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http://en.wikipedia.org/wiki/Poisson_regression
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# Poisson regression
Regression analysis
Models
Estimation
Background
In statistics, Poisson regression is a form of regression analysis used to model count data and contingency tables. Poisson regression assumes the response variable Y has a Poisson distribution, and assumes the logarithm of its expected value can be modeled by a linear combination of unknown parameters. A Poisson regression model is sometimes known as a log-linear model, especially when used to model contingency tables.
Poisson regression models are generalized linear models with the logarithm as the (canonical) link function, and the Poisson distribution function.
## Regression models
If $x \in \mathbb{R}^n$ is a vector of independent variables, then the model takes the form
$\log (\operatorname{E}(Y|x))=a' x + b,\,$
where $a \in \mathbb{R}^n$ and $b \in \mathbb{R}$. Sometimes this is written more compactly as
$\log (\operatorname{E}(Y|x))=\theta' x,\,$
where x is now an (n + 1)-dimensional vector consisting of n independent variables concatenated to some constant, usually 1. Here θ is simply a concatenated to b.
Thus, when given a Poisson regression model θ and an input vector $x$, the predicted mean of the associated Poisson distribution is given by
$\operatorname{E}(Y|x)=e^{\theta' x}.\,$
If Yi are independent observations with corresponding values xi of the predictor variable, then θ can be estimated by maximum likelihood. The maximum-likelihood estimates lack a closed-form expression and must be found by numerical methods. The probability surface for maximum-likelihood Poisson regression is always convex, making Newton–Raphson or other gradient-based methods appropriate estimation techniques.
## Maximum likelihood-based parameter estimation
Given a set of parameters θ and an input vector x, the mean of the predicted Poisson distribution, as stated above, is given by
$\operatorname{E}(Y|x)=e^{\theta' x}\,$,
and thus, the Poisson distribution's probability mass function is given by
$p(y|x;\theta) = \frac{[\operatorname{E}(Y|x)]^y \times e^{-\operatorname{E}(Y|x)}}{y!} = \frac{e^{y \theta' x} e^{-e^{\theta' x}}}{y!}$
Now suppose we are given a data set consisting of m vectors $x_i \in \mathbb{R}^{n+1}, \, i = 1,...,m$, along with a set of m values $y_1,\ldots,y_m \in \mathbb{R}$. Then, for a given set of parameters θ, the probability of attaining this particular set of data is given by
$p(y_1,\ldots,y_m|x_1,\ldots,x_m;\theta) = \prod_{i=1}^m \frac{e^{y_i \theta' x_i} e^{-e^{\theta' x_i}}}{y_i!}.$
By the method of maximum likelihood, we wish to find the set of parameters θ that makes this probability as large as possible. To do this, the equation is first rewritten as a likelihood function in terms of θ:
$L(\theta|X,Y) = \prod_{i=1}^m \frac{e^{y_i \theta' x_i} e^{-e^{\theta' x_i}}}{y_i!}$.
Note that the expression on the right hand side has not actually changed. A formula in this form is typically difficult to work with; instead, one uses the log-likelihood:
$\ell(\theta|X,Y) = \log L(\theta|X,Y) = \sum_{i=1}^m \left( y_i \theta' x_i - e^{\theta' x_i} - \log(y_i!)\right)$.
Notice that the parameters θ only appear in the first two terms of each term in the summation. Therefore, given that we are only interested in finding the best value for θ we may drop the yi! and simply write
$\ell(\theta|X,Y) = \sum_{i=1}^m \left( y_i \theta' x_i - e^{\theta' x_i}\right)$.
To find a maximum, we need to solve an equation $\frac{\partial \ell(\theta|X,Y)}{\partial \theta} = 0$ which has no closed-form solution. However, the negative log-likelihood, $-\ell(\theta|X,Y)$, is a convex function, and so standard convex optimization techniques such as gradient descent can be applied to find the optimal value of θ.
## Poisson regression in practice
Poisson regression may be appropriate when the dependent variable is a count, for instance of events such as the arrival of a telephone call at a call centre.[1]:pp.740-752 The events must be independent in the sense that the arrival of one call will not make another more or less likely, but the probability per unit time of events is understood to be related to covariates such as time of day.
### "Exposure" and offset
Poisson regression may also be appropriate for rate data, where the rate is a count of events occurring to a particular unit of observation, divided by some measure of that unit's exposure. For example, biologists may count the number of tree species in a forest, and the rate would be the number of species per square kilometre. Demographers may model death rates in geographic areas as the count of deaths divided by person−years. More generally, event rates can be calculated as events per unit time, which allows the observation window to vary for each unit. In these examples, exposure is respectively unit area, person−years and unit time. In Poisson regression this is handled as an offset, where the exposure variable enters on the right-hand side of the equation, but with a parameter estimate (for log(exposure)) constrained to 1.
$\log{(\operatorname{E}(Y|x))} = \log{(\text{exposure})} + \theta' x$
which implies
$\log{(\operatorname{E}(Y|x))} - \log{(\text{exposure})} = \log{\left(\frac{\operatorname{E}(Y|x)}{\text{exposure}}\right)} = \theta' x$
Offset in the case of a GLM in R can be achieved using the offset() function:
```glm.fit <- glm(y ~ offset(log(exposure)) + x, family=poisson(link=log) )
```
### Overdispersion
A characteristic of the Poisson distribution is that its mean is equal to its variance. In certain circumstances, it will be found that the observed variance is greater than the mean; this is known as overdispersion and indicates that the model is not appropriate. A common reason is the omission of relevant explanatory variables, or dependent observations. Under some circumstances, the problem of overdispersion can be solved by using a negative binomial distribution instead.[2][3]
Another common problem with Poisson regression is excess zeros: if there are two processes at work, one determining whether there are zero events or any events, and a Poisson process determining how many events there are, there will be more zeros than a Poisson regression would predict. An example would be the distribution of cigarettes smoked in an hour by members of a group where some individuals are non-smokers.
Other generalized linear models such as the negative binomial model or zero-inflated model may function better in these cases.
### Use in survival analysis
Poisson regression creates proportional hazards models, one class of survival analysis: see proportional hazards models for descriptions of Cox models.
### Tests of over-dispersion
One method for testing for over dispersion in the data is to regress a variable zi against the predicted values of t estimated from the Poisson regression.[4] This test has three steps.
1. Estimate a Poisson regression of yi on xi and generate the predicted values ( ti )
2. Calculate the zi variable
$z_i = \frac{ ( y_i - t_i )^2 - y_i } { t_i \sqrt{2} }$
3. Regress zi against ti with ordinary least squares. In symbols
$z_i = a t_i + e_i \,$
where a is a constant and ei is a random variable with an expectation of zero.
The null hypothesis being tested here is that the data are Poisson-distributed: in this case a = 0.
## Extensions
### Regularized Poisson regression
When estimating the parameters for Poisson regression, one typically tries to find values for θ that maximize the likelihood of an expression of the form
$\sum_{i=1}^m \log(p(y_i;e^{\theta' x})),$
where m is the number of examples in the data set, and $p(y_i;e^{\theta' x})$ is the probability mass function of the Poisson distribution with the mean set to $e^{\theta' x}$. Regularization can be added to this optimization problem by instead maximizing
$\sum_{i=1}^m \log(p(y_i;e^{\theta' x})) - \lambda \left\|\theta\right\|_2^2$,
for some positive constant $\lambda$. This technique, similar to ridge regression, can reduce overfitting.
## Implementations
Some statistics packages include implementations of Poisson regression.
• MATLAB Statistics Toolbox: Poisson regression can be performed using the "glmfit" and "glmval" functions.[5]
• Microsoft Excel: Excel is not capable of doing Poisson regression by default. One of the Excel Add-ins for Poisson regression is XPost
• R: The function for fitting a generalized linear model in R is glm(), and can be used for Poisson Regression
• SAS: Poisson regression in SAS is done by using GENMOD
• SPSS: In SPSS, Poisson regression is done by using the GENLIN command
• Stata: Stata has a procedure for Poisson regression named "poisson", and for panel data "xtpoisson"
• mPlus: mPlus allows for Poisson regression using the command COUNT IS when specifying the data
## References
• Cameron, A.C. and P.K. Trivedi (1998). Regression analysis of count data, Cambridge University Press. ISBN 0-521-63201-3
• Christensen, Ronald (1997). Log-linear models and logistic regression. Springer Texts in Statistics (Second ed.). New York: Springer-Verlag. pp. xvi+483. ISBN 0-387-98247-7. MR 1633357.
• Hilbe, J. M. (2007). Negative Binomial Regression, Cambridge University Press. ISBN 978-0-521-85772-7
1. Greene, William H., Econometric Analysis, fifth edition, Prentice-Hall, 2003.
2. Paternoster R, Brame R (1997). "Multiple routes to delinquency? A test of developmental and general theories of crime". Criminology 35: 45–84.
3. Berk R, MacDonald J (2008). "Overdispersion and Poisson regression". Journal of Quantitative Criminology 24: 269–284.
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http://nrich.maths.org/7274
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nrich enriching mathematicsSkip over navigation
### Ball Bearings
If a is the radius of the axle, b the radius of each ball-bearing, and c the radius of the hub, why does the number of ball bearings n determine the ratio c/a? Find a formula for c/a in terms of n.
### Overarch 2
Bricks are 20cm long and 10cm high. How high could an arch be built without mortar on a flat horizontal surface, to overhang by 1 metre? How big an overhang is it possible to make like this?
### Cushion Ball
The shortest path between any two points on a snooker table is the straight line between them but what if the ball must bounce off one wall, or 2 walls, or 3 walls?
# Population Dynamics - Part 5
### The Lotka-Volterra Equations
The Lotka-Volterra equations allow us to model a biological system containing a predator and a prey species. They arose in the 1920s due to the independent work of the mathematicians Alfred Lotka and Vito Volterra.
### Formulating the Equations
Let x be the size of a fish population and y be the size of the shark population. If the prey have an unlimited food supply and no predators, then they grow exponentially according to the equation $\frac {\mathrm{d}x}{\mathrm{d}t}=\lambda x$ , for some constant $\lambda$.
If there are more sharks, more fish will be killed and if there are fewer sharks, then fewer fish will be killed. So the rate of predation upon the prey is proportional to the rate at which the predators and prey meet. We represent this by $\alpha xy$.
Note that if either x or y is zero then there can be no predation.
The equation for prey becomes: $$\frac {\mathrm{d}x}{\mathrm{d}t}=\lambda x-\alpha xy =x(\lambda -\alpha y)$$ Conversely, predators rely on prey to survive, so in the absence of any prey, the predator equation is $\frac {\mathrm{d}y}{\mathrm{d}t}=-\gamma y$. The growth of the predator population will depend on the population sizes of the predators and prey, and the ability of the predators to successfully catch prey. We represent this by $\beta xy$.
The equation for predators becomes: $$\frac {\mathrm{d}y}{\mathrm{d}t}=\beta xy-\gamma y =y(\beta x -\gamma)$$
### Using the Equations
We are interested in population equilibrium, which occurs when neither population is changing: $$\begin{align*} \frac {\mathrm{d}x}{\mathrm{d}t}&=\frac {\mathrm{d}y}{\mathrm{d}t}=0 \\ x(\lambda -\alpha y)&=y(\beta x -\gamma)=0 \\ \Rightarrow x=0,y=0 &\text{ and } x=\frac{\gamma}{\beta}, y=\frac{\lambda}{\alpha} \end{align*}$$ The first solution occurs when both species die out, the second when the sizes of both species reach an equilibrium with each other.
Determining the stability of these equations, requires knowledge of the Jacobian matrix and linearization, which we exclude for complexity.
Question: Consider a species of fish, x, and a species of shark, y, with population equations: $$\frac {\mathrm{d}x}{\mathrm{d}t}=2x-4xy , \frac {\mathrm{d}y}{\mathrm{d}t}=xy-2y$$ a) Where are the equilibrium points?
b) Below is a phasor diagram of the above equations. Try and work out what it is representing. What happens at the red point?
We can also plot the size of the populations against time on the same graph. The graph below clearly shows the dependency of the shark population on a large population of prey, and the oscillatory nature of the two population sizes.
Question: Why do you think the shark population peaks after the fish population has soared? Why do you think the shark population is always less than the fish population?
### Limitations to our model
In reality if two species coexist, their interactions also depend on the environment they inhabit. This is explored further here. Some of the assumptions made in the above equations are:
• The prey always find ample food
• The rate of change of population is proportional to its size
Question: Can you think of any other assumptions we have made in our model?
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://crypto.stackexchange.com/tags/collision-resistance/hot
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# Tag Info
## Hot answers tagged collision-resistance
9
### From hash to Cryptographic hash
Well, first of all, you need to be clear about the meanings of various cryptographical primitives. Cryptographic hash function; this is a function that takes an input string, and generates a hash. The idea is that we don't know how to create two input strings with the same hash, and so the hash can be used as a replacement for the original string. Now, ...
7
### What is pre-image resistance, and how can the lack thereof be exploited?
Preimage resistance is about the most basic property of a hash function which can be thought. It means: For a given $h$ in the output space of the hash function, it is hard to find any message $x$ with $H(x) = h$. (Note that the it is hard here and in the next definitions is not formally defined, but can be formalized by looking at families of hash ...
5
### bcrypt - collision-resistance against chosen salt and work factor?
The answer to the original question would have been: Yes, it would be impossibly difficult to exhibit workfactor, salt, password0, password1 such that bcrypt(workfactor, salt, password0) = bcrypt(workfactor, salt, password1); but even if that was feasible, it would not matter much, because in normal use at least one of the password is unknown to an adversary ...
4
### Real life collision when only using truncated hash
Accidental collisions are interesting for certain applications, and one would expect accidental collisions to occur less frequently in a system than malicious collisions. So, if you are not worried about malicious collisions, only accidental, it is easy to compute how many digests you would need to compute before seeing an accidental collision. If the ...
3
### What is pre-image resistance, and how can the lack thereof be exploited?
Often the hash (iterated and salted mostly) of a password is saved in a database, instead of the password. If a user logs in, the hash is computed and compared against the stored hash value. This way a user that can see the database of hashes does not see the password directly, but this property depends crucially on the hash being resistant to a pre-image ...
3
### What is pre-image resistance, and how can the lack thereof be exploited?
A collision attack is the ability to find two inputs that produce the same result, but that result is not known ahead of time. In a typical case (e.g., the attack on MD5) only a relatively small number of specific inputs are known to produce collisions. Collision resistance obviously means that a collision attack is difficult (for some definition of ...
2
### RSA SHA1 signature from public key
No, it's not possible. RSA based signatures can only be generated if you know the private key. At the RSA primitive level, signing is the same as decrypting a message (where the data being decrypted in this instance is a hash of the message). If you change the hash, then verification of that signature would fail. Without knowledge of the corresponding ...
2
### Hash collision resistance requirements for Lamport signatures
Yes, it makes sense to truncate the hash to 128 bits. The security proof actually says that if finding a preimage for F requires effort 2^n, then breaking the Lamport signature scheme with G having k-bit digests requires effort (2^n)/(2k). So strictly speaking, with F truncated to 128 bits and G having 256 bits (2k=512=2^9), you will have 128-9=119 bits of ...
2
### Real life collision when only using truncated hash
Yes, there have been real life inputs with collisions working on the full length of MD5. There is a pair of X.509 certificates that share an MD5 hash. There is also a pair of PostScript documents that are an MD5 collision. There are also two binary strings a mere 6 bits different that are a collision. The whole length of MD5 has been broken. Attacks that ...
2
### What “Tag Length” should be used for the EAX MAC?
While the EAX mode permits truncating the tag to any length $\tau$ between 0 and $n$ bits, where $n$ is the block size of the underlying block cipher, this should only be taken as a statement that truncating the tag in EAX mode does not introduce any security issues beyond the obvious (an attacker only needs $2^\tau$ attempts to froge a $\tau$-bit tag by ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://physics.stackexchange.com/questions/32592/limits-on-a-gauss-box-of-light/32672
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# limits on a gauss box of light
Consider a wall defined by $w(x,y,z) = \Theta(x-L)$ which is nonzero in the infinite semi-space of $x \ge L$, as well as a coherent planar standing EM wave travelling in the $z$ plane given by its electric field:
$$E_x = \Theta(x - L) \sin(kz) \sin(\omega t), E_y = E_z = 0$$
Consider a complementary standing wave travelling in a opposite plane, in the region $x \le -L$:
$$E_x = - \Theta(L - x) \sin(kz) \sin(\omega t), E_y = E_z = 0$$
now, if i take a small closed box in the region $-L- \epsilon \le x \le L + \epsilon$ and z such that $0 \le kz \le \pi$, the net electric flux over this box at any given moment of time is:
$$\frac{ 8 \Delta y \sin(\omega t) }{k}$$
Naively, it would seem "conceivable" to setup standing waves in this way and produce a net electric flux in a given region of vacuum without any spatial charges anywhere. Now, since this apparent flux increases with the wavelength, i suspect there is some optical bound in the coherence that the walls can sustain due to dispersion, i.e: something like $\Delta k_z \Delta x \ge \hbar$, but i can figure out exactly what is the reason this won't work.
What i'm trying to figure out is how physical this solution is and what limits are imposed by optics in the realization of the charge-violating gauss box
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## 2 Answers
It is easy to see that your light cannot have such a well defined wall without any charges, i.e:
$$\nabla \cdot E = \lbrack \delta (x-L) + \delta (L-X) \rbrack \sin(kz) \sin(\omega t)$$
which means you need surface charges to sustain the wall.
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Oh I see--- the OP was taking the field to vanish inside the wall, and making the box straddle the wall so that part is in and part is out. I didn't understand. +1, you had the right interpretation. The reason I didn't understand is because he said "-(L+\epsilon)<x<L+\epsilon" where the flux is zero, and also -(L-\epsilon)<x<L-epsilon (where it's all plane wave) the flux is zero. He probably meant -L-\epsilon<x<L-\epsilon, where the flux is nonzero and it is what you said. – Ron Maimon Jul 23 '12 at 14:20
The net electric flux over any box in any configuration of electromagnetic waves without charges is zero. The electric field lines that enter the box also leave the box. In your case, it's obvious--- the E field is in the x-direction for both plane waves, and so the flux is only nonzero on the two opposite sides which are in the y-z plane. The flux entering one of the sides is equal point by point to the flux leaving the other side, because the plane waves are translationally invariant.
More generally, you can't violate Gauss's law by superposing plane-waves--- each one already satisfies Gauss's law by itself, and superposing two things that satisfy Gauss's law gives a third thing that also satisfies Gauss's law. The demonstration is by noting that for a plane wave
$$\nabla \cdot E \propto k\cdot E$$
that is, it is zero if E is perpendicular to k. This is just saying that the field lines in a plane wave go from one side of space to the other side of space at any one time, without being created or destroyed.
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http://math.stackexchange.com/questions/305088/proof-that-a-n-hypercube-is-n-vertex-connected
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# Proof that a n-hypercube is n-vertex-connected
I'm new to graph theory, I'm finding it hard to get upon proofs.
To prove: An n-hypercube is n-vertex connected. Approaches I thought:
• It holds true for `n=2`, so assume it holds true for `n=k-1`, and prove it for `n=k`, so its proved by induction.
• Prove that there are `n` vertex disjoint paths between every pair of points (u,v) in the n-hypercube, then it's n-vertex connected. (right?)
Can someone please point me in the right direction? What different approaches are possible for this problem?
I'm unable to get think of solutions myself clearly. But after I know a solution, it seems easy. I want to improve my problem solving skills in graphs. Which types of problems would you recommend me to start with? Any resources would be highly appreciated. Thanks!
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## 3 Answers
To show that your approaches work, let's prove that there are $n$ disjoint path's by induction ;-)
It definitely works for $n = 2$, so assume it holds true for $n = k-1$. Let $u = (u_0,u_1,\ldots,u_{n-1})$ and $v = (v_0,v_1,\ldots,v_{n-1})$. Now, there are two cases:
• Vertices $u$ and $v$ have some common index $i$ such that $u_i = v_i$. By removing this index and applying induction hypothesis we get $n-1$ disjoint paths. The $n$th path $$u \leadsto u[i\to 1-u_i] \leadsto^* v[i\to 1-v_i] \leadsto v$$ is surely disjoint, because it differs at index $i$ all the time.
• Vertices $u$ and $v$ differ on all the indexes. WLOG we can assume that $u_i = 0$ and $v_i = 1$ for all $i$. The $k$th path goes like this: $$0 \leadsto e_k \leadsto e_k+e_{k+1} \leadsto^* e_k+\ldots+e_{n-1}+e_0+e_1+\ldots+e_{k-2} \leadsto 1,$$ or equivalently, $j$th vertex of $k$th path equals $$\sum_{i=k}^{k+j-1}e_{i\ \bmod\ n},$$ where $e_i = (0,0,\ldots,0,1,0,0,\ldots)$ has a single $1$ (one) at $i$th place.
I hope it helps ;-)
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Nice, I just need to ask, does there exist a simpler proof for this? – Bruce Feb 16 at 3:00
Isn't Joe Zeng's proof simpler? If you were to expand his hint, just do induction again, where the hypothesis is "after removing $n-1$ vertices the $n$-cube stays connected". It is true if all the removed vertices have a common neighbor, and if not, then you split your cube into two parts (there must be an index where they differ) that have at most $n-2$ removed vertices and you apply the inductive hypothesis. – dtldarek Feb 16 at 7:56
Hint: Consider that every vertex of an $n$-hypercube is connected to $n$ vertices. What does this mean about deleting any $n - 1$ vertices in the cube?
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Deleting n-1 vertices must result in a bridge in the graph, right? Sorry but I can't figure out how to prove the presence of a bridge with that hint.. – Bruce Feb 15 at 22:09
I was more thinking that the graph will always stay connected after removing $n-1$ vertices. It doesn't necessarily need to be a bridge. – Joe Z. Feb 15 at 23:36
And if you remove all $n$ vertices adjacent to one vertex, that one vertex is then disconnected from the rest of the graph. – Joe Z. Feb 15 at 23:36
Oops I got confused with edge connectivity there. – Bruce Feb 16 at 2:55
I really need a formal explanation/proof, can you give one? How can it be said that the graph will always remain connected after removing n-1 vertices? I'm unable to understand! – Bruce Feb 16 at 2:59
show 1 more comment
I found a proof in Introduction to Graph Theory by Douglas B. West
@Joe Zeng, @dtldarek please have a look at this. What are your thoughts on this?
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Well, this is exactly the proof I've given, not the first time of "someone has been here before". – dtldarek Feb 16 at 7:59
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http://mathoverflow.net/revisions/37778/list
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Return to Question
2 added 206 characters in body
If $A \subseteq \mathcal B(\mathcal H)$ is an algebra of operators that is closed under adjoint, then its bicommutant $A''$ is a von Neumann algebra, and is the ultraweak closure of $A$; this is one version of von Neumann's bicommutant theorem. Does the theorem hold relative to an arbitrary von Neumann algebra $\mathcal M$? Concretely, what is the truth value of the following statement:
Let $\mathcal M$ be a W*-algebra, and $A\subseteq \mathcal M$ be a subalgebra closed under adjoint. Then the relative bicommutant $A'' = \{ m \in \mathcal M | \forall x \in \mathcal M. (\forall a \in A. ax = xa) \implies mx=xm \}$ is a W*-algebra, and is the ultraweak closure of $A$.
A W*-algebra is C*-algebra that is isomorphic to a von Neumann algebra. I use the term W*-algebra to emphasize that the bicommutant is being computed relative to $\mathcal M$ itself rather than relative to a Hilbert space on which $\mathcal M$ is represented.
Edit: As Matthew points out below, $A$ should contain the unit of the ambient algebra, i.e., of $\mathcal B(\mathcal H)$ in the bicommutant theorem, and of $\mathcal M$ in the statement in question.
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Relative Bicommutant
If $A \subseteq \mathcal B(\mathcal H)$ is an algebra of operators that is closed under adjoint, then its bicommutant $A''$ is a von Neumann algebra, and is the ultraweak closure of $A$; this is one version of von Neumann's bicommutant theorem. Does the theorem hold relative to an arbitrary von Neumann algebra $\mathcal M$? Concretely, what is the truth value of the following statement:
Let $\mathcal M$ be a W*-algebra, and $A\subseteq \mathcal M$ be a subalgebra closed under adjoint. Then the relative bicommutant $A'' = \{ m \in \mathcal M | \forall x \in \mathcal M. (\forall a \in A. ax = xa) \implies mx=xm \}$ is a W*-algebra, and is the ultraweak closure of $A$.
A W*-algebra is C*-algebra that is isomorphic to a von Neumann algebra. I use the term W*-algebra to emphasize that the bicommutant is being computed relative to $\mathcal M$ itself rather than relative to a Hilbert space on which $\mathcal M$ is represented.
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http://physics.stackexchange.com/questions/27598/massive-excitations-in-conformal-quantum-field-theory
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# Massive excitations in Conformal Quantum Field Theory
Single particle states in quantum field theory appear as discrete components in the spectrum of the Poincare group's action on the state space (i.e. in the decomposition of the Hilbert space of quantum states into irreducible representations of the Poincare group). Classification of irreducible unitary representations of the Poincare group leads to the notions of mass and spin.
Now, suppose we have a conformal QFT and are doing the same trick with the conformal group. Which irreducible representations do we have?
We still have the massless particles (at least I'm pretty sure we do although I don't immediately see the action of special conformal transformations). However, all representations for a given spin s and any mass m > 0 combine into a single irreducible representation.
What sort of physical object corresponds to this representation? Is it possible to construct a scattering theory for such objects? Is it possible to define unstable objects of this sort?
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Very very naive question: you say there will be (irreducible) representations with a fixed spin $s$ and any mass $m>0$. Since any mass $m$ introduce a length scale $L\sim \frac 1m$, conformal transformations would transform states of different masses into each other. So you would need a theory of uncountable number of particles with any mass $m>0$? If this is correct, doesn't it (naively) seem to be quit hopeless to construct any consistent quantum field theory of this kind? Has such a theory ever been constructed? – Heidar Dec 3 '11 at 1:05
@Heidar, these states would not be particles. This is because the mass spectrum within each such representation is continuous. – Squark Dec 3 '11 at 8:32
## 1 Answer
Representation theory of the conformal group is discussed in the canonical reference by Mack. As for physical interpretation of the theory, the construction of asymptotic states and scattering theory does not work in CFT for the reasons you write. Rather, the basic observables are Euclidean correlation functions, and the operators of the theory can be arranged into Hilbert space. This is explained in the classic paper of Mack and Luscher.
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http://mathoverflow.net/revisions/68428/list
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## Return to Question
2 added math-physics tag, since this more a question of physics than of maths
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# Why is the output of an LTI system the convolution of the input funtion and the impulse response?
I am looking at the description of LTI systems in the time domain.
Intuitively, I'd have guessed it would be the composition of the input function and some "system function". $$y(t) = f(x(t)) = (f\circ x)(t)$$ Where $x(t)$ is the input, $y(t)$ output and $f(x)$ a "system function".
Why is it not that way? Could such a "system function" be found for, say, an R-C-Circuit?
The actual output function y(t), is defined as $$y(t) = (h * x)(t)$$ Where $h(t)$ is the response to a dirac impulse. This is hard to grasp for me. Why is it so? I have looked at various explanations, drawings of rectangles becoming infinitely narrow, which I sort of understood, but it is still "hard to grasp"! I am looking for a simple explanation in one or two sentences here.
http://en.wikipedia.org/wiki/LTI_system_theory
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http://mathoverflow.net/questions/6180/why-are-fusion-categories-interesting/6187
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## Why are fusion categories interesting?
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In the same vein as Kate and Scott's questions, why are fusion categories interesting? I know that given a "suitably nice" fusion category (which probably means adding adjectives such as "unitary," "spherical," and "pivotal"), we get a subfactor planar algebra which, in turn, gives us a subfactor. Also, I vaguely understand that these categories give us Turaev Viro TQFTs.
What else do fusion categories do? What's a good reference for the Turaev Viro stuff?
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## 7 Answers
Fusion categories (over $\mathbb{C}$) are a natural generalization of finite groups and their behavior over $\mathbb{C}$. The complex representation theory of a finite group is a fusion category, but there are many others. In fact, you can think of a fusion category as a non-commutative, non-cocommutative generalization of a finite group. A finite-dimensional Hopf algebra is that too, but they don't have to be semisimple, while the semisimple ones give you many fusion categories, but again not by any means all of them.
Many of the basic results about the structure and representation theory of finite groups generalize, or seem like they could generalize, to fusion categories. This principle has been worked out to a very incomplete but interesting extent by Etingof and others. For instance there is an analogue of the theorem that the dimension of complex irrep of a finite group $G$ divides $|G|$. (Addendum: A qualified analogue, as Scott and Noah point out. If the category is braided, it is a strict analogue; otherwise it is an analogue of dividing $|G|^2$.) There are also semisimple Hopf algebras and other fusion categories that look a lot like $p$-groups.
You can think of the whole theory as a rebooted theory of finite groups. However, we are miles and miles away from any fusion category equivalent of the classification of finite simple groups. It is a struggle to make fusion categories that are not derived very closely from finite groups, or do not come from quantum groups at roots of unity. Only a few types of examples are known, and who knows what else is out there.
One enticing thing that does change is that dimensions of irreducible objects in a fusion category don't have to be integers. For instance, one of the simplest fusion categories is the Fibonacci category. It has two irreducible objects, the trivial one $I$ and the other object $F$. The dimension of $F$ is the golden ratio, as you can infer from the branching equation $F \otimes F \cong F \oplus I$. (But the dimensions are algebraic integers, and even cyclotomic algebraic integers. Hence divisibility is still a sensible question.)
You could also ask, why the semisimple case. As you learn in undergraduate or basic graduate representation theory, the semisimple representation theory of a finite group is much cleaner than the modular representation theory in positive characteristic.
And yes, you also get 3-manifold invariants and subfactors.
For references: Really Turaev and Viro's original paper, state sum invariants of 3-manifolds and 6j symbols, is pretty good. The generalization to spherical categories is due to Barrett and Westbury, Invariants of Piecewise-Linear 3-manifolds. And there is a discussion in Turaev's book.
A sketch: Recall that a basis-independent expression in tensor calculus has the structure of a graph with vertices labelled by tensors and edges labelled by vector spaces. A monoidal category allows the evaluation of similar expressions, except that the graph must be planar and acyclic. In a rigid pivotal category, there are good duals and the graph just needs to be planar. In a spherical category, left trace equals right trace, so a closed graph can be drawn on a sphere. If it is spherical, rigid, and semisimple, then you can use the graph of a tetrahedron to make a local interaction on the tetrahedra of a triangulated 3-manifold, and the result up to normalization is the Turaev-Viro 3-manifold invariant. (In this setting you should dualize the tetrahedra, so that a tensor morphism in the category is associated to a face of the tetrahedron.)
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The relationship between (sufficiently nice) fusion categories and subfactor planar algebras is a little complicated. There are three things to keep in mind.
Almost trivially, a pivotal fusion category gives an unshaded planar algebra. This is just unravelling definitions. This doesn't give a subfactor planar algebra, however.
Alternatively, you can take the "alternating part" of your fusion category $\mathcal{C}$, with respect to your favourite object $V$. This essentially defines a shaded planar algebra with `$\mathcal{P}_k = \operatorname{End}(V \otimes V^* \otimes ... V)$` ($k$ tensor factors). This is a useful and interesting subfactor valued invariant of the pair $(\mathcal{C}, V)$.
Finally, a categorical Morita equivalence between two fusion categories is precisely a finite-depth subfactor (the two fusion categories are the A-A and B-B bimodules, the Morita equivalence and its inverse are the A-B and B-A bimodules).
(Warning, all of the above may require adding adjectives ...)
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You should say "oriented" somewhere above. – Noah Snyder Nov 19 2009 at 23:02
The analogy with finite groups is not quite as strong as Greg suggests. If you're looking at a braided fusion category then the dimensions of objects all divide the the global dimension (the sum of the squares of the dimensions).
The actual theorem (thanks Noah!) is: given $X$ an object of $Z(\mathcal{C})$, the double of a fusion category $\mathcal{C}$, $\operatorname{dim}(X)$ divides $|\mathcal{C}|$. When $\mathcal{C}$ is already braided, $Z(\mathcal{C})$ includes into $\mathcal{C}$, and this gives the statement above.
However, for general fusion categories this is only a conjecture, and indeed a conjecture with a proposed counterexample (we're still working on this one...)
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Noah points out that the "analogue" in Greg's statement that "For instance there is an analogue of the theorem that the dimension of complex irrep of a finite group $G$ divides $|G|$." is doing some work, and perhaps I shouldn't complain. – Scott Morrison♦ Nov 19 2009 at 22:32
"When C is already braided, Z(C) is equivalent to C" ... this statement seems false to me. – pasquale zito Nov 19 2009 at 22:42
Equivalent is too strong. He just meant that there's an inclusion of C into Z(C). In particular, every object in C gives and object in Z(C), so his conclusion follows. – Noah Snyder Nov 19 2009 at 23:01
To be fair, it is a somewhat weakened analogue. The general statement for all fusion categories is that it divides $|G|^2$, the dimension of $Z(C)$. (I thought if $C$ is braided, then $Z(C)$ is two mirror copies of $C$.) – Greg Kuperberg Nov 19 2009 at 23:06
Sorry, if $C$ is modular rather than just braided, then $Z(C)$ is two mirror copies of $C$. – Greg Kuperberg Nov 19 2009 at 23:19
show 2 more comments
I am very far from an expert on the subject, but I have had to answer this question before. I will provide a few of my answers that seem to be omitted from the above discussion.
1. Fusion rings, or fusion rules, appear in certain physical "thought experiments" (though I am not aware that they have yet actually been observed). For instance "anyons" are defined by certain fusion rules,and if they existed would be useful in quantum computation. In the dogma of field theory, for these fusion rules to be physically meaningful, there needs to be a categorification of them, which would be a fusion category. This initiates a broad class of problems: given a fusion ring, decide whether or not it has a categorification, and if so, how many (up to equivalence). By Ocneanu's rigidity theorem, there are only finitely many categorifications of any given fusion ring, so it's possibly a "tame" problem to solve.
2. There is no hope (at present, per Greg's comment =]) to completely classify fusion categories in any sense, so far as I understand. The classification of fusion categories would encompass not only the classification of finite groups, but also of compact Lie groups (via the tilting module construction on the associated quantum group which yields a fusion category). So people classify fusion categories in small classes under the assumption that group theoretical categories (ones defined purely in terms of group theory: representations of groups, cohomology of groups, morita equivalences, etc.) is "easy" and they want to study the difference between the two contexts.
3. Greg mentioned that studying fusion categories is like studying semi-simple Hopf algebras, except that (a) there isn't necessarily a fiber functor to vector spaces, and (b) even if there does abstractly exist one, you don't choose one. If one admits the interest in studying Hopf algebras, then one has to admit the interest in fusion categories as a sort of "basis free" version. A direct application to finite groups is pinning down the precise relation between the groups D_8 and the quaternions. They are obviously not isomorphic; however their group rings are morita equivalent as rings (since they have the same number of irreducibles). Their irreps even have the same dimensions, so one can ask if their fusion categories are equivalent as fusion categories (they are not in this case, but there are some non-isomorphic groups which are so-called "isocategorical" meaning that not only are their group rings isomorphic, but the Hopf algebras are twist equivalent as Hopf algebras. The most sensible way to prove this sort of statement is through fusion categories.
4. For me, I am a fairly concrete-minded person, but someone who nevertheless tries to understand modern algebraic geometry, algebraic topology and category theory as best I can. Fusion categories have been a fantastic discovery for me, because they are in many ways homotopy theoretic/higher category-type constructions, but they are about as simple as one can get (because you basically have constrained the 1-morphisms as much as possible by the semi-simplicity assumption, and just focus on the higher morphisms). So for instance the first 2-groupoid I was ever able to understand in completely concrete terms arises in a paper of Etingof Nikshych and Ostrik about fusion categories. As such they can be viewed as a kindergarten of higher categories.
5. By the way, there is also some interest in "finite tensor categories" which are not semi-simple but satisfy the other finiteness conditions of fusion categories. (so you have finitely many simple objects, and you posit that every object is a finite-length extension of the simples). There's actually a great deal of the theory from fusion categories which generalizes here. So far as I can tell, the only obstacle in developing this notion more completely is that no one has had time to do it yet.
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2. There is no hope at the present of classifying fusion categories, but that's not the same as no hope ever. (At least the "simple" ones, whatever that means.) Finite simple groups already engulf simple Lie algebras, so a mutual generalization is not unfathomable. 3. The fiber functor, when it exists, is probably unique. For group algebras that is a classical Tannakian theorem. (Morita equivalence is of course a different matter.) – Greg Kuperberg Nov 21 2009 at 16:31
Your point about 2 is fair; my original statement was vague and speculative enough not to be evaluated one way or the other very effectively =]. In any case, no one that I know of sees it as a serious goal for the moment. – David Jordan Nov 21 2009 at 17:49
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As to your response to (3), I don't believe that is true. If you fix a symmetry on the category, then a symmetric fiber functor is unique up to isomorphism. However if you forget the symmetry, there are non-isomorphic fiber functors. This is, I believe the point behind <arxiv.org/pdf/math/0007196>;. It is also one way to prove that D_8 and Q_8 aren't isocategorical: they admit different numbers of (non-Tannakian) fiber functors. – David Jordan Nov 21 2009 at 17:52
On (3), you're right and I'm wrong. It's a great sequence of examples. – Greg Kuperberg Nov 21 2009 at 18:43
I also wrote a sequence of blog posts explaining the Turaev-Viro construction from the point of view of planar algebras. It has pretty pictures and might be relevant.
TQFTs via Planar Algebras I
TQFTs via Planar Algebras II
TQFTs via Planar Algebras III
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The following is a nice overview http://arxiv.org/abs/0804.3587 .
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When we were trying to understand Turaev-Viro and how the representation theory of $u_q(sl(2))$ mirrored the classical representation theory, we wrote The classical and Quantum 6j Symbols. This does not help with fusion categories per se, except that the representations of the quantum group must be a fusion category (correct me someone if I mis-stated that). But it does give the algebraic underpinning to the Kauffman Lins approach to TV invariants.
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If you divide by the negligible ideal, then the representation theory of a quantum group at a root of unity indeed becomes a fusion category. As nlab says, fusion = rigid + semisimple + monoidal + finitely many irreducibles. These are also spherical, hence Turaev-Viro. The only twist is that the category is multiplicity-free, so the $6j$-symbol is a scalar rather than a tensor for each six $j$s. ncatlab.org/nlab/show/fusion+category – Greg Kuperberg Nov 20 2009 at 2:09
Greg, Thanks. Scott – Scott Carter Nov 20 2009 at 14:48
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http://skullsinthestars.com/2008/06/19/invisibility-physics-schotts-radiationless-orbits/
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The intersection of physics, optics, history and pulp fiction
## Invisibility Physics: Schott’s radiationless orbits
Posted on June 19, 2008 by
Conventional wisdom, even to this day, dictates that accelerating charges necessarily give off electromagnetic radiation. This is seen, for instance, in large-scale particle accelerators (synchrotrons), such as the Tevatron at Fermilab and the soon-to-be-operational LHC at CERN: the charged particles moving around the ring are constantly shedding radiation over a range of frequencies, including X-rays.
In the first post in my series on the physics of invisibility, however, we discussed a little-known 1910 paper by Paul Ehrenfest, in which he demonstrates theoretically that one can have accelerating extended distributions of charge which produce no radiation fields. Ehrenfest was attempting to explain one of the most vexing problems of physics at the time: the presence of electrons in the atom. The atom was known to have electrons moving about within it, and these electrons should have been radiating constantly, according to the known physics of the time, but were not seen to do so.
Soon after Ehrenfest’s paper, Bohr produced his model of the atom, which eliminated the need for radiationless orbits and ended most speculation on atomic structure. Ehrenfest’s work was mostly forgotten, but other researchers independently discovered other radiationless motions of charges, and this research would lead eventually to more detailed studies of invisibility. One of the most important researchers on radiationless motions was G.A. Schott, who in 1933 produced a beautiful and amazing theoretical result* which we discuss in this post.
We’ve mentioned Schott before, in our discussion of failed atomic models in the period 1903-1913. Schott had volunteered his own speculations on the origins of atomic structure, but they were discarded in the wake of Bohr’s successful explanation. Schott, however, was evidently not convinced. He remained what we might call one of the last respectable ‘anti-quantum’ scientists, and he continued to seek a stable model for the atom which could be explained purely through classical mechanics and Maxwell’s electromagnetic theory.
I’ve quoted Schott’s obituary before, but it’s worth doing so again, as it is directly relevant to his invisibility work (Obituary Notices of Fellows of the Royal Society, Vol. 2, No. 7, (Jan., 1939), pp. 451-454):
The first serious attack on the classical theory was made by Bohr in 1913. Although Planck’s constant appeared in the theory of light in 1900, Bohr was the first to use it to give an electrical explanation of line spectra. There were two violent departures from the classical theory. Firstly, the existence of radiationless electronic orbits, secondly the expression for the change in energy. Schott devoted the most of the remainder of his life attempting to fit both those in the classical theory… The mathematical difficulties are enormous, and the skill showed in getting numerical results shows Schott’s mastery at its highest. It might be called the supreme attack of a heroic defender before his death. Defeated? Who shall say? I like to think that in future years the work of Schott will always be consulted for inspiration to tackle the difficulties which come across the path of all theories.
So what did Schott achieve? He demonstrated mathematically that a uniformly charged spherical shell of radius $a$, undergoing periodic motion of period $T$, will not produce radiation if the following condition is satisfied:
$a = m c T/2$,
where $m$ is an integer (1,2,3,…), and $c$ is the speed of light. For those unfamiliar, the period $T$ refers to the amount of time it takes for the sphere to finish one full period and return to its starting position. A slowed-down movie of Schott’s sphere in motion might look something like this:
This result is astounding for a number of reasons. First, the result is completely independent of the specific path of motion: the sphere can be moved in a circular path, an elliptical path, a square path, or any other shape, as long as the sphere repeats the pattern in the time $T$. Second, unlike the simple examples Ehrenfest provided in his 1910 paper, Schott’s example has no inherent symmetry which prevents radiation. So why doesn’t the sphere radiate?
The answer is, in short, complete destructive interference. The time-dependent electric field produced by a charge on one point of the sphere is canceled out by contributions to the field from all other charges. The net result is that the accelerating sphere produces only a static electric field, and if we were to put an appropriate negative point charge in the center of the sphere’s orbit, we could negate even that.
Before we get too excited, however, and start planning ‘radiationless synchrotrons’ and other unusual applications of this effect, we should make a number of observations. First, we consider Schott’s condition for radiationless motion. Since $cT$ is the distance light travels during the course of one ‘orbit’ of our electrified sphere, this suggests that
$cT= 2a/m$.
In other words, light cannot travel a distance further than the sphere’s diameter. This in turn suggests that the sphere cannot move in a broad orbit; rather, its motion is constrained to be a small ‘wobble’ or ‘vibration’. In simple illustration,
Schott’s theorem allows for WOBBLE:
but not ORBITS:
Schott himself was aware of this when he wrote his paper, as he wrote, “…if one of our spheres were used as a model of the electron, the radiationless orbits of its centre would be far too small, since they would lie entirely inside the electron…”
Another concern with such a model is radiation reaction, the interaction of a charged object with its own electric field. Schott’s model neglects the reality that his sphere must necessarily pass through its own radiation field, and its is unclear whether such a reaction might wreck the radiationless motion. We will see in future posts that Schott later looked at the radiation reaction of his sphere, and found even more unusual results!
Another concern is the inherently relativistic nature of Schott’s model. For a sphere of any practical size, the period of vibration must be exceedingly fast, and this translates into a high velocity or acceleration of the sphere. It is unclear whether relativistic concerns invalidate Schott’s theorem, though one later paper suggests that they do.
In spite of such concerns, Schott tried to ground his thought experiment in reality with the following ‘concrete example’ of his sphere:
imagine a metal sphere suspended by a fine metal wire in such a manner that it can be earthed or insulated at will. Surround it by a closely fitting insulating coating, e.g. two thin hollow hemispheres of ebonite fitted together, and then place around the whole and concentric with it a larger insulated metal sphere made of two hemispheres with a very small hole through which the suspending wire passes without touching the outer sphere. Now connect the outer metal sphere to one pole of a battery and the inner one momentarily to the other pole, and again insulate the latter sphere. This receives a charge which resides on the ebonite in contact with it according to the theory of Maxwell and a well known experiment of Faraday. Remove the outer metal hemispheres and also the ebonite hemispheres. Joining the latter together we obtain a very nearly uniformly charged insulating sphere, and if the ebonite insulated perfectly, the charge would remain uniform however the sphere moved about as a whole. This charged ebonite sphere is a concrete example, realized approximately, of what is meant in the present paper.
By the time of Schott’s paper in 1933, the quantum theory was firmly established and Schott himself seems to have realized that it was futile to attempt to use his sphere to try and construct a classical model of the atom. However, another unexplained phenomenon had just recently been discovered, in 1930: a new type of radiation which was uncharged and penetrated deeply into matter. In 1932, James Chadwick performed a series of experiments elucidating the nature of this new particle, the neutron, and by 1933 physicists were trying to understand how this uncharged particle, with approximately the mass of the proton, fit into the picture.
Schott suggested that the neutron might be a collection of one or more of his radiationless spheres:
Having now established the principal result of our investigation, we may perhaps be permitted to indulge in a little speculation, and, though models of the atom and its constituents, especially classical ones, are out of fashion, enquire whether such models, constructed out of charged sphere, like the one considered above, may not, after all, be of use in the elucidation of atomic problems… Obviously it does not help to account for Bohr’s radiationless electron orbits, for, if one of our spheres were used as a model of the electron, the radiationless orbits of its centre would be far too small, since they would lie entirely inside the electron, as we saw in the last section.
But this very fact suggests that, if two of our spheres were taken as models of the electron and proton, it might prove possible to use them to construct a permanent model of the neutron, possibly also permanent models of atomic nuclei; for we require radiationless orbits of nuclear dimensions, which can be provided by our spheres, if they are of such dimensions, and have no spin.
Schott’s model of the neutron, like his earlier atomic model, was overwhelmed by experimental reality, and the undeniable existence of additional fundamental forces of nature, namely the strong nuclear force and the weak nuclear force. Schott’s beautiful calculation would continue to inspire a number of physicists through the decades to come to perform their own investigations of radiationless motion, and as we will see in future posts these investigations would eventually lead to investigations of the possibility of truly invisible objects.
* G.A. Schott, “The electromagnetic field of a moving uniformly and rigidly electrified sphere and its radiationless orbits,” Phil. Mag. Vol. 15, Ser. 7 (1933), 752-761.
********************************************************************
A Mathematical Appendix: For those of a mathematical inclination who might be interested in trying to reproduce Schott’s results, I give you the following outline:
1. The sphere is moving periodically, and is of finite size; therefore its charge and current densities can be expanded in terms of a mixed Fourier series in time/Fourier transform in space:
${\bf j}({\bf r},t)=\sum_{n=-\infty}^\infty \int {\bf j}_n({\bf K})e^{i({\bf K}\cdot{\bf r}-\omega_n t)}d^3K$.
2. In the Lorentz gauge, one can express the scalar potential $\phi$ and vector potential ${\bf A}$ in terms of an integral over the retarded forms of the charge and current densities.
3. We may also write the charge and current densities as follows:
$\rho({\bf r},t)=e f({\bf r}-{\bf \xi}(t))$,
${\bf j} ({\bf r},t)=e {\bf \dot{\xi}}f({\bf r}-{\bf \xi}(t))$,
where ${\bf \xi}$ is the vector position of the center of the sphere as a function of time.
4. Determine the radiation fields far away from the region of the sphere, i.e. in the ‘far-zone’. If you do everything correctly, you get a frightfully complicated series form which includes, in each term,
$\tilde{f}(k_n{\bf \hat{r}})$,
the three-dimensional Fourier transform of $f$, evaluated at spatial frequencies $k_n=n 2\pi/T$, and ${\bf \hat{r}}$ is the direction of observation.
5. Since we are assuming that our sphere is an infinitely thin shell of charge, its Fourier transform is a Bessel function with evenly-spaced zeros. Requiring the function to be zero for those special spatial frequencies listed above results in Schott’s condition.
6. No, it isn’t easy!
### Like this:
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### 2 Responses to Invisibility Physics: Schott’s radiationless orbits
1. Brett says:
I very much enjoy these posts. You may be interested in the work of Randell Mills, who uses the nonradiation condition as the basis for the stability of the bound and free electron, and as the cause of superconductivity. I also suggest you add to a wikipedia article on the nonradiation condition which I will be posting soon.
2. yoron says:
Yes, I may not understand it all, but at least I found it all. And now onwards
to part three And may I congratulate you on your clarity of thought.
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http://physics.stackexchange.com/questions/tagged/waves?page=5&sort=votes&pagesize=15
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### Percentage increase in the length of pendulum
I'm struggling with a physics question : What should be the percentage increase in the length of the chord of a pendulum for the period increased by 1%? The answer is 2%. I tried with \$w = 2 \pi/T ...
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http://math.stackexchange.com/questions/174316/uniqueness-of-meets-and-joins-in-posets/174317
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# Uniqueness of meets and joins in posets
Exercise 1.2.8 (Part 2), p.8, from Categories for Types by Roy L. Crole.
Definition: Let $X$ be a preordered set and $A \subseteq X$. A join of $A$, if such exists, is a least element in the set of upper bounds for $A$. A meet of $A$, if such exists, is a greatest element in the set of lower bounds for $A$.
Exercise: Suppose that $X$ is a poset (and thus also a preorder). Show that meets and joins in a poset are unique if they exist.
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I have posted my solution as an answer and would appreciate any feedback. Thanks. – Code-Guru Jul 23 '12 at 17:41
1
Please choose titles that describe the problem, not it's location. – Gone Jul 23 '12 at 18:37
@BillDubuque Is that better? – Code-Guru Jul 23 '12 at 18:39
1
Much better, thanks. – Gone Jul 23 '12 at 19:03
1
– Martin Sleziak Jul 24 '12 at 10:41
show 1 more comment
## 1 Answer
Let $A \subseteq X$ and suppose $x$ and $y$ are both joins of $A$. Then $x$ and $y$ are both upper bounds of $A$. So $A \le x$ and $A \le y$. But since $x$ and $y$ are joins of $A$ and $x, y \in X$, $y \le x$ and $x \le y$. Therefore, $x = y$ because because $\le$ is anti-symmetric. Meets are unique by a similar argument.
