text:
# Intuition for groups

This is quite a non-standard question, certainly for mathematics, though I believe it is no less important (for me and my peers, i.e. grads).

The course I am reading so far introduced us to Groups, rings, fields, etc. in the first year, progressing to characters, reps etc in the latter years.

However, even with representation theory I still don't feel like I have a good intuition.

How do you gain intuition in such an area. I know there will be a response of doing examples, but this doesn't really help - I've done lots already.

Am I doomed.

• Could you clarify, are you having trouble gaining an intuitive feeling for groups, rings, fields, etc., just representation theory, or both? Mar 10, 2014 at 1:15
• Hello Callus, I would have to say both. The more I learn, i.e. representation theory, the less I seem to have a grip of what the view is. I can happily follow a proof, remember and understand results, answer questions and so forth, but don't really have a coherent connection between them, or any kind of easy way to see that this result is obviously really important, or this one is patently false. Apologies, I can't be more specific. Mar 10, 2014 at 1:31

I'll take a stab at this. First off, intuition in mathematics is invaluable. That being said, it is often difficult to gain a level of comfort with abstract material on an intuitive level. However, any example/picture/ect. that can give some insight is wonderful. When dealing with groups, the first thing that comes to mind for myself is symmetry. A group structure indicates a degree of symmetry in an underlying object. I am quite certain you have seen groups being described as the symmetries of regular polyhedra ect. As for representations, they are in fact meant to make such symmetries more concrete. Instead of thinking of group elements as abstract symbols, we can obtain a set of matrices which behave the same way. Representations can actually be quite intuitive in some "nice" examples. For instance, a rotation group of a geometric figure can be visualized as matrices that take the figure to itself in the plane. However, when dealing with Hilbert spaces and vector spaces over arbitrary fields, this picture becomes less intuitive. I wish I had more insight as to visualizing such concepts (this lies outside of my typical area) but from experience with other abstract concepts such as higher dimensional manifolds, the best way to gain intuition is working with these things. You don't need to be able to draw what it "looks like" per se, all you need is to be able to draw a picture that means something to you. If a concept has meaning to you then its worth something (provided its accurate of course). An excellent series of videos on Youtube from a conference celebrating the proof of the Poincare conjecture has a lecture by Bill Thurston. It might be worth looking into (even though it doesn't mention representations) because Thurston was a true believer in intuition. Hope this helps a little.

• +1. The fact that group theory is the study of symmetry (to be more precise: invariance under transformation) is crucial for an intuitive understanding, yet almost never receives emphasis in group theory courses, in my experience at least. I like to think of the whole of group theory etc. as one single, very elaborate answer to the single question of "What possible ways are there for something to be symmetric?" Mar 10, 2014 at 1:34

A group is a set $$G$$ with an operation $$\cdot$$ that combines the elements of $$G$$ to give an element that also belongs to $$G$$. In order for $$(G,\cdot)$$ to be a group it must satisfy the following axioms: closure, associativity, identity element, inverse element.

For me the intuitive way to think about groups is to consider the Rubik's Cube group. This group consists of all possible states the cube can be in. All of the states can be reached by performing one of the face rotations: $${F, B, U, D, L, R}$$ (Front, Back, Upper, Down, Left, Right) and these are called generators of the group. The operation $$\cdot$$ is the composition of those rotations. The closure of this group means that after performing a sequence of rotations, we end up with one of the possible Cube's permutations. Associativity is intuitive - if we perform moves $$(F \cdot U)\cdot R$$ we would end up in the same state as if we performed $$F \cdot (U \cdot R)$$. For each move there exists an opposite move (with the same face but in the opposite direction, e.g. $$F$$ and $$F'$$) that brings you back to the initial state (inverse element) and for each state of the cube there exists an identity element $$e$$ that means not moving the cube at all. Now you can abstract away the whole idea of cube and perform mathematical operations on this set using composition $$\cdot$$ and you know that this could be translated back to the physical orientation of the cube.

Sometimes different unrelated structures (atoms, particles) express some symmetry and therefore could be viewed as a subgroup of symmetry groups, e.g the Rubik's Cube group is the subgroup of the symmetric group $$S_{48}$$. This allows us to use common language to solve seemingly unrelated problems.

To be very intuitive, representation theory makes abstract algebraic objects "concrete" by describing its elements by matrices and the algebraic operations in terms of matrix addition and matrix multiplication respectively. Roughly, it makes abstract algebra into linear algebra, which is a concrete subject.


text:
Book:Journal/Journal of Research of the National Bureau of Standards

Journal of Research of the National Bureau of Standards

(J. Res. Natl. Bur. Stand.)

Publisher: National Institute of Standards and Technology

Dates

Started publication: $1934$
(current)

Featured Articles

• 1961: John ToddComputational problems concerning the Hilbert matrixVol. 65: pp. 19 – 22)
• 1969: Edward W. Ng and Murray GellerA table of integrals of the Error functionsVol. 73B: pp. 1 – 20)

format: plain


text:
# Math Help - Proofs of Invertibility

1. ## Proofs of Invertibility

I'm supposed to prove an invertible, but I'm not sure how, as the prof didn't talk about it in class, and the book is rather vague...

