text:
# Example of a bialgebra which is Frobenius but not Hopf

Let $$\Bbbk$$ be a field. A well-known result (see Larson, Sweedler, An Associative Orthogonal Bilinear Form for Hopf Algebras and Pareigis, When Hopf algebras are Frobenius algebras) states that a bialgebra $$B$$ over $$\Bbbk$$ is a finite-dimensional Hopf algebra if and only if it is Frobenius as an algebra and the Frobenius morphism $$\psi:B\to\Bbbk$$ is a (left) integral on $$B$$ (i.e. $$\sum b_1\psi(b_2)=\psi(b)1_B$$ for all $$b\in B$$).

For didactical reasons, I would like to have an example of a (necessarily finite-dimensional) bialgebra $$B$$ which is Frobenius as an algebra but that is not a Hopf algebra. However, I didn't manage to construct one. Does anybody know one? Even a non-elementary one would be fine.

Alternatively, if somebody knows a proof of the fact that a Frobenius bialgebra is automatically a Hopf algebra, without further assumptions on the Frobenius homomorphism, then I would be very glad to see it.

Here is an example. Take k a field and S the multiplicative semigroup {0,1}. For notational convenience denote it S={1,x}. Consider B=k[S] the semigroup algebra.

B is bialgebra and not Hopf because S is a semigroup and nota a group.

$$B\cong k[x]/(x^2-x)\cong k\times k$$ is semisimple. In particular is a Frobenius algebra.

• Thanks, that's a nice example. For personal future reference, an explicit Frobenius coordinate system is given by $e= 1\otimes 1 - 1\otimes x - x\otimes 1$ and $\psi:B\to\Bbbk: 1\mapsto 0, x\mapsto -1$. – Ender Wiggins May 26 '19 at 16:55

Ussually finite dimension is part of the definition of a Frobenius algebra. So, what definition of "Frobenius" algebra do you have in mind?

In wikipedia (https://en.wikipedia.org/wiki/Frobenius_algebra#Category-theoretical_definition) a categorical definition is given. If you take a "Frobenius object" in the category of vector spaces, then Frobenius impies finite dimensionality.

• This is exactly the reason why I stressed finite-dimensional: I was trying to avoid un-helpful (non-)answers because I would like to see a genuine, ring-theoretic example of a bialgebra over a field that (as an algebra) is Frobenius (Casimir+Frobenius morphism or non-degenerate bilinear associative form, choose the description you prefer) – Ender Wiggins May 17 '19 at 7:26
• Just to be sure, when I read your post I thought of an infinite dimensional example, but now I read carefully and I see that what you wrote is "necessarily finite dimensional". But then, what's wrong with Larson-Sweedler-Pareigis result? You are lookin for finite or infinite dimensional examples? – Marco Farinati May 17 '19 at 11:59
• Since the additional condition on the Frobenius morphism is required to have a Hopf algebra structure, I would like to see an example of a bialgebra which is Frobenius but doesn't satisfy that condition, so that it is not a Hopf algebra. In particular, being Frobenius, it has to be finite-dimensional. – Ender Wiggins May 18 '19 at 12:08
• I think you misunderstood the statement. Larson-Sweedler proves that a finitely generated bialgebra over a p.i. ring, (e.g. a finite dimensional bialgebra over a field) a Frobenius structure provides an atipode (and viceversa). So, if a finite dimensional bialgebra $B$ is Frobenius as algebra, then it is necesarily Hopf. – Marco Farinati May 21 '19 at 13:28
• I am missing the converse: why a bialgebra which is Frobenius as an algebra should admit an antipode? Larson-Sweedler proved only one implication, from Hopf to Frobenius. Pareigis proved the converse, but under the additional assumption that the Frobenius morphism is an integral. I would like to see a proof of $B$ Frobenius $\Rightarrow$ $B$ fd Hopf or a counterexample of a Frobenius bialgebra which is not Hopf – Ender Wiggins May 21 '19 at 17:36


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# 61 is 10% of what number?

