\Rightarrow\quad x \:=\:0^n \quad\Rightarrow\quad \boxed{x \:=\:0}$
. . $x^{\frac{n^1-1}{n}} - 1 \:=\:0 \quad\Rightarrow\quad x^{\frac{n^2-1}{n}} \:=\:1 \quad\Rightarrow\quad x \:=\:1^{\frac{n}{n^2-1}} \quad\Rightarrow\quad\boxed{x \:=\:1}\;\;\text{ for }n \neq \pm1$
7. That's just a little insane.
Back to my original question:
What number is alterered identically by a power and a root?
There are not too many candidates.
Only 0 and 1. finis.
Why drag through messy algebra with ambiguous exponents when it's really just a definition / property question? | {"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": 10, "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": 0.9441981911659241, "perplexity_flag": "head"} |
http://www.physicsforums.com/showpost.php?p=3131114&postcount=1 | View Single Post
## Continuous and differentiable points of f(x,y,z) (Existence of multivariable limit?)
1. The problem statement, all variables and given/known data
Find the continuous points P and the differentiable points Q of the function $$f$$ in $${R}^3$$, defined as
$$f(0,0,0) = 0$$
and
$$f(x,y,z) = \frac{xy(1-\cos{z})-z^3}{x^2+y^2+z^2}, (x,y,z) \ne (0,0,0)$$.
2. Relevant equations
3. The attempt at a solution
If you want to look at the limit I'm having trouble with, just skip a few paragraphs. I'm mostly including the rest in case anyone is in the mood to point out flaws in my reasoning.
Differentiating $$f$$ with respect to x, y and z, respectively (when $$(x,y,z) \ne (0,0,0)$$ will make it apparent that all three partials will contain a denominator of $$(x^2+y^2+z^2)^2$$ and a continuous numerator. Thus, these partials are continuous everywhere except in $$(0,0,0)$$, and it follows that $$f$$ is differentiable, and consequently, also continuous in all points $$(x,y,z) \ne (0,0,0)$$.
Investigating if $$f$$ is differentiable at $$(0,0,0)$$, we investigate the limit
$$\lim_{(h_1,h_2,h_3) \to (0,0,0)}{\frac{f(h_1,h_2,h_3) - f(0,0,0) - h_1 f_1(0,0,0) - h_2 f_2(0,0,0) - h_3 f_3(0,0,0)}{\sqrt{{h_1}^2 + {h_2}^2 + {h_3}^2}}} = \lim_{(h_1,h_2,h_3) \to (0,0,0)}{\frac{h_1 h_2 (1-\cos{h_3}) - {h_3}^3}{({h_1}^2 + {h_2}^2 + {h_3}^2)^{3/2}}}.$$
Evaluating along the line $$x = y = z$$, that is, $$h_1 = h_2 = h_3$$, it is found after a bit of work and one application of l'Hôpital's rule that the limit from the right does not equal the limit from the left, and hence, $$f$$ is not differentiable in $$(0,0,0)$$.
To prove continuity of $$f$$, we want to show that $$\lim_{(x,y,z) \to (0,0,0)}f(x,y,z) = 0$$. Since I haven't found any good counter-examples to this, I've tried to prove it with the epsilon-delta definition instead, with little luck.
We see that
$$|f(x,y,z) - 0| = \left|\frac{xy(1-\cos{z})-z^3}{x^2 + y^2 + z^2}\right| \le \left|\frac{xy(1-\cos{z})-z^3}{z^2}\right|,$$
getting me nowhere.
Trying with spherical coordinates instead, we get
$$|f(x,y,z)-0| = \left|\frac{{\rho}^2 {\sin^2 \phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - {\rho}^3 \cos^3 {\phi}}{{\rho}^2 \sin^2 {\phi} \cos^2 {\theta} + {\rho}^2 \sin^2 {\phi} \sin^2 {\theta} + {\rho}^2 \cos^2 {\phi}}\right| = \left|\sin^2 {\phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - \rho \cos^3 {\phi}\right|.$$
I'm not sure how to proceed. Suggestions?
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug | {"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": 21, "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": 0.928421676158905, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/7295/is-there-a-list-of-all-connected-t-0-spaces-with-5-points | # Is there a list of all connected $T_0$-spaces with 5 points?
Is there some place (on the internet or elsewhere) where I can find the number and preferably a list of all (isomorphism classes of) finite connected $T_0$-spaces with, say, 5 points?
In know that a $T_0$-topology on a finite set is equivalent to a partial ordering, and wikipedia tells me that there are, up to isomorphism, 63 partially ordered sets with precisely 5 elements. However, I am only interested in connected spaces, and I'd love to have a list (most preferably in terms of Hasse diagrams).
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## 1 Answer
This question was answered here.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.95074063539505, "perplexity_flag": "head"} |
http://mathhelpforum.com/math-topics/39571-quadratic-not-print.html | # is this a quadratic or not?
Printable View
• May 25th 2008, 10:01 AM
vicereine
is this a quadratic or not?
I'm doing this math course through correspondence and really the only indication they've given to determine a quadratic function is that the first difference numbers are not constant, while the second difference numbers are constant (same #). Well I'm stuck on this question because it doesn't follow the above rule, plus aren't the points always at the same distance on both sides of the parabola (mirrored)?
http://img172.imageshack.us/img172/8...uestiongx4.jpg
Question A says 'confirm' that it is quadratic though, so I'm wondering if it really is, so could someone explain WHY it is, if it is?
• May 25th 2008, 10:14 AM
galactus
Using Excel, I get a quadratic of $y=-0.3167x^{2}+2.3286x+1.5024$
This has an R^2 of .9994. Which is pretty dang good.
Plug in x=7 to find the height at 7 seconds. I get 2.3
It hits the ground when y=0. That occurs when x=7.96 seconds.
• May 25th 2008, 11:11 AM
vicereine
Well no where in the explanations or examples did they show how to form an equation from the chart or turn it into a graph...this is the first unit so maybe I'm not there yet. Like I said in the original post, it just covers the first and second differences. In all four examples given, the second differences were the same number...that's why I'm confused.
• May 25th 2008, 11:52 AM
bobak
Quote:
Originally Posted by vicereine
Well no where in the explanations or examples did they show how to form an equation from the chart or turn it into a graph...this is the first unit so maybe I'm not there yet. Like I said in the original post, it just covers the first and second differences. In all four examples given, the second differences were the same number...that's why I'm confused.
I don't know much about this subject but form what I understand the second derivative of a quadratic is constant but the method of second difference is a numerical approximation of the derivative (so there is bound to be variations). I think what you're looking for is a fairly constant second difference and linear relationship in with first differences.
Out of interest have you been taught to use technology to solve these problems or have you been taught to do them using graphical method. I plotted a graph of the first differences to get an approximations of the linear equation that relates them then integrated and I got an answer similar to what galactus posted.
Bobak
• May 25th 2008, 11:59 AM
ThePerfectHacker
1 Attachment(s)
Go here and download the program and install it.
1)Click F4 to insert a point series. And insert all these points.
2)Right click on the left where it says "Series 1" and select "Insert Tredline".
3)Click on polynomial and set the order be 2, then click okay.
All times are GMT -8. The time now is 05:02 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": 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": 0.939253032207489, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/20403/a-question-on-complex-analysis | # A question on complex analysis
Let $f,g:\mathbb C \to \mathbb C$ be two analytic functions such that $f(z)(g(z)+z^2)=0$ for all $z$ .Then prove that either $f(z)=0$ or $g(z)=-z^2$.
-
6
The $z^2$ is only a distraction. $h(z)=g(z)+z^2$ is analytic if $g$ is. (I assume that you forgot to mention that $f$ and $g$ are analytic.) What do you know about the zeros of analytic functions? – Jonas Meyer Feb 4 '11 at 15:06
yes ,f and g both are analytic functions . – user6618 Feb 4 '11 at 15:40
Please tell me whether i am correct or not: – user6618 Feb 4 '11 at 15:44
Is Uniqueness theorem applicable to f and this g – user6618 Feb 4 '11 at 15:48
@N Jana: Yes, that's the result you want to apply. Do you see how to apply it? – Jonas Meyer Feb 4 '11 at 15:50
show 2 more comments
## 2 Answers
I thought it might be a good idea to attempt to get this off of the Unanswered list by posting what is essentially the solution N Jana gave in the comments.
If there exists $a\in\mathbb C$ such that $f(a)\neq 0$, then by continuity there is an open disk at $a$ where $f$ is nonzero. Then $g(z)+z^2$ is zero on this disk, and by the identity theorem for analytic functions, $g(z)+z^2=0$ for all $z\in\mathbb{C}$.
The $z^2$ here adds nothing essential. The obvious generalization is that if a product of analytic functions on a connected open set is identically $0$, then one of the functions is identically zero. (And this means that the ring of analytic functions on a connected open set is an integral domain with pointwise operations.)
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If $f=0$ you have done, so can assume this is not the case. Can assume there is a point where both $f$ and $g-z^2$ are both different from zero otherwise you have done. Up to traslation can assume this point is the origin. So there exists integers $h,k$ such that $f=z^hf'$ and $g=z^kg'$ with $f',g'$ different from zero at the origin. then you have $f'(z^{h+k}g'-z^{2+k})=0$. Since $f'$ not zero at the origin this means that locally $z^{h+k}g'-z^{2+k}=0$ from which the thesis follows.
- | {"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": 29, "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": 0.9377259612083435, "perplexity_flag": "head"} |
http://cms.math.ca/Events/winter10/abs/cna.html | 2010 CMS Winter Meeting
Coast Plaza Hotel and Suites, Vancouver, December 4 - 6, 2010 www.cms.math.ca//Events/winter10
Convex and Nonsmooth Analysis
Org: Philip Loewen (UBC) and Yves Lucet (UBC-Okanagan)
[PDF]
HEINZ BAUSCHKE, UBC Okanagan
A Demiclosedness Principle and its Applications [PDF]
Browder's Demiclosedness Principle is one of the building blocks of fixed point theory. In this talk, I will report on a new demiclosedness principle and its applications.
(Based on joint work with Patrick Combettes, Paris VI.)
JAMES V. BURKE, University of Washington
On the subdifferential regularity of max root functions for polynomials [PDF]
In 2001, Burke and Overton showed that the abscissa mapping on polynomials is subdifferentially regular on the monic polynomials of degree $n$. In this talk I will discuss an extension of this result to the class of max polynomial root functions which includes both the polynomial abscissa and radius mappings. The approach to the computation of subgradient is somewhat simpler than that given by Burke and Overton and provides new insight into the variational properties of these functions.
This is joint work with Julia Eaton, University of Puget Sound.
MCLEAN EDWARDS, University of British Columbia
Relationships between classes of monotone operators [PDF]
Among monotone operators, subdifferentials of convex functions are particularly well behaved. Generalizations are often defined by selecting a certain desirable property of subdifferentials. We consider the classes of monotone operators defined by five such properties: maximality, strict monotonicity, 3-cyclic monotonicity, 3*-monotonicity, and paramonotonicity. Surprisingly, only three general relationships (two of which are obvious) link these five classes. This talk is a report on a complete catalogue of theorems and examples addressing all possible combinations of these five properties. I will describe the relationships that hold in general, and illustrate some low dimensional and linear operators that explore the boundaries.
NASSIF GHOUSSOUB, University of British Columbia
Modern convexity methods in nonlinear PDEs [PDF]
Convexity is making a comeback in PDE, Analysis, and geometry. I will try to show how.
YASIN GOCGUN, University of Washington
Approximate Dynamic Programming for Dynamic Stochastic Resource Allocation [PDF]
Dynamic allocation of multiple renewable resources to different types of jobs arriving randomly over time is a class of scheduling problems that arise in many fields such as logistics, transportation, and healthcare. Due to stochasticity inherent in these problems, they are generally termed as Dynamic Stochastic Resource Allocation Problems (DSRAPs). In DSRAPs, the challenge faced by the decision-maker in each time-period is how to schedule jobs waiting in queue or which jobs to service for the current period so as to optimize a certain performance metric over a finite/infinite horizon, given resource constraints. These problems are naturally modeled as a Markov Decision Process (MDP), a mathematical framework used to model systems that evolve stochastically over time.
While different classes of DSRAPs have been studied over decades, there is no mathematical technique to solve realistic instances of these problems mainly due to their combinatorial nature. As such, researchers resort to approximation techniques such as Approximate Dynamic Programming (ADP) to tackle these problems, yet they typically ignore certain aspects of dynamic resource allocation because of imposing assumptions such as single resource and identical resource consumption values per job.
In this research, we employ a mathematical-programming-based ADP technique called Lagrangian-relaxation for dealing with a large class of the DSRAPs. Specifically, we develop abstract problem description, MDP models of these problems, and approximately solve them using the Lagrangian approach. The results of our extensive computational experiments reveal that the Lagrangian approach significantly outperforms easy-to-use heuristic decision rules.
RAFAL GOEBEL, Loyola University Chicago
Set-valued Lyapunov functions for pointwise asymptotic stability [PDF]
Pointwise asymptotic stability of a set, for a difference inclusion, requires that each point of the set be Lyapunov stable and that every solution to the inclusion, from a neighborhood of the set, be convergent and have the limit in the set. It is equivalent to the usual asymptotic stability for a single equilibrium, but is different in general, especially for noncompact sets of equilibria. Set-valued Lyapunov functions are set-valued mappings which characterize pointwise asymptotic stability in a way similar to how Lyapunov functions characterize the usual asymptotic stability. The talk will present necessary and sufficient conditions for pointwise asymptotic stability in terms of weak and strict set-valued Lyapunov functions.
ELDAD HABER, UBC
ZHAOSONG LU, Simon Fraser University
First-order Methods for Constrained Nonsmooth Optimization [PDF]
We propose a novel augmented Lagrangian (AL) method for a class of constrained nonsmooth optimization problems. Also, we propose two nonmonotone gradient methods for solving the AL subproblems. Global and local convergence of the methods are established. Finally, we discuss the application of this method to a new formulation of sparse principal component analysis and present some computational results which substantially outperform the other existing approaches.
YVES LUCET, UBC Okanagan
Recent Developments in Computational Convex Analysis: Convex Hulls and Graph-Matrix Calculus [PDF]
We present a couple of algorithms to compute the convex envelope of piecewise linear-quadratic (PLQ) univariate functions. The first is simple to implement but has quadratic complexity while the second is more involved but has linear time complexity. In addition, we present a new family of algorithms for computing convex operators of PLQ functions by manipulating the graph of the subdifferential using Goebel's graph-matrix calculus rules.
TAMON STEPHEN, Simon Fraser University
Minimal Conflicting Sets in ancestral genome reconstruction [PDF]
We consider problems of generating minimal obstacles (Conflicting Sets) to the consecutive-ones property for binary matrices used in ancestral genome reconstruction. We show that this problem can be reduced to a joint generation problem for boolean functions, and that this strategy can be helpful in discriminating between true and false positive ancestral syntenies in simulated and real data sets.
This is joint work with Cedric Chauve, Utz-Uwe Haus and Vivija You.
MOHAMED TAWHID, Thompson Rivers University
Solving Nonlinear Complementarity Problem by a Derivative-Free Descent Method [PDF]
We consider smooth/nonsmooth nonlinear complementarity Problem (NCP). We show under certain assumptions, any stationary point of the unconstrained minimization problem is already a solution of smooth NCP. Also, we suggest a derivative-free descent algorithm and give conditions for its convergence. Furthermore, we present some preliminary numerical results.
SHAWN WANG, University of British Columbia Okanagan
Self-dual Regularization of Monotone Operators via the Resolvent Average [PDF]
We give two self-dual regularization of maximal monotone operators on Hilbert spaces. These regularization and their set-valued inverse are strongly monotone, single-valued and Lipschitz with full domain. Moreover, these regularization graphically converges to the original monotone operator. If a maximal monotone operator has nonempty zeros, these self-dual regularization can be used to find its least norm solution. When the maximal monotone operator is the subdifferential of a proper lower semicontinuous convex function with nonempty minimizers, this translates to find the least norm minimizer.
LIN XIAO, Microsoft Research
Optimizing Orthogonality [PDF]
We consider a new method for the problem of multi-class classification in machine learning. Our formulation involves minimizing the weighted sum of the absolute values of the inner products between a set of vectors, which promotes orthogonality between the vectors. We give a sufficient condition on the weights such that this objective is a convex function of all the vectors. We also propose an efficient dual-averaging method for solving such a nonsmooth convex optimization problem.
