text:
# Citation

Iyengar D, Klawans MR, Stotts AL, Northrup TF (2019) Use of Text Messaging as an Intervention to Promote Weight Loss and Reduce Waist Circumference among an Overweight and Abdominally Obese Patient Population: A Randomized-Controlled Quality Improvement Study in Primary Care. J Obes Weight-Loss Medic 5:029. doi.org/10.23937/2572-4010.1510029

# Use of Text Messaging as an Intervention to Promote Weight Loss and Reduce Waist Circumference among an Overweight and Abdominally Obese Patient Population: A Randomized-Controlled Quality Improvement Study in Primary Care

##### Deepa Iyengar, MD, MPH*, Michelle R Klawans, MPH, Angela L Stotts, PhD and Thomas F Northrup, PhD

Department of Family & Community Medicine, UTHealth, USA

# Abstract

## Purpose

Obesity is a national epidemic resulting in significant morbidity and mortality. Weight loss as modest as 5% of body weight significantly reduces the risk of hypertension, diabetes, and cardiovascular disease. The primary aim of this pilot study was to test the efficacy of a text messaging intervention in a primary care setting to motivate overweight patients to lose weight and reduce their waist circumference.

## Methods

A total of 128 adult primary care patients who were overweight and classified as abdominally obese via waist circumference measurement were randomized to enhanced care (EC, n = 62) or standard care (SC, n = 66). Patients assigned to both groups met with the physician, and the SC received a standard clinic brochure on weight loss. In addition to the brochure, the EC group met with a research assistant to discuss the risks of abdominal obesity and resources to lose weight in addition to receiving 3 motivational text messages per week for 12 weeks and 2 phone calls to check in on their progress. At baseline, three, and six months, all patients had a survey administered and weight, waist circumference and blood pressure were obtained.

## Results

In the EC group, 50% (n = 16) reported reading all of the text messages, 31% (n = 10) found the text messages helpful, and 84% (n = 26) reported using the diet and physical activity resources provided. The average weight loss at six months was 4.3 pounds in the SC group and 3 pounds in the EC group (p = 0.17). The average reduction in waist circumference at six months was 0.5 inches in the SC group and 1.2 inches in the EC group (p = 0.15). The most commonly reported diet change was reducing calories, with 28% (n = 17) and 36% (n = 18) reporting a low-calorie diet at three and six months, respectively. At six months, 26% (n = 13) reported engaging in cardiovascular exercise like jogging aerobics or swimming, while 24% (n = 12) reported engaging in some form of weightlifting as a change from baseline.

## Conclusions

This pilot study demonstrated the potential feasibility of a simple, texting-based intervention designed to motivate overweight individuals to lose weight and reduce their waist circumference in a primary care clinic. Both groups showed modest weight loss, slightly favoring the SC group, and waist circumference reductions favored the intervention group, although neither result was statistically significant. Future studies need creative ways to increase intervention exposure and harness the potential for mobile technology-based health coaching. While constraints of time and infrastructure may limit personalized counselling, mobile technology may serve as an excellent adjunct medium of communicating with patients about resources related to healthy eating and exercise. Novel methods for boosting the effectiveness of technology-based interventions are needed.

# Citation

Iyengar D, Klawans MR, Stotts AL, Northrup TF (2019) Use of Text Messaging as an Intervention to Promote Weight Loss and Reduce Waist Circumference among an Overweight and Abdominally Obese Patient Population: A Randomized-Controlled Quality Improvement Study in Primary Care. J Obes Weight-Loss Medic 5:029. doi.org/10.23937/2572-4010.1510029

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# Coaxial Cylindrical Conductors

1. Feb 5, 2008

### heartofaragorn

1. The problem statement, all variables and given/known data
Two coaxial cylindrical conductors are shown. The inner cylinder has radius a = 2 cm, length 10 m, and carries a total charge of Q inner = +8nC. The outer cylinder has an inner radius b = 6 cm, outer radius c= 7 cm, length 10 m, and carries a total charge of Q outer = -16nC. What is Ex, the x-component of the electric field at point P which is located at the midpoint of the length of the cylinders at a distance r= 4 cm from the origin and makes an angle of 30 degrees with the x-axis?

2. Relevant equations
Surface charge density = 2pi * radius a * length
Linear charge density = 2pi * radius a * surface charge density
Electric field = 2 * k * linear charge denisty / r

3. The attempt at a solution
I drew a Gaussian sphere at the radius r = 4 according to where point P lies and tried to determine the charge inside, which I think may be where I went wrong. There is an enclosed cylinder of +8 nC that has a radius of 2 that falls completely within the Gaussian sphere; however, I cannot see that the outer charge affects this particular sphere since it is within the 2nd cylinder. I calculated the surface charge density to be 2.5133 using the radius of 0.04 m, then plugged that into the linear charge density formula to receive an answer of 0.63165. I then tried plugging that into the formula for the electric field given by my prof and received an answer of 2.84245 e 11. I took that and mulitplied by cos 30 degrees to account for the diagnol line upon which P lies. I cannot get the right answer (interactive online problem) and I'm getting the feeling that I'm either using the wrong charge, wrong radius, or wrong formulae...or a combination of all 3!!!! Please help me!!!! Thanks! :uhh:
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 5, 2008

### Mindscrape

So what direction does the electric field, from Gauss's law, travel in? I'm guessing this is where are getting tripped up.

Also, do you have a picture? I can't tell why they even give you an outer cylinder, something you mentioned also, maybe the problem creator is just trying to throw you off. Could you repost your work but in terms of symbols rather than numbers?

3. Feb 6, 2008

### heartofaragorn

I'll try attaching the image, but as for the symbols, I don't have anything that will display them. My computer is a little too old to have packages like Microsoft Equation or anything like that. If it's really a problem, I'll try seeing if I can download something. Also, I'm really not sure what direction the electric field travels in!

#### Attached Files:

• ###### Cylinder.gif
File size:
2.9 KB
Views:
118
Last edited: Feb 6, 2008
4. Feb 6, 2008

### Staff: Mentor

Your Gaussian cylinder at (radius r, say) totally encloses the inner conductor. What's Gauss's law? Set up the equation and solve for E. (No need to compute surface charge!)

The charge on the outer conductor is irrelevant. The field at r = 4 cm will be radially outward.

You can just type in your equations without using symbols, or you can use Latex. (See Introducing LaTeX Math Typesetting .)

5. Feb 6, 2008

### heartofaragorn

The interactive help that comes with the problem stated that I had to calculate the surface charge in order to calculate the linear charge. Also, do I have the right formula for E? And if I don't need the surface charge, from where do I get the linear charge?

6. Feb 6, 2008

### Staff: Mentor

That makes no sense to me. Using the data given, you can calculate the linear charge density directly.
Yes. (Use Gauss's law to verify it, if you're unsure.)
You are given the total charge and the length. (Assume it's uniformly distributed.)

7. Feb 6, 2008

### heartofaragorn

YEA! It worked!!!!!! Thanks so much!


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Cosets and Lagrange's Theorem

• May 15th 2009, 06:37 PM
Chris L T521
Cosets and Lagrange's Theorem
Hello,

I'm am stuck on this problem (or so I think) XD

Quote:

Let $n$ be a positive integer. Let $H=\{0,\pm n,\pm2n,\pm3n,\dots\}$. Find all left cosets of $H$ in $\mathbb{Z}$. How many are there?
Source: Contemporary Abstract Algebra, 6e, by J. Gallian.

Since
$n\in\mathbb{Z}^+$, I thought that it would be safe to say that the left cosets would be

\begin{aligned}
0+H & = \{0,\pm n,\pm2n,\pm 3n,\dots\}= nk+H,\,k\in\mathbb{Z}^+\\
1+H & = \{\dots,1-2n,1-n,1,1+n,1+2n,\dots\}=nk+1+H,k\in\mathbb{Z}^+\\
2+H & = \{\dots,2-2n,2-n,2,2+n,2+2n,\dots\}=nk+2+H,k\in\mathbb{Z}^+\\
& \,\vdots\\
\left(n-1\right)+H & = \{\dots,-2n-1,-n-1,-1,n-1,2n-1,\dots\}=2nk-1+H,k\in\mathbb{Z}^+
\end{aligned}

From this, it would appear that there are $n$ distinct left cosets. But is there another way to verify that this is the case? Lagrange's Theorem wouldn't apply to this, since $\left|n\mathbb{Z}\right|$ and $\left|\mathbb{Z}\right|$ are infinite.