The question is this:

Let u and v be column vectors in R^n, and let A= I + u(v^T). Show that if (u^T)v =/= -1, then A is invertible and
A^-1 = I - (1/(1+((u^T)v)) (u(v^T))

Sorry about the last equation being so confusing, but that's it. I'm not sure how to prove this, since I don't even know how it works. I'm terrible at this stupid inverse matrix stuff...

2. Originally Posted by Hellreaver

Let u and v be column vectors in R^n, and let A= I + u(v^T). Show that if (u^T)v =/= -1, then A is invertible and A^-1 = I - (1/(1+((u^T)v)) (u(v^T))
this is a nice problem! first we need a little point:

$\boxed{1}$ if $u,v$ are two vectors in $\mathbb{R}^n,$ then $uv^Tuv^T=uv^Tu^Tv.$

Proof. note that $u^Tv$ and $v^Tu$ are both scalars and obviously $u^Tv=v^Tu.$ also a scalar commutes with a vector. thus: $uv^Tu v^T=uu^Tvv^T=uv^Tu^Tv. \ \ \ \Box$

back to your problem: we have $A^2=(I+uv^T)^2=I + 2uv^T+uv^Tuv^T=I+2uv^T+uv^Tu^Tv. \ \ \ \ \ \ \ \text{by} \ \boxed{1}$

hence: $A^2=I+uv^T + uv^T + uv^Tu^Tv=A+uv^T(1+u^Tv)=A+(A-I)(1+u^Tv),$ which gives us: $A(A-I-(1+u^Tv)I)=-(1+u^Tv)I. \ \ \ \ \ \ (*)$

now $1+u^Tv \neq 0.$ so we can divide both sides of $(*)$ by $-(1+u^Tv).$ also we have $A-I=uv^T.$ thus $(*)$ gives us: $A \left(I - \frac{1}{1+u^Tv}uv^T \right)=I.$

3. Friggen complicated enough?
I so don't understand this stuff. I was pretty good at trig proofs, but that was back when I actually had a teacher. This linear algebra stuff just doesn't make sense...

4. Originally Posted by Hellreaver
Friggen complicated enough?
I so don't understand this stuff. I was pretty good at trig proofs, but that was back when I actually had a teacher. This linear algebra stuff just doesn't make sense...
his response is very easy to follow, grab a pen and paper and write it carefully down. You should be able to understand it.

5. I understand his response perfectly enough (thanks, NonCommAlg). I just don't understand the majority of the concept behind it.


text:
# Integers. Integer Storage Since Binary consists only of 0s and 1s, we cant use a negative sign ( - ) for integers. Instead, the Most Significant Bit is.

## Presentation on theme: "Integers. Integer Storage Since Binary consists only of 0s and 1s, we cant use a negative sign ( - ) for integers. Instead, the Most Significant Bit is."— Presentation transcript:

Integers

Integer Storage Since Binary consists only of 0s and 1s, we cant use a negative sign ( - ) for integers. Instead, the Most Significant Bit is used to represent the sign. This way, half the combinations in a fixed length of bits can be used to represent negative values. But which value of the sign bit (0 or 1) will represent a negative number?

Integers 2s Complement Notation

2s Complement Notation (examples in 8 bits to save space) Fixed length notation system. Uses 1 to represent negative values. The largest non-negative value: 01111111 The smallest non-negative value: 00000000 The largest negative value is: 11111111 The smallest negative value is: 10000000

2s Complement Notation The representations of non-negative integers in 2s Complement look the same as they do for Natural numbers. However, negative values look VERY different than we might expect.

2s Complement Notation Complementary numbers sum to 0. Decimal is a Signed Magnitude system so complements have the same magnitude but different signs: 5 and -5, for example. 2s Complement is a Fixed Length system. There are no signs, so to find a numbers complement, another technique is needed.

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1.

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5:

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5: Represent +5 in fixed length

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5: Represent +5 in fixed length00000101

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5: Represent +5 in fixed length00000101 flip the bits ( 1 0, 0 1 )

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5: Represent +5 in fixed length00000101 flip the bits ( 1 0, 0 1 ) 11111010

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5: Represent +5 in fixed length00000101 flip the bits ( 1 0, 0 1 ) 11111010 add 1 to the new pattern+1

2s Complement Notation One such technique is to simply change each bit to its opposite, and then add 1. To find the 2s complement notation for -5: Represent +5 in fixed length00000101 flip the bits ( 1 0, 0 1 ) 11111010 add 1 to the new pattern+1 to produce -511111011

2s Complement Notation Complementary numbers sum to 0.

2s Complement Notation Complementary numbers sum to 0. So if to +5 00000101

2s Complement Notation Complementary numbers sum to 0. So if to +5 we add -5 00000101 +11111011

2s Complement Notation Complementary numbers sum to 0. So if to +5 we add -5 we should get 00000101 +11111011 1 00000000 discard the carry bit to retain the fixed length

Download ppt "Integers. Integer Storage Since Binary consists only of 0s and 1s, we cant use a negative sign ( - ) for integers. Instead, the Most Significant Bit is."