LIANGJIN YAO, UBC Okanagan
The sum of a maximal monotone operator of type (FPV) and a maximal monotone operator with full domain is maximal monotone [PDF]
The most important open problem in Monotone Operator Theory concerns the maximal monotonicity of the sum of two maximal monotone operators provided that Rockafellar's constraint qualification holds.
In this paper, we prove the maximal monotonicity of the sum of $A + B$ provided that $A$ and $B$ are maximal monotone operators such that $\operatorname{dom} A\cap\operatorname{int}\operatorname{dom} B\neq\varnothing$, $A+N_{\overline{\operatorname{dom} B}}$ is of type (FPV), and $\operatorname{dom} A\cap\overline{\operatorname{dom} B}\subseteq\operatorname{dom} B$. The proof utilizes the Fitzpatrick function in an essential way.
JIM ZHU, Western Michigan University
Why bankers need to know convex analysis [PDF]
Maximizing (expected) concave utility functions has been a traditional model in economic and financial decision making. In early 1970s the work by Black and Scholes on option pricing lead to a `revolution'. The paradigm shifted to price financial assets with replicating portfolio. Later Cox and Ross proposed simpler computation method using risk-neutral probability which became a default in the financial industry. However, the result is less than idea. In this talk we argue that the Black-Scholes method is a special case of concave utility maximization and the Cox-Ross formula is the natural result of its dual. Moreover, in the system of Black-Scholes and Cox-Ross, the analysis of sensitivity to model perturbation is dangerously inadequate. After many financial crises, it is perhaps time to return to the time tested method of utility maximization in which convex analysis plays an essential role.
YURIJ ZINCHENKO, University of Calgary
Numerical estimation of the relative entropy of entanglement in Quantum Information Theory [PDF]
We propose a practical algorithm for the calculation of the relative entropy of entanglement (REE), defined as the minimum relative entropy between a state and the set of states with positive partial transpose. Our algorithm is based on a practical semi-definite cutting plane approach. In low dimensions the implementation of the algorithm in MATLAB provides an estimation for the REE with an absolute error smaller than 1e-3. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": 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": 0.8869368433952332, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/51287/what-defines-the-maximum-charge-a-capacitor-can-store/51292 | # What defines the maximum charge a capacitor can store?
The formula for a capacitor discharging is $Q=Q_0e^{-\frac{t}{RC}}$ Where $Q_0$ is the maximum charge. But what property defines the maximum charge a capacitor can store?
If it depends on capacitance then that means it depends on the voltage you put across the capacitor, but how can any capacitor "cope" with any voltage?
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## 1 Answer
The maximum charge a capacitor stores depends on the voltage $V_0$ you've used to charge it according to the formula:
$$Q_0=CV_0$$
However, a real capacitor will only work for voltages up to the breakdown voltage of the dielectric medium in the capacitor. So in reality, for every capacitor there is a maximum possible charge $Q_{max}$ given by:
$$Q_{max}=CV_{max}$$
where $V_{max}$ is the breakdown voltage of the dielectric medium in the capacitor.
-
I don't think this answers the whole question. $C$ may be the constant for a given capacitor but there are factors like the surface area of the plates in the capacitor that affect $C$. – Brandon Enright Apr 4 at 23:23
@BrandonEnright Yes, but I think the question is what the maximum charge is for a given capacitor. There is no maximum charge for an arbitrary capacitor. – jkej Apr 5 at 0:07 | {"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": 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": 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": 0.8840025663375854, "perplexity_flag": "head"} |
http://mathhelpforum.com/trigonometry/92398-re-substituion-problem.html | Thread:
1. Re-substituion problem
this is part of an integration, thats why i posted it here.
the needed substitution here is $u=sin(x)$
i end up with this sometime, which is correct:
$2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C$
now my book says the result should be:
$\frac{2}{5}\sqrt{sin(x)}[4+cos^2(x)]+C$
and wofram alpha says it should be:
$\frac{1}{5}\sqrt{sin(x)}[cos(2x)+9]+C$
to be honest... i tried and i tried, i dont get it how i will get any of those 2 solutions. would be so nice if someone could explain it me reeeeal slow
thx in advance
2. I think there's a typo in there. Did you mean $\sin^5( x)$ instead of $\sin (x^5)$?
In that case:
$2\sqrt{\sin x}-\frac{2\sqrt{\sin^5 x}}{5}+C=2\cdot \frac{5}{5}\sqrt{\sin x}-\frac{2\sqrt{\sin x}\sqrt{\sin^4 x}}{5}+C=$ $\frac{2\sqrt{\sin x}}{5}(5-\sqrt{\sin^4 x})+C=\frac{2\sqrt{\sin x}}{5}(5-\sin^2 x)+C=$ $\frac{2\sqrt{\sin x}}{5}(4 + (1-\sin^2 x))+C=$ $\frac{2\sqrt{\sin x}}{5}(4+\cos^2 x)+C$
3. oh yes that was a typo... thank you very much, all clear now!
need to improve my algebra skills
4. Originally Posted by coobe
this is part of an integration, thats why i posted it here.
the needed substitution here is $u=sin(x)$
i end up with this sometime, which is correct:
$2\sqrt{sin(x)}-\frac{2*\sqrt{sin(x^5)}}{5}+C$ ${\color{red} <font color=$2\sqrt{sin(x)}-\frac{2*\sqrt{(sin(x))^5}}{5}+CMr F says: You mean }" alt="{\color{red} 2\sqrt{sin(x)}-\frac{2*\sqrt{(sin(x))^5}}{5}+CMr F says: You mean }" />.
now my book says the result should be:
$\frac{2}{5}\sqrt{sin(x)}[4+cos^2(x)]+C$
and wofram alpha says it should be:
$\frac{1}{5}\sqrt{sin(x)}[cos(2x)+9]+C$
to be honest... i tried and i tried, i dont get it how i will get any of those 2 solutions. would be so nice if someone could explain it me reeeeal slow
thx in advance
$2\sqrt{\sin(x)}-\frac{2 \sqrt{(\sin(x))^5}}{5}+C = 2\sqrt{\sin(x)}-\frac{2 \sqrt{(\sin(x))^4 \sin x}}{5}+C$
$= 2\sqrt{\sin(x)}-\frac{2 \sin^2 x\sqrt{\sin x}}{5}+C$.
Now take out a factor of $\frac{2 \sqrt{\sin x}}{5}$ and substitute $\sin^2 x = 1 - \cos^2 x$ to get your book's answer.
To show that Wolfram's answer is equivalent to the book's answer you need to recall that $\cos (2x) = 2 \cos^2 (x) - 1$ .... | {"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": 20, "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": 0.9111067652702332, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2007/08/02/free-monoid-objects/ | # The Unapologetic Mathematician
## Free Monoid Objects
When we have an algebraic concept described as a set with extra structure, the morphisms between such structured sets are usually structure-preserving functions between the underlying sets. This gives us a “forgetful” functor which returns the underlying sets and functions. Then as we saw, we often have a left adjoint to this forgetful functor giving the “free” structure generated by a set.
But now that we’re talking about monoid objects we’re trying not to think about sets. A monoid object in $\mathcal{C}$ is a monoidal functor from $\mathrm{Th}(\mathbf{Mon})$ to $\mathcal{C}$, and a “homomorphism” of such monoid objects is a monoidal natural transformation. But the object part of such a functor is specified by one object of $\mathcal{C}$ — the image of $M\in\mathrm{Th}(\mathbf{Mon})$ — which we can reasonably call the “underlying object” of the monoid object. Similarly, a natural transformation will be specified by a morphism between the underlying objects (subject to naturality conditions, of course). That is, we have a “forgetful functor” from monoid objects in $\mathcal{C}$ to $\mathcal{C}$ itself. And a reasonable notion of a “free” monoid object will be a left adjoint to this functor.
Now, if the monoidal category $\mathcal{C}$ has coproducts indexed by the natural numbers, and if the functors $C\otimes\underline{\hphantom{X}}$ and $\underline{\hphantom{X}}\otimes C$ preserve these coproducts for all objects $C\in\mathcal{C}$, then the forgetful functor above will have a left adjoint. To say that the monoidal structure preserves these coproducts is to say that the following “distributive laws” hold:
$\coprod\limits_n(A\otimes B_n)\cong A\otimes\biggl(\coprod\limits_nB_n\biggr)$
$\coprod\limits_n(A_n\otimes B)\cong\biggl(\coprod\limits_nA_n\biggr)\otimes B$
For any object $C\in\mathcal{C}$ we can define the “free monoid object on $C$” to be $\coprod\limits_nC^{\otimes n}$, equipped with certain multiplication and unit morphisms. For the unit, we will use the inclusion morphism $C^{\otimes0}\rightarrow\coprod\limits_nC^{\otimes n}$ that comes for free with the coproduct. The multiplication will take a bit more work.
Given any natural numbers $m$ and $n$, the object $C^{\otimes m}\otimes C^{\otimes n}$ is canonically isomorphic to $C^{\otimes m+n}$, which then includes into $\coprod\limits_kC^{\otimes k}$ using the coproduct morphisms. But this object also includes into $\coprod\limits_{i,j}(C^{\otimes i}\otimes C^{\otimes j})$, which is isomorphic to $\biggl(\coprod\limits_iC^{\otimes i}\biggr)\otimes\biggl(\coprod\limits_jC^{\otimes j}\biggr)$. Thus by the universal property of coproducts there is a unique morphism $\mu:\biggl(\coprod\limits_iC^{\otimes i}\biggr)\otimes\biggl(\coprod\limits_jC^{\otimes j}\biggr)\rightarrow\biggl(\coprod\limits_kC^{\otimes k}\biggr)$. This is our multiplication.
Proving that these two morphisms satisfy the associativity and identity relations is straightforward, though somewhat tedious. Thus we have a monoid object in $\mathcal{C}$. The inclusion $C=C^{\otimes1}\rightarrow\coprod\limits_nC^{\otimes n}$ defines a universal arrow from $C$ to the forgetful functor, and so we have an adjunction.
So what does this look like in $\mathbf{Set}$? The free monoid object on a set $S$ will consist of the coproduct (disjoint union) of the sets of ordered $n$-tuples of elements of $S$. The unit will be the unique ${0}$-tuple $()$, and I’ll leave it to you to verify that the multiplication defined above becomes concatenation in this context. And thus we recover the usual notion of a free monoid.
One thing I slightly glossed over is showing that $\mathbf{Set}$ satisfies the hypotheses of our construction. It works here for the same reason it will work in many other contexts: $\mathbf{Set}$ is a closed category. Given any closed category with countable coproducts, the functor $C\otimes\underline{\hphantom{X}}$ has a right adjoint by definition. And thus it preserves all colimits which might exist. In particular, it preserves the countable coproducts, which is what the construction requires. The other functor preserves these coproducts as well because the category is symmetric — tensoring by $C$ on the left and tensoring by $C$ on the right are naturally isomorphic. Thus we have free monoid objects in any closed category with countable coproducts.
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http://mathhelpforum.com/calculus/51449-exponential-growth.html | Thread:
1. Exponential Growth
The count in a culture of bacteria was 400 after 2 hours and 25,600 after 6 hours.
(a) What is the relative rate of growth of the bacteria population?
(b) What was the initial size of the culture?
My big problem is how to find the rate of growth. Then when I find that I assume I plug in the rate and time=0 to find the initial size of the culture am I correct? I just don't know how to find that blasted rate of growth.
2. Let A be the initial size of the population, R be the growth rate per hour, and t be the time in hours. Let P represent the number of bacteria present. Then $P = AR^t$. We have $400 = AR^2$ and $25600 = AR^6$. Then:
$\frac{25600}{400} = \frac{AR^6}{AR^2} = R^4$.
Solve for R and then you can find A.
3. Hmm, I don't know how to solve for $R^4$ what am I doing with it? Modeling isn't exactly my strong suit, in fact its my worst. I'm sorry.
4. Bump because I don't understand.
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http://mathoverflow.net/questions/64440/suitable-references-for-the-the-stone-von-neumann-theorem/64473 | ## Suitable references for the the Stone-von Neumann Theorem
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Hi all,
I am working on a mathematical physics project now and I need to understand the Stone-von Neumann Theorem properly. Wikipedia says that it is any one of a number of different formulations of the uniqueness of the canonical commutation relations between position and momentum operators'.
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If you are familiar with the representation theory, I will recommend Kirillov : Elements of the theory of representations Mackey : The Theory of Unitary Group Representations I haven't looked, but suspect that Mackey : Unitary group representations in physics, probability, and number theory will have some enlightening comments. A toy model of this is looking at representation theory of $SL(2,\mathbb{R})$ on functions annihilated by the laplace beltrami operator. Any books on spectral theory of automorphic form will talk about it. – isildur May 10 2011 at 1:13
## 4 Answers
von Neumann's original proof is beautiful and quite simple. It deduces the Stone-von Neumann theorem from the Plancherel theorem. This proof can be found in Folland's book "Harmonic analysis in phase space". I have also tried to give an exposition of this proof, along with a proof of a generalization of this theorem due to Mackey, in my paper "An Easy Proof of the Stone-von Neumann-Mackey Theorem" (Expositiones Mathematicae, 24(1):110-118, 2011, also available on the arXiv).
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An excellent reference explaining the history and significance of the Stone von Neumann theorem is Jonathan Rosenberg's
"A Selective History of the Stone-von Neumann Theorem"
available at
http://www-users.math.umd.edu/~jmr/StoneVNart.pdf
Note that one doesn't actually have uniqueness of representations of the position and momentum operators (since they're unbounded). The theorem applies to their exponentiated versions, which give the Heisenberg (mathematician's name) or Weyl (physicist's name) group.
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I first learned about it from the first chapter of Lion and Vergne's book "The Weil representation, Maslov index and theta series", which is really much more down to earth than its title might suggest.
EDIT : It maybe should be emphasized that I am not a representation theorist, and when I read the above book I knew far less about the subject than I do now (which isn't a lot). This seems to me to be a positive thing -- I found the book pretty easy going despite not knowing all that much when I looked at it.
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I would also recommend M. Reed and B. Simon, Methods of modern mathematical physics, vol I, chapter VIII (and maybe vol II, chapter X). The theorem itself is VIII.14 with Corollary after that. The reference is useful also due to counterexample, demonstrating that the idea about “one-to-one correspondence” used in Wikipedia and some other places is not very good one.
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I might be biased, but this is the RIGHT reference. At least for the analysts under us (or better functional analysts or spectral theorists). – Helge May 17 2011 at 16:04 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9292919635772705, "perplexity_flag": "middle"} |
http://nrich.maths.org/6781 | nrich enriching mathematicsSkip over navigation
### 14 Divisors
What is the smallest number with exactly 14 divisors?
### Summing Consecutive Numbers
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Rule of Three
If it takes four men one day to build a wall, how long does it take 60,000 men to build a similar wall?
# Weekly Problem 20 - 2010
##### Stage: 3 Short Challenge Level:
The diagram show the net of a regular octohedron.
In a Magic Octahedron, the four numbers on the faces that meet at a vertex add up to make the same total for every vertex. If the letters $F, G, H, J$ and $K$ are replaced with the numbers $2, 4, 6, 7$ and $8$, in some order, to make a Magic Octohedron, what is the value of $G + J$?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
This problem is taken from the UKMT Mathematical Challenges.
View the previous week's solution
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9044267535209656, "perplexity_flag": "middle"} |
http://quant.stackexchange.com/questions/tagged/itos-lemma+martingale | # Tagged Questions
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### How to use Itô's formula to deduce that a stochastic process is a martingale?
I'm working through different books about financial mathematics and solving some problems I get stuck. Suppose you define an arbitrary stochastic process, for example $X_t := W_t^8-8t$ where \$ W_t ... | {"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": 0.9099560379981995, "perplexity_flag": "middle"} |
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## Intuitive content of Loop Gravity--Rovelli's program
http://arxiv.org/abs/1302.3833
Loop Quantum Cosmology
Ivan Agullo, Alejandro Corichi
(Submitted on 15 Feb 2013)
This Chapter provides an up to date, pedagogical review of some of the most relevant advances in loop quantum cosmology. We review the quantization of homogeneous cosmological models, their singularity resolution and the formulation of effective equations that incorporate the main quantum corrections to the dynamics. We also summarize the theory of quantized metric perturbations propagating in those quantum backgrounds. Finally, we describe how this framework can be applied to obtain a self-consistent extension of the inflationary scenario to incorporate quantum aspects of gravity, and to explore possible phenomenological consequences.
