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# Investigating the Ebola response

Should you donate to efforts to contain the Ebola outbreak in west Africa? With hundreds of millions of dollars coming in from other donors, will your donation make a difference? How does this compare to giving to GiveWell’s top charities?

These are difficult questions. It’s always hard to estimate how much good a donation does; it’s much harder in the midst of a rapidly evolving situation like this one. It requires predicting the future path of the pandemic and the effects of response efforts. New information (and new donations) are constantly changing the picture. Further complicating matters, the people who best understand the situation are extremely busy, and we need to be careful with how we request their time. Even coming up with a rough take on Ebola involves major effort. However, at this point – due to some preliminary analysis and estimates – we are in the midst of conducting a small investigation, and hope to publish our take on donating for Ebola response within the next week or two.

In this post, we lay out the steps we’ve taken and the steps we’re planning next for our investigation. We then discuss what goes into our decisions about how to respond to sudden, prominent donation opportunities like this one, and why we’ve decided to do an investigation in this case.

Our investigation
The basic question we’re trying to answer is: what is the cost-effectiveness (in terms of lives saved and similar benefits per dollar spent) of additional donations to the Ebola response (beyond what’s already been raised, and including factors such as the risk that Ebola might spread to more countries in Africa or become endemic if not contained)?

Unfortunately, we don’t know of any published efforts to answer this question. We also don’t know of efforts to answer related questions such as “What is the expected death toll of the Ebola outbreak conditional on the current planned response, and how would this change if the response were better-funded?” The information and analysis we do have that seems most relevant is:

• A CDC model that projects Ebola deaths under different assumptions about what proportion of cases are effectively isolated. The projection goes only through January 20, and covers only Liberia and Sierra Leone. There are also some other models with broadly similar properties. We initially tried using these models, but now provisionally believe they cannot be used for our purposes (more below).
• Some basic information on the status of the UN fundraising appeal. As of now, $988 million has been requested;$486 million has been raised and an additional $233 million has been pledged. • Some basic information on the World Health Organization (WHO)’s hopes for the containment effort. A recent press briefing with a WHO representative states: “…the numbers we need to get behind are 70:70:60; that number is 70% safe burials, 70% cases being managed and cared for properly; and within 60 days of our start date which for UNMEER we’re taking as 1st October. So, our goal is to have that in place by 60 days which would be 1st December.” Initially, we tried to focus on using the CDC model to forecast Ebola cases at higher and lower levels of response efforts, which we tried to map to higher or lower levels of funding. However, we ran into several issues here. • One fundamental issue is that we know too little about the relationship between “how much money is raised” and “what sort of response is possible”: it might be that the activities most crucial to containing the epidemic can already be funded at current levels, and that additional donations would do relatively little. • Another major issue turned out to be that the CDC model already appears to be out of date (and specifically, overly pessimistic). The model incorporates data on cases through late August; reported Ebola cases since then are lower than the model predicted even in the maximal “strong response effort” scenario. It is possible that the recent reports of Ebola cases reflect issues with data collection (for example, perhaps people with Ebola are now avoiding care or healthcare workers are too overwhelmed to report data); but based purely on the numbers, we don’t feel we can use the CDC model to make good forecasts for cost-effectiveness analysis. • Even if we resolved the above two issues, there would be major questions remaining. The CDC model covers only two countries, and only through January 20; it does not address cases in Guinea, the possibility that Ebola becomes endemic, or the possibility that Ebola spreads to other countries. We know little about the organizations involved in the response effort and how well they’re performing, and it’s unlikely that we’ll be able to find out much about this question while the epidemic is ongoing. We have also experimented with using a model published more recently by the Virginia Bioinformatics Institute, but we haven’t yet determined whether this model could be useful. We haven’t been able to compare this model’s predictions to recent reports directly, but it appears to make similar projections to the CDC’s for Ebola cases conditional on strong control as of December 31. We would need a better understanding of the model, and more discussion, in order to determine whether it might be used for a cost-effectiveness estimate (but even if we did use it for such an estimate, the estimate would remain problematic for many of the reasons listed above). At this point, we’re focusing on trying to set up conversations to gain more information about the following questions: • If we were to recommend donations to the response effort, how quickly could donations be utilized on the ground? Would they make a difference to the response effort? • What would these donations allow that could not be funded otherwise? Would they expand the most important response activities? Should we think of additional donations as having similar impact to the average dollar in the response effort? • How significant is the risk that Ebola spreads to other countries and/or becomes endemic? How should we think about the likely longer-term death toll, factoring in unlikely but extremely bad scenarios? • Should we infer from recent data that the CDC model was overly pessimistic, or is there another explanation for the low (relative to the CDC model’s projections) reports of further cases? • If one donates to the response effort, whom specifically should they donate to? We’re first trying to see whether we can gain information by speaking with people who aren’t directly involved with the effort, and who can therefore take time to speak with us in a low-stakes way. If necessary, we may need to create an estimate of how much money we might be able to raise for the response, in order to give people more information about whether talking to us is worth their time. How we decide which crises to investigate When a humanitarian crisis hits the headlines, we usually get a lot of questions along the lines of “How can I help and where should I give?” At the same time, there are several reasons that headline-dominating-crises tend not to make for the best giving opportunities, and particularly tend to be a poor fit for our work. • The people best positioned to understand, and help with, Ebola response are probably the people who have been working on pandemic containment, developing-world health systems, and other related areas for years before this crisis emerged. The best opportunities to prevent or contain the epidemic were probably before it was widely recognized as a crisis (and perhaps before Ebola had broken out at all – more funding for preventive surveillance could have made a big difference). We’d guess that a similar dynamic holds in general: it takes years to build expertise and context in an area, and the most crucial opportunities to make a difference will often be before the issue is getting widespread attention. In general, we think we’ll find the best giving opportunities by picking good causes to focus on and working on them for years, not by scrambling to catch up on the state of knowledge about an urgent and chaotic situation. As it happens, biosecurity is one of our leading contenders for a focus area, and we have been actively investigating the area for a few months. One of our main focuses is on strengthening routine preventive surveillance. However, we are far from having the network and knowledge needed for a rapid diagnosis of the Ebola outbreak. • When an issue is getting a lot of media coverage, it often attracts a lot of funding. All else equal, this makes giving less attractive, since we emphasize room for more funding. In past investigations (2010 Haiti earthquake, Japan tsunami), we found evidence that money was not the limiting factor for the relief effort. • Urgent issues also tend to be particularly difficult to investigate. The people who know the most about them tend to be extremely busy, and issues tend to be more newsworthy when they are more unprecedented and chaotic. • If we do choose to investigate a crisis, we generally need to make the investigation an urgent top priority in order to keep up with developing news. That means high involvement from senior staff and major disruptions to our workflow. It can be worth it, but the costs are high. In some past crises, we have made major efforts to put out helpful content – particularly the 2010 Haiti earthquake and 2011 Japan tsunami. Our work attracted a fair amount of media coverage, and helped us formulate general principles for disaster relief giving, but it also took a lot of time and did not result in large amounts of donations (in 2011, when we covered both the Japan tsunami and the Somalia famine and recommended Doctors Without Borders for both, we tracked ~$50,000 in money moved to Doctors Without Borders; note that in these cases, we also stated that we did not feel the giving opportunity was as strong as giving to our top charities). We provided more limited coverage for the 2011 Somalia famine and choose only to provide general tips in response to the 2013 Philippines earthquake.

When a crisis starts getting coverage, we weigh factors such as (a) how many people are asking for our views and (b) how much capacity we have for an investigation, as well as (c) the likely “cost per life saved” (or similar metric) for donating to the relief effort.

In the case of the Ebola outbreak, we initially guessed that the outbreak would remain relatively contained, and that ample funding for the relief effort would come in. (High-profile donations from individuals and significant attention from governments both contributed to this view.) Recently, several things have changed:

• Over the past week, we’ve heard from more people – particularly people who follow GiveWell closely – than we had in previous weeks.
• The crisis has now been attracting significant attention, yet funding remains substantially below what has been requested.
• The crisis appears quite relevant to our ongoing investigation of preventive surveillance. Many of the people we are speaking to about surveillance are heavily involved in the Ebola response.
• In light of the above factors, we decided to put some time into a very rough estimate of what the “cost per life saved” might look like for the Ebola response. Some initial calculations indicated that the cost-effectiveness could be quite strong, consistent with the idea that containing a small number of cases now could prevent a large number of cases later. However, in light of our questions about the CDC model (among other issues), we don’t think our estimate is usable, and decided to gather more information along the lines described above.