Similar presentations


text:
# Showing $\lim_{x \rightarrow \infty}\int_{4}^{x}\frac{1}{x\log(x)\log \log(x)}$ converges or divegences

In the text "Elementary Theory of Calculus" I'm having trouble applying the integral test to show that the series in $(1.)$ converges.

$(1.)$

$$\sum_{}^{}\frac{1}{n\log(n)\log \log(n)}$$

My attack on $(1.)$ can be seen within $(2.)$ yielded the following observations:

$(2.)$ $$\sum_{}^{}\frac{1}{n\log(n)\log \log(n)} <=> \int_{4}^{\infty}\frac{1}{x\log(x)\log \log(x)}$$

$$\lim_{x \rightarrow \infty}\int_{4}^{x}\frac{1}{x\log(x)\log \log(x)} <=> \int_{4}^{x}(\lim_{x \rightarrow \infty}\frac{1}{x\log(x)\log \log(x)})$$

On the RHS side $(2.)$ is appropriate to pass the limit inside the integral or is their a better way to approach proving the integral converges or diverges ?

• You have several very deep errors here: it's not true that the sum and integral are equal to each other (but it is true that they'll converge or diverge together - try and understand why!), and your attempt to interchange limit and order-of-integration makes no sense at all (and above and beyond that, you really shouldn't use the same variable name for your (formal) integration variable and the limit of the integral, because it leads to exactly this sort of confusion.) – Steven Stadnicki Jun 5 '17 at 19:29
• Look for how to compute improper integrals. Limit cannot be simply taken inside. – Anurag A Jun 5 '17 at 19:30
• i think your integral doesn't converge on the given interval – Dr. Sonnhard Graubner Jun 5 '17 at 19:30
• Thanks for the notification The integral of our given function whether it converges or diverges would apply convergence or divergence for our series it was my mistake to think that the sum and integral were equivalent to each other. – Zophikel Jun 5 '17 at 19:35
• Let $x=e^{e^u}$ and compare to $\int_1^\infty\frac1x~\mathrm dx$. – Simply Beautiful Art Jun 6 '17 at 1:26

We have $$\frac{d}{dx} \log\log\log{x} = \frac{1}{x\log{x}\log{\log{x}}}.$$ It should be fairly clear what happens to this antiderivative for large $x$.
And no, you certainly can't pass the limit inside the integral: $$\int_4^X \frac{dx}{x\log{x}\log{\log{x}}}$$ is a function of one variable: $X$. This is but one reason that it's so important to use different letters for the integration variables and the limits. See also this post.
There seem to be a few misunderstandings here. First, the integral test tells you that for a decreasing function $f(x)$ such that $\lim_{x\to\infty} f(x) = 0$, we have that $$\int^\infty_k f(x) dx \,\,\,\,\,\, \text{ and } \,\,\,\,\,\, \sum^\infty_{n=k} f(n)$$ either both converge or both diverge. It does not say that the two are equal. Secondly, $$\lim_{x\to \infty} \int^x_{4} f(x) dx \neq \int^x_4 \left[\lim_{x\to \infty} f(x) \right]dx.$$ Instead, you need to find an anti-derivative $F$ of $f$ and then $$\lim_{x\to\infty}\int^x_{4} f(x) dx = \lim_{x\to\infty} (F(x) - F(4));$$ i.e., you need to take the anti-derivative before considering the limit.
In your case, note that you can make the substitution $u = \log\log x$ and see that \begin{align*} \int_4^\infty \frac{dx}{x \log x \log \log x} &= \int^\infty_{\log\log4} \frac{du}{u} \\&= \log(u)\bigg|^\infty_{\log\log 4} \\&= \log\log\log x\bigg|^\infty_{4}\\&= \lim_{t\to\infty} \log\log\log(t) - \log\log\log(4) = \infty.\end{align*} Thus the integral diverges and so does the sum.


text:
# mages' blog

We released googleVis version 0.5.8 on CRAN last week. The update is a maintenance release for the forthcoming release of R 3.2.0.

 Screen shot of some of the Google Charts
New to googleVis? The package provides an interface between R and the Google Charts Tools, allowing you to create interactive web charts from R without uploading your data to Google. The charts are displayed by default via the R internal help browser.

Visit the examples of all googleVis charts on CRAN and review the vignettes.

## Communicating Risk and Uncertainty

David Spiegelhalter gave a fascinating talk on Communicating Risk and Uncertainty to the Public & Policymakers at the Grantham Institute of the Imperial College in London last Tuesday.

In a very engaging way David gave many examples and anecdotes from his career in academia and advisory. I believe his talk will be published on the Grantham Institute's YouTuble channel, so I will only share a few highlights and thoughts that stuck in my mind here.

### Framing: Opportunity vs. Risk

Do you prefer survival ratios or mortality rates? The way you present probabilities/frequencies will make a difference in the way the audience perceives them and therefore how they will make decisions. David suggested one should aim to give a balanced view to mitigate the framing effect; e.g. show the opportunity and risk, the survival ratio and mortality rate. Interestingly, that is exactly how Wolfram Alpha displays life expectancies.