52 pages, 5 figures. To appear as a Chapter of "The Springer Handbook of Spacetime," edited by A. Ashtekar and V. Petkov. (Springer-Verlag, at Press).
http://arxiv.org/abs/1302.1496 Standard Model Higgs field and energy scale of gravity F.R. Klinkhamer (Submitted on 6 Feb 2013 (v1), last revised 14 Feb 2013 (this version, v3)) The effective potential of the Higgs scalar field in the Standard Model may have a second degenerate minimum at an ultrahigh vacuum expectation value. This second minimum then determines, by radiative corrections, the values of the top-quark and Higgs-boson masses at the standard minimum corresponding to the electroweak energy scale. An argument is presented that this ultrahigh vacuum expectation value is proportional to the energy scale of gravity, E_{Planck} \equiv \sqrt{\hbar c^5/G_N}, considered to be characteristic of a spacetime foam. In the context of a simple model, the existence of kink-type wormhole solutions places a lower bound on the ultrahigh vacuum expectation value and this lower bound is of the order of E_{Planck}. http://arxiv.org/abs/1302.3680 Quantum Gravity on a Quantum Computer? Achim Kempf (Submitted on 15 Feb 2013) EPR-type measurements on spatially separated entangled spin qubits allow one, in principle, to detect curvature. Also the entanglement of the vacuum state is affected by curvature. Here, we ask if the curvature of spacetime can be expressed entirely in terms of the spatial entanglement structure of the vacuum. This would open up the prospect that quantum gravity could be simulated on a quantum computer and that quantum information techniques could be fully employed in the study of quantum gravity. http://arxiv.org/abs/1302.3648 Causality and non-equilibrium second-order phase transitions in inhomogeneous systems A. del Campo, T. W. B. Kibble, W. H. Zurek (Submitted on 14 Feb 2013) When a second-order phase transition is crossed at fine rate, the evolution of the system stops being adiabatic as a result of the critical slowing down in the neighborhood of the critical point. In systems with a topologically nontrivial vacuum manifold, disparate local choices of the ground state lead to the formation of topological defects. The universality class of the transition imprints a signature on the resulting density of topological defects: It obeys a power law in the quench rate, with an exponent dictated by a combination of the critical exponents of the transition. In inhomogeneous systems the situation is more complicated, as the spontaneous symmetry breaking competes with bias caused by the influence of the nearby regions that already chose the new vacuum. As a result, the choice of the broken symmetry vacuum may be inherited from the neighboring regions that have already entered the new phase. This competition between the inherited and spontaneous symmetry breaking enhances the role of causality, as the defect formation is restricted to a fraction of the system where the front velocity surpasses the relevant sound velocity and phase transition remains effectively homogeneous. As a consequence, the overall number of topological defects can be substantially suppressed. When the fraction of the system is small, the resulting total number of defects is still given by a power law related to the universality class of the transition, but exhibits a more pronounced dependence on the quench rate. This enhanced dependence complicates the analysis but may also facilitate experimental test of defect formation theories.
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1302.5265 The loop quantum gravity black hole Rodolfo Gambini, Jorge Pullin (Submitted on 21 Feb 2013) We quantize spherically symmetric vacuum gravity without gauge fixing the diffeomorphism constraint. Through a rescaling, we make the algebra of Hamiltonian constraints Abelian and therefore the constraint algebra is a true Lie algebra. This allows the completion of the Dirac quantization procedure using loop quantum gravity techniques. We can construct explicitly the exact solutions of the physical Hilbert space annihilated by all constraints. New observables living in the bulk appear at the quantum level (analogous to spin in quantum mechanics) that are not present at the classical level and are associated with the discrete nature of the spin network states of loop quantum gravity. The resulting quantum space-times resolve the singularity present in the classical theory inside black holes. The new observables that arise suggest a possible resolution for the "firewall" problem of evaporating black holes. Comments: 4 pages,
http://arxiv.org/abs/1302.5273 There exist no 4-dimensional geodesically equivalent metrics with the same stress-energy tensor Volodymir Kiosak, Vladimir S. Matveev (Submitted on 21 Feb 2013) We show that if two 4-dimensional metrics of arbitrary signature on one manifold are geodesically equivalent (i.e., have the same geodesics considered as unparameterized curves) and are solutions of the Einstein field equation with the same stress-energy tensor, then they are affinely equivalent or flat. Under the additional assumption that the metrics are complete or the manifold is closed, the result survives in all dimensions >2. http://arxiv.org/abs/1302.5162 On CCC-predicted concentric low-variance circles in the CMB sky V. G. Gurzadyan, R. Penrose (Submitted on 21 Feb 2013) A new analysis of the CMB, using WMAP data, supports earlier indications of non-Gaussian features of concentric circles of low temperature variance. Conformal cyclic cosmology (CCC) predicts such features from supermassive black-hole encounters in an aeon preceding our Big Bang. The significance of individual low-variance circles in the true data has been disputed; yet a recent independent analysis has confirmed CCC's expectation that CMB circles have a non-Gaussian temperature distribution. Here we examine concentric sets of low-variance circular rings in the WMAP data, finding a highly non-isotropic distribution. A new "sky-twist" procedure, directly analysing WMAP data, without appeal to simulations, shows that the prevalence of these concentric sets depends on the rings being circular, rather than even slightly elliptical, numbers dropping off dramatically with increasing ellipticity. This is consistent with CCC's expectations; so also is the crucial fact that whereas some of the rings' radii are found to reach around $15^\circ$, none exceed $20^\circ$. The non-isotropic distribution of the concentric sets may be linked to previously known anomalous and non-Gaussian CMB features.
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1302.5695 Quantum matter in quantum space-time Martin Bojowald, Golam Mortuza Hossain, Mikhail Kagan, Casey Tomlin (Submitted on 22 Feb 2013) Quantum matter in quantum space-time is discussed using general properties of energy-conservation laws. As a rather radical conclusion, it is found that standard methods of differential geometry and quantum field theory on curved space-time are inapplicable in canonical quantum gravity, even at the level of effective equations. 16 pages http://arxiv.org/abs/1302.5564 Spin-cube Models of Quantum Gravity A. Mikovic (Submitted on 22 Feb 2013) We study the state-sum models of quantum gravity based on a representation 2-category of the Poincare 2-group. We call them spin-cube models, since they are categorical generalizations of spin-foam models. A spin-cube state sum can be considered as a path integral for a constrained 2-BF theory, and depending on how the constraints are imposed, a spin-cube state sum can be reduced to a path integral for the area-Regge model with the edge-length constraints, or to a path integral for the Regge model. We also show that the effective actions for these spin-cube models have the correct classical limit. 16 pages brief mention (Shapo is always interesting): http://arxiv.org/abs/1302.5619 Spontaneously Broken Conformal Symmetry: Dealing with the Trace Anomaly Roberta Armillis, Alexander Monin, Mikhail Shaposhnikov (Submitted on 22 Feb 2013) The majority of renormalizable field theories possessing the scale invariance at the classical level exhibits the trace anomaly once quantum corrections are taken into account. This leads to the breaking of scale and conformal invariance. At the same time any realistic theory must contain gravity and is thus non-renormalizable. We show that discarding the renormalizability it is possible to construct viable models allowing to preserve the scale invariance at the quantum level. We present explicit one-loop computations for two toy models to demonstrate the main idea of the approach. Constructing the renormalized energy momentum tensor we show that it is traceless, meaning that the conformal invariance is also preserved. 20 pages, 5 figures
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1302.6264 The Solution to the Problem of Time in Shape Dynamics Julian Barbour, Tim Koslowski, Flavio Mercati (Submitted on 25 Feb 2013) The absence of unique time evolution in Einstein's spacetime description of gravity leads to the hitherto unresolved 'problem of time' in quantum gravity. Shape Dynamics is an objectively equivalent representation of gravity that trades spacetime refoliation invariance for three-dimensional conformal invariance. Its logical completion presented here gives a dimensionless description of gravitational dynamics. We show that in this framework the classical problem of time is completely solved. Since a comparable definitive solution is impossible within the spacetime description, we believe Shape Dynamics provides a key ingredient for the creation of quantum gravity. 14 pages http://arxiv.org/abs/1302.6566 Singularity resolution from polymer quantum matter Andreas Kreienbuehl, Tomasz Pawlowski (Submitted on 26 Feb 2013) We study the polymeric nature of quantum matter fields using the example of a Friedmann-Lemaitre-Robertson-Walker universe sourced by a minimally coupled massless scalar field. The model is treated in the symmetry reduced regime via deparametrization techniques, with the scale factor playing the role of time. Subsequently the remaining dynamic degrees of freedom are polymer quantized. The analysis of the resulting dynamic shows that the big bang singularity is resolved, although with the form of the resolution differing significantly from that of the models with matter clocks: dynamically, the singularity is made passable rather than avoided. Furthermore, the results of the genuine quantum analysis expose crucial limitations to the so-called effective dynamics in loop quantum cosmology when applied outside of the simplest isotropic settings. 12 pages and 4 figures
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1302.7142 Holonomy Operator and Quantization Ambiguities on Spinor Space Etera R. Livine (Submitted on 28 Feb 2013) We construct the holonomy-flux operator algebra in the recently developed spinor formulation of loop gravity. We show that, when restricting to SU(2)-gauge invariant operators, the familiar grasping and Wilson loop operators are written as composite operators built from the gauge-invariant 'generalized ladder operators' recently introduced in the U(N) approach to intertwiners and spin networks. We comment on quantization ambiguities that appear in the definition of the holonomy operator and use these ambiguities as a toy model to test a class of quantization ambiguities which is present in the standard regularization and definition of the Hamiltonian constraint operator in loop quantum gravity. 14 pages http://arxiv.org/abs/1302.7135 Fields and Laplacians on Quantum Geometries Johannes Thürigen (Submitted on 28 Feb 2013) In fundamentally discrete approaches to quantum gravity such as loop quantum gravity, spin-foam models, group field theories or Regge calculus observables are functions on discrete geometries. We present a bra-ket formalism of function spaces and discrete calculus on abstract simplicial complexes equipped with geometry and apply it to the mentioned theories of quantum gravity. In particular we focus on the quantum geometric Laplacian and discuss as an example the expectation value of the heat kernel trace from which the spectral dimension follows. 3 pages, submitted to the Proceedings of the 13th Marcel Grossmann Meeting (MG13), Stockholm, July 1-7, 2012 http://arxiv.org/abs/1302.7037 Loop Quantization of Shape Dynamics Tim Koslowski (Submitted on 28 Feb 2013) Loop Quantum Gravity (LQG) is a promising approach to quantum gravity, in particular because it is based on a rigorous quantization of the kinematics of gravity. A difficult and still open problem in the LQG program is the construction of the physical Hilbert space for pure quantum gravity. This is due to the complicated nature of the Hamilton constraints. The Shape Dynamics description of General Relativity (GR) replaces the Hamilton constraints with spatial Weyl constraints, so the problem of finding the physical Hilbert space reduces to the problem of quantizing the Weyl constraints. Unfortunately, it turns out that a loop quantization of Weyl constraints is far from trivial despite their intuitive physical interpretation. A tentative quantization proposal and interpretation proposal is given in this contribution. 3 pages, talk given at the 13th Marcel Grossmann Meeting, Stockholm 1-7 July 2012 I don't normally list online seminar talks but these might be of particular interest to people following QG research: http://pirsa.org/13020132/ Quantum Gravity as Random Geometry Vincent Rivasseau Abstract: Matrix models, random maps and Liouville field theory are prime tools which connect random geometry and quantum gravity in two dimensions. The tensor track is a new program to extend this connection to higher dimensions through the corresponding notions of tensor models, colored triangulations and tensor group field theories. 27/02/2013 http://pirsa.org/13020146/ The universe as a process of unique events Lee Smolin 26/02/2013 (Covers material from a new book "Time Reborn" listed on Amazon to appear April 2013) also of interest, possibly related: http://arxiv.org/abs/1302.7291 General Relativity and Quantum Cosmology The arrow of time and the nature of spacetime George F R Ellis (Submitted on 28 Feb 2013) This paper extends the work of a previous paper [arXiv:1108.5261] on top-down causation and quantum physics, to consider the origin of the arrow of time. It proposes that a past condition' cascades down from cosmological to micro scales, being realized in many microstructures and setting the arrow of time at the quantum level by top-down causation. This physics arrow of time then propagates up, through underlying emergence of higher level structures, to geology, astronomy, engineering, and biology. The appropriate space-time picture to view all this is an emergent block universe (EBU'), that recognizes the way the present is different from both the past and the future. This essential difference is the ultimate reason the arrow of time has to be the way it is. 56 pages, 6 figures
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1303.0195 Living in Curved Momentum Space J. Kowalski-Glikman (Submitted on 1 Mar 2013) In this paper we review some aspects of relativistic particles' mechanics in the case of a non-trivial geometry of momentum space. We start with showing how the curved momentum space arises in the theory of gravity in 2+1 dimensions coupled to particles, when (topological) degrees of freedom of gravity are solved for. We argue that there might exist a similar topological phase of quantum gravity in 3+1 dimensions. Then we characterize the main properties of the theory of interacting particles with curved momentum space and the symmetries of the action. We discuss the spacetime picture and the emergence of the principle of relative locality, according to which locality of events is not absolute but becomes observer dependent, in the controllable, relativistic way. We conclude with the detailed review of the most studied kappa-Poincare framework, which corresponds to the de Sitter momentum space. 23 pages http://arxiv.org/abs/1303.0196 Inhomogeneous Universe in Loop Quantum Gravity Francesco Cianfrani (Submitted on 1 Mar 2013) It is discussed a truncation of the kinematical Hilbert space of Loop Quantum Gravity, which describes the dynamical system associated with an inhomogeneous cosmological model. 3 pages, contribution to the Proceedings of the 13th Marcel Grossman Meeting (Stockholm, Sweden, July 1-7 2012) http://arxiv.org/abs/1303.0060 Differences and similarities between Shape Dynamics and General Relativity Henrique Gomes, Tim Koslowski (Submitted on 1 Mar 2013) The purpose of this contribution is to elucidate some of the properties of Shape Dynamics (SD) and is largely based on a recent longer article. We shall point out some of the key differences between SD and related theoretical constructions, illustrate the central mechanism of symmetry trading in electromagnetism and finally point out some new quantization strategies inspired by SD. We refrain from describing mathematical detail and from citing literature. For both we refer to the longer article. 3 pages, talk given at the 13th Marcel Grossmann Meeting in Stockholm, July 2012 http://arxiv.org/abs/1303.0174 MG13 Proceedings: A lattice Universe as a toy-model for inhomogeneous cosmology Jean-Philippe Bruneton (Submitted on 1 Mar 2013) We briefly report on a previously found new, approximate, solution to Einstein field equations, describing a cubic lattice of spherical masses. This model mimics in a satisfactory way a Universe which can be strongly inhomogeneous at small scales, but quite homogeneous at large ones. As a consequence of field equations, the lattice Universe is found to expand or contract in the same way as the solution of a Friedmann Universe filled with dust having the same average density. The study of observables indicates however the possible existence of a fitting problem, i.e. the fact that the Friedmann model obtained from past-lightcone observables does not match with the one obtained by smoothing the matter content of the Universe. 3 pages. Prepared for MG13 conference
http://arxiv.org/abs/1303.0762 A new perspective on early cosmology Emanuele Alesci (Submitted on 4 Mar 2013) We present a new perspective on early cosmology based on Loop Quantum Gravity. We use projected spinnetworks, coherent states and spinfoam techniques, to implement a quantum reduction of the full Kinematical Hilbert space of LQG, suitable to describe inhomogeneous cosmological models. Some preliminary results on the solutions of the Scalar constraint of the reduced theory are also presented. http://arxiv.org/abs/1303.0433 A Dynamics for Discrete Quantum Gravity Stan Gudder (Submitted on 2 Mar 2013) This paper is based on the causal set approach to discrete quantum gravity. We first describe a classical sequential growth process (CSGP) in which the universe grows one element at a time in discrete steps. At each step the process has the form of a causal set (causet) and the "completed" universe is given by a path through a discretely growing chain of causets. We then quantize the CSGP by forming a Hilbert space $H$ on the set of paths. The quantum dynamics is governed by a sequence of positive operators $\rho_n$ on $H$ that satisfy normalization and consistency conditions. The pair $(H,\brac{\rho_n})$ is called a quantum sequential growth process (QSGP). We next discuss a concrete realization of a QSGP in terms of a natural quantum action. This gives an amplitude process related to the sum over histories" approach to quantum mechanics. Finally, we briefly discuss a discrete form of Einstein's field equation and speculate how this may be employed to compare the present framework with classical general relativity theory.