Does this look right, or am I way over my head? XD

I would appreciate any suggestions! :D
• May 15th 2009, 07:33 PM
PaulRS

Okay, let us see: $
aH = bH \Leftrightarrow H = \left( {a^{ - 1} b} \right)H \Leftrightarrow a^{ - 1} b \in H
$
or equivalently: $
n\left| {\left( {b - a} \right)} \right.
$
( written in another way: $
b \equiv a\left( {\bmod .n} \right)
$
)

... ;)

There are $n$ distinct left-cosets indeed.
• May 15th 2009, 07:36 PM
Gamma
Yeah, that is exactly right. $H=n\mathbb{Z}$. It is a normal subgroup as $\mathbb{Z}$ is abelian. You will notice that

$\mathbb{Z}/H = \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$

This is precisely how you define the modular numbers. So you can say $[G:H]=n=|\mathbb{Z}_n|$


text:
# ARest: Estimation of Autoregressive (AR) Parameters In funtimes: Functions for Time Series Analysis

## Description

Estimate parameters φ of autoregressive time series model

X_t = ∑_{i=1}^pφ_iX_{t-i} + e_t,

by default using robust difference-based estimator and Bayesian information criterion (BIC) to select the order p. This function is employed for time series filtering in functions sync_test and wavk_test.

## Usage

 1 ARest(x, ar.order = NULL, ar.method = "HVK", BIC = TRUE)

## Arguments

 x a vector containing a univariate time series. Missing values are not allowed. ar.order order of autoregressive model when BIC = FALSE, or the maximal order for BIC-based filtering. Default is round(10*log10(length(x))), where x is the time series. ar.method method of estimating autoregression coefficients. Default "HVK" delivers robust difference-based estimates by \insertCiteHall_VanKeilegom_2003;textualfuntimes. Alternatively, options of ar function can be used, such as "burg", "ols", "mle", and "yw". BIC logical value indicates whether the order of autoregressive filter should be selected by Bayesian information criterion (BIC). If TRUE (default), models of orders p= 0,1,...,ar.order or p= 0,1,...,round(10*log10(length(x))) are considered, depending on whether ar.order is defined or not (x is the time series).

## Details

The same formula for BIC is used consistently for all methods:

BIC=n\ln(\hat{σ}^2) + k\ln(n),

where n = length(x), k=p+1.

## Value

A vector of estimated AR coefficients. Returns numeric(0) if the final p=0.

## Author(s)

Vyacheslav Lyubchich

## References

\insertAllCited

ar, HVK, sync_test, wavk_test
  1 2 3 4 5 6 7 8 9 10 11 12 # Fix seed for reproducible simulations: set.seed(1) #Simulate some time series, possibly with trend: n <- 100 Y <- arima.sim(n = n, list(order = c(2, 0, 0), ar = c(-0.7, -0.1))) plot.ts(Y) #Estimate the coefficients: ARest(Y) #HVK by default ARest(Y, ar.method = "yw") #Yule--Walker ARest(Y, ar.method = "burg") #Burg


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< Previous Section Home Next Section >

# Section 4.7Wrinkly Functions

## Summation Notation

In this section we will speak of adding large numbers of terms. It will be convenient to represent a sum of many terms without having to write it out. Summation notation was designed to do just that. Instead of writing

1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39

we could instead use a notation that says, in effect, "add up the first 20 odd numbers".

Every odd number can be represented as $2k-1$ for some natural number k. So we will write the above sum as in Figure 4.7.1 and mean the same thing as "add up the first 20 odd numbers".

Read the statement in Figure 4.7.1 as "The sum from k equals 1 to 20 of the numbers represented by $2k-1$." Check that this verbal description matches the long expression given above.

$$\sum_{k=1}^{20} (2k-1)$$ Figure 4.7.1. Summation notation that represents the sum 1+3+…+39.

The summation notation in Figure 4.7.1 expands to $(2\cdot1-1) + (2\cdot2-1) + (2\cdot3-1) + … + (2\cdot20-1)$ as the value of k varies from 1 to 20 in increments of 1.

GC understands summation notation, so using it can be very practical.

Reflection 4.7. Write the following sums using summation notation:
• 12 + 32 + 52 + 72 +…+ 992;
• 3 + 6 + 9 + 12 + 15 + … + 600
• 1 + 2 + 4 + 8 + 16 + 32 + 64 + … + 32768;
• 1 + 3 + 4 + 9 + 16 + 15 + 64 + 21 + 256 + 27 + … + 65536 + 51.
• Enter your summation notations on separate lines in GC (type ctrl-shift-S to get the summation symbol). GC should display 25100 for the first, 60300 for the second, 65535 for the third, and 87624 for the fourth.

## A Function That Is a Sum of Many Functions

Sound travels through air as waves of energy formed by variations in pressure caused by something vibrating. Simple sounds are sine waves.

Complex sounds are made by a jumble of simple sounds happening together. When sounds happened together, their pressure waves add to make a complex, bumpy wave.

When graphed relative to time, these complex sound waves have very wrinkly graphs. In this section we will not investigate properties of sound, but we will examine functions that have wrinkly graphs so that you have a better understanding of sound if you study it.

Start with $f(x) = \sin(x)$ and its graph $y = \sin(x)$. The function g defined as $g(x) = (\frac{1}{2})\sin(2x)$ will have half the amplitude of f and twice the frequency—g’s period will be $\frac{1}{2}$ the period of sin(x). Functions f and g are like two simple sounds; the function $(f+g)(x) = f(x)+g(x)$ is like the two sounds happening together for some period of time.

Suppose we combine N sounds $w_0$, $w_1$, …, $w_N$, where sound $w_{k+1}$ has half the amplitude and twice the frequency as sound $w_k$. Suppose also that sound $w_1$ is modeled by sin(x). The function w, below, models the combined sound over time:$$w(x,N) = \sum_{k=0}^{N} {2^{-k}\sin(2^k x)}$$

where x is a measure of time and N is the number of sounds.

Figure 4.7.2 gives a visual perspective on how the function w is made for any value of N. The first animation shows $y = w(x,1)$ being made by adding the next sound wave to $y = w(x,0)$. The second animation shows $y = w(x,2)$ being made by adding the next sound wave to $y = w(x,1)$. The third animation shows $y = w(x,3)$ being made by adding the next sound wave to $y = w(x,2)$.

Figure 4.7.2. The next iteration of w(x,N) comes from adding $2^{-(N+1)}\sin(2^{N+1}x)$ to w(x,N)

Put another way, the definition of w builds in stages. When $N = 0$, w is $w(x,0) = 2^0 \sin(2^0x)$, or just sin(x). When $N = 1$, w is $w(x,1) = \sin(x) + \frac{1}{2}\sin(2x)$. When $N = 3$, w is:

## Exercise Set 4.7

1. First do the following steps:

• Enter the definition of w in GC. Type ctrl-shift-S to get the summation sign: $w(x,N) = \sum_{k=0}^{N} {2^{-k}\sin(2^k x)}$
• Enter $y = w(x, 15)$ on a second line.
• Enter $x = 1.1$ on a third line.
• Zoom in repeatedly around the point where the graphs of $x = 1.1$ and $y = w(x,15)$ intersect. Keep zooming until w appears to have a constant rate of change over the visible x-axis.

Now call the width of the visible x-axis ∆x ("delta x"). What is the value of ∆x? (Trace the graph to see values of x.)

2. Suppose that $dx$ varies from 0 to ∆x. Let $dy = mdx$. The value of m is w’s essential constant rate of change at the moment that x = 1.1. What is your estimate of the value of m?

 < Previous Section Home Next Section >


text:
## 17Calculus - Invalid Derivative Quotient Rule

##### 17Calculus

As you know, the quotient rule for taking the derivative is $\left[ \frac{f(x)}{g(x)} \right]' = \frac{f'g-fg'}{g^2}$ One of the most common mistakes students do when first learning this rule is to take the derivative of each function and then divide the results. The correct equation is based on limits.

When does $$\displaystyle{ \left[ \frac{f(x)}{g(x)} \right]' = \frac{f'(x)}{g'(x)} }$$ hold?

It would be nice if we could take the derivative of each function and then just divide $$g'$$ into $$f'$$. But this will not work nearly all the time. This video shows that this invalid quotient rule works only in a couple of special cases.

### Michael Penn - Let's change the quotient rule!

video by Michael Penn

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.


text:
Getting annotated and normalized data.
2
0
Entering edit mode
jslow • 0
@jslow-7582
Last seen 5 months ago
United States

Hi,

I am trying to analyze microarray data with R and am stuck at annotation step. I was wondering if anyone could help.