Nov 28, 2016

If 61 = 10% of X, then $X = 610$

#### Explanation:

Proof:

61 = 10% of $X$ means 61 = X *10% = X *10/100 = X/10

or:

$61 \cdot 10 = X$ that is $X = 610$.

format: plain


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# Factor the polynomial

• Jan 20th 2011, 03:10 PM
Sapphire19
Factor the polynomial
6 3
x + 16X + 64 =

3 3 2 2
This formula: x
+ a = (X+a)(x - ax + a )

• Jan 20th 2011, 03:11 PM
Sapphire19
oops the problem is
x^6 +16x^3_64 and the formula is X^3+a^3= (x^2-ax + a^2)
• Jan 20th 2011, 03:26 PM
Quote:

Originally Posted by Sapphire19
oops the problem is
x^6 +16x^3_64 and the formula is X^3+a^3= (x^2-ax + a^2)

This is a quadratic in $x^3$

$\left(x^3\right)^2+16\left(x^3\right)+64=\left(x^3 +8\right)\left(x^3+8\right)$
• Jan 20th 2011, 03:34 PM
Sapphire19
That's not the right answer. This is regarding homework on mymathlab. If you write the wrong answer it tells you right away if it is wrong. The answer should look exactly like (except with different numbers)
(x+3)^2 (x^2-3x+9)^2

It says "notice that the given polynomial is a perfect square and can be factored by (x+a)^2.
For example, x^6+54x^3+729= (x^3+27)^2. "Now notice that this answer is the sum of 2 cubes. Now you use the formula that I previously posted to get the answer.
Even with this help that mymathlab gives me I still can't manage to complete this question. After literally over 45 minutes of trying I am giving up because i have A LOT of math homework to do and simply cannot spend 45 minutes on one question!
• Jan 20th 2011, 03:41 PM
e^(i*pi)
Archie Meade's factored form is the sum of two cubes

$x^3+8 = x^3+2^3$
• Jan 20th 2011, 07:23 PM
Quote:

Originally Posted by Sapphire19
That's not the right answer. This is regarding homework on mymathlab. If you write the wrong answer it tells you right away if it is wrong. The answer should look exactly like (except with different numbers)
(x+3)^2 (x^2-3x+9)^2

It says "notice that the given polynomial is a perfect square and can be factored by (x+a)^2.
For example, x^6+54x^3+729= (x^3+27)^2. "Now notice that this answer is the sum of 2 cubes. Now you use the formula that I previously posted to get the answer.
Even with this help that mymathlab gives me I still can't manage to complete this question. After literally over 45 minutes of trying I am giving up because i have A LOT of math homework to do and simply cannot spend 45 minutes on one question!

You don't sound like you've got too much patience!

$x^3+2^3=(x+2)\left(x^2-2x+2^2\right)$

Does that help ?
• Jan 20th 2011, 08:07 PM
mirajshah
Related question
Hi MHFers! I've got what's supposed to be a relatively easy problem here... can anyone solve?
$0.00278=p^{4}\times(1-p)^{3}$

Thanks!
• Jan 21st 2011, 04:17 AM
$0.00278=p^{4}\times(1-p)^{3}$


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# Tag Info

24

In order to understand Russian Roulette, let's look at a very basic backward path tracer: void RenderPixel(uint x, uint y, UniformSampler *sampler) { Ray ray = m_scene->Camera.CalculateRayFromPixel(x, y, sampler); float3 color(0.0f); float3 throughput(1.0f); // Bounce the ray around the scene for (uint bounces = 0; bounces < 10; +...

17

There are multiple areas in path tracing that can be importance sampled. In addition, each of those areas can also use Multiple Importance Sampling, first proposed in Veach and Guibas's 1995 paper. To better explain, let's look at a backwards path tracer: void RenderPixel(uint x, uint y, UniformSampler *sampler) { Ray ray = m_scene->Camera->...

13

I am posting this for anyone wondering about the confusion between the terms $\frac{1}{\pi}$ and $\frac{1}{4}$. The term $\frac{1}{\pi}$ is an error from the original Cook-Torrance reference. In fact, the whole term $\frac{1}{4(n \cdot \omega_i)}$ comes from the Jacobian of the transformation from reflected solid angle to normal solid angle. According to ...

12

According to this paper, the $\frac{1}{\pi}$ in your $f_r$ should be $\frac{1}{4}$: $$f_r = \frac{DFG}{4(n\cdot w_i)(n \cdot w_o)},$$ so you would end up with $$\frac{\pi}{2}L_i(p,w_k)\left(\frac{DFG}{n\cdot w_o}\right).$$

12

The theoretical ideal antialiasing filter for discretely sampled data is a sinc filter, because it perfectly removes all frequencies higher than the Nyquist frequency, while leaving alone all the lower ones. So, to some extent, we can expect antialiasing filters that more closely resemble the sinc filter to produce better-quality images. The tent filter (...

12

This question is somewhat related to this one. As Alan has already said, following the actual light path through each layer leads to more physically accurate results. I will base my answer on a paper by Andrea Weidlich and Alexander Wilkie ("Arbitrarily Layered Micro-Facet Surfaces") that I have read and partially implemented. In the paper the authors ...