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1303.0752 Inclusion of matter in inhomogeneous loop quantum cosmology Daniel Martín-de Blas, Mercedes Martín-Benito, Guillermo A. Mena Marugán (Submitted on 4 Mar 2013) We study the hybrid quantization of the linearly polarized Gowdy $T^3$ model with a massless scalar field with the same symmetries as the metric. For simplicity, we quantize its restriction to the model with local rotational symmetry. Using this hybrid approach, the homogeneous degrees of freedom of the geometry are quantized \`a la loop, leading to the resolution of the cosmological singularity. A Fock quantization is employed both for the matter and the gravitational inhomogeneities. Owing to the inclusion of the massless scalar field this system allows us to modelize flat Friedmann-Robertson-Walker cosmologies filled with inhomogeneities propagating in one direction, providing a perfect scenario to study the quantum back-reaction of the inhomogeneities on the polymeric homogeneous and isotropic background. 4 pages
Quote by marcus I told you I would make several false starts. Eventually there should be a non-technical description of loop gravity in only one to ten pages. Let's keep this thread going until we have one, or find one in the literature.
Hi, sorry for the interruption, I quoted the above from post #17 of this thread.
I'm just wondering if a non-technical discussion of loop gravity in only one to ten pages has ever been achieved?
I would love to have a nice concise and refined synopsis of the main ideas of LQG along with perhaps a well-organized outline of what types of topics are most required.
My specific interests in LQG is to try to understand how and why it is 'background independent", and also any information that might be known on how it might relate to the entropy associated with the horizon area of a black hole. Or actually any information on how a spin network itself relates to the concept of entropy.
Thanks.
Recognitions: Gold Member Science Advisor Hi Leucippus, this thread started out with discussion (in 2003) but evolved into a Loop and allied QG bibliography. We can start separate discussion threads. You have some good questions so I gathered excerpts (including from this latest) and started a new thread. I hope the new one will prove satisfactory. http://www.physicsforums.com/showthr...=1#post4298277
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1303.1687 Quantum states of the bouncing universe Jean Pierre Gazeau, Jakub Mielczarek, Wlodzimierz Piechocki (Submitted on 7 Mar 2013) In this paper we study quantum dynamics of the bouncing cosmological model. We focus on the model of the flat Friedman-Robertson-Walker universe with a free scalar field. The bouncing behavior, which replaces classical singularity, appears due to the modification of general relativity along the methods of loop quantum cosmology. We show that there exist a unitary transformation that enables to describe the system as a free particle with Hamiltonian equal to canonical momentum. We examine properties of the various quantum states of the Universe: boxcar state, standard coherent state, and soliton-like state, as well as Schrödinger's cat states constructed from these states. Characteristics of the states such as quantum moments and Wigner functions are investigated. We show that each of these states have, for some range of parameters, a proper semiclassical limit fulfilling the correspondence principle. Decoherence of the superposition of two universes is described and possible interpretations in terms of triad orientation and Belinsky-Khalatnikov-Lifshitz conjecture are given. Some interesting features regarding the area of the negative part of the Wigner function have emerged. 18 pages, 19 figures brief mention: http://arxiv.org/abs/1303.1535 The Structure of the Gravitational Action and its relation with Horizon Thermodynamics and Emergent Gravity Paradigm Krishnamohan Parattu, Bibhas Ranjan Majhi, T. Padmanabhan (Submitted on 6 Mar 2013) http://arxiv.org/abs/1303.1782 Non-Associative Geometry and the Spectral Action Principle Shane Farnsworth, Latham Boyle (Submitted on 7 Mar 2013) Chamseddine and Connes have shown how the action for Einstein gravity, coupled to the SU(3) × SU(2) × U(1) standard model of particle physics, may be elegantly recast as the "spectral action" on a certain "non-commutative geometry." In this paper, we show how this formalism may be extended to "non-associative geometries," and explain the motivations for doing so. As a guiding illustration, we present the simplest non-associative geometry (based on the octonions) and evaluate its spectral action: it describes Einstein gravity coupled to a G2 gauge theory, with 8 Dirac fermions (which transform as a singlet and a septuplet under G2). We use this example to illustrate how non-associative geometries may be naturally linked to ordinary (associative) geometries by a certain twisting procedure. This is just the simplest example: in a forthcoming paper we show how to construct realistic models that include Higgs fields, spontaneous symmetry breaking and fermion masses.
Don't miss these http://arxiv.org/abs/1303.1537 On the theory of composition in physics Lucien Hardy (Submitted on 6 Mar 2013) We develop a theory for describing composite objects in physics. These can be static objects, such as tables, or things that happen in spacetime (such as a region of spacetime with fields on it regarded as being composed of smaller such regions joined together). We propose certain fundamental axioms which, it seems, should be satisfied in any theory of composition. A key axiom is the order independence axiom which says we can describe the composition of a composite object in any order. Then we provide a notation for describing composite objects that naturally leads to these axioms being satisfied. In any given physical context we are interested in the value of certain properties for the objects (such as whether the object is possible, what probability it has, how wide it is, and so on). We associate a generalized state with an object. This can be used to calculate the value of those properties we are interested in for for this object. We then propose a certain principle, the composition principle, which says that we can determine the generalized state of a composite object from the generalized states for the components by means of a calculation having the same structure as the description of the generalized state. The composition principle provides a link between description and prediction. http://arxiv.org/abs/1303.1538 Reconstructing quantum theory Lucien Hardy (Submitted on 6 Mar 2013) We discuss how to reconstruct quantum theory from operational postulates. In particular, the following postulates are consistent only with for classical probability theory and quantum theory. Logical Sharpness: There is a one-to-one map between pure states and maximal effects such that we get unit probability. This maximal effect does not give probability equal to one for any other pure state. Information Locality: A maximal measurement is effected on a composite system if we perform maximal measurements on each of the components. Tomographic Locality: The state of a composite system can be determined from the statistics collected by making measurements on the components. Permutability: There exists a reversible transformation on any system effecting any given permutation of any given maximal set of distinguishable states for that system. Sturdiness: Filters are non-flattening. To single out quantum theory we need only add any requirement that is inconsistent with classical probability theory and consistent with quantum theory. http://arxiv.org/abs/1303.1632 Higgs potential and confinement in Yang-Mills theory on exotic R^4 Torsten Asselmeyer-Maluga, Jerzy Król (Submitted on 7 Mar 2013) We show that pure SU(2) Yang-Mills theory formulated on certain exotic R^4 from the radial family shows confinement. The condensation of magnetic monopoles and the qualitative form of the Higgs potential are derived from the exotic R^4, e. A relation between the Higgs potential and inflation is discussed. Then we obtain a formula for the Higgs mass and discuss a particular smoothness structure so that the Higgs mass agrees with the experimental value. The singularity in the effective dual U(1) potential has its cause by the exotic 4-geometry and agrees with the singularity in the maximal abelian gauge scenario. We will describe the Yang-Mills theory on e in some limit as the abelian-projected effective gauge theory on the standard R^4. Similar results can be derived for SU(3) Yang-Mills theory on an exotic R^4 provided dual diagonal effective gauge bosons propagate in the exotic 4-geometry. http://arxiv.org/abs/1303.1803 Classifying gauge anomalies through SPT orders and classifying anomalies through topological orders Xiao-Gang Wen (Submitted on 7 Mar 2013) In this paper, we systematically study gauge anomalies in bosonic and fermionic weak-coupling gauge theories with gauge group G (which can be continuous or discrete). We argue that, in d space-time dimensions, the gauge anomalies are described by the elements in Free[H^{d+1}(G,R/Z)]\oplus H_\pi^{d+1}(BG,R/Z). The well known Adler-Bell-Jackiw anomalies are classified by the free part of the group cohomology class H^{d+1}(G,R/Z) of the gauge group G (denoted as Free[H^{d+1}(G,\R/\Z)]). We refer other kinds of gauge anomalies beyond Adler-Bell-Jackiw anomalies as nonABJ gauge anomalies, which include Witten SU(2) global gauge anomaly. We introduce a notion of \pi-cohomology group, H_\pi^{d+1}(BG,R/Z), for the classifying space BG, which is an Abelian group and include Tor[H^{d+1}(G,R/Z)] and topological cohomology group H^{d+1}(BG,\R/\Z) as subgroups. We argue that H_\pi^{d+1}(BG,R/Z) classifies the bosonic nonABJ gauge anomalies, and partially classifies fermionic nonABJ anomalies. We also show a very close relation between gauge anomalies and symmetry-protected trivial (SPT) orders [also known as symmetry-protected topological (SPT) orders] in one-higher dimensions. Such a connection will allow us to use many well known results and well developed methods for gauge anomalies to study SPT states. In particular, the \pi-cohomology theory may give a more general description of SPT states than the group cohomology theory. http://arxiv.org/abs/1212.4863 Boundary Degeneracy of Topological Order Juven Wang, Xiao-Gang Wen (Submitted on 19 Dec 2012 (v1), last revised 23 Jan 2013 (this version, v2)) We introduce the notion of boundary degeneracy of topologically ordered states on a compact orientable spatial manifold with boundaries, and emphasize that it provides richer information than the bulk degeneracy. Beyond the bulk-edge correspondence, we find the ground state degeneracy of fully gapped edge states depends on boundary gapping conditions. We develop a quantitative description of different types of boundary gapping conditions by viewing them as different ways of non-fractionalized particle condensation on the boundary. Via Chern-Simons theory, this allows us to derive the ground state degeneracy formula in terms of boundary gapping conditions, which reveals more than the fusion algebra of fractionalized quasiparticles. We apply our results to Toric code and Levin-Wen string-net models. By measuring the boundary degeneracy on a cylinder, we predict Z_k gauge theory and U(1)_k x U(1)_k non-chiral fractional quantum hall state at even integer k can be experimentally distinguished. Our work refines definitions of symmetry protected topological order and intrinsic topological order.
Recognitions: Gold Member Science Advisor http://arxiv.org/abs/1303.2773 BTZ Black Hole Entropy in Loop Quantum Gravity and in Spin Foam Models J.Manuel Garcia-Islas (Submitted on 12 Mar 2013) We present a comparison of the calculation of BTZ black hole entropy in loop quantum gravity and in spin foam models. We see that both give the same answer. 6 pages, 3 figures brief mention: http://arxiv.org/abs/1303.2719 Another Survey of Foundational Attitudes Towards Quantum Mechanics Christoph Sommer (Submitted on 11 Mar 2013) Although it has been almost 100 years since the beginnings of quantum mechanics, the discussions about its interpretation still do not cease. Therefore, a survey of opinions regarding this matter is of particular interest. This poll was conducted following an idea and using the methodology of Schlosshauer et al. (arXiv:1301.1069 [quant-ph]), but among a slightly different group. It is supposed to give another snapshot of attitudes towards the interpretation of quantum mechanics and keep discourse about this topic alive. 10 pages, 18 figures, 1 table
Recognitions: Science Advisor There is one problem with these results, namely that some people may agree that they believe in interpretation x, but that a careful analysis may also show, that they do NOT agree what this interpretation x MEANS. So without such an analysis one always misses the fact that people agreeing on Copenhagen do unfortunately not agree on the same Copenhagen :-)
http://arxiv.org/abs/1303.3576 Cosmology from Group Field Theory Steffen Gielen, Daniele Oriti, Lorenzo Sindoni (Submitted on 14 Mar 2013) We identify a class of condensate states in the group field theory (GFT) approach to quantum gravity that can be interpreted as macroscopic homogeneous spatial geometries. We then extract the dynamics of such condensate states directly from the fundamental quantum GFT dynamics, following the procedure used in ordinary quantum fluids. The effective dynamics is a non-linear and non-local extension of quantum cosmology. We also show that any GFT model with a kinetic term of Laplacian type gives rise, in a semi-classical (WKB) approximation and in the isotropic case, to a modified Friedmann equation. This is the first concrete, general procedure for extracting an effective cosmological dynamics directly from a fundamental theory of quantum geometry. http://arxiv.org/abs/1303.3497 The DeWitt Equation in Quantum Field Theory Parikshit Dutta, Krzysztof A. Meissner, Hermann Nicolai (Submitted on 14 Mar 2013) We take a new look at the DeWitt equation, a defining equation for the effective action functional in quantum field theory. We present a formal solution to this equation, and discuss the equation in various contexts, and in particular for models where it can be made completely well defined, such as the Wess-Zumino model in two dimensions.
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http://physics.stackexchange.com/questions/10731/primordial-black-hole-detection?answertab=active | # Primordial Black Hole Detection
There's ample direct evidence for the existence of galactic and stellar mass black holes. However, there is no such direct evidence of primordial black holes, those formed after the Big Bang. A recent paper [http://arxiv.org/abs/1106.0011] describes oscillations the Sun might undergo if it encountered a primordial black hole. The theory that primordial black holes exist hasn't had experimental backing. Why should one believe they actually exist? Is this science or "guesstimating?"
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Sorry if my disbelief is so open. – Michael Luciuk Jun 3 '11 at 14:52
Since my early reading of Stephen Hawking books I've been fascinated by the idea that interstellar space is littered with mountain-sized black holes. I'm interested to hear some updates on the current state of thinking about these. – AlanSE Jun 3 '11 at 23:27
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The title is about formation, but the question is about detection. Maybe you could change the title? – Ben Crowell Aug 9 '11 at 14:22
## 2 Answers
The question might be asked as to why the universe did not produce a gas of black holes? The universe started out at a very low entropy, and black holes have entropy proportional to their horizon areas. So there could not have been a lot of primordial black holes in the early universe.
The lifetime of a black hole is $$t~=~\frac{5120\pi G^2M^3}{\hbar c^4}.$$ A black hole generated at the start of the universe and a mass of $1.7\times 10^{11}$ kg would Hawking evaporate about now. The final explosion would involve the quick conversion of mass into energy, where $10^5$ kg would be converted into energy in around a second. This is a considerable explosion, but it is less than conversion rate of matter by a star. So this is not as easy as looking at supernovae if primordial black holes are very distant. Also the energy would largely be in the gamma ray domain. The energy output would be a chirp function in its frequency increase and energy. The primordial black hole needs to explode in our local neighborhood. The actual dynamics of the explosion needs further examination, which I am sure is in the literature.
The Chandra satellite could pick up primordial black hole explosion, but so far nothing like this has been found. The lack of such detection gives weight to the estimate that the universe had a very low entropy at the start.
LC
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Chandra has been active for a dozen years without detecting a primordial BH radiation chirp signal. I'm curious what the mass distribution of these theoretical items was expected to be in their BB formation. Is it possible that the creation probability of less massive BHs, say less than 1.7E11 kg, was much smaller than for more massive BHs? – Michael Luciuk Jun 4 '11 at 3:22
Lawrence, Out of topic, but have you noticed that you lost your reputation once again? tsk tsk :) – anna v Jun 4 '11 at 3:51
""The primordial black hole needs to explode in our local neighborhood."" A black hole living somewhere in a galaxy, has a lot of chances to grow sucking in interstellar material, hasn't it? – Georg Jun 4 '11 at 8:53
In terms of entropy, the statement I've heard is that the most likely state for the early universe is one dominated by gravitational waves (like the mixmaster universe), not by black holes. I don't know this based on any personal understanding of the calculations. But if we were going to maximize entropy by having black holes, we would certainly do that by having big ones, not little ones. – Ben Crowell Aug 9 '11 at 14:12
– Benjamin Horowitz Aug 10 '11 at 3:21
Is this science or "guesstimating?"
It's theory.
• General relativity allows for black holes at essentially any mass.