Here's the code i have so far

library(affycoretools)

library(oligo)

OligoRaw<-read.celfiles(filenames=list.celfiles()) OligoEset<-rma(OligoRaw) # 35556 features, 24 samples  data.oligo<-exprs(OligoEset)

library(mogene10sttranscriptcluster.db) library(pd.mogene.1.0.st.v1)

ID <- getMainProbes(OligoEset) annot <- select(mogene10sttranscriptcluster.db, featureNames(ID),                 c("SYMBOL","GENENAME","ENTREZID")) # 36631

I am stuck at trying to merge the "annot" with "OligoEset". I would like to have annotated and normalized data set in a dataframe/.txt/.xls files to analyze.

I'd very much appreciate any help.

Thanks,

Jun

microarray rstudio annotation • 756 views
0
Entering edit mode
@james-w-macdonald-5106
Last seen 58 minutes ago
United States

You actually don't want the annotated and normalized data any of those forms. If you are going to use Bioconductor to analyze, then you need to learn to use the tools that are supplied.

The ExpressionSet containing your data is a perfect input to say, the limma package. So you now need to define what comparisons you want to make, and express that as a design matrix. See the limma user's guide.

What you would tend to do is something like

design <- model.matrix(~<args go here>)

fit <- lmFit(data.oligo, design)

fit2 <- eBayes(fit)

You will have duplicates in your annot data.frame, so you have to deal with that. The most naive thing you could do is choose the first one:

annot <- annot[!duplicated(annot[,1]),]

fit2\$genes <- annot

Now your topTable() output will have annotations, as well as statistics.

topTable(fit2, coef = 2)

0
Entering edit mode

Thanks James,

but assuming I want a collated file so that non-R users can read and understand it, Im thinking a file that contains normalized raw datum such that layman can compare and make their own analysis.

Is there a way I can get this done?

0
Entering edit mode
@james-w-macdonald-5106
Last seen 58 minutes ago
United States

Sure. If you are already using affycoretools, see ?writeFit.

I am not in general enthused with giving normalized data to 'laymen' so they can make their own analyses. In other words, generating summarized data from raw celfiles is not usually the part of the analysis that requires the most sophistication (although the QC part does take some base knowledge). Instead, fitting models to the data and ensuring that statistically unsophisticated collaborators understand what was done and why is the main deliverable for my line of work.

Because of that, I much prefer giving people either HTML or Excel spreadsheets that already contain the comparisons they wanted. The ReportingTools package makes it very easy to generate HTML tables that are easy to work with. The openxlsx package makes it easy to output Excel spreadsheets directly, which allows you to circumvent Excel's tendency to convert gene symbols that look like dates into actual dates, when people import data incorrectly (as an example, SEPT1 is helpfully converted to 9/1/2015, because obviously).


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## Book tickets knowing flight numbers

I’m trying to book a round trip tickets KRR -> SVO -> AMS -> SEA.

I know there is a flight KRR -> SVO departing at 2:30pm however all search engines then end up with results that give me overnight layover in AMS. If I choose later departing flight they do have me staying overnight in SVO but I need to get there earlier in a day.

Any ideas on how I can go about booking the right combination of flights?

Thank you.

## How to present a book online?

Hello everyone! First time poster here.

I am looking for some advice on how to "web-ify" a How-To book that I wrote and put it online.

The problem is that it's not practical to have one webpage that is the entire book, but I'm not sure how to split the book up.

This may sound elementary, but I am selling the book and giving people access, so I don't want to split up the book just any way as that will make the web development part difficult.

Does that make sense?

## Missing a connection and don’t have money to book next flight

What happens if I miss a connection flight and it’s my fault? Let’s say i got distracted or something. The airline may not book me onto the next flight because it wasn’t their fault. And if I don’t have the money to book another flight i’m stuck in the airport.

What would be the best course of action under these circumstances?

## Can I use Accommodation or Hotel Schema markup if customers can’t book it

We sell ferry travel, and want to capture customers browsing holidays to the destinations in order to raise awareness of going by ferry.

On this page we mention the different types of holiday accomm you could book on the islands and give 3 examples of each. 3 hotels, 3 resorts/holiday parks & 3 self catering apartments.

Can we use the accommodation markup schema, or as we don’t mention price and you can’t book would this be pointless?

Thanks!

## Why can I not book these flights on Air France?

I’m looking to book a multi-city itinerary on Air France, specifically CDG–>KIX (Paris CDG to Kansai International) and KIX–>LHR (Kansai International to London Heathrow).

Air France operates flights for both these pairs, and it seems I would be able to book them separately without any issue (though very expensive, of course). However, I can’t book them together (for any amount of money). When I search through ITA Matrix, this combination on Air France does not show up (even though both one-ways show up).

When I search on the Air France website, I get a little bit of hope, since on the first page after the search it indeed lists many flights in both directions. However when I select which flights I want, it gives me a strange error message “There are no flights available. Please modify your travel dates.” (even though it just displayed available flights!).

Is there any hope of booking this combination, perhaps through a SkyTeam partner, or does Air France simply disallow this sort of flight combination?

By the way, the combination CDG–>KIX and NRT–>CDG seems to be easily bookable.

## I want to buy a book review.

I'm in College and I've been assigned to write a book review. I have no idea how to do it and want to ask for help in a professional service. I think it can help me to write reviews further with my mind. What do you recommend?

## What is the compressibility of this simple “book”?

Compressibility is defined as

$$C=\frac{2^{HN}}{2^{H_{max}N}}$$

The book is made up of a simple alphabet of only {a,b,c,d} which occur with probabilities $$P(a)=0.2, P(b)=0.4, P(c)=0.1, P(d)=0.3$$

In class we were given the example of calculating the entropy of a simple string of coin tosses, where $$N$$ is the number of coin tosses.

So, for example, we could say that $$P(heads)=0.1$$ and $$P(tails)=0.9$$. The entropy of each coin toss is therefore $$H(p=0.9)=0.469$$ If we toss the coin 4 times then we end up with a compressibility of: $$C=\frac{2^{HN}}{2^{H_{max}N}}=\frac{2^{0.469\cdot4}}{2^{1\cdot 4}}=22.3\%$$ How do we extend this to the book case?

The entropy of the letter occurance is $$H(p=0.2,p=0.4,p=0.1,p=0.3)=1.85$$

and the maximum entropy is just 2 bits (i.e. we require 2 bits to descibe four characters)

Since this is all the information given in the question, I presume that we just set $$N=4$$ and plug in the values for $$H$$ and $$H_{max}$$ to get the entropy. However, I am struggeling to see why. Why is $$N$$ not the number of letters that make up this book?

## Confusion about proof of undecidability of REGULAR TM in Sipser’s book

in the book “Introduction to the Theory of Computation” by Michael Sipser there is an example of undecidable languages in which there is a language REGULR_TM which is described as follows :

REGULAR_TM = { <M> | M is a Turing machine and L(M) is regular language }. Well, Sipser says that this is an undecidable language since we cannot have a decider to decide this language. Because if we could, we could create a TM that decides ATM which we know previously that is undecidable. so this could cause a contradiction. Here is the proof from this book :

Proof. We let R be a TM that decides REGULARtm and construct TM S to decide Atm. Then S works in the following manner.

S = “On input , where M is a TM and w is a string:

1. Construct the following TM M2.

M2 = “On input x:

  1. If x has the form 0^n 1^n, accept.    2. If x does not have this form, run M on input w and accept if M accepts w."
2. Run R on input <M2>.

3. If R accepts, accept; if R rejects, reject.”

Now my question is, is this kind of constrcuting a TM S correct? the part that is really confusing me is the part that says “run M on input w and accept if M accepts w”. this is the whole question for Atm. If we could answer that question why we need a machine M2 to feed it into the R and output the result of R. We just “run M on w”.

Also for some languages that [we know] are decidable we can have the same argument and conclude that it is not decidable.

## Which book is the most definitive, yet elementary, discussion of C? C++? Python?

I in October 2018 completed the MITx course INTRODUCTION TO COMPUTER SCIENCE AND PROGRAMMING USING PYTHON on the edx.org web platform by scoring 97% marks.

Besides that, I’ve also read the first few chapters of the book SCHAUM’S OUTLINES OF THEORY & PROBLEMS OF PROGRAMMING WITH C by Byron S. Gottfried back in 2003, but I don’t remember much from that particular book.

For C, I’ve got the book THE C PROGRAMMING LANGUAGE by Kermighan & Richie, 2nd edition 1988. Is this book still current for learning C? If so, is the discussion in it elementary enough for a learner like me? If not, then which is the most up-to-date book on C?

I also have got the book THE C++ PROGRAMMING LANGUAGE by Bjourn Stroustup, 4th edition. Should I read this book for learning C++? I mean is it going to be accessible enough for me?