11

Without explicit light sampling, I'd expect very slow convergence. Any path that doesn't happen to hit the light source before being terminated will be useless (contributes zero). Since you have a small light source, the vast majority of paths will not hit it. It would be interesting to add some statistics to your tracer to see how many paths return zero vs ...

10

Generally speaking, path tracing removes a number of assumptions that ray tracing makes. Ray tracing usually assumes that there is no indirect lighting (or that indirect lighting can be approximated by a constant function), because handling indirect lighting would require casting many additional rays whenever you shade an intersection point. Ray tracing ...

9

The Russian roulette technique itself is a way of terminating paths without introducing systemic bias. The principle is fairly straightforward: if at a particular vertex you have a 10% chance of arbitrarily replacing the energy with 0, and if you do that an infinite number of times, you will see 10% less energy. The energy boost just compensates for that. If ...

9

There is a great paper from 2006 on this topic, Filter Importance Sampling. They propose your method 2, study the properties, and come out generally in favor of it. They claim that this method gives smoother rendering results because it weights all samples that contribute to a pixel equally, thereby reducing variance in the final pixel values. This makes ...

9

The box.obj file has no vertex normals, and by default Mitsuba will generate smooth normals for OBJ files that don't specify their own normals. This creates the magnification effect: the box with smooth normals forms a convex lens! By adding this line to the box object in the scene file: <boolean name="faceNormals" value="true"/> I got results that ...

8

As mentioned in the comments, I would highly suggest starting with Full Volumetric Scattering. This is two fold: Since you are doing path tracing, adding volumetrics isn't super difficult. Fully understanding how full volumetric scattering works will be a great basis for understanding the estimations. In addition, it can provide great "references" to see if ...

7

First of all, a good reference for Monte Carlo path tracing in participating media is these course notes from Steve Marschner. The way I like to think about volume scattering is that a photon traveling through a medium has a certain probability per unit length of interacting (getting scattered or absorbed). As long as it doesn't interact, it just goes in a ...

7

Just to expand on some of the other answers, the proof that Russian Roulette does not give a biassed result is very simple. Suppose that you have some random variable $F$ which is the sum of several terms: $$F = F_1 + \cdots + F_N$$ Replace each term with: $$F'_i = \left\{ \begin{array}{ll} \frac{1}{p_i} F_i & \hbox{with probability } p_i \\ ... 7 Let us take one step back. When you do path tracing, you are doing a Monte Carlo integration. What does this mean? You try to solve \int f(x)\mathrm{d}x by sampling. The Monte Carlo integration says for a sufficiently big n: \frac{1}{n}\sum_{i=0}^n f(x_i) \rightarrow \int f(x)p(x) \mathrm{d}x If we instead sum \frac{f(x_i)}{p(x_i)} we get \frac{1}{... 7 Wenzel Jakob et al presented a framework for layered Materials at SIGGRAPH 2014. Section 6.2 explains importance sampling. If you prefer code over equations, the method is implemented in the Mitsuba renderer. 7 Assuming that you are familiar with the concept of BSDFs, the usual way of modelling rough dielectric surfaces (i.e. glass, water, plastics) is to use microfacet-based models like Microfacet Models for Refraction through Rough Surfaces. To make it work efficiently in a path tracer you will need a good sampling strategy, like Importance Sampling Microfacet-... 7 Overview Here is a short overview of the most used space representations, MLT variants and mutation strategies for these MLT variants. As you can see, there are quite some papers dating back to 2017 (e.g., three papers explore combining the Path Space and the Primary Sample Space by jumping back and forth between the two). Path Space (PS) representation, ... 6 When you perform regular Monte Carlo integration over a hemisphere using N samples, each sample represents \frac{2\pi}{N} steradians. So the Monte Carlo integration for Lambertian BRDF is:$$\frac{2\pi}{N}\sum_{i=1}^N\frac{\rho}{\pi}L_i*Cos\theta_i For path tracing, you only take one sample per path segment, so because $N$=1, the above sum becomes: \$...

6

TL;DR Yes, you can do it like that, you just have to divide the result by the probability of choosing the direction. Full Answer The topic of sampling in path tracers allowing materials with both reflection and refraction is actually a little bit more complex. Let's start with some background first. If you allow BSDFs - not just BRDFs - in your path ...

6

This is not a full answer, I would just like to share the knowledge I obtained by studying two of the papers mentioned in the question: Steerable Importance Sampling and Practical Product Importance Sampling for Direct Illumination. Steerable Importance Sampling In this paper they propose a method for sampling the product of the clamped cosine component ...

6

The problem lies mainly in CIE1931XYZ::tristimulusValues() function, where you normalize the resulting color to the luminance of your illuminant which causes that directly observed light source has luminance 1, but everything else is much darker. That is a nice thing to do if you just want to visualize colours of various reflectance spectra under a given ...