Ebola response may be an outstanding use of funds, largely because the right preventive measures could stop the problem from becoming much larger and more costly to contain. The same logic would apply at an even earlier stage – to the strengthening of everyday preventive surveillance, of the kind that could have led to much earlier detection and containment of this epidemic. If that’s right, surveillance could turn out to be an outstanding cause to specialize in, under the heading of the Open Philanthropy Project.


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# Please Show me how to find the intercompany transactions: On October 1, 2010, Madison Ltd. acquir...

Please Show me how to find the intercompany transactions:

On October 1, 2010, Madison Ltd. acquired all the shares of Dobson Ltd. for $849,600. On that date, Dobson’s statement of financial position showed share capital of$540,000 and retained earnings of $273,600. In addition, at the acquisition date, all of Dobson’s identifiable assets and liabilities had carrying values that equaled their fair values. Madison and Dobson’s financial statements for September 30, 2014 are presented below: Statement of Financial Position As of September 30, 2014 Madison Ltd. Dobson Ltd. Assets: Current assets: Cash$    144,000               $131,400 Short-term investments 27,000 122,400 Accounts receivable 18,000 540,000 Inventory 302,400 64,800 491,400 858,600 Non-current assets: Land 126,000 216,000 Equipment, net 75,600 27,000 Investment in Dobson 849,600 ___-___ 1,051,200 243,000 1,542,600 1,101,600 Liabilities and shareholders’ equity: Current liabilities: Accounts payable 9,000 23,400 Non-current liabilities: Deferred income taxes 93,600 54,000 102,600 77,400 Shareholders’ equity: Share capital 900,000 540,000 Retained earnings 540,000 484,200 1,440,000 1,024,200$1,542,600               $1,101,600 Statement of Income For the year ended September 30, 2014 Madison Ltd. Dobson Ltd. Sales revenue$ 2,152,500              $1,670,400 Cost of sales 1,598,400 1,207,225 Gross profit 554,100 463,175 Expenses: Salaries and benefits 103,500 57,600 Amortization 9,360 8,640 Other 7,200 __-___ 120,060 66,240 Other revenues and expenses: Investment income 300 1,225 Loss on disposal of asset (1,800) __-___ 432,540 398,160 Income tax expense 173,016 213,264 Net income$ 259,524                $184,896 Statement of Changes in Equity For the year ended September 30, 2014 Madison Ltd. Dobson Ltd. Share capital, October 1, 2013$   900,000               $540,000 Changes during the year ___-___ ___-___ Share capital, September 30, 2014 900,000 540,000 Retained earnings, October 1, 2013 424,476 299,304 Net income 259,524 184,896 Dividends declared (144,000) ______ Retained earnings, September 30, 2014 540,000 484,200$ 1,440,000             $1,024,200 Additional information: Both companies use a perpetual inventory system, have a September 30 year-end, and a 30% tax rate. Madison uses the entity theory method for consolidation. On June 30, 2014, Madison sold some equipment to Dobson for$10,800. At that date, the net book value of the equipment to Madison was $12,600. The equipment is expected to have a remaining useful life of 10 years. On April 1, 2014, Madison purchased$90,000 of merchandise from Dobson. Dobson had acquired the goods for $54,000. On July 15, Madison sold half of the goods to a customer for$50,400. The remaining goods were still in Madison’s inventory at its 2014 fiscal year-end.

At October 1, 2013, Madison had some goods in inventory that it had purchased from Dobson at May 25, 2013. The profit on these goods was $10,800. These goods were sold by December 31, 2013. In 2011, Madison sold a tract of land to Dobson for an accounting gain of$36,000. Dobson plans to build a warehouse and office complex on the land in 2015.