And, for learning Python, is the book BEGIN TO CODE WITH PYTHON by Rob Miles going to be a good choice? Or, how about Prof. John V. Guttag’s INTRODUCTION TO COMPUTATION AND PROGRAMMING USING PYTHON WITH APPLICATIONS TO DATA SCIENCE, 3rd edition?

One of my aims in learning Python, especially, is to be able to extend the features of the Non-Visual Desktop Access (NVDA) screen-reading program, which has been coded in Python. Please visit nvaccess.org.


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# Conditional expectation on an indicator

## Homework Statement

Let $X$ and $Y$ be independent Bernoulli RV's with parameter $p$. Find,
$$\mathbb{E}[X\vert 1_{\{X+Y=0\}}]$$ and $$\mathbb{E}[Y\vert 1_{\{X+Y=0\}}]$$

## The Attempt at a Solution

I'm trying to show that,

$$\mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] = 0$$

by,

\begin{align*} \mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] &= \frac{\mathbb{E}[(X+Y)1_{\{X+Y=0\}}]}{\mathbb{P}[X+Y=0]} \\ &= \frac{0}{(1-p)^2} \\ &= 0 \end{align*}

haruspex
Homework Helper
Gold Member
2020 Award
$$\mathbb{E}[X\vert 1_{\{X+Y=0\}}]$$
I'm not familiar with the notation for the condition ($1_{\{X+Y=0\}}$). How does one read it? Can you post a link?

I'm not familiar with the notation for the condition ($1_{\{X+Y=0\}}$). How does one read it? Can you post a link?

It's the indicator of the event $\{X+Y=0\}$.

$$1_{\{X+Y=0\}}=\begin{cases}1, \qquad \text{if }X+Y=0; \\ 0, \qquad \text{otherwise.}\end{cases}$$

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

Let $X$ and $Y$ be independent Bernoulli RV's with parameter $p$. Find,
$$\mathbb{E}[X\vert 1_{\{X+Y=0\}}]$$ and $$\mathbb{E}[Y\vert 1_{\{X+Y=0\}}]$$

## The Attempt at a Solution

I'm trying to show that,

$$\mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] = 0$$

by,

\begin{align*} \mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] &= \frac{\mathbb{E}[(X+Y)1_{\{X+Y=0\}}]}{\mathbb{P}[X+Y=0]} \\ &= \frac{0}{(1-p)^2} \\ &= 0 \end{align*}

Basically, you are trying to show that the conditional distribution ##f(k) \equiv P(X= k|X+Y=0), \; k=0,1,2, \ldots## has mean zero; that is, that ##\sum_{k=0}^{\infty} k f(k) = 0##. How do you suppose that could happen?

Basically, you are trying to show that the conditional distribution ##f(k) \equiv P(X= k|X+Y=0), \; k=0,1,2, \ldots## has mean zero; that is, that ##\sum_{k=0}^{\infty} k f(k) = 0##. How do you suppose that could happen?

I think it is possible since both ##X## and ##Y## are Bernoulli, if their sum is 0, then ##\mathbb{P}[X=0|X+Y=0]=1##. Then, ##\sum_{k=0}^1kf(k)=0*1##.

Ray Vickson


text:
Free Version
Easy

# Same Base Exponential Equations

SATSTM-OEJXRN

If ${a}^{3b-5}={a}^{4b+3}$ for all values of $a$, then what is the value of $b$?

A

$8$

B

$-8$

C

$-2$

D

$2$

E

$4$


text:
# Zeta function universality: How to compute the shift parameter for simple functions?

I've come across Zeta function universality. For a nice function $f$ in a nice subset $U$ of the complex strip between real $0$ and $1$, one can find a real $t$, such the zeta function $\zeta$ shifted by $it$ approximates $f$:

I've plugged in $|\zeta(s+it)-f(s)|$ to an computer algebra system and varied $t$. For the $t$ I've tried (some smaller values), that absolute never got small at all $s$ in some interval I tried at once. Analytically, I doN't quite know how to find the $t$, even for simple $f$ like $\frac{1}{1-s}$ or polyomials in $s$, because sums over running $n^{-it}$ are rather mystical to me.

How to compute $t$ for $\frac{1}{1-s}$ or $2+s-3s^2$? Any finite $U$ you choose.


text:
# Generalizations of Poisson structures related to rational Gaudin model

Abstract : The Poisson structure arising in the Hamiltonian approach to the rational Gaudin model looks very similar to the so-called modi?ed Reflection Equation Algebra. Motivated by this analogy, we realize a braiding of the mentioned Poisson structure, i.e. we introduce a "braided Poisson" algebra associated with an involutive solution to the quantum Yang-Baxter equation. Also, we exhibit another generalization of the Gaudin type Poisson structure by replacing the ?rst derivative in the current parameter, entering the so-called local form of this structure, by a higher order derivative. Finally, we introduce a structure, which combines both generalizations. Some commutative families in the corresponding braided Poisson algebra are found.
Type de document :
Pré-publication, Document de travail
LATEX, 16 pp. 2013

https://hal.archives-ouvertes.fr/hal-00931978
Soumis le : jeudi 16 janvier 2014 - 10:34:04
Dernière modification le : mercredi 19 décembre 2018 - 14:08:04

### Identifiants

• HAL Id : hal-00931978, version 1
• ARXIV : 1312.7813

### Citation

Dimitri Gurevich, Vladimir Rubtsov, Pavel Saponov, Zoran Skoda. Generalizations of Poisson structures related to rational Gaudin model. LATEX, 16 pp. 2013. 〈hal-00931978〉

### Métriques

Consultations de la notice


text:
# Scaling with Combinators and the Static Modeling Language

Up until this point, we have been using Gen’s generic built-in modeling language, which is a very flexible modeling language that is shallowly embedded in Julia. However, better performance and scaling characteristics can be obtained using specialized modeling languages or modeling constructs. This notebook introduces two built-in features of Gen:

These features provide both constant-factor speedups, as well as improvements in asymptotic orders of growth, over the generic built-in modeling language.

## Outline

Section 2: Introducing the map combinator

Section 5: Checking the inference programs

using Gen

## 1. Studying the scaling behavior of an inference program

Recall the regression model used to introduce iterative inference in an earlier tutorial:

@gen function model(xs::Vector{Float64})
slope = @trace(normal(0, 2), :slope)
intercept = @trace(normal(0, 2), :intercept)
noise = @trace(gamma(1, 1), :noise)
prob_outlier = @trace(uniform(0, 1), :prob_outlier)
n = length(xs)
ys = Vector{Float64}(undef, n)
for i = 1:n
if @trace(bernoulli(prob_outlier), :data => i => :is_outlier)
(mu, std) = (0., 10.)
else
(mu, std) = (xs[i] * slope + intercept, noise)
end
ys[i] = @trace(normal(mu, std), :data => i => :y)
end
ys
end;

We wrote a Markov chain Monte Carlo inference update for this model that performs updates on each of the ‘global’ parameters (noise, slope, intercept, and prob_outlier), as well as the ‘local’ is_outlier variable associated with each data point. The update takes a trace as input, and returns the new trace as output. We reproduce this here:

function block_resimulation_update(tr)

# Block 1: Update the line's parameters
line_params = select(:noise, :slope, :intercept)
(tr, _) = mh(tr, line_params)

# Blocks 2-N+1: Update the outlier classifications
(xs,) = get_args(tr)
n = length(xs)
for i=1:n
(tr, _) = mh(tr, select(:data => i => :is_outlier))
end

# Block N+2: Update the prob_outlier parameter
(tr, _) = mh(tr, select(:prob_outlier))

# Return the updated trace
tr
end;

We write a helper function that takes a vector of y-coordinates and populates a constraints choice map:

function make_constraints(ys::Vector{Float64})
constraints = choicemap()
for i=1:length(ys)
constraints[:data => i => :y] = ys[i]
end
constraints
end;

Finally, we package this into an inference program that takes the data set of all x- and y-coordinates ,and returns a trace. We will be experimenting with different variants of the model, so we make the model an argument to this function:

function block_resimulation_inference(model, xs, ys)
observations = make_constraints(ys)
(tr, _) = generate(model, (xs,), observations)
for iter=1:500
tr = block_resimulation_update(tr)
end
tr
end;

Let’s see how the running time of this inference program changes as we increase the number of data points. We don’t expect the running time to depend too much on the actual values of the data points, so we just construct a random data set for each run:

ns = [1, 3, 7, 10, 30, 70, 100]
times = []
for n in ns
xs = rand(n)
ys = rand(n)
start = time_ns()
tr = block_resimulation_inference(model, xs, ys)
push!(times, (time_ns() - start) / 1e9)
end

We now plot the running time versus the number of data points:

using PyPlot

plot(ns, times)
xlabel("number of data points")
ylabel("running time (seconds)");

The inference program seems to scale quadratically in the number of data points.