 Life expectance: UK, male, 62 years old (Source: Wolfram Alpha)

### Absolute and relative risk

What did you have for breakfast this morning? Fried up bacon? Perhaps you have fried up bacon every day? It could increase your chances of developing pancreatic cancer by almost a fifth, as an article by the Daily Express on 16th January 2012 suggested.

 Daily Express, 16th January 2012

Well, this might be true, but fortunately pancreatic cancer is quite rare. About 5 people out of 400 will develop this cancer even without eating fried up bacon every day (absolute risk). Eating fry ups every morning will increase your chances by 20% (relative risk) to 6 out of 400 - still pretty low. Less of a scary headline.

Small increases in absolute risks can result in large increases in relative risk. Depending on how you choose to frame the story you can generate very different reactions. David mentioned also the work of Gerd Gigerenzer, who is promoting the use of natural frequencies in the context of medical conversations, e.g. say '1 out of 200 people will experience a certain effect', instead of 'the probability for this effect is 0.5%'.

### Conditional probabilities

One of the big challenges in communicating risk are conditional probabilities. I believe, conditional probabilities are not intuitive for most of us. As an example David talked about breast cancer screening. The test itself is not a 100% reliable. No test is. There are always results which are false positives (test predicts cancer when there is none) and false negatives (test predicts no cancer when there is one). Tests are usually designed to have a higher false alarm rate than no alarm at all. Hence, even when the test result is positive the likelihood of cancer could be low, because cancer is rare and the test is not that good. David presented a great diagram of the NHS breast screening - Helping you decide leaflet to illustrate the conditional probabilities for breast screening.

 NHS breast screening - Helping you decide
I think this is a wonderful example of trying to give a balanced view of the underlying risk, communicated in a very clear and understandable way.

### Translating probabilities into words

The use of natural frequencies is not always appropriate and sometimes probabilities are more useful. However, instead of saying that something has a 50% chance of occurring, you might also say it is equally likely to happen than not. David mentioned the work of the Intergovernmental Panel on Climate Change (IPCC) that aims to use a consistent language in all their reports. To achieve that, they publish guidance on consistent treatments of uncertainty. This is their translation of probabilities into words:

 Source: IPCC Uncertainty guidance note
I think this table will become handy in the future when I have to elicit expert opinions.

### Speaking with confidence

The IPCC uncertainty guidance also talks about confidence and interestingly they correlate the level of agreement with the level of evidence. Note, even with a lot of evidence, without agreement the confidence is relatively low.

 Source: IPCC Uncertainty guidance note
I wonder, what low agreement with robust evidence could mean. Perhaps, even if there is strong evidence that I should change my diet, e.g. drink less coffee, I might not agree with it and as a result wouldn't take any actions.

### Fan charts

The last example from David's talk I will mention here is the Bank of England's fan chart of GDP projection. The Bank of England has been using this chart now for a number of years and tweaked it only a little over that time. Until the financial crisis the Bank of England showed bands of confidence within the 5-95% interval. The financial crisis happened outside that interval, which of course doesn't mean that it couldn't happen, it was just very unlikely. Since then they illustrate the 0-100% confidence interval as a grey background. The other interesting aspect is, that they don't show the mean projection line initially to overcome the anchoring effect. Note also that the term projection is used and not prediction.

 November 2014 GDP Fan chart (Source: Bank of England)

Finally, David announced that he has a new book in the pipeline. I am sure it will sell well, as it has the catchy title Sex by numbers.

## Extended Kalman filter example in R

Last week's post about the Kalman filter focused on the derivation of the algorithm. Today I will continue with the extended Kalman filter (EKF) that can deal also with nonlinearities. According to Wikipedia the EKF has been considered the de facto standard in the theory of nonlinear state estimation, navigation systems and GPS.

### Kalman filter

I had the following dynamic linear model for the Kalman filter last week:
\begin{align} x_{t+1} & = A x_t + w_t,\quad w_t \sim N(0,Q)\\ y_t &=G x_t + \nu_t, \quad \nu_t \sim N(0,R)\\ x_1 & \sim N(\hat{x}_0, \Sigma_0) \end{align}With $$x_t$$ describing the state space evolution, $$y_t$$ the observations, $$A, Q, G, R, \Sigma_0$$ matrices of appropriate dimensions, $$w_t$$ the evolution error and $$\nu_t$$ the observation error. The Kalman filter provides recursive estimators for $$x_t$$ via:
\begin{align} K_{t} &= A \Sigma_t G' (G \Sigma_t G' + R)^{-1}\\ \hat{x}_{t+1} &= A \hat{x_t} + K_{t} (y_t - G \hat{x}_t) \\ \Sigma_{t+1} &= A \Sigma_t A' - K_{t} G \Sigma_t A' + Q \end{align}In the case of nonlinearities on the right hand side of either the state ($$x_t$$) or observation ($$y_t$$) equation the extended Kalman filter uses a simple and elegant trick: Taylor series of the first order, or in other words, I simply linearise the right hand side. The matrices $$A$$ and $$G$$ will be the Jacobian matrices of the respected vector functions.