Required:

Prepare Madison’s consolidated financial statements for the year ended September 30, 2014. (Round numbers to the nearest dollar, and show all your calculations.)

format: plain


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R survival analysis
1
3
Entering edit mode
8.6 years ago
hdy ▴ 160

I am now learning how to do survival analysis in R and using COX proportional hazards model, which can be referred to the 'coxph' function under package 'survival'. When I looked some online examples, eg, here, I found sometimes the code looks like

Surv(time, censor) ~ factor(hercoc)

But I can also do

Surv(time, censor) ~ hercoc

I was wondering what is the difference between these two and when shall I use which one. And how should I interpret the result from the one used 'factor'

R survival • 14k views
0
Entering edit mode

In R the factor data format should be used for categorical data. For example, if you were doing survival analysis for three different treatments

treatments<- c(1,2,3)

Then you should pass this vector as a factor because the data are categorical. If you did not do this then R would assume the data are continuous and might cause misinterpretations of the results.

On the other hand, if the treatment was of one drug but at different concentrations such as

treatment<- c(0,1,1.5,2)

Then you should not factor these data because they are continuous.

At least that's my understanding, others please chime in

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Entering edit mode
8.6 years ago
russhh 5.7k

If your variable hercoc has only two levels then there is no difference. However, if it has 3 or more levels then there is a difference. You haven't provided any example data, and I am assuming that hercoc is numeric.

Using a more concrete example:

library(survival)

attach(lung)

#     inst time status age sex ph.ecog ph.karno pat.karno meal.cal wt.loss
#1      3  306      2  74   1       1       90       100     1175      NA
#2      3  455      2  68   1       0       90        90     1225      15
#3      3 1010      1  56   1       0       90        90       NA      15
#4      5  210      2  57   1       1       90        60     1150      11

The variable sex has two levels but is coded as a numeric variable here (1 for male, 2 for female or vice versa).

summary(as.factor(sex))
1   2
138  90

There's only one coefficient fitted by the model regardless of how you write it:

coxph(Surv(time, status) ~ sex)

# Call:
coxph(formula = Surv(time, status) ~ sex)

coef exp(coef) se(coef)     z      p
sex -0.531     0.588    0.167 -3.18 0.0015

Likelihood ratio test=10.6  on 1 df, p=0.00111  n= 228, number of events= 165

coxph(Surv(time, status) ~ as.factor(sex))
Call:
coxph(formula = Surv(time, status) ~ as.factor(sex))

coef exp(coef) se(coef)     z      p
as.factor(sex)2 -0.531     0.588    0.167 -3.18 0.0015

Likelihood ratio test=10.6  on 1 df, p=0.00111  n= 228, number of events= 165

For the variable ph.ecog there are 4 levels

summary(as.factor(ph.ecog))
#   0    1    2    3 NA's
#  63  113   50    1    1

On fitting the survival model against ph.ecog it really does make a difference whether the variable enters as a numeric or a factor. If treated numerically, only a single coefficient is fitted (for a given individual, the value for ecog is multiplied by this coefficient before entering into the coxph calculation);

coxph(Surv(time, status) ~ ph.ecog)
Call:
coxph(formula = Surv(time, status) ~ ph.ecog)

coef exp(coef) se(coef)   z       p
ph.ecog 0.476      1.61    0.113 4.2 2.7e-05

Likelihood ratio test=17.6  on 1 df, p=2.77e-05  n= 227, number of events= 164
(1 observation deleted due to missingness)

However, treated as a factor, three different coefficients will be fitted, one for each non-reference level (ie, levels 1 2 and 3 each have a coefficient) and for a given individual you would look up the coefficient corresponding to the level of the ecog factor.

> coxph(Surv(time, status) ~ as.factor(ph.ecog))
Call:
coxph(formula = Surv(time, status) ~ as.factor(ph.ecog))

coef exp(coef) se(coef)    z       p
as.factor(ph.ecog)1 0.369      1.45    0.199 1.86 6.3e-02
as.factor(ph.ecog)2 0.916      2.50    0.225 4.08 4.5e-05
as.factor(ph.ecog)3 2.208      9.10    1.026 2.15 3.1e-02

Likelihood ratio test=18.4  on 3 df, p=0.000356  n= 227, number of events= 164
(1 observation deleted due to missingness)

Look into how the coefficients enter the survival model in a good Generalised linear model book (I really can't explain that quickly for you)

0
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I also found the "strata" function in survival analysis. What is the different between this "strata" and "factor"？

0
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no idea I'm afraid. Might be better as another question

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For strata, it means stratification. If you stratify on ph.ecog, you will have no coef estimation on that variable, but that variable has been adjusted in the model. I think you'd better check the distribution on that variable to see how many samples in each stratum of that variable first.