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1
Your solution seems fine. Just a remark about notation - I am not sure whether $A\le x$ is frequently used as a shortcut for $(\forall a\in A) a\le x$. (But if it used this way in your course/book, then it is certainly ok for you to use this notation.) – Martin Sleziak Jul 24 '12 at 10:44
@MartinSleziak That is the notation used in the textbook I am reading. When i typed this up, I wondered if it was a standard notation or not. – Code-Guru Jul 24 '12 at 13:25
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http://mathhelpforum.com/trigonometry/58079-tan-cot-terms-p-q.html
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# Thread:
1. ## tan, cot in terms of p,q
Given that sin a/sin b = p and cos a/cos b = q, express tan a and cot a in terms of p and q. (a and b are acute angles). I have no idea where to start on this one. I've tried a few things but always get stuck.
Thanks much.
2. Originally Posted by daydreembelievr
Given that sin a/sin b = p and cos a/cos b = q, express tan a and cot a in terms of p and q. (a and b are acute angles). I have no idea where to start on this one. I've tried a few things but always get stuck.
Thanks much.
Here's a big start:
$\frac{\sin a}{\sin b} = p \Rightarrow \sin b = \frac{\sin a}{p}$ .... (1)
$\frac{\cos a}{\cos b} = q \Rightarrow \cos b = \frac{\cos a}{q}$ .... (2)
Square both equations and add:
$1 = \frac{\sin^2 a}{p^2} + \frac{\cos^2 a}{q^2} \Rightarrow p^2 q^2 = q^2 \sin^2 a + p^2 \cos^2 a$
$\Rightarrow \frac{p^2 q^2 }{\cos^2 a} = q^2 \tan^2 a + p^2$ .... (3)
Substitute the identity $1 + \tan^2 a = \frac{1}{\cos^2 a}$ into equation (3) and solve for $\tan^2 a$ and hence $\tan a$.
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http://physics.stackexchange.com/questions/21654/how-to-find-the-value-of-the-parameter-a-in-this-transfer-function/21655
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# How to find the value of the parameter $a$ in this transfer function?
I am given a transfer function of a second-order system as: $$G(s)=\frac{a}{s^{2}+4s+a}$$
and I need to find the value of the parameter $a$ that will make the damping coefficient $\zeta=.7$. I am not sure how to do this but I might have found something that might have helped so I am going to take a stab at it. I found a transfer function in the book of a second order spring-mass-damper system with an external applied force in the book as: $$G(s)=\frac{a}{m\omega_{n}^{2}}(\frac{\omega_{n}^{2}}{s^{2}+2\zeta \omega_{n}+\omega_{n}^{2}})$$
I was thinking that I could just write $\omega_{n}^{2}=a$ and $2\omega_{n}\zeta=4$. Then I could just solve for $a$. Is this possible?
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Is this really a homework question? It doesn't read like one at all. – David Zaslavsky♦ Mar 1 '12 at 1:35
Yes it's a homework question. I am given a transfer function and then asked to find the value of $a$ that would make $\zeta=.7$ – Greg Harrington Mar 1 '12 at 2:05
I guess you're just really good at writing questions then ;-) – David Zaslavsky♦ Mar 1 '12 at 2:19
Well i gave the question and then gave my solution and asked if it was right. Maybe it didn't look like a question to the common reader. Oops. Do you have any knowledge about this question though? – Greg Harrington Mar 1 '12 at 2:28
No "oops" needed, you certainly haven't done anything wrong here. I guess it's just different enough from the usual "how do I find velocity" type stuff that I wouldn't peg it as a homework question if you hadn't included the tag. I don't have any particular knowledge about this stuff, though. – David Zaslavsky♦ Mar 1 '12 at 2:49
## 1 Answer
You have the answer.
Consider $2*\zeta*\omega_n = 4$. $\zeta = 0.7$. $\omega_n^2 = a$ What value of $\omega_n$ (or $a$ in your case) satisfies this?
It wouldn't be 2.86 because $\omega_{n}=\sqrt{a}$. I got $a=8.163$. But thank you for letting me know that I did it right. I didnt think that I could equate it like that – Greg Harrington Mar 1 '12 at 2:05
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http://mathoverflow.net/questions/28079?sort=votes
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Points on binary hemispheres of the n-sphere
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Let $\mathbb{S}^{n-1}=${$x\in \mathbb{R}^n| \sum_{k=1}^n x_k^2 =1$} be the $n-1$ sphere and $n_i\in\mathbb{R}^n$ with components $n_{ij}\in${$-1,1$}$\ \forall\ j=1,2,\dots,n$. There are obviously $2^n$ such distinct "binary" vectors $n_i$. Now define for every binary vector $n_i$ a Hemisphere $H_i=${$x \in \mathbb{S}^{n-1}\ |\ (x,n_i)\geq0$} where $(.,.)$ means the usual inner product $(x,y)=\sum_{k=1}^n x_k y_k$.
What is the Probability that $p$ uniformly distributed vectors on $\mathbb{S}^{n-1}$ are all on the same of those $H_i$?
EDIT, Will Jagy: evidently these really are hemispheres, Tom Goodwillie pointed out that for $n=3$ they are $$x + y + z \geq 0, \; \; \; x + y -z \geq 0, \; \; \; x - y + z \geq 0, \; \; \; x - y - z \geq 0 ,$$ and their complements. $$-x - y - z \geq 0, \; \; \; -x - y +z \geq 0, \; \; \; -x + y - z \geq 0, \; \; \; -x + y + z \geq 0 .$$
EDIT,problem poser:
Here is a solution for $n=3$:
The problem is much easier, if you see the similarity of the sphere with the great arcs that define the $8$ hemispheres and a cuboctahedron (www.wikipedia.org/wiki/Cuboctahedron).
As you can see the cuboctahedron actually consists only of $2$ distinct kinds of faces, these are $8$ triangles and $6$ squares. It is not difficult to see, that the intersections of the hemispheres consists of only these triangles and squares projected on the sphere and that these spherical triangles and “spherical” squares are the smallest parts you can get by the intersection of some hemispheres (except the empty set of course). Unfortunately the area is not preserved under this projection. But what is preserved is of course the fact that one hemisphere consists of $3$ spherical triangles and $4$ “spherical” squares (Lets call the triangles $T$ and the squares $S$ from now on).
It is also pretty obvious that the intersection of two hemispheres with hamming distance $2$ (that means two hemispheres with binary vectors with only two different components) consists of two $T$ and one $S$. That gives us the two equations $2T+S=I_2$ and $4T+3S=2\pi$ where $I_2$ is the intersection of two hemispheres with hamming-distance $2$.
To calculate $I_2$ we use the fact that the angle between two binary vectors of hamming distance $2$ is given by $\varphi_2=arccos−\frac{1}{3}$ and therefor $I_2=\frac{\pi-\varphi_2}{2\pi}4\pi=2(\pi-\varphi_2)$.
Now we have from the equations above $$T=\frac{3I_2-2\pi}{2}=2\pi-3\varphi_2$$ and $$S=2(\pi-I_2))=4\varphi_2-2\pi.$$Now, that we know the smallest parts $T$ and $S$ we have to understand what kinds of intersections we have and how they consist of $T$ and $S$. To do that we only need to care wlog about the hemisphere $H_1$ with binary vector $n_1=(1,1,1)^T$. Lets call $I_{ijk}$ the intersection of $H_1$ and hemispheres with hamming distance $i,j,k$(relatively to $n_1$). It is obvious, that there cant be more than intersections of $4$ hemispheres. (If there would be more, there would be a pair of disjoint hemispheres with vanishing intersection.) So there are the non vanishing intersections $I_1,I_2,I_{11},I_{12},I_{22},I_{111},I_{211},I_{221}$ (note that i mean here that you can find hemispheres with hamming distances $i,j,k$ so that $I_{ijk}$ doesnt vanish. For $n=3$ I think it is obvious, that in this case the value of the intersection is either $0$ or $I_{ijk}$ depending on which hemispheres of the given hamming distances you chose.). As you can easily check we have $I_{221}=I_{111}=I_{22}=T$, $I_{211}=S$, $I_{12}=I_{11}=S+T$, $I_2=2T+S$ and $I_1=2T+2S$. I think everyone sees now, that we just have to use the inclusion exclusion principle to get the desired probability $P(3,p)$. I won't do this now because i think that this doesnt give more insights.
The big question that arises now is, whether there are similarities in higher dimensions. If we could find the smallest parts of intersections on $\mathbb{S}^3$ without too much effort (like having to integrate) we could use a similar strategy I used for $n=3$. Then, if we could find some further symmetries, we may even be able to get some kind of recursion and solve it for arbitrary dimensions.
EDIT, problem poser:
Here is a little summary of the progress I made for n>3:
My aim is to find the smallest pieces of intersections of the hemispheres. If you look at the case n=3, you recognize that one of the smallest pieces is a spherical triangle with equal angles (the equal angles come from the symmetry of the problem). The good thing is, as Will Jagy pointed out, that you can get the area by the method Marco Radeschi describes in his comment of http://mathoverflow.net/questions/8846/proofs-without-words . Even more great is the fact, that we can always apply this method if we have a set of hemispheres, that all have the same distance to each other (that means that all have binary vectors with same hamming distance to each other) and for that the intersections of all these hemispheres do not vanish. For example for n=3 the intersection of the hemispheres with binary vectors (1,1,1), (1,-1,-1) and (-1,1,-1) is such a spherical triangle mentioned above. If you add the hemisphere (-1,-1,1) now to that set, all binary vectors still have the same distance, but the intersection of all four vanishes. That shows that the vanishing property is important to apply the method.
Now it seems logical to search for such a set for n=4 and to hope that the intersection (that you could easily calculate by the method mentioned above) is a smallest piece. What really concerns me is the fact that there is no such set for n=4. For example a set with the maximum number of hemispheres that all have hamming distance 2 to each other and for that the intersection of all hemispheres does not vanish is {(1,1,1,1), (1,-1,-1,1), (1,1,-1,-1), (1,-1,1,-1)}. (It is easy to prove this, because there are only 3 possible hemispheres left and each of them is the negative one of a hemisphere in the set. ) Now you can add for example the hemisphere (1,-1,-1,-1) and you will still have a non vanishing intersection of all hemispheres. This proves that for n=4 the intersection of hemispheres that all have hamming distance 2 to each other and that have a non vanishing intersection is not a smallest piece. The same argumentation proves that there also can't be such sets with hamming distance 1 and 3.
But I also found a positive result: For all dimensions the intersection of hemispheres with the same given component, is a smallest piece.(To avoid confusion I mean hemispheres that for example all have the same first component) For example for n=3 these intersections are the spherical squares. The problem for n>3 is to find the volume of such intersection.
Any ideas?
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This looks like a basic probability homework problem. – Victor Protsak Jun 14 2010 at 7:53
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No Victor Protsak, i dont think so. I have already spent a lot of time on that problem, and the only thing i recognized was that it is extremely complicated and unsolvable if you cant find any symmetry. The experts to whom i talked to adviced me not to waste more time on this because they are sceptical about the existence of a solution (that means an explicit formula f(n,p) ) and told me that the solution will probably be a byproduct of a bigger theory. – unknown Jun 14 2010 at 8:54
From the geometric point of view using the inclusion-exclusion principle, you will have to find the area of all kinds of intersections of the hemispheres. For n=3 this is not really a problem because of the gauss-bonnet theorem. Is there a generalized theorem so that i can easily get these intersections in higher dimensions? I thought about maybe finding a recursion from the results i have of small dimensions n<4, but this seems nearly impossible, because i cant find any usefull symmetry. Other approaches lead to a huge amount of distinction of cases you cant control without symmetry. – unknown Jun 14 2010 at 9:00
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The following question is perhaps simpler and gives at least an upper bound: What is the probability that $p$ uniformly distributed points of $\mathcal S^{d−1}$ belong to some hemisphere (given by the intersection of $\mathcal S^{d−1}$ with some half-space? Since hemispheres are in bijection with elements of $\mathcal S^{d−1}$ there exists an expectancy $A$ of the area corresponding to all hemispheres which work. A first order approximation is then given by $2^{-p}A/v(\mathcal S^{d-1})$ with $v(\mathcal S^{d-1})$ denoting the volume of $\mathcal S^{d-1}$. – Roland Bacher Jun 14 2010 at 16:56
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Hi there. I have been fiddling with it. I have become convinced that there is no "closed form" way to evaluate the volume of a tetrahedron in $\mathbb{S}^3 ,$ even one with all "dihedral" angles equal. See pages 141-142 of "Regular Polytopes" by H. S. M. Coxeter, where instead he does a tetrahedron with three right angles and three others. Here he refers to work of Schlafli, and later in English by Richmond in 1903 "The volume of a tetrahedron in elliptic space" and himself in 1935 "The functions of Schlafli and Lobatschefsky." In particular, I finally realized why my easiest trick fails. – Will Jagy Jun 21 2010 at 19:14
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1 Answer
A good way to think about a uniformely random point on the sphere is to think of it as the normalization of a vector of normally distributed i.i.d random variables. In fact, you don't even need to normalize. Consider hemisphere $v_i$ and $v_j$, form the $2 \times N$ matrix $V$, $v_i$ and $v_j$ being the rows of $V$. Let $X$ be a random Gaussian vector with covariance matrix $I$ (identity). $V.X$ is a vector with two components, whose sign represent membership in the hemisphere $i$ and $j$... $V.X$ follows a multivariate normal distribution with mean $\left(\begin{array}{cc}0\\0\end{array}\right)$ and covariance matrix $n \left[\begin{array}{cc}1 & 1-2h/n\\1-2h/n & 1\end{array}\right]$ where $h$ is the hamming distance. Notice that any covariance matrix can be approximated, up to a multiplicative factor, for $n$ large enough. The measure of the intersection of the two hemispheres is the integral of the the probability density over an orthant.
For the intersection of two hemispheres, the orthant is a quartant and the intersection is given as
$$\frac{1}{4} + \frac{\sin^{-1}(1-2h/n)}{2\pi}$$
(note that this is not the area but the fraction of the $(n-1)$ sphere area that is in the intersection)
There is also a formula for the trivariate case, i.e. if you're considering the intersection of three hemispheres
$$\frac{1}{8} + \frac{1}{4\pi}( \sin^{-1}(1-2h_{12}/n) + \sin^{-1}(1-2h_{23}/n) + \sin^{-1}(1-2h_{13}/n) )$$
Unfortunately, it is known that there is no closed-form expression of the probability of the orthants of a general multivariate normal distribution above the trivariate case. In your case, the covariance matrix have a certain structure, so it is not definite evidence that there is no formula for the area of the intersection of more than three hemispheres, but it is pretty good evidence.
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http://physics.stackexchange.com/questions/51852/if-hydrogen-and-helium-are-lighter-than-air-why-wont-liquid-hydrogen-and-liqui/51854
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# If hydrogen and helium are lighter than air, why won't liquid hydrogen and liquid helium defy gravity?
Title says it all. If hydrogen and helium are lighter than air, why won't liquid hydrogen and liquid helium defy gravity?
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## 3 Answers
Gaseous hydrogen and helium are lighter than air. Hydrogen, helium and air are close approximations to ideal gases, and for an ideal gas the volume of one mole of gas is about 22.4 litres. That means the density of an ideal gas is inversely proportional to its molecular weight, so hydrogen (Mw = 2) and helium (Mw = 4) are lighter than air (average Mw = 28.8).
However you're asking about liquid hydrogen and helium, and liquids are much denser than gases because the molecules are much more tightly packed. For example the density of liquid hydrogen is around 68 kg/m$^3$ compared to air at about 1.3 kg/m$^3$. That's why liquid hydrogen doesn't float in air.
Incidentally, the density of liquid nitrogen (a close approximation to liquid air) is about 800 kg/m$^3$ so liquid hydrogen would float on liquid air.
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Because liquid hydrogen and liquid helium are much heaver than air.
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please explain why. – think123 Jan 22 at 7:22
Well, this is an experimental fact - it is easy to look up their densities and the density of air. Or do you want some theoretical explanation why liquids typically have higher densities than gases? – akhmeteli Jan 22 at 7:28
First of all, no mass 'defies' gravity. Second, the normal state of these two is air so they cannot be 'lighter than air' which would in effect be lighter than themselves. If you were meaning to ask why the liquid forms of these elements won't float, the answer is simply because liquids don't float. This is one of the properties that defines a gas. The molecules must overcome the IMF's and have a high enough KE but liquids do not. The density of the liquid would have to be lower than that of the gas and no liquid has a lower density than a gas. Although hydrogen has a low gaseous density, it's liquid density is still considerably greater than any gas.
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Actually, hydrogen and helium are lighter than air. Air is made up of other gases as well, and H and He will sit at the top of a room filled with air. – Kitchi Feb 24 at 13:25
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http://math.stackexchange.com/questions/115917/represent-mathbbr3-as-an-union-of-disjoint-circles-using-stereographic-pro
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# Represent $\mathbb{R}^3$ as an union of disjoint circles using stereographic projection
I have begun to learn complex analysis and have solved a few problems on stereographic projection and Riemann sphere but can't solve the problem in the subject.
Could you help please?
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"few" means "not many", "a few" means "several" -- the "but" seems to imply that you meant "a few", since "few" doesn't form a contrast to not being able to solve a problem. – joriki Mar 3 '12 at 11:52
@joriki thanks, english is not my native language. – Sergey Filkin Mar 3 '12 at 13:24
Maybe you could see the stereographic projection as a diffeo from the 3-sphere (minus a point) to $\mathbb{R}^3$, and then transport the Hopf fibration to $\mathbb{R}^3$. That still leaves one manifold homeomorphic to $\mathbb{R}$ and not to $\mathbb{S}_1$, because of the point we eliminated at the beginning. – D. Thomine Mar 3 '12 at 14:02
I've found a way to represent $\mathbb{R}^3$ as a disjoint union of manifolds diffeomorphic to $\mathbb{S}_1$, but it is not very elegant and it makes no use of the stereographic projection. Are you interested? – D. Thomine Mar 7 '12 at 15:30
@D.Thomine sure! – Sergey Filkin Mar 8 '12 at 1:32
## 1 Answer
Since you said you're interested in D. Thomine's answer (and s/he seems to have forgotten about it), here's an answer using no complex analysis or stereographic projection. We will use transfinite induction. We will show $\mathbb{R}^3$ can be partitioned into circles of radius $1$. This idea actually given as an exercise in Ciesielski's book Set Theory for the Working Mathematician. (In the same book, he proves that $\mathbb{R}^2$ can't be partitioned as a union of circles of positive radius.)
To that end, choose a bijection with $\mathbb{R}^3$ and $\mathfrak{c}$, the cardinality of the continuum. (Equivalently, well order $\mathbb{R}^3$ with its minimal well ordering.) The important thing about $\mathfrak{c}$ is that it is a cardinal number, meaning for any ordinal number $\alpha < \mathfrak{c}$, we have $|\alpha| < |\mathfrak{c}| = \mathfrak{c}$ where $|\cdot |$ denotes cardinality.
To begin the induction, let $r_0$ denote the "first" real number. Choose any circle $C_0$ of unit radius which goes through $r_0$.
Now, assume inductively that for all $\beta < \alpha$, we have chosen pairwise disjoint circles of radius $1$ so that all of the $r_\beta$ lie on 1. We wish to extend the induction to $r_\alpha$.
First, if $r_\alpha$ already lies on a prechosen circle, we are done. So, we may assume $r_\alpha$ does not lie on any previously chosen circle of radius 1.
Now, consider all the planes in $\mathbb{R}^3$ passing through $r_\alpha$. It is easy to see that there are $\mathfrak{c}$ such planes. Since, at this point, we have chosen at most $|\alpha|$ circles and $|\alpha|< \mathfrak{c}$, and each circle lies in one plane, there must be a plane $P_\alpha$ which doesn't contain any of our previously chosen circles (though, of course, they may intersect it in $1$ or $2$ points).
Let's just focus on $P_\alpha$ for now. Consider all the circles of radius $1$ contained in $P_\alpha$ passing through $r_\alpha$, which I'll call candidate circles. Again, a not-too-hard counting argument shows there are $\mathfrak{c}$ candidate circles. Each of our previously chosen circles intersects $P_\alpha$ in at most 2 points so there are at most $2|\alpha| = |\alpha| < \mathfrak{c}$ "bad" points in $P_\alpha$ which we must avoid. Any point in $P_\alpha$ is on at most 2 candidate circles, so all "bad" points remove at most $2|\alpha| < \mathfrak{c}$ candidate circles from consideration. But since there are $\mathfrak{c}$ candidate circles, there must be at least one left over. Let $C_\alpha$ be one of these left overs candidates.
By "candidateness", $r_\alpha \in C_\alpha$. Further, $C_\alpha$ cannot intersect any of our previously chosen circles because those intersection points would correspond to "bad" points, which we avoided. Thus, we have continued the induction.
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http://en.wikipedia.org/wiki/Magnetic_flux
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# Magnetic flux
Electromagnetism
• Magnetization
• Magnetic flux
Scientists
In physics, specifically electromagnetism, the magnetic flux (often denoted Φ or ΦB) through a surface is the component of the magnetic B field passing through that surface. The SI unit of magnetic flux is the weber (Wb) (in derived units: volt-seconds), and the CGS unit is the maxwell. Magnetic flux is usually measured with a fluxmeter, which contains measuring coils and electronics, that evaluates the change of voltage in the measuring coils to calculate the magnetic flux.
## Description
The magnetic flux through a surface when the magnetic field is variable relies on splitting the surface into small surface elements, over which the magnetic field can be considered to be locally constant. The total flux is then a formal summation of these surface elements (see surface integration).
Each point on a surface is associated with a direction, called the surface normal; the magnetic flux through a point is then the component of the magnetic field along this direction.
The magnetic interaction is described in terms of a vector field, where each point in space (and time) is associated with a vector that determines what force a moving charge would experience at that point (see Lorentz force). Since a vector field is quite difficult to visualize at first, in elementary physics one may instead visualize this field with field lines. The magnetic flux through some surface, in this simplified picture, is proportional to the number of field lines passing through that surface (in some contexts, the flux may be defined to be precisely the number of field lines passing through that surface; although technically misleading, this distinction is not important). Note that the magnetic flux is the net number of field lines passing through that surface; that is, the number passing through in one direction minus the number passing through in the other direction (see below for deciding in which direction the field lines carry a positive sign and in which they carry a negative sign). In more advanced physics, the field line analogy is dropped and the magnetic flux is properly defined as the component of the magnetic field passing through a surface. If the magnetic field is constant, the magnetic flux passing through a surface of vector area S is
$\Phi_B = \mathbf{B} \cdot \mathbf{S} = BS \cos \theta,$
where B is the magnitude of the magnetic field (the magnetic flux density) having the unit of Wb/m2 (tesla), S is the area of the surface, and θ is the angle between the magnetic field lines and the normal (perpendicular) to S. For a varying magnetic field, we first consider the magnetic flux through an infinitesimal area element dS, where we may consider the field to be constant:
$d\Phi_B = \mathbf{B} \cdot d\mathbf{S}.$
A generic surface, S, can then be broken into infinitesimal elements and the total magnetic flux through the surface is then the surface integral
$\Phi_B = \iint_S \mathbf{B} \cdot d\mathbf S.$
From the definition of the magnetic vector potential A and the fundamental theorem of the curl the magnetic flux may also be defined as:
$\Phi_B = \oint_{\partial S} \mathbf{A} \cdot d\boldsymbol{\ell},$
where the line integral is taken over the boundary of the surface S, which is denoted ∂S.
## Magnetic flux through a closed surface
Some examples of closed surfaces (left) and open surfaces (right). Left: Surface of a sphere, surface of a torus, surface of a cube. Right: Disk surface, square surface, surface of a hemisphere. (The surface is blue, the boundary is red.)
Main article: Gauss's law for magnetism
Gauss's law for magnetism, which is one of the four Maxwell's equations, states that the total magnetic flux through a closed surface is equal to zero. (A "closed surface" is a surface that completely encloses a volume(s) with no holes.) This law is a consequence of the empirical observation that magnetic monopoles have never been found.
In other words, Gauss's law for magnetism is the statement:
$\Phi_B=\,\!$$\scriptstyle S$$\mathbf{B} \cdot d\mathbf S = 0$
for any closed surface S.
## Magnetic flux through an open surface
Main article: Faraday's law of induction
For an open surface Σ, the electromotive force along the surface boundary, ∂Σ, is a combination of the boundary's motion, with velocity v, through a magnetic field B (illustrated by the generic F field in the diagram) and the induced electric field caused by the changing magnetic field.
While the magnetic flux through a closed surface is always zero, the magnetic flux through an open surface need not be zero and is an important quantity in electromagnetism. For example, a change in the magnetic flux passing through a loop of conductive wire will cause an electromotive force, and therefore an electric current, in the loop. The relationship is given by Faraday's law:
$\mathcal{E} = \oint_{\partial \Sigma}\left( \mathbf{E} +\mathbf{ v \times B}\right) \cdot d\boldsymbol{\ell} = -{d\Phi_B \over dt},$
where
$\mathcal{E}$ is the electromotive force (EMF),
ΦB is the magnetic flux through the open surface Σ,
∂Σ is the boundary of the open surface Σ; note that the surface, in general, may be in motion and deforming, and so is generally a function of time. The electromotive force is induced along this boundary.
dℓ is an infinitesimal vector element of the contour ∂Σ,
v is the velocity of the boundary ∂Σ,
E is the electric field,
B is the magnetic field.
The two equations for the EMF are, firstly, the work per unit charge done against the Lorentz force in moving a test charge around the (possibly moving) surface boundary ∂Σ and, secondly, as the change of magnetic flux through the open surface Σ. This equation is the principle behind an electrical generator.
Area defined by an electric coil with three turns.
## Comparison with electric flux
Main articles: Electric flux and Gauss's law
By way of contrast, Gauss's law for electric fields, another of Maxwell's equations, is
$\Phi_E =\,\!$$\scriptstyle S$$\mathbf{E}\cdot d\mathbf{S} = \frac{Q}{\epsilon_0}\,\!$
where
E is the electric field,
S is any closed surface,
Q is the total electric charge inside the surface S,
ε0 is the electric constant (a universal constant, also called the "permittivity of free space").
Note that the flux of E through a closed surface is not always zero; this indicates the presence of "electric monopoles", that is, free positive or negative charges.
Magnetic Circuits
Conventional Magnetic Circuits
• Magnetomotive force $\mathcal F$
• Magnetic flux $\Phi$
• Magnetic reluctance $\mathcal R$
Phasor Magnetic Circuits
• Complex reluctance $Z_\mu$
Related Concepts
• Magnetic permeability $\mu$
Gyrator-capacitor model variables
• Magnetic impedance $z_M$
• Effective resistance $r_M$
• Magnetic inductivity $L_M$
• Magnetic capacitivity $C_M$
This box:
## See also
• Magnetic field
• Flux
• Maxwell's equations (sometimes called the Maxwell equations) are the set of four equations, attributed to James Clerk Maxwell, that describe the behavior of both the electric and magnetic fields, as well as their interactions with matter.
• Gauss's law gives the relation between the electric flux flowing out a closed surface and the electric charge enclosed in the surface.
• Magnetic circuit is a method using an analogy with electric circuits to calculate the flux of complex systems of magnetic components.
• Magnetic monopole is a hypothetical particle that may loosely be described as "a magnet with only one pole".
• Magnetic flux quantum is the quantum of magnetic flux passing through a superconductor.
• Carl Friedrich Gauss developed a fruitful collaboration with the physics professor Wilhelm Weber; it led to new knowledge in the field of magnetism.
• James Clerk Maxwell demonstrated that electric and magnetic forces are two complementary aspects of electromagnetism.
## External articles
Patents
• Vicci, , Magnetic-flux conduits
• Magnetic Flux through a Loop of Wire by Ernest Lee, Wolfram Demonstrations Project.
• Conversion Magnetic flux Φ in nWb per meter track width to flux level in dB - Tape Operating Levels and Tape Alignment Levels
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http://mathoverflow.net/questions/9311/cubic-spline-of-a-two-variable-function
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## Cubic spline of a two-variable function
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
So, I am aware of how to (both iteratively and using a linear equation) compute the cubic spline of a one-variable function with $m$ control points. However, I am not sure how to do any type of spline on a two-variable function with $mn$ control points in a square grid. Is this even possible? Is there a simple algorithm? I've tried to see if the process is separable, but so far I can't seem to prove it one way or the other.
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Carl de Boor has elaborated on the theory of multivariable splines. See here: pages.cs.wisc.edu/~deboor/ftpreadme.html – Steve Huntsman Dec 18 2009 at 21:04
## 1 Answer
There is a nice book by Les A. Piegl and Wayne Tiller, which is intended as an introduction to NURBS, but in the first chapters also provides theoretical background for uni- and bivariate Beziers und Splines.
The NURBS Book Les A. Piegl, Wayne Tiller Springer ISBN-10: 3540615458 ISBN-13: 978-3540615453
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Thanks, will see if I can get my hands on it. – rlbond Dec 26 2009 at 1:49
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http://crypto.stackexchange.com/questions/5279/carmichael-number-factoring?answertab=votes
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# Carmichael number factoring
Unsure whether this is the right forum for this question, worth a try.
The task im faced with is to implement a poly-time algorithm that finds a nontrivial factor of a carmichael number. Many resources on the web states that this is easy, however, without further explanation why that is?
Furthermore, since miller-rabbin exits when a nontrivial square root of 1 is found, this can be used to find a factor to the carmichael number: $x^2 \equiv 1 = (x+1)(x-1)\equiv0\ mod\ N$, where N is the carmichael number we want to factor. Hence factors must be found using $\gcd(x+1,N)$ and $\gcd(x-1, N)$, correct?
Due to problems with strong liars, in some cases we will miss out on factors, is this a major problem? Since miller-rabbin tests only passes composites with a probability 1/4 is it correct to say that the chances of finding a factor is > 0.5?
Kind regards!
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– mikeazo♦ Nov 6 '12 at 14:28
Thanks, ill have a look at it:) – Nyfiken Nov 6 '12 at 14:37
I think that you have the right sketch: find a witness to the Miller-Rabbin test; since that's a liar to the Fermat test, you can efficiently exhibit a non-trivial factor of $N$ as $\gcd(x-1,N)$ or $\gcd(x+1,N)$, where $x$ is the butlast result in the Miller-Rabbin test. I see no reason why odds of success would not be $\ge3/4$ for each random base tested. – fgrieu Nov 6 '12 at 15:57
@fgrieu: Thank you for answering. Could you perhaps elaborate a bit more on why you think the success rate should be >= 3/4 – Nyfiken Nov 6 '12 at 16:19
## 1 Answer
Yes, using Miller-Rabin with a random witness does give a practical factoring method. When you run the Miller-Rabin algorithm, it can end in one of three ways:
1. The final value is not 1; this case causes Miller-Rabin to output "Composite"
2. An intermediate value was not 1 or N-1, but the next value was 1; this causes Miller-Rabin to output "Composite"
3. The initial value was 1; or an intermediate value was N-1; this causes Miller-Rabin to output "Probably Prime".
Case 1 will happen only if $W^{N-1} \neq 1 \bmod N$, however, that we get that inequality for Carmichael numbers only if $W$ is a multiple of a factor of $N$; in this case, $gcd(W, N)$ gives us a nontrivial factor.
For Case 2, well, that gives us a nontrivial value $X$ for which $X^2 = 1 \bmod N$; as you correctly point out, a nontrivial square root of $N$ immediately gives us the nontrivial factors $gcd(X+1, N), gcd(X-1, N)$
Hence, in the specific case of Carmichael numbers, if the Miller Rabin test outputs "Composite", we always have enough information to immediately factor. And, when Miller-Rabin is given a composite number, then it will output "Composite" with probability > 3/4; hence, a single iteration will allow us to factor with high probability.
In fact, for Carmichael numbers, the probability of success with Miller Rabin is actually > 7/8; Carmichael numbers are not the worse case for Miller Rabin.
This gives us a practical factorization method; however for pedantic reasons, it doesn't actually answer the question. It's not a poly-time algorithm; it is a probabilistic poly-time algorithm which, per iteration, has a good probability of yielding a factor, but it also has a probability of not finding a factor.
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http://physics.stackexchange.com/questions/7922/in-quantum-mechanics-why-do-the-probabilities-of-the-possible-outcomes-of-a-mea/7968
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# In quantum mechanics, why do the probabilities of the possible outcomes of a measurement add up to 1?
The question assumes the standard formalism with projector-valued measures rather than POVMs. Suppose a measurement has two possible outcomes, and the corresponding probabilities are greater than 0 and less than 1. Neither outcome is therefore certain. Then why is it certain that either outcome is obtained (as it seems, if the probabilities add up to 1)?
Added after four answers: All the answers provided so far elaborate on the comment by @Vladimir: "It is not a 'quantum mechanical' feature but a consequence of probability definition." @Lubos and @Mark cast the question into a quantum-mechanical form, e.g., why do the absolute squares of the amplitudes associated with the possible outcomes of a measurement add up to 1? They also explain why the sum remains equal to 1. (However, for a decaying particle the probability of finding it decreases, while the probability of finding its decay products increases. So the "conservation of probability" has something to do with the proper conservation laws.) @David makes it clearest why these answers are insufficient.
Keep in mind that no actual measurement is perfect. While theorists may ignore this, experimenters know well enough that in many runs of a given experiment no outcome is obtained. (The efficiency of many real-world detectors is rather low.) This means that in order to make the probabilities add up to 1, one discards (does not consider) all those experiments in which no outcome is obtained.
So let me follow up with another question.
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7
It is not a "quantum mechanical" feature but a consequence of probability definition. – Vladimir Kalitvianski Apr 2 '11 at 9:12
2
@Koantum Since you have a new question - or have clarified the old one to the extent it appears new - you might want to start a new thread so that the answers stay relevant to the question at hand. – Mark Eichenlaub Apr 3 '11 at 3:15
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What could you possibly mean by "no outcome"? The experimenter is transported backwards in time and the experiment never happened? I think this is what Vladimir was alluding to above. At any rate try to define "no outcome" – sigoldberg1 Apr 3 '11 at 4:00
@sigoldberg1, a successful measurement has an outcome. An attempted measurement may or may not be successful. An unsuccessful measurement has no outcome. – Koantum Apr 3 '11 at 4:58
@Mark, as you suggested, I have posted the follow-up question separately. – Koantum Apr 3 '11 at 5:08
show 1 more comment
## 6 Answers
Suppose we have a quantum state that is normalized at some time. Then it will remain normalized if the Hamiltonian is Hermitian.
$$\frac{\partial}{\partial t}\langle \Psi \mid \Psi \rangle = \left(\frac{\partial}{\partial t} \langle \Psi \right) \mid \Psi \rangle + \langle \Psi \mid \left(\frac{\partial}{\partial t} \Psi \rangle \right)$$
Schrodinger's equations says
$$\frac{\partial}{\partial t}\mid \Psi \rangle = \frac{-i}{h} H \mid \Psi \rangle$$
and
$$\frac{\partial}{\partial t} \langle \Psi \mid = \frac{i}{h} \langle H \Psi \mid = \frac{i}{h} \langle \Psi \mid H^*$$
Substituting in gives
$$\frac{\partial}{\partial t}\langle \Psi \mid \Psi \rangle = \frac{i}{h}\left( -\langle \Psi \mid H \Psi \rangle + \langle \Psi \mid H^* \Psi \rangle\right)$$
As long as $H = H^*$, this is zero. The initial normalization is a postulate, as David said.
This answer is pretty much an explicit way of saying what Lubos wrote - I was half way done when he posted, so check there for more detail.
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Thanks, Mark, +1 for the relevant math supplement. ;-) – Luboš Motl Apr 2 '11 at 7:47
Suppose a measurement has two possible outcomes
That's why. One of the possible outcomes has to occur - it's the definition of "possible outcomes." Probability theory takes this as an axiom, the axiom of unit measure according to Wikipedia.
Note that we often have to normalize the probabilities so that they add up to 1.
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The probabilities of individual outcomes are given by the squared absolute values of the complex probability amplitudes $a_i$ associated with the individual outcomes. Their sum $${\rm Total\,\,probability} = \sum_{i} |a_i|^2$$ is therefore nothing else than a formula for the squared length of the state vector $|\psi\rangle$: note that it is a complexified version of the Pythagorean theorem.
In quantum mechanics, if the state vector (wave function) has the length equal to one at the beginning, it will have the length equal to one at all times - because of the so-called "unitarity". Unitarity means that the evolution according to Schrödinger's equation is essentially just a rotation around some "axes" in the Hilbert space - an element of $U(N)$ or $U(\infty)$, a complex and/or infinite-dimensional generalization of $O(N)$.
Unitarity means that the evolution of $|\psi\rangle$ - by Schrödinger's equation - preserves the length of the vector, and the squared length of the vector is nothing else than the sum of probabilities of all mutually exclusive outcomes of experiments, regardless of the orthogonal basis of the Hilbert space that we choose.
The unitarity condition is $U U^\dagger={\bf 1}$ for the evolution operators $U$ and may be reduced to $H=H^\dagger$, the hermiticity of the Hamiltonian, whenever the evolution is encoded in a Hamiltonian via $U=\exp(Ht/i\hbar)$, as Mark's answer shows explicitly.
It is important that in quantum mechanics, we don't have to normalize the probabilities in a way that would depend on the evolution: the preservation of the total probability is guaranteed by the equations of motion, namely by the hermiticity of the Hamiltonian that enters these equations of motion.
If an ad hoc normalization were required, quantum mechanics would be spoiled by a new source of non-locality and the "wave function collapse" would become observable. Both of these problems would lead to inconsistencies with the observations - and, from a theorist's viewpoint, internal logical inconsistencies in the theory.
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An outcome is anything that can happen. You might not want to fail to detect a particle, but if you do, it's an outcome. It still causes quantum decoherence (and waveform collapse if you believe the Copenhagen interpretation), in that there will not be interference between waves that are there if the particle is detected and waves that are there if it isn't.
Also, it's entirely possible to predict the probability of not detecting the particle when it's there, and accidentally detecting the particle when it isn't. In case you're wondering, the decoherence is essentially from the detector. There will be interference between part of a waveform that missed the detector and part that just didn't manage to set it off.
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@user2898, one needs to distinguish two cases. (i) If a 100% efficient detector doesn't detect a particle, there is no particle. (ii) If a real-world detector (which never is 100% efficient) fails to detect a particle, there may or may not be a particle. You could mock up a 100% efficient detector by ignoring all instances in which it failed to work, but this would beg the question. – Koantum Apr 3 '11 at 5:17
Lets look at simple probability theory.
Suppose you toss a coin $N$ times. And you get heads $H$ times and tails $T$ times. When you don't get heads you get tails.
This means $T = N - H$.
Now, probability of getting heads $P_h = \frac{H}{N}$ (by definition)
and probability of getting tails $P_t = \frac{T}{N} = \frac{N-H}{N}$
$P_h + P_t = \frac{H}{N} + \frac{N-H}{N} = \frac{H + N - H}{N} = \frac{N}{N} = 1$
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What operationally is "no outcome"? As mentioned above this has nothing to to with quantum mechanics, since every quantum experiment must have a classical reading (Bohr's dictum). Perhaps this is the part of which you are unaware.
I have thought about it myself since yesterday. One definition of the absence of outcome is "no stability" or reproducibility. Something like, you think you see blue. You blink or wish or something and the outcome changes, you see green. Blink again and it changes again. Another possibility might be that the experimental apparatus is destroyed.
Also, merely apparent measurements of unphysical quantities may have no outcome, like trying to measure our absolute x,y,z coordinates in the universe, but even this is not operationally so well defined. Some purely statistical experiments like telepathy appear to have no outcome, as they change on reinterpretation.
Do you have other operational possibilities in mind for "no outcome"? Please focus on the meaning of "operational". This was the giant advance in 20th century physics.
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Perhaps it is best to focus on a position measurement using an array of detectors. To simplify further, just two detectors and a state assigning to them a total probability of 1. Since no detector is perfect, no every one of these measurements will have an outcome. All Bohr would require in this case that the unsuccessful measurements be discarded or ignored, if we want to compare the predicted probabilities with the measured ones. – Koantum Apr 4 '11 at 1:42
Do this first with ants, in a Y shaped maze with cameras as detectors. There is the possibility that the ant a) "gets reflected", i.e. turns around, or b) "gets absorbed" in the apparatus, i.e. stops or gets eaten by a spider. To you both of these are "no outcome" results. Similarly, for atoms in a Stern-Gerlach apparatus, there can be poor beam focusing, or poor vacuum, etc. However, to me, in both these circumstances, there was indeed an outcome, namely that neither detector registers an ant or an atom (in the delta t assumed). Another outcome is that both detectors could register. – sigoldberg1 Apr 4 '11 at 2:30
Sorry, ran out of space. My basic criticism is that there is way too much theory in your definition of the "outcome" of an experiment. As an experimentalist, to me, the outcome is whatever comes out, which can more or less be anything. Your question has to do more with the rescalings of probabilities which occur all the time in real experiments, as you indicated. So to me you want to know whether there could be theories which always predict missing probability of detection, like missing energy or spin before neutrinos were discovered? But wouldn't we just call that particle decay? – sigoldberg1 Apr 4 '11 at 2:43
the outcomes isn't whatever "comes out." It's what is represented by a projector. My question is not: does QM "always predict missing probability of detection" but: can QM predict with certainty that a measurement will have one of its possible outcomes (which are represented by orthogonal projectors on an HS)? – Koantum Apr 5 '11 at 1:56
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http://mathoverflow.net/questions/2293?sort=newest
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## Elliptic Curves, Lattices, Lie Algebras
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I've recently started to look at elliptic curves and have three basic questions:
1. Is it correct to say that elliptic curves $E$ in the projective plane are in bijective correspondence with lattices $L$ in the complex plane via $E$ <--> $C/L$.
2. If so, is there an explicit expression of the lattice generators in terms of the equation defining the curve? Or, at least, is there a simple example of a curve and its corresponding lattice?
3. Since every elliptic curve is a Lie group, it must have a corresponding Lie algebra. Is there an explicit expression of the Lie algebra in terms of the equation or lattice? Or, again, a simple example of a curve and its Lie algebra (or, even better, an example of a curve, its lattice, and its Lie algebra).
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## 4 Answers
What you want in terms of the relation between lattices and elliptic curves over C is proposition I.4.4 of Silverman's Advanced topics in the arithmetic of elliptic curves. Additionally, to go from a lattice to the equation of the elliptic curve (explicitly), you use Eisenstein series as in Corollary I.4.3 of that book. To go from the elliptic curve to the lattice is described in the proof of proposition I.4.4: you look at the homology of the curve and compute period integrals (incidentally, this is how you go from an abelian variety over C to a complex torus).
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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.
For a bit more info on question 3: if you are interested in the elliptic curve only as a complex Lie group, then when you identify it with C/L for C the complex plane and L a lattice, the Lie algebra is canonically C and the exponential map is the reduction mod L.
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1. No, for any cubic curve in the plane, there is a family of cubic plane curves (3 or 4 dimensional - I forget Edit: 8-dimensional, with a transitive action of PGL3) that are isomorphic as curves. For any lattice in C, there is a 1-dimensional family of lattices that form isomorphic curves. Over C, the curves are classified by their j-invariant (a complex number).
2. You can fix the lattice and choose generators so that one of the generators is 1 and the other lies in the upper half plane. There is then an action of SL2(Z) on the choices of generators, yielding an action on the upper half plane. If you choose a fundamental domain for this action, you get a "canonical" choice of lattice for each elliptic curve.
3. The Lie algebra is the unique 1-dimensional Lie algebra, whose bracket is zero. One can sometimes find more interesting information using the formal group law, but that mostly applies when you work in characteristic p.
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Thanks alot for your answers. One more question then: How much of the structure of an elliptic curve does this correspondence preserve? That is, is it homeomorphic, diffeomorphic, or biholomorphic? – John McCarthy Oct 24 2009 at 15:55
Corollary I.4.3 of Silverman that I speak of in my answer tells you that the map from C/lattice to E is a complex analytic isomorphism (it's given by the Weierstrass P-function and its derivative). – Rob Harron Oct 24 2009 at 16:10
Great, that's what I wanted to hear. Thanks alot. – John McCarthy Oct 24 2009 at 16:15
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Scott, I am confused by point 1. There is definitely a typo here, but I think there may be a larger error. {Smooth cubic curves}/PGL_3 is in bijection with {genus one curves}/isomorphism. To see this, notice that, given a genus one curve C, the set of embeddings of C into P^2, modulo PGL_3, is in bijection Pic^3(C). The automorphisms of C act transitively on Pic^3(C). I can think of various ways to vary my answer by talking about genus 1 curves with a specified origin or by not quotienting by PGL_3, but I can't think of any question whose answer should be "3 or 4 dimensional". – David Speyer Nov 3 2009 at 20:17
I was looking at free parameters in the full Weierstrass cubic form, but you're right that it is a poor choice. I will try to change it appropriately. – S. Carnahan♦ Nov 3 2009 at 21:39
Depending on what you mean, then no. Elliptic curves (abstractly) are in bijection with lattices modulo homothety (also, the upper half plane modulo SL(2,Z)). As for the Lie algebra of an elliptic curve, an elliptic curve is abelian, so the lie algebra is a one dimensional abelian complex Lie algebra, so it has trivial bracket.
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you mean "...with lattices modulo homothety" or "...with the upper half plane modulo SL(2,Z)" in that first sentence. – Kevin Buzzard Nov 3 2009 at 21:16
Yeah, good point. I don't do much with elliptic curves (most of the problems I think of are only interesting in the hyperbolic case) so I mix the language up at times. Corrected. – Charles Siegel Nov 3 2009 at 22:22
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http://mathforum.org/mathimages/index.php/Image_Convolution
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# Image Convolution
### From Math Images
Image Convolution
Field: Other
Image Created By: [[Author:| ]]
Image Convolution
Image Convolution is the process of applying a filter to images. Clockwise from top left, this images shows an original image, a Gaussian Blur filter, a Poster Edges filter, and a Sharpen filter. The filters were applied in Photoshop.
# Basic Description
Images can be convolved by applying a function to each pixel of the image. Usually, this function is precalculated inside a small two dimensional array called a kernel.
# A More Mathematical Explanation
[Click to view A More Mathematical Explanation]
Most generally, the convolution of two functions f and g is defined as the following:
UNIQ3d834044 [...]
[Click to hide A More Mathematical Explanation]
Most generally, the convolution of two functions f and g is defined as the following:
$(f * g)(x,y) = \sum_{v=-\infty}^{\infty} \sum_{u=-\infty}^{\infty} f(u,v) g(x - u, y - v)$
In this formula $f(x,y)$ is a function that represents the image, and $g(x,y)$ is the kernel. In practical situations, the kernel is only defined over a finite set of points, so we can modify our definition as follows:
$(f * g)(x,y) = \sum_{v=y-h}^{y+h} \sum_{u=x-w}^{x+w} f(u,v) g(x - u, y - v)$
Where $2w + 1$ is the width of the kernel and $2h + 1$ is the height of the kernel. In this example g is only defined over the points $[-w, w] \times [-h, h]$. To convolve an image, this formula is evaluated at every point in the image. In the following pseudocode to convolve an image, f(x, y) is the original image, g(x, y) is the kernel, and h(x, y) is the new image.
```for y from 0 to imageHeight
for x from 0 to imageWidth
sum := 0;
for v from y - h to y + h
for u from x - w to x + w
sum := sum + f( u, v ) * g( x - u, y - v );
h( x, y ) := sum;
```
## Examples
The most common type of kernel is a gaussian which acts as a lowpass filter, suppressing high frequency data in the signal. The most common example of the gaussian for a 3x3 kernel is the following:
$\frac{1}{16} \left[ {\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \\ \end{array} } \right]$
Other types of filters include edge detectors, which are essentially high pass filters. One type of edge detectors is called the Sobel operator. Mathematically, it computes the first derivative. The first derivative is large when the image greatly increases in intensity between two adjacent points which is what an edge looks like. The sobel operator has the form:
$\left[ {\begin{array}{ccc} 1 & 0 & -1 \\ 2 & 0 & -2 \\ 1 & 0 & -1 \\ \end{array} } \right]$
$\left[ {\begin{array}{ccc} 1 & 2 & 1 \\ 0 & 0 & 0 \\ -1 & -2 & -1 \\ \end{array} } \right]$
## Applet
[Click to show applet]
[{{{HideMessage}}}]
If you can see this message, you do not have the Java software required to view the applet.
# Teaching Materials
There are currently no teaching materials for this page. Add teaching materials.
If you are able, please consider adding to or editing this page!
Have questions about the image or the explanations on this page?
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.
[[Category:]]
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http://www.mathplanet.com/education/algebra-2/exponential-and-logarithmic-functions/exponential-functions
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Exponential functions
Until now we have dealt with various calculations of functions and equations where x is either in the base or the exponent. When x is the exponent the function is known as an exponential function. We will now see how an exponential function appears graphically. As with any function whatsoever, an exponential function may be correspondingly represented on a graph. We will begin with two functions as examples - one where the base is greater than 1 and the other where the base is smaller than is smaller than 1.
In this function the base is 2. The function is inclining.
In this function the base is 0.1. The function is declining.
An exponential function may generally be written as:
$y=C\cdot a^{x}$
The following graph depicts four different curves indicating how an exponential function behaves depending on whether C and a are positive or negative. The curves are identical but inverted in the various quadrants:
No curve ever intersects the x-axis but only approaches it infinitely.