To understand why, consider the block of code inside block_resimulation_update that loops over the data points:

# Blocks 2-N+1: Update the outlier classifications
(xs,) = get_args(tr)
n = length(xs)
for i=1:n
(tr, _) = mh(tr, select(:data => i => :is_outlier))
end

The reason for the quadratic scaling is that the running time of the call to mh inside this loop also grows in proportion to the number of data points. This is because the updates to a trace of a model written the generic built-in modeling language always involve re-running the entire model generative function.

However, it should be possible for the algorithm to scale linearly in the number of data points. Briefly, deciding whether to update a given is_outlier variable can be done without referencing the other data points. This is because each is_outiler variable is conditionally independent of the outlier variables and y-coordinates of the other data points, conditioned on the parameters.

We can make this conditional independence structure explicit using the Map generative function combinator. Combinators like map encapsulate common modeling patterns (e.g., a loop in which each iteration is making independent choices), and when you use them, Gen can take advantage of the restrictions they enforce to implement performance optimizations automatically during inference. The Map combinator, like the map function in a functional programming language, helps to execute the same generative code repeatedly.

## 2. Introducing the map combinator

To use the map combinator to express the conditional independences in our model, we first write a generative function to generate the is_outlier variable and the y-coordinate for a single data point:

@gen function generate_single_point(x::Float64, prob_outlier::Float64, noise::Float64,
slope::Float64, intercept::Float64)
is_outlier = @trace(bernoulli(prob_outlier), :is_outlier)
mu = is_outlier ? 0. : x * slope + intercept
std = is_outlier ? 10. : noise
y = @trace(normal(mu, std), :y)
return y
end;

We then apply the Map, which is a Julia function, to this generative function, to obtain a new generative function:

generate_all_points = Map(generate_single_point);

This new generative function has one argument for each argument of generate_single_point, except that these arguments are now vector-valued instead of scalar-valued. We can run the generative function on some fake data to test this:

xs = [0, 1, 2, 3, 4]
prob_outliers = fill(0.5, 5)
noises = fill(0.2, 5)
slopes = fill(0.7, 5)
intercepts = fill(-2, 5)
trace = simulate(generate_all_points, (xs, prob_outliers, noises, slopes, intercepts));

We see that the generate_all_points function has traced 5 calls to generate_single_point, under namespaces 1 through 5. The Map combinator automatically adds these indices to the trace address.

println(get_choices(trace))

│
├── 1
│   │
│   ├── :y : -9.299459431407582
│   │
│   └── :is_outlier : true
│
├── 2
│   │
│   ├── :y : -1.321273012367464
│   │
│   └── :is_outlier : false
│
├── 3
│   │
│   ├── :y : 0.520205709733691
│   │
│   └── :is_outlier : true
│
├── 4
│   │
│   ├── :y : -0.2497222484225019
│   │
│   └── :is_outlier : false
│
└── 5
│
├── :y : 0.4961726596283081
│
└── :is_outlier : true

Now, let’s replace the Julia for loop in our model with a call to this new function:

@gen function model_with_map(xs::Vector{Float64})
slope = @trace(normal(0, 2), :slope)
intercept = @trace(normal(0, 2), :intercept)
noise = @trace(gamma(1, 1), :noise)
prob_outlier = @trace(uniform(0, 1), :prob_outlier)
n = length(xs)
ys = @trace(generate_all_points(xs, fill(prob_outlier, n), fill(noise, n), fill(slope, n), fill(intercept, n)), :data)
ys
end;

Note that this new model has the same address structure as our original model had, so our inference code will not need to change. For example, the 5th data point’s $y$ coordinate will be stored at the address :data => 5 => :y, just as before. (The :data comes from our @trace invocation in the better_model definition, and the :y comes from generate_point; only the 5 has been inserted automatically by Map.)

trace = simulate(model_with_map, (xs,));
println(get_choices(trace))

│
├── :intercept : -5.258730415956281
│
├── :slope : -0.35831621634793925
│
├── :prob_outlier : 0.6504269974681032
│
├── :noise : 0.873700919240994
│
└── :data
│
├── 1
│   │
│   ├── :y : -3.6313467104186437
│   │
│   └── :is_outlier : true
│
├── 2
│   │
│   ├── :y : 10.914595507340668
│   │
│   └── :is_outlier : true
│
├── 3
│   │
│   ├── :y : -6.0345837282037635
│   │
│   └── :is_outlier : false
│
├── 4
│   │
│   ├── :y : -13.647061274089857
│   │
│   └── :is_outlier : true
│
└── 5
│
├── :y : 10.76592726168024
│
└── :is_outlier : true

Let’s test the running time of the inference program, applied to this new model:

with_map_times = []
for n in ns
xs = rand(n)
ys = rand(n)
start = time_ns()
tr = block_resimulation_inference(model_with_map, xs, ys)
push!(with_map_times, (time_ns() - start) / 1e9)
end

We plot the results and compare them to the original model, which used the Julia for loop:

plot(ns, times, label="original")
plot(ns, with_map_times, label="with map")
xlabel("number of data points")
ylabel("running time (seconds)")
legend();

We see that the quadratic scaling did not improve. In fact, we actually got a that happed was a constant factor slowdown.

We can understand why we still have quadratic scaling, by examining the call to generate_single_point:

ys = @trace(generate_all_points(xs, fill(prob_outlier, n), fill(noise, n), fill(slope, n), fill(intercept, n)), :data)

Even though the function generate_all_points knows that each of the calls to generate_single_point is conditionally independent, and even it knows that each update to is_outlier only involves a single application of generate_single_point, it does not know that none of its arguments change within an update to is_outlier. Therefore, it needs to visit each call to generate_single_point. The generic built-in modeling language does not provide this information the generative functions that it invokes.

## 3.Combining the map combinator with the static modeling language

In order to provide generate_all_points with the knowledge that its arguments do not change during an update to the is_outlier variable, we need to write the top-level model generative function that calls generate_all_points in the Static Modeling Language, which is a restricted variant of the built-in modeling language that uses static analysis of the computation graph to generate specialized trace data structures and specialized implementations of trace operations. We indicate that a function is to be interpreted using the static language using the static annotation:

@gen (static) function static_model_with_map(xs::Vector{Float64})
slope = @trace(normal(0, 2), :slope)
intercept = @trace(normal(0, 2), :intercept)
noise = @trace(gamma(1, 1), :noise)
prob_outlier = @trace(uniform(0, 1), :prob_outlier)
n = length(xs)
ys = @trace(generate_all_points(xs, fill(prob_outlier, n), fill(noise, n), fill(slope, n), fill(intercept, n)), :data)
return ys
end;

The static language has a number of restrictions that make it more amenable to static analysis than the unrestricted modeling language. For example, we cannot use Julia for loops, and the return value needs to explicitly use the return keyword, followed by a symbol (e.g. ys). Also, each symbol used on the left-hand side of an assignment must be unique. A more complete list of restrictions is given in the documentation.

Below, we show the static dependency graph that Gen builds for this function. Arguments are shown as diamonds, Julia computations are shown as squares, random choices are shown as circles, and calls to other generative function are shown as stars. The call that produces the return value of the function is shaded in blue.

Now, consider the update to the is_outlier variable:

(tr, _) = mh(tr, select(:data => i => :is_outlier))

Because this update only causes values under address :data to change, the static_model_with_map function can use the graph above to infer that none of the arguments to generate_all_point could have possibly changed. This will allow us to obtain the linear scaling we expected.

However, before we can use a function written in the static modeling language, we need to run the following function (this is required for technical reasons, because functions written in the static modeling language use a staged programming feature of Julia called generated functions).

Gen.load_generated_functions()

Finally, we can re-run the experiment with our model that combines the map combinator with the static language:

static_with_map_times = []
for n in ns
xs = rand(n)
ys = rand(n)
start = time_ns()
tr = block_resimulation_inference(static_model_with_map, xs, ys)
push!(static_with_map_times, (time_ns() - start) / 1e9)
end

We compare the results to the results for the earlier models:

plot(ns, times, label="original")
plot(ns, with_map_times, label="with map")
plot(ns, static_with_map_times, label="with map and static outer fn")
xlabel("number of data points")
ylabel("running time (seconds)")
legend();

We see that we now have the linear running time that we expected.