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# Timothée Poisot

## Computational Ecologist

2012/01/01Goodbye Wordpress, hello Jekyll!

A few weeks ago, I discovered Jekyll. In a nutshell, Jekyll is a static site generator who promised me I’ll be able to blog like a hacker. This site is now hosted on GitHub, and both the template and the content can be forked, so everything is basically open source.

Jekyll is a lot of fun to use, especially because all of the content reside in static text files instead of a database. It’s also fully compatible with the markdown format, which is a lot easier to write than html. For example, writing a simple text in bold in markdown is done through

**text in bold**

rather than the equivalent html

text in bold

So a less lot keys to press (especially on my new QWERTY keyboard). The fact that Jekyll uses the Liquid templating language also make the overall design process fun. In addition, the whole system is working on GitHub, so I have complete version control on everything. This is so awesome!


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# Homework Help: Variation of Parameter Problem

1. Feb 20, 2009

### bengaltiger14

1. The problem statement, all variables and given/known data

Find the general solution using the method of variation of parameters of:

y''-6y'+9y=(x^-3)(e^3x)

I found the roots of the corresponding homogeneous equation to be lamba = 3. So there are repeated roots. My question is, how do I solve a variation of parameter question with repeated roots? I know how to do it using reduction of order but confused on variation of parameters

2. Feb 20, 2009

### HallsofIvy

The same way you solve it without repeated roots! Since x= -3 is a double root of the characteristic equation, two independent solutions to the associated homogeneous equation are $e^{-3x}$ and $xe^{-3x}$.

Look for a solution to the entire equation of the form $y(x)= u(x)e^{-3x}+ v(x)xe^{-3x}$. Now just follow the usual procedure for variation of parameters.


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# Help with this operator

eljose79
let be the non-linear operator acting this way:

K(f)=g K(f)=(Int(c,x)f(t)dt)**r

where c and r are real constant and f is unknown.

in fact can we relate K(f) and solve it by applying the fractional derivatives?..in fact can be K(f) expressed as combination of fractional derivatives?..

nbo10
Originally posted by eljose79
let be the non-linear operator acting this way:

K(f)=g K(f)=(Int(c,x)f(t)dt)**r

where c and r are real constant and f is unknown.

in fact can we relate K(f) and solve it by applying the fractional derivatives?..in fact can be K(f) expressed as combination of fractional derivatives?..

And what are fractional derivatives?

JMD

You mean partial derivatives?


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# BrokenPowerLaw1D¶

class astropy.modeling.powerlaws.BrokenPowerLaw1D(amplitude=1, x_break=1, alpha_1=1, alpha_2=1, **kwargs)[source]

One dimensional power law model with a break.

Parameters: amplitude : float Model amplitude at the break point. x_break : float Break point. alpha_1 : float Power law index for x < x_break. alpha_2 : float Power law index for x > x_break.

Notes

Model formula (with $$A$$ for amplitude and $$\alpha_1$$ for alpha_1 and $$\alpha_2$$ for alpha_2):

$\begin{split}f(x) = \left \{ \begin{array}{ll} A (x / x_{break}) ^ {-\alpha_1} & : x < x_{break} \\ A (x / x_{break}) ^ {-\alpha_2} & : x > x_{break} \\ \end{array} \right.\end{split}$

Attributes Summary

 alpha_1 alpha_2 amplitude input_units This property is used to indicate what units or sets of units the evaluate method expects, and returns a dictionary mapping inputs to units (or None if any units are accepted). param_names x_break

Methods Summary

 evaluate(x, amplitude, x_break, alpha_1, alpha_2) One dimensional broken power law model function fit_deriv(x, amplitude, x_break, alpha_1, …) One dimensional broken power law derivative with respect to parameters

Attributes Documentation

alpha_1
alpha_2
amplitude
input_units

This property is used to indicate what units or sets of units the evaluate method expects, and returns a dictionary mapping inputs to units (or None if any units are accepted).