Videolesson: show how to rewrite the given expression
$0.25^{x}=(\frac{1}{4})^{x}=4^{-x}$
Next Class: Exponential and logarithmic functions, Logarithm and logarithm functions
• Pre-Algebra
• Algebra 1
• Algebra 2
• Geometry
• Sat
• Act
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http://mathoverflow.net/revisions/6533/list
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## Return to Answer
3 oops, this is a presentation of monoids, not groups
Many standard examples of algebraic "forgetful" functors $U : C \to \mathrm{Set}$ have the following form:
• $C$ is a presentable category, i.e., there is a small category $I$ and a collection $S$ of cones of $I$ such that $C$ is equivalent to the full subcategory of functors $I \to \mathrm{Set}$ consisting of those functors which send the cones of $S$ to limit diagrams in $\mathrm{Set}$;
• $U$ is evaluation at an object $u \in I$.
For example, if $C$ is the category of groupsmonoids, take $I = \Delta^{\mathrm{op}}$ so that functors $I \to \mathrm{Set}$ are simplicial sets and choose $S$ so that the objects of $C$ are those simplicial sets $X$ such that $X_0 = \ast$ and $X_{i+j} \to X_{i} \times X_{j}$ is an isomorphism (where this map is induced by the inclusions of the first $i+1$ and last $j+1$ elements of an ordered $i+j+1$ element set). The object $u$ is the two-element set $[1]$. (One actually needs only the full subcategory of $\Delta^\mathrm{op}$ on the objects $[0]$, $[1]$, $[2]$, $[3]$, and the cones involving these objects; expanding this gives a possibly more familiar presentation of the notion of group.monoid.)
In these cases (which include models of any essentially algebraic theory) the existence of a left adjoint is guaranteed by the theory of presentable categories. Indeed, the inclusion of $C$ into $\mathrm{Set}^I$ has a left adjoint which we compose with the constant diagram functor $\mathrm{Set} \to \mathrm{Set}^I$ to obtain a left adjoint to $U$. See Adamek and Rosicky, Locally presentable and accessible categories, for an excellent introduction to the subject.
2 added 225 characters in body; added 12 characters in body
Many standard examples of algebraic "forgetful" functors $U : C \to \mathrm{Set}$ have the following form:
• $C$ is a presentable category, i.e., there is a small category $I$ and a collection $S$ of cones of $I$ such that $C$ is equivalent to the full subcategory of functors $I \to \mathrm{Set}$ consisting of those functors which send the cones of $S$ to limit diagrams in $\mathrm{Set}$;
• $U$ is evaluation at an object $u \in I$.
For example, if $C$ is the category of groups, take $I = \Delta^{\mathrm{op}}$ so that functors $I \to \mathrm{Set}$ are simplicial sets and choose $S$ so that the objects of $C$ are those simplicial sets $X$ such that $X_0 = \ast$ and $X_{i+j} \to X_{i} \times X_{j}$ is an isomorphism (where this map is induced by the inclusions of the first $i+1$ and last $j+1$ elements of an ordered $i+j+1$ element set). The object $u$ is the two-element set $[1]$. (One actually needs only the full subcategory of $\Delta^\mathrm{op}$ on the objects $[0]$, $[1]$, $[2]$, $[3]$, and the cones involving these objects; expanding this gives a possibly more familiar presentation of the notion of group.)
In these cases (which include models of any essentially algebraic theory) the existence of a left adjoint is guaranteed by the theory of presentable categories. Indeed, the inclusion of $C$ into $\mathrm{Set}^I$ has a left adjoint which we compose with the constant diagram functor $\mathrm{Set} \to \mathrm{Set}^I$ to obtain a left adjoint to $U$. See Adamek and Rosicky, Locally presentable and accessible categories, for an excellent introduction to the subject.
1
Many standard examples of algebraic "forgetful" functors $U : C \to \mathrm{Set}$ have the following form:
• $C$ is a presentable category, i.e., there is a small category $I$ and a collection $S$ of cones of $I$ such that $C$ is equivalent to the full subcategory of functors $I \to \mathrm{Set}$ consisting of those functors which send the cones of $S$ to limit diagrams in $\mathrm{Set}$;
• $U$ is evaluation at an object $u \in I$.
For example, if $C$ is the category of groups, take $I = \Delta^{\mathrm{op}}$ so that functors $I \to \mathrm{Set}$ are simplicial sets and choose $S$ so that the objects of $C$ are those simplicial sets $X$ such that $X_0 = \ast$ and $X_{i+j} \to X_{i} \times X_{j}$ is an isomorphism (where this map is induced by the inclusions of the first $i+1$ and last $j+1$ elements of an ordered $i+j+1$ element set). The object $u$ is the two-element set $[1]$.
In these cases (which include models of any essentially algebraic theory) the existence of a left adjoint is guaranteed by the theory of presentable categories. Indeed, the inclusion of $C$ into $\mathrm{Set}^I$ has a left adjoint which we compose with the constant diagram functor $\mathrm{Set} \to \mathrm{Set}^I$ to obtain a left adjoint to $U$. See Adamek and Rosicky, Locally presentable and accessible categories, for an excellent introduction to the subject.
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http://quant.stackexchange.com/questions/47/how-to-calculate-future-distribution-of-price-using-volatility?answertab=votes
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# How to calculate future distribution of price using volatility?
I want to create a lognormal distribution of future stock prices. Using a monte carlo simulation I came up with the standard deviation as being $\sqrt{(days/252)}$ $*volatility*mean*$ $\log(mean)$. Is this correct?
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Thanks for the help, I figured it was a simple mistake. Actual formula for stdev is sqrt(days/252)*volatility*mean. The mistake I was making is days doesn't include first day. So if you have 5 days of random walk, days=4 in above formula. – CptanPanic Feb 3 '11 at 1:34
## 3 Answers
I'm not sure I understand, but if you want to compute the variance of $exp(X)$, where $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$, that variance is (from Wikipedia): $$\left(\exp{(\sigma^2)} - 1\right) \exp{(2\mu + \sigma^2)}$$
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I don't think I asked my question correctly. Assuming a random walk, starting at $\mu$ and yearly volatility of $\sigma$ . What is $\sigma$ for the distribution (expected prices) in like 1 month? I must be overthinking this, but it doesn't seem like \$\sigma / 12 works either. – CptanPanic Feb 1 '11 at 1:47
1
The volatility scales as the square root of time. So in one month, you would have $\sigma / \sqrt{12}$, not $\sigma / 12$. This has nothing to do with being log normal, though. – shabbychef Feb 1 '11 at 3:29
The distribution of the log of a stock price in n days is a normal distribution with mean of $\log(current_price)$ and standard deviation of $volatility*\sqrt(n/365.2425)$ if you're using calendar days, and assuming no dividends and 0% risk-free interest rate.
Note that the standard deviation is independent of the current_price: if $\log(current_price)$ increases by 0.3 (for example), the stock has increased by 35%, regardless of its current_price.
To include dividends and the risk-free interest rate, see:
http://en.wikipedia.org/wiki/Black-Scholes
which models future stock prices w/ an eye towards pricing options.
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To create a lognormal distribution (that is, to generate values from it), you need to start with normally distributed numbers and then exponentiate them.
That is to say, take a sample $z$ from the standard normal distribution, and form the lognormally distributed underlying value
$$U_T = U_0 \exp\left( (r-q-\sigma^2/2)T + \sigma \sqrt{T} z \right)$$
The probability density function of $U_T$ is formed from solving this for $z$ and then applying the normal PDF.
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http://everydayecon.wordpress.com/2012/03/
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Monthly Archives: March 2012
On Fiat Money
Posted on March 23, 2012
Why do people accept fiat money? It is, after all, intrinsically useless. David Glasner has written a thoughtful post on fiat money. Glasner writes:
Why should a fiat money not be able to retain value? Well, consider the following thought experiment. For a pure medium of exchange, a fiat money, to have value, there must be an expectation that it will be accepted in exchange by someone else. Without that expectation, a fiat money could not, by definition, have value. But at some point, before the world comes to its end, it will be clear that there will be no one who will accept the money because there will be no one left with whom to exchange it. But if it is clear that at some time in the future, no one will accept fiat money and will then lose its value, a logical process of backward induction implies that it must lose its value now.
So why are people willing to accept it? Glasner suggests that the reason is because the government accepts money as payment for taxes. This is the chartalist view and has been around for some time. Glasner traces this back to Wicksteed. Ross Starr provides an analysis within the context of general equilibrium theory. However, I would like to suggest another proposition. To do so, we need to talk about two different questions: (a) why does fiat money exist?, and (b) why do people accept it?
There are a number of ways that have been suggested to explain the use of money. However, I would like to focus on one in particular. As emphasized by Brunner and Meltzer (1971) and Ostroy (1973), money is a substitute for information. To illustrate this point, consider an example. Suppose that there are three people, three goods, and three time periods. Person 1 produces Good 1 in Period 1, but wants to consume Good 2. Person 2 produces Good 2 in Period 2, but wants to consume Good 3. Person 3 produces Good 3 in Period 3, but wants to consume Good 1. Notice that there is a basic double coincidence of wants problem. This problem, however, does not necessarily require money. For example, the three individuals could get together and make a multilateral arrangement in which each individual promises to deliver the desired good to the appropriate trader in each period. No money is necessary.
The example above, however, requires that the individuals can perfectly commit to those actions. In reality, the individuals have an incentive to renege on their promises. For example, Person 3 could promise to give Good 3 to Person 2 in Period 3 so long as Person 1 supplies Good 1. However, if there is disutility associated with production, Person 3 has no incentive to produce anything given that they have already received their consumption good. But even this isn’t sufficient to require money. If trading histories are perfectly monitored and costly available to all parties, this would provide an incentive for individuals to behave honestly.
This latter assertion, however, bears little resemblance to the world in which we live. We do not have costless access to the trading histories of every possible counterparty. As such, when individuals cannot perfectly commit and there is imperfect monitoring, money is essential in the language of Hahn in that it expands the possible allocations available to economic agents. The role of money in this context is as a substitute for information. Money is memory.
The informational role of money makes money essential and is therefore preferable to other arrangements. However, this does not resolve the solution by backward induction that Glasner suggested above. Thus, we need to answer question (b) above.
In search models of money it is standard to denote the price of money as $\phi_t$. The price of money refers to the goods price of money rather than the money price of goods as we are usually accustomed to thinking. It is important to think about the price of money because it is possible that this price could be zero (i.e. we have a non-monetary economy). A condition for a monetary equilibrium in these models is that
$\phi_t \geq \beta \phi_{t + 1}$
This implies that it must be true that the price of money today is greater than or equal to the present discounted value of the price of money tomorrow. In more familiar language to monetary theorists, it must be true that the inflation rate is greater than or equal to minus the rate of time preference.
According to Glasner, we know that at some date, T, the world ends and therefore nobody would accept money. Through backward induction, nobody would accept money today. However, the solution by backward induction is contingent upon knowing the date at which the world ends. For example, suppose that the probability of the world ending tomorrow is $p(\phi_{t+1} = 0 | \Omega_t)$ where $\Omega_t$ is the information available at time t. Thus, the expected future value of the price of money at time t is:
$E_t \phi_{t + 1} = [1 - p(\phi_{t+1} = 0 | \Omega_t)] \phi_{t + 1} + p(\phi_{t+1} = 0 | \Omega_t) * 0$
Thus, we can re-write the necessary condition for equilibrium as:
$\phi_t \geq \beta [1 - p(\phi_{t+1} = 0 | \Omega_t)] \phi_{t + 1}$
So long as the probability of the world ending is not equal to one, a monetary equilibrium obtains and is therefore not subject to the backward induction critique. Fiat money can therefore be thought of as a rational bubble. We do not need to appeal to irrationality as Glasner suggests. Rather as a trader I know that money increases the set of feasible allocations in trade and therefore I have an incentive to use it and accept it so long as I anticipate that others will accept it in the future. In addition, I know that it is intrinsically worthless, but so long as the future is not certain (or time is considered infinite) it can have positive value because of its role as medium of exchange. As a result, fiat money trades above its fundamental value.
Health Care and Health Insurance Are Not the Same
Posted on March 22, 2012
Jeffrey Toobin writing in the New Yorker:
The main argument that opponents of the health-care law have come up with is that the mandate regulates economic inactivity—i.e., not buying insurance—and the Commerce Clause allows only the regulation of economic activity. In the first appellate review of the law, last summer, the Sixth Circuit demolished that argument. The court pointed out that there are two unique characteristics of the market for health care: “(1) virtually everyone requires health care services at some unpredictable point; and (2) individuals receive health care services regardless of ability to pay.” Thus, there was no such thing as “inactivity” in the health-care market; everyone participates, even if he or she chooses not to buy insurance.
I don’t understand this argument. Health care is not health insurance. There is no inactivity in the health care market, but there is inactivity in the health insurance market. The law mandates health insurance purchases, not health care purchases. This does not seem to be a valid comparison.
Also, Ezra Klein writes regarding Paul Ryan’s budget plan:
It’s Medicaid and other health spending, which includes the Affordable Care Act, where Ryan really brings down the hammer: That category falls by 1.25 percent of GDP. So Ryan’s cuts to health care for the poor are almost twice the size of his cuts to health care for the old.
The plan cuts government-provided health insurance to the poor, not health care to the poor. The distinction is important because there is evidence that in some cases being uninsured is comparable to being on Medicare (see here, here, and here).
→ 3 Comments
Posted in Economic News, Politics
Schools of Thought, Rules, Discretion, and Rational Expectations
Posted on March 20, 2012
Arnold Kling has written a piece in which he implies that Keynesians and Monetarists are all victims of confirmation bias. They see things around them that reinforces their priors and they view this as evidence that their worldview is correct. He also concludes by suggesting that he thinks that both schools of thought are wrong. There are a couple of points that I want to assess.
First, I have pointed out confirmation bias before when Paul Krugman described why he is a Keynesian. However, as I discuss in that post, the problem is not that Krugman is wrong, but rather that there are other theories that predict observationally equivalent observations. The key is to develop a null hypothesis to differentiate between the two. Confirmation bias is only a bias if you are incorrect, otherwise it is evidence. (I am reminded of the idea that the first stage of being an alcoholic is the same as that of someone who is not alcoholic: denial.)
The problem, I think, is with falsifying schools of thought. You cannot falsify Keynesianism, Monetarism, or any other school of thought. Each of these schools — to the extent to which they even exist today in the profession — is made up of diverse members. There are a broad range of issues in which there is agreement. Some of these positions might be well-founded, others not. However, the idea is to develop testable hypotheses. If my model is correct, I would expect to observe X. Do I observe X? Are there other possible explanations? If so, can I differentiate between these explanations?
One of the most dubious aspects of modern macro, in my view, is the role of sticky prices. One can track prices and determine that they are slow to adjust, but does that mean that they are important from a macroeconomic perspective? Not necessarily. In order to test this, one needs a better way to test this hypothesis than simply observing that prices adjust slowly or asking business people how, when, and why they adjust prices. If sticky prices don’t have macroeconomic effects, this is an IO story. The conventional way to test the hypothesis is to write down a model without sticky prices and the identical model with sticky prices, simulate the models and see which one better predicts what we observe in reality. However, does this necessarily tell us that sticky prices are important? No. It could be that something is missing from the model and that sticky prices are picking up that effect. For example, we know that monetary policy shocks have persistent effects on real output. A basic RBC model cannot replicate this observation. Modifying the RBC model to have sticky prices (i.e. a New Keynesian model) is better able to explain the persistence. But so what? The RBC model cannot pick up this characteristic because all markets are centralized. In reality, the true data generating process might have segmented markets, decentralized trading, or a variety of other features. The fact that models with sticky prices perform better than models without does not necessarily tell us anything.
The point that I want to make is not that we should forsake sticky price models or a belief in the macroeconomic implications thereof. Rather, my point is that we need to generate better testable hypotheses to examine these questions. In other words, the reason that I am skeptical of the role of sticky prices is not because I believe that there is no evidence, but rather that it is not clear how this evidence stacks up against possible alternatives, which have not been examined. The evidence in favor of these views lies in the confirmation of a basic hypothesis. It does not, however, mean that those who accept the important macroeconomic role of sticky prices have committed confirmation bias, but rather that they have tested the hypothesis against insufficient alternatives.
What’s more, this is only one issue. Keynesians believe in a lot more than sticky prices. As such, it should take a lot more evidence to overturn an entire school of thought. And examining each issue is hard. Economics don’t get to use natural experiments. As such, one needs to develop clear, testable null hypotheses against reasonable alternatives.
This brings me to my second point. Who are the stubborn Monetarists and the stubborn Keynesians? The view of stubborn Monetarists and Keynesians suggests that those who see the world through a similar lens to these historical schools of thought are somehow standard bearers for this line of thinking and that nothing much has changed. New Keynesianism is different from Old. New Keynesians use microfoundations, they accept rational expectations, etc. Modern monetarists are largely divided into two categories: Market Monetarists and New Monetarists. If we take out nominal income targeting, there is very much agreement among the two groups. The blend of modern Monetarists is different from Old Monetarism, although both advocate the importance of money, both from a micro and a macro perspective. Nonetheless, there is a difference between modern Monetarism and Old Monetarism. Gone are the days of advocating constant rates of money growth, for example.
Arnold Kling seems to suggest that the primary difference between Monetarists and Keynesians is one of policy, with Keynesians favoring fiscal policy and discretion and Monetarists favor monetary policy and rules. This is a mischaracterization and is, in fact, undermined by Kling’s own article as he implies that John Taylor is a stubborn Monetarist advocating the Taylor rule. But there is nothing Monetarist about John Taylor or his monetary policy rule. Taylor is a New Keynesian and his monetary rule is of central use in New Keynesian models. While it is true that the most vocal advocates of fiscal stabilization are what one would consider Keynesians, the consensus of the profession prior to the recession was not one in favor of fiscal stabilization. See, for example, James Bullard’s “Death of a Theory” paper.
The rational expectations hypothesis virtually ensures that macroeconomic performance will be better under rules than under discretion. If policy follows predictable rules, then individuals in the private sector will not make mistakes that come from guessing incorrectly about the path of policy. If such mistaken guesses are the main cause of recessions, then the use of rules could eliminate recessions.
First, the literature on rules versus discretion with rational expectations was really about the inflationary bias of the central bank, not about recessions. This mixes the view of Friedman and other Old Monetarists that the Fed tended to either stimulate too much, causing inflation, or contract too much, causing a recession with the rational expectations literature that focused on inflation bias.
Second, Kling seems to offer a veiled criticism of rational expectations. What rational expectations suggests is that people do not make systematic errors. As such the literature on rules versus discretion suggests that if the central bank is able of systematically misleading the public, then rules are better than discretion. In that sense the rational expectations hypothesis is important to the debate, but what is the alternative? Even under adaptive learning, the solution will converge to the rational expectations solution.
Overall, I don’t see schools of thought dominating the literature. I see a lot of loud voices in the blogosphere and I think that Arnold Kling views these voices as a representative sample of the profession. I think this is very misleading.
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Posted in Economic News, Macroeconomic Theory
Forecasts and Standard Errors
Posted on March 19, 2012
Via John Whitehead, I was led to this discussion of climate change by the EPA. The EPA repeats the following claim from the IPCC:
The average surface temperature of the Earth is likely to increase by 2 to 11.5°F (1.1-6.4°C) by the end of the 21st century, relative to 1980-1990, with a best estimate of 3.2 to 7.2°F (1.8-4.0°C) (see Figure 1). The average rate of warming over each inhabited continent is very likely to be at least twice as large as that experienced during the 20th century.
Does the standard error on this century-long forecast seem small to anyone else? (I’m not being facetious, I’m genuinely asking in the hopes that those will greater knowledge of the issue will answer.)
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Posted in Uncategorized
Pet Peeves: Monetary Policy Edition
Posted on March 7, 2012
A recent op-ed in the WSJ posits the following claim:
A few critics of quantitative easing (QE) and the zero interest rate (ZIRP) have correctly pointed out that these policies weaken the dollar and thereby reduce the purchasing power of American paychecks. They increase the risk of future inflation, obscure the true cost of the unsustainable fiscal policy the federal government is running, and transfer wealth from savers to debtors.
But QE and ZIRP also reduce long-term economic growth by punishing savers, reducing saving and investment over the long run. They encourage the misallocation of resources that at a minimum is preventing the natural rebalancing of our economy and could sow the seeds of another painful boom-bust.
There is a lot going on in these statements. Also, I don’t like thinking about monetary policy in terms of interest rates. Nonetheless, here are my main points of rebuttal:
(a) Does the Fed control real interest rates in the long run? In other words, the author seems to think that by having the Fed target a federal funds rate near zero that this effects savings and investment (i.e. lower interest rates lead to lower savings and since savings=investment this means lower investment and lower growth). But don’t people make savings and investment decisions based on real interest rates whereas the Fed is targeting a nominal interest rate? Consider an example to illustrate my point. Suppose that the Fed came out tomorrow and said that they wanted to promote savings and investment and therefore were raising their target of the federal funds to 5%. Holding inflation expectations constant, the Fisher equation implies that the real interest rate would rise substantially. However, we cannot hold inflation expectations constant in this example because that policy would necessarily have an effect on inflation expectations. In fact, it is very likely that this policy would induce expectations of deflation. The policy would do nothing to raise real interest rates.
(b) The use of “boom-bust” is clearly a reference to the Wicksell-Hayek business cycle model. However, in that model the boom (which leads to a subsequent bust) is caused by malinvestment due to the fact that the market rate of interest is set below the natural rate of interest. Assuming that this view is empirically valide, what is the current natural rate of interest? One empirically testable hypothesis from this view is that when the market rate of interest is held below the natural interest rate, this results in ever-accelerating inflation. Do we observe ever-accelerating inflation? Do we at least observe rapidly growing productivity in the face of stable inflation? I think that the answer to those questions is no.
(c) Does inflation reduce the purchasing power of American paychecks? First, this depends on how we measure things and the time horizon. If American paychecks have less purchasing power, then we would expect to see real compensation declining. We don’t. Perhaps the author is referring to the purchasing power relative to other currencies. Have we seen a marked depreciation in the USD relative to the Euro or the Pound sterling? No.
Do not mistake this analysis for downplaying the costs of inflation. The costs associated with inflation can be significant. However, just because something costs more in nominal or real terms does not necessarily mean that individuals are worse off. For example, we have seen a large increase in the price level since 1970. Nevertheless, given that an 18 cubic foot refrigerator cost about \$400 in 1971 and the same size refrigerator costs about \$450 today, the average worker in 1971 would have had to work about 107 hours to purchase that refrigerator (based on average wages) whereas the worker in 2011 would only have had to work about 23 hours.
Rising prices aren’t a cost of inflation, they are the definition of inflation.
(The costs of inflation are the things that would have been avoided in the absence of inflation. For example, the inflation tax on money holdings, “shoe leather” costs, costs associated with shorter contract durations and negotiation, etc. Read Axel Leijonhufvud, for example.)
(d) Inflation reallocates wealth from savers to borrowers only when it is unexpected. Has inflation been higher than expected?
Money and Measurement
Posted on March 5, 2012
My colleague Michael Belongia and his co-author Peter Ireland have new NBER working paper entitled, “The Barnett Critique After Three Decades: A New Keynesian Analysis.” Here is the abstract:
This paper extends a New Keynesian model to include roles for currency and deposits as competing sources of liquidity services demanded by households. It shows that, both qualitatively and quantitatively, the Barnett critique applies: While a Divisia aggregate of monetary services tracks the true monetary aggregate almost perfectly, a simple-sum measure often behaves quite differently. The model also shows that movements in both quantity and price indices for monetary services correlate strongly with movements in output following a variety of shocks. Finally, the analysis characterizes the optimal monetary policy response to disturbances that originate in the financial sector
Measurement matters.
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http://physics.stackexchange.com/questions/39148/open-boundary-condition-and-glauber-dynamics/39179
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# Open boundary condition and Glauber Dynamics
Warning: by background is in math, not physics. I've just recently started working with things that are close to theoretical physics. So please note that
1. I'm still very confused by the jargon.
2. Maybe I'm asking something trivial, in which case please just give me some reference and I'll be happy to read them.
I am reading some notes about Glauber Dynamics for discrete spin models (mainly stuff from Martinelli, Olivieri and Schonmann), and I'm surprised by how little space is given to models with "open boundary conditions" (which as far as I understood is the current terminology for no boundary conditions).
Googling the web didn't helped me understand if this is a choice of the authors, or there is some more stringent motivation beyond.
Let's suppose that we are working with a regular lattice (i.e. $\mathbb{Z}^d$), discrete spin space ($\{\pm 1\}$ is enough), and finite range interactions ($H(\sigma) = \sum_{A} J_A(\sigma_A)$, where the sum is taken over finite sets $A$ of diameter smaller than the interaction length).
1. Are there any problems with defining the Gibbs measure for finite volumes $V \subset \mathbb Z^d$ without boundary conditions, i.e. $$\mu_V(\sigma) = Z_V^{-1} \exp \left( \sum_{A \subset V} J_A(\sigma_A) \right)$$ where by "problem" I think of things like "is not a measure on the right space of configurations", or "does not verify some important compatibility rule" (DLR?), or "is not unique", etc.
2. Are known conditions for which Glauber Dynamics on this finite region $V$ will converge quickly to the above-defined distribution will be uniformly gapped?
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– Yvan Velenik Oct 6 '12 at 11:37
I knew the Martinelli's notes, but I overlooked at the chapter over phase coexistence. Thanks a lot for pointing it out! – Angelo Lucia Oct 8 '12 at 11:01
## 1 Answer
There are no problems--- I don't know a proof of the convergence of the Glauber dynamics to the Boltzmann distribution which doesn't work independent of boundary conditions. You just need to check two things:
1. The Glauber dynamics preserves the Boltzmann distribution (this is easy to check by starting with the Boltzmann distribution, and showing it is obeys detailed balance, i.e. that $\rho_i K_{i\rightarrow j} = \rho_j K_{j\rightarrow i}$, in words: the rate of the i->j transition in equilibrium is balanced by the rate of the j->i transition. This shows that Glauber dynamics has a stationary distribution which is the Boltzmann distribution.
2. Show ergodicity: this says that starting from any two configurations at time zero, there is a path with total nonzero Glauber probability which ends at the same state, i.e. the two paths collide with nonzero probability after finite time.
Under these conditions, it is easy to prove by a coupling argument that the stationary distribution is unique. The proof goes as follows: you consider the coupled walk, where the two walks are independent until they happen to be the same, and then stay the same forever. When there is a finite probability for combining, this will guarantee eventual convergence of the two walks, and therefore the final distribution is independent of the starting distribution, and must be the Boltzmann distribution (this is covered in one of my answers here in more detail, but I can't remember which).
The estimates from this argument are usually not optimal. The mixing time of the Glauber dynamics is usually reasonably fast away from a critical point, where there is critical slowing down in large systems. To get around this, you should use block updates.
-
I know the coupling method to prove convergence, but thanks for explaining it. What I really meant with "converge quickly" is whether the Glauber dynamics is gapped (i.e. the spectral gap is uniform in the system size) or not. – Angelo Lucia Oct 9 '12 at 15:19
This is false, for example, just a diffusing particle has no gap for infinite size. This means that a number-conserving Glauber dynamics on a finite number of diffusing spins is ungapped. The issue of gap uniformity is not possible to adress independent of the model. – Ron Maimon Nov 8 '12 at 17:19
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http://mathhelpforum.com/calculus/161270-complex-number-help.html
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# Thread:
1. ## complex number help
I have a couple of questions that i cant figure out to save myself =/
1.these are expressing in the form a+ib $\frac{1}{(2+i)(z+1)} = 5-3i$
i got z= -1/14 + i/14
2. $2z^2=6iz + 3$
didnt know how to do this.
3. $(z)(zcon)+2(z-zcon)=10 +2i$ (where zcon is the conjugate of z)
okay and theres:
" $1-zi$ is the root of the equation
$z^2+(i+1)z+k=0$
find the other root and the value of k."
now i assume you use the $\Delta$ but im not sure.
im sorry theres so many but i need the help.
also if this isnt the correct forum also sorry as there was a similar problem listed here too
The first one is incorrect.
$\frac{1}{(2+i)(z+1)}=5-3i$
$1=(5-3i)(2+i)(z+1))$
$1=(13-i)(z+1)$
$z+1=\frac{1}{(13-i)}\times\frac{13+i}{13+i}$
Hope you will be able to continue.
The second one is a quadratic equation of z.
$2z^2=6iz + 3\Rightarrow{2z^2-6iz - 3=0}$
$z^2-3iz-\frac{3}{2}=0$
$(z-\frac{3i}{2})^2+\frac{9}{4}-\frac{3}{2}=0$
$(z-\frac{3i}{2})^2+\frac{9}{4}-\frac{3}{2}=0$
$(z-\frac{3i}{2})^2+\frac{3}{4}=0$
Can you continue from here??
3. ok thanks i understand the first one
is the second
+/- $\frac{3i}{2}+\frac{9}{4}$?
i think i got my last posted one as well.
its a 2 not a z
do you use the fact that the sum of the roots would be $let x=second root<br /> x+(1-2i)=i+1 ?$
therfore figuring out k?
4. that didnt post as expected
should be
"letx = the second root
then the eqn starting with x
5. is the first one $z=\frac{-157}{170}+\frac{i}{170}$??
is the first one $z=\frac{-157}{170}+\frac{i}{170}$??
Correct.
ok thanks i understand the first one
is the second
+/- $\frac{3i}{2}+\frac{9}{4}$ ?
Incorrect.
$(z-\frac{3i}{2})^2=-\frac{3}{4}$
$(z-\frac{3i}{2})=\pm{\frac{i\sqrt{3}}{2}}$
$z=\frac{3i}{2}\pm{\frac{i\sqrt{3}}{2}}$
$z=(\frac{3}{2}\pm\frac{\sqrt{3}}{2})i$
3. $(z)(zcon)+2(z-zcon)=10 +2i$ (where zcon is the conjugate of z)
Use $z=x+iy$ and simplify. You would end up with two equations with x and y; hence you would be able to solve them simultaneously.
okay and theres: $1-zi$" is the root of the equation $z^2+(i+1)z+k=0$ find the other root and the value of k."
You could use the sum and product of roots of a quadratic equation. That is if $\alpha~and~\beta$ are the roots of the quadratic, $ax^2+bx+c=0$; then, $\alpha+\beta=-\frac{b}{a}~and~\alpha\beta=\frac{c}{a}$
1.these are expressing in the form a+ib $\frac{1}{(2+i)(z+1)} = 5-3i$
i got z= -1/14 + i/14
There is a much easier way to do this one.
Recall that $\dfrac{1}{z} = \dfrac{{\overline z }}{{\left| z \right|^2 }}$
So $z + 1 = \dfrac{1}{{2 + i}}\dfrac{1}{{5 - 3i}} = \dfrac{{2 - i}}{5}\dfrac{{5 + 3i}}{{34}}$.
8. would my way of working out k in question 4 be right?
If the two roots of this equation are $\alpha~and~\beta$ then, $\alpha+(1-2i)=-(1+i)~and~\alpha(1-2i)=k$. From these two equations you could find k.
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http://math.stackexchange.com/questions/101797/topology-of-almost-convex-sets
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# Topology of “almost convex” sets
Let $C$ be a compact set in $\mathbb{R}^n$. If $C$ is convex, it must be homeomorphic to a closed ball. Now suppose that instead of convexity we require the intersection of $C$ with any line to have at most $m$ connected components. What topological restrictions does this condition place on $C$ and its embedding in $\mathbb{R}^n$?
For example, the result at http://www.ics.uci.edu/~eppstein/junkyard/knot-colinear.html shows that for $n=3$ and $m=3$, the complement of $C$ can't be knotted.
I would be interested in results for specific small cases as well as in the following question: For fixed $n$, does there exist $m$ so that $C$ can come from any "reasonable" isotopy class? (Define reasonable reasonably.)
-
The empty set or a singleton are compact convex subsets of $\mathbb R^n$ which are not homeomorphic to a closed ball. – Rasmus Jan 23 '12 at 22:49
@Rasmus A singleton of $\mathbb{R}^n$ is homeomorphic to a closed ball of dimension $0$ and the empty set is homeomorphic to a closed ball of dimension $-1$ (by convention). We have a result: if $U\subseteq \mathbb{R}^n$ is a bounded convex open set, then $\overline{U}$ is homeomorphic to the closed ball in $\mathbb{R}^n$. – Amitesh Datta Jan 24 '12 at 1:33
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http://mathhelpforum.com/calculus/28415-trigonometric-integrals.html
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# Thread:
1. ## Trigonometric Integrals
Can anyone help me or give me a good idea of how to approach the definite integral ([sinx]^2)*([cosx]^2) with the limits of integration from 0 to pi/2? I'm pretty sure I should somehow be able to simplify it using half-angle identities but I'm stuck. Thanks
2. $<br /> \sin ^2 x\cos ^2 x = \frac{1}<br /> {4}\sin ^2 2x = \frac{1}<br /> {4}\left( {\frac{{1 - \cos 4x}}<br /> {2}} \right) = \frac{{1 - \cos 4x}}<br /> {8}$
$<br /> <br /> ^* \sin 2x = 2\sin x\cos x<br /> <br /> <br />$
3. Use the identities $sin^{2}(x)=\frac{1-cos(2x)}{2}$ and $cos^{2}(x)=\frac{(1+cos(2x)}{2}$
Then, $\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos(2x))(1+cos(2x))dx$
$=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos^{2}(2x))dx$ $=\frac{1}{4}\int_{0}^{\frac{\pi}{2}}sin^{2}(2x)dx$ $=\frac{1}{8}\int_{0}^{\frac{\pi}{2}}(1-cos(4x))dx$
Now continue.
4. Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
5. Yes, that is an identity. Try deriving it.
6. Originally Posted by thejabronisayz
Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
You know that $1 - cos^2(y) = sin^2(y)$ right? Well, let y = 2x ...
-Dan
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http://math.stackexchange.com/questions/147120/simplifying-trigonometric-expression?answertab=active
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# Simplifying trigonometric expression
I'm sorry, if this question seems too simple..
Is it possible to reduce `sin 8x/cos 4x` it like this:
````4 (sin 2x)/ 4(cos x)
sin 2x/cos x
````
?
Or this trig. expression cannot be simplified any further?
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## 1 Answer
We have $\sin(8x)=2\sin(4x)\cos(4x)$. This is a particular case of the double-angle identity $\sin 2u=2\sin u\cos u$ (we just take $u=4x$).
So our numerator can be written as $2\sin(4x)\cos(4x)$, and our denominator is $\cos(4x)$. There is cancellation, and we obtain $2\sin(4x)$.
Remark: Your proposed simplification was not correct. It seems to assume, for example, that $\cos 4x=4\cos x$. That is not true in general. Take almost any specific value of $x$, like $x=\pi/2$ ($90$ degrees). We have $\cos 4x=1$ and $4\cos x=0$. There was a similar incorrect assertion about $\sin 8x$.
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I have written out a simplified form $2\sin(4x)$. There is no reason to fool around with the denominator, since it cancels anyway. – André Nicolas May 19 '12 at 18:37
At first time, i didnt notice.. :( Thanks a lot!!) – DrStrangeLove May 19 '12 at 18:39
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http://math.stackexchange.com/questions/202450/could-someone-verify-that-i-got-this-negation-correctly-in-discrete-mathematics
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# Could someone verify that I got this negation correctly in Discrete Mathematics?
~(∃ x ∈ Z | ∀ y ∈ Z, y/x ∈ Z)
= ∀ x ∈ Z | ∀ y ∈ Z, y/x ∈ Z
I'm just not 100% sure. I'm under the impression that if expressions following "such that" are not included in the quantifier they do not get negated. Is that correct?
And if that is true, how does one read the bottom expression? It seems bizarre to me.
-
## 2 Answers
Hint:
Assuming Z refers to the integers, your original un-negated statement reads:
There exists an integer $x$ so that for any integer $y$, the value of $\frac{y}{x}$ is an integer.
(It's referring to the integers $1$ and $−1$.)
Your (incorrectly) negated statement is read:
For any integers $x$ and $y$, the value of $\frac{y}{x}$ is an integer.
Correctly negated, it should read:
For any integer $x$, there is an integer $y$ so that $\frac{y}{x}$ is not an integer.
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Thanks for the beautiful explanation. I understand it conceptually (like an "English" translation of negation, if you will) but mathematically converting is where I'm having trouble. Could you see the comment on the other answer? – Doug Smith Sep 25 '12 at 21:37
The negation is
$$\forall x \in \mathbb{Z}, \exists y \in \mathbb{Z} \,\,\,\,\,\, \frac{y}{x} \notin \mathbb{Z}$$
When you have a statement of the form
$$\forall \exists \exists \forall \exists \cdots P$$
that is, a bunch of quantifiers before a quantifier-free formula, you negate by turning $\forall$ into $\exists$ and vice versa, and then negating $P$ to get
$$\neg (\forall \exists \exists \forall \exists \cdots P) = \exists \forall \forall \exists \forall \cdots \neg P$$
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Okay, but how does the such that (|) come into play? In the original equation there is one expression before the | and in the negation there is two. What's the conversion rule for that? – Doug Smith Sep 25 '12 at 21:36
@DougSmith: The "such that" is just there to make it read nicely in English. Most people leave it out, I believe. For example, I would have written $\exists x \in \mathbb{Z} \; \forall y \in \mathbb{Z} \; \frac{y}{x} \in \mathbb{Z}$ for the original expression. – Snowball Sep 25 '12 at 21:41
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http://physics.stackexchange.com/questions/45112/velocity-of-real-gas-molecules/45115
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# Velocity of real gas molecules?
It is known that the velocity of ideal gas molecules can be computed using Maxwell-Boltzmann law of distribution of molecular velocities, with average velocity given as:
$\overline{v}=\sqrt{\frac{8RT}{\pi M}}$
What would be the corresponding velocity for a real gas (which are not at STP)? Thanks!
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At STP everyday gasses are pretty close to ideal. – dmckee♦ Nov 26 '12 at 2:58
## 1 Answer
There is an important thing to note about your question, which is the understanding the relationship between math and physics. Physics uses mathematical equations to describe reality. Physics equations are approximations and may assume certain conditions (pressure, temperature, etc.) to offer an equation that is strictly true under those certain conditions. Thus, physics equations like the one you provided has boundaries in where its true.
In order to find the average velocity for real gases, you'll need to strictly define the conditions your gas is under. Although the average velocity of gases given by Maxwell-Boltzman distribution may be accurate, it may still deviate from a real gas you are asking about because it may be under different conditions. You may need to add or subtract some term to the equation you provided in order for it to match your real gas.
In short, there is no corresponding velocity for a real gas in the sense that "no physics equation to date is perfectly accurate in describing reality."
You may have a very accurate and precise equation that corresponds to reality, but it still serves as an approximation. What's great is that depending upon the system you are studying, sometimes you don't need such a large accuracy and your physics equation may have an accuracy that may exceed the system in question. This would be the case where your physics equation would be equivalent to describing reality, but you really need to specify your system first.
Moreover, since the equation is describing an average velocity, it means that you could have a distribution of velocities that has a large spread or a small spread.
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The current model I use is already based on the assumption of no major deviation in ideal and real gas velocity. However, I want to improve on that ideal gas assumption I am using in my model to the real gas. – S_H Nov 26 '12 at 6:05
In that case, like I've stated, you'll need to find some extra term that's added or subtracted to the ideal gas law that applies to the real gas you are talking about. That extra term is currently not there and physicists don't bother to find an extra correction because the ideal gas law works pretty well already. However, this gives more reason for you to try to find it! Perhaps you'll rewrite textbooks! :D – QEntanglement Nov 30 '12 at 17:25
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http://mathhelpforum.com/number-theory/160910-number-exponentiation.html
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# Thread:
1. ## number exponentiation
"Is there an exponent of number $3 ( 3^x )$ that ends with 0001 in decimal system?" Hope you understood.
It can be written like this :
$3^x=10000*k+1$. Find x
2. Originally Posted by teps
"Is there an exponent of number $3 ( 3^x )$ that ends with 0001 in decimal system?" Hope you understood.
It can be written like this :
$3^x=10000*k+1$. Find x
Hint: Use Euler's theorem. If you want the actual value of $x$, you will need to compute $\varphi(10000)$. See here for how to do that.
3. This can be done by a pigeon hole argument.
Choose 10001 different integers $\{x_1,x_2,\ldots,x_{10001}\}$. Since there are 10000 remainders upon division by 10000, two of the numbers $x$ and $y$ in our set must have the property that $3^x$ and $3^y$ have the same remainder when divided by 10000. Say $x>y$.
Hence $3^x-3^y = 10000k$ for some integer $k$, whereas $3^y(3^{x-y}-1) = 10000k$. But 10000 and $3^y$ are relatively prime, hence $3^{x-y}-1 = 10000l$ for some integer $l$, like you asked for.
4. Thank you. Got it
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http://mathhelpforum.com/algebra/117756-graphing-log-function.html
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# Thread:
1. ## Graphing log Function
Graph
y=log base 3 (x+7)
It asks to graph, but i am just having problem solving for points(Y)
I know that you can also graph this function in exponential form
3^y=x+7
however, i tried to substitute in for x in the log function, but getting wrong answers
x=-7
y=log base 3(-7+7) = 1
expo
x=-6
3^y=-6+7
3^y=1
y=1
I am substituting different numbers for x but getting same answer, in fact the -6 is correct not the -7.
How do i find y in a using the log function?
Thank You
2. Originally Posted by hovermet
Graph
y=log base 3 (x+7)
It asks to graph, but i am just having problem solving for points(Y)
I know that you can also graph this function in exponential form
3^y=x+7
however, i tried to substitute in for x in the log function, but getting wrong answers
x=-7
y=log base 3(-7+7) = 1 <<<<<< log(0) is not defined!
...
If you don't want to use a calculator you should get powers of 3 as argument of the log function:
$y = \log_3(x-7)$
Therefore:
$y = \log_3(8-7)~\implies~y = 0$
$y = \log_3(10-7)~\implies~y = 1$
$y = \log_3(16-7)~\implies~y = 2$
$y = \log_3\left(\frac{22}3-7\right)~\implies~y = -1$
If you are allowed to use a calculator use the base-change-formula to get the y-values for any x > 7:
$y = \log_3(x)~\implies~y=\dfrac{\ln(x)}{\ln(3)}$
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http://mathhelpforum.com/differential-geometry/168761-arcwise-connected-set.html
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# Thread:
1. ## arcwise connected set
Taylor proves the following (with one incomprehensible step), and Rosenlicht gives it as a problem, a problem!! Is the answer given? Do bears poop in the woods?
Any connected open subset of E^n is arcwise connected.
Definitions from Rosenlicht:
"A metric space E is connected if the only subsets of E which are both open and closed are E and 0. A subset S of a metric space is a connected subset if the subspace S is connected."
"A metric space E is said to be arcwise connected if, given any p,q eE, there is a continuous function f:[0,1] -> E such that f(0)=p , f(1)=q."
Rudin doesn't touch this.
2. What is the question in all of that?
3. Originally Posted by Hartlw
Taylor proves the following (with one incomprehensible step), and Rosenlicht gives it as a problem, a problem!! Is the answer given? Do bears poop in the woods?
Any connected open subset of E^n is arcwise connected.
Definitions from Rosenlicht:
"A metric space E is connected if the only subsets of E which are both open and closed are E and 0. A subset S of a metric space is a connected subset if the subspace S is connected."
"A metric space E is said to be arcwise connected if, given any p,q eE, there is a continuous function f:[0,1] -> E such that f(0)=p , f(1)=q."
Rudin doesn't touch this.
This seems to be true. What's the "incomprehensible step" in Taylor's (Taylor who?) proof?
Tonio
4. The question is how do you prove the statement about arcwise connectedness.
Taylor, "General Theory of Functions and Integration"
Taylor gives a long and involved proof, most of which can be followed.
From Taylor:
"Theorem: Let S be an open and connected set in R^k. Then, given any two distinct points x,y in S, there exists a polygonal arc lying in S and having x and y as its end points."
"Let A consist of x and all points a in s such that there exists a polygonal arc lying in S with x and a as its end points. We have to prove that y is in A. We shall assume the contrary and deduce a contradiction. Let B=S-A. Evidentally A unequal 0, B unequal 0, S = AUB, AXB = 0."
To be disconnected, AXB must equal zero (just proved) and AXB' = 0 and A'XB = 0. (Prime means the set of all accumulation points). Then
He proves A is open so that AXB'=0 (no accumulation points in A).
That leaves A'XB =0. That's hard. OK, I'll copy the damn thing:
"If we now prove that A'XB ==0, S will be expressed as the union of the separated sets A, B, contrary to the fact that S is connected, and so we shall have completed our proof. Suppose then that A'XB unequal zero, say z eA'XB. Then zeS, and there exists a spherical neighborhood of z lying entirely in S.This neighborhood must contain a point w in A because zeA'. There is then a polygonal arc L in S joining x to w. Let M be the line segment from w to z, (ends included). Now z is not on L because L is a subset of A and zeB."
Now I get stuck. Continuing with Taylor:
"The set LXM is compact, and hence there is a point y of L at minimum distance from z. We then obtain a polygonal arc joining x to z by proceeding from x along L until we come to y, and then following the straight line segment from y to z."
The rest then is easy:
"But this means that zeA, which is a contradiction. Hence we must conclude that A'XB=0, and the proof is complete."
comment: Not only is this an excercise in Rosenlicht, it is also an excercise in Buck, "Advanced Calculus," with of course, no answer. Are these people serious?
What I was really looking for is an understandable proof, or a reference to one. If someone could explain the underlined sentence, that would work and be greatly appreciated.
5. Originally Posted by Hartlw
The question is how do you prove the statement about arcwise connectedness.
Taylor, "General Theory of Functions and Integration"
Taylor gives a long and involved proof, most of which can be followed.
From Taylor:
"Theorem: Let S be an open and connected set in R^k. Then, given any two distinct points x,y in S, there exists a polygonal arc lying in S and having x and y as its end points."
"Let A consist of x and all points a in s such that there exists a polygonal arc lying in S with x and a as its end points. We have to prove that y is in A. We shall assume the contrary and deduce a contradiction. Let B=S-A. Evidentally A unequal 0, B unequal 0, S = AUB, AXB = 0."
To be disconnected, AXB must equal zero (just proved) and AXB' = 0 and A'XB = 0. (Prime means the set of all accumulation points). Then
He proves A is open so that AXB'=0 (no accumulation points in A).
That leaves A'XB =0. That's hard. OK, I'll copy the damn thing:
"If we now prove that A'XB ==0, S will be expressed as the union of the separated sets A, B, contrary to the fact that S is connected, and so we shall have completed our proof. Suppose then that A'XB unequal zero, say z eA'XB. Then zeS, and there exists a spherical neighborhood of z lying entirely in S.This neighborhood must contain a point w in A because zeA'. There is then a polygonal arc L in S joining x to w. Let M be the line segment from w to z, (ends included). Now z is not on L because L is a subset of A and zeB."
Now I get stuck. Continuing with Taylor:
"The set LXM is compact, and hence there is a point y of L at minimum distance from z. We then obtain a polygonal arc joining x to z by proceeding from x along L until we come to y, and then following the straight line segment from y to z."
The rest then is easy:
"But this means that zeA, which is a contradiction. Hence we must conclude that A'XB=0, and the proof is complete."
comment: Not only is this an excercise in Rosenlicht, it is also an excercise in Buck, "Advanced Calculus," with of course, no answer. Are these people serious?
What I was really looking for is an understandable proof, or a reference to one. If someone could explain the underlined sentence, that would work and be greatly appreciated.
Are you using "X" for intersection?? I know a proof of this theorem, but it's hard to comment on this one with the formatting. Is there any way you could tex it?
6. Originally Posted by Hartlw
Taylor proves the following (with one incomprehensible step), and Rosenlicht gives it as a problem, a problem!! Is the answer given? Do bears poop in the woods?
Any connected open subset of E^n is arcwise connected.
This ought to be a fairly straightforward result, quite reasonable to leave as a problem in a graduate text, and Taylor's proof seems unnecessarily complicated. (I find that surprising, because Taylor is usually an excellent author. His Introduction to functional analysis is still one of the best, 50+ years after its first publication.)
Start as Taylor does, by fixing a point $x\in S$, and defining A to be the set of points in S that can be reached by a path from x. Let B be the complement of A in S. Then A is open. Reason: suppose that $y\in A$; since S is open, there is a ball B(y,e) centred at y and contained in S. Then every point in that ball belongs to A, because you can connect it to y by a straight line which lies within the ball, and then from y there is a path to x.
A very similar argument shows that B is open: suppose that $z\in B$; there is a ball centred at z and contained in S. If that ball contains a point $w\in A$ then there is a path in S consisting of a straight line from z to w and then from w to x. So $z\in A$ – contradiction. Therefore the ball contains no points in A and hence lies entirely in B, showing that B is open.
Thus S is the disjoint union of the open sets A and B. But if S is connected then one of those sets must be empty. It can't be A, because $x\in A$. Therefore B is empty, which means that S is arcwise-connected.
The basic idea in all of this is that the composition of two paths is a path. If there is a path from p to q, and a path from q to r, then you can stitch them together to get a path from p to r.
7. Drexel28: X is intersection and e is epsilon (belongs to).
Opalg: Thanks so much for reply and very impeccable and transparent proof. The problem with clarity is that it enables someone to see a possible fault, a problem that dosn't occur with Taylor's proof.
If B is the complement of A in S, then B must contain accumulation points of A. In that case, a ball in B centered on an accumulation point can contain points of A without a contradiction.
8. Originally Posted by Hartlw
If B is the complement of A in S, then B must contain accumulation points of A. In that case, a ball in B centered on an accumulation point can contain points of A without a contradiction.
The point is that B cannot contain accumulation points of A, because that would give a contradiction. The sets A and B are both open and closed in S. Therefore either A and B are completely separate, with no accumulation points in common (which would mean that S is not connected) or B is empty (and therefore S is arcwise-connected).
9. Originally Posted by Opalg
The point is that B cannot contain accumulation points of A, because that would give a contradiction. The sets A and B are both open and closed in S. Therefore either A and B are completely separate, with no accumulation points in common (which would mean that S is not connected) or B is empty (and therefore S is arcwise-connected).
From Buck, Advanced Calculus: "Definition 4. A set in E^n is said to be connected if it is impossible to split S into two disjoint sets A and B, neither one empty, without having one of the sets contain a boundary point of the other."
So A and B cannot both be open and the argument for B being open fails because a ball exists in B (the boundary point) which can contain a point of A without a contradiction.