6

The problem appears to be unintentionally transparent surfaces Although the image is grainy, it is sufficiently clear to estimate that all of the darker regions are due to surfaces facing away from the light, rather than due to shadows cast onto surfaces facing towards the light. So it does seem that there is a problem, and the lack of shadows is not just ...

6

There are, and I am looking forward to seeing the specifics of other answers, but one way to deal with this is to not have the noise (or as much noise) in the source data to begin with. The noise is coming from the fact that there is high variance in the rendering - the number of samples you've taken haven't converged enough to the actual right answer of ...

6

I don't know if this is exactly what you're looking for. I work tangentially in the film and TV industry. I don't work for a studio, but I work on the software that studios use for their productions. It's software that any user could buy if they wanted to, but it has been used in feature films, television shows, commercials, music videos, and more. About ...

5

is it correct having such a big light? I don't see any problem with having a big area light. That said, it also depends on the scale of your scene. If the light is large compared to it, shadows will tend to be more diffuse, like under an overcast sky. shadows are missing.. [...] Can you help me understanding why? I haven't found any blatant mistake, but ...

5

Firstly, as @trichoplax correctly pointed out, your randomPoint function calculates a point in a cube, then uses rejection sampling to return all points that are inside a unit sphere. In order to return points on a sphere, you would need to change the greater than to an equals. That said, rejection sampling is very inefficient. A better way to sample a ...

5

Yes, this is to be expected. The difference between your two images is the difference between using and not using importance sampling. The first image shows noticeably more noise than the second. This is because instead of biasing the hemisphere samples towards the normal by using cosine weighting, the code instead samples uniformly over the hemisphere and ...

5

Most renderers allow you to set a flag on an object to make shadow rays ignore it (so it won't cast shadows). You could set this flag on your glazing. That said, it's not unusual to make it so that refraction is disabled on shadow rays: that is, if they would be transmitted by an object, a child ray is traced in the same direction, ignoring any refraction ...

5

First of all, you can get caustics just by Path Tracing. Caustics isn't a difficult phenomenon in computer graphics. Almost any global illumination algorithm can render it. It's just a matter of the converging speed. While (naive) Path Tracing can render caustics, it takes a long time to clean the rendered image due to how the algorithm works. Photon ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Bodhee Prep-CAT Online Preparation

| Best Online CAT PreparationFor Enquiry CALL @ +91-95189-40261

# CAT 2017 [slot 2] Question with solution 13

Question 47:
Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal's share, in rupees, is
1. 100
2. 200
3. 300
4. 400
Option: 1
Explanation:

Let the time take by kamal to complete the task be x days.
Hence we have $\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$
=> x = 40 days.
Ratio of the work done by them = $\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$ = 4 : 5 : 1
Hence the wage earned by Kamal = 1/10 × 1000 = 100

Previous QuestionNext Question

### CAT 2023Classroom Course

We are starting classroom course for CAT 2023 in Gurugram from the month of December.


text:
# [R-sig-ME] GLMM-Implementation question

Douglas Bates bates at stat.wisc.edu
Thu Jun 11 18:06:28 CEST 2009

On Thu, Jun 11, 2009 at 10:38 AM, Fabian
Scheipl<Fabian.Scheipl at stat.uni-muenchen.de> wrote:
> Never mind me, the answer to the second question is:
> update_u (called by update_dev) iteratively updates the orthonormalized
> random effects until convergence each time before S_nlminb_iterate is
> called.

Yes.  I was about to write that.

> On Thu, Jun 11, 2009 at 5:17 PM, Fabian Scheipl <
> Fabian.Scheipl at stat.uni-muenchen.de> wrote:

>> Dear List,
>>
>> Preparing the slides for a lecture on GLMMs I'm giving next week I noticed
>> that I don't quite understand eq. 40 (Laplace Approximation of a
>> GLMM-Likelihood) in the Implementation-vignette for lme4.
>> I would be extremely grateful  (my students as well, of course, but
>> probably significiantly less so ;) ) if somebody would find the time to
>> offer his/her thoughts on some of the following points:
>>
>> I do not understand how the expression that is exponentiated  in the second
>> line is equivalent to the quadratic Taylor-approximation of the penalized
>> log-likelihood around the conditional mode \tilde b.
>>
>> AFAIU the first two terms in the sum is just the penalized log-likelihood
>> evaluated at the conditional modes \tilde b. The first term is the
>> likelihood of y conditional on fixed and random effects, the second is
>> equivalent to \tilde b ' G^{-1} \tilde b.
>> [should be \tilde b^\star ' \tilde b^\star instead of  \tilde b ' \tilde
>> b^\star, I think? Also, there are two plus signs after that].