Model sub-classes can also use function annotations in evaluate to indicate valid input units, in which case this property should not be overridden since it will return the input units based on the annotations.

param_names = ('amplitude', 'x_break', 'alpha_1', 'alpha_2')
x_break

Methods Documentation

static evaluate(x, amplitude, x_break, alpha_1, alpha_2)[source]

One dimensional broken power law model function

static fit_deriv(x, amplitude, x_break, alpha_1, alpha_2)[source]

One dimensional broken power law derivative with respect to parameters


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rethinking programming language tags

This question asks whether programming language tags such as [java], [c++], or [python] are on topic at CS stack exchange. The answer says that questions about the programming languages themselves may be on topic but questions about writing in those languages aren't.

That answer is 5 years old and while I agree with it in theory, looking at questions with those tags reveals a graveyard of programming questions that should not have been asked here (and generally are very low quality). The top questions tagged [java], [c++], and [python] are generally not specific to one particular language. In the few cases that they are, they would also be fine under more general tags like [programming-languages] or [type-theory] and the language-specific tag seems unnecessary. For each of these tags, the majority of questions have 0 or fewer votes and a large number have negative votes.

I posit that the presence of those tags gives newbies the impression that their programming questions are on topic here. Even if there are theoretically good questions for these tags, in practice these tags are magnets for bad questions. Given the data, do we really think these tags are a net positive for the site?

• Please flag all programming questions for closing, and edit away programming-language tags if they are irrelevant. – Raphael Oct 25 '18 at 16:36
• "Given the data, do we really think these tags are a net positive for the site?" -- no, certainly not. But they make offtopic questions easier to find, or so the reasoning goes. We can never prevent people from asking crappy questions, and people posting programming questions have already shown they haven't looked at tags, tag descriptions, questions, or the help section of this site. They won't be deterred by not finding their tags; they'd just use others. – Raphael Oct 25 '18 at 16:38
• – Raphael Oct 30 '18 at 22:49
• I think they are beneficial, even if they only help marking likely off-topic content as such. They could be even more beneficial if they triggered a warning. – chi Jan 28 '19 at 21:26

We do have some related data that points to the answer being “no”. Even on Stack Overflow where the tags exists and is obviously a core tag for the site's topic, a sizable minority of askers don't pick a language tag, but some other tag that is less relevant or totally unrelated. For example, as I write, there are 2 non-deleted questions in the question2answer tag — those people have a question to answer. That's just from the past 9 days (older questions are likely to be automatically deleted and doesn't include manually-deleted questions. Back when there was a homework tag, a lot of questions got posted with just this tag. Tags like question had to be blacklisted (even though that tag name could make sense for NLP). Here on CS.SE, we get programming questions with tags like , or just some choice of tag that makes absolutely no sense.
• I completely agree. Further, the programming-languages tag is already abused too much, for off-topic programming questions. I'm subscribed to the tag, since I enjoy PL theory, and it can be a bit annoying to find highlighted off-topic questions. Retagging helps, but does not solve the issue. Removing the tags for specific PLs would only make programming-languages be abused even more, without preventing the issue. – chi Jan 28 '19 at 21:19


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## Glassy dynamics in the East model

### Abstract

We study the dynamics of the East model, comprising a chain of uncoupled spins in a downward-pointing field. Glassy effects arise at low temperatures $T$ from the kinetic constraint that spins can only flip if their left neighbour is up. We give details of our previous solution of the non-equilibrium coarsening dynamics after a quench to low $T$ (Phys. Rev.\ Lett. 83:3238, 1999), including the anomalous coarsening of down-spin domains with typical size $\bar{d} \sim t^{T \ln 2}$, and the pronounced fragile glass'-divergence of equilibration times as $t_*=\exp(1/T^2\ln 2)$. We also link the model to the paste-all coarsening model, defining a family of interpolating models that all have the same scaling distribution of domain sizes. We then proceed to the problem of equilibrium dynamics at low $T$. Based on a scaling hypothesis for the relation between timescales and lengthscales, we propose a model for the dynamics of superdomains' which are bounded by up-spins that are frozen on long timescales. From this we deduce that the equilibrium spin correlation and persistence functions should exhibit identical scaling behaviour for low $T$, decaying as $g(\tilde{t})$. The scaling variable is $\tilde{t}=(t/t_*)^{T\ln 2}$, giving strongly stretched behaviour for low $T$. The scaling function $g(\cdot)$ decays faster than exponential, however, and in the limit $T\to 0$ at fixed $\tilde{t}$ reaches zero at a {\em finite} value of $\tilde{t}$.Comment: 17 pages, revtex4, 8 figures. Minor clarifications, plus new figure 6 to provide simulation support for the notion of superdomain dynamic