### Logistic growth

As an example I will use a logistic growth model, inspired by the Hakell example given by Dominic Steinitz.

The logistic growth model can be written as a time-invariant dynamical system with growth rate $$r$$ and carrying capacity $$k$$:
\begin{aligned} \dot{p} & = r p\Big(1 - \frac{p}{k}\Big) \end{aligned}The above ordinary differential equation has the well known analytical solution:
$p = \frac{kp_0\exp(r\,t)}{k + p_0(\exp(r\,t) - 1)}$Suppose I observe data of a population for which I know the carrying capacity $$k$$, but where the growth rate $$r$$ is noisy.

### Extended Kalman filter

The state space and observation model can then be written as:
\begin{aligned} r_i &= r_{i-1} \\ p_i &= \frac{kp_{i-1}\exp(r_{i-1}\Delta T)}{k + p_{i-1}(\exp(r_{i-1}\Delta T) - 1)} \\ y_i &= \begin{bmatrix}0 & 1\end{bmatrix} \begin{bmatrix}r_i \\ p_i\end{bmatrix} + \nu \end{aligned}Or with $$x_i:=\begin{bmatrix}r_i & p_i\end{bmatrix}'$$ as:
\begin{aligned} x_i &= a(x_i)\\ y_i &= G x_i + \nu_i, \quad \nu_i \sim N(0,R) \end{aligned}In my example the state space model is purely deterministic, so there isn't any evolution noise and hence $$Q=0$$.

With an initial prior guess for $$x_0$$ and $$\Sigma_0$$ and I am ready to go. The R code below shows my implementation with the algorithm above. Note that I use the jacobian function of the numDeriv package.
After a few time steps the extended Kalman filter does a fantastic job in reducing the noise. Perhaps this shouldn't be too surprising as a local linearisation of the logistic growth function will give a good fit. The situation might be different for highly nonlinear functions. The Wikipedia article about the Kalman filter suggests the unscented version in those cases.

### Session Info

R version 3.1.2 (2014-10-31)
Platform: x86_64-apple-darwin13.4.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets methods base

other attached packages:
[1] numDeriv_2012.9-1

loaded via a namespace (and not attached):
[1] tools_3.1.2

## Kalman filter example visualised with R

At the last Cologne R user meeting Holger Zien gave a great introduction to dynamic linear models (dlm). One special case of a dlm is the Kalman filter, which I will discuss in this post in more detail. I kind of used it earlier when I measured the temperature with my Arduino at home.

Over the last week I came across the wonderful quantitative economic modelling site quant-econ.net, designed and written by Thomas J. Sargent and John Stachurski. The site not only provides access to their lecture notes, including the Kalman fitler, but also code in Python and Julia.

I will take their example of the Kalman filter and go through it with R. I particularly liked their visuals of the various steps of the Kalman filter.

### Motivation

Suppose I have a little robot that moves autonomously over my desk. How would the robot know where it is? Well, suppose the robot can calculate from its wheels' movements how far it went and an additional sensor (e.g. GPS-like) provides also information about its location. Of course both pieces of information will be imprecise. How can the various signals be combined?

This is a classic scenario for the Kalman filter. Its key assumptions are that the errors/noise are Gaussian and that the state space evolution $$x_t$$ from one time step to the next is linear, so is the mapping to the sensor signals $$y_t$$. For the example I will use it reads:
\begin{align} x_{t+1} & = A x_t + w,\quad w \sim N(0,Q)\\ y_t &=G x_t + \nu, \quad \nu \sim N(0,R)\\ x_1 & \sim N(\hat{x}_0, \Sigma_0) \end{align}With $$A, Q, G, R, \Sigma$$ matrices of appropriate dimensions. The Kalman filter provides recursive estimators for $$x_t$$:
\begin{align} K_{t} &= A \Sigma_t G' (G \Sigma_t G' + R)^{-1}\\ \hat{x}_{t+1} &= A \hat{x_t} + K_{t} (y_t - G \hat{x}_t) \\ \Sigma_{t+1} &= A \Sigma_t A' - K_{t} G \Sigma_t A' + Q \end{align}

Let's say at time $$t_0$$ the robot has the expected position $$\hat{x} = (0.2, -0.2)$$. That means, the robot doesn't know its location on the table with certainty. Further I shall assume that the probability density function of the position follows a Normal distribution with covariance matrix $$\Sigma = \left[\begin{array}{cc} 0.4 & 0.3\\ 0.3 & 0.45 \end{array} \right]$$
The prior distribution of the robot's position can be visualised in R with a contour plot.

library(mnormt)
xhat <- c(0.2, -0.2)
Sigma <- matrix(c(0.4, 0.3,
0.3, 0.45), ncol=2)
x1 <- seq(-2, 4,length=151)
x2 <- seq(-4, 2,length=151)
f <- function(x1,x2, mean=xhat, varcov=Sigma)
dmnorm(cbind(x1,x2), mean,varcov)
z <- outer(x1,x2, f)
mycols <- topo.colors(100,0.5)
image(x1,x2,z, col=mycols, main="Prior density",
xlab=expression('x'[1]), ylab=expression('x'[2]))
points(0.2, -0.2, pch=19)
text(0.1, -0.2, labels = expression(hat(x)), adj = 1)

### Sensor information

The 'GPS' sensor provides also data about the location of the robot, again the sensor signal can be regarded as the expected position with some noise. Suppose the reading of the sensor is $$y=(2.3, -1.9)$$. I will assume that the sensor signal has an error following a Normal distribution, with a covariance matrix $$R=0.5 \Sigma$$.