I would have to go back and look at that document more carefully to be
able to answer these questions.  Unfortunately I am in "crunch mode"
on another project right now and I don't think I will be able to free
up the time today.

>> The next term should probably read (b^- \tilde b) ' D^-1 (b - \tilde b),
>> the quadratic term in the Taylor-approximation.
>> What bothers me is that, as D is defined in eq. 39 [ which should define
>> Var(b^\star|...), not Var(b|...) ],
>> it is the inverse of the expected Fisher-Info for b^\star, not the observed
>> - e.g. we are not using an expression for the second derivative but for its
>> expectation -  doesn't that make a difference and is what we are doing still
>> a Laplace-Approximation in the conventional sense?
>>
>> My second question: I got lost in the source for lmer - in which function
>> called by ST_setPars  do the PIRLS-updates for b happen and do we actually
>> do Fisher-scoring until convergence for b every time we update \beta and
>> \theta or is it just a single fisher-scoring step on b before each call to
>> S_nlminb_iterate?
>>
>> Best Wishes,
>> Fabian Scheipl
>>
>>
>>
>
>        [[alternative HTML version deleted]]
>
> _______________________________________________
> R-sig-mixed-models at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-sig-mixed-models
>


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Function forecasting::ExponentialSmoothingTrendTune(dataValues, noObservations, alpha, beta, alphaLow, alphaUpp, betaLow, betaUpp)

# forecasting::ExponentialSmoothingTrendTune

The forecasting::ExponentialSmoothingTrendTune() procedure is a time series forecasting helper procedure of forecasting::ExponentialSmoothingTrend() by computing the $$\alpha$$ and $$\beta$$ for which the mean squared error is minimized.

## Function Prototype

forecasting::ExponentialSmoothingTrendTune(
! Provides the alpha for which the mean squared error is minimized.
dataValues,      ! Input, parameter indexed over time set
noObservations,  ! Scalar input, length history
alpha,           ! Scalar output,
beta,            ! Scalar output,
alphaLow,        ! Optional input, default 0.01
alphaUpp,        ! Optional input, default 0.99
betaLow,         ! Optional input, default 0.01
betaUpp)         ! Optional input, default 0.99

## Arguments

dataValues

A one dimensional parameter containing the observations for the first $$T$$ elements of the time set.

noObservations

Specifies the number of elements that belong to the history of the time set. This parameter corresponds to $$T$$ in the notation presented in Time Series Forecasting Notation.

alpha,

beta, i.e. $$\alpha$$ and $$\beta$$ are scalar output parameters of this procedure. The values for $$\alpha$$ and $$\beta$$ are such that the mean squared error of the estimates returned by forecasting::ExponentialSmoothingTrend() are minimized.

alphaLow

Lowerbound on $$\alpha$$, default 0.01.

alphaUpp

Upperbound on $$\alpha$$, default 0.99.

betaLow

Lowerbound on $$\beta$$, default 0.01.

betaUpp

Upperbound on $$\beta$$, default 0.99.

Note

In order to use this function, the Forecasting system library needs to be added to the application.

Please note that this function performs an optimization step; a nonlinear programming solver should be available and, in an AIMMS PRO environment, it should be run server side.


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## The Balian-Low theorem for a new kind of Gabor system.(English)Zbl 1266.42075

The Balian-Low theorem is a very important result in time-frequency analysis. The authors establish the Balian-Low theorem for a new kind of Gabor systems. The proposed Gabor systems are a generalization of already existing cases. Moreover, the authors give some properties of the corresponding generalized Zak transform and provide some examples of the generalized Gabor systems. The topic of this paper is interesting and important to researchers working on both theoretical and practical aspects of time-frequency analysis, Fourier analysis and sampling theorems.

### MSC:

 42C15 General harmonic expansions, frames 33C90 Applications of hypergeometric functions 94A12 Signal theory (characterization, reconstruction, filtering, etc.)
Full Text:

### References:

 [1] DOI: 10.1007/s00041-001-4016-5 · Zbl 0887.42026 [2] DOI: 10.1007/BF02930696 · Zbl 1037.42030 [3] DOI: 10.1063/1.1559415 · Zbl 1062.42026 [4] Benedetto JJ, Math. Res. Lett. 13 pp 467– (2006) · Zbl 1226.42022 [5] DOI: 10.1137/050634104 · Zbl 1142.42013 [6] Czaja, W and Powell, AM. 2006.Recent Developments in the Balian–Low theorem, Harmonic Analysis and Applications, 79–100. Boston: Birkhäuser. · Zbl 1129.42410 [7] Daubechies, I.Ten Lectures on Wavelets, CBMS-NSF, Regional Conference Series in Applied Mathematics, SIAM, Philadelphia, PA, 1992 · Zbl 0776.42018 [8] Gröchenig K, Foundations of Time–frequency Analysis (2001) [9] DOI: 10.1063/1.1768621 · Zbl 1071.42022 [10] Heil C, J. Math. Phys. 47 (11) pp 13506-1– (2006) · Zbl 1112.42004 [11] Janssen AJEM, Philips J. Res. 43 pp 23– (1988) [12] DOI: 10.1103/PhysRevLett.19.1385 [13] Zayed AI, Methods Appl. Anal. 2 (2) pp 160– (1995) [14] Cohen L, Time-Frequency Analysis (1995) [15] Gabor D, J. IEEE, Part III 93 pp 429– (1946) [16] DOI: 10.1109/78.558469 [17] DOI: 10.1016/j.physd.2006.06.013 · Zbl 1104.94004 [18] Chen QH, Int. J. Wavelets, Multiresolution Inform. Proces. 3 (4) pp 465– (2005) · Zbl 1086.42018 [19] DOI: 10.1016/j.na.2007.02.001 · Zbl 1275.94014 [20] DOI: 10.1016/j.physd.2005.03.005 · Zbl 1070.94504 [21] DOI: 10.1216/jiea/1181075323 · Zbl 1086.30035 [22] DOI: 10.1016/j.jmaa.2010.07.037 · Zbl 1200.42026 [23] Young RM, An Introduction to Non-Harmonic Fourier Series, 1. ed. (2001) [24] Dahlke S, Proc. Lond. Math. Soc. 96 (2) pp 464– (2008) · Zbl 1215.42035 [25] Dahlke, S and Teschke, G. and Stingl,Coorbit theory, multi-{$$\alpha$$}-modulation frames, and the concept of joint sparsity for medical multichannel data analysis, EURASIP J. Adv. Sig. Proc. (Special Issue: Signal Processing for Applications in Healthcare Systems) (2008). Article ID 471601, 19 pages · Zbl 1184.94195 [26] DOI: 10.1137/S0036144501386986 · Zbl 0995.42022 [27] Auslander L, IEEE Trans. Signal Proces. 39 (4) pp 825– (1991) [28] Polyak N, IEEE Trans. Signal Proces. 46 (4) pp 857– (1998)
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Center - Maple Help

GroupTheory

 Center
 construct the center of a group

 Calling Sequence Center( G ) Centre( G )

Parameters

 G - a permutation group

Description

 • The center of a group $G$ is the set of elements of $G$ that commute with all elements of $G$. That is, an element $g$ of $G$ belongs to the center of $G$ if, and only if, $g·x=x·g$, for all $x$ in $G$.
 • The Center( G ) command constructs the center of a group G. The group G must be an instance of a permutation group, a group defined by a Cayley table, or a custom group that defines its own center method.
 • The Centre command is provided as an alias.

Examples

 > $\mathrm{with}\left(\mathrm{GroupTheory}\right):$
 > $G≔\mathrm{PermutationGroup}\left(\left\{\left[\left[1,2\right]\right],\left[\left[1,2,3\right],\left[4,5\right]\right]\right\}\right)$
 ${G}{≔}⟨\left({1}{,}{2}\right){,}\left({1}{,}{2}{,}{3}\right)\left({4}{,}{5}\right)⟩$ (1)
 > $\mathrm{Center}\left(G\right)$
 ${Z}{}\left(⟨\left({1}{,}{2}\right){,}\left({1}{,}{2}{,}{3}\right)\left({4}{,}{5}\right)⟩\right)$ (2)
 > $\mathrm{Center}\left(\mathrm{AlternatingGroup}\left(4\right)\right)$
 ${Z}{}\left({{\mathbf{A}}}_{{4}}\right)$ (3)
 > $G≔\mathrm{MetacyclicGroup}\left(3,4,2\right)$
 ${G}{≔}⟨\left({1}{,}{2}{,}{3}\right)\left({4}{,}{6}{,}{8}\right)\left({5}{,}{7}{,}{9}\right)\left({10}{,}{11}{,}{12}\right){,}\left({1}{,}{4}{,}{10}{,}{5}\right)\left({2}{,}{6}{,}{11}{,}{7}\right)\left({3}{,}{8}{,}{12}{,}{9}\right)⟩$ (4)
 > $\mathrm{IsAbelian}\left(\mathrm{Center}\left(G\right)\right)$
 ${\mathrm{true}}$ (5)
 > $\mathrm{GroupOrder}\left(\mathrm{Center}\left(G\right)\right)$
 ${12}$ (6)
 > $\mathrm{IsNormal}\left(\mathrm{Center}\left(G\right),G\right)$
 ${\mathrm{true}}$ (7)

Compatibility

 • The GroupTheory[Center] command was introduced in Maple 17.


text:
# Correct Rational Expression?