## 4. Constant-factor performance gains from the static modeling language

Note that in our latest model above, generate_single_point was still written in the generic modeling language. It is not necessary to write generate_single_point in the static language, but doing so can provide modest constant-factor performance improvements. Here we rewrite this function in the static language. Note that we had to make some changes to fit into the more restrictive syntax of the static language:

@gen (static) function static_generate_single_point(x::Float64, prob_outlier::Float64, noise::Float64,
slope::Float64, intercept::Float64)
is_outlier = @trace(bernoulli(prob_outlier), :is_outlier)
mu = is_outlier ? 0. : x * slope + intercept
std = is_outlier ? 10. : noise
y = @trace(normal(mu, std), :y)
return y
end;

static_generate_all_points = Map(static_generate_single_point);

@gen (static) function fully_static_model_with_map(xs::Vector{Float64})
slope = @trace(normal(0, 2), :slope)
intercept = @trace(normal(0, 2), :intercept)
noise = @trace(gamma(1, 1), :noise)
prob_outlier = @trace(uniform(0, 1), :prob_outlier)
n = length(xs)
ys = @trace(static_generate_all_points(xs, fill(prob_outlier, n), fill(noise, n), fill(slope, n), fill(intercept, n)), :data)
return ys
end;

Gen.load_generated_functions()

Now, we re-run the experiment with our new model:

fully_static_with_map_times = []
for n in ns
xs = rand(n)
ys = rand(n)
start = time_ns()
tr = block_resimulation_inference(fully_static_model_with_map, xs, ys)
push!(fully_static_with_map_times, (time_ns() - start) / 1e9)
end

We see that we get a modest improvement in running time:

plot(ns, times, label="original")
plot(ns, with_map_times, label="with map")
plot(ns, static_with_map_times, label="with map and static outer fn")
plot(ns, fully_static_with_map_times, label="with map and static outer and inner fns")
xlabel("number of data points")
ylabel("running time (seconds)")
legend();

## 5. Checking the inference programs

Before wrapping up, let’s confirm that all of our models are giving good results:

Let’s use a synthetic data set:

true_inlier_noise = 0.5
true_outlier_noise = 10.
prob_outlier = 0.1
true_slope = -1
true_intercept = 2
xs = collect(range(-5, stop=5, length=50))
ys = Float64[]
for (i, x) in enumerate(xs)
if rand() < prob_outlier
y = 0. + randn() * true_outlier_noise
else
y = true_slope * x + true_intercept + randn() * true_inlier_noise
end
push!(ys, y)
end
ys[end-3] = 14
ys[end-5] = 13;

figure(figsize=(4,4))
scatter(xs, ys);
ax = gca()
ax[:set_xlim](-7, 7)
ax[:set_ylim](-7, 15);

We write a trace rendering function that shows the inferred line on top of the observed data set:

function render_trace(trace)
xs,  = get_args(trace)
choices = get_choices(trace)
xlim = [-5, 5]
slope = choices[:slope]
intercept = choices[:intercept]
plot(xlim, slope * xlim .+ intercept, color="black")
ys = [choices[:data => i => :y] for i=1:length(xs)]
scatter(xs, ys)
ax = gca()
ax[:set_xlim](-7, 7)
ax[:set_ylim](-7, 15)
end;

Finally, we run the experiment. We will visualize just one trace produced by applying our inference program to each of the four variants of our model:

figure(figsize=(16, 4))

subplot(1, 4, 1)
tr = block_resimulation_inference(model, xs, ys)
render_trace(tr)
title("model")

subplot(1, 4, 2)
tr = block_resimulation_inference(model_with_map, xs, ys)
render_trace(tr)
title("model with map")

subplot(1, 4, 3)
tr = block_resimulation_inference(static_model_with_map, xs, ys)
render_trace(tr)
title("static model with map")

subplot(1, 4, 4)
tr = block_resimulation_inference(fully_static_model_with_map, xs, ys)
render_trace(tr)
title("fully static model with map");

It looks like inference in all the models seems to be working reasonably.


text:
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Unitary p-wave interactions between fermions in an optical lattice

Abstract

Exchange-antisymmetric pair wavefunctions in fermionic systems can give rise to unconventional superconductors and superfluids1,2,3. The realization of these states in controllable quantum systems, such as ultracold gases, could enable new types of quantum simulations4,5,6,7,8, topological quantum gates9,10,11 and exotic few-body states12,13,14,15. However, p-wave and other antisymmetric interactions are weak in naturally occurring systems16,17, and their enhancement via Feshbach resonances in ultracold systems has been limited by three-body loss18,19,20,21,22,23,24. Here we create isolated pairs of spin-polarized fermionic atoms in a multiorbital three-dimensional optical lattice. We spectroscopically measure elastic p-wave interaction energies of strongly interacting pairs of atoms near a magnetic Feshbach resonance. The interaction strengths are widely tunable by the magnetic field and confinement strength, and yet collapse onto a universal curve when rescaled by the harmonic energy and length scales of a single lattice site. The absence of three-body processes enables the observation of elastic unitary p-wave interactions, as well as coherent oscillations between free-atom and interacting-pair states. All observations are compared both to an exact solution using a p-wave pseudopotential and to numerical solutions using an ab initio interaction potential. The understanding and control of on-site p-wave interactions provides a necessary component for the assembly of multiorbital lattice models25,26 and a starting point for investigations of how to protect such systems from three-body recombination in the presence of tunnelling, for instance using Pauli blocking and lattice engineering27,28.

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3. Mizushima, T. et al. Symmetry-protected topological superfluids and superconductors: from the basics to 3He. J. Phys. Soc. Jpn 85, 022001 (2016).

4. Botelho, S. S. & Sá de Melo, C. A. R. Quantum phase transition in the BCS-to-BEC evolution of p-wave Fermi gases. J. Low Temp. Phys. 140, 409–428 (2005).

5. Gurarie, V., Radzihovsky, L. & Andreev, A. V. Quantum phase transitions across a p-wave Feshbach resonance. Phys. Rev. Lett. 94, 230403 (2005).

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10. Zhang, C., Tewari, S. & Das Sarma, S. Bell’s inequality and universal quantum gates in a cold-atom chiral fermionic p-wave superfluid. Phys. Rev. Lett. 99, 220502 (2007).

11. Nayak, C., Simon, S. H., Stern, A., Freedman, M. & Das Sarma, S. Non-Abelian anyons and topological quantum computation. Rev. Mod. Phys. 80, 1083–1159 (2008).

12. Jona-Lasinio, M., Pricoupenko, L. & Castin, Y. Three fully polarized fermions close to a p-wave Feshbach resonance. Phys. Rev. A 77, 043611 (2008).

13. D’Incao, J. P., Esry, B. D. & Greene, C. H. Ultracold atom-molecule collisions with fermionic atoms. Phys. Rev. A 77, 052709 (2008).

14. Nishida, Y., Moroz, S. & Son, D. T. Super Efimov effect of resonantly interacting fermions in two dimensions. Phys. Rev. Lett. 110, 235301 (2013).

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Acknowledgements

We acknowledge insightful discussions with F. Chevy and S. Zhang and helpful manuscript comments from J. Bohn, A. Kaufman and R. Learn. This work is supported by the AFOSR grants FA9550-19-1-0275, FA9550-19-1-7044 and FA9550-19-1-0365, by ARO W911NF-15-1-0603, by the NSF’s JILA-PFC PHY-1734006 and PHY-2012125 grants, by NIST and by NSERC.

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V.V., P.X., F.C., C.J.F. and J.H.T. developed the experimental techniques and collected the data. V.V., P.X., M.M., F.C. and C.J.F. performed the data analysis. M.M., T.B., C.J.F., J.P.D’I. and A.M.R. developed the theoretical models. All authors contributed to writing the manuscript.

Corresponding authors

Correspondence to Cora J. Fujiwara, Ana Maria Rey or Joseph H. Thywissen.

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Venu, V., Xu, P., Mamaev, M. et al. Unitary p-wave interactions between fermions in an optical lattice. Nature 613, 262–267 (2023). https://doi.org/10.1038/s41586-022-05405-6

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## On the ranks of certain finite semigroups of order-decreasing transformations.(English)Zbl 0844.20050

Let $$X$$ be a totally ordered finite set with $$n$$ elements and let $$K^- (n,r)$$ be the semigroup, under composition, of all selfmaps $$\alpha$$ of $$X$$ such that $$x\alpha \leq x$$ and $$|\text{Im } \alpha |\leq r$$. Let $$L^-(n,r)$$ be the semigroup consisting of the empty map together with all partial one-to-one maps $$\alpha$$ on $$X$$ such that $$x\alpha \leq x$$ and $$|\text{Im } \alpha|\leq r$$. Let $$P^-_r = K^-(n,r)/K^-(n,r-1)$$ for $$n \geq 3$$ and $$r \geq 2$$ and let $$Q_r^- = L^-(n,r)/L^-(n,r - 1)$$ for $$n \geq 2$$ and $$r \geq 1$$. The author shows that both $$P^-_r$$ and $$Q^-_r$$ have a unique generating set. He further shows that $$\text{rank } P^-_r = S(n,r)$$ the Stirling number of the second kind where the rank, $$\text{rank }S$$, of a semigroup $$S$$ is defined to be the least of the cardinal numbers of all the generating sets of $$S$$. Finally, he determines the rank of $$Q^-_r$$ as well.