10. Originally Posted by Hartlw
From Buck, Advanced Calculus: "Definition 4. A set in E^n is said to be connected if it is impossible to split S into two disjoint sets A and B, neither one empty, without having one of the sets contain a boundary point of the other."
So A and B cannot both be open and the argument for B being open fails because a ball exists in B (the boundary point) which can contain a point of A without a contradiction.
In this problem they are both open, and the conclusion is that B must therefore be empty.
11. Originally Posted by Opalg
In this problem they are both open, and the conclusion is that B must therefore be empty.
You haven't proved B is open because B contains a boundary point of A, with a ball which can have a point in A without creating a contradiction.
12. Originally Posted by Hartlw
You haven't proved B is open because B contains a boundary point of A, with a ball which can have a point in A without creating a contradiction.
I don't want to continue saying the same thing much more. But for the last time, B does not contain a boundary point of A. All the boundary points of A lie in A itself, and therefore do not belong to B.
13. Originally Posted by Opalg
I don't want to continue saying the same thing much more. But for the last time, B does not contain a boundary point of A. All the boundary points of A lie in A itself, and therefore do not belong to B.
But you proved A was open, so its boundary points have to be in B.
From Wolfram Math World:
A connected set is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.
I suspect that's why Taylor is so obstruse: You have to get past the boundary points.
14. Originally Posted by Hartlw
But you proved A was open, so its boundary points have to be in B.
From Wolfram Math World:
A connected set is a set which cannot be partitioned into two nonempty subsets such that each subset has no points in common with the set closure of the other.
I suspect that's why Taylor is so obstruse: You have to get past the boundary points.
To get really nasty about this, how do you know that A isn't empty? Let x be a point in S with a spherical neighborhood. How do you know you can connect from x to any point in the spherical neighborhood without assuming what you are trying to prove?
15. Originally Posted by Hartlw
To get really nasty about this, how do you know that A isn't empty? Let x be a point in S with a spherical neighborhood. How do you know you can connect from x to any point in the spherical neighborhood without assuming what you are trying to prove?
Taylor helps us out here:
Two distinct points in R^k determine a unique straight line- the line joining these two points. If x1 and x2 are the points, the parametric equation of the line is: x=(1-t)x1+x2, t in [0,1]. So if x is any point in S and w is a point in its spherical neighborhood, there is a line from x to w and A is not empty.
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http://mathoverflow.net/questions/111121/a-too-easy-normalization-of-a-lax-funtor-between-2-categories/111657
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## A (too easy) normalization of a lax-funtor between 2-categories ?
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Let $(F, \phi): \mathscr{A}\to \mathscr{B}$ a lax-functor between 2-categories. In the setting of 2-categories the axiom of lax-functor become:
Ul) $1: F(f)=F(f)\circ 1_{F(X)}\xrightarrow{1\circ \phi_X}F(f)\circ F(1_X)\xrightarrow{\phi_{f, 1}} F(f\circ 1_X)= F(f)$.
For $f: X \to Y$ in $\mathscr{A}$.
Ur) $1: F(f)= 1_{F(Y)}\circ F(f)\xrightarrow{\phi_X\circ 1} F(1_Y)\circ F(f)\xrightarrow{\phi_{1, f}} F( 1_Y \circ f)= F(f)$.
For $f: X \to Y$ in $\mathscr{A}$.
UA) The compositions $F(h)\circ F(g)\circ F(f) \xrightarrow{\phi_{h, g}\circ 1} F(h\circ g)\circ F(f) \xrightarrow{\phi_{h\circ g, f}\circ 1} F(h\circ g\circ f)$ and
$F(h)\circ F(g)\circ F(f) \xrightarrow{1\circ \phi_{g, f}} F(h)\circ F(g\circ f) \xrightarrow{\phi_{h, g\circ f}} F(h\circ g\circ f)$ are equal.
For componibile morphisms $h, g, f$.
.
I call $(F, \phi)$ normal (or unitary) if in the axioms $(Ul)$ and $(Ur)$ above all arrows are identities.
Gived $(F, \phi)$ (general) I define a normal lax.functor $(\tilde{F}, \tilde{\phi})$ that is the some of $F$ on objects and on non-identity morphisms, with of course $\tilde{F}(1_X)=1_{F(X)}$, $\widetilde{\phi}_{X}= 1: \tilde{F}(1_X)\to 1_{F(X)}$, and with $\tilde{\phi}_{g, f}$ defined as:
$\phi_{g, f}: F(g)\circ F(f)\to F(g\circ f)$ if $f$ and $g$ arent identity, and the obvious identity if $f$ or $g$ is a identity. I checked (easly) that axiom $(UA)$ is true for $(\tilde{F}, \tilde{\phi})$, then $(\tilde{F}, \tilde{\phi})$ is a lax.funtor.
I ask if this (very easy normalization) is just know in literature ( I dont know), and if is right (I'm sure its right, but I'm no too sure of myself).
Edit: If in $(Ul)$ all arrows are identies, this imply the some in $(Ur)$ (i.e. $(F, \phi)$ is normal)?
My motivations is a generalization of the J.W. Gray concept of quasi-functors to lax.functors (Gray gived this definition for 2-functors).
EDit As Jonathan Chiche observed, these is the obvious condition than a lax.functor induce hom-functors between hom-categories (I am ashamed for this rough oversight). Anyway if the canonical morphism $\phi_X: 1_{FX}\Rightarrow F(1_X)$ is a isomorphism then the mine definition preserving this funtorialiy, only need a easy correction based on the following observation: the cells of type $\sigma: 1_{FX}\Rightarrow W$ are in bijections with the cells of type $\sigma': F(1_X)\Rightarrow W$ and the cells like $\tau: W\Rightarrow 1_{FX}$ are in bijections with the cells of type $\tau': W\Rightarrow F(1_X)$. With this correction I seems that also the naturality of $\tilde{\phi}$ work well.
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## 2 Answers
It seems to me that your question woud be clearer if you stated more precisely the axioms for lax functors. Anyway, while I am unsure as to what you have in mind, I guess that "lax functors" and "normal lax functors" have the same meaning for you and me. You may therefore be interested in Lemma 4.2. and the beginning of the proof of Theorem 6.3. of the paper Nerves and classifying spaces for bicategories by Carrasco, Cegarra and Garzón. Hope this helps.
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Of course the axioms are the classical: J.Benabou, Introduction to bicategories, (M1) and (M2) p. 30. Thank your for the reference. A lax functors (for me) is what J.Benabou call morphisms (of bicategories) or what J.W. Gray call "pseudo.functors" in its "Formal category theory" LNM 391, 1974. A lax.functor $F$ (for me) is normal if the canonical cell $1_{F(X)}\Rightarrow F(1_X)$ is the identity. – Buschi Sergio Nov 1 at 10:55
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Well, I know what the axioms are, but part of my point is that some of these axioms are missing in your question. For instance, you don't mention the fact that lax functors induce functors between the categories of 1-cells. I don't think this condition is automatically satisfied as soon as the ones you mention are, even in the realm of (strict) 2-categories. – Jonathan Chiche Nov 1 at 11:50
THank you very much Jonatan (I'm no a researcher, I study alone after job, and sometime I'm tired and I take misunderstandings for stress or enthusiasm). Anyway for pseudonfunctors I seems that is possible correct the HOm-functors (functors inducted between Hom-categories) – Buschi Sergio Nov 1 at 17:37
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I think Jonathan is right: your "normalized" $\tilde{F}$ may not be a lax functor, because you may not be able to extend it functorially on the 2-cells. Suppose that $\mathcal{A}$ contains a 1-cell $f:X\to X$ with a 2-cell $\alpha: f\to 1_X$, whereas $\mathcal{B}$ contains no 2-cells from $F(f)$ to $1_{F(X)}$.
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THis is no possible, consider $\phi_X^{-1}\ast F(\alpha): F(f) \to F(1_X)\to 1_{F(X)}$, in the last Edit on my inital question I admit that the "Lax" condiction is too strong, then I limitate the conjecture on pseudo-functors (its enought pseudo only about unity condiction). LEt me know if languaage is not understable. – Buschi Sergio Nov 6 at 16:12
For a pseudofunctor this normalization is right (by the a well know coherence theorem), I used this in a my old answere to a your question abot Gray tensor product (i'm still studing about..). But I think that condiction is right also for lax functors pseudo on unity $\phi_X: 1_{F(AX)}\cong F(1_X)$. – Buschi Sergio Nov 6 at 16:17
Yes, for a normal lax functor it may work. – Mike Shulman Nov 7 at 21:09
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http://mathoverflow.net/questions/110784/computing-the-limit-of-a-certain-recursively-defined-sequence/110789
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## Computing the limit of a certain recursively defined sequence
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The following is not exactly a research question (it was originated from manufacturing of exercises for calculus), and has no other motivation than explaining a phenomenon. I apologize if it is inappropriate (and will quickly remove it).
Consider the sequence of real numbers defined recursively as follows $$u_0:=\lambda > -1\ ,$$ $$u_{n+1}=\frac{2u_n}{1+\sqrt{1+2^{-n}u_n}}\ .$$ Numerical evidence suggests that $u_n$ always converges to $\log(1+\lambda)$. I imagine there should be a simple explanation. How can one prove it?
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u_(n+1) is very close to the harmonic mean of 2^n and u_n when u_n/2^n is small. Perhaps that helps for another derivation? Gerhard "Ask Me About System Design" Paseman, 2012.10.26 – Gerhard Paseman Oct 26 at 22:51
## 1 Answer
Multiplying top and bottom by the conjugate and simplifying (assuming $u_n \neq 0$) we get: $$u_{n+1}=2^{n+1}(\sqrt{1+2^{-n}u_n}-1).$$ Calling $v_n:=\frac{u_n}{2^n}+1$ we have the recursion: $v_{n+1}=\sqrt{v_n}$ and therefore $v_n=v_0^{2^{-n}}$. Going back to $u$´s we have $u_n=2^n[(1+\lambda)^{2^{-n}}-1]$. Finally: $$\lim_{n\to\infty} u_n=\lim_{h\to 0}\frac{(1+\lambda)^h-(1+\lambda)^0}{h}= \ln (1+\lambda).$$
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Your first display is valid even when $u_n=0$, and this case occurs if and only if $\lambda=0$. – GH Oct 26 at 22:20
@CH: That´s right, thanks. – Ramiro de la Vega Oct 26 at 22:24
that's very nice – Pietro Majer Oct 26 at 22:37
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So $x=u_n$ is a solution of $(1+\frac{x}{N})^N=1+\lambda$ (with $N:=2^n$), which we may see as an approximated problem for $e^x=1+\lambda$... – Pietro Majer Oct 27 at 7:47
@Pietro: that´s a nice way to look at it. – Ramiro de la Vega Oct 27 at 18:46
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http://mathoverflow.net/questions/100552/do-cartesian-fibrations-preserve-pullbacks
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## Do Cartesian fibrations preserve pullbacks?
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If $F:C \to D$ is a Cartesian fibration of $\infty$-categories, I would like to show that $F$ preserves pullbacks. This seems intuitively clear, but I haven't found it in HTT (but perhaps I missed it).
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## 1 Answer
This is false already for ordinary fibrations of 1-categories. It is a theorem that if $D$ has pullbacks and $F:C\to D$ is a fibration, then $C$ has and $F$ preserves pullbacks if and only if each fiber of $F$ has pullbacks and the reindexing functors preserve them. Thus, the Grothendieck construction of any functor $D^{\mathrm{op}}\to \mathrm{Cat}$ which does not factor through $\mathrm{Cat}_{\mathrm{pullbacks}}$ will give a counterexample.
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Thanks Mike! Not only does this explain why I shouldn't find it in HTT, it also tells me the modified statement I should prove :). – David Carchedi Jun 27 at 6:31
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http://mathoverflow.net/questions/37214?sort=votes
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## Why aren’t there more classifying spaces in number theory?
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Much of modern algebraic number theory can be phrased in the framework of group cohomology. (Okay, this is a bit of a stretch -- much of the part of algebraic number theory that I'm interested in...). As examples, Cornell and Rosen develop basically all of genus theory from cohomological point of view, a significant chunk of class field theory is encoded as a very elegant statement about a cup product in the Tate cohomology of the formation module, and Neukirch-Schmidt-Wingberg's fantastic tome "Cohomology of Number Fields" convincingly shows that cohomology is the principal beacon we have to shine light on prescribed-ramification Galois groups.
Of course, we also know that group cohomology can be studied via topological methods via the (topological) group's classifying space. My question is:
Question: Why doesn't this actually happen?
More elaborately: I'm fairly well-acquainted with the "Galois cohomology for number theory" literature, and not once have I come across an argument that passes to the classifying space to use a slick topological trick for a cohomological argument or computation (though I'd love to be enlightened). On the other hand, for example, are things like Tyler's answer to my question
http://mathoverflow.net/questions/15375/coboundary-representations-for-trivial-cup-products
which strikes me as saying that there may be plenty of opportunities to carry over interesting constructions and/or lines of reasoning from the topological side to the number-theoretic one.
Maybe the classifying spaces for gigantic profinite groups are too hideous to think about? (Though there's plenty of interesting Galois cohomology going on for finite Galois groups...). Or maybe I'm just ignorant to the history, and that indeed the topological viewpoint guided the development of group cohomology and was so fantastically successful at setting up a good theory (definition of differentials, cup/Massey products, spectral sequences, etc.) that the setup and proofs could be recast entirely without reference to the original topological arguments?
(Edit: This apparently is indeed the case. In a comment, Richard Borcherds gives the link http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.bams/1183537593 and JS Milne suggests MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280)., both of which look like good reads.)
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5
My impression is that many automorphic forms describe cohomology classes on arithmetic manifolds or orbifolds, which are classifying spaces for their fundamental groups. Also, the class number of a quadratic imaginary number field is the number of cusps of the corresponding Bianchi orbifold, which could be turned into a cohomological statement. But I'm not sure this is related to your question. – Agol Aug 31 2010 at 4:32
You might be slightly interested in the answer by Josh Roberts in this thread: mathoverflow.net/questions/10879/… – David Corwin Aug 31 2010 at 4:36
@Agol: Well, perhaps not directly, but it sounds fascinating regardless, so +1. If there's an answer there to be elaborated on, I'd love to see it. This will save me from asking the new question "What was Agol talking about when he said..." :) – Cam McLeman Aug 31 2010 at 12:29
1
Related: Trees and more general buildings can give you the total spaces of the universal bundle for the group they were constructed from. In any case their homology is a G-module. And buildings are used in arithmetic, not directly for Galois groups though, as far as I know, you rather look e.g. at SL_n(L) and then get an action of Gal(L/k) on the result... – Peter Arndt Sep 1 2010 at 13:09
## 3 Answers
Classifying spaces are widely used in algebraic number theory, but in slightly disguised form. A classifying space is really just an approximation to the classifying topos of a group. However the classifying topos is just the category of G-sets, which is exactly what one uses in defining group cohomology and so on. Or to put it another way, all the useful information in the classifying space is already contained in the category of G-sets.
The comment at the end of the question is correct: group cohomology was discovered as the cohomology of the classifying space, and the topological constructions were then turned into algebraic constructions and removed from the theory. So in some sense all the group cohomology calculations are implicitly using the classifying space.
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+1: Thanks for this interpretation -- I think this makes a lot of sense, though I'm going to spend some time working it out more carefully. Do you have a reference for the historical comment? I'd like to understand the historical development better. – Cam McLeman Aug 31 2010 at 12:43
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From my Class Field Theory Notes p86. In the mid-1930s, Hurewicz showed that the homology groups of an "aspherical space" $X$ depend only on the fundamental group $\pi$ of the space. Thus one could think of the homology groups $H_{r}(X,\mathbb{Z})$ of the space as being the homology groups $H_{r}(\pi,\mathbb{Z})$ of the group $\pi$. It was only in the mid-1940s that Hopf, Eckmann, Eilenberg, MacLane, Freudenthal and others gave purely algebraic definitions of the homology and cohomology groups of a group $G$. – JS Milne Aug 31 2010 at 14:27
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Continued: It was then found that $H^{1}$ coincided with the group of crossed homomorphisms modulo principal crossed homomorphisms, and $H^{2}$ with the group of equivalence classes of "factor sets", which had been introduced much earlier (e.g., I. Schur, \"{U}ber die Darstellung der endlichen$\ldots$ , 1904; O. Schreier, \"{U}ber die Erweiterungen von Gruppen, 1926; R. Brauer, \"{U}ber Zusammenh\"{a}nge$\ldots$ , 1926). For more on the history, see MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280). – JS Milne Aug 31 2010 at 14:29
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projecteuclid.org/euclid.bams/1183537593 – Richard Borcherds Aug 31 2010 at 14:32
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As a topologist, my view is that group cohomology of interest to number theorests seems to generally be with non-trivial module coefficients. Many of the tricks topologists employ to study spaces do not apply in this setting, and indeed there are some topologists whose view is that once the coefficients are non-trivial then one is "just doing algebra." But attitudes can change: cohomology of profinite groups was perhaps viewed similarly, but in the "chromatic" (number theoretic) framework for stable homotopy, cohomology of profinite groups figures prominently. For this reason, some topologists work hard to show that techniques which work for finite groups pass sensibly to the profinite settings.
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One interesting special case: if I remember correctly, the $E^2$-term of the Adams-Novikov spectral sequence is isomorphic to the continuous cohomology of the Morava stabilizer group (the automorphism group of the Honda formal group law) with coefficients the Lubin-Tate universal deformation ring. For the relevance of Lubin-Tate theory in class field theory, this might be called a number-theoretic object. But the usual methods for computation are of a more topological kind. See math.rochester.edu/people/faculty/doug/… , especially chapter 7. – Lennart Meier Aug 31 2010 at 11:33
This answer agrees very much with my impression: it's not so much finite or pro-finite that's the issue, but the focus on trivial or non-trivial coefficients. I do wish topologists would think more about non-trivial coefficients, since they're so good at coming up with ideas useful for arithmeticians. – Minhyong Kim Aug 31 2010 at 12:06
@Dev Sinha: Okay, fair point. On the other hand, there are a lot of important statements in group cohomology which involve trivial coefficients: The Schur multiplier (or Hopf's Integral Homology Formula) comes to mind, as does the interpretation of the relation rank of a pro-$p$-group $G$ as $\operatorname{dim}H^2(G,\mathbb{F}_p)$. – Cam McLeman Aug 31 2010 at 12:25
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I must admit, I twisted something in my head. The cohomology of the Morava stabilizer group is the $E^2$ term of a spectral sequence computing the stable homotopy groups of the $K(n)$-localized sphere, while in the book by Ravenel the Adams-Novikov $E^2$-term for the ($p$-localized) sphere is computed. But I think, my point remains probably valid. – Lennart Meier Aug 31 2010 at 12:37
@Cam: certainly there are plenty of important statements about trivial coefficients, and for some it is helpful to think about the classifying space as a topological object. I myself have been doing computations in cohomology of symmetric groups (and others) thinking geometrically. – Dev Sinha Aug 31 2010 at 17:30
show 2 more comments
I suppose we should also mention algebraic k-theory. Quillen defined the k-groups as the homotopy groups of certain classifying spaces. For a unital, associative ring $R$, $$K_n(R):=\pi_n(BGL(R)^+),$$ where $GL(R)$ is the direct limit of the general linear groups and $^+$ is Quillen's plus-construction on spaces whose fundamental groups have perfect subgroups. Now I'm not sure how useful this has been for computation (these are homotopy groups, after all), but the classifying space is used. And there are number theory applications of algebraic k-theory.
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## Is Soergel’s proof of Kazhdan-Lusztig positivity for Weyl groups independent of other proofs?
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Let $(W, S)$ be a Coxeter system. Soergel defined a category of bimodules $B$ over a polynomial ring whose split Grothendieck group is isomorphic to the Hecke algebra $H$ of $W$. Conjecturally, the image of certain indecomposable (projective?) bimodules in $B$ is the well-known Kazhdan-Lusztig basis of $H$. Assuming the conjecture, Soergel showed that the coefficients of the Kazhdan-Lusztig polynomials of $W$ are given by the dimensions of certain Hom-spaces in $B$. It follows that these coefficients are non-negative, which was already known by work of Kazhdan-Lusztig in the Weyl group case by linking these coefficients to intersection cohomology of the corresponding Schubert varieties.
Soergel proved this conjecture in 1992 for $W$ a Weyl group, and Härterich proved it in 1999 for $W$ an affine Weyl group. Unfortunately, I can't access the first paper, and the second paper is in German, so I don't know anything about either of these proofs.
Question: Do these proofs depend on the relationship of the coefficients of the K-L polynomials to intersection cohomology, or are they independent of the corresponding machinery?
(The reason I ask is that I am potentially interested in relating known combinatorial proofs of positivity to Soergel's work, and I want to get an idea of how much machinery I would need to learn to do this.)
Edit: Soergel's 1992 paper is here, if only I had the appropriate journal access. If anybody does and would like to send me this paper, that would be excellent - my contact information is at a link on my profile.
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It seems that you can find Soergel's paper at an open access site called DigiZeitschriften. A link to the journal containing this paper is digizeitschriften.de/main/dms/toc/?IDDOC=502899. Issue number 429 has the relevant paper. – Mike Skirvin Jun 24 2010 at 21:06
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Thanks! Unfortunately, the paper only brings me more bad news; the proof depends on results from another paper of Soergel's (jstor.org/stable/1990960) which is in German... – Qiaochu Yuan Jun 24 2010 at 21:57
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@Mike: This is a good site for online viewing, though U.S. users might want to read the copyright notice there before photocopying. (That's like illegal downloads of music, videos.) On his Webpage, Wolfgang suggests that reprints might be sent by his secretary: home.mathematik.uni-freiburg.de/soergel/ (Both reprints and secretaries are however vanishing breeds.) @Qiaochu: In any language you'll find Soergel's papers challenging to read, if that's any consolation. But they are often important papers, as are those of his adviser Jantzen in Hamburg. And some papers are intertwined. – Jim Humphreys Jun 24 2010 at 22:12
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Qiaochu: 1 As someone who had to learn mathematical German in order to understand Jantzen's book on primitive ideals (which is still the best source for Harish-Chandra bimodules), I feel for you. At the same time, it's well worth it! 2 Can you, please, give a reference for "known combinatorial proofs of positivity"? I am not aware of any that work for general $y,w$ even in the case of the $A$ series. – Victor Protsak Jun 25 2010 at 5:00
Neither am I! The best result I can track down is dmtcs.org/dmtcs-ojs/index.php/proceedings/article/… . – Qiaochu Yuan Jun 25 2010 at 15:13
## 2 Answers
My understanding is that Soergel's approach applies just to finite Weyl groups and not directly to other finite Coxeter groups (or more generally), since what he can actually prove depends on some of the geometric machinery used to prove the Kazhdan-Lusztig Conjecture. The same must be true of the 1999 thesis work of his student Martin Harterich involving affine Weyl groups, which doesn't seem to have been formally published. In those situations the coefficients of KL polynomials were seen to be nonnegative in the early steps taken by Kazhdan and Lusztig toward understanding their conjecture via Schubert varieties: they occur as dimensions of certain cohomology groups.
Later on, Soergel made his program more explicit for proving the nonnegativity for arbitrary Coxeter groups using his more algebraic/categorical setting of bimodules: MR2329762 (2009c:20009) 20C08 (20F55) Soergel,Wolfgang (D-FRBG), Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln ¨uber Polynomringen. (German. English, German summaries) [Kazhdan-Lusztig polynomials and indecomposable bimodules over polynomial rings] J. Inst. Math. Jussieu 6 (2007), no. 3, 501–525. This is in a French journal but written in German; the helpful review by Ulrich Goertz is however in English if you have access to MathSciNet. (In any case, J. Reine Angew. Math. has become super-expensive for libraries, so print or online access gets tricky.)
A helpful follow-up paper (in English) by Soergel's later student Peter Fiebig (now at Erlangen) should also be consulted, though it is still unclear to me how far one can get with Soergel's conjectural approach in this spirit: MR2395170 (2009g:20087) 20F55 (20C08) Fiebig, Peter (D-FRBG), The combinatorics of Coxeter categories. Trans. Amer. Math. Soc. 360 (2008), no. 8, 4211–4233. (Fiebig's papers are on arXiv, by the way.)
I'll have to take another look at this literature, but in any case the nonnegativity of coefficients of KL polynomials for arbitrary Coxeter groups (predicted in 1979 by Kazhdan and Lusztig) remains an intriguing question. The general setting is far from the kind of representation theory or geometry one encounters in Lie theory, but a purely combinatorial approach seems at the moment unlikely to succeed.
ADDED: Special cases where Kazhdan-Lusztig polynomials have been computed are discussed in section 7.12 of my 1990/1992 book on reflection groups and Coxeter groups. In particular, noncrystallographic finite Coxeter groups all yield nonnegative coefficients. For dihedral groups, the polynomials are all 1, while for type `$H_3$` the computer tables found by Mark Goresky are still on his Webpage at IAS. The 1987 paper in J. Algebra by Dean Alvis which I cited involved his unpublished computer results on the polynomials for `$H_4$`, for which his current Webpage gives details: http://mypage.iusb.edu/~dalvis/h4data/index.html
These polynomials were later recovered by Fokko du Cloux using his computer system Coxeter: see his last published paper MR2255133 (2007e:20010) 20C08 (20F55) du Cloux, Fokko (F-LYON-ICJ), Positivity results for the Hecke algebras of noncrystallographic finite Coxeter groups. J. Algebra 303 (2006), no. 2, 731–741.
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Perhaps I can supplement Jim's answer a little.
In the paper "Kazhdan-Lusztig-Polynome und unzerlegbare Bimoduln uber Polynomringen" Soergel shows that there are certain graded indecomposable bimodules over a polynomial ring (now known as Soergel bimodules) which categorify the Hecke algebra. (Note that these are not projective!!)
That is, the indecomposable objects are classified (up to shifts and isomorphism) by the Weyl group, and one has an isomorphism between the Hecke algebra and the split Grothendieck group of the category of Soergel bimodules.
As a consequence, one obtains a basis for the Hecke algebra which is positive in the standard basis (as follows from the construction of the isomorphism of the Grothendieck group with the Hecke algebra) and has positive structure constants (because it is a categorification).
Soergel conjectures that this basis is in fact the Kazhdan-Lusztig basis, which would imply positivity in general.
Up until now there are only two cases when one can verify Soergel's conjecture:
• when one has some sort of geometry (in which case one can show that the indecomposable Soergel bimodules are the equivariant intersection cohomology of Schubert varieties, and then use old results of Kazhdan and Lusztig). This shows Soergel's conjecture for Coxeter groups associated to Kac-Moody groups (in particular finite and affine Weyl groups).
• when the combinatorics is very simple (i.e. for dihedral groups (Soergel) or universal Coxeter groups (Fiebig, Libedinsky)).
Hence, up until now there are no examples where Soergel's conjecture has yielded positivity when it was not known by other means. Also note that in the vast majority of cases, the proof using Soergel bimodules is strictly more complicated than the geometry proof, as one needs an extra step to get from geometry to Soergel bimodules.
Soergel's conjecture would however have more far reaching consequences than a proof that Kazhdan-Lusztig polynomials have positive coefficients. For example it provides a natural "geometry" for arbitrary Coxeter groups. For example, generalising some sort of Soergel bimodules to complex reflection groups would yield a natural setting for the study of "spetses" (unipotent characters associated to complex reflection groups).
One should also note that Dyer has developed a very similar conjectural world associating commutative algebra categories to Coxeter groups. He instead considers modules over the dual nil Hecke ring (which is the analogue of the cohomology of a flag variety), and has many nice results and conjectures. (Much of his work considers more general orders than the Bruhat order, and so will probably come in handy soon ...!)
While I am at it I should mention Peter Fiebig's theory of Braden-MacPherson sheaves on moment graphs. This is (in a sense made precise in one of Peter's papers) a local version of Soergel bimodules, and hence many questions become more natural on the moment graph.
Finally, one should mention the recent work of Elias-Khovanov and Libedinsky, which give generators and relations for the monoidal category of Soergel bimodules for certain Coxeter groups. (Elias-Khovanov in type A and Libedinsky in right-angled type.) These are very interesting results, but it is unclear to what extent they can be used to attack Soergel's conjecture.
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# Find the smallest value of $a$ that a solution for this equation exists?
How to find the smallest value of $a$ that a solution for this equation exists? $2\ln x + \frac{5}{x} < a$
That are the steps I've done, and have no idea what to do next:
$2\ln x < -\frac{5}{x} + a$
$\ln x^2 < -\frac{5}{x} + a$
$e^{-\frac{5}{x} + a} < x^2$
$e^{-5} \cdot e^{\frac{1}{x}} \cdot e^a < x^2$
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$u<v$ if and only if $e^u<e^v$. You've instead concluded $e^v<e^u$. – Thomas Andrews Jan 23 at 12:52
## 1 Answer
As Thomas Andrews has pointed out, the inequalities in your third and fourth lines are the wrong way around. Having said that, I'm not sure whether your approach will work even if you make these corrections. Instead I present an alternative method which uses a little bit of differential calculus.
Let $f(x) = 2\ln x + \frac{5}{x}$ and note that it has domain $(0, \infty)$. The only way the inequality $f(x) < a$ can't have solutions is if $f(x) \geq a$ for all $x \in (0, \infty)$. In particular, if we know that $f$ has a global minimum (it need not have one), then for any $a$ less than or equal to that, the inequality will have no solutions.
You should be able to use calculus to determine whether $f$ has a global minimum, and if so, what it is.
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http://math.stackexchange.com/questions/236632/related-rates-problem-with-an-absolute-answer
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# Related rates, problem with an absolute answer
A round shaped athletic track (centre in a point C) has a radius of $50.0$ $mts$. An athlete runs around the track and a referee is going to time from a starting point A. If the athlete is in a point B, 'how fast does the area of the triangle ABC increases when the central angle is $\pi/4$?
I started by stating that $h=r\sin \alpha$ and $b=r$
Then I said that $A={r^2 \over 2}\sin \alpha$
Implicitly deriving...
$${dA \over dt}={r^2 \over 2} \cos \alpha {d\alpha \over dt}$$ Replacing:
$${dA \over dt}=625 \sqrt2 {d\alpha \over dt}$$
But I don't know how to get ${d\alpha \over dt}$? I even tried $l=r*\alpha$
$${dl \over dt} = r{d\alpha \over dt}$$
And then replacing again:
$${dA \over dt}={25 \over 2} \sqrt2 {dl \over dt}$$
But then again I have no way out of there.
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## 1 Answer
You can’t really say more than $${dA \over dt}=625 \sqrt2 ~\frac{d\alpha}{dt}\tag{1}$$ unless you have some information on the runner’s speed. The most you can do is rewrite $(1)$ directly in terms of his speed. If he’s running at a speed of $v$ metres per second, then $$v=r\frac{d\alpha}{dt}=50\frac{d\alpha}{dt}\;,$$ and $$\frac{dA}{dt}=\frac{625\sqrt2}{50}v=\frac{25\sqrt2 v}2\;.\tag{2}$$ Although they say essentially the same thing, I slightly prefer $(2)$ to $(1)$, since linear units are a little more familiar in this context than angular units.
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Why did you write that $v=r {d\alpha \over dt}?$ – ChairOTP Nov 13 '12 at 20:07
@ChairOTP: Because it’s true. The arc of the circle that subtends an angle $\alpha$ has length $r\alpha$, and $v$ is the rate of change of that length. – Brian M. Scott Nov 13 '12 at 20:13
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http://www.nag.com/numeric/cl/nagdoc_cl23/html/G07/g07eac.html
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# NAG Library Function Documentnag_rank_ci_1var (g07eac)
## 1 Purpose
nag_rank_ci_1var (g07eac) computes a rank based (nonparametric) estimate and confidence interval for the location argument of a single population.
## 2 Specification
#include <nag.h>
#include <nagg07.h>
void nag_rank_ci_1var (Nag_RCIMethod method, Integer n, const double x[], double clevel, double *theta, double *thetal, double *thetau, double *estcl, double *wlower, double *wupper, NagError *fail)
## 3 Description
Consider a vector of independent observations, $x={\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)}^{\mathrm{T}}$ with unknown common symmetric density $f\left({x}_{i}-\theta \right)$. nag_rank_ci_1var (g07eac) computes the Hodges–Lehmann location estimator (see Lehmann (1975)) of the centre of symmetry $\theta $, together with an associated confidence interval. The Hodges–Lehmann estimate is defined as
$θ^=median xi+xj2,1≤i≤j≤n .$
Let $m=\left(n\left(n+1\right)\right)/2$ and let ${a}_{\mathit{k}}$, for $\mathit{k}=1,2,\dots ,m$ denote the $m$ ordered averages $\left({x}_{i}+{x}_{j}\right)/2$ for $1\le i\le j\le n$. Then
• if $m$ is odd, $\stackrel{^}{\theta }={a}_{k}$ where $k=\left(m+1\right)/2$;
• if $m$ is even, $\stackrel{^}{\theta }=\left({a}_{k}+{a}_{k+1}\right)/2$ where $k=m/2$.
This estimator arises from inverting the one-sample Wilcoxon signed-rank test statistic, $W\left(x-{\theta }_{0}\right)$, for testing the hypothesis that $\theta ={\theta }_{0}$. Effectively $W\left(x-{\theta }_{0}\right)$ is a monotonically decreasing step function of ${\theta }_{0}$ with
$mean W=μ= nn+14, varW=σ2= nn+12n+124.$
The estimate $\stackrel{^}{\theta }$ is the solution to the equation $W\left(x-\stackrel{^}{\theta }\right)=\mu $; two methods are available for solving this equation. These methods avoid the computation of all the ordered averages ${a}_{k}$; this is because for large $n$ both the storage requirements and the computation time would be excessive.
The first is an exact method based on a set partitioning procedure on the set of all ordered averages $\left({x}_{i}+{x}_{j}\right)/2$ for $i\le j$. This is based on the algorithm proposed by Monahan (1984).
The second is an iterative algorithm, based on the Illinois method which is a modification of the regula falsi method, see McKean and Ryan (1977). This algorithm has proved suitable for the function $W\left(x-{\theta }_{0}\right)$ which is asymptotically linear as a function of ${\theta }_{0}$.
The confidence interval limits are also based on the inversion of the Wilcoxon test statistic.
Given a desired percentage for the confidence interval, $1-\alpha $, expressed as a proportion between $0$ and $1$, initial estimates for the lower and upper confidence limits of the Wilcoxon statistic are found from
$Wl=μ-0.5+σΦ-1α/2$
and
$Wu=μ+ 0.5+σ Φ-11-α /2,$
where ${\Phi }^{-1}$ is the inverse cumulative Normal distribution function.
${W}_{l}$ and ${W}_{u}$ are rounded to the nearest integer values. These estimates are then refined using an exact method if $n\le 80$, and a Normal approximation otherwise, to find ${W}_{l}$ and ${W}_{u}$ satisfying
$PW≤Wl≤α/2 PW≤Wl+1>α/2$
and
$PW≥Wu≤α /2 PW≥Wu- 1>α /2.$
Let ${W}_{u}=m-k$; then ${\theta }_{l}={a}_{k+1}$. This is the largest value ${\theta }_{l}$ such that $W\left(x-{\theta }_{l}\right)={W}_{u}$.
Let ${W}_{l}=k$; then ${\theta }_{u}={a}_{m-k}$. This is the smallest value ${\theta }_{u}$ such that $W\left(x-{\theta }_{u}\right)={W}_{l}$.
As in the case of $\stackrel{^}{\theta }$, these equations may be solved using either the exact or the iterative methods to find the values ${\theta }_{l}$ and ${\theta }_{u}$.
Then $\left({\theta }_{l},{\theta }_{u}\right)$ is the confidence interval for $\theta $. The confidence interval is thus defined by those values of ${\theta }_{0}$ such that the null hypothesis, $\theta ={\theta }_{0}$, is not rejected by the Wilcoxon signed-rank test at the $\left(100×\alpha \right)%$ level.
## 4 References
Lehmann E L (1975) Nonparametrics: Statistical Methods Based on Ranks Holden–Day
Marazzi A (1987) Subroutines for robust estimation of location and scale in ROBETH Cah. Rech. Doc. IUMSP, No. 3 ROB 1 Institut Universitaire de Médecine Sociale et Préventive, Lausanne
McKean J W and Ryan T A (1977) Algorithm 516: An algorithm for obtaining confidence intervals and point estimates based on ranks in the two-sample location problem ACM Trans. Math. Software 10 183–185
Monahan J F (1984) Algorithm 616: Fast computation of the Hodges–Lehman location estimator ACM Trans. Math. Software 10 265–270
## 5 Arguments
1: method – Nag_RCIMethodInput
On entry: specifies the method to be used.
${\mathbf{method}}=\mathrm{Nag_RCI_Exact}$
The exact algorithm is used.
${\mathbf{method}}=\mathrm{Nag_RCI_Approx}$
The iterative algorithm is used.
Constraint: ${\mathbf{method}}=\mathrm{Nag_RCI_Exact}$ or $\mathrm{Nag_RCI_Approx}$.
2: n – IntegerInput
On entry: $n$, the sample size.
Constraint: ${\mathbf{n}}\ge 2$.
3: x[n] – const doubleInput
On entry: the sample observations, ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
4: clevel – doubleInput
On entry: the confidence interval desired.
For example, for a $95%$ confidence interval set ${\mathbf{clevel}}=0.95$.
Constraint: $0.0<{\mathbf{clevel}}<1.0$.
5: theta – double *Output
On exit: the estimate of the location, $\stackrel{^}{\theta }$.
6: thetal – double *Output
On exit: the estimate of the lower limit of the confidence interval, ${\theta }_{l}$.
7: thetau – double *Output
On exit: the estimate of the upper limit of the confidence interval, ${\theta }_{u}$.
8: estcl – double *Output
On exit: an estimate of the actual percentage confidence of the interval found, as a proportion between $\left(0.0,1.0\right)$.
9: wlower – double *Output
On exit: the upper value of the Wilcoxon test statistic, ${W}_{u}$, corresponding to the lower limit of the confidence interval.
10: wupper – double *Output
On exit: the lower value of the Wilcoxon test statistic, ${W}_{l}$, corresponding to the upper limit of the confidence interval.
11: fail – NagError *Input/Output
The NAG error argument (see Section 3.6 in the Essential Introduction).
## 6 Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
NE_BAD_PARAM
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_CONVERGENCE
Warning. The iterative procedure to find an estimate of the lower confidence point had not converged in $100$ iterations.
Warning. The iterative procedure to find an estimate of Theta had not converged in $100$ iterations.
Warning. The iterative procedure to find an estimate of the upper confidence point had not converged in $100$ iterations.
NE_INT
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 2$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
NE_REAL
On entry, clevel is out of range: ${\mathbf{clevel}}=〈\mathit{\text{value}}〉$.
NE_SAMPLE_IDEN
Not enough information to compute an interval estimate since the whole sample is identical. The common value is returned in theta, thetal and thetau.
## 7 Accuracy
nag_rank_ci_1var (g07eac) should produce results accurate to five significant figures in the width of the confidence interval; that is the error for any one of the three estimates should be less than $0.00001×\left({\mathbf{thetau}}-{\mathbf{thetal}}\right)$.
## 8 Further Comments
The time taken increases with the sample size $n$.
## 9 Example
The following program calculates a 95% confidence interval for $\theta $, a measure of symmetry of the sample of $50$ observations.
### 9.1 Program Text
Program Text (g07eace.c)
### 9.2 Program Data
Program Data (g07eace.d)
### 9.3 Program Results
Program Results (g07eace.r)
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http://mathoverflow.net/revisions/82849/list
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## Return to Answer
2 added 141 characters in body
I'm a little confused by your augmentation map.
Do you mean the augmentation map on the group ring $\mathbb{Z}[F]$? \mathbb{Z}[F]$(And and in the pro-$p$case, the completed group ring$\mathbb{Z}_p[[F]]\$?) I ask only because this augmentation map comes up frequently and significantly in the study of large number-theoretic Galois groups.
Assuming this is the case (and apologies for misinterpreting if not)not -- hopefully the answer will still be of some use to you), there is a tremendous amount of machinery set up for dealing exactly with questions of this sort -- probably the best starting place is the phrase "pro-p Fox Differential Calculus." (And so, indeed, your intuition that solving the discrete problem turns out to provide the correct pro-$p$ analog is correct. It was Iwasawa who carefully established the fundamental analogy here. In fact, thanks to the topology of $\mathbb{Z}_p$, in some ways the pro-p Fox calculus is nicer than the discrete version.) In particular, if you filter the group ring $\mathbb{Z}_p[[F]]$ by powers of the augmentation ideal (the group-ring version of your $A$), you land upon the sequence of "dimension subgroups" of F.
These subgroups have shown up repeatedly in the analysis of pro-$p$-groups arising in the study of large Galois group groups arising from restricted ramification questions (as appears to be the case for you). A couple of the highlights of the theory are the work of Vogel and Morishita interpreting number-theoretic analogs of the a priori knot-theoretic notion of Milnor invariants, refined versions of Golod-Shafarevich-type inequalities, and perhaps most relevant for your question, work of Arrigoni (e.g., "On Schur $\sigma$-groups") which I think explicitly answers questions of your type. For a more fundamental reference, see Koch's "Galois theory of $p$-extensions.")
Sorry to be mostly hand-wavey -- I'm away from good references at the moment.
1
I'm a little confused by your augmentation map. Do you mean the augmentation map on the group ring $\mathbb{Z}[F]$? (And in the pro-$p$ case, the completed group ring $\mathbb{Z}_p[[F]]$?)
Assuming this is the case (and apologies for misinterpreting if not), there is a tremendous amount of machinery set up for dealing exactly with questions of this sort -- probably the best starting place is the phrase "pro-p Fox Differential Calculus." (And so, indeed, your intuition that solving the discrete problem turns out to provide the correct pro-$p$ analog is correct. It was Iwasawa who carefully established the fundamental analogy here. In fact, thanks to the topology of $\mathbb{Z}_p$, in some ways the pro-p Fox calculus is nicer than the discrete version.) In particular, if you filter the group ring $\mathbb{Z}_p[[F]]$ by powers of the augmentation ideal (the group-ring version of your $A$), you land upon the sequence of "dimension subgroups" of F.
These subgroups have shown up repeatedly in the analysis of pro-$p$-groups arising in the study of large Galois group arising from restricted ramification questions (as appears to be the case for you). A couple of the highlights of the theory are the work of Vogel and Morishita interpreting number-theoretic analogs of the a priori knot-theoretic notion of Milnor invariants, refined versions of Golod-Shafarevich-type inequalities, and perhaps most relevant for your question, work of Arrigoni (e.g., "On Schur $\sigma$-groups") which I think explicitly answers questions of your type. For a more fundamental reference, see Koch's "Galois theory of $p$-extensions.")
Sorry to be mostly hand-wavey -- I'm away from good references at the moment.
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http://mathinsight.org/product_rule_refresher
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# Math Insight
• Top
• In threads
When printing from Firefox, select the SVG renderer in the MathJax contextual menu (right click any math equation) for better printing results
### A refresher on the product rule
Not only will the product rule be of use in general and later on, but it's already helpful in perhaps unexpected ways in dealing with polynomials. Anyway, the general rule is $${d\over dx}(fg)=f'g+fg'$$ While this is certainly not as awful as the quotient rule, it is not as simple as the rule for sums, which was the good-sounding slogan that the derivative of the sum is the sum of the derivatives. It is not true that the derivative of the product is the product of the derivatives. Too bad. Still, it's not as bad as the quotient rule.
One way that the product rule can be useful is in postponing or eliminating a lot of algebra. For example, to evaluate $${d\over dx}\left((x^3+x^2+x+1)(x^4+x^3+2x+1)\right)$$ we could multiply out and then take the derivative term-by-term as we did with several polynomials above. This would be at least mildly irritating because we'd have to do a bit of algebra. Rather, just apply the product rule without feeling compelled first to do any algebra: $${d\over dx}\left((x^3+x^2+x+1)(x^4+x^3+2x+1)\right)$$ $$=(x^3+x^2+x+1)'(x^4+x^3+2x+1)+(x^3+x^2+x+1)(x^4+x^3+2x+1)'$$ $$=(3x^2+2x+1)(x^4+x^3+2x+1)+(x^3+x^2+x+1)(4x^3+3x^2+2)$$ Now if we were somehow still obliged to multiply out, then we'd still have to do some algebra. But we can take the derivative without multiplying out, if we want to, by using the product rule.
For that matter, once we see that there is a choice about doing algebra either before or after we take the derivative, it might be possible to make a choice which minimizes our computational labor. This could matter.
#### Exercises
1. Find ${ d \over dx }(x^3-1)(x^6+x^3+1))$
2. Find ${ d \over dx }(x^2+x+1)(x^4-x^2+1)$.
3. Find ${ d \over dx }(x^3+x^2+x+1)(x^4+x^2+1))$
4. Find ${ d \over dx }(x^3+x^2+x+1)(2x + \sqrt{x}))$
#### Thread navigation
##### Calculus Refresher
• Previous: The quotient rule
• Next: The chain rule
#### Cite this as
Garrett P, “A refresher on the product rule.” From Math Insight. http://mathinsight.org/product_rule_refresher
Keywords: derivative, derivative properties, ordinary derivative, product rule
#### Credits
The page is based off the Calculus Refresher by Paul Garrett.
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http://en.wikipedia.org/wiki/Deterministic_pushdown_automaton
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# Deterministic pushdown automaton
In automata theory, a deterministic pushdown automaton (DPDA or DPA) is a variation of the pushdown automaton . The DPDA accepts the deterministic context-free languages, a proper subset of context-free languages.[1]
Machine transitions are based on the current state and input symbol, and also the current topmost symbol of the stack. Symbols lower in the stack are not visible and have no immediate effect. Machine actions include pushing, popping, or replacing the stack top. A deterministic pushdown automaton has at most one legal transition for the same combination of input symbol, state, and top stack symbol. This is where it differs from the nondeterministic pushdown automaton.
## Formal definition
A (not necessarily deterministic) PDA M can be defined as a 7-tuple:
$M=(Q\,, \Sigma\,, \Gamma\,, q_0\,, Z_0\,, A\,, \delta\,)$
where
• $Q\,$ is a finite set of states
• $\Sigma\,$ is a finite set of input symbols
• $\Gamma\,$ is a finite set of stack symbols
• $q_0\,\in Q\,$ is the start state
• $Z_0\,\in\Gamma\,$ is the starting stack symbol
• $A\,\subseteq Q\,$, where $A$ is the set of accepting states
• $\delta\,$ is a transition function, where
$\delta\colon(Q\, \times ( \Sigma\, \cup \left \{ \varepsilon\, \right \} ) \times \Gamma\,) \longrightarrow \mathcal{P}(Q \times \Gamma ^{*})$
where $\Gamma^{*}$ means $\{a_{1}a_{2}...a_{n}|n\ge0\wedge\forall i.a_{i}\in\Gamma\}$, "the set of all finite strings (including the empty string $\varepsilon$) of elements of $\Gamma$", $\varepsilon$ denotes the empty string, and $\mathcal{P}(X)$ is the power set of a set $X$.
M is deterministic if it satisfies both the following conditions:
• For any $q \in Q, a \in \Sigma \cup \left \{ \varepsilon \right \}, x \in \Gamma$, the set $\delta(q,a,x)\,$ has at most one element.
• For any $q \in Q, x \in \Gamma$, if $\delta(q, \varepsilon, x) \not= \emptyset\,$, then $\delta\left( q,a,x \right) = \emptyset$ for every $a \in \Sigma.$
There are two possible acceptance criteria: acceptance by empty stack and acceptance by final state. The two are not equivalent for the deterministic pushdown automaton (although they are for the non-deterministic pushdown automaton). The languages accepted by empty stack are the languages that are accepted by final state, as well as have no word in the language that is the prefix of another word in the language.
## Languages recognized
If $L(A)$ is a language accepted by a PDA $A$ it can also be accepted by a DPDA if and only if there is a single computation from the initial configuration until an accepting one for all strings belonging to $L(A)$. If $L(A)$ can be accepted by a PDA it is a context free language and if it can be accepted by a DPDA it is a deterministic context-free language.
Not all context-free languages are deterministic. This makes the DPDA a strictly weaker device than the PDA. For example, the language of even-length palindromes on the alphabet of 0 and 1 has the context-free grammar S → 0S0 | 1S1 | ε. An arbitrary string of this language cannot be parsed without reading all its letters first which means that a pushdown automaton has to try alternative state transitions to accommodate for the different possible lengths of a semi-parsed string.[2]
Restricting the DPDA to a single state reduces the class of languages accepted to the LL(1) languages.[3] In the case of a PDA, this restriction has no effect on the class of languages accepted.
## Properties
### Closure
Closure properties of deterministic context-free languages (accepted by deterministic PDA by final state) are drastically different from the context-free languages. As an example they are (effectively) closed under complementation, but not closed under union. To prove that the complement of a language accepted by a deterministic PDA is also accepted by a deterministic PDA is tricky. In principle one has to avoid infinite computations.
As a consequence of the complementation it is decidable whether a deterministic PDA accepts all words over its input alphabet, by testing its complement for emptiness. This is not possible for context-free grammars (hence not for general PDA).
### Equivalence problem
Geraud Senizergues (1997) proved that the equivalence problem for deterministic PDA (i.e. given two deterministic PDA A and B, is L(A)=L(B)?) is decidable,[4] a proof that earned him the 2002 Gödel Prize. For nondeterministic PDA, equivalence is undecidable.
## Notes
1. Michael Sipser (1997). Introduction to the Theory of Computation. PWS Publishing. p. 102. ISBN 0-534-94728-X.
2. Hopcroft, John; Rajeev Motwani & Jeffrey Ullman (2001). (2 ed.). Addison-Wesley. pp. 249–253.
3. Kurki-Suonio, R. (1969). "Notes on top-down languages". BIT 9 (3): 225–238. doi:10.1007/BF01946814.
4. Sénizergues, Géraud (1997). "The equivalence problem for deterministic pushdown automata is decidable". AUTOMATA, LANGUAGES AND PROGRAMMING. 1256/1997: 671–681. doi:10.1007/3-540-63165-8_221.