• January 26th 2010, 12:34 AM
Noegddgeon
Correct Rational Expression?
Hello, everybody. I did an exercise from a textbook I'm learning from and here is the rational expression which I'm required to add:

$

\frac {6}{x + 3} - \frac {x + 4}{9 - x^2} + \frac {2x - 3}{9 - 6x + x^2}

$

Here is the answer which I got:

$

\frac {9x^2 - 32x + 33}{(x + 3)(x - 3)^2}

$

I would very much appreciate if anyone can tell me whether or not I did this problem correctly, as there is no answer for it in the book. Thank you very much for your time. :]

Colton
• January 26th 2010, 12:45 AM
Noegddgeon
Just discovered a beautiful thing called mathway.com and my question was answered with the power of technology. :] Thanks to all anyways who may have helped me.
• January 26th 2010, 12:49 AM
Sudharaka
Dear Noegddgeon,


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# Tag Info

0

After several hours of frustration I finally decided to put the file "ectaart.cls" in the same folder of my LaTeX file, and it worked! I know it is not the standard way to do this, but I really tried all other options.

4

I don't know about a class for this. However, it would not make much sense, since many in-document elements are a bit class-dependent, as well as many class options wouldn't work. Still, there comes the package savetrees as a rescue. You simply add \usepackage{savetrees} in your preamble, and voilà! the document has less pages! :) To typeset in ...

2

Under the line class option for res, \\ is redefined to be , . This can be seen when viewing the definition of \@linename, used to print the resume title under the line option: \def\@linename{\begingroup \def\\{, } {\namefont\@name} \vskip 2pt \fullline \vskip 2pt % where do you live? \@ifundefined{@addressone}{% % do nothing }{% ...

3

Placing the content of the address in a \parbox{<width>}{<content>} can be a quick hack. Something like the following may help: \address{\parbox{0.4\linewidth}{% 1985 Storm Lane, Troy, NY 12180\\ % 1985 Storm Lane, Troy, NY 12180} }

5

You don't need to add any \@ifstar to your definitions, because \@startsection already takes care of that. However, some package you are using redefines \@ifstar. For some reason this redefinition uses \new@ifnextchar rather than the standard \@ifnextchar. The old definition would ignore spaces, but this new one does not. You have introduces a space between ...

1

Note how all the default sectional units define their fonts as switches (from article.cls: \newcommand\section{\@startsection {section}{1}{\z@}% {-3.5ex \@plus -1ex \@minus -.2ex}% {2.3ex \@plus.2ex}% {\normalfont\Large\bfseries}} ...

1

The sbl-latex repo on github is owned by Robert Fisher. It uses a .bst file to compile the bibliography. it is very sophisticated and outputs really nice documents. I don't think it is still under active development. https://github.com/RobertFischer/sbl-latex However, when I wrote my thesis, I found a few things about it that I couldn't change easily, so I ...

1

The editor (TeXMaker) is irrelevant to the question. In MiKTeX settings, if you chose "Install missing packages on the fly", it means that MiKTeX will do automatic installation for any file requested during compilation that is in its file name database. More clearly, if you're running (pdf|xe|lua)(la)tex etc., and the engine tries to load a file that is ...

1

Go to the link: Latex styles for the Society of Biblical Literature Handbook of Style (updated class and style files) Click the Download Zip icon on the lower right corner as shown in the image below: This will ensure that you have all of the .cls files. Open the sbl-latex-master.zip file and extract all. Search in the folder: ...

0

I googled it by looking for the actual name of the class rcfsblpaper.cls This seems to be the github location now

2

Try the following: Open Lyx Click on Tools Menu/Reconfigure Lyx will be unresponsive until reconfigure completes restart Lyx

Top 50 recent answers are included


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# Exam P Practice Problem 22 – Poisson Number of Claims

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Problem 22A
Ten percent of the policyholders for an auto insurance company are considered high risk and ninety percent of its policyholders are considered low risk. The number of claims made by a policyholder in a calendar year follows a Poisson distribution with mean $\lambda$.

For high risk policyholders, $\lambda=0.9$. For low risk policyholders, $\lambda=0.1$. An actuary selects one policyholder at random. What is the expected number of claims made by this policyholder in the next calendar year?