And the theory is strongly supported on several fronts, including the apparent observation of large and huge compact object which fit the bill a black holes.
• Big Bang cosmology allows for the conditions necessary to produce "small" ones.
Again the theory has some experimental evidence to back it up, though in this case some of the fine detail needed to be convinced that primordial black hole did or did not form is lacking.
Why do you imagine that every reasonable theory will have or not have experimental justification right away? Some experiments or observations are hard or just take a lot of time.
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I really don't believe that every possible theory needs instant justification to be correct. Perhaps observational evidence for primordial BHs is too faint to be detected. However, to be fair, GR is almost a century old, and the Big Bang over half that. If they existed in quantity one might expect the heavens to be lit up with x-ray/gamma radiation. – Michael Luciuk Jun 3 '11 at 16:17
It's possible to determine the mass of primordial BHs that would evaporate via Hawking radiation over the past and present. Could the paucity of radiation give hints on the mechanism of their theoretical formation? – Michael Luciuk Jun 3 '11 at 16:31
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@Michael: It can put a upper bound on the density now and from that contribute to understanding the very early universe. But you'd have to ask a cosmologist what the state of these bounds are. Last I heard it wasn't very strong. – dmckee♦ Jun 3 '11 at 17:57 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "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": 0.9432063698768616, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/17786/why-are-local-systems-and-representations-of-the-fundamental-group-equivalent/18589 | ## Why are local systems and representations of the fundamental group equivalent
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My question: Let X be a sufficiently 'nice' topological space. Then there is an equivalence between representations of the fundamental group of X and local systems on X, i.e. sheaves on X locally isomorphic to a constant sheaf. Does anyone know of a self contained, detailed treatment of this suitable for my background? I've looked at the first few pages of Delignes "Équations différentielles à points singuliers réguliers" (which my advisor suggested I take a look at) but here it just says that the equivalence is "well known", giving no reference. Neither googling ("local systems representations fundamental group") (nothing usable comes up), wiki nor the nLab entry (not detailed anough and more interested in generalisation) on local systems were of much help to me. I apologise in case the equivalence should obvious once one knows about universal covering spaces/deck transformations. I haven't learned those yet. If so, please let me know.
Why I care: I am trying to read Simpsons "Higgs bundles and local systems",. Publ. Math. I. H. E. S. 75 (1992) 5–95". Simpson assumes this equivalence but gives no references.
My background: I am a Diplom (roughly equivalent to MSc) student at a german university. For my Diplom thesis I am aiming to understand Narasimhan and Seshadris Theorem (from "Stable and unitary vector bundles on a compact Riemann surface") which I think roughly states that a holomorphic vector bundle of degree zero on a compact Riemann surface X of genus g ≥ 2 is stable if and only if it arises from an irreducible unitary representation of the fundamental group of X. I also hope to read and understand parts of Hitchins paper "The Self-Duality Equations on a Riemann Surface" (Proc. London Math. Soc. 1987 s3-55: 59-126) and as mentioned above parts of Simpsons article "Higgs bundles and local systems".
If there is any way this question could be improved upon, please let me know.
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This is the standard relationship between covering spaces and the fundamental group covered in most basic algebraic topology textbooks, no? – Ryan Budney Mar 11 2010 at 0:43
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Judging from the comments it seems to me that the only reason why this doesn't appear to be well covered in the literature is that you're looking in the wrong places (or insisting on keywords that aren't used in most intro algtop books). In books like Hatcher's, they use the word "bundle", not "locally constant sheaf". Instead of "local system" words like "bundle of groups" are used. Moreover it looks like you prefer not to think about bundles of groups, but the induced vector bundles from the construction Arapura describes below. – Ryan Budney Mar 11 2010 at 3:55
## 6 Answers
I agree that the correspondence between representations of the fundamental group(oid) and locally constant sheaves is not very well documented in the basic literature. Whenever it comes up with my students, I end up having to sketch it out on the blackboard. However, my recollection is that Spanier's Algebraic Topology gives the correspondence as a set of exercises with hints. In any case, one direction is easy to describe as follows. Suppose that $X$ is a good connected space X (e.g. a manifold). Let $\tilde X\to X$ denote its universal cover. Given a representation of its fundamental $\rho:\pi_1(X)\to GL(V)$, one can form the sheaf of sections of the bundle $(\tilde X\times V)/\pi_1(X)\to X$. More explicitly, the sections of the sheaf over U can be identified with the continuous functions $f:\tilde U\to V$ satisfying $$f(\gamma x) = \rho(\gamma) f(x)$$ for $\gamma\in \pi_1(X)$. This sheaf can be checked to be locally constant. Essentially the same procedure produces a flat vector bundle, i.e. a vector bundle with locally constant transition functions. This is yet another object equivalent to a representation of the fundamental group.
With regard to your other comments, perhaps I should emphasize that the Narasimhan-Seshadri correspondence is between stable vector bundles of degree 0 and irreducible unitary representations of the fundamental group. The bundle is constructed as indicated above. Anyway, this sounds like a good Diplom thesis problem. Have fun.
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To give an alternative formulation of the same thing: One could consider the universal cover as a principal $\pi_1(X)$-bundle on X, then there exists a cover such that the transition functions $\gamma_{ij}$ (which are elements of $\pi_1$) of this bundle generate $\pi_1$, and one then gets transition functions on the associated bundle/local system by $\theta_{ij}:=\rho(\gamma_{ij})$. (I believe this works, but correct me if I'm wrong.) – Ketil Tveiten Mar 12 2010 at 15:20
Yes, absolutely right. – Donu Arapura Mar 13 2010 at 0:30
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Tamas Szamuely's new book "Galois Groups and Fundamental Groups"
http://www.ams.org/mathscinet-getitem?mr=2548205
contains a proof of this result and is excellently written, starting from the ground up.
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Pramod Achar's notes (from a lecture in an course he taught on perverse sheaves) are two pages.
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This is an old correspondence by deligne. you find it in books like:
Voisin: Hodge theory and complex algebraic geometry I Sabbah: Isomonodromic deformations and Frobenius manifolds Kobayashi: Geometry of complex vector bundles
It is true for local systems with complex coefficients. The rough picture is this:
A flat connections is equivalent to a local system and the parallel transport of this connection in a loop only depends on the loop, so gives you a represenation of the fundamental group.
A representation rep of the fundamental group of X defines a complex vector bundle of rank r via X x C^{r}/~ (fibers identified by rep). This carries again a flat connection.
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I'd strongly recommend having a look at Atiyah & Bott's paper The Yang-Mills Equations over Riemann Surfaces. They explain this pretty well, and embed it in a beautiful larger picture.
The basic idea is that a locally constant sheaf can be viewed as the horizontal sections of a bundle with respect to a flat connection. The holonomy of a flat connection along a curve only depends on the homotopy class of the curve, hence gives a representation of the fundamental group.
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http://mathoverflow.net/questions/46303/definition-of-the-kervaire-invariant-for-normal-maps-as-in-browders-book | ## Definition of the Kervaire invariant for normal maps (as in Browder’s book)
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Browder's book "Surgery on simply-connected manifolds" defines the Kervaire invariant in a very general setting. My question is: how does one get the more usual definition of the invariant for a framed (or Wu-orinented) manifold from Browder's general definition?
Browder's setting: For Poincare pairs $(X,Y)$ and $(A,B)$ with specified Spivak normal bundles, a normal map is a map $f: (X,Y) \to (A,B)$ that has degree 1 and is covered by a bundle map $b$ from one Spivak normal bundle to the other. Browder manages to define the $\mathbb Z /2$-valued Kervaire invariant $\sigma$ for any such map (no additional information needed), at least as long as $f^*$ takes the Wu class of $A$ to the Wu class of $X$. I find it hard to understand his definition.
Some simplification: For the sake of understanding, it should be safe to replace all the Poincare pairs with closed manifolds, and Spivak normal bundles with normal bundles. Then, a normal map becomes a degree-1 map of manifolds embedded in a high-dimensional Euclidean space that is covered by a bundle isomorphism of normal bundles. If the manifolds can be framed (or Wu-oriented), the Wu classes will be zero, so there's no reason to worry about them.
Question: How does this situation relate to the more usual situation of having a manifold with a framing or Wu orientation (which are, of course, necessary to define the Kervaire invariant in the usual way)? I'd imagine that there is some standard thing I could fix $(A,B)$ to be so that Browder's version of Kervaire invariant computes the usual Kervaire invariant of $(X,Y)$. I couldn't figure out what it should be. This choice of $(A,B)$ should somehow encode the Wu orientation on $(X,Y)$.
Please tell me if there are any mistakes in the above. Thank you!
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I'm not familiar with the term "Wu orientation" (or "Wu oriented"). If $X$ is framed then surely $(A,B)$ can be taken to be $(D^n,S^{n-1})$. – Tom Goodwillie Nov 17 2010 at 4:19
@Tom: What would the map of degree 1 be? Is there some obvious way to construct one map for every framing of X? (As for Wu orientations, I think you can safely ignore them). – Ilya Grigoriev Nov 17 2010 at 7:50
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I would have thought that you take $(A,B) = (D^n, \partial D^n)$. Then for any closed framed manifold $X$ take $f$ to be the map that collapses the complement of a ball, with the isomorphism $TX \cong f^* TD^n$ determined by the framing of $X$. – Oscar Randal-Williams Nov 17 2010 at 9:06
@Oscar: Yeah, that's probably it! I was confused about whether we can get any framing in this manner, but this seems clear to me now (as $f^∗ TD^n$ is a trivial bundle on X, the bundle isomorphism is precisely the same as a framing); thanks for straightening me out! Also, is there some deep meaning to the fact that this doesn't seem to work for manifolds with boundary, or am I missing something simple again? – Ilya Grigoriev Nov 17 2010 at 16:32
## 2 Answers
I found a useful reference on Doug Ravenel's website that first explains Browder's definition and then relates it to Kervaire's. Here it is (pp. 142-143; the file's a whopping 5MB, but just because it's scanned in).
Also, hi, and looking forward to your presentation!
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Thanks, Elizabeth! I like this reference a lot, although I don't think it answers this specific question (Oscar did, though). – Ilya Grigoriev Nov 17 2010 at 16:29
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My algebraic theory of surgery gives the following approach to the definition of the Kervaire invariant. Let $Q_n(C,\gamma)$ be the Weiss twisted quadratic $Q$-group defined for any chain bundle $(C,\gamma)$. A spherical fibration $\nu:X \to BG(k)$ determines a chain bundle $(C(X),\gamma(\nu))$ with a Hurewicz-style group morphism $$h~:~\pi^S_{n+k}(T(\nu)) \to Q_n(C(X),\gamma(\nu))$$ from the stable homotopy groups of the Thom space $T(\nu)$. The image $h(\rho)$ of a stable homotopy class $\rho$ relates the evaluations on the Hurewicz-Thom image fundamental homology class $$[X]~=~[\rho] \in H_{n+k}(T(\nu))~=~H_n(X)$$ of the Steenrod squares of $X$ and the cup products with the Wu classes $v_r(\nu) \in H^r(X)$, verifying on the chain level the formula of Wu and Thom $$\langle v_r(\nu) \cup y,[X] \rangle = \langle Sq^r(y),[X] \rangle~(y\in H^{n-r}(X))~.$$ An $n$-dimensional geometric Poincare complex $X$ (e.g. an $n$-dimensional manifold) has a canonical class of pairs $(\nu_X:X \to BG(k),\rho_X:S^{n+k} \to T(\nu_X))$ with $\nu_X$ the Spivak normal fibration (= sphere bundle of the normal bundle of an embedding $X \subset S^{n+k}$ for a manifold $X$). A fibre homotopy trivialization $b:\nu_X \simeq *:X \to BG(k)$ (e.g. one determined by a framing of a manifold) determines a morphism $Q_n(C(X),\gamma(\nu_X)) \to {\mathbb Z}_2$ such that the image of $h(\rho_X)$ is the Kervaire invariant $K(X,b)\in {\mathbb Z}_2$. More generally, a Wu-orientation $b$ of $X$ (for which Browder's 1969 Annals paper The Kervaire invariant of framed manifolds and its generalization is a good reference) determines a morphism $Q_n(C(X),\gamma(\nu_X)) \to {\mathbb Z}_8$ such that the image of $h(\rho_X)$ is the Brown generalized Kervaire invariant $K(X,b)\in {\mathbb Z}_8$. Most of this is already explained in my paper Algebraic Poincare cobordism.
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As it has been done for the Intersection of conics using matrix representation the aim of this page is providing an exaustive and clear numerical example that describe the math behind the ... | {"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": 5, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9263177514076233, "perplexity_flag": "head"} |
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## Continuous and nowhere differentiable
Yah, that's pretty nifty, isn't it?
Actually (at least in the current context), there is a viewpoint from which one can say that everything is, in fact, describable. However, that's a far cry from, for example, being able to evaluate any real function at 0.
Anyways, back to the problem at hand... your problem is that you swapped the order of limits, an operation that isn't generally allowed. Specifically, there's a limit operation implicit in differentiating, and also with summation. So, in generally, summing then differentiating won't be the same as differentiating then summing.
Well, swapping them is ok if you have uniform convergence of the differentiated series (which he does~). ie. if $$f(x) = \sum_{n=0}^\infty f_n(x)$$ on a closed interval $I$ and $$\sum_{n=0}^\infty f^\prime_n(x)$$ is uniformly convergent on $I$ and each $f^\prime_n$ is continuous on $I$ then $$f^\prime(x) = \sum_{n=0}^\infty f^\prime_n(x)$$ on $I$. Of course, that doesn't mean that the function isn't differentiable outside of the interval.
Recognitions: Gold Member Science Advisor Staff Emeritus Only when AB < 1, though. The fact the series diverges almost everywhere for AB > 1 doesn't guarantee the derivative doesn't exist.
yep, that's definitely true~
Recognitions: Homework Help Science Advisor Thanks guys. I need to review some Analysis. So, as I see it I can swap the order of summation and differentiation as long as the series converges uniformly to a function which it does if AB<1. This I can prove via the Weierstrass M-test for both the function and the sum of the derivatives of $f_n(x)$. So, however, when AB becomes greater than 1 this no longer applies and another approach must be used. Is that correct? I found a good reference proving "a" function exists which is nowhere differentiable which is under the title of "The Weierstrass Function": Link to proof Think I'll get a hardcopy and spend some time going over this proof. I suspect it can be applied to prove nowhere differentiable of the cosine series above.
...I can swap the order of summation and differentiation as long as the series converges uniformly...
as long as the series of derivatives converges uniformly, yes
Recognitions: Homework Help Science Advisor In an effort to approach this problem, I've chosen to fall back to an easier one first: $$f(x)=\sum_{n=0}^\infty \frac{GetDistance[4^n x]}{4^n}$$ Where the GetDistance[x] function returns the distance from x to the integer nearest x. This is the saw-tooth function commonly used to demonstrate continuous and nowhere differentiable. I've attached a plot superimposing $f_0$,$f_0+f_1$ and$f_0+f_1+f_2$. My Analysis book has a proof I can follow of it's nowhere differentiable behavior. Essentially, we seek to show: $\lim_{n\rightarrow\infty}{} \frac{f(a+h)-f(a)}{h}$ does not have a limit. My interpretation of this is that it has a "corner" at all points in its domain. Next, I'll pursue the following: $$f(x)=\sum_{n=0}^\infty \frac{Sin[(n!)^2 t]}{n!}$$ I'm told that T.W. Koerner in his book "Fourier Analysis" gives a detailed proof that this function is nowhere differentiable. Again, this involves an analysis of the difference quotient. Apparently using the same technique, one can prove the non-differentiability of the Weierstrass function. So, I'll work with them a bit first myself and if nothing pans out, well, it's a pleasant ride to the University library from my home. Attached Thumbnails
Recognitions: Homework Help Science Advisor I've retrieved the reference cited above from T.W.Koerner's book, "Fourier Analysis". I wish to focus on reviewing the proof here of nowhere differentiable for the function below (reviewing/reporting here will help me better understand it and perhaps will help others who also wish to learn more about these functions). I'll report it in parts. I should add that he states after thoroughly investigating this proof, one should be able to apply it to prove nowhere differentiable of the cosine series of the original post which is my ultimate goal. $$f(t)=\sum_{r=0}^\infty \frac{1}{r!}Sin[(r!)^2 t]$$ In general, we seek to investigate the difference quotient: $$\mathop {\lim }\limits_{x_n \to x} \frac{f(x)-f(x_n)}{x-x_n}$$ In order to investigate this limit, Koerner breaks the sum into three parts which turn out to be the key to the proof: $$h_n(t)=\sum_{r=0}^{n-1}\frac{1}{r!}Sin[(r!)^2 t]$$ $$k_n(t)=\frac{Sin[(n!)^2t]}{n!}$$ $$I_n(t)=\sum_{r=n+1}^{\infty}\frac{1}{r!}Sin[(r!)^2 t]$$ And thus: $$f(t)=h_n(t)+k_n(t)+I_n(t)$$ These function components will be used to evaluate the difference quotient. However, some Lemmas are needed in order to set bounds on the various quantities. That's next.