## References
• G. Sénizergues: L(A)=L(B)? decidability results from complete formal systems. Theoretical Computer Science 251(1-2): 1-166 (2001)
• G. Sénizergues: L(A)=L(B)? A simplified decidability proof. Theoretical Computer Science 281(1-2): 555-608 (2002)
## Further reading
• Hamburger, Henry; Dana Richards (2002). . Upper Saddle River, NJ 07458: Prentice Hall. pp. 284–331. ISBN 0-13-065487-6.
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http://mathoverflow.net/questions/41837?sort=oldest
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## 1..n game, how to analyze?
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I came up with a simple game.
A permutation of 1..n is available for purchase in that order. 2 players each have m in money each to bid for one number at a time in the permutation in order and will get that score for purchasing it. The player who won the previous auction starts bidding for the next number A player may bid zero and then wins if the other player does not bid. In the end, only the score counts, not remaining money.
Naturally, by a simple strategy-stealing argument, there has to be a winning move for the first player, and there can only be one such winning move.
E.g. the case n=5, m=8
Each example shows only one line of play:
54321: 5 wins // (bid for) 5: P1=5, P2 pass 4: P1=3, P2=4 3: P2=3 2: P2=1, P1=2 1: P1=1 wins
54321: 6 loses // 5: P1=6, P2 pass 4: P1=2, P2 = 3 3: P2=2 2: P2=2 wins
52431: 4 wins // 5: P1=4, P2 pass 2: P1=1, P2=2 4: P2=4 3: P2=2, P1=3 wins
Becase a player needs to win 8 points to win, and there are a total of 8 money per player, it seems natural that the number n is worth about n. My conjecture is that a winning bid in the special case if the first number in the permutation is p, is always p or p-1
Has this game been considered before?
What is the natural way to proceed with this problem? (except exhaustive search)
Edit: It seems like a fairly interesting game, but after a bit more thought, I guess that the fact that the players are bidding for a permutation of 1..n does not make the game any simpler to analyze than an arbitrary set. Because, the central point of the analysis will be: Which possible subsets of 1..n do I have to win to win the game? For example, to win the game 1..5 you will need to win {5 4}, {5 3}, {5 2 1}, {4 3 2} or {4 3 1} or some super-set of one of those sets.
This means that the simplest case to analyze will instead be a game consisting of n 1's to bid for (where n is odd and you need to win ceil(n/2) auctions to win). That game is fairly simple to analyze, but I haven't been able to generalize further than that.
What I find interesting is how the order in which the auctions proceed will affect the best bid. This is a little surprising in such a simple setting.
I have seen a blind bidding variant similar to the one suggested in comments played also (and I believe it is quite possible to get really good at it), but that is a completely different game.
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I'm afraid I don't understand your notation for the example. Could you expand a little? – jc Oct 12 2010 at 0:11
I have played such a game as a young child with a deck of cards: the permutation consisted of one shuffled suit, and each player had a suit from which to bid, but bidding was secret. Unfortunately, I don't remember its name :( – JBL Oct 12 2010 at 12:46
The open information isn't clear. Do both players know the permutation before the game begins? If not, do players learn of the number if they don't win the bid? What happens if the bids are tied? If both players are out of money? – Kevin O'Bryant Oct 12 2010 at 14:48
Both players know the permutation before the game begins. You can't make a bid that would tie. If you are out of money then you have to either bid 0 or lose the bid. – ohai Oct 12 2010 at 16:33
## 2 Answers
"My conjecture is that a winning bid in the special case if the first number in the permutation is p, is always p or p-1"
Did you mean n instead of p here? For the 5 and 6 case I think this is true, if my minimax codes are right, but I don't think this is the case for n=10 m=28 when the initial permutation is (10,1,2,3,4,5,6,7,8,9) for example. In this case the winning initial bid is n-2=8 whereas n-1=9 and n=10 both lose.
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I was thinking that if the sum of the numbers is odd, perhaps the case with m = ceil(n(n+1)/4) would be simpler. However, with the evidence on bids I have seen so far, I don't have that much support for that. E.g. n=5, m=8, and n=10, m=23. But it is of course possible that the game just needs an exhaustive search to beat. – Carl Johan Oct 12 2010 at 7:07
Do you have some code for how you implemented this minimax function? For the case n=10, m=28, branching seems rather high and the number of options very large unless you impose some restrictions. – Carl Johan Oct 13 2010 at 13:21
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I will be assuming that money starts at 1 for each player and players can bid any real number. I'll also assume that prizes are real numbers.
First, it should be noted that in each position with prizes $p_1 , \dots , p_n$ and accumulated scores $s_A, s_B$, if one player wins with money $m$ then he can win with any amount $m'>m$ and if that player loses with money $m$ he also loses with $m' < m$.
Let's assume that there is no partition of the prizes into two classes which yield the same sum; the above argument shows that there is a "singular value" which is the separator of the classes of winning and losing position for a player (such existence is guaranteed in $\mathbb{R}$).
There is a practical rule to find out what that value is in general, but let's start with simple positions: if player $A$ has to win $n$ times in a row in order to win, it is easy to see that the winning condition is $\frac{1}{n} \cdot m_A \geq m_B$.
To analyse a general position, given that the win-the-prize option has a winning condition of the form $\chi_W \cdot m_A \geq m_B$ and the lose-the-prize option has a winning condition of the form $\chi_L \cdot m_A \geq m_B$, the position the player $A$ is in has a winning condition of $\frac{1+\chi_L}{1+ \frac{1}{\chi_W}} m_A \geq m_B$.
To see why this happens, just consider the associated system of inequalities:
$$\chi_W \cdot (m_A - p) \geq m_B$$
$$\chi_L \cdot m_A \geq m_B - p$$
then deduce
$$m_A - \frac{ m_B}{\chi_W} \geq p \geq m_B - \chi_L \cdot m_A$$
Taking equalities yields the relation stated above.
This way, starting form the end of the auction tree you can go backward and find out all the $\chi$ for each position, until you get to the first one. Note that exacxtly meeting the inequality does not guarantee a win, it depends on who is going first on that auction (i.e. who won the last auction).
I'll add an example:
Prizes: 10 6 4 8 5
Minimal winning sequences: 10 8, 10 4 5, 10 6 4, 10 6 5, 6 8 5, 6 4 8, 4 8 5.
Results: In the starting position, $\chi=1$. After the first auction, the losing and winning $\chi$ are respectively $\frac{4}{7}$ and $\frac{7}{4}$; after the second auction the $\chi$ are $\frac{3}{1}$ if you won both 10 and 6, $\frac{4}{3}$ if you won 10 and lost 6, $\frac{3}{4}$ if you lost 10 and won 6, $\frac{1}{3}$ if you lost both.
Note that in the general case where there could be ties, the $\chi$s may not be one the inverse of the other, resulting in a $\chi>1$ for the starting position, which means the first player has no way to win (but still has a way to draw). In fact, I think that every time there is a partition in two subsets so that each yields the same sum, it is not possible for the first player to win (he can just draw).
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http://physics.stackexchange.com/questions/18847/on-electromagnetic-self-energy/18849
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On Electromagnetic Self Energy
In the process of pair annihilation an electron and a positron annihilate each other to produce a pair of photons, conserving momentum and energy. As the oppositely charged particles approach each other, the field around them decreases. We may consider them as a dipole with a decreasing dipole moment and radiating energy. As the field decreases the value of E changes. Consequently, B should change: $$\nabla \times \mathbf{B}=\epsilon_0\mu_0\frac{\partial \mathbf{E}}{\partial t}$$ and $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}.$$ So we have a Poynting vector that drives energy in the outward direction (If it was in the inward direction the field would grow stronger with the approach of the particles which is not the case). By the time the electron and the positron are close enough to clash the classical self energy (electromagnetic self energy) has moved out to an infinite distance.
Quantum Mechanical Picture:Virtual photons emitted by the electron (or the positron) hit back the particle increasing its KE.This extra energy gets included into the mass of the particle leading to the dressed mass of the particle. In pair annihilation this energy is included in $E=mc^2$.And we get it after the collision of the particles. But the classical electromagnetic self energy, as we have seen in the previous paragraph, flows out to infinity before the collision of the particles.
So the classical self energy of the electron and its QM self energy must be physically different. Is this true?
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2 Answers
The quantum electrodynamics process includes the distant field of an electron only in a crude way, by additional photon emissions from the electron line. The mass, however, includes the self-field contribution, in a physical renormalization, because the mass of the electron is fixed at its experimental value. The tree-level QED description is not of the far-region outside the compton wavelength of the electron, but of the behavior in the near region, close to and inside the Compton wavelength.
The field of the electron and positron is described by adding photon lines to the simple tree level annihilation diagram, and making these lines have low momentum, soft photons. The soft photon amplitude is always divergent, and this divergence is only made sensible by doing a summation of infinitely many soft photon emissions, with the classical field as a guide for the summation. This type of thing is very involved, but since the physical situation is well understood classically, along the lines you suggest, people generally do not worry about it. This line of research is generally classified as infraparticle analysis, which tries to take into account the infrared behavior of QED.
Classical electron radius vs. Compton wavelength
The classical electron radius is defined, up to small factors of order unity, as the radius of a sphere of total charge e whose self-energy equals the mass of the electron. This distance is about 2.5 fermis. If the electron were a classical sphere of this size, nearly all its mass would be self-mass, and the annihilation process would proceed by the process you describe, gradual classical dipole moment relaxation, and outward Poynting flow.
But the Compton wavelength of the electron is about a thousand times larger, or 2500 fermis (the ratio of the two is the perturbtion parameter $\alpha\over 2\pi$). The annihilation process in quantum electrodynamics is not causal within the Compton wavelength, the intermediate electron line between the two photons is executing a mostly space-like motion.
So there is a separation of scales--- for scales significantly larger than the Compton wavelength, the classical dipole field picture applies, as you describe, and leads to a shower of coherent soft photons during the annihilation process whose total energy is approximately equal to the fraction of the mass-energy of the electron contained in the field at a distance larger than the Compton wavelength. This is about a tenth of a percent of the mass of the electron.
The rest of the process is the emission of two (or less often three) hard photons described by a QED tree diagram, and this happens when the electron and positron are within the compton wavelength. The hard process is not related to the classical field relaxation, it is a pure quantum effect, and it is the thing calculated in field theory books.
Classical vs. Quantum self energy
The behavior of the self-energy in quantum electrodynamics is completely different from the behavior in classical electrodynamics. The self-energy diagram for the electron is only log divergent, while the classical sphere of radius r has a self-energy which diverges as 1/r. Both calculations are well known, the classical one is just from the field of a spherical source, while the quantum one is from the Feynman diagram where an electron emits and reabsorbs a photon along its worldline, where the integrand is (Wick rotated, with factors absorbed in dk, schematic $\gamma$ structure, and ignoring the external wavefunction parts)
$$\int {\gamma^\mu (\gamma\cdot k + m ) \gamma_\mu\over (k+p)^2(k^2 + m^2)} dk$$
Which is log divergent because the linear-in-k divergence cancels by symmetry after combining denominators.
This cancellation of the classical linear divergence is nearly self-evident in the modern Feynman formalism, which is why people often neglect to mention that this was originally a non-trivial result, due to Victor Weisskopf. Weisskopf's calculation uses old fasioned perturbation theory and is unreadable today. Old fasioned perturbation theory consists of putting all the intermediate lines on shell and summing over the intermediate momentum, which reproduces the Feynman propagator, but in a way that separates time and space. This separates out positron and electron contributions on intermediate states. The classical linear in k divergence is still present in each part alone, but cancels (up to a relatively negligible log divergence) at large k between the two parts.
There is a simple heuristic to understand the cancellation. The electron is propagating back and forth in time by a particle path integral, and when it is going backwards, it is a positron. The positron segments have positive charge, and the electron segments have negative charge. When you slice this at any time, you always have one more electron line than positron on each slice (because overall the electron is moving from past to future), but if you look close to one time slice, each individual path zig-zags a diverging number of times, making the effective instantaneous distribution a smeared out mess of positive and negative charges, which is approximately as wide as the Compton wavelength (which is the scale at which the electron path stops being strictly future oriented). This makes an effective fractal smeared out charge distribution which adds up to e, and the self-energy of a smeared out smooth charge distribution would be finite, except if irregular, it can still be divergent.
The cancellation of the positron and electron classical divergences would be suspect if you didn't have a covariant regulator, which is why the Feynman style regulator is so important. In the Pauli-Villars picture, you add additional heavy electrons with wrong sign loop contributions to cancel out the high-k divergences in the diagram. In general, this requires two regulator fields. The regulator fields allow the same formal manipulations on their divergent integrals as the physical fields, so that the end result just subtracts out the same diagram at a higher mass. The subtraction process gives a finite integral, which justifies the intermediate manipulations, and shows that, in a covariant regulator, the electron-positron cancellation is really happening.
If you just cut off positron and electron at high k, you need to make sure that the cutoff respects the cancellation. If you include more high-k electrons than positrons, you get the classical linear in the cutoff divergence again.
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+1 : Very interesting "bringing together" of pieces of the picture of the annihilation process. – twistor59 Dec 29 '11 at 9:11
The classical annihilation process described in the original post is a falling-at-the-center problem with releasing the potential energy difference as radiation due to time-dependent dipole moment (Anamitra, you missed the current $\vec{j}$ in the first equation). So the total radiated energy is essentially determined with the radius at which the falling stops. If we use the classical electron radius $r_0=\frac{e^2}{mc^2}$, the released energy will be $E=\frac{e^2}{2r_0}=\frac{mc^2}{2}$. If we take into account the magnetic field contribution (an electron is a magnet), we may derive another "classical electron radius" $r_1>r_0$, so the released energy will be even less. What I wanted to underline is that it is the transformation of the interaction potential energy into radiation energy, which is not a self-energy transformation. As soon as the electron and positron remain slow (non relativistic, due to loss of energy into the radiation), the final state in CED is some sort of a still dipole "molecule" of some small size and new total mass (if they do not merge). If they merge, we can (theoretically) get a neutral system with zero rest mass and zero rest energy (the mass defect being $2mc^2$).
The notion of self-energy is rather strange. It is a theoretical difficulty of our describing interactions via fields that is resolved by a brute force - with discarding the self-energy contributions (they are not necessary, to say the least). The more important thing is the total electron energy that includes the famous $mc^2$ in the relativistic mechanics with its new conservation laws.
In QED, when a non-relativistic positronium annihilates, we use the energy conservation law that gives us the total radiated energy providing the electron and positron disappear. The difference between CED and QED in an essentially "discontinuous" spectrum of the radiation (transitions from one level to another) and in admitting the possibility for particles to disappear, the energy conservation law still holds. The interaction manifests itself here in attracting the charges in the initial state and in disappearing the charges in the final state.
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The soft photons of annihilation should lead the hard photons to always add up to less than 2mc^2 by a certain small fraction of order $\alpha$. Do you know what these corrections are experimentally or theoretically? Just curious. – Ron Maimon Dec 30 '11 at 9:50
@RonMaimon: No, I do not know exactly, but I suspect that the only certain thing is an average value. One can get two less hard photons with more energetic soft contributions within the same initial energy of the electron-positron pair. – Vladimir Kalitvianski Dec 30 '11 at 12:20
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http://en.wikiversity.org/wiki/Advanced_elasticity/Polar_decomposition
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# Advanced elasticity/Polar decomposition
From Wikiversity
## Polar decomposition
The w:Polar decomposition theorem states that any second order tensor whose determinant is positive can be decomposed uniquely into a symmetric part and an orthogonal part.
In continuum mechanics, the deformation gradient $\boldsymbol{F}$ is such a tensor because $\det(\mathbf{F}) > 0$. Therefore we can write
$\boldsymbol{F} = \boldsymbol{R}\cdot\boldsymbol{U} = \boldsymbol{V}\cdot\boldsymbol{R}$
where $\boldsymbol{R}$ is an orthogonal tensor ($\boldsymbol{R}\cdot\boldsymbol{R}^T = \boldsymbol{\mathit{1}}$) and $\boldsymbol{U}, \boldsymbol{V}$ are symmetric tensors ($\boldsymbol{U} = \boldsymbol{U}^T$ and $\boldsymbol{V} = \boldsymbol{V}^T$) called the right stretch tensor and the left stretch tensor, respectively. This decomposition is called the polar decomposition of $\boldsymbol{F}$.
Recall that the right Cauchy-Green deformation tensor is defined as
$\boldsymbol{C} = \boldsymbol{F}^T\cdot\boldsymbol{F}$
Clearly this is a symmetric tensor. From the polar decomposition of $\boldsymbol{F}$ we have
$\boldsymbol{C} = \boldsymbol{U}^T\cdot\boldsymbol{R}^T\cdot\boldsymbol{R}\cdot\boldsymbol{U} = \boldsymbol{U}\cdot\boldsymbol{U} = \boldsymbol{U}^2$
If you know $\boldsymbol{C}$ then you can calculate $\boldsymbol{U}$ and hence $\boldsymbol{R}$ using $\boldsymbol{R} = \boldsymbol{F}\cdot\boldsymbol{U}^{-1}$.
### How do you find the square root of a tensor?
If you want to find $\boldsymbol{U}$ given $\boldsymbol{C}$ you will need to take the square root of $\boldsymbol{C}$. How does one do that?
We use what is called the spectral decomposition or eigenprojection of $\boldsymbol{C}$. The spectral decomposition involves expressing $\boldsymbol{C}$ in terms of its eigenvalues and eigenvectors. The tensor product of the eigenvectors acts as a basis while the eigenvalues give the magnitude of the projection.
Thus,
$\boldsymbol{C} = \sum_{i=1}^3 \lambda_i^2~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$
where $\lambda_i^2$ are the principal values (eigenvalues) of $\boldsymbol{C}$ and $\boldsymbol{N}_i$ are the principal directions (eigenvectors) of $\boldsymbol{C}$.
Therefore,
$\boldsymbol{U}^2 = \sum_{i=1}^3 \lambda_i^2~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$
Since the basis does not change, we then have
$\boldsymbol{U} = \sum_{i=1}^3 \lambda_i~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$
Therefore the $\lambda_i$ can be interpreted as principal stretches and the vectors $\boldsymbol{N}_i$ are the directions of the principal stretches.
#### Exercise:
If
$\boldsymbol{U} = \sum_{i=1}^3 \lambda_i~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i$
show that
$\boldsymbol{U}^2 = \boldsymbol{U}\cdot\boldsymbol{U} = \sum_{i=1}^3 \lambda_i^2~ \boldsymbol{N}_i\otimes\boldsymbol{N}_i ~.$
### Example of polar decomposition
Let us assume that the motion is given by
$\begin{align} x_1 &= \cfrac{1}{4} \left[4~X_1 + (9 - 3~X_1 - 5~X_2 - X_1~X_2)~t\right] \\ x_2 &= X_2 + (4 + 2~X_1)~t \end{align}$
The adjacent figure shows how a unit square subjected to this motion evolves over time.
An example of a motion.
#### Deformation gradient
The deformation gradient is given by
$\boldsymbol{F} = \frac{\partial \mathbf{x}}{\partial \boldsymbol{X}} \quad \implies \quad F_{ij} = \frac{\partial x_i}{\partial X_j}$
Therefore
$\begin{align} F_{11} &= \frac{\partial x_1}{\partial X_1} = \cfrac{1}{4}\left[4 + (- 3 - X_2)~t\right] \\ F_{12} &= \frac{\partial x_1}{\partial X_2} = \cfrac{1}{4}\left[(- 5 - X_1)~t\right] \\ F_{21} &= \frac{\partial x_2}{\partial X_1} = 2~t \\ F_{22} &= \frac{\partial x_2}{\partial X_2} = 1 \end{align}$
At $t = 1$ at the position $\boldsymbol{X} = (0,0)$ we have
$\mathbf{F} = \begin{bmatrix} \frac{\partial x_1}{\partial X_1} & \frac{\partial x_1}{\partial X_2} \\ \frac{\partial x_2}{\partial X_1} & \frac{\partial x_2}{\partial X_2} \end{bmatrix} = \cfrac{1}{4} \begin{bmatrix} 1 & -5 \\ 8 & 4 \end{bmatrix}$
You can calculate the deformation gradient at other points in a similar manner.
#### Right Cauchy-Green deformation tensor
We have
$\boldsymbol{C} = \boldsymbol{F}^T\cdot\boldsymbol{F}$
Therefore,
$\mathbf{C} = \mathbf{F}^T~\mathbf{F} = \cfrac{1}{16} \begin{bmatrix} 65 & 27 \\ 27 & 41 \end{bmatrix}$
To compute $\boldsymbol{U}$ we have to find the eigenvalues and eigenvectors of $\boldsymbol{C}$. The eigenvalue problem is
$(\mathbf{C} - \lambda^2~\mathbf{I})\mathbf{N} = \mathbf{0}$
where
$\mathbf{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
To find the eigenvalues we solve the characteristic equation
$\det(\mathbf{C} - \lambda^2~\mathbf{I}) = 0$
Plugging in the numbers, we get
$\det\begin{bmatrix} \cfrac{65}{16} - \lambda^2 & \cfrac{27}{16} \\ \cfrac{27}{16} & \cfrac{41}{16} - \lambda^2 \end{bmatrix} = 0$
or
$\lambda^4 - \cfrac{53}{8}~\lambda^2 + \cfrac{121}{16} = 0$
This equation has two solutions
$\begin{align} \lambda_1^2 & = \cfrac{53}{16} + \cfrac{3}{16}~\sqrt{97} = 5.159 \\ \lambda_2^2 & = \cfrac{53}{16} - \cfrac{3}{16}~\sqrt{97} = 1.466 \end{align}$
Taking the square roots we get the values of the principal stretches
$\lambda_1 = 2.2714 \qquad \lambda_2 = 1.2107$
To compute the eigenvectors we plug into the eigenvalues into the eigenvalue problem to get
$\left\{ \begin{bmatrix} 65 & 27 \\ 27 & 41 \end{bmatrix} - \lambda^2_1 ~ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right\}~\begin{bmatrix} N_1^{(1)} \\ N_2^{(1)} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Because this system of equations is not linearly independent, we need another equation to solve this system of equations for $N^{(1)}_1$ and $N^{(1)}_2$. This problem is eliminated by using the following equation (which implies that $\mathbf{N}$ is a unit vector)
$(N^{(1)}_2)^2 = \sqrt{1 - (N^{(1)}_1)^2}$
Solving, we get
$\mathbf{N}_1 = \begin{bmatrix} N_1^{(1)} \\ N_2^{(1)} \end{bmatrix} = \begin{bmatrix} 0.8385 \\ 0.5449 \end{bmatrix}$
We can do the same thing for the other eigenvector $\mathbf{N}_2$ to get
$\mathbf{N}_2 = \begin{bmatrix} N_1^{(2)} \\ N_2^{(2)} \end{bmatrix} = \begin{bmatrix} -0.5449 \\0.8385 \end{bmatrix}$
Therefore,
$\boldsymbol{N}_1\otimes\boldsymbol{N}_1 = \mathbf{N}_1^T~\mathbf{N}_1 = \begin{bmatrix} 0.8385 & 0.5449 \end{bmatrix} \begin{bmatrix} 0.8385 \\ 0.5449 \end{bmatrix} = \begin{bmatrix} 0.7031 & 0.4569 \\ 0.4569 & 0.2969 \end{bmatrix}$
and
$\boldsymbol{N}_2\otimes\boldsymbol{N}_2 = \mathbf{N}_2^T~\mathbf{N}_2 = \begin{bmatrix} -0.5449 & 0.8385 \end{bmatrix} \begin{bmatrix} -0.5449 \\0.8385 \end{bmatrix} = \begin{bmatrix} 0.2969 & -0.4569 \\ -0.4569 & 0.7031 \end{bmatrix}$
Therefore,
$\boldsymbol{C} = \lambda_1^2~\boldsymbol{N}_1\otimes\boldsymbol{N}_1 + \lambda_2^2~\boldsymbol{N}_2\otimes\boldsymbol{N}_2 \quad \implies \quad \mathbf{C} = 5.159~ \begin{bmatrix} 0.7031 & 0.4569 \\ 0.4569 & 0.2969 \end{bmatrix} + 1.466~ \begin{bmatrix} 0.2969 & -0.4569 \\ -0.4569 & 0.7031 \end{bmatrix}$
We usually don't see any problem to calculate $\boldsymbol{C}$ at this point and go straight to the right stretch tensor.
#### Right stretch
The right stretch tensor $\boldsymbol{U}$ is given by
$\boldsymbol{U} = \lambda_1~\boldsymbol{N}_1\otimes\boldsymbol{N}_1 + \lambda_2~\boldsymbol{N}_2\otimes\boldsymbol{N}_2 \quad \implies \quad \mathbf{U} = 2.2714~ \begin{bmatrix} 0.7031 & 0.4569 \\ 0.4569 & 0.2969 \end{bmatrix} + 1.2107~ \begin{bmatrix} 0.2969 & -0.4569 \\ -0.4569 & 0.7031 \end{bmatrix}$
or
$\mathbf{U} = \begin{bmatrix} 1.9565 & 0.4846 \\ 0.4846 & 1.5256 \end{bmatrix}$
We can invert this matrix to get
$\mathbf{U}^{-1} = \begin{bmatrix} 0.5548 & -0.1762 \\ -0.1762 & 0.7114 \end{bmatrix}$
#### Rotation
We can now find the rotation matrix by using th relation
$\boldsymbol{R} = \boldsymbol{F}\cdot\boldsymbol{U}^{-1}$
In matrix form,
$\mathbf{R} = \cfrac{1}{4} \begin{bmatrix} 1 & -5 \\ 8 & 4 \end{bmatrix} \begin{bmatrix} 0.5548 & -0.1762 \\ -0.1762 & 0.7114 \end{bmatrix} = \begin{bmatrix} 0.3590 & -0.9334 \\ 0.9334 & 0.3590 \end{bmatrix}$
You can check whether this matrix is orthogonal by seeing whether $\mathbf{R}~\mathbf{R}^T = \mathbf{R}^T~\mathbf{R} = \mathbf{I}$.
You thus get the polar decomposition of $\boldsymbol{F}$. In an actual calculation you have to be careful about floating point errors. Otherwise you might not get a matrix that is orthogonal.
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http://mathforum.org/mathimages/index.php?title=Coefficients&oldid=34991
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Coefficients
From Math Images
Revision as of 17:14, 15 November 2012 by Marhot (Talk | contribs)
Quadratic
Field: Algebra
Created By: [[Author:| ]]
Website: [ ]
Quadratic
Just a quadratic function.
Basic Description
Every graphical expression has variables, numbers that change dependent upon the location of the point on the graph, and has coefficients and constants, numbers that do not change, but still influence the shape of the graph. The following paragraphs will be examining how these coefficients affect various types of graphical functions.
Linear Functions
A single line is a representation of a linear expression shown by the general equation $y = ax + b$. $x$ and $y$ are the variables (the changing values described above); And $a$ and $b$ are the coefficients. $a$ is a coefficient with an $x$ in its term, this means that when $a$ is changed, it will affect both $x$ and $y$. In this case, $a$ affects how quickly one changes with respect to the other. When a is one, these two variables change the same amount in comparison to other, therefore the steepness of the line will be a one-to-one ratio. When $x$ is 5, $y$ is 5. When $x$ is -1.672, $y$ is also -1.672. If $a$ is greater than 1, than $y$ is increasing more rapidly than $x$ and the line becomes steeper, ($y$'s value will be greater than its respective $x$ value). If it is less than 1, but greater than zero, $x$ has greater respective values, therefore the line becomes less steep. If $a$ is zero, $y$ is never changing, it remains at a constant value throughout. Therefore, the line is horizontal. To make the line vertical, $a$ would have to be very very large, at infinity. When $a$ is negative, the line will tilt in the other direction. When $a$ is positive the line goes from the bottom left to the upper right, so when it is negative it goes from the bottom right to the upper left. And as it becomes more negative, it will increase the steepness in this direction.
The coefficient $b$ is specifically a constant (a term without variables). Constants affect where the linear function intersects the y-axis (typically the vertical axis). For instance if $b$ is 1, the line will pass through the part of the y-axis where $y$'s value would correlate to 1. As can be inferred, $b$ only changes the y values. It raises or lowers the line by the value it is equal to.
Polynomial Functions
For other polynomials (functions that use addition, subtraction, and multiplication, and contain terms with integer exponents that are not negative) of higher degrees (the value of the highest power), the graph will be a curve. If the degree is an even number, the graph will be symmetrical, if not it will will have its ends going in opposite directions. The leading coefficient (coefficient of the term with the highest power) will affect this curve. The larger it is, the thinner the curve will appear because it is scaling the entire graph, increasing the y values more and more in comparison to the x value that was initially used. As the coefficient decreases, the curve becomes wider. Whether it is positive or negative also affects the graph. Even degrees, for a positive coefficient, will go up towards positive infinity on either side and if negative, both ends will go towards negative infinity. If the degree is odd, and the leading coefficient positive, the right side will go to positive infinity and the left towards negative infinity, when the leading coefficient is negative, the end behavior of each side will reverse.
Although the shape of polynomials stays essentially the same regardless of the other coefficients, the remaining terms do affect the shape close to the vertex. This is because the first term becomes significantly larger as the values of x become large positive or negative numbers. Since this value grows larger and larger, the other terms affect the graphs less and less until they are often negligible. In most cases, if one "zooms out" on any polynomial, the little bumps and curves that may be created by other terms in the middle are difficult to notice, because they are so small in comparison to the overall graph, which is infinite in size. To see how the other terms do affect behavior closer to the vertex, see a more mathematical explanation.
Power Functions
Power functions are functions in which the variable is the base of a number with a constant exponent. Therefore, when interpreting the graph, the base changes while the exponent stays the same. Therefore, depending on the exponent, the graph could a various collection of curves.
If the exponent is an even number, the graph will be symmetrical about the y-axis. This is because even when x is negative, raising it to an even power will cause it to be positive. Two negatives always cancel each other and make a positive number. If the exponent is odd, the graph will not be symmetrical since when x is negative and raised to an odd power, there will be negative y values on one side of the graph and positive values on the other.
The shape of the graph is also influenced by a few other factors. If the exponent is greater than zero, the graph is increasing and if it is less than zero, it is decreasing.
If the exponent is a fraction $\frac{m}{n}$, both $m$ and $n$ must be examined and their relationship to one another in order to understand what shape they may make. If $n$ is even, then it is undefined for values less than zero. When $n$ is even, there are no negative y values produced. If $n$ is odd than it is defined for values greater than and less than zero (and at zero). If $m$ is even, the function is even (symmetrical) and if $m$ is odd, the function is odd, meaning it is symmetrical about the origin. If $m$ is less than $n$, the function is increasing at a decreasing rate, and if $m$ is greater than $n$, the function is increasing at an increasing rate.
There are other coefficients to consider in a power function. When looking at the general equation, $y = ax^{b}$ we have so far, only discussed $b$. The coefficient $a$ however, is just a scaling factor and therefore makes the graph appear thinner if larger and fatter is smaller. If it is negative, then the graph will be flipped.
Exponential Functions
Exponential equations are equations where the variable is the exponent, therefore the constant is the base. Although the base isn't necessarily a coefficient, what number it is, changes the shape of the graph. In an exponential function, this base is always a positive number. If the number is between 0 and 1, the graph is decreasing, because one is raising a fraction to higher and higher powers, which makes the number smaller and smaller. If the number is greater than one, the graph is increasing because it is raising the base to higher and higher powers which will make the values increase. If the base is 1, the graph will make a straight line due to the fact that one raised to any power is always one. A coefficient can be multiplied to this base and its variable exponent and will scale the graph.
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http://mathhelpforum.com/advanced-statistics/127510-conditional-probability-question.html
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# Thread:
1. ## Conditional Probability Question
I was wondering if I could get some help with a problem. I have thought about this for a long time and can't figure it out. I don't need to solve the problems (at least not yet), I just need to know if they have different answers and if so, why. I thought they should all have the same answer, but my calculations are not working out that way :-(
Imagine a glass with 6 white balls numbered 1-6 and 4 black balls numbered 1-4. 3 balls are chosen at random.
A) What is the probability you choose 3 white balls given you choose at least 1 white ball.
B) What is the probability you choose 3 white balls given you choose at least 1 even numbered white ball?
C) What is the probability you choose 3 white balls given you choose white ball #2?
2. I'm not entirely confident with my answers.
A) $\frac{\binom{6}{3} \binom{4}{0}}{\binom{6}{1} \binom{4}{2} +\ \binom{6}{2} \binom{4}{1} + \binom{6}{3} \binom{4}{0}}$
B) $\frac{\binom{3}{1} \binom{5}{2} \binom{4}{0}}{\binom{3}{1} \binom{5}{2} \binom{4}{0} + \binom{3}{1} \binom{5}{1} \binom{4}{1} + \binom{3}{1} \binom{5}{0} \binom{4}{2}}$ because you want to choose at least one of the 3 even numbered white balls
C) $\frac{1 \binom{5}{2} \binom{4}{0}}{1 \binom{5}{2} \binom{4}{0} + 1 \binom{5}{1} \binom{4}{1} + 1 \binom{5}{0} \binom{4}{2}}$
3. I don't trust (b), since you only have 10 balls, the sum of the number of balls doesn't add up to 10 here.
4. Originally Posted by matheagle
I don't trust (b), since you only have 10 balls, the sum of the number of balls doesn't add up to 10 here.
Yeah, that answer doesn't make any sense.
By splitting the white balls into two groups, I get the following answer which makes a bit more sense:
$\frac{\binom{3}{1} \binom{3}{2} \binom{4}{0} + \binom{3}{2} \<br /> \binom{3}{1} \binom{4}{0} + \binom{3}{3}\binom{3}{0}\binom{4}{0}}{\binom{3}{1} \binom{3}{2} \binom{4}{0} + \binom{3}{1} \binom{3}{1} \binom{4}{1} + \binom{3}{1} \binom{3}{0} \binom{4}{2} + \binom{3}{2} \binom{3}{1} \binom{4}{0} + \binom{3}{2} \binom{3}{0} \binom{4}{1} + \binom{3}{3} \binom{3}{0} \binom{4}{0}}$
5. I get the same denominator, but I would only consider TWO groups.
I lump the odd whites with the black balls.
${3\choose 1}{7\choose 2}+{3\choose 2}{7\choose 1}+{3\choose 3}{7\choose 0}=85$
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http://mathoverflow.net/questions/76415/kahler-manifolds-with-special-submanifolds
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## Kahler manifolds with special submanifolds
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
This question is related to another question of mine. Let $X$ be a kahler manifold with $\dim_{\mathbb{C}}(X)=n$, let $\pi:E\rightarrow M$ be a holomorphic vector bundle of $rank_{\mathbb{C}}(E)=n-k$ over a kahler manifold $M$ with $\dim(M)=k$. Suppose there is a holomorphic embedding $\Phi$ of a neighborhood $N$ of the zero section of $E$ in $X$ that is $$\Phi:N\rightarrow X$$
What can be said about $X$? If $M$ is a point nothing, but what if $M$ is a higher dimensional manifold?
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Dear Italo, it would be nice if you can give a hint on the "type" of statement that you would like to have, i.e. give a slightly more specific question. At least, do you encounter such situations naturally? – Dmitri Sep 26 2011 at 15:26
I know that the question is rather vague, i imagined a situation like this thinking about my last question and your answer, in particular van de ven theorem. I was wondering if situations like the one described above are "natural" or conditions above are quite restrictive (as i suspect). And as consequence of the rigidity there is a statement like this: if you have a kahler manifold $X$ a vector bundle $E$ as above necessarily X is isomorphic $\mathbb{P}(E\osum \mathcal{L})$ with $\mathcal{L}$ a line bundle. – Italo Sep 26 2011 at 16:02
## 2 Answers
If your $X$ is projective and $E$ is an ample vector bundle over $M$ then the field of meromorphic functions of $X$ is a finite extension of the field of meromorphic functions of $\mathbb P(E \oplus \mathcal O_M )$. This follows from Corollary 6.8 of Hartshorne's Cohomological dimension of algebraic varieties.
There is also an analytic version by Andreotti, which precedes Hartshorne's paper, implying the same result.
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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
I would like to make one guess concerning one particular situation.
Correction In fact, the following guess is wrong (thanks to jvp)! But I don't want to delete this answer, since jvp left a comment to it with a cool link, that shows why the guess is wrong
Guess. Suppose that $M\cong \mathbb CP^1$, and $E=O(1)\oplus...\oplus O(1)$. In this case $X$ is rational, i.e. birational to $\mathbb CP^n$.
I have to say that I don't know how to prove this guess and even not 100% sure that this guess is correct. But first of all, this statement is obviously true if $n=2$ and I rather belive it should be possible to prove this for $n=3$ using twistor theory. The reason to make this guess is that if you have such a line in $X$ you can start to deform it and it deforms (at least locally) in exactly the same way as a line in $\mathbb CP^n$. Also, it is obvious that $X$ is rationally connected in this case.
Of course, this guess is a pure speculation. But maybe someone can prove it? :)
Remark 1. You can always blow up something in $X$ that does not touch the image of the zero section of $N$, so one can only say something (if one can at all) about birational type of $X$.
Remark 2. Note that if you don't assume $X$ to be Kaehler, but only ask it to be complex, then the guess is completely wrong. There exist huge amount of complex 3-folds containing lines with neighborhoods biholomorphic to a neighbourhood of a line in $\mathbb CP^3$. Such examples come from twistor theory -- one should take the twistor space of a conformally flat 4-manifold.
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1
Related paper arxiv.org/PS_cache/math/pdf/0304/0304006v1.pdf – jvp Sep 26 2011 at 17:28
Thank you for this link, this is an interesting paper! – Dmitri Sep 26 2011 at 18:26
You are welcome Dmitri. – jvp Sep 26 2011 at 20:02
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http://mathhelpforum.com/algebra/1607-proof-correct.html
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# Thread:
1. ## Is proof correct?
I have to prove that $a^5-a$ is dividable with 30 for every integer value of $a$.
My proof is:
$a^5-a=a(a^2-1)(a^2+1)$ and $30 = 2(2^2-1)(2^2+1)$.
So if we put $a$ instead of 2 (or 2 instead of a) the proof is proven.
I have found in book much complicated solution but this seems ok to me.
Am I right?
P.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.
2. No, you've proven it for a = 2 but not for all a.
3. Originally Posted by DenMac21
I have to prove that $a^5-a$ is dividable with 30 for every integer value of $a$.
My proof is:
$a^5-a=a(a^2-1)(a^2+1)$ and $30 = 2(2^2-1)(2^2+1)$.
So if we put $a$ instead of 2 (or 2 instead of a) the proof is proven.
I have found in book much complicated solution but this seems ok to me.
Am I right?
It would help if you told us what level course this
is related to, otherwise you risk getting a totally
incomprehensible explanation of how to prove the
result.
P.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.
What you want is the LaTeX guide, its in the following
thread:
http://www.mathhelpforum.com/math-he...read.php?t=266
RonL
4. I am learning Algebra for high-school.
5. [QUOTE=DenMac21]I have to prove that $a^5-a$ is dividable with 30 for every integer value of $a$.
/[QUOTE]
I'm not sure if this qualifies as high school maths but here goes:
(read $|$ as divides so $a|b$ means $a$ divides $b$.
if $30|(a^5-a)$ then $5|(a^5-a)$ and $6|(a^5-a)$, also if
$5|(a^5-a)$ and $6|(a^5-a)$ then $30|(a^5-a)$ because $5$
and $6$ have no common factor.
So we need only show that for any positive integer $a$
that $5|(a^5-a)$ and $6|(a^5-a)$.
Now any positive integer $a$ may be written
$a=k.5+r$,
for some positive integer $k$ and remainder $r=0, 2,3, \mbox{ or }4$.
Now:
$a^5-a=5.K+r^5 - 5.k-r$,
for some positive integer $K$. so if:
$5|(a^5-a)$
then:
$5|(r^5-r)$
where $r=0, 2,3, \mbox{ or }4$, and these five cases can be
checked by hand.
Now repeat the same type of argument for $6$ in place
of $5$ to complete the proof.
RonL
6. Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for $n$ then it must work for $n+1$. Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).
Thus, let $n^5-n$ be divisible by 30.
Then show that $(n+1)^5-(n+1)$ is divisible by 30. Use the binomial theorem
$(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1$
Thus,
$(n^5-n)+5n^4+10n^3+10n^2+5n$
But, $n^5-n$ is divisible by 30,
thus,
$5n^4+10n^3+10n^2+5n$ if divisible by 30 then the whole expression is divisible by 30.
But
$5n^4+10n^3+10n^2+5n$ is divisible by 5
$n^4+2n^3+2n^3+n$. Factor,
$n(n^3+2n^2+2n+1)$.
Notice if this is true for $n=1,2,3,4,5,6$ thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.
In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
7. Originally Posted by ThePerfectHacker
Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for $n$ then it must work for $n+1$. Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).
I was deliberately avoiding induction, as I suspect its
not high school maths
Fermat's little theorem is ruled out for the same reason
RonL
8. This is the proof from the book which is (at the moment) most understable to me:
$a^5-a=(a-1)(a)(a+1)(a^2+1)=$
$(a-1)a(a+1)(a^2-4)+5)=$
$(a-2)(a-1)a(a+1)(a+2) + 5(a-1)a(a+1)$
$(a-2)(a-1)a(a+1)(a+2)$ are five successive integer numbers and they are divisible with 30.
$5(a-1)a(a+1)$ have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.
So, both expressions are divisible with 30 and their sum is then also divisible with 30.
9. Originally Posted by ThePerfectHacker
Just like CaptainBlack said I do not know what high school math is but try it like this:
Show that if it works for $n$ then it must work for $n+1$. Meaning if it works for 1 it must work for 2 but if it works for 2 then it works for 3 but if it works for 3 it must work for 4.... (This is called induction).
Thus, let $n^5-n$ be divisible by 30.
Then show that $(n+1)^5-(n+1)$ is divisible by 30. Use the binomial theorem
$(n+1)^5-(n+1)=n^5+5n^4+10n^3+10n^2+5n+1-n-1$
Thus,
$(n^5-n)+5n^4+10n^3+10n^2+5n$
But, $n^5-n$ is divisible by 30,
thus,
$5n^4+10n^3+10n^2+5n$ if divisible by 30 then the whole expression is divisible by 30.
But
$5n^4+10n^3+10n^2+5n$ is divisible by 5
$n^4+2n^3+2n^3+n$. Factor,
$n(n^3+2n^2+2n+1)$.
You have lost me around here, presumably you are trying to show that
$n(n^3+2n^2+2n+1)$.
is divisible by 6 (or 2 and 3), but I can't follow how you are doing so
Notice if this is true for $n=1,2,3,4,5,6$ thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.
In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
RonL
10. Originally Posted by CaptainBlack
You have lost me around here, presumably you are trying to show that
$n(n^3+2n^2+2n+1)$.
is divisible by 6 (or 2 and 3), but I can't follow how you are doing so
RonL
Just manually filling in n = 1 -> 6.
11. Originally Posted by DenMac21
This is the proof from the book which is (at the moment) most understable to me:
$a^5-a=(a-1)(a)(a+1)(a^2+1)=$
$(a-1)a(a+1)(a^2-4)+5)=$
$(a-2)(a-1)a(a+1)(a+2) + 5(a-1)a(a+1)$
$(a-2)(a-1)a(a+1)(a+2)$ are five successive integer numbers and they are divisible with 30.
$5(a-1)a(a+1)$ have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.
So, both expressions are divisible with 30 and their sum is then also divisible with 30.
The proof is fine but you have to come up with the little 'trick' in the second step of course
12. Originally Posted by TD!
Just manually filling in n = 1 -> 6.
Well that would do it as it's true, but the words don't seem to make that
clear.
RonL
13. Originally Posted by CaptainBlack
Well that would do it as it's true, but the words don't seem to make that
clear.
Perhaps "check" would've been better than "notice", but he meant to say that here. It was also implied a bit in his last paragraph.
Originally Posted by ThePerfectHacker
Notice if this is true for $n=1,2,3,4,5,6$ thus, it is true for all numbers. Because 7 is considered as 1 because it leaves the same remainder. 8 is considered as 2 because it leaves the same remainder....
Q.E.D.
In a similar way you could have checked all numbers from 1 to 30 in the problem by the same reason that 31 is considered as 1 in the last paragraph. But I did it this way because it is faster and leaves less computation.
14. Originally Posted by DenMac21
This is the proof from the book which is (at the moment) most understable to me:
$a^5-a=(a-1)(a)(a+1)(a^2+1)=$
$(a-1)a(a+1)(a^2-4)+5)=$
$(a-2)(a-1)a(a+1)(a+2) + 5(a-1)a(a+1)$
$(a-2)(a-1)a(a+1)(a+2)$ are five successive integer numbers and they are divisible with 30.
Seems to me that you might have to prove this last assertion (not too
difficult) unless its a result you are expected to know. Same comment
applies to similar assertion below.
$5(a-1)a(a+1)$ have as factors 5 and three successive integer numbers which product is 6 so also this expression is divisible with 30.
So, both expressions are divisible with 30 and their sum is then also divisible with 30.
RonL
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# Topic: Coordinate rotation
###### In the News (Wed 15 May 13)
Rotation (disambiguation) - Wikipedia, the free encyclopedia
Generally, it is used to denote (in 3D) the rotation of movement of a rigid body in such a way that any given point of that body remains at a constant distance from some fixed line (in 2D: point).
The retail practice of rotating stock (especially fresh produce/bread) to ensure the stock with the soonest sell by date is at the front of the shelf - meaning it will be sold first.
In baseball pitching, the rotation is the group of starting pitchers for a team, and the order in which they pitch.
en.wikipedia.org /wiki/Rotation_(disambiguation) (297 words)
Coordinates (elementary mathematics) (Site not responding. Last check: 2007-11-05)
The coordinates of a point are the components of a tuple of numbers used to represent the location of the point in the plane or space.
The polar coordinate systems are coordinate systems in which a point is identified by a distance from some fixed feature in space and one or more subtended angles.
In terms of the Cartesian coordinate system, one usually picks O to be the origin (0,0) and L to be the positive x-axis (the right half of the x-axis).
hallencyclopedia.com /Coordinates_(elementary_mathematics) (942 words)
Encyclopedia: Coordinate rotation
In linear algebra and geometry, a coordinate rotation is a type of transformation from one system of coordinates to another system of coordinates such that distance between any two points remains invariant under the transformation.
On the other hand, the composition of a reflection and a rotation, or of a rotation and a reflection (composition is not commutative), will be equivalent to a reflection.
Rotation matrices have a determinant of +1, and reflection matrices have a determinant of -1.
www.nationmaster.com /encyclopedia/Coordinate-rotation (395 words)
Station Information - Coordinate rotation
In linear algebra and geometry, a coordinate rotation is a transformation from one system of coordinates to another system of coordinates, such that distance between any two points remains invariant undr the transformation.
It can also be described by means of quaternions (see quaternions and spatial rotation), an approach which is similar to the use of vector calculus.
More generally, coordinate rotations are represented by orthogonal matrices.
www.stationinformation.com /encyclopedia/c/co/coordinate_rotation.html (345 words)
Optimal Complex-Coordinate Rotation Method for the Calculation of Resonance States (Site not responding. Last check: 2007-11-05)
Complex-coordinate rotation method is useful for the calculation of resonance states encountered in atomic and molecular physics.
In principle, as long as the hidden resonance pole is exposed by the complex-coordinate rotation, the actual angle of rotation used does not matter for exact calculation.
The resulting optimal complex-coordinate rotation method, which has a wide range of applicability, is applied here to the Stark effect for illustration.
flux.aps.org /meetings/BAPSMAY96/abs/S220015.html (163 words)
CSPICE Routines: EUL2M_C
r is a rotation matrix representing the composition of the rotations defined by the input angle-axis pairs.
The resulting matrix r may be thought of as a coordinate transformation; applying it to a vector yields the vector's coordinates in the rotated system.
Viewing r as a coordinate transformation matrix, the basis that r transforms vectors to is created by rotating the original coordinate axes first by angle1 radians about the coordinate axis indexed by axis1, next by angle2 radians about the coordinate axis indexed by axis2, and finally by angle3 radians about coordinate axis indexed by axis3.
www.gps.caltech.edu /~marsdata/cspice/eul2m_c.html (985 words)
The reserved FITS coordinate system keywords
Axis Rotation: the rotation angle between the pixel axis and the physical coordinate axis in degrees.
Rotation Matrix: the rotation (and skew) terms needed to convert from the pixel coordinate system to the physical coordinate system.
Coordinate Rotation: the longitude and latitude in the native coordinate system of the standard system's north pole.
heasarc.gsfc.nasa.gov /docs/heasarc/ofwg/docs/general/wcs_keywords/node6.html (691 words)
Projections and Coordinate Systems
This section regards the simple case of rectangular coordinate systems in which x and y axes are perpendicular to eachother (orthagonal), and the units on each axis are the same and constant.
Techniques for transformation of rectangular coordinate systems come into play when you have two maps that have been projected with the same method, or when the projection for one of the maps is unknown or irrelevant.
Rotation of coordinate systems is simply illustrated as the pinning of the coordinate origin, and a radial motion of the axes while maintaining their orthagonal relationship.
www.geog.ubc.ca /courses/geog516/notes/coordinates/Rect_CoordsLect.html (1733 words)
Rotations (Site not responding. Last check: 2007-11-05)
In the polar coordinate system however, the ordered pair is not represented by x or y but of r and angle originating from the origin or the pole, which is the center of the coordinate system.
Now doing these rotations are pretty straightforward, all we have to do is smack the needed values on our 2d rotation equation for each axis and we're good to go.
Which means for a full rotation on all the axes, do not directly put values until they are fully rotated on the axis that they are rotated.
qbnz.com /pages/tutorials/relsoft/Chapter2 (1662 words)
Matrix and Quaternion FAQ (Site not responding. Last check: 2007-11-05)
Rotation in X transforms Y and Z Rotation in Y transforms X and Z Rotation in Z transforms X and Y The argument to this goes as follows: Given a vertex V = (x,y,z), rotation angles (A,B and C) and translation (D,E,F).
Negating the rotation angle is equivalent to generating the transpose of the matrix.
Because the rotation axis is specifed as a unit direction vector, it may also be calculated through vector mathematics or from spherical coordinates ie (longitude/latitude).
www.flipcode.com /documents/matrfaq.html (7478 words)
Coordinate rotation - (Site not responding. Last check: 2007-11-05)
In other words, a rotation is a type of isometry – note however that there are isometries other than rotations, such as translations, reflections, and glide reflections.