Topics: Condensed Matter - Statistical Mechanics, Condensed Matter - Disordered Systems and Neural Networks
Publisher: 'American Physical Society (APS)'
Year: 2003
DOI identifier: 10.1103/PhysRevE.68.031504
OAI identifier: oai:arXiv.org:cond-mat/0303318

### Suggested articles

To submit an update or takedown request for this paper, please submit an Update/Correction/Removal Request.


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Please show work 5a and 5b .Thanks

Image text transcribed for accessibility: The half rectified cosine wave x(t) = max{/4cos( omega t).0} and its time-shifted version z(t) = x(t - t0) are shown below What is the fundamental period omega 0 and the amplitude A. and also the time-delay . The trigonometric Fourier series for x(t) can be shown to be Using this and the properties of the Fourier series provided, find the trigonometric Fourier series for z(t). The steps to do so are follows: Compute the coefficients of the exponential series for x(t)from the trigonometric series. Use Fourier series properties to compute the exponential coefficient of the shifted signal z(t), and Compute the trigonometric coefficients from the exponential coefficients for z(t).


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# Discrete Fourier

For natural number $$N>1$$, the Discrete Fourier is operator $$\hat ふ_N$$ acting on $$N$$-vector $$A$$ in such a way that

$$\!\!\!\!\!\!\!\!\!\!\! (1) ~ ~ ~ \displaystyle (\hat ふ_{N} A)_n = \sum_{m=0}^{N-1} ふ_{n,m} A_m$$

While deal with Fourier operators, it is convenient to begin numeration with zero. Elements of matrix $$ふ$$ are

$$\!\!\!\!\!\!\!\!\!\!\! (2) ~ ~ ~ \displaystyle ふ_{m,n}=\frac{1}{\sqrt{N}} \exp\! \left( \frac{- \mathrm i ~2 ~\pi}{N}~m~n\right)$$

The Fourier matrix $$ふ$$ is symmetric and unitary;

$$\!\!\!\!\!\!\!\!\!\!\! (3) ~ ~ ~ \displaystyle ふ_{m,n}= ふ_{n,m}$$
$$\!\!\!\!\!\!\!\!\!\!\! (4) ~ ~ ~ \displaystyle {\hat ふ}^*= {\hat ふ}^{-1}$$

in such a way, if $$A=\hat ふ B$$, then $$B={\hat ふ}^* A$$; so, $${\hat ふ}^\dagger={\hat ふ}^*$$.

## Different notations

Sometimes, the operator that deviate from $$\hat ふ$$ with factor $$\sqrt{N}$$ is considered; for such modified operator, the equation (4) does not hold, but the mofified operator is still called "Discrete Fourier" [1]. Often, such an operator is denoted with latin letter $$F$$. (Sometimes the special font is used to avoid the confusion with parameters and functions.)

In order to avoid confusions, in TORI, the hiragana [2] character ふ is used for the (normalized) Fourier operator, so, here, $$ふ_N$$ character denotes "DiscreteFourier with $$N$$ points" and nothing else.