### MSC:

 20M20 Semigroups of transformations, relations, partitions, etc. 20M05 Free semigroups, generators and relations, word problems
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# Colección

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## Something's Got a Hold on Me

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Rendering a subset in DX11

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How do I render only a select subset of a mesh in DX11?

I'm setting up the render like this:

void EFFECTMESH::Render(void){
unsigned int stride;
unsigned int offset;

// Set vertex buffer stride and offset.
stride = sizeof(NMVertex);
offset = 0;

// Set the vertex buffer to active in the input assembler so it can be rendered.
GE->devcon->IASetVertexBuffers(0, 1, &VB, &stride, &offset);

// Set the index buffer to active in the input assembler so it can be rendered.
GE->devcon->IASetIndexBuffer(IB, DXGI_FORMAT_R32_UINT, 0);

// Set the type of primitive that should be rendered from this vertex buffer, in this case triangles.
GE->devcon->IASetPrimitiveTopology(D3D11_PRIMITIVE_TOPOLOGY_TRIANGLELIST);
}

and then I'm calling the shader. The entire mesh renders fine, but I want to render different parts with different shaders......

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You just have to call DrawIndexed with the correct IndexCount (the numbers of indices in the index buffer of the subset you want to draw) and StartIndexLocation (the position of the first index in the index buffer of the subset you want to draw). The other draw calls have similar arguments.

struct Subset
{
uint index_count;
uint start_index;
Texture* textures;
CBuffer* cbuffers;
};

void EFFECTMESH::Render(uint num_subsets, Subset* subsets){
unsigned int stride;
unsigned int offset;

// Set vertex buffer stride and offset.
stride = sizeof(NMVertex);
offset = 0;

// Set the vertex buffer to active in the input assembler so it can be rendered.
GE->devcon->IASetVertexBuffers(0, 1, &VB, &stride, &offset);

// Set the index buffer to active in the input assembler so it can be rendered.
GE->devcon->IASetIndexBuffer(IB, DXGI_FORMAT_R32_UINT, 0);

// Set the type of primitive that should be rendered from this vertex buffer, in this case triangles.
GE->devcon->IASetPrimitiveTopology(D3D11_PRIMITIVE_TOPOLOGY_TRIANGLELIST);

for(uint i = 0; i < num_subsets; i++)
{
//Bind correct textures (subsets[i].textures)
//Bind correct cbuffers (subsets[i].cbuffers)

//Draw current subset
GE->devcon->DrawIndexed(subsets[i].index_count, subsets[i].start_index, 0);
}
}

Edited by TiagoCosta

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Thanks, I was thinking I might have to create a separate index and vertex buffer for each subset. It's not exactly how you showed, but I got it to work for me. I'll post my finished code when I can make it look better (right now it's a mess).

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Ok. It looks a little more intelligent now:

void EFFECTMESH::PrimeRenderVB(void){
unsigned int stride;
unsigned int offset;
// Set vertex buffer stride and offset.
stride = sizeof(NMVertex);
offset = 0;
// Set the vertex buffer to active in the input assembler so it can be rendered.
GE->devcon->IASetVertexBuffers(0, 1, &VB, &stride, &offset);
// Set the index buffer to active in the input assembler so it can be rendered.
//	GE->devcon->IASetIndexBuffer(IB, DXGI_FORMAT_R32_UINT, 0);//this will be done in PrimeRenderIB(..)
// Set the type of primitive that should be rendered from this vertex buffer, in this case triangles.
GE->devcon->IASetPrimitiveTopology(D3D11_PRIMITIVE_TOPOLOGY_TRIANGLELIST);
}
void EFFECTMESH::PrimeRenderIB(int index){
if (index>=NumSubsets) return;
GE->devcon->IASetIndexBuffer(IB, DXGI_FORMAT_R32_UINT, indexInfo[index].iStart*sizeof(UINT));
GE->devcon->DrawIndexed(indexInfo[index].iCount,indexInfo[index].iStart,0);
}

and to render the two subsets:

	EffectMesh[testMesh].PrimeRenderVB();

EffectMesh[testMesh].PrimeRenderIB(0);
worldMatrix=EffectMesh[testMesh].indexInfo[0].KeyframeMatrix[int(float((GE->Timer.TotalTime/1000)%500)/100.0f)];
worldMatrix(3,2)+=190-camloc.z;
worldMatrix(3,1)+=-40-camloc.y;
worldMatrix(3,0)+=-camloc.x;
if (!NMShader.Render(GE->devcon,EffectMesh[testMesh].indexInfo[0].iCount, worldMatrix, m_Camera->m_viewMatrix, projectionMatrix,TA)) tp1++;

EffectMesh[testMesh].PrimeRenderIB(1);
worldMatrix=EffectMesh[testMesh].indexInfo[1].KeyframeMatrix[int(float((GE->Timer.TotalTime/1000)%500)/100.0f)];
worldMatrix(3,2)+=190-camloc.z;
worldMatrix(3,1)+=-40-camloc.y;
worldMatrix(3,0)+=-camloc.x;
if (!NMShader.Render(GE->devcon,EffectMesh[testMesh].indexInfo[1].iCount, worldMatrix, m_Camera->m_viewMatrix, projectionMatrix,TA)) tp1++;

This works nicely for my purposes. I only "setup" the common stuff once with PrimeRenderVB() and then I render each subset with PrimeRenderIB() and the appropriate shader.

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# Comparative balance sheets for 2021 and 2020 and a statement of income for 2021 are given...

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Comparative balance sheets for 2021 and 2020 and a statement of income for 2021 are given below for Metagrobolize Industries. Additional information from the accounting records of Metagrobolize also is provided.

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We are arranging the numbers from 1 to 8 in an order so that three consecutive terms cannot be increasing. For example, 12345678 isn’t allowed but 81436572 is. How many ways are there to do it? Please don’t use computers.

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You can try this with pencil and paper, or make it a physical puzzle you can try out with your kids if you have one of those large 6x3 egg cartons laying around. Imagine you have 3 rows of 6 spots. Fill the first row with 6 red colored eggs, the second with 6 green colored eggs, and the last with...

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@Randal'Thor But Deus stays in first place lol :p

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Feb 8 at 23:03, by bobble
Quick public service announcement for the Low Quality Posts Queue:
(again)
Feb 8 at 23:03, by bobble
> Edit if the post is good quality, but you can improve the post to make it better. Avoid editing posts that should be closed or deleted, as it will remove the post from review.
Feb 8 at 23:04, by bobble
If a post is low-quality and should be deleted (e.g. breaks rules or is a blatantly worse duplicate answer), please don't just edit in LQP to add spoilers. It doesn't fix the fundamental issues.
Clicking the "Edit" button means you're saying "I can make this a completely acceptable answer, by our standards, with an edit". If the answer is unacceptable for its content, then adding spoilers does nothing to fix the content

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@msh210 yes
crossword progress: 103/180

I have a quarter-baked idea about the C4 and would like at least a few minutes to contemplate it futilely before getting a further hint.

What is the idea baking in?

(For the avoidance of doubt, almost certainly there's nothing to it.)
@bobble Cerebrospinal fluid, I guess?
Well, if my idea had any actual merit I've been unable to find it. No longer contemplating it now.

9:19 PM
@Jafe exciting!

10:04 PM
C4 hint 2: "King" appears at the start of the clue for a good reason.
@Jafe Looking forward to it.

That hint kinda fits with some of my quarter-baked pseudo-semi-pre-ideas.

11:11 PM
0

In multiplayer Tetris, there are a number of opening setups that let you send powerful attacks very quickly. One popular opening is the DT Cannon, which allows the player to very quickly send a T-Spin Double followed by a T-Spin Triple. The standard DT Cannon setup looks like this: Here's a vide...

11:36 PM
0

Someone told me that I should put a link to Fortnightly Topic Challenge #50: Escape Rooms... whatever that is, I guess. PSE, I need your help. I just got back from the worst escape room experience I've ever had in my life. Not only was the solution incredibly obvious, but it was also wrong for ...