Recognitions: Homework Help Science Advisor In order to prove nowhere differentiable, we must show that the difference quotient diverges at all x. Thus we seek to analyze: $$\mathop {\lim }\limits_{a \to x}\frac{f(x)-f(a)}{x-a}$$ If we can show that for any such x, this limit increases without bound as additional elements of the function sequence are added, then we will have shown the derivative is undefined at x. It suffices therefore, to consider the absolute value of the quotient: $$\mathop{\lim}\limits_{a\to x}\frac{|f(x)-f(a)|}{|x-a|}$$ In order to evaluate this limit, f(x) is decomposed into the three functions stated earlier which are dependent on the summation index n. Later, we will evaluate the limit as n approaches infinity. Thus we analyze: $$|k_n(x)-k_n(a)|$$ $$|h_n(x)-h_n(a)|$$ $$|L_n(x)-L_n(a)|$$ Lemma A: If$K\geq 3$ is an integer and $x\in D_f$ (x is in the domain of the function), then there exists an $a\in D_f$, such that: $$\frac{\pi}{K}<|x-a|\leq\frac{3\pi}{K}$$ and: $$|Sin[Kx]-Sin[Ka]|\geq 1$$ This is the central idea in the whole proof and needs to be fully understood in order to proceed further. Proof: Consider the plot of the function Sin[Kx] and an arbitrary point, x, along it as well as the following interval I'll call I: $$I=(x+\frac{\pi}{K},x+\frac{3\pi}{K}]$$ An attached plot shows this setup for K=6 and x=1.2. Note that within I, Sin[Kx] takes on all values between -1 and 1. Any value, a, within this interval will be in absolute value: $$\frac{\pi}{K}<|x-a|\leq\frac{3\pi}{K}$$ But since Sin[Kx] takes on all values between -1 and 1 in this interval, I can choose a such that Sin[Ka] is either -1 or 1 so that: $$|Sin[Kx]-Sin[Ka]|\geq 1$$ QED Using this Lemma, we can now evaluate $k_n(x)$ which I'll do next. Attached Thumbnails
Recognitions: Homework Help Science Advisor Using Lemma A, for any $x\in D_f$, we can find an "a" in the neighborhood of x such that: $$\frac{\pi}{(n!)^2}<|x-a|\leq\frac{3\pi}{(n!)^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ Note that as n goes to infinity, a approaches x. Thus, the final limit will be evaluated as n goes to infinity. Using the definition of k(x) above, we have: $$|k_n(x)-k_n(a)|=\frac{1}{n!}|Sin[(n!)^2 x]-Sin[(n!)^2 a]|$$ which, according to Lemma A is bounded from below, thus giving:: $$|k_n(x)-k_n(a)|\geq\frac{1}{n!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ The determination of a bound for h(x) is a very interesting problem and deserves its own post:
Recognitions: Homework Help Science Advisor In order to analyze $h_n(x)$, the following Lemma is needed: Lemma B: For all $x,a\in D_f$: $$|Sin[x]-Sin[a]|\leq|x-a|$$ Proof: Using the Mean Value Theorem we have: $$Sin[x]-Sin[a]=-Cos[c](x-a)$$ Thus: $$|Sin[x]-Sin[a]|\leq|x-a|$$ QED The following is as messy as the proof gets: Using the definition of $k_n(x)$ above, we have: $$|h_n(x)-h_n(a)|\leq\sum_{r=0}^{n-1}\frac{1}{r!}|Sin[(r!)^2x]-Sin[(r!)^2a]|$$ (The triangle inequality) Using Lemma A: $$\sum_{r=0}^{n-1}\frac{1}{r!}|Sin[(r!)^2x]-Sin[(r!)^2a]|\leq\sum_{r=0}^{n-1}\frac{1}{r!}|(r!)^2x-(r!)^2a|$$ Now: $$\sum_{r=0}^{n-1}\frac{1}{r!}|(r!)^2x-(r!)^2a|=\sum_{r=0}^{n-1}(r!)|x-a|$$ and: $$\sum_{r=0}^{n-1}(r!)|x-a|=[1+2!+3!+ . . . +(n-1)!]|x-a|$$ $$=[(n-1)!+\sum_{r=0}^{n-2}r!]|x-a|$$ Now: $$\sum_{r=0}^{n-2}r!\leq\sum_{r=0}^{n-2}(n-2)!$$ Thus we have: $$|h_n(x)-h_n(a)|\leq[(n-1)!+\sum_{n=0}^{n-2}(n-2)!]|x-a|$$ Since: $$\sum_{r=0}^{n-2}(n-1)!=(n-1)(n-2)!$$ This reduces to: $$|h_n(x)-h_n(a)|\leq 2(n-1)!|x-a|$$ Since: $$|x-a|\leq\frac{3\pi}{(n!)^2}$$ Finally then: $$|h_n(x)-h_n(a)|\leq\frac{6\pi}{n(n!)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ I know, this is messy but it involves a lot of good work with summations, absolute values, and inequalities. The rest is less-complicated.
Recognitions: Homework Help Science Advisor Lemma C: $$\sum_{r=n}^{\infty}\frac{1}{r!}\leq\frac{2}{n!}$$ For $n\geq 2$ Proof: $$\sum_{r=n}^{\infty}\frac{1}{r!}=\frac{1}{n!}+\frac{1}{(n+1)!}+\frac{1}{ (n+2)!}+. . .$$ $$=\frac{1}{n!}[1+\frac{1}{n+1}+\frac{1}{(n+2)(n+1)}+\frac{1}{(n+3)(n+2)(n+1)}+. . .]$$ $$\leq\frac{1}{n!}[1+\frac{1}{2}+(\frac{1}{2})^2+(\frac{1}{2})^3+. . .]$$ Since the last term is a geometric series, we have: $$\sum_{r=n}^{\infty}\frac{1}{r!}\leq\frac{2}{n!}$$ QED Analysis of $L_n(x)$ Since: $$|L_n(x)|\leq\sum_{r=n+1}^{\infty}\frac{1}{r!}$$ Using Lemma C we have: $$|L_n(x)|\leq\frac{2}{(n+1)!}$$ Thus: $$|L_n(x)-L_n(a)|\leq|L_n(x)|+|L_n(a)|\leq\frac{4}{(n+1)!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$ We now have all the expressions needed to analyze the limit of the difference quotient as n goes to infinity. I'll complete the proof next.
Recognitions: Homework Help Science Advisor We now have: $$|x-a|\leq\frac{3\pi}{(n!)^2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ $$|k_n(x)-k_n(a)|\geq\frac{1}{n!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$ $$|h_n(x)-h_n(a)|\leq\fraq{6\pi}{n(n!)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$ $$|L_n(x)-L_n(a)|\leq\fraq{4){(n+1)!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$ Recalling: $$f(x)=k(x)+h(x)+L(x)$$ and noting that we have expressed each term in terms of n, we wish to evaluate: $$\mathop{\lim}\limits_{n\to \infty}\frac{|(k_n(x)+h_n(x)+L_n(x))-(k_n(a)+h_n(a)+L_n(a))|}{|x-a|}$$ A short Lemma to assist with evaluating this absolute value: Lemma D: $$|a+b+c|\geq|a|-|b|-|c|$$ Proof: $$|a|=|(a+b+c)-(b+c)|\leq|a+b+c|+|b+c|$$ $$|a|-|b+c|\leq|a+b+c|$$ Since: $$|b+c|\leq|b|+|c|$$ we have: $$|a|-|b|-|c|\leq|a|-|b+c|\leq|a+b+c|$$ QED Thus: $$|f(x)-f(a)|\geq|k_n(x)-k_n(a)|-|h_n(x)-h_n(a)|-|L_n(x)-L_n(a)|$$ $$\ \ \ \ \ \ \geq\frac{1}{n!}-\frac{6\pi}{n(n!)}-\frac{4}{(n+1)!}$$ Since: $$\frac{1}{n!}-\frac{6\pi}{n(n!)}-\frac{4}{(n+1)!}=\frac{1}{n!}(1-\frac{6\pi}{n}-\frac{4}{n+1})$$ And: $$\frac{1}{n!}(1-\frac{6\pi}{n}-\frac{4}{n+1})\geq\frac{1}{n!}(1-\frac{30}{n!})$$ Finally, for n>60, $$\frac{1}{n!}(1-\frac{30}{n!})\geq\frac{1}{2(n!)}$$ Substituting this expression and the results from equation (1) above, $$\mathop{\lim}\limits_{n\to\infty}\frac{|(k_n(x)+h_n(x)+L_n(x))-(k_n(a)+h_n(a)+L_n(a))|}{|x-a|}\geq\frac{\frac{1}{2(n!)}}{\frac{3\pi}{(n!)^2}}=\frac{n!}{6\pi}$$ Thus for $n\geq 60$, the limit goes to infinity proving that the derivative does not exist for arbitrary point x in the domain. QED Initially, I was overwhelmed with this proof and was having a difficult time. The author made an interesting point: "reflection and the passage of time will help a great deal". He was right. Now I understand it fully and am comfortable with its construction. It's now a relatively simple matter to prove the same for the Cosine series which I'll do in a brief summary report next time.
Recognitions: Homework Help Science Advisor "relatively simple matter" . . . . . . yea, right . . . . . I'm wrong and I ain't proud. Can someone show me how to prove: $$f(x)=\sum_{r=0}^{\infty} A^r Sin[B^r x]$$ Is nowhere differentiable. Even with all I did above I can't apply it to this equation with A<1 and B an integer: I end up getting the difference quotient is larger than minus infinity which is meaningless.
Recognitions: Homework Help Science Advisor I was incorrectly interpreting the partial sums of a geometric series. It should be: $$\sum_{r=0}^{n-1}(\frac{1}{A})^r+(\frac{1}{A})^n+\sum_{r=n+1}^{\infty}(\frac{1}{A})^r=$$ $$\frac{1-(\frac{1}{A})^n}{1-\frac{1}{A}}+(\frac{1}{A})^n+\frac{(\frac{1}{A})^{n+1}}{1-\frac{1}{A}}$$ The important point is to devise sizes for A and B such that the following partial differences are less than 1. The most difficult is for $h_n(x)$. I rationalized that since factorials squared were being used above, then I should try to make the B term quite large with respect to A. In the following example, I used $B=A^2$. $$|k_n(x)-k_n(x_0)|\geq(\frac{1}{A})^n$$ $$|I_n(x)-I_n(x_0)|\leq\frac{2}{A-1}(\frac{1}{A})^n$$ $$|h_n(x)-h_n(x_0)|\leq(\frac{A^n-1}{A-1})\frac{3\pi}{A^nA^n}$$ Plugging this into the differential difference quotient leads to the following expression: $$\mathop{\lim}\limits_{n\to\infty}\frac{(\frac{1}{A})^n[1-\frac{2}{a-1}-\frac{3\pi(A^n-1)}{A^n(A-1)}]}{\frac{3\pi}{B^n}}$$ With $B=A^2$ This limit tends to infinity if $A\geq 13$ Thus, $$f(x)=\sum_{r=0}^\infty(\frac{1}{13})^r Sin[(13)^{2r}x]$$ is nowhere differentiable. (this is not a proof)
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http://mathoverflow.net/questions/23992/question-about-computing-group-cohomology-using-cochains/23996 | ## Question about computing group cohomology using cochains
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In Milne's notes on Class Field Theory (http://www.jmilne.org/math/CourseNotes/CFT.pdf), he initially defines group cohomology in terms of injective resolutions, then he talks about computing cohomology using cochains. I don't see him mention anywhere that the group has to be finite in order for cochains to work, but this seems to be the case?
Later, he discusses profinite groups, in which he says that cohomology of profinite groups can be computed using continuous cochains. What isn't clear is the following: is the cohomology using continuous cochains a modified cohomology theory, different from the one using injective resolutions? In this case, then, do we get the cohomology theory using injective resolutions if we use all cochains, not just continuous ones? Or do the continuous cochains give the same cohomology as injective resolutions, and cochains which are not necessarily continuous only give cohomology in the case of finite groups? I.e., is there only one such cohomology theory? At the very least, using general cochains versus continuous cochains in the case of infinite profinite groups is different, for in one case $H^1$ is $\mathrm{Hom}(G,M)$ when M is trivial, and in the other case $H^1$ is $\mathrm{Hom}_{\mathrm{cts}}(G,M)$.
Assuming that set-theoretic cochains only work for finite groups, why is it the case? It seems that the proof that cochains compute cohomology (i.e. by looking at a projective resolution of $\mathbb{Z}$) fails because the modules used in the case of finite groups, i.e. tensor powers of $\mathbb{Z}[G]$, aren't necessarily projective when $G$ is infinite (in the case when $G$ is finite, they are free). Is this correct?
-
1
Uniform answer (including finite groups with discrete topology as profinite groups) is to work in category of discrete $G$-modules, so take injective resolutions there. In general, for $G$-module $M$ its "discretization" is maximal discrete $G$-submodule (i.e., elts with open stabilizer). This has univ. mapping property, so discretization of injective $G$-mod is inj. in category of discrete $G$-mods. So latter category has enough injectives. For finite discrete $G$ it's the old thing. For general $G$ recovers "cont. cochain" construction, via work; dropping continuity does matter. – BCnrd May 9 2010 at 8:12
1
For a group (no topology), I define the cohomology using injective resolutions in the category of all G-modules, and I prove that it can be computed using cochains (no conditions on G). For a profinite group, I define the cohomology using injective resolutions in the category of all discrete G-modules, and state that it can be computed using continuous cochains. A finite group can be regarded as a profinite group with the discrete topology, in which case the two definitions coincide (obviously). – JS Milne May 9 2010 at 19:53
Yes, that make sense. My main problem is that I didn't realize that a module could be injective in the category of discrete G-modules but not in the category of G-modules, which is what lead me to wonder whether cochains only worked for finite groups. This last wonder was in part justified by the fact that I hadn't looked in detail at the proof that cochains worked, which is only in the appendix. Thanks for clearing up the confusion! – David Corwin May 10 2010 at 3:31
## 2 Answers
You should take a look at the beginning of chapter 2 of Serre's Galois Cohomology. He explains there that if G is a profinite group, then the category of discrete abelian groups with a continuous action of G has enough injectives (but not enough projectives in general), and that cohomology can be "computed" as a direct limit of cohomology of a finite group. To sum up:
-if G is discrete (i.e. no topology), then the category has enough injectives and projectives (it is the category of left $\mathbb{Z}[G]$-modules), and using a projective resolution for $\mathbb{Z}$ gives you the equivalence between the derived functor definition and the cochains definition (using the fact that Ext can be computed two ways). You can find this in Serre's Local Fields.
-if G is profinite, and we consider the category of discrete modules with a continuous action of G, then there are enough injectives (this can be seen quite easily from the discrete case), but not enough projectives. Luckily though, the two definitions agree (thanks to the "direct limit computability"). You can find this in Serre's Galois Cohomology.
-if G is an arbitrary topological group, there is not much left. There aren't enough injectives nor projectives in general, and if you define cohomology with cochains, you don't get an homological functor (only the beginning of the long exact sequence exists). However, see the end of J.-M. Fontaine and Yi Ouyang's book (it's a pdf, I found it on Fontaine's web page) about p-adic representations, they mention that if you have a continuous set-theoretic section in your short exact sequence, you get a long exact sequence. I haven't read the reference they provide, though.