Conceptually I think about it like this, given that the robot is at position $$x$$, what would be the likelihood that the sensor reading is $$y\,|\,x\,$$? The Kalman filter assumes a linear mapping from $$x$$ to $$y$$:
$y = G x + \nu \mbox{ with } \nu \sim N(0,R)$In my case $$G$$ will be the identity matrix. Let's add the sensor information to the plot.

R <- 0.5 * Sigma
z2 <- outer(x1,x2, f, mean=c(2.3, -1.9), varcov=R)
image(x1, x2, z2, col=mycols, main="Sensor density")
points(2.3, -1.9, pch=19)
text(2.2, -1.9, labels = "y", adj = 1)
points(0.2, -0.2, pch=19)
text(0.1, -0.2, labels = expression(hat(x)), adj = 1)

### Filtering

I have two pieces of information about the location of the robot, both with a Normal distribution, my prior is $$x \sim N(\hat{x},\Sigma)$$ and for the distribution of my sensor I have $$y\,|\,x \sim N(Gx, R)$$.

What I would like to know is the location of the robot given the sensor reading:
$p(x \,|\, y) = \frac{p(y \,|\, x) \, p(x)} {p(y)}$Now, because $$x, y$$ are Normal and $$G$$ is the identity matrix, I know that the posterior distribution of $$x \,|\, y$$ is Normal again, with parameters (see conjugate prior):
$\hat{x}_f = (\Sigma^{-1} + R^{-1})^{-1} (\Sigma^{-1} \hat{x} + R^{-1} y) \\ \Sigma_f = (\Sigma^{-1} + R^{-1})^{-1}$Using the matrix inversion identity $(A^{-1} + B^{-1})^{-1} = A - A (A + B)^{-1}A = A (A + B)^{-1} B$ I can write the above as:
\begin{align} \hat{x}_f & = (\Sigma - \Sigma (\Sigma + R)^{-1}\Sigma)(\Sigma^{-1} \hat{x} + R^{-1}y)\\ & =\hat{x} - \Sigma (\Sigma + R)^{-1} \hat{x} + \Sigma R^{-1} y -\Sigma (\Sigma + R)^{-1}\Sigma R^{-1}y\\ & = \hat{x} + \Sigma (\Sigma + R)^{-1} (y - \hat{x})\\ & = (1.667, -1.333)\\ \Sigma_f &= \Sigma - \Sigma (\Sigma + R)^{-1} \Sigma\\ &=\left[\begin{array}{lll} 0.133 & 0.10\\ 0.100 & 0.15 \end{array} \right] \end{align}In the more general case when $$G$$ is not the identity matrix I have
\begin{align} \hat{x}_f & = \hat{x} + \Sigma G' (G \Sigma G' + R)^{-1} (y - G \hat{x})\\ \Sigma_f &= \Sigma - \Sigma G' (G \Sigma G' + R)^{-1} G \Sigma \end{align}The updated posterior probability density function $$p(x\,|\,y)=N(\hat{x}_f, \Sigma_f)$$ is visualised in the chart below.
I can see how the prior and likelihood distributions have influenced the posterior distribution of the robot's location. The sensor information is regarded more credible than the prior information and hence the posterior is closer to the sensor data.

G = diag(2)
y <- c(2.4, -1.9)
xhatf <- xhat + Sigma %*% t(G) %*% solve(G %*% Sigma %*% t(G) + R) %*% (y - G %*% xhat)
Sigmaf <- Sigma - Sigma %*% t(G) %*% solve(G %*% Sigma %*% t(G) + R) %*% G %*% Sigma
z3 <- outer(x1, x2, f, mean=c(xhatf), varcov=Sigmaf)
image(x1, x2, z3, col=mycols,
xlab=expression('x'[1]), ylab=expression('x'[2]),
main="Filtered density")
points(xhatf[1], xhatf[2], pch=19)
text(xhatf[1]-0.1, xhatf[2],
labels = expression(hat(x)[f]), adj = 1)
points(0.2, -0.2, pch=19, col=lb)
text(0.1, -0.2, labels = expression(hat(x)), adj = 1, col=lb)
points(2.3, -1.9, pch=19, col=lb)
text(2.2, -1.9,labels = "y", adj = 1, col=lb)

### Forecasting

Ok, I have a better understanding of the robot's position at time $$t_0$$, but the robot is moving and what I'd like to know is where it will be after one time step.