@Sphinx Phew, got this just before the end of the challenge period


text:
# How to represent constraints for optimization problems in a data model?

I am at the moment writing a program focusing on asset allocation and I am thinking about how I should represent my constraints in the data model.

The first approach that came to mind was to define some categories to classify the constraints so that they could be stored in a table according to their "category" (for example, unary constraints x>=y, binary constraints y

I then came up with another idea which is to define my own "constraint language" with its own grammar and to store it as a string in the database (like Sum("Equities")<Percent(20,Portfolio)). This would imply writing a parser. I could also opt to use an XML representation of the constraints to use one of the many XML parsers.

I wanted to know if anyone had another potential solution and if you knew about some papers discussing this subject?

• What kind of restrictions are you talking about? For the linear restrictions you mention a matrix is the way to go. – Bob Jansen Jul 5 '11 at 17:45
• yeah they're not all linear – SRKX Jul 7 '11 at 7:04


text:
What is Root Mean Square Error RMSE in GIS?

Root Mean Square Error (RMSE) (also known as Root Mean Square Deviation) is one of the most widely used statistics in GIS. RMSE can be used for a variety of geostatistical applications.

RMSE measures how much error there is between two datasets. RMSE usually compares a predicted value and an observed value. For example, a LiDAR elevation point (predicted value) might be compared with a surveyed ground measurement (observed value).

Predicted value: LiDAR elevation value
Observed value: Surveyed elevation value

Root mean square error takes the difference for each LiDAR value and surveyed value. You can swap the order of subtraction because the next step is to take the square of the difference. (The square of a negative or positive value will always be a positive value).

After that, divide the sum of all values by the number of observations. This is how RMSE is calculated.

RMSE Formula:

$RMSE = {\sqrt {\frac{1} {N}{\sum\limits_{i = 1}^N {(x_{i} - \hat{x}_{i} } })^{2} } }$

How to calculate RMSE in Excel?

Here is a quick and easy guide to calculate RMSE in Excel. You will need a set of observed and predicted values:

1. In cell A1, type “observed value” as a title. In B1, type “predicted value”. In C2, type “difference”.

2. If you have 10 observations, place observed elevation values in A2 to A11. Place predicted values in B2 to B11.

3. In column C2, subtract observed value and predicted value: =A2-B2. Repeat for all rows below where predicted and observed values exist.

4. In cell D2, use the following formula to calculate RMSE: =SQRT(SUMSQ(C2:C11)/COUNTA(C2:C11))

Cell D2 is the root mean square error value.

RMSE quantifies how different a set of values are. Give this quick RMSE guide a try and master one of the most widely used statistics in GIS.


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# Is there a better way to factor $375007$ with out testing first $612$ primes? No calculators please

Is there a better way to factor $375007$ with out testing first $612$ primes ?

I know this factors to $31\times 12097$ by testing the primes $2,3,5,\ldots,31$. Is there any other clever way to work this ? I have tried Fermat's factorization by writing the number as $x^2-y^2$ but it is also taking too many iterations because the factors differ by large magnitude.

Also I have been trying to factor it by changing the base to 10^2 : $37x^2 + 50x+7 = (ax+b)(cx+d)$ and other bases but no success yet.

• because it is a dumb method - I wouldn't know in advance how small/large the least prime factor of a given number can be. I am looking for alternate ways to factor numbers whose factors differ by large magnitudes – pooja Jan 6 '15 at 4:35
• Not dumb, but heuristic. A dumb method wouldn't really lead you to any sort of solution. – daOnlyBG Jan 6 '15 at 4:40
• There are a lot of ways. See Wikipedia's article on integer factorization, which has a list of methods. The Pollard rho algorithm is a good method of intermediate complexity that is not too hard to implement. – MJD Jan 6 '15 at 4:41
• Well that method works always. My question was if we can do better than that and yes I am also going through wiki and getting myself confused – pooja Jan 6 '15 at 4:46
• @MJD The other nice thing about Pollard rho is that it runs (on average) faster if the least prime factor is small. In effect it beats trial division except in small cases where both run extremely quickly regardless. – Erick Wong Jan 6 '15 at 4:51

The last prime I need to know to factorize that number is $113$, the $30$th prime number. This is because $\sqrt{12097}\approx110$. After $110$, I know that any higher value to divide $12097$ would force its counterpart to be smaller: $\frac{12097}{113}\approx107$. Thus, if a number higher than $110$, like $113$, divided $12097$, I would have already found its counterpart. Once I hit $113$, I know that $12097$ is prime, because I've crossed the square root barrier.
• This is, of course, only using the naive trial division method. Both this and, if I understand correctly, Pollard rho, will not do nearly as well with $617\times619=381923$; in this case, trial division will take 113 primes (this by the way is coincidence, don't take it as meaning anything) to get anywhere. – Dan Uznanski Jan 6 '15 at 5:18


text:
2 Put in details of local calculations.

Q1) I am going to go for "no". I think it's precisely Tate's thesis that shows that the gamma factor for Riemann zeta can be interpreted as a local factor analogous to the usual local factors at the finite primes. However let me absolutely stress that you do not need to read all the technical details of Tate's thesis to understand the analogue completely. Any function on a local field (including $\mathbf{Q}_p$ and $\mathbf{R}$) has a Fourier transform, which is another such function. If you normalise things in a sane way, then the characteristic function of $\mathbf{Z}_p$ is its own Fourier transform, and there's a standard function on $\mathbf{R}$ that is its own Fourier transform. Now do a certain explicit integral to these functions---the same sort in both cases---this is in Tate's thesis. On the $p$-adic side you get $(1-p^{-s})^{-1}$ and on the real side you get the correct Gamma factor. None of this is a mystery really and is really just scratching the surface of Tate's thesis (the meat of which is the functional equation, not the definition!). I would give details In fact, let's do it now.

So $k$ is either $\mathbf{Q}_p$ or $\mathbf{R}$, and $\mu^*$ is a Haar measure on $k^*$. Now if $f$ a function on $k$, let's define$$\zeta(f,s)=\int_{k^\times}f(t)|t|^s d\mu^*$$(the integral should be over $k^\times$)Let's now compute this integral for some choices of $f$, $k$. If $k=\mathbf{Q}_p$and $f$ is the integrals, characteristic function of $\mathbf{Z}_p$ (which are also easy if you know Haar turns out to beits own Fourier transform) and $\mu^*$ is normalised to make $\mathbf{Z}_p^*$ havemeasure 1, but I'm in then the integral is (breaking up $\mathbf{Z}_p\backslash\{0\}$ into a hurry sum of $p^m\mathbf{Z}_p^\times$ for $m\geq0$) $$\zeta(f,s)=\sum_{m\geq0}p^{-ms}=(1-p^{-s})^{-1}.$$

Now let's try another example: -/let's let $f$ be $e^{-\pi x^2}$, which is its own Fourier transform, let's let Haar measure on $\mathbf{R}^\times$ be $dx/|x|$, and let's compute the integral. It's done on p317 of Cassels-Froehlich---Tate's thesis: we need to compute$$\int_{x\in\mathbf{R}}e^{-\pi x^2}|x|^{s-1} dx$$which is readily checked to be $\pi^{-s/2}\Gamma(s/2)$.

1

I am no expert, but let me give you some guesses as to the answers.

Q1) I am going to go for "no". I think it's precisely Tate's thesis that shows that the gamma factor for Riemann zeta can be interpreted as a local factor analogous to the usual local factors at the finite primes. However let me absolutely stress that you do not need to read all the technical details of Tate's thesis to understand the analogue completely. Any function on a local field (including $\mathbf{Q}_p$ and $\mathbf{R}$) has a Fourier transform, which is another such function. If you normalise things in a sane way, then the characteristic function of $\mathbf{Z}_p$ is its own Fourier transform, and there's a standard function on $\mathbf{R}$ that is its own Fourier transform. Now do a certain explicit integral to these functions---the same sort in both cases---this is in Tate's thesis. On the $p$-adic side you get $(1-p^{-s})^{-1}$ and on the real side you get the correct Gamma factor. None of this is a mystery really and is really just scratching the surface of Tate's thesis (the meat of which is the functional equation, not the definition!). I would give details of the integrals, which are also easy if you know Haar measure, but I'm in a hurry :-/

Q2) I think there is a misunderstanding here (perhaps mine, perhaps yours). The global functional equation reflects Poincare duality in the function field case. I don't really see a local functional equation for $(1-p^{-s})^{-1}$ so I don't really know what you're asking.