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Problem 22B
Use the same information as in Problem 22A. What is the variance of the number of claims made by this randomly selected policyholder?

___________________________________________________________________________

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___________________________________________________________________________

$\copyright \ 2013$


text:
Scalable Bayesian Deep Learning with Modern Laplace Approximations

The Laplace approximation is a particular case of variational inference where one aims to approximate the posterior with a Gaussian. More precisely, a Laplace approximation uses a MAP estimate for the mean of the Gaussian posterior and derives the covariance matrix from a second-order expansion of the loss function around the MAP estimate. This method usually provides posterior approximations superior to a mean field Gaussian approximation, which is known to systematically underestimate the variance. However, the Laplace approximation is challenging to apply in the setting of Bayesian deep learning because estimating the full covariance matrix scales quadratically in the number of parameters.

[Dax22L] combines Laplace approximations with sub-network selection guided by Wasserstein distance. For a small sub-network they are able to improve the uncertainty estimates using the Laplace approximation and still maintain essentially the same accuracy as the original network.

It should be mentioned that there are still a few computational challenges open that currently have to be tackled with heuristics, most prominently the sub-network selection.

There is also a good talk by one of the authors.

# References

• Laplace Redux - Effortless Bayesian Deep Learning, Erik Daxberger, Agustinus Kristiadi, Alexander Immer, Runa Eschenhagen, Matthias Bauer, Philipp Hennig. (2022)


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# Accelerations (kinematic analysis)

1. May 17, 2009

### disclaimer

Hi there;

I guess this must be really easy and obvious but I just can't seem to be able to figure it out right now. I need to find (grapho-analytically) the acceleration of B (which is obviously the same as of C) in the position show in the picture. $$v_1=2 m/s$$ is the input velocity.

Data:
$$|AB|=0.068 [m], |BC|=0.06 [m], |CS_3 |=0.088 [m], |S_3 D|=0.088 [m], v_1=2 [m/s]$$

I calculated all the velocities:

$$v_{B1}=2 [m/s], v_{B2}=3.111 [m/s], v_{B2/B1}=2.384 [m/s], v_C=3.111 [m/s], v_{S3}=1.555 [m/s],$$

$$\omega_3=\frac{v_C}{|CD|}=\frac{3.111}{0.176}=17.676 [1/s]$$

As for the accelerations, certainly the acceleration of A is zero ($$v_1=const$$). Also, the normal component of the acceleration of C:

$$a_C^n=\omega_3^2\cdot{|CD|}=54.989 [m/s^2]$$

And here I am a bit lost. It looks like slider#2 is accelerating with respect to slideway#1. Any thoughts? I analyzed the structure in SAM, and the absolute acceleration of B (and C) is $$85.574 m/s^2$$. What other accelerations should I consider? Thanks in advance for taking your time.

2. May 18, 2009

### disclaimer

Okay, I already figured it out....


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# paley_graph#

paley_graph(p, create_using=None)[source]#

Returns the Paley $$\frac{(p-1)}{2}$$ -regular graph on $$p$$ nodes.

The returned graph is a graph on $$\mathbb{Z}/p\mathbb{Z}$$ with edges between $$x$$ and $$y$$ if and only if $$x-y$$ is a nonzero square in $$\mathbb{Z}/p\mathbb{Z}$$.

If $$p \equiv 1 \pmod 4$$, $$-1$$ is a square in $$\mathbb{Z}/p\mathbb{Z}$$ and therefore $$x-y$$ is a square if and only if $$y-x$$ is also a square, i.e the edges in the Paley graph are symmetric.

If $$p \equiv 3 \pmod 4$$, $$-1$$ is not a square in $$\mathbb{Z}/p\mathbb{Z}$$ and therefore either $$x-y$$ or $$y-x$$ is a square in $$\mathbb{Z}/p\mathbb{Z}$$ but not both.

Note that a more general definition of Paley graphs extends this construction to graphs over $$q=p^n$$ vertices, by using the finite field $$F_q$$ instead of $$\mathbb{Z}/p\mathbb{Z}$$. This construction requires to compute squares in general finite fields and is not what is implemented here (i.e paley_graph(25) does not return the true Paley graph associated with $$5^2$$).

Parameters:
pint, an odd prime number.
create_usingNetworkX graph constructor, optional (default=nx.Graph)

Graph type to create. If graph instance, then cleared before populated.

Returns:
Ggraph

The constructed directed graph.

Raises:
NetworkXError

If the graph is a multigraph.

References

Chapter 13 in B. Bollobas, Random Graphs. Second edition. Cambridge Studies in Advanced Mathematics, 73. Cambridge University Press, Cambridge (2001).