Suppose I have a linear model that explains how the state evolves over time, e.g. based on wheel spin. Again, I will assume a Gaussian process. In this toy example I follow the assumptions of Sargent and Stachurski, although this doesn't make much sense for the robot:
$x_{t+1} = A x_t + w_{t+1} \mbox{, where } w_t \sim N(0, Q)$with$\begin{split}A = \left( \begin{array}{cc} 1.2 & 0.0 \\ 0.0 & -0.2 \end{array} \right), \qquad Q = 0.3 \Sigma\end{split}$As all the processes are linear and Normal the calculations are straightforward:
\begin{align} \mathbb{E} [A x_f + w] &= A \mathbb{E} [x_f] + \mathbb{E} [w]\\ &= A \hat{x}_f\\ &= A \hat{x} + A \Sigma G' (G \Sigma G' + R)^{-1}(y - G \hat x) \end{align}\begin{align} \operatorname{Var} [A x_f + w] &= A \operatorname{Var}[x_f] A' + Q\\ &= A \Sigma_f A' + Q\\ &= A \Sigma A' - A \Sigma G' (G \Sigma G' + R)^{-1} G \Sigma A' + Q \end{align}The matrix $$A \Sigma G' (G \Sigma G' + R)^{-1}$$ is often written as $$K_\Sigma$$ and called the Kalman gain. Using this notation, I can summarise the results as follows:\begin{split}\begin{align} \hat{x}_{new} &:= A \hat{x} + K_{\Sigma} (y - G \hat{x}) \\ \Sigma_{new} &:= A \Sigma A' - K_{\Sigma} G \Sigma A' + Q \nonumber \end{align}\end{split}Adding the prediction to my chart gives me:

A <- matrix(c(1.2, 0,
0, -0.2), ncol=2)
Q <- 0.3 * Sigma
K <- A %*% Sigma %*% t(G) %*% solve(G%*% Sigma %*% t(G) + R)
xhatnew <- A %*% xhat + K %*% (y - G %*% xhat)
Sigmanew <- A %*% Sigma %*% t(A) - K %*% G %*% Sigma %*% t(A) + Q
z4 <- outer(x1,x2, f, mean=c(xhatnew), varcov=Sigmanew)
image(x1, x2, z4, col=mycols,
xlab=expression('x'[1]), ylab=expression('x'[2]),
main="Predictive density")
points(xhatnew[1], xhatnew[2], pch=19)
text(xhatnew[1]-0.1, xhatnew[2],
labels = expression(hat(x)[new]), adj = 1)
points(xhatf[1], xhatf[2], pch=19, col=lb)
text(xhatf[1]-0.1, xhatf[2], col=lb,
labels = expression(hat(x)[f]), adj = 1)
points(0.2, -0.2, pch=19, col=lb)
text(0.1, -0.2, labels = expression(hat(x)), adj = 1, col=lb)
points(2.3, -1.9, pch=19, col=lb)
text(2.2, -1.9,labels = "y", adj = 1, col=lb)
That's it, having gone through the updating process for one time step gives me the recursive Kalman filter iteration mentioned above.

Another way to visualise the four steps can be achieved with the lattice package:
library(lattice)
grid <- expand.grid(x=x1,y=x2)
grid$Prior <- as.vector(z) grid$Likelihood <- as.vector(z2)
grid$Posterior <- as.vector(z3) grid$Predictive <- as.vector(z4)
contourplot(Prior + Likelihood + Posterior + Predictive ~ x*y,
data=grid, col.regions=mycols, region=TRUE,
as.table=TRUE,
xlab=expression(x[1]),
ylab=expression(x[2]),
main="Kalman Filter",
panel=function(x,y,...){
panel.grid(h=-1, v=-1)
panel.contourplot(x,y,...)
})

You can find the code of this post on Github.

### Session Info

R version 3.1.2 (2014-10-31)
Platform: x86_64-apple-darwin13.4.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets methods base

other attached packages:
[1] lattice_0.20-29 mnormt_1.5-1

loaded via a namespace (and not attached):
[1] grid_3.1.2  tools_3.1.2


text:
💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

WZ

# Determine whether each integral is convergent or divergent. Evaluate those that are convergent.$\displaystyle \int_0^\infty \sin^2 \alpha\ d \alpha$

## $\infty$, integral is divergent

#### Topics

Integration Techniques

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

problem is determined why this integral is Cordant or the verdant. This problem we need to use the trick in my trick. Identity Bye. Off a square is equal to assign to off er minus one over to now. This integral If you got to integral from zero to infinity. Fine. Teo, Might this wand over too? Yeah, by definition of proper into girl, This is equal to the limit. He goes to infinity integral from their own tea. And Kasai Tio, huh? Minus one over two the And this is equal to Lim. He goes Tio infinity. It's anti derogative off. Assigned to Alpha as they call to sign Tio, huh? Over to and want untidy ritual. One days are past minus off. Um huh. Off from zero. Sure. This is Nico too. They make he goes to infinity. Sign twenty over four, minus he over too. On dh minus zero. Thank you. Auntie goes to infinity. Sign Tootie, That's not consist. There's this limit that's now that exist. Is the intern integral off? This function is that wouldn't

WZ

#### Topics

Integration Techniques

Lectures

Join Bootcamp


text:
## Pacific Journal of Mathematics

### Quasi-rotation $C^*$-algebras.

H. Rouhani

#### Article information

Source
Pacific J. Math., Volume 148, Number 1 (1991), 131-151.