In two dimensions, a counterclockwise rotation of the plane about the origin, where $\left(x,y\right)$ is mapped to $\left(x\text{'},y\text{'}\right)$, is given by the same formulas as a coordinate transformation with a clockwise rotation of the coordinate axes, resulting in a change of coordinates $\left(x,y\right)$ into $\left(x\text{'},y\text{'}\right)$:
This rotation is similar to a two dimensional rotation, except that instead of x and y axes, there are $\mathbf\left\{u_\perp\right\}$ and $\mathbf\left\{v\right\} \times \mathbf\left\{u_\perp\right\}$ axes, both of which are perpendicular to v.
www.grohol.com /psypsych/Coordinate_rotation (993 words)
Coordinate rotation digital computer processor (cordic processor) for vector rotations in carry-save architecture - ...
A coordinate rotation digital computer processor for vector rotations, particularly for solving real-time processing of a vector of the vector components X.sub.0, Y.sub.0, by an angle Z.sub.0, where X.sub.0, Y.sub.0 and Z.sub.0 are each digital words having a plurality of data bits, said data bits having particular significances relative to one another, said processor comprising:
The rotation of a vector is therefore realized by a plurality of identical stages that are composed only of adder/subtractor circuits, devices for realizing shift operations and a sign detection of Z.sub.i (sign (Z.sub.i)).
Given a rotational angle Z.sub.o between +pi/2 and -pi/2 and an accuracy requirement of g=10.sup.-3, two additional bits for places preceding the decimal point are required, so that the input word width of the rotational angle amounts to 13 bits.
www.freepatentsonline.com /5317753.html (6627 words)
OGIP/94-006
Note that this image representation is qualitatively different from the others in that the pixel coordinates are explicitly given in the table whereas in the first 2 cases the pixel coordinates are only implied by the location of the pixel within the array.
Axis Rotation: the rotation angle between the pixel axis and the physical coordinate axis.
Coordinate Rotation: the longitude in the native coordinate system of the standard system's north pole; default value = 180 degrees.
heasarc.gsfc.nasa.gov /docs/heasarc/ofwg/docs/general/ogip_94_006/ogip_94_006.html (2084 words)
BASIC Stamp: CORDIC math (Site not responding. Last check: 2007-11-05)
The next rotation is (tan H = +/- 1/2) in the direction that rotates toward the given angle.
The action taken in sequence is to rotate the coordinates in progressively smaller steps until the transformed x axis lines up with the point P, that is, the y coordinate finally is reduced to zero.
In the calculation of the arctangent and length of the vector, it is the coordinate y that is minimized, while in the calculation of the sine and cosine, it is the difference between the given angle and the rotated angle that is minimized.
www.emesystems.com /BS2mathC.htm (3270 words)
CS184 Lecture 6 summary (Site not responding. Last check: 2007-11-05)
So R is an explicit representation for a coordinate rotation, and vice versa.
That follows because applying R to both the p and the coordinates [X' Y' Z'] moves the latter to normal coordinates [X Y Z], from which point the coordinates of p can be read off.
A general 3D coordinate frame can be described by the directions of its axes X' Y' Z' and the position of its origin t.
www.cs.berkeley.edu /~jfc/cs184f98/lec6/lec6.html (398 words)
Geographic Coordinate System Transformations
One such application is the coordinate transformation introduced to enable the conversion of coordinates expressed in the North American Datum of 1927 (including the Clarke 1866 ellipsoid) to coordinates expressed in the new North American Datum of 1983 which takes the GRS 1980 ellipsoid.
The sign convention is such that a positive rotation about an axis is defined as a clockwise rotation of the position vector when viewed from the origin of the Cartesian coordinate system in the positive direction of that axis; e.g.
Because the coordinate differences are small and the coordinates in the source and target systems are very similar in magnitude, in reversible polynomial transformations it is possible to adopt the same coordinate values for the evaluation point in both source and target systems.
www.posc.org /Epicentre.2_2/DataModel/ExamplesofUsage/eu_cs35.html (5780 words)
Rotation
Given a rotation R and an orthonormal basis B, the matrix representation of R relative to B is a rotation matrix.
For example, if M is the matrix that transforms vectors from J2000 coordinates to body equator and prime meridian coordinates, then the first row is the vector, expressed in J2000 coordinates, that points from the body center to the intersection of the prime meridian and body equator.
The canonical form we've found shows why three-dimensional rotations are very much like two-dimensional rotations: The effect of a three-dimensional rotation on any vector is to rotate the component of that vector that is normal to the rotation axis, and leave the component parallel to the rotation axis fixed.
www.gps.caltech.edu /~marsdata/req/rotation.html (10322 words)
[No title] (Site not responding. Last check: 2007-11-05)
Likewise, if your molecule has D2d symmetry, you will be given the option to rotate the coordinates of your molecule by 45 degrees in the XY plane, which generally should not be done unless GRABFF has the rotation matrix.
The former refers to the set of coordinates generated by the Gxx Z-matrix input routine; the latter is obtained by a center-of-mass translation and principal moments of rotation transformation, and in most cases is identical to the default "master frame" coordinate system used in GAMESS.
Therefore, if the rotation matrix is unknown, one has to use the Z-matrix coordinate system in GAMESS, which in turn generally means that one has to specify the molecular symmetry as C1.
www.osc.edu /PET/CCM/software/tested/source/grabff/doc.html (2508 words)
Modern Physics:Math:Vectors - Wikibooks, collection of open-content textbooks
The direction of a vector in two dimensions is generally represented by the counterclockwise angle of the vector relative to the x axis, as shown in figure 2.
All that remains to be proven for equation (2.6) to hold in general is to show that it yields the same answer regardless of how the Cartesian coordinate system is oriented relative to the vectors.
Since the laws of physics cannot depend on the choice of coordinate system being used, we insist that physical laws be expressed in terms of scalars and vectors, but not in terms of the components of vectors.
en.wikibooks.org /wiki/Modern_Physics:Math:Vectors (1161 words)
Coordinate rotation for numerical control system - Patent 4370720
Multi-dimensional coordinate translation, rotation, and scaling for a machine control system provides translation, rotation, and scaling of coordinates such as for alignment, inch and metric, scaling, and expanded and contracted dimensions.
Data can be entered in either absolute or incremental coordinates, G code selectable, with the capability to mix both coordinate schemes in the same block of commands.
Position offset and axis rotation parameters are commanded either from the tape or with the manual data input (MDI) keyboard.
www.freepatentsonline.com /4370720.html (15749 words)
CORDIC (COordinate Rotation DIgital Computer)
CORDIC works by rotating the coordinate system through constant angles until the angle is reduced to zero.
Let's start with some coordinates (X,Y) which we want to rotate by an angle 'a'.
At each step, it also increments or decrements the X and Y coordinate register by the appropriate value (i.e.
www.ee.ualberta.ca /courses/ee401/microboard/cordic_CCink.html (815 words)
SCATMECH: BRDF_Model
The variable rotation is the rotation angle of the sample in radians about the surface normal.
This variable stores the coordinate system for which the scattering is defined in the code.
Near the surface normal, this coordinate system is close to the directions x and y, where x is a vector in the plane of incidence and the plane of the sample, and y is a vector perpendicular to the plane of incidence.
physics.nist.gov /Divisions/Div844/facilities/scatmech/html/BRDF_Model.htm (1186 words)
Untitled Document (Site not responding. Last check: 2007-11-05)
CORDIC (COordinate Rotation DIgital Computer) algorithms are a family of iterative shift-add algorithms for computing a wide range of functions including certain trigonometric, hyperbolic, linear, and logarithmic functions.
Trigonometric CORDIC algorithms were invented in 1956 as a digital solution for real-time navigation problems, and the theory was later extended to provide solutions to a broader class of functions.
This presentation aims to introduce the main idea used in this family of algorithms with a mathematical formulation, and then show how it can be used to compute various functions.
www.ee.bilkent.edu.tr /~eee591/abslevent.html (105 words)
Coordinate rotation (Site not responding. Last check: 2007-11-05)
In two dimensions, a counterclockwise coordinate rotation from a coordinate system
Then z can be rotated counterclockwise by an angle θ by pre-multiplying it with
Another way is to multiply by a matrix M, which will rotate by an angle
www.sciencedaily.com /encyclopedia/coordinate_rotation (638 words)
Coordinate rotation (Site not responding. Last check: 2007-11-05)
Prior to calculating any fluxes and statistics, gen2 always rotates the two horizontal wind components into along-wind and cross-wind components based on the mean wind direction for the record.
The record mean cross-wind component will be zero by definition.
These estimates of wind speed are used in the third generation program in the bulk flux formula and in all calculations that require an estimate of mean wind speed.
blg.oce.orst.edu /Software/2nd_and_3rdgen/node2.html (152 words)
Citations: Performance Degradation of Genetic Algorithms under Coordinate Rotation - Salomon (ResearchIndex) (Site not responding. Last check: 2007-11-05)
Previously, researchers have examined problems for which coordinate transformations were shown to provide measurable changes in the performances of EAs (e.g.
The reason for this is that three of the five crossover methods used by GADO are rotation independent.
A rotation induces epistasis, which describes a nonlinear interaction between the parameters with respect to the tness function.
citeseer.ifi.unizh.ch /context/272952/0 (1087 words)
U.S. Pregrant 20030227324 - COORDINATE ROTATION OF PRE-DISTORTION VECTOR IN FEEDFORWARD LINEARIZATION AMPLIFICATION ... (Site not responding. Last check: 2007-11-05)
An amplifier system (120, 820) for radio frequency signals comprises a phase and gain adjuster (122) which receives an input signal and a control vector (C) for producing a distortion-adjusted input signal.
The composite control vector is applied to the phase and gain adjuster, thereby enabling the phase and gain adjuster to produce the distortion-adjusted input signal.
In essence, the modified pre-distortion vector generator performs a vector multiplication (coordinate rotation) of a pre-distortion vector so that the resultant modified pre-distortion vector has a proper direction relative to the control vector.
cxp.paterra.com /uspregrant20030227324.html (258 words)
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http://mathoverflow.net/revisions/84650/list
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## Return to Answer
2 added 5 characters in body
You can prove it by using the fact that the following holds always:
$\max_{1 \le k \le n}|S_k| \le \max_{1 \le k \le n/2}|S_k| + \max_{1 \le k \le n/2}|M_k|$
If the left hand side is larger than $\epsilon$ then one of the right hand terms is larger than $\epsilon/2$.
This also shows that the inequality is valid under absolutely no assumptions on the joint distribution of the variables $X_1,\ldots,X_n$.
1
You can prove it by using the fact that the following holds always:
$\max_{1 \le k \le n}|S_k| \le \max_{1 \le k \le n/2}|S_k| + \max_{1 \le k \le n/2}|M_k|$
If the left hand is larger than $\epsilon$ then one of the right hand terms is larger than $\epsilon/2$.
This also shows that the inequality is valid under absolutely no assumptions on the joint distribution of the variables $X_1,\ldots,X_n$.
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http://mathhelpforum.com/differential-geometry/209485-cauchy-sequence.html
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# Thread:
1. ## Cauchy sequence
Would someone help me with this problem?
Define the sequence $\{a_n\}$ in a metric space $(X,d)$ such that $d(a_{n+2},a_{n+1})\leq cd(a_{n+1},a_n)$ for $c\in (0,1)$. Show that $\{a_n\}$ is Cauchy.
2. ## Re: Cauchy sequence
See the proof of Banach fixed-point theorem.
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http://math.stackexchange.com/questions/165090/riemann-sum-estimate-of-cauchy-residue-formula?answertab=active
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# Riemann sum estimate of Cauchy residue formula
Let $e(k) = \exp \left(\frac{2 \pi i k}{N}\right)$ be a root of unity.
I wanted to numerically verify the Cauchy residue formula in Mathematica.
$$\lim_{N \to \infty}\frac{1}{N}\sum_{k=0}^{N-1} \frac{e(k) }{e(k) - a} = \frac{1}{2\pi i} \oint_{|z|=1} \frac{dz}{z - a} = \mathbf{1}( |a| < 1)$$
This computes the winding number of the curve $|z|=1$ (counter-clockwise) around $a \in \mathbb{C}$.
Can this Riemann sum be evaluated exactly? I would like to know the leading-order correction of the Riemann sum to the integral .
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## 1 Answer
Denote the lhs as $f_N(a)$. It can be regarded as a rational function of complex variable $a$. It's straightforward to check that $f_N(a e(i))=f_N(a)$, $i=0,\ldots,N-1$. Which means that $f_N$ is a rational function of $a^N$. From the other hand $\lim_{a\to\infty}f_N(a)=0\;$. This leaves the only possibility $$f_N(a)=\frac c{\prod_{i=0}^{N-1}(e(i)-a)}=\frac c{1-a^N}.$$ Plugging $a=0$ gives $c=1$ so $f_N(a)=\frac1{1-a^N}$. Thus the difference between the sum and the integral is $f_N(a)-1=\frac {a^N}{1-a^N}$.
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This is more clever than the way I was going. Very nice. (+1) – oen Jul 9 '12 at 5:25
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http://math.stackexchange.com/questions/178909/bilinear-optimization-problem
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# Bilinear Optimization Problem
How could I solve the following optimization problem using MATLAB or an other way?
Given ${E}^{1}, {A}^{21}, {A}^{22}, {C}^{1}, {A}^{12}$
$\underset{{C}^{2}, {E}^{2}}{min} {\left \| {C}^{2}{E}^{1} - {A}^{21} \right \|}_{F}^{2} + {\left \| {C}^{2}{E}^{2} - {A}^{22} \right \|}_{F}^{2} + {\left \| {C}^{1}{E}^{2} - {A}^{12} \right \|}_{F}^{2}$
I know the approximated linear solution, assuming ${A}^{22}$ is negligible is given by:
${C}^{2} = {A}^{21} {{E}^{1}}^{T} {\left( {E}^{1} {{E}^{1}}^{T} \right)}^{-1}, {E}^{2} = {\left( {\begin{bmatrix} {C}^{1}\\ {C}^{2} \end{bmatrix}}^{T} \begin{bmatrix} {C}^{1}\\ {C}^{2} \end{bmatrix} \right)}^{-1} {\begin{bmatrix} {C}^{1}\\ {C}^{2} \end{bmatrix}}^{T} \begin{bmatrix} {A}^{12}\\ {A}^{22} \end{bmatrix}$
Yet it would be nice to know how to solve it accurately. Thanks.
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Note that , if you fix $E^2$(or $C^2$), you can exactly solve for $C^2$(or $E^2$). You can take an iterative approach to solving this. 1. Initialize $E^2$ arbitrarily 2. Solve for $C^2$, say $C^{'}_2$ 3. Set $C^2$ to $C^{'}_2$ 4. Solve for $E^2$, say $E^{'}_2$ 5. Set $E^2$ to $E^{'}_2$ 6. Goto Step 2 – TenaliRaman Aug 9 '12 at 6:57
– Drazick Aug 9 '12 at 7:06
– TenaliRaman Aug 9 '12 at 7:18
The problem is when I use 'lsnonlin' in MATLAB it asks for the Jacobian where the input is a vector. Meaning the expression should be derived by every single item in C2 and E2. For instance, how do you derive the expression by ${C}^{2}_{1, 1}$? – Drazick Aug 9 '12 at 7:46
Derivative of a function always lies in the domain of the function. This means that the derivative of your objective has to be a matrix. If you want to translate this into a vector, then rewrite your objective function in terms of the individual matrix components. Then take derivative w.r.t each component. – TenaliRaman Aug 9 '12 at 8:14
## 2 Answers
Since your problem is unconstrained you can us `fminunc` to solve it in MATLAB (given an initial starting point $x_0$). Perhaps, you can use your approximate solution as $x_0$.
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Yet it is a function of 2 variables - C2 and E2. – Drazick Aug 5 '12 at 5:57
OK, it turns out the best way to solve it is using MATLAB's 'lsnonlin' function. Now, I just need to compute the Jacobain of the function relative to C2, E2 and the Frobenius Norm.
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http://www.bioscience.ws/encyclopedia/index.php?title=Ionization_energy
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Ionization energy
Ionization energy - Wikipedia, the free encyclopedia
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# Ionization energy
Periodic trend for ionization energy. Each period begins at a minimum for the alkali metals, and ends at a maximum for the noble gases.
The ionization energy of an atom or molecule describes the amount of energy required to remove an electron from the atom or molecule in the gaseous state.
X → X+ + e-
The units for ionization energy vary from discipline to discipline. In physics, the ionization energy is typically specified in electron volts (eV) and refers to the energy required to remove a single electron from a single atom or molecule. In chemistry, the ionization energy is typically specified as a molar quantity (molar ionization energy or enthalpy) and is reported in units of kJ/mol or kcal/mol (the energy needed to remove one mole of electrons from one mole of atoms or molecules).
The nth ionization energy refers to the amount of energy required to remove an electron from the species with a charge of (n-1). For example, the first three ionization energies are defined as follows:
1st ionization energy
X → X+ + e-
2nd ionization energy
X+ → X2+ + e-
3rd ionization energy
X2+ → X3+ + e-
## Values and trends
Main article: Molar ionization energies of the elements
Generally the (n+1)th ionization energy is larger than the nth ionization energy. Always, the next ionization energy involves removing an electron from an orbital closer to the nucleus. Electrons in the closer orbitals experience greater forces of electrostatic attraction; thus, their removal requires increasingly more energy. Ionization energy becomes greater up and to the right of the periodic table.
Some values for elements of the third period are given in the following table:
Successive molar ionization energies in kJ/mol
(96.485 kJ/mol = 1 eV/particle)
Element First Second Third Fourth Fifth Sixth Seventh
Na 496 4,560
Mg 738 1,450 7,730
Al 577 1,816 2,881 11,600
Si 786 1,577 3,228 4,354 16,100
P 1,060 1,890 2,905 4,950 6,270 21,200
S 999.6 2,260 3,375 4,565 6,950 8,490 27,107
Cl 1,256 2,295 3,850 5,160 6,560 9,360 11,000
Ar 1,520 2,665 3,945 5,770 7,230 8,780 12,000
Large jumps in the successive molar ionization energies occur when passing noble gas configurations. For example, as can be seen in the table above, the first two molar ionization energies of magnesium (stripping the two 3s electrons from a magnesium atom) are much smaller than the third, which requires stripping off a 2p electron from the very stable neon configuration of Mg2+.
Ionization energy is also a periodic trend within the periodic table organization. Moving left to right within a period or upward within a group, the first ionization energy generally increases. As the atomic radius decreases, it becomes harder to remove an electron that is closer to a more positively charged nucleus. Ionization energy increases from left to right in a period and decreases from top to bottom in a group.
## Electrostatic explanation
Atomic ionization energy can be predicted by an analysis using electrostatic potential and the Bohr model of the atom, as follows (note that the derivation uses Gaussian units).
Consider an electron of charge -e and an atomic nucleus with charge +Ze, where Z is the number of protons in the nucleus. According to the Bohr model, if the electron were to approach and bind with the atom, it would come to rest at a certain radius a. The electrostatic potential V at distance a from the ionic nucleus, referenced to a point infinitely far away, is:
$V = \frac{Ze}{a} \,\!$
Since the electron is negatively charged, it is drawn inwards by this positive electrostatic potential. The energy required for the electron to "climb out" and leave the atom is:
$E = eV = \frac{Ze^2}{a} \,\!$
This analysis is incomplete, as it leaves the distance a as an unknown variable. It can be made more rigorous by assigning to each electron of every chemical element a characteristic distance, chosen so that this relation agrees with experimental data.
It is possible to expand this model considerably by taking a semi-classical approach, in which momentum is quantized. This approach works very well for the hydrogen atom, which only has one electron. The magnitude of the angular momentum for a circular orbit is:
$L = |\mathbf r \times \mathbf p| = rmv = n\hbar$
The total energy of the atom is the sum of the kinetic and potential energies, that is:
$E = T + U = \frac{p^2}{2m_e} - \frac{Ze^2}{r} = \frac{m_e v^2}{2} - \frac{Ze^2}{r}$
Velocity can be eliminated from the kinetic energy term by setting the Coulomb attraction equal to the centripetal force, giving:
$T = \frac{Ze^2}{2r}$
Solving the angular momentum for v and substituting this into the expression for kinetic energy, we have:
$\frac{n^2 \hbar^2}{rm_e} = Ze^2$
This establishes the dependence of the radius on n. That is:
$r(n) = \frac{n^2 \hbar^2}{Zm_e e^2}$
Now the energy can be found in terms of Z, e, and r. Using the new value for the kinetic energy in the total energy equation above, it is found that:
$E = - \frac{Ze^2}{2r}$
At its smallest value, n is equal to 1 and r is the Bohr radius a0. Now, the equation for the energy can be established in terms of the Bohr radius. Doing so gives the result:
$E = - \frac{1}{n^2} \frac{Z^2e^2}{2a_0} = - \frac{Z^213.6eV}{n^2}$
## Quantum-mechanical explanation
According to the more complete theory of quantum mechanics, the location of an electron is best described as a probability distribution. The energy can be calculated by integrating over this cloud. The cloud's underlying mathematical representation is the wavefunction which is built from Slater determinants consisting of molecular spin orbitals. These are related by Pauli's exclusion principle to the antisymmetrized products of the atomic or molecular orbitals.
In general, calculating the nth ionization energy requires calculating the energies of $Z-n+1$ and $Z-n$ electron systems. Calculating these energies exactly is not possible except for the simplest systems (i.e. hydrogen), primarily because of difficulties in integrating the electron correlation terms. Therefore, approximation methods are routinely employed, with different methods varying in complexity (computational time) and in accuracy compared to empirical data. This has become a well-studied problem and is routinely done in computational chemistry. At the lowest level of approximation, the ionization energy is provided by Koopmans' theorem.
## Vertical and adiabatic ionization energy in molecules
Figure 1. Franck–Condon principle energy diagram. For ionization of a diatomic molecule the only nuclear coordinate is the bond length. The lower curve is the potential energy curve of the neutral molecule, and the upper curve is for the positive ion with a longer bond length. The blue arrow is vertical ionization, here from the ground state of the molecule to the v=2 level of the ion.
Ionization of molecules often leads to changes in molecular geometry, and two types of (first) ionization energy are defined – adiabatic and vertical.1
Adiabatic ionization energy: The adiabatic ionization energy of a molecule is the minimum amount of energy required to remove an electron from a neutral molecule, i.e. the difference between the energy of the vibrational ground state of the neutral species and that of the positive ion. The specific equilibrium geometry of each species does not affect this value.
Vertical ionization energy: Due to the possible changes in molecular geometry that may result from ionization, additional transitions may exist between the vibrational ground state of the neutral species and vibrational excited states of the positive ion. In other words, ionization is accompanied by vibrational excitation. The intensity of such transitions are explained by the Franck–Condon principle, which predicts that the most probable and intense transition corresponds to the vibrational excited state of the positive ion that has the same geometry as the neutral molecule. This transition is referred to as the "vertical" ionization energy since it is represented by a completely vertical line on a potential energy diagram (see Figure).
For a diatomic molecule, the geometry is defined by the length of a single bond. The removal of an electron from a bonding molecular orbital weakens the bond and increases the bond length. In Figure 1, the lower potential energy curve is for the neutral molecule and the upper surface is for the positive ion. Both curves plot the potential energy as a function of bond length. The horizontal lines correspond to vibrational levels with their associated vibrational wave functions. Since the ion has a weaker bond, it will have a longer bond length. This effect is represented by shifting the minimum of the potential energy curve to the right of the neutral species. The adiabatic ionization is the diagonal transition to the vibrational ground state of the ion. Vertical ionization involves vibrational excitation of the ionic state and therefore requires greater energy.
In many circumstances, the adiabatic ionization energy is often a more desirable physical quantity since it describes the difference in energy between the two potential energy surfaces. However, due to experimental limitations, the adiabatic ionization energy is often difficult to determine, whereas the vertical detachment energy is easily identifiable and measurable.
## Analogs of Ionization Energy to Other Systems
While the term ionization energy is largely used only for gas-phase atomic or molecular species, there are a number of analogous quantities that consider the amount of energy required to remove an electron from other physical systems.
Electron binding energy: A generic term for the ionization energy that can be used for species with any charge state. For example, the electron binding energy for the chloride ion is the minimum amount of energy required to remove an electron from the chlorine atom when it has a charge of -1. In this particular example, the electron binding energy has the same magnitude as the electron affinity for the neutral chlorine atom. In another example, the electron binding energy refers the minimum amount of energy required to remove an electron from the dicarboxylate dianion -O2C(CH2)8CO2-.
Work Function: The minimum amount of energy required to remove an electron from a solid surface.
## See also
• Bragg-Gray Cavity Theory
• Electronegativity
• The work function is the energy required to strip an electron from a solid.
• Koopmans' theorem
• Di-tungsten tetra(hpp) has the lowest recorded ionization energy for a stable chemical compound.
• Electron affinity
• Debye length
• Ionization energies of the elements (data page)
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http://mathhelpforum.com/algebra/205257-write-complex-number-form-bi.html
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# Thread:
1. ## Write this complex number on the form a + bi
I'm dealing with this problem where I'm to write the complex number below on the form a + bi.
$\frac {\sqrt{2+i}}{\sqrt{2-i}}$
I've tried writing it as
$\sqrt {\frac {2+i}{2-i}}$
Then I tried multiplying by the conjugate, writing it using polar coordinates, considering the squre root as an exponent of 0.5 etc. However, I just kept reaching a dead end. The expression is supposedly equal to
$\fr {2\sqrt{5}}{5} + \fr {i\sqrt{5}}{5}$
2. ## Re: Write this complex number on the form a + bi
You suspect have misread the answer you are supposed to get. It should be $2\sqrt{5}/5+ i\sqrt{5}/5$ or $2\frac{\sqrt{5}}{5}+ i\frac{\sqrt{5}}{5}$ either of which is equal to $\frac{2}{\sqrt{5}}+ \frac{i}{\sqrt{5}}$.
You get that by, as you say, multiplying by the conjugate. Then "rationalize the denominator".
3. ## Re: Write this complex number on the form a + bi
Originally Posted by MathCrusader
I'm dealing with this problem where I'm to write the complex number below on the form a + bi.
$\frac {\sqrt{2+i}}{\sqrt{2-i}}$
I've tried writing it as
$\frac {2\sqrt{5}}{5} + \frac {i\sqrt{5}}{5}$
$\frac {\sqrt{2+i}}{\sqrt{2-i}}\frac {\sqrt{2-i}}{\sqrt{2-i}}=\frac {\sqrt{5}}{2-i}=\frac {2\sqrt{5}+\sqrt{5}i}{5}$
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http://mathhelpforum.com/differential-geometry/147930-calculation-lie-brackets.html
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# Thread:
1. ## calculation of lie brackets
Hello!
We have two vectorfields on $\mathbb{R}^n$ X(x)=Ax+a, Y(x)=Bx+b, where A,B are nxn-matrices and $a, b \in \mathbb{R}^n$.
Is it true that the lie bracket [X,Y]=0?
2. What happens when you plug your vector fields into the definition of the Lie bracket?
3. Thanks for your answer. Thats what i did. I just want to be sure if my calculation is right. The lie bracket of X and Y is zero since the Hessian Matrix of a smooth function is symmetric.
Is the result correct?
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http://physics.stackexchange.com/questions/41437/do-christoffel-symbols-commute?answertab=active
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# Do Christoffel symbols commute?
Do Christoffel symbols commute? For example, does $\Gamma^{e}_{db}\Gamma^{c}_{ea} = \Gamma^{c}_{ea}\Gamma^{e}_{db}$?
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3
They are just numbers; so yes, they commute. – user10001 Oct 22 '12 at 17:15
no they don't, he showed you what he meant by commute with the pattern of index contraction. – Ron Maimon Oct 22 '12 at 17:15
Yes, to clarify, I am specifically referring to the 'commutator' above. – user12345 Oct 22 '12 at 17:17
3
You didn't do anything with the indices above rather than swap the order of the terms. If you take the indices as abstract indices, then yes, they commute by the commutation of ordinary numbers. – Jerry Schirmer Oct 22 '12 at 17:19
3
Do Christoffel symbols commute? As matrices they do not. See @Ron's answer. does $\Gamma^{e}_{db}\Gamma^{c}_{ea} = \Gamma^{c}_{ea}\Gamma^{e}_{db}$? Yes, because they are just numbers. – user10001 Oct 22 '12 at 18:22
show 3 more comments
## 3 Answers
In classical theory, all observables commute. The components $\Gamma^a_{bc}$ are just real numbers so of course that they commute.
In quantum theory, they don't commute. It's probably a bit laborious to calculate the commutator.
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Sorry, @Ron, but $\Gamma^a_{bc}$ is no "linear transformation" or any other transformation. It's a dynamical variable. Classically, it's a pure number. Quantum mechanically, it's a $q$-number, an operator. What you may have computed is the "commutator" in which the indices $b,c$ of $\Gamma^a_{bc}$ are treated as two indices of a matrix and one multiplies two Gammas as matrices. Then they don't commute. But in the products, the indices would be written differently. The original question writes a clear formula which proves that the OP is asking whether the components commute, and they do. – Luboš Motl Oct 25 '12 at 7:58
The two matrix indices are a and c, not b and c, and treating them as a matrix is the definition of connection. You are doing rhetoric now, not honest science. I will delete my comment above yours. – Ron Maimon Oct 25 '12 at 12:58
At any rate, your answer is wrong and my answer is right. – Luboš Motl Oct 25 '12 at 21:07
yes, in the technical sense that I interpreted "commutator of connection coefficients" in a non-vacuous way. You were right about the OP's confusion, so +1. That's more rhetoric by the way, since we all agree on what happened. By the way, Lubos, you might be interested in Thomas Gold's book "The Deep Hot Biosphere", which makes the case (completely persuasively IMO) that we will never run out of oil, since oil is abiogenic and produced deep inside the Earth. While, unlike you, I believe in global warming and think this is a bad thing, it is still good science by Gold. – Ron Maimon Oct 25 '12 at 23:21
This might be overkill, but here it goes:
Let $\pi:E\rightarrow B$ be an arbitrary fiber bundle with typical fiber $F$ and $\Phi=\tau\times\varphi$ be a local trivialization over $U\subset B$ $$\Phi:\pi^{-1}(U)\rightarrow U\times F$$
Geometrically, the Christoffel symbol $\Gamma$ of a connection is an element $$\Gamma\in\mathrm{Hom}(\tau^*(\mathrm TU),\varphi^*(\mathrm TF))$$ which is a fancy way of saying that for each $e\in E$, there is a linear map $$\Gamma(e):\mathrm T_{\tau(e)}U\rightarrow\mathrm T_{\varphi(e)}F$$
In general, talking about the composition $$\Gamma(e)\circ\Gamma(e')$$ makes no sense as domain and codomain don't agree.
In relativity however, $E=\mathrm TB$ which allows us to identify these spaces and we end up with $$\Gamma(V):\mathrm T_bU\rightarrow\mathrm T_bU$$ where $V\in\mathrm T_bU$ is a tangent vector, eg a 4-velocity.
As $\Gamma(V)$ is linear, it can be expressed in local coordinates via matrix multiplication $$W^i\mapsto\Gamma(V)^i{}_j W^j$$ In case of the Levi-Civita connection, the map $$V\mapsto\Gamma(V)$$ is linear as well and we arrive at $$W^i\mapsto(\Gamma_k V^k)^i{}_j W^j = \Gamma_k{}^i{}_j V^k W^j$$
The compositions read $$(\Gamma(V)\circ\Gamma(W))^i{}_j = \Gamma_k{}^i{}_a \Gamma_l{}^a{}_j V^k W^l \\ (\Gamma(W)\circ\Gamma(V))^i{}_j = \Gamma_l{}^i{}_a \Gamma_k{}^a{}_j V^k W^l$$ which in general do not commute.
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hopefully correct – Christoph Oct 22 '12 at 23:52
Correct, but I said the same thing without obfuscation. – Ron Maimon Oct 23 '12 at 2:40
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@Ron: which is why I wasn't sure if I should add this answer; in the end, I decided that explicitly mentioning the domains of the map and showing where the indices come from had sufficient surplus value – Christoph Oct 23 '12 at 6:03
I see, yes, it does add value, +1. – Ron Maimon Oct 23 '12 at 15:23
No they don't commute in this sense, except you screwed up the commutator. It should be
$$\Gamma^{e}_{db}\Gamma^c_{ea} - \Gamma^c_{eb}\Gamma^{e}_{da}$$
Which is
$$\Gamma_b \Gamma_a - \Gamma_a \Gamma_b$$
in a matrix form, where I have taken one upper and one lower index and suppressed them in the Christoffel symbol (which lower index doesn't matter, becuase of the symmetry on the lower indices). This has the interpretation I give below.
The Christoffel symbols are the coordinate form of the infinitesimal rotations associated with parallel transporting a vector a ways along a short distance. If all the frames were orthonormal, the parallel transport would be SO rotations, and the infinitesimal form would be a bunch of stuff in the Lie Algebra of SO, and these are antisymmetric matrices (or the upper index representation of antisymmetric forms for the Lorentzian case). Their commutators tell you when the rotations corresponding to moving in a certain direction is noncommutative with the rotation corresponding to moving in another direction.
For a coordinate basis for the tangent space, the basis vectors are not orthonormal, so the connection coefficients don't obey the Lie algebra conditions, but they still stay noncommutative. You can easily verify this in a generic example by direct computation, but it is also obvious from those cases where you happen to choose coordinates where the coordinate basis is orthogonal, like spherical coordinates, and think about the different coordinate rotations associated with parallel transport in different infinitesimal directions.
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Look at the index contractions he's made. This isn't the curvature term. – Jerry Schirmer Oct 22 '12 at 17:24
@JerrySchirmer: Yeah, I didn't see is screw up. I think he meant what I say above--- it's easy to screw up index stuff like this. – Ron Maimon Oct 22 '12 at 17:26
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No, it wasn't a screw up. I meant exactly what I'd written in the question. – user12345 Oct 22 '12 at 17:35
Yet, my answer is still correct, this is the proper interpretation of non-commutativity of Christoffel symbols, and this is what people mean when they usually say it. I believe the answer is superior to others, and does not deserve downvotes. It doesn't matter that I misunderstood what user16307's screw up was, the question is best answered as above. – Ron Maimon Oct 22 '12 at 18:07
Ron's answer has quite useful information, and moreover its not irrelevant to OP's question. +1 – user10001 Oct 22 '12 at 18:09
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http://math.stackexchange.com/questions/53969/what-does-formal-mean/168730
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# What does “formal” mean?
I know the definition of formal power series, power series and polynomials. But what does the adjective "formal" mean? In google English dictionary, does it mean "9. Of or relating to linguistic or logical form as opposed to function or meaning" or maybe another one in the link?
Or does "formal" have some mathematical meaning which is other than usual dictionary meaning?
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## 5 Answers
I see formal used in at least two senses in mathematics.
• Rigorous, i.e. "here is a formal proof" as opposed to "here is an informal demonstration."
• "Formal manipulation," that is, manipulating expressions according to certain rules without caring about convergence, etc.
Confusingly they can mean opposite things in certain contexts, although "formal manipulations" can be made rigorous in many cases.
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Isn't there a notion of "formal" in algebraic geometry? – Damien Jul 27 '11 at 1:44
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– Qiaochu Yuan Jul 27 '11 at 1:54
I think the etymology of the word shows how the senses are related. ‘Formal’ comes from ‘form’; the association with rigour is via Hilbert's formalist school. By ‘rigour’ what is really meant is formal manipulations of logical propositions in accordance to the rules of inference, rather than following ‘intuition’. – Zhen Lin Jul 27 '11 at 2:00
– Bruno Stonek Jul 27 '11 at 3:17
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I've always thought of the latter meaning as the meaning "of or related to form" — a formal power series is something that has the form of a power series, formal manipulations are those that work on the form directly (without caring about what the expression may "mean" in the analysis sense), etc. – ShreevatsaR Jul 27 '11 at 4:42
As an example, formal power series is analyzed without regard to convergence. Really, what is of interest is the sequence of coefficients.
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And don't forget the notion of formal space arising in rational homotopy theory.
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The word "formal" in "formal power series" is indicating that you are considering all objects that are algebraically "like a power series". This is opposed to its use in analysis where you spend a lot of time figuring out for which $x$ the series converges.
Basic analysis goes like this:
"$\displaystyle\sum_{n=1}^{\infty} x^n$ is a series which converges for $|x|<1$ and therefore the function $f(x) = \displaystyle\sum_{n=1}^{\infty} x^n$ has the domain $|x| < 1$".
You then proceed to use the function and talk about derivatives and integrals on the restricted domain. If the series has very few points of convergence such as $\displaystyle\sum_{n=1}^{\infty} n!x^n$ which converges only for $x=0$, then casting it as the function $g(x) = \displaystyle\sum_{n=1}^{\infty} n!x^n$ can only have domain $x=0$ and its value is $g(0)=0$. Pretty boring function when it comes to derivatives and integrals!
When you study formal power series, you ignore the consideration of convergence and use the series as it is presented as an algebraic entity, so even though $g$ only converges at $x=0$, you ignore that and focus on other properties of the series.
Another common use of the word "formal" is with a "formal system" which is basically a big rulebook for an artificial language comprised of an alphabet (a list of symbols), a grammar (a way of arranging those symbols), and axioms (initial lists of symbols to start from). The word "formal" here is needed because it is very prim and proper and only allows manipulations according to the grammar and axioms; you can't combine symbols in any way like you can in English (for example this ee cummings poem is an "acceptable" combination of the symbols of English, but is also seemingly "wrong" according to our standard grammar).
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The first series certainly converges for $|x| < 1$; why wouldn't you consider it a power series with a finite radius of convergence? (I would have chosen an example with zero radius of convergence, such as $\sum n! x^n$.) – Qiaochu Yuan Jul 27 '11 at 1:39
Yes, @Qiaochu's series is an excellent example; even with the zero radius of convergence, it can be manipulated formally to produce... interesting and useful identities. – J. M. Jul 27 '11 at 2:23
Thank you for the suggestion! – tomcuchta Jul 27 '11 at 4:39
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"not considered as a power series in analysis since it does not converge for any $x\in \mathbf R$" - actually, it does... if $x=0$ that is. – J. M. Jul 27 '11 at 4:45
When I was learning about logic as an undergraduate, I recall being told that the word "formal", with respect to "formal languages" meant that the "form" of expressions written in that language had primacy.
In other words, rules for manipulating expressions in a formal language could be given in terms of the form of the expression only, without needing to know to what values the variables in the expression were bound.
So a formal language permits us to use relatively simple pattern-matching algorithms to decide which transformations of an expression are valid at any given time.
In this context, formality is linked to the simplicity of the rules that define the set of valid transformations of an expression.
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http://en.wikipedia.org/wiki/Indeterminate_equation
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# Indeterminate equation
An indeterminate equation, in mathematics, is an equation for which there is more than one solution; for example, 2x = y is a simple indeterminate equation. Indeterminate equations cannot be solved uniquely. For example, the equations
$\ ax + by = c,$
$\ x^2 - Py^2 = 1,$
$x^2=1$
where a, b, c, and P are given integers (provided that P is not a square number), are indeterminate equations. Equations of the second form are named Pell's equations.
## References
This algebra-related article is a stub. You can help Wikipedia by expanding it.
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http://quant.stackexchange.com/tags/risk-neutral-measure/hot
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# Tag Info
## Hot answers tagged risk-neutral-measure
10
### Black-Scholes and Fundamentals
I think you are interpreting too much into the matter. The $-\frac12\sigma^2$ is just a correction term that comes from Jensen's inequality. You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution). I think there are no deeper truths to be found here.
7
### Formal proof for risk-neutral pricing formula
Note first that this key equation is only assumed to hold true under some extra assumptions. Typically those assumptions are taken to be about absence of arbitrage, though it is possible to weaken them somewhat if you are willing to consider portfolio arguments or collectively agreeable objective function. Anyway, the argument is this: if all the risk can ...
6
### Financial Mathematics - Martingales example
The classic argument using risk-neutral pricing is to assume that discounted stock prices are $\tilde{P}$-martingales where $\tilde{P}$ is the risk-neutral probability measure. Then, you know that $$\frac{S_t}{(1+r)^t}=\tilde{E}[\frac{S_T}{(1+r)^T} | \mathcal{F}_t]$$ by definition of a martingale process. As the discounts are non-stochastic, you can ...
6
### Formal proof for risk-neutral pricing formula
You can find a simple proof in the discrete time case at http://kalx.net/ftapd.pdf. I'm not sure what you are trying to derive with your Ito calculus, but here is a rigourous derivation of the Black-Sholes/Merton PDE: http://kalx.net/dsS2011/bms.pdf. The Black-Scholes '73 derivation is not mathematically correct. The modern approach does not use so called ...
4
### Risk Neutral Probability and invariant measure
No, you obtain a risk-neutral measure by any change of measure; invariance is far more restrictive. Because in your formula $\mu\circ f^{-1} (A)=\mu(A)$, it has to be for any $A$. Risk-neutrality can be seen as a way to inject into your model a list of market prices you really want to not be exposed to: once they are taken into account (i.e. once you made ...
4
### How to transform process to risk-neutral measure for Monte Carlo option pricing?
The use of risk-neutral measure is based on the ability to arbitrage away the instantaneous risk of contingent claims. Although for forward contracts the hedge quantity is 1.0, in the general contingent claims case we must assume it varies instantaneously with the market state. The Girsanov Theorem tells us what the difference is, instantaneously, between ...
3
### Black-Scholes and Fundamentals
One thing to keep in mind here is that the world of risk-free/arbitrage-free models is not necessarily the real world. Specifically, this equation $$\mu = r - \frac{1}{2}\sigma^2$$ occurs not because this is the way stocks behave in reality (they don't! For S&P 500, long-run $\mu$ is closer to 6-9%, if I recall correctly), but because using any ...
3
### St Petersburg lottery pricing & short investing horizons
What a great question -- it touches on many issues at the core of quantitative finance. This answer might be a lot more than you bargained for, but it's too interesting to pass up. References Mostly, this subject falls somewhere at the intersection of these three highly-interrelated topics: risk-neutral valuation, rational pricing and the fundamental ...
3
### Is drift rate the same as interest rate in risk-neutral random walk when using Monte Carlo for option pricing?
Yes. The risk neutral and the real path share the same volatility, so the difference is in the drift rate, where the risk-neutral path drifts with the risk-free rate r. You may want to check out Paul Willmots book, esp. ch. 26, for applications.
3
### How does one go from measure P to Q(risk-neutral) when modeling an asset paying dividends?
Just following Musiela Rutkowski (the link redirects to Amazon). The risk neutral measure is derived form imposing that the present value of a self financed portfolio (i.e.; no infusion or withdraw of money) is a martingale. A portfolio can be seen as a stochastic process where its value at time $t$ is given by $$V_t = \phi^0_tP_t + \phi^1_tS_t\ ,$$ ...
3
### Formal proof for risk-neutral pricing formula
Since I did not get any comments to my latest update, and since I find it quite convincing, I hereby post my solution as an answer. maybe I can prove that Q exists assuming a lognormal distribution of $S_t$. Assuming $dS_t = \mu S_t dt + \sigma S_t dW_t$ By Itô, $d(e^{-rt} S_t) = -r e^{-rt} S_t dt + e^{-rt} S_t dS_t$. Replacing with the definition of ...
3
### Formal proof for risk-neutral pricing formula
The only requirement if you are risk neutral is the property of martingale on your discounted stock price $M_t=e^{-rt} S_t$. But if you apply Itô $d( S_t\cdot e^{-rt} e^{rt})=d(M_t\cdot e^{rt})=r_tM_te^{rt}dt + ..dW_t=r_tS_tdt+..dW_t$ you see see that under the risk free probability, the asset price must have $r_t$ as yield and to answer to your question, ...
2
### Formal proof for risk-neutral pricing formula
The first think you have to ask is ¿¿What price??? Monetary price or equity price?? All answers,the ones I read, related to monetary price, but are equity price really risk free???? One of the biggest problem with Black Scholes (personal opinion) is that they consider the behave of equity price as monetary price: Solve this ODE: S(t)'/dt= r*S(0), this tell ...
2
### How does one go from measure P to Q(risk-neutral) when modeling an asset paying dividends?
ad) "Is it normal to assume no other drift?" Under measure P you might have drift. You could use it as a working assumption, but in general indices drift every now and then. So, no, usually you do not assume away the drift. "The index is described as "following a geometric Brownian motion", which to me says that the there is no other drift going on" ...
2
### American Option price formula assuming a logLaplace distribution?
Have you looked at using Laplace in a Monte Carlo simulation? Here is how you price American style options within a MC framework: http://www2.math.uu.se/research/pub/Jia1.pdf and the Longstaff, Schwartz paper: http://escholarship.org/uc/item/43n1k4jb#page-1 Regarding the discretization of a process that draws its random variables from a Laplace ...
2
### How to choose model parameters?
Yes, you do really use market prices to calibrate models derived under the risk-neutral measure. That is the whole reason why risk-neutral measures are utilized, to a) ease the calculations but mostly b) because under no arbitrage and one price for each security assumptions (among couple other other assumptions) the price derived under the risk neutral ...
2
### Pricing forward contract on a stock
In my mind you are simply right: you arrive at $$f(t,S) = S(t) - K e^{-r(T-t)}.$$ Assume that $t=0$, so we are at the inception of the contract, then $$f(0,S) = S(0) - Ke^{-r T}.$$ If you choose $K = S(0) e^{r T}$ then the contract value at inception is zero. This simply means that the fair price for the forward is given by $K= S(0) e^{r T}$ which is ...
2
### What mathematical characteristics are required from the asset price process in order to stay within the RNP framework?
This depends. I am not aware of a general risk neutral pricing framework applying to all asset classes and/or stochastic processes. In order to reach more general statements about risk neutral pricing you need to consider jumps and autocorrelation depending on the asset regarded, maybe stochastic interest rates and/or volatility. If I remember correctly, in ...
1
### St Petersburg lottery pricing & short investing horizons
This may or may not be helpful, since I don't have anything to point you to that specifically addresses the high skewness of the distribution you mention. However, this sounds like it is probably an idiosyncratic risk, and that certainly has bearing on whether or not it would be priced. In the standard capital asset pricing model, the marginal investor ...
1
### Version of Girsanov theorem with changing volatility
I don't think Girsanov's formula works when the volatilities are different between the P measure and P* measure. P and P* will be singular with respect to each other. Please see Prof. Goodman's class notes on page 11 at http://www.math.nyu.edu/faculty/goodman/teaching/StochCalc2012/notes/Week10.pdf . Also, from [ ...
1
### Version of Girsanov theorem with changing volatility
The Girsanov theorem applies to any compatible change of measure, including a volatility change. The version you have written above is a simplified version for drift changes only, but if you look in any good stochastic calculus book, you will see that full version just requires that you be able to compute the cross-variation of the two processes.
1
### Risk neutral probability in binomial short rate model assumed to be 0.5?
I'll take a stab. In short rate modelling we start by postulating dynamics under pricing measure $Q$ and then calibrating our model to market prices. General short rate dynamics under $Q$ can be described as $dr(t)=b(t_0+t,r(t))dt+\sigma(t_0+t,r(t))dW(t)$ , where $W$ is $Q$ Brownian motion. Note that the short rate process itself is not a $Q$-martingale ...
1
### What mathematical characteristics are required from the asset price process in order to stay within the RNP framework?
I assume that by "this machinery breaks down" you mean when it breaks down as theory, but not as a practical tool. I would say that the exact point where risk neutral pricing approach fails is when the payoff is no more attainable. There exist a precise mathematical characterization for attainable payoffs (see the book of Hans Föllmer, Alexander Schied, ...
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http://math.stackexchange.com/questions/119511/odd-powers-in-the-taylor-series-expansion-of-operatornamelog1ez
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# Odd powers in the Taylor series expansion of $\operatorname{Log}(1+e^z)$.
I noticed a question while reviewing Taylor series expansions that has been bugging me. The question is: How many nonzero terms with odd exponents does the Taylor series $\operatorname{Log} (1 + e^z)$ about $z = 0$ have?
I'm having trouble figuring out how to do this. Can anyone help?
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## 2 Answers
Let $\displaystyle f(z) = \log (1 + e^z)$.
Now
$$f'(z) = \frac{e^z}{1+e^z} = \frac{e^z - 1}{2(e^z + 1)} + \frac{1}{2} = g(z) + \frac{1}{2}$$
Now we have that
$$g(-z) = \frac{e^{-z} - 1}{2(e^{-z} + 1)} = \frac{1 - e^z}{2(1 + e^z)} = -g(z)$$
Thus $\displaystyle g(z)$ is an odd function and its power series will only contain odd terms.
Since $\displaystyle f(z) = C + \frac{z}{2} + \int g(z)$, the only odd power in the series for $\displaystyle f(z)$ is $\displaystyle z$, and its coefficient is $\displaystyle \frac{1}{2}$.
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$$\log(1 + e^z) = \log(e^{z/2}(e^{z \over 2} + e^{-{z \over 2}}))$$ $$=\log(e^{z \over 2}) + \log(e^{z \over 2} + e^{-{z \over 2}})$$ $$= {z \over 2} + \log(2\cosh({z \over 2}))$$ Since $\cosh({z \over 2})$ is even, so is any function of $\cosh({z \over 2})$. So all terms of the Taylor series of $\log(2\cosh({z \over 2}))$ have even powers. Thus ${z \over 2}$ is the only term of $\log(1 + e^z)$'s Taylor series with odd powers of $z$ showing up.
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http://quant.stackexchange.com/tags/collective-risk-model/hot
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# Tag Info
## Hot answers tagged collective-risk-model
5
### Copula models and the distribution of the sum of random variables without Monte Carlo
If the density of $(X,Y)$ is known, then you may obtain the density of the sum $X+Y$ simply by applying the Jacobi's transformation formula, which describes the density of the transformed random variable $g(X,Y)$ for $g(x,y) = (x+y, x)$. Integrating out the $x$-component yields the density of $X+Y$. See Jacod/Protter Probability Essentials ch. 12 for ...