-
At first, I was confused by the following: If we compute the cohomology of profinite groups using injective modules in the category of discrete modules in order to get the theory using continuous cochains, then how is this theory any different from the normal cohomology theory? (But the two clearly are different, for example in the case of $H^1$). – David Corwin May 9 2010 at 15:16
But I think the answer to my concerns is the following: An injective object in the category of discrete $G$-modules is not necessarily injective in the category of $G$-modules. Hence, when we compute cohomology of profinite groups in the category of discrete modules, the injective resolution isn't injective in the category of $G$-modules, so the cohomology we get isn't the same as normal group cohomology. At the very least, we can conclude that the category of discrete $G$-modules doesn't have enough injectives-which-are-injective-in-the-category-of-$G$-modules-as-well. Is all this true? – David Corwin May 9 2010 at 15:20
I disagree with the last part of the answer ("if G is an arbitrary topological group, there is not much left"): there is a very developed theory of continuous cohomology for topological groups and, in particular, Lie groups and p-adic Lie groups. See the books of Guichardet and Borel-Wallach. The key technical notion in Guichardet is a "strong relatively injective resolution". Of course, it would be facile to expect that the cohomology of discrete and t.d. groups had an exact analogue for arbitrary topological groups, but much is known for the Lie case. – Victor Protsak May 9 2010 at 19:13
yes, I should have written "there is not much left that I know of" ;) @Davidac: you are right, injective is relative to the category you're working with. – Homology May 9 2010 at 20:11
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You are basically correct. The point is that for applications to class field theory one does use a modified theory, the "cohomology of profinite groups", which is not the same as plain cohomology of groups.
Cochains do work for cohomology of arbitrary groups. Tensor powers of $\mathbb{Z}[G]$ are free $\mathbb{Z}[G]$-modules on the basis $1\otimes g_2\otimes\cdots\otimes g_n$. But in class field theory we don't use the plain theory for infinite Galois group. The theory we do use is either defined by taking direct limits of the theory for finite groups, or by using continuous cochains.
-
Actually, I see you're right - you don't need that $G$ is finite to show that tensor powers of $\mathbb{Z}[G]$ are free. – David Corwin May 9 2010 at 15:07 | {"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": 29, "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": 0.9345066547393799, "perplexity_flag": "head"} |
http://en.wikipedia.org/wiki/Truth_function | # Truth function
In mathematical logic, a truth function is a function from a set of truth values[clarification needed] to truth values. Classically the domain and range of a truth function are {truth, falsehood}, but they may have any number of truth values, including an infinity of these[citation needed].
A sentence is truth-functional if the truth-value of the sentence is a function of the truth-value of its sub-sentences. A class of sentences is truth-functional if each of its members is. For example, the sentence "Apples are fruits and carrots are vegetables" is truth-functional since it is true just in case each of its sub-sentences "apples are fruits" and "carrots are vegetables" is true, and it is false otherwise. Not all sentences of a natural language, such as English, are truth-functional.
Sentences of the form "x believes that..." are typical examples of sentences that are not truth-functional. Let us say that Mary mistakenly believes that Al Gore was President of the USA on April 20, 2000, but she does not believe that the moon is made of green cheese. Then the sentence
"Mary believes that Al Gore was President of the USA on April 20, 2000"
is true while
"Mary believes that the moon is made of green cheese"
is false. In both cases, each component sentence (i.e. "Al Gore was president of the USA on April 20, 2000" and "the moon is made of green cheese") is false, but each compound sentence formed by prefixing the phrase "Mary believes that" differs in truth-value. That is, the truth-value of a sentence of the form "Mary believes that..." is not determined solely by the truth-value of its component sentence, and hence the (unary) connective (or simply operator since it is unary) is non-truth-functional.
In classical logic a truth function is a compound proposition whose truth or falsity is unequivocally determined by the truth or falsity of its components for all cases, the class of its formulas (including sentences) is truth-functional since every sentential connective (e.g. &, →, etc.) used in the construction of formulas is truth-functional. Their values for various truth-values as argument are usually given by truth tables. Truth-functional propositional calculus is a formal system whose formulas may be interpreted as either true or false.
## Table of binary truth functions
There are sixteen Boolean functions associating the inputs P and Q with four-digit binary outputs. Any of these functions is identified as a truth table of a certain logical connective in classical logic, including several degenerate cases such as a function not depending on one or both of its arguments.
Contradiction/False
Notation Equivalent
formulas
Truth table Venn diagram
$\bot$
"bottom"
P $\wedge$ ¬P
Opq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 0 |
| 1 | 0 | 0 | |
Tautology/True
Notation Equivalent
formulas
Truth table Venn diagram
$\top$
"top"
P $\vee$ ¬P
Vpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 1 |
| 1 | 1 | 1 | |
Proposition P
Notation Equivalent
formulas
Truth table Venn diagram
P p
Ipq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 0 |
| 1 | 1 | 1 | |
Negation of P
Notation Equivalent
formulas
Truth table Venn diagram
¬P
~P
Np
Fpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 1 |
| 1 | 0 | 0 | |
Proposition Q
Notation Equivalent
formulas
Truth table Venn diagram
Q q
Hpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 1 |
| 1 | 0 | 1 | |
Negation of Q
Notation Equivalent
formulas
Truth table Venn diagram
¬Q
~Q
Nq
Gpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 0 |
| 1 | 1 | 0 | |
Conjunction
Notation Equivalent
formulas
Truth table Venn diagram
P $\wedge$ Q
P & Q
P · Q
P AND Q
P $\not\rightarrow$¬Q
¬P $\not\leftarrow$ Q
¬P $\downarrow$ ¬Q
Kpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 0 |
| 1 | 0 | 1 | |
Alternative denial
Notation Equivalent
formulas
Truth table Venn diagram
P ↑ Q
P | Q
P NAND Q
P → ¬Q
¬P ← Q
¬P $\lor$ ¬Q
Dpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 1 |
| 1 | 1 | 0 | |
Disjunction
Notation Equivalent
formulas
Truth table Venn diagram
P $\lor$ Q
P OR Q
P $\leftarrow$ ¬Q
¬P → Q
¬P ↑ ¬Q
¬(¬P $\wedge$ ¬Q)
Apq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 1 |
| 1 | 1 | 1 | |
Joint denial
Notation Equivalent
formulas
Truth table Venn diagram
P ↓ Q
P NOR Q
P $\not\leftarrow$ ¬Q
¬P $\not\rightarrow$ Q
¬P $\wedge$ ¬Q
Xpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 0 |
| 1 | 0 | 0 | |
Material nonimplication
Notation Equivalent
formulas
Truth table Venn diagram
P $\not\rightarrow$ Q
P $\not\supset$ Q
P $\wedge$ ¬Q
¬P ↓ Q
¬P $\not\leftarrow$ ¬Q
Lpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 0 |
| 1 | 1 | 0 | |
Material implication
Notation Equivalent
formulas
Truth table Venn diagram
P → Q
P $\supset$ Q
P ↑ ¬Q
¬P $\lor$ Q
¬P ← ¬Q
Cpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 1 |
| 1 | 0 | 1 | |
Converse nonimplication
Notation Equivalent
formulas
Truth table Venn diagram
P $\not\leftarrow$ Q
P $\not\subset$ Q
P ↓ ¬Q
¬P $\wedge$ Q
¬P $\not\rightarrow$ ¬Q
Mpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 1 |
| 1 | 0 | 0 | |
Converse implication
Notation Equivalent
formulas
Truth table Venn diagram
P $\leftarrow$ Q
P $\subset$ Q
P $\lor$ ¬Q
¬P ↑ Q
¬P → ¬Q
Bpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 0 |
| 1 | 1 | 1 | |
Exclusive disjunction
Notation Equivalent
formulas
Truth table Venn diagram
P $\not\leftrightarrow$ Q
P $\not\equiv$ Q
P $\oplus$ Q
P XOR Q
P $\leftrightarrow$ ¬Q
¬P $\leftrightarrow$ Q
¬P $\not\leftrightarrow$ ¬Q
Jpq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 0 | 1 |
| 1 | 1 | 0 | |
Biconditional
Notation Equivalent
formulas
Truth table Venn diagram
P $\leftrightarrow$ Q
P ≡ Q
P XNOR Q
P IFF Q
P $\not\leftrightarrow$ ¬Q
¬P $\not\leftrightarrow$ Q
¬P $\leftrightarrow$ ¬Q
Epq
| | | | |
|----|----|----|----|
| | | Q | |
| 0 | 1 | | |
| P | 0 | 1 | 0 |
| 1 | 0 | 1 | |
## Functional completeness
See also: Functional completeness
Because a function may be expressed as a composition, a truth-functional logical calculus does not need to have dedicated symbols for all of the above-mentioned functions to be functionally complete. This is expressed in a propositional calculus as logical equivalence of certain compound statements. For example, classical logic has ¬P ∨ Q equivalent to P → Q. The conditional operator "→" is therefore not necessary for a classical-based logical system if "¬" (not) and "∨" (or) are already in use.
A minimal set of operators that can express every statement expressible in the propositional calculus is called a minimal functionally complete set. A minimally complete set of operators is achieved by NAND alone {↑} and NOR alone {↓}.
The following are the minimal functionally complete sets of operators whose arities do not exceed 2:
One element
{↑}, {↓}.
Two elements
{$\vee$, ¬}, {$\wedge$, ¬}, {→, ¬}, {←, ¬}, {→, $\bot$}, {←, $\bot$}, {→, $\not\leftrightarrow$}, {←, $\not\leftrightarrow$}, {→, $\not\to$}, {→, $\not\leftarrow$}, {←, $\not\to$}, {←, $\not\leftarrow$}, {$\not\to$, ¬}, {$\not\leftarrow$, ¬}, {$\not\to$, $\top$}, {$\not\leftarrow$, $\top$}, {$\not\to$, $\leftrightarrow$}, {$\not\leftarrow$, $\leftrightarrow$}.
Three elements
{$\lor$, $\leftrightarrow$, $\bot$}, {$\lor$, $\leftrightarrow$, $\not\leftrightarrow$}, {$\lor$, $\not\leftrightarrow$, $\top$}, {$\land$, $\leftrightarrow$, $\bot$}, {$\land$, $\leftrightarrow$, $\not\leftrightarrow$}, {$\land$, $\not\leftrightarrow$, $\top$}.
## Algebraic properties
Some truth functions possess properties which may be expressed in the theorems containing the corresponding connective. Some of those properties that a binary truth function (or a corresponding logical connective) may have are:
• Associativity: Within an expression containing two or more of the same associative connectives in a row, the order of the operations does not matter as long as the sequence of the operands is not changed.
• Commutativity: The operands of the connective may be swapped without affecting the truth-value of the expression.
• Distributivity: A connective denoted by · distributes over another connective denoted by +, if a · (b + c) = (a · b) + (a · c) for all operands a, b, c.
• Idempotence: Whenever the operands of the operation are the same, the connective gives the operand as the result.
• Absorption: A pair of connectives $\land$, $\lor$ satisfies the absorption law if $a\land(a\lor b)=a$ for all operands a, b.
A set of truth functions is functionally complete if and only if for each of the following five properties it contains at least one member lacking it:
• monotonic: If f(a1, ..., an) ≤ f(b1, ..., bn) for all a1, ..., an, b1, ..., bn ∈ {0,1} such that a1 ≤ b1, a2 ≤ b2, ..., an ≤ bn. E.g., $\vee$, $\wedge$, $\top$, $\bot$.
• affine: Each variable always makes a difference in the truth-value of the operation or it never makes a difference. E.g., $\neg$, $\leftrightarrow$, $\not\leftrightarrow$, $\top$, $\bot$.
• self dual: To read the truth-value assignments for the operation from top to bottom on its truth table is the same as taking the complement of reading it from bottom to top; in other words, f(¬a1, ..., ¬an) = ¬f(a1, ..., an). E.g., $\neg$.
• truth-preserving: The interpretation under which all variables are assigned a truth value of 'true' produces a truth value of 'true' as a result of these operations. E.g., $\vee$, $\wedge$, $\top$, $\rightarrow$, $\leftrightarrow$, ⊂. (see validity)
• falsehood-preserving: The interpretation under which all variables are assigned a truth value of 'false' produces a truth value of 'false' as a result of these operations. E.g., $\vee$, $\wedge$, $\not\leftrightarrow$, $\bot$, ⊄, ⊅. (see validity)
### Arity
See also: arity
A concrete function may be also referred to as an operator. In two-valued logic there are 2 nullary operators (constants), 4 unary operators, 16 binary operators, 256 ternary operators, and $2^{2^n}$ n-ary operators. In three-valued logic there are 3 nullary operators (constants), 27 unary operators, 19683 binary operators, 7625597484987 ternary operators, and $3^{3^n}$ n-ary operators. In k-valued logic, there are k nullary operators, $k^k$ unary operators, $k^{k^2}$ binary operators, $k^{k^3}$ ternary operators, and $k^{k^n}$ n-ary operators. An n-ary operator in k-valued logic is a function from $\mathbb{Z}_k^n \to \mathbb{Z}_k$. Therefore the number of such operators is $|\mathbb{Z}_k|^{|\mathbb{Z}_k^n|} = k^{k^n}$, which is how the above numbers were derived.
However, some of the operators of a particular arity are actually degenerate forms that perform a lower-arity operation on some of the inputs and ignores the rest of the inputs. Out of the 256 ternary boolean operators cited above, $\binom{3}{2}\cdot 16 - \binom{3}{1}\cdot 4 + \binom{3}{0}\cdot 2$ of them are such degenerate forms of binary or lower-arity operators, using the inclusion-exclusion principle. The ternary operator $f(x,y,z)=\lnot x$ is one such operator which is actually a unary operator applied to one input, and ignoring the other two inputs.
"Not" is a unary operator, it takes a single term (¬P). The rest are binary operators, taking two terms to make a compound statement (P $\wedge$ Q, P $\vee$ Q, P → Q, P ↔ Q).
The set of logical operators $\Omega\!$ may be partitioned into disjoint subsets as follows:
$\Omega = \Omega_0 \cup \Omega_1 \cup \ldots \cup \Omega_j \cup \ldots \cup \Omega_m \,.$
In this partition, $\Omega_j\!$ is the set of operator symbols of arity $j\!$.
In the more familiar propositional calculi, $\Omega\!$ is typically partitioned as follows:
nullary operators: $\Omega_0 = \{\bot, \top \} \,$
unary operators: $\Omega_1 = \{ \lnot \} \,$
binary operators: $\Omega_2 \subseteq \{ \land, \lor, \rightarrow, \leftrightarrow \} \,$
## Principle of compositionality
Instead of using truth tables, logical connective symbols can be interpreted by means of an interpretation function and a functionally complete set of truth-functions (Gamut 1991), as detailed by the principle of compositionality of meaning. Let I be an interpretation function, let Φ, Ψ be any two sentences and let the truth function fnand be defined as:
• fnand(T,T)=F; fnand(T,F)=fnand(F,T)=fnand(F,F)=T
Then, for convenience, fnot, for fand and so on are defined by means of fnand:
• fnot(x)=fnand(x,x)
• for(x,y)= fnand(fnot(x), fnot(y))
• fand(x,y)=fnot(fnand(x,y))
or, alternatively fnot, for fand and so on are defined directly:
• fnot(T)=F; fnot(F)=T;
• for(T,T)=for(T,F)=for(F,T)=T;for(F,F)=F
• fand(T,T)=T; fand(T,F)=fand(F,T)=fand(F,F)=F
Then
• I(~)=I(¬)=fnot
• I(&)=I(^)=I(&)=fand
• I(v)=I($\lor$)= for
• I(~Φ)=I(¬Φ)=I(¬)(I(Φ))=fnot(I(Φ))
• I(Φ&Ψ) = I(&)(I(Φ), I(Ψ))= fand(I(Φ), I(Ψ))
etc.
Thus if S is a sentence that is a string of symbols consisting of logical symbols v1...vn representing logical connectives, and non-logical symbols c1...cn, then if and only if I(v1)...I(vn) have been provided interpreting v1 to vn by means of fnand (or any other set of functional complete truth-functions) then the truth-value of I(s) is determined entirely by the truth-values of c1...cn, i.e. of I(c1)...I(cn). In other words, as expected and required, S is true or false only under an interpretation of all its non-logical symbols.