Dates
First available in Project Euclid: 8 December 2004

https://projecteuclid.org/euclid.pjm/1102644786

Mathematical Reviews number (MathSciNet)
MR1091534

Zentralblatt MATH identifier
0744.46059

#### Citation

Rouhani, H. Quasi-rotation $C^*$-algebras. Pacific J. Math. 148 (1991), no. 1, 131--151. https://projecteuclid.org/euclid.pjm/1102644786

#### References

• [I] R. Bott and L. Tu, Differential Forms in Algebraic Topology,Graduate Texts in Mathematics No. 82, Springer Verlag, 1982.
• [2] L. Brown, P. Green, and M. Rieffel, Stable isomorphism and strong Morita equivalenceof C*-algebras, Pacific J. Math., 71 (1977), 349-363.
• [3] G. Elliott, On the K-theory of the C*-algebra generated by a projective repre- sentation of a torsion-free abelian group, Operator Algebras and GroupRepre- sentations, Vol. 1, Pitman, London (1984), 157-184.
• [4] H. Furstenberg, Strict ergodicityand transformation of the torus,Amer. J.Math., 83 No. 4(1961), 573-601.
• [5] R. Ji, On the crossedproduct C*-algebras associated with Furstenbergtrans- formations on tori, Ph.D. dissertation, State University of New York at Stony Brook, 1986.
• [6] I. Megory-Cohen, Properties of hyperbolic crossedproduct algebras,Ph.D. dis- sertation, University of California at Berkeley, 1985.
• [7] J. Packer, K-theoretic invariantsfor C*-algebras associated to transformations and induced flows, J. Funct. Anal., 67 (1986), 25-59.
• [8] W. Parry, Topics in Ergodic Theory, Cambridge University Press, 1981.
• [9] G. Pederson, C*-algebras and Their Automorphism Groups, London Math. Soc. Monographs 14, Academic Press, London-New York, 1979.
• [10] M. Pimsner, Ranges of Traces on KQ of Reduced Crossed Products by Free Groups,Lecture Notes in Mathematics, Vol. 1132, Springer-Verlag, 1985.
• [II] M. Pimsner, and D. Voiculescu, Exact sequencesfor K-groups and Ext-groups of certain crossed product C*-algebras, J. Operator Theory, 4 (1980), 93-118.
• [12] S. C. Power, Simplicity of C*-algebras of minimal dynamical systems, J. London Math. Soc, 18 (1978), 534-538.
• [13] N. Riedel, Classification of the C*-algebras associated with minimal rotations, Pacific J. Math., 101 (1982), 153-161.
• [14] M. Rieffel, C*-algebras associated with irrational rotations, Pacific J. Math., 93 (1981), 415-429.
• [15] H. Rouhani, Classification of certain non-commutative three-tori, Ph.D. disser- tation, Dalhousie University, 1988.
• [16] H. Rouhani, A Furstenberg transformation of the 2-torus without quasi-discrete spec- trum, Canad. Math. Bull., 33 (3) (1990), 316-322.
• [17] H. Rouhani, Strong Morita equivalencefor the quasi-rotation C*-algebras, preprint, 1989.
• [18] P. Walters, Ergodic Theory--Introductory Lectures, Lecture Notes in Mathe- matics 458, Springer-Verlag, 1975.
• [19] H.-S. Yin, A simple proof of the classification of rational rotation C* -algebras, Proc. Amer. Math. Soc, 98 No. 3 (1986), 469-470.


text:
Navigation timing makes it easy to measure the real-world speed and performance of websites and locate problem areas that need tuning.

For example, navigation timing can help you locate latency problems by helping you precisely monitor customer webpage navigation and track the timing of user activities.

performance.timing is generated based on all requests and responses (but not including the ajax ones).

console.log(window.performance);

Will show you the timing object:

connectEnd                 // Time when server connection is finished.
connectStart               // Time just before server connection begins.
domComplete                // Time just before document readiness completes.
domInteractive             // Time just before readiness set to interactive.
domainLookupEnd            // Time after domain name lookup.
domainLookupStart          // Time just before domain name lookup.
fetchStart                 // Time when the resource starts being fetched.


text:
Archives

Slightly $m$-continuous multifunctions
by Takashi Noiri   Valeriu Popa

Vol. 1 No. 4 (2006) P.485~P.505

ABSTRACT

We introduce a new class of multifunctions called a slightly $m$-continuous multifunction which is a generalization of both slightly continuous functions [10] and $m$-continuous multifunctions [33]. In this paper we obtain several properties of such a multifunction.

KEYWORDS
$m$-continuous, multifunction, slightly continuous, slightly $m$-continuous

MATHEMATICAL SUBJECT CLASSIFICATION 2010
Primary: 54C08, 54C60

MILESTONES