3
### Copula models and the distribution of the sum of random variables without Monte Carlo
In general setting this is quite a tough problem and it looks like just switching from regular multivariate probability to copulas doesn't make it easier. In general case you need to rely on numerical methods for integration. There is a nice overview of the problem in Copula Theory and Its Applications: Proceedings of the Workshop Held in Warsaw, 25-26 ...
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http://mathhelpforum.com/pre-calculus/211845-x-x-2lnx.html
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# Thread:
1. ## x/x^2lnx
All!
Please, how do I get f'(x) and f''(x) if I have f(x)=x/(x^lnx)?
Many thanks!
2. ## Re: x/x^2lnx
Originally Posted by Boo
All!
Please, how do I get f'(x) and f''(x) if I have f(x)=x/(x^lnx)?
Many thanks!
Which do you have? $f(x) = \frac{x}{x^2 ln(x)}$
or
$f(x) = \frac{x}{x^{ln(x)}}$
-Dan
Or maybe:
$f(x) = \frac{x}{x^{2ln(x)}}$
3. ## Re: x/x^2lnx
Hello, Boo!
$\text{Given: }\:f(x) \:=\:\dfrac{x}{x^{\ln x}}. \quad \text{Find }f'(x)\text{ and }f''(x).$
You need Logarithmic Differentiation.
We have: . $y \;=\;\dfrac{x}{x^{\ln x}}$
Take logs: . $\ln(y) \;=\;\ln\left(\frac{x}{x^{\ln x}}\right)$
. . . . . . . . $\ln(y) \;=\;\ln(x) - \ln\left(x^{\ln x}\right)$
. . . . . . . . $\ln(y) \;=\;\ln(x) - \ln x\ln x$
. . . . . . . . $\ln(y) \;=\;\ln(x) - [\ln(x)]^2$
Differentiate implicitly:
. . . . . . . . $\frac{y'}{y} \;=\;\frac{1}{x} - 2\ln(x)\!\cdot\!\frac{1}{x}$
. . . . . . . . $\frac{y'}{y} \;=\;\frac{1}{x}\left[1 - 2\ln(x)\right]$
. . . . . . . . $y' \;=\;y\cdot \frac{1-2\ln(x)}{x}$
. . . . . . . . $y' \;=\;\frac{x}{x^{\ln x}}\cdot\frac{1-2\ln(x)}{x}$
Hence: . . $y' \;=\;\frac{1-2\ln(x)}{x^{\ln x}}$
Now repeat the process to find the second derivative, $y''.$
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http://stats.stackexchange.com/questions/24022/why-levene-test-of-equality-of-variances-rather-than-f-ratio
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Why Levene test of equality of variances rather than F ratio?
SPSS uses the Levene test to evaluate homogeneity of variances in the independent group t-test procedure.
Why is the Levene test better than a simple F ratio of the ratio of the variances of the two groups?
-
– chl♦ Mar 2 '12 at 22:45
1 Answer
You could use an F test to assess the variance of two groups, but the using F to test for differences in variance strictly requires that the distributions are normal. Using Levene's test (i.e., absolute values of the deviations from the mean) is more robust, and using the Brown-Forsythe test (i.e., absolute values of the deviations from the median) is even more robust. SPSS is using a good approach here.
Update In response to the comment below, I want to clarify what I'm trying to say here. The question asks about using "a simple F ratio of the ratio of the variances of the two groups". From this, I understood the alternative to be what is sometimes known as Hartley's test, which is a very intuitive approach to assessing heterogeneity of variance. Although this does use a ratio of variances, it is not the same as that used in Levene's test. Because sometimes it is hard to understand what is meant when it is only stated in words, I will give equations to make this clearer.
Hartley's test: $$F=\frac{s^2_2}{s^2_1}$$ Levene's test / Brown-Forsythe test: $$F=\frac{MS_{b/t-levels}}{MS_{w/i-levels}}$$
In all three cases, we have ratios of variances, but the specific variances used differ between them. What makes Levene's test and the Brown-Forsythe test more robust (and also distinct from any other ANOVA), is that they are performed over transformed data, whereas the F ratio of group variances (Hartley's test) uses the raw data. The transformed data in question are the absolute values of the deviations (from the mean, in the case of Levene's test, and from the median, in the case of the Brown-Forsythe test).
There are other tests for heterogeneity of variance, but I'm restricting my discussion to these, as I understood them to be focus of the original question. The rationale for choosing amongst them is based on their performance if the original data are not truly normal; with the F test being sufficiently non-robust that it is not recommended; Levene's test being slightly more powerful than BF if the data really are normal, but not quite as robust if they aren't. The key citation here is O'Brien (1981), although I could not find an available version on the internet. I apologize if I misunderstood the question or was unclear.
-
1
Because Levene's statistic is a ratio of squares constructed from those absolute residuals, and is referred to an F distribution, it is not immediately apparent that it should be any more robust than other tests based on ratios of squares! You may be thinking of more robust variants, such as the Brown-Forsythe test. See a good discussion by @chl at stats.stackexchange.com/questions/2591/…. – whuber♦ Mar 2 '12 at 22:54
@whuber, thanks for the comment & link. There is too much to respond to in a comment, so I edited my answer. I believe what I am trying to get at should be clearer now. However, if I misunderstood, or am simply wrong, I can delete this answer. – gung Mar 3 '12 at 7:20
The (new) last paragraph makes your point well (+1). – whuber♦ Mar 3 '12 at 15:39
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http://math.stackexchange.com/questions/tagged/philosophy?sort=votes&pagesize=15
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http://physics.stackexchange.com/questions/53110/generic-born-stability-criteria
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# Generic Born stability criteria
The tensorial form of Hooke's law for the strain-stress relationship in a crystal is (in the Voigt notation):
where $\sigma$ is the strain, $\epsilon$ is the stress and C is the stiffness tensor:
For a crystalline system of the cubic symmetry class, the stiffness tensor reduces to:
The Born criterion for the stability of an unstrained crystal is that free energy must be represented by a positive defined quadratic form. In the case of a cubic crystal, it is known that this is equivalent to the following three conditions on the elastic constants:
$$C_{11} - C_{12} > 0$$ $$C_{44} > 0$$ $$C_{11} + 2 C_{12} > 0$$
But what about lower symmetry classes? What is the generic Born criterion for stability of a crystal? I have quite convinced myself that all the eigenvalues of $C$ must be positive, but I cannot find confirmation of that anywhere. Is it right? Is there a reference on that topic?
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## 4 Answers
This derivation is correct for cubic crystals without external pressure.
A fresh review can be found in Rev. Mod. Phys. 84, 945 (2012) Lattice instabilities in metallic elements http://rmp.aps.org/abstract/RMP/v84/i2/p945_1
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While I can't find a reference for this, I think your criterion is correct. Here's the argument I would use - if it's the same as yours, then maybe it's right! The elastic energy is (http://ciks.cbt.nist.gov/garbocz/manual/node8.html) $$E = \frac{1}{2}\int d^d r \epsilon_i C_{ij} \epsilon_j$$ where $C_{ij}$ is symmetric. Because $C_{ij}$ is symmetric and real, it can be diagonalized, and its eigenvectors are complete and orthogonal (https://en.wikipedia.org/wiki/Hermitian_matrix). Then you can expand $\epsilon_j$ in terms of the eigenvectors of $C_{ij}$, $\bf{v}^{(k)}$, i.e. $C_{ij} v_j^{(k)} = \lambda_k v_i^{(k)}$. $$\epsilon_j = \sum_k (\bf{v}^{(k)} \cdot \bf{\epsilon}) v^{(k)}_j$$ (This assumes that $\bf{v}^{(k)}$ is orthonormal, which we can always choose without loss of generality.) We then find that $$E = \frac{1}{2} \int d^d r \lambda_k (\bf{v}^{(k)} \cdot \bf{\epsilon})^2$$.
Stability means that there are no modes that will lead to an unbounded decrease in energy, and no marginal modes - i.e. that $\lambda_k > 0$ for all $k$ - your positive definite quadratic form.
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Yes, that's how I figured it too… – F'x Apr 2 at 7:43
I've found a good analysis of the stability conditions for a crystal's elastic constants, both unstrained and under stress, in:
J. W. Morris Jr and C. R. Krenn, Philos. Mag. A 2000, 12, 2827–2840
To quote them:
In the linear elastic limit the conditions of internal stability reduce to the familiar condition that the 6 x 6 matrix $C_{ij}$ of elastic moduli have no negative eigenvalues.
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A detailed analysis may be found also in arXiv:1104.0173 [astro-ph.SR], D. A. Baiko "Shear modulus of neutron star crust", Eqs. (29)-(33). http://arxiv.org/abs/1104.0173
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That paper is, at best, only very marginally related to my question :( – F'x Apr 2 at 7:43
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# Tag Info
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5
### What does the * mean in spherical harmonics?
The superscript $*$ is a common notation for complex conjugate. Going back to check, (3.53) in the blue English edition states $$Y_{l,m} = \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}P^m_l(\cos\theta)e^{im\phi}$$ which is followed by (3.54) $$Y_{l,-m}(\theta,\phi) = (-1)^m Y^*_{l,m}(\theta,\phi),$$ making is clear that it has to be complex conjugation.
4
### What is the Lagrangian for a relativistic charge that includes the self-force?
classical electrodynamics mainly deals with two kinds of proplems: a) The action of a field on a charged particle and b) the fields arising from the motion of such a field. Of course, this can only be approximative but it turns out that a lot of phenomena can be described in this way. However, you are right, an entire treatment would include a) and b) ...
4
### Is it true that any system of accelerating charges will radiate?
The claim that accelerated charges must radiate is simply false. There are very many simple situations in which they do, but in general things should be examined on a case-by-case basis; there is not simple thumb rule like "acceleration yields radiation." The simplest way to see this is to consider a wire carrying a constant current. This situation is ...
4
### Does GR provide a maximum electric field limit?
The answer is: the Motion Mountain book is wrong, your TA and John Baez are correct A couple answers here are discussing specific solutions in GR. I think there is a much easier and more general way to answer this. In special relativity, we can choose to use a different coordinate system (we don't need to physically change our motion as some posters seem ...
4
### Why do electrons around nucleus radiate light according to classical physics
Because there is a changing configuration of charges with time, a time-dependent dipole moment. The electron's field is coming from a different center at different times, so that the field is oscillating in magnitude with a period the orbital period of the electron. When you have an oscillation of electric fields, it sets up oscillations of the entire ...
3
### Einstein's Field equations and impulse-energy tensor
The stress-energy tensor $T_{\mu\nu}$ on the right side of the Einstein equation characterizes all of the various forms of "stuff" in the spacetime. If there are electromagnetic fields in the spacetime, then the stress-energy tensor of the electromagnetic field is part of that $T_{\mu\nu}$, along with contributions from other forms of energy, mass, etc.
3
### Trouble with the Lorentz law of force: Incompatibility with special relativity and momentum conservation?
Whoever the PRL referee(s) was/were, they should have sent it back to the author to put the argument into a manifestly covariant formalism. The editors should have done the same before the paper got to a referee. As it is, everybody has to waste time unpicking the 3-d vector mess. 3-d vectors have a perfectly legitimate place in Physics, but not if one is ...
3
### Non-linear dynamics of classical hydrogen atom
No, it is not possible, and the argument is simple--- there is no dimensional parameter with unit of length, so if there were a stable equilibrium at one radius, there would be many such equilibria obtained by rescaling the original solution to a one-parameter family of solutions. In fact, it is easier to see that the stable solution is for the electron to ...
3
### Classical (or semi-classical) interpretation of photoelectric effect?
It may be a reference to the fact that you can reproduce the characteristics of the photoelectron production in a model which treats the incident light classically, but treats the matter in the target quantum mechanically. This is explained in Mandel and Wolf's book (chapter 9), which explains how a simple semiclassical calculation can be used to derive the ...
3
### Noether theorem and classical proof of electric charge conservation
By the word classical we will mean $\hbar=0$, and we will use the conventions of Ref. 1. The Lagrangian density for Maxwell theory with various matter content is $$\tag{1} {\cal L} ~=~{\cal L}_{\rm Maxwell} + {\cal L}_{\rm matter} ,$$ $$\tag{2} {\cal L}_{\rm Maxwell}~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu},$$ \tag{3} {\cal L}_{\rm matter}~=~{\cal L}_{\rm ...
2
### What is the conserved canonical momentum for a relativistically moving charge in a static Coulomb electric field?
Put speed of light $c=1$ and use sign convention $(-,+,+,+)$. To have that the canonical momentum $$\vec{p}~=~\frac{\partial L}{\partial d_t \vec{x}}~=~ \gamma m_{0}d_t \vec{x} +q\vec{A}$$ is conserved $d_t \vec{p}=0$, and a $\vec{x}$-translational symmetry of the Lagrangian L~=~T-U, \qquad T~=~-\frac{m_0}{\gamma}, \qquad U~=~ q (\phi -\vec{A}\cdot ...
2
### Does a static electric field and the conservation of momentum give rise to a relationship between $E$, $t$, and some path $s$?
We have a thermodynamic understanding of energy violation, in terms of perpetual motion, but momentum violation is somewhat less immediately paradoxical in a changing background, maybe because we have intuition that the field has momentum in this case. But the field momentum is irrelevant for this question. Mechanical momentum should be conserved in a ...
2
### What is the Lagrangian for a relativistic charge that includes the self-force?
Solving both, Lorentz force and Maxwell's equation, does include radiation reaction. Radiation reaction emerges from the interaction of the charge with the field it has emitted itself, the self-field. If you do not want to solve Maxwell's equations and treat the fields as external, you can use Lorentz-Abraham-Dirac equation with some known inconsistencies ...
2
### How do we visualise antenna reception of individua radiowave photons building up to a resonant AC current on the antenna?
Here is an experimentalist's view of the question: 1) one photon hits the antenna and raises a molecular electron band to a higher energy level, and it will fall back to its lower one, with the characteristic electromagnetic transition time of the order of 10^-16sec, giving the energy to the antenna grid of molecules. One photon will just disappear. 2)a ...
2
### Interaction of matter with EM fields
Generally speaking, the answer to both questions is linked to some number becoming increasingly large so that, for atoms you have a large density of higher excited states (think to Rydberg atoms as an example) or for electromagnetic field one has such a large number of photons that a coherent state is a good description of it and an average field can be ...
2
### Interaction of matter with EM fields
I'm not sure there is a generic answer to your questions other than the trivial "don't bother including the quantization when the accuracy of your result isn't compromised by making this approximation". I know that doesn't really help much, because you may not be able to verify this until you've done the calculation including the quantization anyway. You ...
2
### Explanation for speed of an electrical impulse
I think i came to the origins of this equation. In all likelihood, this equation describes not a speed of an electrical impulse but a direct current power transmitted via a superconducting coaxial cable. A proof: Consider a simple transmission DC coaxial cable. To eliminate the energy losses due to Joule heating in the cable, the inner(of radius $r$) ...
2
### Trouble with the Lorentz law of force: Incompatibility with special relativity and momentum conservation?
There is a comment by Daniel A. T. Vanzella with a counter argument that essentially removes the paradox. He uses the natural covariant formulation of the problem. In this, you can see that the lorentz force has no spatial component in the charge/dipole rest frame, but the four force is not null. The dipole develops a time dependent angular momentum ...
2
### What is the answer to Feynman's Disc Paradox?
Chapter 17 precedes chapter 27 which covers field momentum and so he's looking for a simple explanation involving mechanical angular momentum. The initial angular momentum of the system is carried by the initial current in the coil, and so there's no paradox. Note also that the magnetic field can't collapse immediately, but has to disspiate the stored ...
2
### What happens to electrons in an open circuit?
The circuit likely was closed by his body, or by a grounding wire he was holding. At least that's one way the demonstration has been done. I assume the other end was in contact with the generator. Another way is to suspend the tube so it is not in electrical contact with anything, and swing it around, so that you observe a momentary discharge when the ...
Only top voted, non community-wiki answers of a minimum length are eligible
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http://mathoverflow.net/questions/91161/eigenvectors-and-eigenvalues-of-tridiagonal-matrix
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## Eigenvectors and eigenvalues of Tridiagonal matrix
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Hi, is it possible to analytically evaluate the eigenvectors and the eigenvalues of a tridiagonal matrix of the form :
$$\mathcal{T}^{a}_n(p,q) = \begin{pmatrix} 0 & q & 0 & 0 &\cdots & 0 & 0 & 0 \\ p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\ 0 & p & 0 & q &\cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\ 0 & 0 & 0 & 0 &\cdots & p & 0 & q \\ 0 & 0 & 0 & 0 & \cdots & 0 & p & 0 \end{pmatrix}$$
where $p>0\ \ \& \ \ q > 0$ and where there are $n$ rows and $n$ columns in the matrix above?
Furthermore is it possible to do the same for the equivalent matrix where periodic boundary conditions are implemented? i.e.
$$\mathcal{T}^{b}_n(p,q) = \begin{pmatrix} 0 & q & 0 & 0 &\cdots & 0 & 0 & p \\ p & 0 & q & 0 &\cdots & 0 & 0 & 0 \\ 0 & p & 0 & q &\cdots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots& \vdots \\ 0 & 0 & 0 & 0 &\cdots & p & 0 & q \\ q & 0 & 0 & 0 & \cdots & 0 & p & 0 \end{pmatrix}$$
Thanks
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See my edits, which fix the second case. – Denis Serre Mar 16 2012 at 6:47
I wrote my answer before Denis had fixed the second case. Apologies for any redundancy. – Yemon Choi Mar 16 2012 at 7:54
## 4 Answers
Here is the calculation of the spectrum of the first matrix, which I write $pJ+qK$ with $K=J^T$. Define $D={\rm diag}(1,a,a^2,\ldots,a^{n-1})$. Then $D^{-1}JD=a^{-1}J$ and $D^{-1}KD=aK$. Thus, taking $a=\sqrt{p/q}$, one sees that you matrix is similar to $\sqrt{pq}(J+K)$. Its eigenvalues are $\sqrt{pq}$ times those of $J+K$. The spectrum of the latter matrix is made of numbers $2\cos\frac{k\pi}{n+1}$ for $k=1,\ldots,n$.
The second case is easy too. Eigenvectors are $n$-periodic solutions of the recursion $qu_{j+1}+pu_{j-1}=\lambda u_j$. This means that some power of $\omega=\exp\frac{2i\pi}n$ is a root of the characteristic equation $qr^2+p=\lambda r$. Whence the spectrum $\lambda_1,\ldots,\lambda_n$ $$\lambda_j=p\omega^{-j}+q\omega^j.$$
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2
I don't think the eigenvalues of the second matrix given here are correct. The answer should be real when $p=q$, which the answer supplied here is not. – alex o. Mar 15 2012 at 0:40
@alex. You're right. See my edit. – Denis Serre Mar 15 2012 at 9:53
2
@Denis Serre - I think its still not right. It isn't true that the spectrum will be real - for example, if $p=2,q=1,n=4$ then $-i$ is an eigenvalue with eigenvector $[1,i,-1,-i]$. – alex o. Mar 15 2012 at 20:43
@Alex. Yes, my trick does not work for the second matrix. I withdraw this point. – Denis Serre Mar 15 2012 at 22:13
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Another way to look at this problem, from the ground up, is to expand the characteristic polynomial of your first matrix, $\mathcal{T}_n(p,q)$, along the last row and then along the last column, from which you'll quickly get `$$\det(\mathcal{T}_n(p,q)-\lambda)=-\lambda \det(\mathcal{T}_{n-1}(p,q)-\lambda)-pq \det(\mathcal{T}_{n-2}(p,q)-\lambda).$$` You can then compare this with the recurrence relations for the standard orthogonal polynomials; it is then rather easy to match it to that for the Chebyshev polynomials of the first kind, `$$T_{n+1}(x)=2x T_n(x)-T_{n-1}(x),$$` with $T_0(x)=1$, $T_1(x)=x$,which should make it clear that the identification is `$$T_n(\lambda)=\frac{1}{2(\sqrt{pq})^n}\det\left(\mathcal{T}_n(p,q)+2\sqrt{pq}\lambda\right),$$` or the equivalent $\det(\mathcal{T}_n(p,q)-\lambda)=2(\sqrt{pq})^n T_n\left(\frac{-\lambda}{2\sqrt{pq}}\right)$. Since the Chebyshev polynomials are given by $T_n(\cos(\theta))=\cos(n\theta)$, this gives all the eigenvalues. The reason this happens is that (Denis' symmetric version of) your matrix is the Jacobi matrix for the Chebyshev polynomials. You can exploit this to get the eigenvectors in terms of lower order polynomials $T_m$, $m\leq n$, evaluated at the eigenvalue; the construction is in
Gautschi, Walter. Orthogonal Polynomials, Computation and Approximation. Numerical Mathematics and Scientic Computation, Oxford University Press, 2004..
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Thanks. Just remembered this morning I had read a derivation in the context of field theory several years back, in "Field theory, a path integral approach" by Ashok Das, chapter 3.3 Harmonic Oscillator, Matrix Method. This approach parallels the one above. – surfstack Mar 15 2012 at 20:51
@ Emilio: Could you please show how you get $\det(\mathcal{T}_n(p,q)-\lambda)=2(pq)^n T_n(-\lambda/2)$? – Johann Cigler Mar 16 2012 at 9:02
@Johann: You can't - I made an error with the normalizations, it's corrected now. I think the easiest way is to say $\det(\mathcal{T}_n(p,q)-\lambda)=a b^n T_n\left(c\lambda\right)$ for some constants $a, b, c$. Comparing the recursion relations gives constraints for $b$ and $c$, and the first polynomial, for $n=1$, gives the overall normalization $a$. – Emilio Pisanty Mar 16 2012 at 19:29
@Emilio: I think that the correct formula is as in my answer $\det(\mathcal{T}_n(p,q)-\lambda)=(\sqrt{pq})^n U_n\left(\frac{-\lambda}{2\sqrt{pq}}\right)$ with the Chebyshev polynomials of the second kind. – Johann Cigler Mar 16 2012 at 20:37
@Emilio: Perhaps I should add a proof. Let $a(n, x)=\det(\mathcal{T}_n(p,q)-x)$. As you have shown this sequence satisfies the recurrence $a(n, x)=-x a(n-1, x)-pqa(n-2,x)$ with initial values $a(1, x)=-x$ and $a(2, x)= x^2-pq.$ Therefore $b(n,x)= (1/{\sqrt {pq}) ^n}a(n, - 2 \sqrt {pq} x)$ satisfies $b(n,x)=2xb(n-1,x)-b(n-2,x)$ with initial values $b(1,x)=2x$ and $b(2,x)=4x^2-1$ and therefore $b(n,x) = {U_n}(x).$ – Johann Cigler Mar 17 2012 at 7:35
show 1 more comment
If I have read your question correctly, the second matrix is a so-called circulant matrix, and so one can read off the spectrum using known methods. Wikipedia gives you a formula that can be used.
That said, I prefer to approach things from scratch. A circulant $n\times n$ matrix can always be written as $f(S)$ where $f$ is a polynomial of degree $\leq n-1$ and $S$ is a cyclic shift matrix of order $n$. If the first column of the matrix reads $a_0, \dots, a_{n-1}$, then take $S$ to be the shift matrix $e_1\mapsto e_2 \mapsto \dots \mapsto e_n \mapsto e_1$, and take $f(z)=a_0+a_1z+ \dots + a_{n-1}z^{n-1}$.
So in your case, the matrix is just $A=pS+ qS^{n-1} = pS + qS^{-1}$, and since we know the eigenvalues of $S$ (they are the $n$ distinct complex $n$th roots of unity) and corresponding one-dimensional eigenspaces, this allows us to write down the eigenvalues of $A$
```$$\{ p\omega^j + q\omega^{-j} : j=0,1,\dots, n-1\} \quad\quad(\omega=\exp(2\pi i/n)
$$```
with corresponding eigenvectors (depending on the values of $p$ and $q$ some of the eigenvalues may have non-trivial multiplicity).
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Note: at the time I wrote this answer, Denis Serre's answer was temporarily missing an answer to the second part. – Yemon Choi Mar 16 2012 at 7:52
@ Emilio: Could you please show how you get $\det(\mathcal{T}_n(p,q)-\lambda)=2(pq)^n T_n(-\lambda/2)$?
Sorry I wanted to comment another answer. I do not know how to delete this post.
Edit
I think the first problem can be reduced to Chebyshev polynomials of the second kind ${U_n}(x)$ because
$$\det \left( {{T_n}(p,q) - \lambda } \right) = - \lambda \det \left( {{T_{n - 1}}(p,q) - \lambda } \right) - pq\det \left( {{T_{n - 2}}(p,q) - \lambda } \right).$$
implies
$$\det \left( {{T_n}(p,q) - \lambda } \right) ={(\sqrt {pq} )^n}{U_n}\left( {-\frac{\lambda }{{2\sqrt {pq} }}} \right).$$
From the fact that the zeros of ${U_n}(x)$ are $\cos \frac{{k\pi }}{{n + 1}}$ the eigenvalues are
$2\sqrt {pq} \cos \frac{{k\pi }}{{n + 1}}.$
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http://math.stackexchange.com/questions/16783/show-fx-leq-a-2-if-fx-leq-a
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# Show $|f'(x)| \leq (A/2)$ if $|f''(x)|\leq A$
Problem: Given that $f$ is differentiable at $[0,1]$ and $f(0)=f(1)=0$. If $\forall x\in (0,1)$ $|f''(x)|\leq A$ show that $\forall x \in [0,1]$ $|f'(x)| \leq (A/2)$.
My attempt was to to develop a Taylor series for $f(x)$ and $f'(x)$ around point $c \in (0,1)$ where $f'(c)=0$.
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## 1 Answer
A hint: For a given $c\in[0,1]$ compute $f(0)$ and $f(1)$ by means of a Taylor expansion at $c$ (with Lagrange remainder term) and draw conclusions.
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Actually I found a proof without using Taylor expansion. MVT theorem is enough to prove the above statement. I just assume $c$ (where $f'(c) = 0$) is $\leq 1/2$ and go from there. It also works in the case $\geq 1/2$ from symmetry. – Ma.H Jan 8 '11 at 20:00
1
How did you use the assumption $f(0)=f(1)=0\thinspace$? – Christian Blatter Jan 9 '11 at 12:31
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http://physics.stackexchange.com/questions/tagged/laws-of-physics?sort=faq&pagesize=15
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# Tagged Questions
DO NOT USE THIS TAG just because the question deals with a law of physics!
learn more… | top users | synonyms
2answers
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### What happens to matter in a standard model with zero Higgs VEV?
Suppose you reset the parameters of the standard model so that the Higgs field average value is zero in the vacuum, what would happen to standard matter? If the fundamental fermions go from a finite ...
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### Why is it thought that normal physics doesn't exist inside the event horizon of a black hole?
A black hole is so dense that a sphere around it called the event horizon has a greater escape velocity than the speed of light, making it black. So why do astronomers think that there is anything ...
2answers
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### Why didn't Newton just propose the 2nd Law and leave it at that?
Why didn't Newton just propose the 2nd Law ($F=\dot{p}$) and leave it at that? The 2nd Law implicitly contains the first, doesn't it? If so, it seems he wasn't following his own Rule #1 of Book 3 of ...
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http://mathhelpforum.com/algebra/140007-express-single-logarithm-please-help-few-questions.html
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# Thread:
1. ## express as a single logarithm: PLEASE HELP with a few questions?
hello everyone, i am new here and plan to use this forum often, you guys seem very friendly. I am in college algebra right now and im having a huge problem with a few questions on my review packet for the exam coming up, the test is in a few days and these questions are going to be similar to the exam. i have been looking at my math book for the last 3-4 hours and i just cant find any similar questions, im stuck. very frustrated & stressed from this as well.
i scanned the math problem packet my teacher gave me to my pc, i mostly need help on "Express as a single logarithm" problems, they are the last 4 questions. but if you got extra time i would really appreciate it if you can help me on the top questions as well, but i know that might be asking to much so if anything just an answer to atleast one question from problems 11,12,13, or 14 would be great. thank you so much
here is the link of my problem packet, its a .jpg
if possible show your work to how you solved the problem, because i really would like to know how to do these. but if you cant i understand, it would be hard to type it out.
Thanks again everyone, this really is going to help me understand it a lot better.
2. Originally Posted by murkr
hello everyone, i am new here and plan to use this forum often, you guys seem very friendly. I am in college algebra right now and im having a huge problem with a few questions on my review packet for the exam coming up, the test is in a few days and these questions are going to be similar to the exam. i have been looking at my math book for the last 3-4 hours and i just cant find any similar questions, im stuck. very frustrated & stressed from this as well.
i scanned the math problem packet my teacher gave me to my pc, i mostly need help on "Express as a single logarithm" problems, they are the last 4 questions. but if you got extra time i would really appreciate it if you can help me on the top questions as well, but i know that might be asking to much so if anything just an answer to atleast one question from problems 11,12,13, or 14 would be great. thank you so much
here is the link of my problem packet, its a .jpg
http://img197.imageshack.us/img197/5791/page1math.jpg
if possible show your work to how you solved the problem, because i really would like to know how to do these. but if you cant i understand, it would be hard to type it out.
Thanks again everyone, this really is going to help me understand it a lot better.
Dear murkr,
To answer these questions you only need to know some laws of logarithms. Basically,
$log_{a}A+log_{a}B=log_{a}(AB)$
$log_{a}A-log_{a}B=log_{a}\left(\frac{A}{B}\right)$
$c~log_{a}A=log_{a}A^{c}$
For a more detailed explanation please refer List of logarithmic identities - Wikipedia, the free encyclopedia
3. Originally Posted by Sudharaka
Dear murkr,
To answer these questions you only need to know some laws of logarithms. Basically,
$log_{a}A+log_{a}B=log_{a}(AB)$
$log_{a}A-log_{a}B=log_{a}\left(\frac{A}{B}\right)$
$c~log_{a}A=log_{a}A^{c}$
For a more detailed explanation please refer List of logarithmic identities - Wikipedia, the free encyclopedia
i have a problem i got two different answers for question 8
is this right
$%288%29%5C;%5C;%5Clog_%7B17%7D%5C,%5Cleft%28%5Cfrac%7B%5Csqrt[4]%7B5%7D%7D%7Bn%5E2m%5E2%7D%5Cright%29$
$%5Clog_%7B17%7D%5Cleft%28%5Cfrac%7B%5Csqrt[4]%7B5%7D%7D%7Bn%5E2m%5E2%7D%5Cright%29%20%5C;=%5C;%20%5Clog_%7B17%7D%5Cleft%28%5Csqrt[4]%7B5%7D%5Cright%29%5C,-%5C,%5Clog_%7B17%7D%5Cleft%28n%5E2m%5E2%5Cright%29$
. . . . . . . . . .$=%5C;%5Clog_%7B17%7D%5Cleft%285%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5Cright%29%20%5C,-%5C,%20%5Cbigg[%5Clog_%7B17%7D%28n%5E2%29%20%5C,+%5C,%5Clog_%7B17%7D%28m%5E2%29%5Cbigg]$
. . . . . . . . . .$=%5C;%5Cfrac%7B1%7D%7B4%7D%5C,%5Clog_%7B17%7D%285%29%20%5C,-%5C,2%5C,%5Clog_%7B17%7D%28n%29%20%5C,-%5C,2%5C,%5Clog_%7B17%7D%28m%29%5C;%5C;%5C;%20%28A%29$
i got C for #8
the problem above says A for #8
4. Originally Posted by murkr
i have a problem i got two different answers for question 8
is this right
$%288%29%5C;%5C;%5Clog_%7B17%7D%5C,%5Cleft%28%5Cfrac%7B%5Csqrt[4]%7B5%7D%7D%7Bn%5E2m%5E2%7D%5Cright%29$
$%5Clog_%7B17%7D%5Cleft%28%5Cfrac%7B%5Csqrt[4]%7B5%7D%7D%7Bn%5E2m%5E2%7D%5Cright%29%20%5C;=%5C;%20%5Clog_%7B17%7D%5Cleft%28%5Csqrt[4]%7B5%7D%5Cright%29%5C,-%5C,%5Clog_%7B17%7D%5Cleft%28n%5E2m%5E2%5Cright%29$
. . . . . . . . . .$=%5C;%5Clog_%7B17%7D%5Cleft%285%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5Cright%29%20%5C,-%5C,%20%5Cbigg[%5Clog_%7B17%7D%28n%5E2%29%20%5C,+%5C,%5Clog_%7B17%7D%28m%5E2%29%5Cbigg]$
. . . . . . . . . .$=%5C;%5Cfrac%7B1%7D%7B4%7D%5C,%5Clog_%7B17%7D%285%29%20%5C,-%5C,2%5C,%5Clog_%7B17%7D%28n%29%20%5C,-%5C,2%5C,%5Clog_%7B17%7D%28m%29%5C;%5C;%5C;%20%28A%29$
i got C for #8
the problem above says A for #8
Hi murkr,
"A" is the correct answer for #8.
5. Originally Posted by masters
Hi murkr,
"A" is the correct answer for #8.
thank you for varying that for me.
i just got one last question, i understand all the problems expect problems #10 and #14
if you can do just one or both of these i would really appreciate it, im having such a hard time with these my books have no examples of these. without you guys i wouldnt know what to do thank you.
6. ## express as a single log
Originally Posted by murkr
hello everyone, i am new here and plan to use this forum often, you guys seem very friendly. I am in college algebra right now and im having a huge problem with a few questions on my review packet for the exam coming up, the test is in a few days and these questions are going to be similar to the exam. i have been looking at my math book for the last 3-4 hours and i just cant find any similar questions, im stuck. very frustrated & stressed from this as well.
i scanned the math problem packet my teacher gave me to my pc, i mostly need help on "Express as a single logarithm" problems, they are the last 4 questions. but if you got extra time i would really appreciate it if you can help me on the top questions as well, but i know that might be asking to much so if anything just an answer to atleast one question from problems 11,12,13, or 14 would be great. thank you so much
here is the link of my problem packet, its a .jpg
http://img197.imageshack.us/img197/5791/page1math.jpg
if possible show your work to how you solved the problem, because i really would like to know how to do these. but if you cant i understand, it would be hard to type it out.
Thanks again everyone, this really is going to help me understand it a lot better.
You've gotten helpful info already. Here is another demo
no 12
3logB6(x) + 5 logB6(x-6)
LogB6 (x^3) +logB6(x-6)^5
logB6( X^3) (x-6)^5 Answer is A
bjh
7. Originally Posted by murkr
thank you for varying that for me.
i just got one last question, i understand all the problems expect problems #10 and #14
if you can do just one or both of these i would really appreciate it, im having such a hard time with these my books have no examples of these. without you guys i wouldnt know what to do thank you.
Dear murkr,
They are done using the same identities I have given you. However you must do some factorization to get the answer of question 14.
8. ## express as a single log
Originally Posted by murkr
hello everyone, i am new here and plan to use this forum often, you guys seem very friendly. I am in college algebra right now and im having a huge problem with a few questions on my review packet for the exam coming up, the test is in a few days and these questions are going to be similar to the exam. i have been looking at my math book for the last 3-4 hours and i just cant find any similar questions, im stuck. very frustrated & stressed from this as well.
i scanned the math problem packet my teacher gave me to my pc, i mostly need help on "Express as a single logarithm" problems, they are the last 4 questions. but if you got extra time i would really appreciate it if you can help me on the top questions as well, but i know that might be asking to much so if anything just an answer to atleast one question from problems 11,12,13, or 14 would be great. thank you so much
here is the link of my problem packet, its a .jpg
if possible show your work to how you solved the problem, because i really would like to know how to do these. but if you cant i understand, it would be hard to type it out.
Thanks again everyone, this really is going to help me understand it a lot better.
Your latest question help for 10 and 14.
you were told by others that you needed to factor the quadradics.When you do this and follow the log rules you get answer B.Each quadradic is easily factored.
bjh
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http://unapologetic.wordpress.com/2010/12/29/the-sign-lemma/
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# The Unapologetic Mathematician
## The Sign Lemma
As we move towards proving the useful properties of Specht modules, we will find the following collection of results helpful. Through them all, let $H\subseteq S_n$ be a subgroup, and also consider the $S_n$-invariant inner product on $M^\lambda$ for which the distinct Young tabloids form an orthonormal basis.
First, if $\pi\in H$, then
$\displaystyle\pi H^-=H^-\pi=\mathrm{sgn}(\pi)H^-$
where $H^-$ is the alternating sum of the elements of $H$. The proof basically runs the same as when we showed that $\pi e_t=\mathrm{sgn}(\pi)e_t$ where $t$ has shape $(1^n)$.
Next, for any vectors $u,v\in M^\lambda$ we have
$\displaystyle\langle H^-u,v\rangle=\langle u,H^-v\rangle$
Indeed, we can calculate
$\displaystyle\begin{aligned}\langle H^-u,v\rangle&=\sum\limits_{\pi\in H}\langle\mathrm{sgn}(\pi)\pi u,v\rangle\\&=\sum\limits_{\pi\in H}\langle u,\mathrm{sgn}(\pi)\pi^{-1}v\rangle\\&=\sum\limits_{\pi\in H}\langle u,\mathrm{sgn}(\pi^{-1})\pi^{-1}v\rangle\\&=\sum\limits_{\tau\in H}\langle u,\mathrm{sgn}(\tau)\tau v\rangle\\&=\langle u,H^-v\rangle\end{aligned}$
where we have used the facts that $\mathrm{sgn}(\pi)=\mathrm{sgn}(\pi^{-1})$, and that as $\pi$ runs over a group, so does $\tau=\pi^{-1}$.
Next, if the swap $(b\,c)\in H$, then we have the factorization
$\displaystyle H^-=k(1-(b\,c))$
for some $k\in\mathbb{C}[S_n]$. To see this, consider the subgroup $K=\{1,(b\,c)\}\subseteq H$, and pick a transversal. That is, write $H$ as a disjoint union:
$\displaystyle H=\biguplus\limits_ik_iK$
but then we can write the alternating sum
$\displaystyle\begin{aligned}H^-&=\sum\limits_{\pi\in H}\mathrm{sgn}(\pi)\pi\\&=\sum\limits_i\left(\mathrm{sgn}(k_i)k_i+\mathrm{sgn}(k_i(b\,c))k_i(b\,c)\right)\\&=\sum\limits_i\left(\mathrm{sgn}(k_i)k_i-\mathrm{sgn}(k_i)k_i(b\,c)\right)\\&=\sum\limits_i\mathrm{sgn}(k_i)k_i\left(1-(b\,c)\right)\\&=\left(\sum\limits_i\mathrm{sgn}(k_i)k_i\right)\left(1-(b\,c)\right)\end{aligned}$
as we stated.
Finally, if $t$ is some tableau with $b$ and $c$ in the same row, and if the swap $(b\,c)\in H$, then
$\displaystyle H^-\{t\}=0$
Our hypothesis tells us that $(b\,c)\{t\}=\{t\}$. We can thus use the above factorization to write
$\displaystyle\begin{aligned}H^-\{t\}&=k(1-(b\,c))\{t\}\\&=k\{t\}-k(b\,c)\{t\}\\&=k\{t\}-k\{t\}\\&=0\end{aligned}$
## 4 Comments »
1. [...] of the Sign Lemma The results we showed last time have a few immediate consequences we will have use [...]
Pingback by | December 31, 2010 | Reply
2. [...] entries in tells us that we must have some pair of and in the same row of . Thus the swap . The sign lemma then tells us that . Since this is true for every summand of , it is true for [...]
Pingback by | January 18, 2011 | Reply
3. [...] we’ve used the sign lemma. So any two polytabloids coming from tableaux in the same column equivalence class are scalar [...]
Pingback by | January 21, 2011 | Reply
4. [...] the other hand, assume in the same column of . Then . But then the sign lemma tells us that is a factor of , and thus [...]
Pingback by | February 8, 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.
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http://crypto.stackexchange.com/questions/5851/what-do-0n-and-1n-mean-in-cryptography/5877
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# What do $0^n$ and $1^n$ mean in cryptography?
EDIT : I added more context.
Let $f$ = {$f_k$} be a pseudo random function family.
Let $G(x)$ be a pseudo-random generator such that: $G(x)$ = $f_x$($0^k$)$f_x$($1^k$) where k=$|x|$ .
I don't understand the meaning of $1^n$ , $0^n$ and the differences between them within that context. What do they represent?
What is the special role / affect they have on the above context? and how?
Why $1^n$ , $0^n$ and not other combinations?
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## 3 Answers
Without seeing the entire formal construction: It seems like they wanted different strings. Meaning they needed $f_x(a)||f_x(b)$ where $a≠b$. The easiest way to express this is using the all $0$ and all $1$ strings, but any other pair of distinct strings of that length would yield the same effect.
As to why they wanted this: They're using a PRF twice to construct a PRG. Consider what would happen if they used the same string both times. You'd get $$G(x)=f_x(0^k)||f_x(0^k)$$
And the output string would have the property that the first half of the bits are same as the second half of the bits and this would not be pseudorandom(a distinguisher can just check if the the first half and second half of the bits are the same), so $G$ would definitely not be a PRG.
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Typically that means a string of either $n$ zeros or $n$ ones.
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yeah but what is the meaning of those strings inside the above equation? why $0^k$ and not $0^{k-1}$1 ? – Yoni Hassin Dec 28 '12 at 23:34
Why always in cryptography $1^k$ and $0^k$ are used? do they have a symbolic value? that the part that confuses me. – Yoni Hassin Dec 28 '12 at 23:40
@YoniHassin: what are you actually asking? What $0^k$ means, or why that was selected rather than $0^{k-1}1$? As for the former, it (in context) means a bitstring consisting of k zero bits. As for the latter, well, the reason the author of the primitive selected that specific choice probably depends a lot on the context of how the primitive works; without knowing the context, it is unlikely anyone could guess. – poncho Dec 29 '12 at 3:49
I will try to rewrite the question so it will be more clear. – Yoni Hassin Dec 29 '12 at 3:57
1
In light of the above edit, it seems like they wanted different strings. Meaning they needed $f_x(a)||f_x(b)$ where $a \neq b$. The easiest way to express this is using the all 0 and all 1 strings, but any other unique pair would yield the same effect. – AFS Dec 29 '12 at 23:04
show 1 more comment
$0^n$ means a string of $n$ zeros (the $n$-bit string that is all zeros). $1^n$ means a string of $n$ ones.
Why were these used? There's nothing special about $0^n$ or $1^n$, in this context. They could have used any pair of two constant $n$-bit strings, as long as the two strings were not the same. $0^n$ and $1^n$ is a convenient choice of two strings that are different, but they could have chosen any other pair (as long as they're not both the same) and that would have worked fine, too.
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http://mathoverflow.net/questions/52186/self-dual-complete-category
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## Self-dual Complete Category
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is there an example of a self-dual complete category that is not a partially-ordered set?
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15
$C\times C^\mathit{op}$ for $C$ complete and cocomplete. – Denis-Charles Cisinski Jan 15 2011 at 22:54
1
@Denis-Charles: Nice! – Martin Brandenburg Jan 15 2011 at 23:02
1
Any abelian monoid. – Buschi Sergio Jan 16 2011 at 9:16
1
I don't understand the abelian monoid example. – arsmath Jan 16 2011 at 13:28
1
But is that category complete? I would think it's only complete when the monoid only has one element. – arsmath Jan 17 2011 at 18:48
show 5 more comments
## 3 Answers
The category of suplattices (complete join-semilattices) is complete and self-dual. It is complete because it is monadic over Set (the monad is the covariant powerset monad). The duality $Sup^{\mathrm{op}}\to Sup$ sends a suplattice $X$ to the opposite poset $X^{\mathrm{op}}$, which is again a suplattice since every suplattice is also an inflattice (define $\bigwedge S = \bigvee { x | x \le S}$). It sends a sup-preserving map $f:X\to Y$ to $(f^\ast)^{\mathrm{op}}:Y^{\mathrm{op}}\to X^{\mathrm{op}}$, where $f^\ast:Y\to X$ is the right adjoint of $f$—this exists by the adjoint functor theorem for posets, and preserves infs since it is a right adjoint.
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Thanks! I had to think about it for a little bit to make sure I understood it. – arsmath Jan 17 2011 at 21:14
Yes, it can be a little mind-bending at first. – Mike Shulman Jan 18 2011 at 5:04
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Mike has already answered the question, but I thought I would add some assorted remarks on some of the discussion above.
Mike gave an example of what is called a $\ast$-autonomous category, as first defined by Michael Barr in Springer LNM 752 (1979). The field has been very heavily investigated, partly because $\ast$-autonomous categories are the categorical models of Girard's celebrated linear logic. In brief, a $\ast$-autonomous category is a symmetric monoidal closed category $V$ equipped with a "dualizing object" $D$, which means, letting $\hom$ denote the internal hom of $V$, that the canonical natural transformation
$$A \to \hom(\hom(A, D), D)$$
is a natural isomorphism. This implies in particular that
$$\hom(-, D): V^{op} \to V$$
is an equivalence, so that $V$ is self-dual. Complete $\ast$-autonomous categories are certainly of considerable interest, and many examples are known. Some (including Mike's example) are given in this nLab article.
There are also various constructions which allow one to embed sufficiently nice instances of duality, such as the Pontryagin duality alluded to by Martin, into complete $\ast$-autonomous categories. One of the most potent general constructions is called the Chu construction, which takes an input any symmetric monoidal closed category with pullbacks and a preassigned object $d$, and produces as output a $\ast$-autonomous category $Chu(C, d)$ together with a coreflective (in particular full) symmetric monoidal embedding
$$i: C \to Chu(C, d)$$
whose dualizing object is manufactured from $d$ in a canonical way. If $C$ is complete and cocomplete, then so is $Chu(C, d)$. For a sample theorem which shows how certain nice topological dualities embed into Chu-type dualities, including the case of Pontryagin duality, see this paper by Barr.
Incidentally, with regard to Martin's answer, it is indeed the case that the category $LCHAb$ of locally compact Hausdorff abelian groups is not complete. We know for example that the topological product $\mathbb{R}^{\mathbb{N}}$ of countably many copies of $\mathbb{R}$ is not locally compact; it remains to see that there is no such product in $LCHAb$ (i.e., even if we try to strengthen the product topology on the set $\mathbb{R}^\mathbb{N}$ to some locally compact topology in some way, there still exists no solution to the universal mapping problem). If there were, then using the universal property, we could show that the scalar product $\mathbb{R} \times \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}$ is continuous, making $\mathbb{R}^\mathbb{N}$ a Hausdorff topological vector space. But it is well-known that a locally compact Hausdorff TVS over the real numbers is finite-dimensional (reference; see theorem 4.3 on page 5). As arsmath suggests, this implies that the category of locally compact Hausdorff spaces cannot be complete (if it were, then so would be the category of abelian group objects, which we just proved is not the case).
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Thanks for the link to the Chu construction, Todd. It looks interesting. – arsmath Jan 17 2011 at 21:15
Thanks for adding the more general context! – Mike Shulman Jan 18 2011 at 5:03
If $LCA$ denotes the category of locally compact topological abelian groups, then Pontryagin duality implies that $LCA$ is self-dual. To see that this category is complete, note that the forgetful functor to abelian groups creates limits.
EDIT: The category is not complete. See the answer by Todd. I won't delete my answer because Todd refers to it.
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Really? Can you give an explicit description of the topology on the product (it can't be the product topology)? – Theo Buehler Jan 15 2011 at 23:05
You're right, this is not clear at all. Does some abstract-nonsense produce infinite products? I mean something like the statement that the category of compact hausdorff spaces is cocomplete: the coproduct of infinitely many $X_i$ is the stone-cech-comp. of the coproduct of the underlying spaces. – Martin Brandenburg Jan 15 2011 at 23:10
If there's an argument along those abstract-nonsense lines, then it would suggest that the category of locally compact topological spaces has infinite products, which I would find very surprising. – arsmath Jan 16 2011 at 13:32
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http://mathoverflow.net/questions/42201/what-are-the-zero-entropy-invariant-measures-for-an-anosov-geodesic-flow/85345
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## What are the zero entropy invariant measures for an Anosov geodesic flow?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$ be the double-torus with a hyperbolic Riemannian metric. The geodesic flow on the unit tangent bundle $T^1M$ has many invariant Borel probability measures. In particular there are closed geodesics projecting to each non-trivial homotopy class of $M$, and one can support an invariant probability measure on each of these. Also one can take convex linear combinations of these invariant measures.
My question is the following: Are these the only invariant measures of zero metric (Kolmogorov-Sinai) entropy?
More generally, what are the zero entropy invariant probability measures of an Anosov geodesic flow?
Also I'm interested in the same question for shifts on finitely many symbols (i.e. What are the zero entropy invariant measures?).
Besides references giving an answer, other related references are of course very welcome.
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Added ergodic theory tag. – Ian Morris Oct 20 2010 at 9:37
## 5 Answers
There are lots of zero entropy invariant probability measures, many more than just the obvious ones supported on periodic orbits. As you suggest in the question, one can understand the general case by just considering what happens for symbolic systems.
Explicit example: Let $\alpha$ be irrational and let $a_n$ denote the fractional part of $n\alpha$. Consider the sequence in $\Sigma_2 = \{0,1\}^\mathbb{Z}$ given by $x_n = 0$ if $0 \leq a_n < 1/2$, and $x_n=1$ if $1/2\leq a_n<1$. Let $X\subset \Sigma_2$ be the orbit closure of $x=(x_n)$; then there is an entropy-preserving isomorphism between the space of invariant measures for the shift map $\sigma\colon X\to X$ and for the irrational rotation $R_\alpha\colon S^1 \to S^1$. The latter preserves Lebesgue measure on the circle and is uniquely ergodic with zero entropy, so $X$ supports exactly one invariant probability measure $\mu$, which comes from Lebesgue and has zero entropy. Now $\mu$ is a shift-invariant probability measure on $\Sigma_2$ that has zero entropy but is not supported on a periodic orbit.
General result: In fact, the above construction is representative of a general phenomenon. As RW points out in his answer, you can get lots of zero entropy measures on shift spaces by taking generating partitions for zero entropy transformations. You can even get more, using the Jewett-Krieger embedding theorem (see Petersen's "Ergodic Theory" or Denker, Grillenberger, and Sigmund's "Ergodic Theory on Compact Spaces"), which lets you find a closed shift-invariant subset of the shift space that has the desired measure as its only shift-invariant probability measure. So there's a lot there.