## Computer science
Logical operators are implemented as logic gates in digital circuits. Practically all digital circuits (the major exception is DRAM) are built up from NAND, NOR, NOT, and transmission gates. NAND and NOR gates with 3 or more inputs rather than the usual 2 inputs are fairly common, although they are logically equivalent to a cascade of 2-input gates. All other operators are implemented by breaking them down into a logically equivalent combination of 2 or more of the above logic gates.
The "logical equivalence" of "NAND alone", "NOR alone", and "NOT and AND" is similar to Turing equivalence.
That fact that all truth functions can be expressed with NOR alone is demonstrated by the Apollo guidance computer.
## Further reading
• Józef Maria Bocheński (1959), A Précis of Mathematical Logic, translated from the French and German versions by Otto Bird, Dordrecht, South Holland: D. Reidel.
• Alonzo Church (1944), Introduction to Mathematical Logic, Princeton, NJ: Princeton University Press. See the Introduction for a history of the truth function concept. | {"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": 119, "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": 0.8554813861846924, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/97847-plane-tangent-paraboloid.html | # Thread:
1. ## Plane Tangent to a Paraboloid.
- I need to find at what point on te paraboloid $y=x^2+z^2$ is the tangent plane parallel to the plane $x+2y+3z=1$.
I can handle most of this problem myself, but I need to know if this is the correct equation to start with:
$f(x,y,z)=x^2-y+z^2$
Then I would just apply:
$f_x(x_o,y_o,z_o)(x-x_o) + f_y(x_o,y_o,z_o)(y-y_o) +f_z(x_o,y_o,z_o)(z-z_o)=0$
Is this correct?
2. Originally Posted by adkinsjr
- I need to find at what point on te paraboloid $y=x^2+z^2$ is the tangent plane parallel to the plane $x+2y+3z=1$.
I can handle most of this problem myself, but I need to know if this is the correct equation to start with:
$f(x,y,z)=x^2-y+z^2$
Then I would just apply:
$f_x(x_o,y_o,z_o)(x-x_o) + f_y(x_o,y_o,z_o)(y-y_o) +f_z(x_o,y_o,z_o)(z-z_o)=0$
Is this correct?
Should be very simple and direct. While this is correct, but not necessary to apply this. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 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": 0.9497888684272766, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/20263?sort=votes | ## Software for computing multi-graded Hilbert series
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The ring of invariants $S^T$ of $k[a,b,c,d]$ under the algebraic torus action $T = k^{*}$ with weights (1,1,-1,-1) is $S = k[ac,ad,bc,bd]$ which has multigraded Hilbert series
$\frac{1 - abcd}{(1-ac)(1-ad)(1-bc)(1-bd)}$
Is there a software package that can compute multigraded Hilbert series? Can it be computed using Macaulay2?
Alternatively, is there software that can compute the multigraded Hilbert series of a toric variety, specified by its fan?
For this example $v_1 = (0,0,1), v_2 = (1,0,1), v_3 = (1,1,1), v_4 = (0,1,1)$ specify the vertices of the toric fan. The multigraded Hiblert series is given by the index which counts points in the dual cone $S_{C^{*}}$ $\sum_{m \in S_{C^{*}}} q^m = \frac{(1 - q_1)}{ (1 - q_2)(1 - q_3)(1 - q_1 q_2^{-1}) (1 - q_1 q_3^{-1}) }$
References:
"Linear diophantine equations and local cohomology," R.P. Stanley, 1982.
"Combinatorial commutative algebra," E. Miller and B. Sturmfels, 2005.
"Sasaki-Einstein manifolds and volume minimisation," Martelli, Sparks, Yau, 2006.
-
## 2 Answers
Macaulay 2 can do multigraded Hilbert series. Let's first assume that you have a presentation of your multigraded ring. I'll mention how to calculate this below. So for your $S = k[ac,ad,bc,bd]$, we'll write it as $S = k[x,y,z,w] / (xz - yw)$.
Assuming that each of $a,b,c,d$ has its own degree direction (so the grading is by ${\bf Z}^4$), we input $k[x,y,z,w]$ as
````S = QQ[x,y,z,w, Degrees=> {{1,0,1,0}, {1,0,0,1}, {0,1,1,0}, {0,1,0,1}}]
````
where here QQ means the rationals. Then we want the Hilbert series of the ideal $(xz - yw)$, so we put
````i8 : hilbertSeries ideal(y*z-x*w)
````
and the answer is:
```` 1 - T T T T
0 1 2 3
o8 = ----------------------------------------
(1 - T T )(1 - T T )(1 - T T )(1 - T T )
1 3 1 2 0 3 0 2
````
If you also need to get the presentation, we can do this as follows. First, download normaliz: http://www.mathematik.uni-osnabrueck.de/normaliz/
and the Macaulay 2 interface to normaliz (I think this is automatically there in Macaulay 2 1.3.1): http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Normaliz/html/
````loadPackage "Normaliz"
````
Set the path to normaliz in Macaulay 2 using the command
````setNmzExecPath("path to the executables norm32 and norm64");
````
The torusInvariants command will give you generators for the subring of invariants. Finally, we can use the ringmap command to define a surjection from a polynomial ring onto the subring to get the desired ideal.
-
### 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.
Piggyback question: What if the grading is by a f.g. abelian 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": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8607865571975708, "perplexity_flag": "middle"} |
http://mathhelpforum.com/statistics/180356-probability-distribution-functions.html | # Thread:
1. ## Probability Distribution Functions
Hello Friends
I need help with a past exam paper question for statistics; it's from a first-year undergraduate paper. It goes like this:
The probability distribution of a random variable has the following density function:
(x) = cx^2(x-1) if 0=<x=<1
= o otherwise
and we are given a probability graph with a normal distribution with f(x) on the y-axis. x: 0=<x=<1 and f(x): 0=<x=<1 .
i) Show that c=30.
Can anyone please advise how I would go about solving this? I don't want the question solved for me, just a couple of hints.
Hope this is okay. Muchos gracios.
Best,
ThetaPhi
[/FONT]
2. solve this
$\displaystyle \int_0^1 cx^2(x-1) ~dx = 1$
3. Originally Posted by pickslides
solve this
$\displaystyle \int_0^1 cx^2(x-1) ~dx = 1$
Yep, done that. I moved c in front of the integral and then solved to find that the integral is 1/12. Hence, as per the integral in your quote, c=12. But this is my point, why c=30?
4. Originally Posted by ThetaPhi
Yep, done that. I moved c in front of the integral and then solved to find that the integral is 1/12. Hence, as per the integral in your quote, c=12. But this is my point, why c=30?
Whoever said c = 30 is obviously wrong and you will need to discuss it with them. By the way, c = 12 is wrong too. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9462786316871643, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/108296-witch-maria-agnesi.html | # Thread:
1. ## witch of Maria Agnesi
The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).
y =
It looks so weird
2. Originally Posted by miss.strw
The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).
slope, $m = y'(-2)$
now use the point-slope form of a linear equation ...
$y - y_1 = m(x - x_1)$ where $(x_1,y_1)$ is the given point.
3. Originally Posted by skeeter
slope, $m = y'(-2)$
now use the point-slope form of a linear equation ...
$y - y_1 = m(x - x_1)$ where $(x_1,y_1)$ is the given point.
Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! please reply soon!
i do not know how to get the y` like i don't know how to do tat.
4. Originally Posted by miss.strw
Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! please reply soon!
i do not know how to get the y` like i don't know how to do tat.
the slope is the derivative of the function evaluated at x = -2
what does this mean? ...
like i don't know how to do tat.
you don't know how to find a derivative?
5. Originally Posted by skeeter
the slope is the derivative of the function evaluated at x = -2
what does this mean? ...
you don't know how to find a derivative?
Not for those one. i don't know is it like 1/1+x^2 = 2x?
so it's like
y-(1/5)=(2x)(x-(-2)) ?
thank you so much for helping!
6. Originally Posted by miss.strw
Not for those one. i don't know is it like 1/1+x^2 = 2x? no
so it's like
y-(1/5)=(2x)(x-(-2)) ? no
$y = \frac{1}{1+x^2}$
use the quotient rule to find y' , or ...
$y = (1+x^2)^{-1}$
... take the derivative using the chain rule.
7. Originally Posted by skeeter
$y = \frac{1}{1+x^2}$
use the quotient rule to find y' , or ...
$y = (1+x^2)^{-1}$
... take the derivative using the chain rule.
g`(-2)h(-2)-g(-2)h`(-2)
---------------------------
[h(-2)]^2
this doesn't make sense to me
if i do the derivative
1
-------
(1+x^2)
is derivative is x^2 = 2x
8. you need to find the derivative first, then evaluate it at x = -2 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "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": 0.93644118309021, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/794?sort=votes | ## Examples of rational families of abelian varieties.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I'd like to know examples of non-trivial families of abelian varieties over rational bases (e.g. open subschemes of the projective line P^1).
One can generate many examples as Jacobians of rational families of curves (e.g. the hyperellitpic family, plane curves, complete intersections). Prym varieties are another example.
Are there any examples which are not obviously Jacobians of a family of curves? I would like to know both principally polarized and non principally polarized examples.
-
David, can you explain the question a bit? Thanks. – shenghao Oct 16 2009 at 20:56
By "rational base" you mean a rational variety like affine spaces, not Spec of the rational numbers, right? So you want a non-constant k-morphism from some rational k-variety to like the moduli space A_{g,d} of dimension g degree d polarized abelian varieties? – shenghao Oct 17 2009 at 0:49
## 6 Answers
David, I don't know if you are still interested in this, it's been over a year. I just stumbled upon your question in the depths of MO. I often found that Weil restrictions of elliptic curves give nice families of examples on which you can test things.
E.g. take a family of elliptic curves $y^2=x^3+t$ and Weil restrict it from ${\mathbb Q}(i)$ to ${\mathbb Q}$. Writing $x=x_1+x_2i$ and similarly for $y$ and $t$, expanding the equation and breaking it into real and imaginary parts, you get a family of 2-dimensional abelian varieties over ${\mathbb Q}(t_1,t_2)$ given by two equations in a 4-dimensional space, $$y_1^2-y_2^2 = x_1^3 - 3x_1 x_2^2 + t_1, \qquad 2y_1y_2 = x_1^2x_2 - 3x_2^3 + t_2.$$ Alternatively, you fix the elliptic curve, but you let the extension vary with $t$ (e.g. ${\mathbb Q}(t^{1/3})$), or both, and you also get interesting families.
The really nice thing is that as opposed to Jacobians, Weil restrictions are trivial to write down in terms of equations. Over the algebraic closure they are isogenous isomorphic to products of elliptic curves (making them boring), but for arithmetic applications they are interesting. There is a small extension of this construction, when you do not base change the elliptic curve but you "tensor it with a ${\mathbb Z}-$module with a Galois action", which is not necessarily a permutation module. This is explained in Milne's paper "On the arithmetic of abelian varieties" (Invent. Math. 1972) section 2, and it is useful if you want to write down non-principally polarised examples.
-
1
Actually, over the algebraic closure, a Weil restriction of an elliptic curve becomes isomorphic (not just isogenous) to a product of elliptic curves (that are Galois-conjugates of the original one). – Yuri Zarhin Feb 5 2011 at 22:33
You are absolutely right. Thank you, fixed! – Tim Dokchitser Feb 5 2011 at 23:17
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Section 5 of Faltings: Arakelov's theorem for abelian varieties is dedicated to the construction of an elaborate example of a family of abelian varieties that do not satisfy a condition Faltings calls $(*)$. I believe the condtion implies that the abelian varieties in the family, at least the general fiber, cannot be a Jacobian. The base of the family is $\mathbb H$ the upper-half plane, so it won't give you a family over a rational curve, but the construction itself is pretty involved so undertsanding it might help with understanding families of abelian varieties in general and might also give you some ideas for this problem.
-
2
There is an explicit example of a certain family of abelian eightfolds at the very end of my paper (Proc. London Math. Soc. 96 (2008), no. 2, 312--334) where the general fiber is the square of a Jacobian while the condition (*) does not hold. – Yuri Zarhin Feb 5 2011 at 21:52
Using a fact that every abelian variety is isogenous to a quotient of a Jacobian, one may construct families of Jacobians that do not satisfy the condition (*). – Yuri Zarhin Feb 6 2011 at 4:18
It is known that $A_g$ is unirational for $g\leq 5$, so at least for these g, we have some nontrivial families over open subsets of projective spaces. For $g> 3$, $\dim M_g < \dim A_g$, so these families are not Jacobians.
-
Do these all come from Prym's? – David Zureick-Brown♦ Oct 17 2009 at 20:43
I don't know what are Prym varieties...but you can read papers proving A_g are unirational, and see the construction there. – shenghao Oct 18 2009 at 0:26
They don't all come from Pryms -- Pryms arise from Jacobians of lower-genus curves, so the locus of Pryms in A_g has dimension strictly smaller than 3g-3. Once g > 3, the Jacobians and Pryms form a finite union of proper subvarieties, so there are certainly rational curves not contained in this locus if A_g is unirational. (If you think of families of abelian varieties ISOGENOUS to Jacobians as also "coming from Jacobians," then you have a countable union of proper subvarieties you have to miss; it's still clear you can do it with a rational curve over C, not so easy over Qbar..) – JSE Oct 19 2009 at 16:14
4
In fact they do all come from Pryms. The point is that to get a g dimensional abelian variety as a Prym one has to consider etale double covers of curves of genus g+1; one gets all ppavs of dimension g as Pryms exactly when g <= 5. – ulrich Oct 20 2009 at 6:32
unknown above is right, sorry for the mistake! – JSE Oct 24 2009 at 1:53
One can construct some families over rational bases which are not Jacobians by taking quotients:
For example, let A be a fixed abelian variety of dimension > 1 and let S be the space of all smooth complete interesection curves for some very ample line bundle on A. For any s in S, let C_s denote the corresponding curve in A. The inclusion of C_s in A induces a surjective morphism from J(C_s) (the Jacobian of C_s) to A and so by duality a morphism A^ to J(C_s) where A^ is the dual abelian variety of A. The quotient of J(C_s) by the image of A^ gives a family of abelian varieties over S. It can be shown using monodromy that this family is non-trivial.
-
"..let S be the space of all smooth complete interesection curves for some very ample line bundle on A." What does this mean? – David Zureick-Brown♦ Oct 19 2009 at 16:34
Let L be a very ample line bundle on A, let T = P(H^0(A,L)) and S be the open subset of T^{d-1}, d = dim(A), corresponding to those intersections of d-1 hypersurfaces which are smooth. – ulrich Oct 20 2009 at 6:27
How about intermediate Jacobians of cubic threefolds? It's easy to write down rational families of cubic hypersurfaces in P^4; it might be much harder to say anything about the corresponding family of abelian varieties, depending on what features you're looking for.
-
1
There are different ways to make intermediate Jacobians. But as far as I know, either they produce abelian varieties, that is they come with a polarization (the Weil ones) or deform holomorphically in families (the Griffiths ones), but not both at the same time. – Abdó Roig-Maranges Oct 19 2009 at 16:23
Hmm. So for this family of intermediate Jacobians of cubic hypersurfaces, can one calculate the monodromy? – David Zureick-Brown♦ Oct 19 2009 at 17:24
These intermediate Jacobians are 5 dimensional ppavs. There are general results about the monodromy action on the middle cohomology of families of hypersurfaces of any degree (for example, by Beauville). The monodromy is usually as big as possible; in the cubic threefold case I think it might be all of Sp(10,Z). – ulrich Oct 20 2009 at 6:40
1
@Abdó: Cubic threefolds have $h^{0,3}=h^{3,0}=0$ so the Hodge structure of $H^3$ is a Tate twist of an abelian variety and there are no problems with getting both a polarisation and holomorphic variation. I think (but am not totally sure) that in general when there are only two non-zero Hodge numbers, the intermediate Jacobians are both polarised and deform holomorphically. It is only when there are at least four non-zero Hodge numbers that one has to involve complex conjugation in the Weil construction. – Torsten Ekedahl Nov 26 2010 at 6:14
Just for the curiosity, into which category falls the fibration which has an elliptic curve with given j-invariant over a point `j\in P^1`?
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An elliptic curve is its own Jacobian, if that helps. – Anweshi Feb 5 2010 at 13:19 | {"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": 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": 0.9282105565071106, "perplexity_flag": "middle"} |
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