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Thank you! This answer is very clearly explained and convincing. I still feel it would be nice if there was some explicit reference or construction explaining how to pass these measures from the shifts over to a geodesic flow, however I've decided to accept this. – Pablo Lessa Oct 15 2010 at 21:48
@Pablo, it's a general theorem (the buzz-word here is Adler-Weiss, see for example this survey by Adler - citeseerx.ist.psu.edu/viewdoc/… ) which tells you you can "encode" such systems and achieve a measurable isomorphism with such Bernoulli systems. From there, it is not hard to construct zero entropy measures as averaging over "thin" orbits (obviously those are shifts of finite type, so you need to do some massaging to get those "thin" sets then the full shift, but it's possible). – Asaf Jan 19 2012 at 18:28
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
Another type of answer. It can be shown that generically (in the sense of Baire), an invariant probability measure for the geodesic flow is ergodic, of full support, with entropy zero, and not mixing. See the article: K. Sigmund. On the space of invariant measures for hyperbolic flows, Amer. J.Math. 94 (1972), 31–37
For shifts, it is due to Oxtoby J. C. Oxtoby. On two theorems of Parthasarathy and Kakutani concerning the shift transformation, (1963) Ergodic Theory (Proc. Internat. Sympos., Tulane Univ., New Orleans, La., 1961) pp. 203–215 Academic Press, New York.
These results can be generalized to any nonelementary negatively curved manifold, or shifts with infinitely many symbols, see for example " Generic measures for hyperbolic flows on non compact spaces", written with Yves Coudene , Israël Journal of Maths vol 179 (2010). (The proof that in this situation, generic measures have entropy zero is not written in this article, and still unpublished, but works without any difficulty.)
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By taking a generating partition you can realize an arbitrary finite entropy transformation as the full shift on a finite alphabet with a certain invariant measure. In particular, this observation provides non-trivial zero entropy invariant measures for the full shift. This can also be done for subshifts of finite type, and therefore (in view of the symbolic business) for Anosov diffeomorphisms - there is an old paper by Lind and Thouvenot where they consider a similar realization problem for arbitrary finite entropy transformations.
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I just saw this... I have recently finished a paper with Terry Soo in which we prove a universality property for geodesic flows: for any measure preserving transformation $T\colon X\to X$ with (measure-theoretic) entropy less than the topological entropy of the time one map of the geodesic flow, there's an injective map from $X$ into the geodesic flow equipped with the time 1 map.
This gives a very rich family of invariant measures for the time 1 map of the geodesic flow. Averaging these measures over times from 0 to 1 gives a large collection of invariant measures for the full flow. In particular if you start off with a zero entropy transformation, the measure you obtain this way is a zero entropy measure for the flow.
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OK, I am sure that there must be something easier for the shifts but... In my paper with Glendinning we briefly mention a construction of a subshift on two symbols of zero topological entropy. Namely, let $\sigma$ denote the shifted Thue-Morse sequence, i.e., $$\sigma = 1101001100101101001011001101001 \dots$$ Define $X$ to be the set of all two-sided 0-1 sequences $(x_n)$ such that $$(x_n, x_{n+1},\dots) \prec \sigma \ \ \ \forall n\in\mathbb Z$$ and $$(1-x_n, 1-x_{n+1},\dots) \prec \sigma \ \ \ \forall n\in\mathbb Z,$$ where $\prec$ stands for lexicographically smaller''. The set $X$ has zero Hausdorff dimension in the standard metric on ${0,1}^{\mathbb Z}$. In fact, the number of admissible words of length $n$ in $X$ grows approximately as $n^{\log n}$.
To endow $X$ with a shift-invariant measure, one can take the sequence of subshifts of finite type converging to $X$ (by truncating $\sigma$) and $\mu$ to be the weak limit of the corresponding measures of maximal entropy, say.
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http://mathhelpforum.com/number-theory/140928-sum-3-squares.html
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# Thread:
1. ## Sum of 3 squares
I apologize for re-posting this question, but someone answered yesterday without reading the entire problem and nobody else seems to be looking at it. Thanks.
Prove that if a prime number is a sum of three squares of different primes, then one of the primes must be equal to 3.
2. Here's a thought:
Try showing $p^2+q^2+r^2$ is composite when $p,q,r>3$ are prime.
3. So we have $p=a^2+b^2+c^2$ where $p,a,b,c$ are all distinct primes.
Assume $a,b,c\neq3 \implies (a,3)=(b,3)=(c,3)=1\implies a^2\equiv b^2\equiv c^2\equiv1\bmod{3}$
This means $p\equiv 1+1+1\equiv0\bmod{3}\implies 3\mid p$, which is a contradiction.
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http://math.stackexchange.com/questions/187420/possible-invariant-subspaces-of-mathbbq2
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Possible invariant subspaces of $\mathbb{Q}^2$
Consider the linear map $T : \mathbb{Q}^2 \rightarrow \mathbb{Q}^2$, associated with the following matrix $$T = \begin{pmatrix}2 & 1\\1&-2\end{pmatrix}$$
I want to show that $\bf 0$ and $\mathbb{Q}^2$ are the only invariant subspaces under $T$. How does this change when $T$ is a real map instead of a rational?
I can't see how being a rational map has influence on the invariant subspaces.
This is an exercise for a course, but is not to be handed in.
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1 Answer
Any proper invariant subspace would be a $1$ - dimensional $\Bbb{Q}$ - subspace. In other words, it must be spanned by an eigenvector with rational eigenvalue. But the characteristic polynomial of your matrix above is irreducible over $\Bbb{Q}$, so what can you deduce?
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http://math.stackexchange.com/questions/116465/probability-combinatorial-related-rolling-dice/116467
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# Probability Combinatorial related; rolling dice
when 3 fair dices roll, let Y be a Random Variable of which value is the minimum result of all 3 dices results. for example if the results are (2,3,2) then Y=2.
how do i find, in the most elegant and reasonable way the P(Y) function?
i tried tackle this for a while, so the more explanations the better! thanks!
this is my professor answer (which i don't get).
(ignore the related foreign letters)
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## 2 Answers
He is calculating the cumulative distribution first. That is, he is finding, for instance, the probability that the minimum die roll is greater than or equal to 2. This is an easier quantity to calculate: the minimum die roll is greater than or equal to 2 if and only if all three die rolls were greater than or equal to 2. So, the probability that the minimum die roll is greater than or equal to 2 is $p(Y\ge 2)=({5\over6})^3$ (we know each die roll was 2 or greater. The rolls are presumed independent and the probability that a particular die is roll is equal to 2 or greater is $5/6$).
Then the probability that the minimum die roll is equal to $i$ is calculated using the cumulative distribution. So, for instance $p(Y=1)=p(Y\ge 1)-p(Y\ge 2)$.
The last formula can be thought of as follows: take the set of outcomes $A=[Y\ge 1]$. Then this set of outcomes differs from the set of outcomes $B=[Y=1]$ by exactly the set of outcomes $C=P[Y\ge 2]$. That is, $A=B\cup C$ and $B\cap C=\emptyset$. We thus have $P(A)=P(B)+P(C)$. Rearranging this gives $P(B)=P(A)-P(C)$.
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that is a well explained answer, thanks a lot! could you reffer to my edit too? – Ofek Ron Mar 4 '12 at 22:45
A=B∪C and B∩C=∅ what sentence\property were you laying on for this to be true? – Ofek Ron Mar 4 '12 at 22:51
@OfekRon It should be fairly clear from the definitions of $A$, $B$, and $C$, that $A=B\cup C$ and $B\cap C=\emptyset$. Take the set where $Y\ge 1$, throw away all the elements with $Y\ge 2$. The elements that remain are those with $Y=1$. – David Mitra Mar 4 '12 at 22:58
$P(X=1)=P(X\leq1)=1-P(X >1)=1-P(X\geq2)$ this means that the possibility of X=1 is the possibility of not being greater than 2. Due to the fact that the 3 throws are independent $P(X\geq2)=\prod\limits_{i=1}^3 P_i(X\geq2)=(\frac{5}{6})^3$ because 5 out of 6 throws a dice will roll a number greater or equal to 2. So you got:$P(X=1)=1-(\frac{5}{6})^3=\frac{91}{216}$. That's the principle for the other calculations too.
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it should be multipication instead of a summation is it not? – Ofek Ron Mar 4 '12 at 22:32
Yeah,sorry it was a typo.I corrected it. – chemeng Mar 4 '12 at 22:35
plz reffer to my edit... – Ofek Ron Mar 4 '12 at 22:35
I didn't quite understand your formula but I think the Inclusion-Exclusion Principle states: $$|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|B\cap C|-|A\cap C|+|A\cap B\cap C|$$ So we got $P(X=1)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}-\frac{1}{6}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{91}{216}$ – chemeng Mar 4 '12 at 22:53
isnt X~B(3,1/6)? – Ofek Ron Mar 4 '12 at 22:53
show 3 more comments
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http://stats.stackexchange.com/questions/32060/sheppards-correction
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# Sheppard's correction
Is there a good expository account of Sheppard's correction, written in a way that any ordinary mathematician can readily follow?
http://mathworld.wolfram.com/SheppardsCorrection.html
(I've thought of writing a Wikipedia article titled "Sheppard's correction", but I haven't done enough homework on it yet.)
Sheppard's correction can be done for various functionals of a probability distribution. The Wolfram article treats cumulants. Curiously, each correction is $-1$ times the correction I'd expect to do for a uniform distribution. I'm not sure whether to believe this.
Think of this: split the interval $(0,\theta)$ into bins of equal length. For a sample from the uniform distribution on the interval, suppose only the midpoint of the bin is reported. Then the variance of the reported data would be too small, since it would fail to report variability within bins. The variability within bins would be just $1/12$ of the square of the bin length. But Sheppard's correction for normally distributed data tells you to add $-1/12$ of the square of the bin length. With the normal distribution, the error from binning is negatively correlated with the observation; with the uniform distribution it is uncorrelated with the observation. If we believe Wolfram, this multiplication by $-1$ seems to apply to all cumulants, not just the second cumulant. If so, that seems like a mystery to be investigated. I hesitate to believe Wolfram on this.
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The software told me that this question "doesn't meet our quality standards". So I added some invisible $\TeX$. Then it apparently met those standards. – Michael Hardy Jul 10 '12 at 22:29
I imagine that it would also pass the quality standards if the question were extended slightly to briefly mention what Sheppard's correction is. Often this can elicit additional interest from "browsers" who may not otherwise know what it is. :) – cardinal Jul 10 '12 at 23:11
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@cardinal : OK, I've added a few things. – Michael Hardy Jul 11 '12 at 0:41
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http://mathoverflow.net/questions/60437/final-steps-for-a-proof-that-a-function-is-not-primitive-recursive
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## final step(s) for a proof that a function is not primitive recursive
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
My function is $f:\mathbb{N} \rightarrow \mathbb{N},\ f(n)=2\uparrow ^n 3$ , the Ackermann(-Péter) function, with the second argument fixed to 3 (and "$\uparrow$" the Knuth up-arrow), which I believe is not primitive recursive, but which I could not prove - and that is what this question is about. Any ideas for the proof are very welcome.
Explanation:
Idea 1 I have tried to prove that it is not primitive recursive in the lines of the proof that the Ackermann function is not primitive recursive - meaning I tried to prove, that my function $f$ bounds in some way all the other primitive recursive functions, so that I arrive at a contradiction, if I assume, that $f$ would be primitive recursive, because than it would bound itself. The way in which I tried to prove, that $f$ should bound all the other functions was: If $g$ is an arbitrary primitive recursive $k$-ary function $g:\mathbb{N}^k \rightarrow \mathbb{N}$, than there exists an natural number $N$, such that for all $n_1 , ..., n_k > N$ we get $g(n_1, ... ,n_k) < f(n_1 + ... +n_k)$.
I thought I could try to prove this by trying to prove, that the set of all primitive recursive functions, which are bounded by f like described above, fulfills the properties which the primitive functions have to fulfill (the "standart" functions belong to this set, the set is closed under composition, the set is closed under primitive recursion). Since the primitive recursive functions are the smallest set, that fulfills these properties, I would obtain that this set has to equal the set of all the primitive recursive functions, which would the give me my contradiction. But this proof failed as well, because I can't show, that this set is closed under composition - that is, I can't show that given functions $h:\mathbb{N}^l \rightarrow \mathbb{N}$ and $g:\mathbb{N}^k \rightarrow \mathbb{N}$, which have the property, that:
$\exists N_1 \ \forall n_1,\ldots,n_{l}>N_1$ such that $h(n_1 ,\ldots ,n_{l})<2 \uparrow^{n_1 + \ldots + n_{l}} 3$
and
$\forall i \in$ {$1,\ldots,l$}$\ \exists N_{i+1} \ \forall n_1,\ldots, n_{k} > N_{i+1}$ such that $g_i(n_1, \ldots ,n_{k})<2 \uparrow^{n_1 + \ldots + n_{k}} 3$
then there should exists an $N$ such that $\forall n_1,\ldots,n_{k}>N$ we should also have, that $t(n_1,\ldots,n_k):=f(g_1(n_1,\ldots ,n_k),\ldots ,g_l(n_1,\ldots,n_k))<2 \uparrow^{n_1 + \ldots + n_{k}} 3$.
In trying to prove the this, I somehow have to make use of the above properties of the functions $g_i$ and $h$. But I can't use the "boundedness" of $h$ because I can't control how much the functions $g_i$ grow - if there aren't values $n_1,\ldots ,n_k$ such that each of the $g_i$'s grow above $N_1$, then I can't make use of $h$'s properties. Maybe there is some other way to prove the closedness under composition, but I can't see it.
Idea 2
I have tried, to proceed indirectly and to assume, that my $f$ would be primitive recursive - the idea was to construct out of this now primitive recursive $f$ another primitive recursive function, that would violate the "Ackermann"-bound - in the sense of the original proof, where one proves that the Ackermann function is not primitive recursive, by first establishing this "Ackermann" bound which all primitive recursive functions have to obey and then showing, that if the Ackermann function is assumend to be primitive recursive one obatains a contradiction.
Could anyone please give me an idea how to finish one of this proof ?
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## 1 Answer
I believe one proof technique was to use a primitive recursive encoding of multiple argument functions, so that it would suffice to show a bound on a composition of unary primitive recursive functions. You might try first showing that your function bounds any unary primitive recursive function first.
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http://nrich.maths.org/6411/note
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# What's Your Mean?
### Why do this problem?
This problem gives an opportunity to practise numerical integration in the context of probability distributions. It will really allow students to get into the meaning of probability density functions in terms of areas and probabilities. Instead of simply requiring an explicit calculation, students will need to engage with decisions concerning limits and integration.
### Possible approach
The first stage of the problem is to realize that a numerical integration is needed to calculate the mean. Once the class has realised that this is the case, they will need to start to perform the integrations. This will require various choices as there are many ways in which this can be done. To facilitate this, you might like to print off copies of the graph for students to draw on.
### Key questions
How do we relate a probability density function to a probability?
How do the two graphs relate to each other?
What is the graphical interpretation of an integral?
How important will the effect of the second graph be?
What happens for values larger than $20$? Are these values relevant?
### Possible extension
How might you try to estimate the variance for these distributions numerically?
### Possible support
First try to show that numerically the area under the red curve is 1. You can then use the decomposition into rectangles and trapezia to try to work out the mean.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.
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http://mathoverflow.net/questions/56025/doing-explicit-computations-with-coordinate-rings/56026
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## Doing explicit computations with coordinate rings
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose that we are given an integral $k$-algebra $A$ of finite type explicitly, by which I mean that we are given the generators of the defining ideal $J$ where $A = k[x_1,...,x_n]/J$. What kinds of tools are out there to compute the integral closure of $A$? I would like the answer as explicitly as possible i.e. generators of the defining ideal.
While suggestions of computer programs are welcome, I want to be able to do these on my own, so I am looking for results which let me prove the answer. I'm asking from the perspective of someone who knows very little computational commutative algebra.
As a related question, if I have two such rings $A$, $B$ given explicitly as above, together with an explicit homomorphism between them, how can I go about determining the kernel and cokernel explicitly? Also, how about if we localize everything at a maximal ideal?
Broadly speaking, I would like to know about what kinds of computational methods are available for rings which arise from studying complex algebraic varieties. Have people out there settled these kinds of computations completely, or is this a hard question in general?
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1
Related: mathoverflow.net/questions/11709/… – David Speyer Feb 20 2011 at 15:03
## 2 Answers
This should be what you're looking for.
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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.
Have you looked at the Swanson-Huneke book on integral closure? Especially chapter 15 which is a discussion of various methods of computing integral closure (including Stolzenberg's method mentioned in Steven Landsberg's answer above).
It is available online HERE
Another recent algorithm is due to A. Singh and I. Swanson, Click HERE. This also has some history discussion, and it is apparently implemented as well.
Since you did ask for computer implementations, please see HERE, an algorithm implemented in Macaulay2 apparently based off T. de Jong's algorithm (mentioned in the sources above).
Finally, I should note that sometimes blowing-up the conductor helps, see section 7 of Greco-Traverso, On seminormal schemes''.
EDIT: With regards to your other questions finding kernels and cokernels of maps of rings, you should see a book on computational commutative algebra. For example, you could try the section of Eisenbud's book "Commutative Algebra with a view towards algebraic geometry'' on Groebner bases, another common source is Ideals, Varieties and Algorithms" by Cox, Litlle and O'Shea. Again, these things are also implemented in Macaulay2 amoung other places.
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http://physics.stackexchange.com/questions/22460/the-quantum-mechanical-description-of-an-electron-motion-in-a-magnetic-monopole
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# The quantum-mechanical description of an electron motion in a magnetic monopole field
The quantum-mechanical motion problem of an electron in electric field of the nucleus is well known. The quantum-mechanical description of electron motion in a magnetic field is also not difficult, since it needs to solve the Schrödinger equation of the form: $$\frac{(\hat p + eA)^2} {2m} \psi = E \psi$$ But if we want to consider the motion of an electron in a magnetic monopole field, the difficulty arises because the definition of the vector potential in the whole space. See, for example. Was this problem solved? What interesting consequences derived from this task? (for energy levels, angular momentum etc.)
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This gives you coherent state representation for angular momentum. – Slaviks Mar 16 '12 at 21:54
The difficulty is not in the vector potential--- you can just use a Dirac string. The difficulty is getting the actual solution. – Ron Maimon Apr 15 '12 at 7:42
## 1 Answer
The classical version of this problem was solved by Henri Poincaré way back in 1896. This is also problem 5.43 in Electrodynamics by Griffiths. The classical trajectories are geodesics on the surface of a cone. A recent treatment of the classical version of this problem is here.
The quantum mechanical version was also solved long back by Igor Tamm in 1931. This is discussed in section 2.3 of the book Magnetic monopoles by Y M Shnir, who follows the treatment in Charge quantization and nonintegrable Lie algebras by Hurst.
The quantum mechanical version of the problem turns out to be separable in spherical polar coordinates. The angular part has the generalized spherical harmonics as its eigenvalues, while the radial solution is the same as the radial wave function of the standard Schroedinger equation. The centrifugal potential in the Schroedinger equation turns out to be always repulsive which implies that there are no bound states for this system of an electron in a magnetic monopole field. However a dyon field does have bound state solutions.
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http://simple.wikipedia.org/wiki/Module_arithmetic
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# Modular arithmetic
(Redirected from Module arithmetic)
Time-keeping on a clock gives an example of modular arithmetic.
Modular arithmetic, sometimes also called clock arithmetic, is a way of doing arithmetic with integers. Much like hours on a clock, which repeat every twelve hours, once the numbers reach a certain value, called the modulus, they go back to zero.
People talked about modular arithmetic in many ancient cultures. For instance, the Chinese remainder theorem is many centuries old. The modern notation and exact definition of modular arithmetic were first described by Carl Friedrich Gauss.[1]
## Congruence
Modular arithmetic can be used to show the idea of congruence. Two integers, a and b, are congruent modulo n if they have the same remainder when both are divided by the positive integer n. Congruence can be written this way:
$a \equiv b \pmod n\,$
The number n is called the modulus. Another definition of congruence, that means the same thing but is sometimes more useful, is that the two integers are congruent modulo n if the difference (a - b) is an integer multiple of n. That is, if n is a factor of (a - b), then a and b are congruent mod n.
For example:
$32 \equiv 11 \pmod 7,\,$
The remainder when 32 is divided by 7 is 4, and the remainder when 11 is divided by 7 is also 4. This tell us they're congruent, mod 7.
We can use this example it with the other definition too. The difference, (a - b), is 32 - 11 = 21. This shows the two numbers are congruent because 21 = 3 * 7, and 7 is a factor of 21.
## Uses
Modular arithmetic is useful in many fields, such as cryptography, computer science, and music. It is often used in calculating checksums and check digits.
## References
1. Richard Taylor (2012). "Modular Arithmetic: Driven by Inherent Beauty and Human Curiosity". Institute for Advanced Study. Retrieved 7 March 2013.
This short article about mathematics can be made longer. You can help Wikipedia by .
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http://math.stackexchange.com/questions/175258/a-difference-equation-related-to-rw-on-hypercube?answertab=oldest
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# A difference equation related to RW on Hypercube
I am trying to solve the following recurrent relation $$T(n)=\frac{n}{m}T(n-1)+\frac{m-n}{m}T(n+1)+1, \,\, \text{subject to } T(m)=0$$
Where $0\leq n\leq m$ and $m$ is a fixed integer. I have written the relation as a difference equation as it is often a way to solve these, and so for $D(n)=T(n)-T(n+1)$ I get $$D(n)=\frac{n}{m-n}D(n-1)+\frac{m}{m-n}.$$ Now I want to sum on both side to telescop the sum and use the fact that $T(m)=0$ but my problem is that on the right side, $D(-1)$ is not defined.
The goal of this problem is to find the hitting time from (0,...0) to (1,...,1) for a simple random walk on the $m$-dimensional hypercube and so $T(n)$ represents this (expected) hitting time.
Thanks
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– Jean-Sébastien Jul 26 '12 at 2:54
## 1 Answer
If you just want the answer, your problem is Exercise 10.5 in Markov Chains and Mixing Times by Levin, Peres, and Wilmer. The book is available online here.
The expected time to go from $(0,\dots,0)$ to $(1,\dots,1)$ is
$$\sum_{k=1}^m \left[k{m\choose k}\right]^{-1} \cdot \sum_{k=1}^m k{m\choose k} =2^{m-1} \sum_{k=1}^m {m-1\choose k-1}^{-1}={m\over 2}\sum_{k=1}^m{2^k\over k} .$$
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I'm not too familiar with network interpretations, but that's a start, Thanks – Jean-Sébastien Jul 26 '12 at 3:29
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http://math.stackexchange.com/questions/tagged/problem-solving?page=2&sort=newest&pagesize=30
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# Tagged Questions
The problem-solving tag has no wiki summary.
1answer
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1answer
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### Worded problems.
There are 9 chocolates in a selection box: 3 white, 3 milk & 3 dark chocolates. There are 3 soft centres, 3 truffles & 3 pralines. Each chocolate is different. Using the clues below, arrange ...
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### asymptotics of a solution for equation
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### rewrite $2ie^{i\pi}+i^3$
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### Solving equations that contain summations
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### Combinatorial restriction on choosing adjacent objects
I am having trouble understanding an approach to the following problem. Suppose 14 books are aligned in a bookcase. How many ways can we choose 5 books so that no adjacent books are chosen? I think my ...
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### Counting Card hands with various restrictions
I would like to know if my solutions are correct for the following three combinatorial card questions. In each question, assume we have a standard deck of cards (13 ranks, and 4 suits). How many ...
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1answer
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### Finding the general term of a sequence
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3answers
429 views
### Arc sums for a circle of $k$ positive integers whose total sum is $n$
This problem got me thinking about the following more general scenario: Suppose you have $k$ positive integers with total sum $n$, and you arrange them in a circle. Given such an arrangement, you ...
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64 views
### How many students like all the three subjects?
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3answers
56 views
### what is the value of the digit in the ones place of the following?
1×3×5×7×9×11×13×...×2007×2009 what is the value of the digit in the ones place of the following? I can't find the solution for this problem. Please give me some hints
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http://unapologetic.wordpress.com/2007/11/29/complete-uniform-spaces/?like=1&source=post_flair&_wpnonce=b0679ce6ea
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# The Unapologetic Mathematician
## Complete Uniform Spaces
Okay, in a uniform space we have these things called “Cauchy nets”, which are ones where the points of the net are getting closer and closer to each other. If our space is sequential — usually a result of assuming it to be first- or second-countable — then we can forget the more complicated nets and just consider Cauchy sequences. In fact, let’s talk as if we’re looking at a sequence to build up an intuition here.
Okay, so a sequence is Cauchy if no matter what entourage we pick to give a scale of closeness, there’s some point along our sequence where all of the remaining points are at least that close to each other. If we pick a smaller entourage we might have to walk further out the sequence, but eventually every point will be at least that close to all the points beyond it. So clearly they’re all getting pressed together towards a limit, right?
Unfortunately, no. And we have an example at hand of where it can go horribly, horribly wrong. The rational numbers $\mathbb{Q}$ are an ordered topological group, and so they have a uniform structure. We can give a base for this topology consisting of all the rays $(a,\infty)=\{x\in\mathbb{Q}|a<x\}$, the rays $(-\infty,a)=\{x\in\mathbb{Q}|x<a\}$, and the intervals $(a,b)=\{x\in\mathbb{Q}|a<x<b\}$, which is clearly countable and thus makes $\mathbb{Q}$ second-countable, and thus sequential.
Okay, I’ll take part of that back. This is only “clear” if you know a few things about cardinalities which I’d thought I’d mentioned but it turns out I haven’t. It was also pointed out that I never said how to generate an equivalence relation from a simpler relation in a comment earlier. I’ll wrap up those loose ends shortly, probably tomorrow.
Back to the business at hand: we can now just consider Cauchy sequences, instead of more general Cauchy nets. Also we can explicitly give entourages that comprise a base for the uniform structure, which is all we really need to check the Cauchy condition: $E_a=\{(x,y)\in\mathbb{Q}\times\mathbb{Q}|\left|x-y\right|<a\}$. I did do absolute values, didn’t I? So a sequence $x_i$ is Cauchy if for every rational number $a$ there is an index $N$ so that for all $i\geq N$ and $j\geq N$ we have $\left|x_i-x_j\right|<a$.
We also have a neighborhood base $\mathcal{B}(q)$ for each rational number $q$ given by the basic entourages. For each rational number $r$ we have the neighborhood $\{x\in\mathbb{Q}|\left|x-q\right|<r\}$. These are all we need to check convergence. That is, a sequence $x_i$ of rational numbers converges to $q$ if for all rational $r$ there is an index $N$ so that for all $i\geq N$ we have $\left|x_i-q\right|<r$.
And finally: for each natural number $n\in\mathbb{N}$ there are only finitely many square numbers less than $2n^2$. We’ll let $a_n^2$ be the largest such number, and consider the rational number $x_n=\frac{a_n}{n}$. We can show that this sequence is Cauchy, but it cannot converge to any rational number. In fact, if we had such a thing this sequence would be trying to converge to the square root of two.
The uniform space $\mathbb{Q}$ is shot through with holes like this, making tons of examples of Cauchy sequences which “should” converge, but don’t. And this is all just in one little uniform space! Clearly Cauchy nets don’t converge in general. But we dearly want them to. If we have a uniform space in which every Cauchy sequence does converge, we call it “complete”.
Categorically, a complete uniform space is sort of alike an abelian group. The additional assumption is an extra property which we may forget when convenient. That is, we have a category $\mathbf{Unif}$ of uniform spaces and a full subcategory $\mathbf{CUnif}$ of complete uniform spaces. The inclusion functor of the subcategory is our forgetful functor, and we’d like an adjoint to this functor which assigns to each uniform space $X$ its “completion” $\overline{X}$. This will contain $X$ as a dense subspace — the closure $\mathrm{Cl}(X)$ in $\overline{X}$ is the whole of $\overline{X}$ — and will satisfy the universal property that if $Y$ is any other complete uniform space and $f:X\rightarrow Y$ is a uniformly continuous map, then there is a unique uniformly continuous $\bar{f}:\overline{X}\rightarrow Y$ extending $f$.
To construct such a completion, we’ll throw in the additional assumption that $X$ is second-countable so that we only have to consider Cauchy sequences. This isn’t strictly necessary, but it’s convenient and gets the major ideas across. I’ll leave you to extend the construction to more general uniform spaces if you’re interested.
What we want to do is identify Cauchy sequences in $X$ — those which should converge to something in the completion — with their limit points in the completion. But more than one sequence might be trying to converge to the same point, so we can’t just take all Cauchy sequences as points. So how do we pick out which Cauchy sequences should correspond to the same point? We’ll get at this by defining what the uniform structure (and thus the topology) should be, and then see which points have the same neighborhoods.
Given an entourage $E$ of $X$ we can define an entourage $\overline{E}$ as the set of those pairs of sequences $(x_i,y_j)$ where there exists some $N$ so that for all $i\geq N$ and $j\geq N$ we have $(x_i,y_j)\in E$. That is, the sequences which get eventually $E$-close to each other are considered $\overline{E}$-close.
Now two sequences will be equivalent if they are $\overline{E}$-close for all entourages $E$ of $X$. We can identify these sequences and define the points of $\overline{X}$ to be these equivalence classes of Cauchy sequences. The entourages $\overline{E}$ descend to define entourages on $\overline{X}$, thus defining it as a uniform space. It contains $X$ as a uniform subspace if we identify $x\in X$ with (the equivalence class of) the constant sequence $x, x, x, ...$. It’s straightforward to show that this inclusion map is uniformly continuous. We can also verify that the second-countability of $X$ lifts up to $\overline{X}$.
Now it also turns out that $\overline{X}$ is complete. Let’s consider a sequence of Cauchy sequences $(x_k)_i$. This will be Cauchy if for all entourages $\overline{E}$ there is an $\bar{N}$ so that if $i\geq\bar{N}$ and $j\geq\bar{N}$ the pair $((x_k)_i,(x_k)_j)$ is in $\overline{E}$. That is, there is an $N_{i,j}$ so that for $k\geq N_{i,j}$ and $l\geq N_{i,j}$ we have $((x_k)_i,(x_l)_j)\in E$. We can’t take the limits in $X$ of the individual Cauchy sequences $(x_k)_i$ — the limits along $k$ — but we can take the limits along $i$! This will give us another Cauchy sequence, which will then give a limit point in $\overline{X}$.
As for the universal property, consider a uniformly continuous map $f:X\rightarrow Y$ to a complete uniform space $Y$. Then every point $\bar{x}$ in $\overline{X}$ comes from a Cauchy sequence $x_i$ in $X$. Being uniformly continuous, $f$ will send this to a Cauchy sequence $f(x_i)$ in $Y$, which must then converge to some limit $\bar{f}(\bar{x})\in Y$ since $Y$ is complete. On the other hand, if $x_i'$ is another representative of $\bar{x}$ then the uniform continuity of $f$ will force $\lim f(x_i)=\lim f(x_i')$, so $\bar{f}$ is well-defined. It is unique because there can be only one continuous function on $\overline{X}$ which agrees with $f$ on the dense subspace $X$.
So what happens when we apply this construction to the rational numbers $\mathbb{Q}$ in an attempt to patch up all those holes and make all the Cauchy sequences converge? At long last we have the real numbers $\mathbb{R}$! Or, at least, we have the underlying complete uniform space. What we don’t have is any of the field properties we’ll want for the real numbers, but we’re getting close to what every freshman in calculus thinks they understand.
### Like this:
Posted by John Armstrong | Point-Set Topology, Topology
## 13 Comments »
1. [...] Miscellany Well, yesterday was given over to exam-writing, so today I’ll pick up a few scraps I mentioned in passing on Thursday. [...]
Pingback by | December 1, 2007 | Reply
2. [...] Order on the Real Numbers We’ve defined the real numbers as a topological field by completing the rational numbers as a uniform space, and then extending the field operations to the new points [...]
Pingback by | December 5, 2007 | Reply
3. [...] functions are defined they agree. Since the rationals are dense in the reals (the latter being the uniform completion of the former) there can be only one continuous extension of to the whole real line. We’ll [...]
Pingback by | April 11, 2008 | Reply
4. [...] We defined the real numbers to be a complete uniform space, meaning that limits of sequences are convergent if and only if they are Cauchy. Let’s write [...]
Pingback by | April 23, 2008 | Reply
5. [...] Now we’ve gone through a lot of work to just add one little extra element to our field, but it turns out this is all we need. Luckily enough, the complex numbers are already algebraically complete! This is very much not the case if we were to try to algebraically complete other fields (like the rational numbers). Unfortunately, the proof really is essentially analytic. It seems to be a completely algebraic statement, but remember all the messy analysis and topology that went into defining the real numbers. [...]
Pingback by | August 7, 2008 | Reply
6. [...] we had sequences of rational numbers which didn’t converge to a rational number. Then we completed the topology to give us the real numbers. Well here we’re just doing the same thing! It turns out that the [...]
Pingback by | August 18, 2008 | Reply
7. [...] integral does not exist. Yes, is countable, and so it has measure zero. However, it’s also dense, which means is discontinuous everywhere in the unit [...]
Pingback by | December 18, 2009 | Reply
8. [...] a lemma: if is irrational, then the set of all numbers of the form with and any integers is dense in the real line. That is, every open interval contains at least one point of . The same is true [...]
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9. [...] got a normed vector space, it’s a natural question to ask whether or not this vector space is complete or not. That is, we have all the pieces in place to define Cauchy sequences in our vector space, [...]
Pingback by | May 12, 2010 | Reply
10. [...] the real numbers form a complete metric space, being Cauchy and being convergent are equivalent — a sequence of finite real numbers is [...]
Pingback by | May 14, 2010 | Reply
11. [...] We won’t talk quite yet about convergence because our situation is sort of like the one with rational numbers; we have a sense of when functions are getting close to each other, but most of these mean Cauchy [...]
Pingback by | May 28, 2010 | Reply
12. [...] We have a notion of “completeness” for boolean rings, which is related to the one for uniform spaces (and metric spaces), but which isn’t quite the same thing. We say that a Boolean ring is [...]
Pingback by | August 10, 2010 | Reply
13. [...] old, awkward approach and has nothing to do with category theory — which tells us that every complete metric space is a Baire space. Let be a countable collection of open dense subsets of . We will show that [...]
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## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
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http://physics.stackexchange.com/questions/tagged/electron?sort=votes&pagesize=50
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# Tagged Questions
Negatively charged particle with spin 1/2. A component of mundane terrestrial matter, and part of all neutral atoms and molecules. It has a mass about 1/1800 that of a proton. Its antiparticle is the positron.
6answers
1k views
### What would happen if Large Hadron Collider would collide electrons?
After some reading about the Large Hadron Collider and it's very impressive instruments to detect and investigate the collision results, there is a remaining question. What would happen if the ...
1answer
2k views
### Are all electrons identical?
Why should two sub-atomic (or elementary particle) - say electrons need to have identical static properties - identical mass, identical charge? Why can't they differ between each other by a very ...
3answers
1k views
### Electrons faster than speed of light
While looking at some exercises in my physics textbook, I came across the following problem which I thought was quite interesting: It is possible for the electron beam in a television picture tube ...
4answers
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### Why is the charge naming convention wrong?
I recently came to know about the Conventional Current vs. Electron Flow issue. Doing some search I found that the reason for this is that Benjamin Franklin made a mistake when naming positive and ...
10answers
3k views
### Does 'electricity' have mass? Is 'electricity' tangible?
Background: I'm in a legal academic discussion about the status of electronic 'goods' and whether they qualify as 'goods' in the same way a chair and a pen do. In this context (and specifically at the ...
2answers
392 views
### What is the mass density distribution of an electron?
I am wondering if the mass density profile $\rho(\vec{r})$ has been characterized for atomic particles such as quarks and electrons. I am currently taking an intro class in quantum mechanics, and I ...
4answers
2k views
### Why do electron and proton have the same but opposite electric charge?
What is the explanation between equality of proton and electron charges (up to a sign)? This is connected to the gauge invariance and renormalization of charge is connected to the renormalization of ...
3answers
301 views
### Current in a simple circuit
I was going over my notes for an introductory course to electricity and magnetism and was intrigued by something I don't have an answer to. I remember my professor mentioning, to the best I can ...
2answers
129 views
### What happens in electron-electron collisions?
What are the results of high energy electron electron collisions? Are other particles created?
1answer
446 views
### Why are noble gases used for lights?
I know that neon is used in advert signs due to its inertness. However, I am not entirely sure how the inertness is exploited. I think it is because Ne being inert means that after electricity frees ...
4answers
3k views
### How fast do electrons travel in an atomic orbital?
I am wondering how fast electrons travel inside of atomic electron orbitals. Surely there is a range of speeds? Is there a minimum speed? I am not asking about electron movement through a conductor.
3answers
449 views
### What was missing in Dirac's argument to come up with the modern interpretation of the positron?
When Dirac found his equation for the electron $(-i\gamma^\mu\partial_\mu+m)\psi=0$ he famously discovered that it had negative energy solutions. In order to solve the problem of the stability of the ...
3answers
432 views
### Electrons in CRT
In a CRT, where do the ejected electrons go after they cause fluorescence on the screen, have they lost most of their energy, or do they actually go through the glass?
4answers
1k views
### Spontaneous pair production?
So I've been looking into particle-antiparticle pair production from a gamma ray and don't understand one thing. Let's say I have a 1,1 MeV photon and it hits a nucleus - electron-positron pair with ...
1answer
606 views
### Why doesn't orbital electron fall into the nucleus of Rb85, but falls into the nucleus of Rb83?
Rb83 is unstable and decays to Kr-83. Mode of the decay is electron capture. Rb85 is stable. The nuclei Rb83 and Rb85 have the same charge. Rb85 is heavier than Rb85, but gravitation is too weak to ...
6answers
421 views
### How low can an electron go?
Title says it all - recently I encountered a strange homework exercise on de Broglie dual theory with an electron wavelength of few millimeters - which implies the velocity lower than 1 m/s. I ...
2answers
282 views
### Is there a published upper limit on the electron's electric quadrupole moment?
I understand an electric quadrupole moment is forbidden in the standard electron theory. In this paper considering general relativistic corrections (Kerr-Newman metric around the electron), however, ...
3answers
509 views
### Measuring the spin of a single electron
Is it possible to measure the spin of a single electron? What papers have been published on answering this question? Would the measurement require a super sensitive SQUID, Superconductive Quantum ...
2answers
143 views
### Do the energy levels of electron orbitals change relativistically?
When an electron emits a photon from changing energy levels, the frequency of the photon depends on the difference between the energy levels. But if someone is moving with respect to the atom, the ...
1answer
186 views
### drift velocity of electrons in a superconductor
is there a formula for the effective speed of electron currents inside superconductors? The formula for normal conductors is: $$V = \frac{I}{nAq}$$ I wonder if there are any changes to this ...
2answers
1k views
### How does electricity propagate in a conductor?
On a systems level, I understand that as electrons are pushed into a wire, there is a net field and a net electron velocity. And I've read that the net electron drift is slow. But electricity ...
3answers
520 views
### Electrons - What is Waving?
If an electron is a wave, what is waving? So many answers on the internet say "the probability that a particle will be at a particular location"... so... the electron is a physical manifestation of ...
4answers
1k views
### How do electrons “know” to share their voltage between two resistors?
My physics teacher explained the difference between voltage and current using sandwiches. Each person gets a bag full of sandwiches when they pass through the battery. Current = the number of people ...
4answers
672 views
### Why photons transfer to electrons perpendicular momentum?
Linear antenna directed along z, photons (EM waves) propagate along x. Momentum of photons have only x component. Why electrons in antenna have z component of momentum?
1answer
132 views
### What happens to 5 electrons on a sphere?
Let's suppose we put 5 electrons on a perfectly conducting (no resistance at all) sphere. There's no equilibrium configuration with 5 (though there is with 2, 3, 4 or 6). So will they keep moving on ...
2answers
2k views
### How does electron move around nucleus?
I need to get a nice picture about how electron moves around nucleus? I find concept of probability and orbitals quite difficult to understand?
1answer
60 views
### Do electrons need specific energies to excite electrons
Photons need specific energy levels, equal to the difference between two energy levels to excite an electron in an atom. Is this the same case with electrons that collide with atoms?
3answers
951 views
### What is the difference between a neutron and hydrogen?
Differences? They are both an electron and a proton, since the neutron decays to a proton and an electron, what's the difference between a neutron and proton + electron? so is it just a higher binding ...
1answer
331 views
### Can electrons move through vacuum?
I studied that electronics is the branch of physics which deals with the practical applications of moving electrons through vacuum, semi-conductors and gasses. Can electrons move through vacuum?
2answers
256 views
### What happens where an electron is annihilated by a spontaneously generated positron-electron pair?
I was visiting the Australian Synchrotron earlier today as part of a tour group; as the guide was going over the booster and storage rings I was reminded of something I learnt of quantum. If I know ...
3answers
4k views
### What happens when we connect a metal wire between the 2 poles of a battery?
As I remembered, at the 2 poles of a battery, positive or negative electric charges are gathered. So there'll be electric field existing within the battery. This filed is neutralized by the chemical ...
3answers
136 views
### Do electrons in multi-electron atoms really have definite angular momenta?
Since the mutual repulsion term between electrons orbiting the same nucleus does not commute with either electron's angular momentum operator (but only with their sum), I'd assume that the electrons ...
1answer
182 views
### How many states can a n qubit quantum computer store?
A classical computer composed of '0' or '1' transistors stores $2^n$ states. Is it true that a quantum computer composed of '0' or '1' or '0 & 1' qubits stores $3^n$ states?
1answer
912 views
### How to know what materials are good conductors of electricity?
I'm not asking a question like "Is the wood conductive?". No. I'm asking what properties do they have to have to be good conductors. Theoretically I mean. Thanks.
2answers
164 views
### Do protons exchange photons with electrons?
I'm sorry for this question but, I just don't get it. According to the electromagnetic field theory, electrons repel each other by exchanging photons. How do protons attract electrons, by photon ...
1answer
276 views
### How would Kohn-Sham orbitals differ from 'true' elecron wavefunctions?
How would the non-interacting electron orbitals from a perfect DFT solution for a given potential shape differ from the 'true' electron wavefunctions? Or can you only really talk about the total ...
1answer
150 views
### Stability of a rotating ring of multiple electrons at relativistic speeds
There was a time when physicists where concerned about electron internal structure. The rotating ring model was one of the proposals to explain how a charge density could become stable against ...
1answer
280 views
### Why can free lithium atoms not take part in an Auger process?
Shouldn't it be possible for an incoming photon to excite one of the 1s electrons to a 2p state (or one of even higher energy) and then for the excited electron to drop back to 1s and kick out the 2s ...
2answers
149 views
### Why the K shell only contains 2 electrons?
It is written in my quantum physics book that the K shell contains only 2 electrons due to the Pauli principle. I know that if $n = 1, l = 0, m = 0$, then the Hilbert space associated to the spin is ...
4answers
669 views
### Where is spin in the Schroedinger equation of an electron in the hydrogen atom?
In my current quantum mechanics, course, we have derived in full (I believe?) the wave equations for the time-independent stationary states of the hydrogen atom. We are told that the Pauli Exclusion ...
4answers
471 views
### How do electrons jump orbitals?
My question isn't how they receive the energy to jump, but why. When someone views an element's emission spectrum, we see a line spectrum which proves that they don't exist outside of their orbitals ...
3answers
1k views
### Electron Positron annihilation Feynman Diagram
I am having some trouble understanding this fenyman diagram, it seems to indicate that the electron produces the positron, as the arrow of the positron is pointing from the electron. Additionally ...
4answers
956 views
### Bohr's model of an atom doesn't seem to have overcome the drawback of Rutherford's model
We, as high school students have been taught that-because Bohr's model of an atom assigns specific orbits for electrons-that it is better than Rutherford's model. But what Rutherford failed to explain ...
2answers
363 views
### Is the free electron wavefunction stable?
The wavefunction of a free electrons is variously described as a plane wave or a wave packet. I am fairly happy with the wave packet, as it is localised. But if we change to the electron's rest ...
2answers
382 views
### How do electrons know that?
The current is maximum through those segments of a circuit that offer the least resistance. But how do electrons know beforehand that which path will resist their drift the least?
2answers
73 views
### Regarding the free electrons on the conductor
In a metal, why don’t the free electrons fall to the bottom of the metal due to gravity? Also, charges in a conductor are supposed to reside on the surface so why don’t the free electrons all go to ...
3answers
159 views
### What if $\gamma$-rays in Electron microscope?
I was referring Electron microscopes and read that the electrons have wavelength way less than that of visible light. But, the question I can't find an answer was that, If gamma radiation has the ...
2answers
298 views
### Why do electrons around nucleus radiate light according to classical physics
As I navigate through physics stackexchange, I noticed Electron model under Maxwell's theory. Electrons radiate light when revolving around nucleus? Why is it so obvious? Note that I do not know ...
2answers
1k views
### In electromagnetic radiation, how do electrons actually “move”?
I've always pictured EM radiation as a wave, in common drawings of radiation you would see it as a wave beam and that had clouded my understanding recently. Illustration on the simplest level: ...
2answers
142 views
### Thought experiment with entangled electrons
Suppose we start out by having two entangled electrons. We separate them by some distance and we put one electron inside a thin loop of wire connected to an extremely sensitive voltage measuring ...
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http://physics.stackexchange.com/questions/45077/direct-nuclear-reaction-in-nuclear-physics
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# Direct nuclear reaction in nuclear physics
Time taken to occur a direct nuclear reaction is very low $10^{-22}$sec . I want to know the Importance of direct nuclear reactions.
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1
– dmckee♦ Nov 25 '12 at 18:21
## 1 Answer
from the wiki article on nuclear reactions:
Direct reactions
An intermediate energy projectile transfers energy or picks up or loses nucleons to the nucleus in a single quick (10−21 second) event. Energy and momentum transfer are relatively small. These are particularly useful in experimental nuclear physics, because the reaction mechanisms are often simple enough to calculate with sufficient accuracy to probe the structure of the target nucleus.
So their importance lies in their use as a probe to study a nucleus, i.e determine its energy levels, quadrupole moment etc. The short lifetime is indicative of a strong interaction lifetime ( in contrast to weak decays or electromagnetic interactions which are much slower)
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http://mathhelpforum.com/geometry/201548-find-equation-lines-print.html
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# find the equation of the lines.
Printable View
• July 30th 2012, 11:26 PM
rcs
find the equation of the lines.
there are two lines through the given point, which are tangent to the given curve. Find the equation each of these lines.
4x^2- 5xy + 2y^2 + 3x - 2y = 0 , P ( 2,3)
thank you
• August 1st 2012, 10:03 AM
earboth
1 Attachment(s)
Re: find the equation of the lines.
Quote:
Originally Posted by rcs
there are two lines through the given point, which are tangent to the given curve. Find the equation each of these lines.
4x^2- 5xy + 2y^2 + 3x - 2y = 0 , P ( 2,3)
thank you
1. All lines passing through P have the equation: $y = m(x-2)+3$
2. Replace the term y in your equation of the ellipse by m(x-2)+3. You'll get a quadratic in x:
$x^2(2m^2 - 5m + 4) - x(8m^2 - 20·m + 12) + 4(2m^2 - 5·m + 3) = 0$
Solve for x.
3. Usually a straight line intercepts an ellipse in 2 or in 1 point or it is a passante that means there are no common points. The case that there exists only one point of interception occurs if the straight line is a tangent to the ellipse. This will happen if the discriminant equals zero:
$\sqrt{- 2·m^2 + 5·m - 3 - 2·m^2 + 5·m - 3)} = 0$
Solve for m. You should come out with $m = 1~\vee~m=\frac32$
4. Plug in these values into the equation of the line.
• August 2nd 2012, 01:53 AM
rcs
Re: find the equation of the lines.
thank you for enlightening my brain... millions of thanks sir.
God Bless
• September 3rd 2012, 07:04 PM
rcs
Re: find the equation of the lines.
Quote:
Originally Posted by earboth
1. All lines passing through P have the equation: $y = m(x-2)+3$
2. Replace the term y in your equation of the ellipse by m(x-2)+3. You'll get a quadratic in x:
$x^2(2m^2 - 5m + 4) - x(8m^2 - 20·m + 12) + 4(2m^2 - 5·m + 3) = 0$
Solve for x.
3. Usually a straight line intercepts an ellipse in 2 or in 1 point or it is a passante that means there are no common points. The case that there exists only one point of interception occurs if the straight line is a tangent to the ellipse. This will happen if the discriminant equals zero:
$\sqrt{- 2·m^2 + 5·m - 3 - 2·m^2 + 5·m - 3)} = 0$
Solve for m. You should come out with $m = 1~\vee~m=\frac32$
4. Plug in these values into the equation of the line.
sir i just dont get when you said to use the y as m(x-2)+3 to ellipse equation?... but what is the ellipse formula sir?
• September 3rd 2012, 10:34 PM
earboth
Re: find the equation of the lines.
Quote:
Originally Posted by rcs
there are two lines through the given point, which are tangent to the given curve. Find the equation each of these lines.
4x^2- 5xy + 2y^2 + 3x - 2y = 0 , P ( 2,3)
thank you
Quote:
Originally Posted by rcs
sir i just dont get when you said to use the y as m(x-2)+3 to ellipse equation?... but what is the ellipse formula sir?
The marked equation describes an ellipse, not the ellipse.
From the attachment of my first post you can see that you have a "tilted" ellipse, that means the axes are not parallel to the coordinate axes, but the curve is still an ellipse.
• September 4th 2012, 04:16 AM
rcs
Re: find the equation of the lines.
Quote:
Originally Posted by earboth
The marked equation describes an ellipse, not the ellipse.
From the attachment of my first post you can see that you have a "tilted" ellipse, that means the axes are not parallel to the coordinate axes, but the curve is still an ellipse.
i got it sir... i solved it manually and i was able to get that equation equal to 0. something that stuck me how you got square root (-2m^2 +5m -3 -2m^2 +5m - 3) = 0... is that the discriminant sir? which being plugged in by the values of the equation above?
thanks
All times are GMT -8. The time now is 05:53 AM.
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