## Numeric implementation of the discrete Fourier

The numeric implementation of the Discrete Fourier is especially efficient when $$N=2^k$$ for some natural number $$k$$. For this case, the example of implementation is suggested below:

/* //Example of headers
#include<math.h>
#include<stdio.h>
#include <stdlib.h>
#include <complex>
using namespace std;
#define z_type complex<double>
*/
// The "miminal" Fast Fourier Transform by Dmitrii Kouznetsov
// n should be 2^m for entire m ; o should be 1 or -1 ;

void fft(z_type *a, int N, int o) // Arry is FIRST!
{z_type u,w,t;  int i,j,k,l,e=1,L,p,q,m;
q=N/2;  p=2; for(m=1;p<N;m++) p*=2;
p=N-1;  z_type y=z_type(0.,M_PI*o); j=0;
for(i=0;i<p;i++){ if(i<j) { t=a[j]; a[j]=a[i]; a[i]=t;}
k=q; while(k<=j){ j-=k; k/=2;}
j+=k; }
for(l=0;l<m;l++)
{ L=e; e*=2; u=1.; w=exp(y/((double)L));
for(j=0;j<L;j++)
{  for(i=j;i<N;i+=e){k=i+L; t=a[k]*u; a[k]=a[i]-t; a[i]+=t;}
u*=w;
} }
}

## Approximation of the Fourier operator

Numerical approximation of the Fourier transform of function $$~f(x)=\exp(−x^2/2)x^2(−3\!+\!x^2)~$$ with $$N\!=\!16$$

For the continuous square-integrable functions, the discrete approximation with $$N$$ points can be written with uniform mesh at pointd $$x_n$$, $$n=0..N$$, such that

$$\displaystyle x_N=(n-N/2) \sqrt{\frac {2\pi}{N}}$$

Let $$f_n=f(x_n)$$ then at the points $$x_n$$, the Fourier-transform $$g=\mathcal F f$$ can be approximated with the sum:

$$\displaystyle g(x_n)= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \exp(\mathrm i x_n y) f(y) \mathrm d y \approx g_n= \frac{1}{\sqrt{N}} \sum_{n=0}^{N-1} \exp(\mathrm i x_n x_m) f_m$$

Then, at large $$N\gg 1$$, the array $$g_n$$ approximates $$g(x_n)$$. Using the routine fft for the discrete Fourier above, the numerical implementation of Fourier operator with the formula above may have the following form:

void fafo(z_type *b,int N,int o)
{ int j; z_type c; double q=sqrt(1./N);
for(j=0;j<N/2;j++) {c=b[j];b[j]=b[j+N/2];b[j+N/2]=c;}
fft(b,N,o);
for(j=0;j<N/2;j++) {c=b[j];b[j]=b[j+N/2]*q;b[j+N/2]=c*q;}
}

where $$b$$ is array to store the function and its Fourier, $$N$$ is number of points, and $$o=1$$ for the direct Fouried and $$o=-1$$ for the conjugated (inverse) Fourier.

As an example, the figure at right shows discrete approximation of the self-Fourier function $$~f(x)=\exp(−x^2/2)x^2(−3\!+\!x^2)~$$ at the mesh with $$N\!=\!16$$. Values $$f(x_n)$$ and $$g(x_n)$$ almost coincide; they are shown with the segmented line. The difference scaled with factor $$100$$ is shown with the saw-like line that oscillates in vicinity of zero.

## Approximation of the Fourier-series

Consider, for continuous function $$f$$ and given $$N$$, array

$$F_n=f\left(\frac{2 \pi}{N} n\right)$$

Then, at large $$N$$ the interpolation of array $$F$$ can be considered as approximation of function $$f$$, and its Discrete Fourier gives an approximation of the Fourier coefficients

$$\!\!\!\!\!\!\!\!\!\! \displaystyle g_n=\frac{1}{2 \pi} \int_0^{2\pi} \exp\left( \mathrm i n x\right) f(x) \mathrm d x \approx \displaystyle \frac{1}{2 \pi} \sum_{m=0}^{N-1} \exp\left(- \mathrm i n \frac{2 \pi}{N} m \right) \frac{2 \pi}{N}= \frac{1}{N} \sum_{m=0}^{N-1} \exp\left(\frac{2 \pi \mathrm i}{N} mn \right) = \frac{1}{\sqrt{N}} (\hat ふ_N F)_n$$

in the Fourier series

$$\!\!\!\!\!\!\!\!\!\!\displaystyle f(x)= \sum_{n=0}^{\infty} ~ g_n \exp( i n x)$$

In such a way, the discrete Fourier operator can be used for approximation of both, the Fourier transform and the coefficients of the Fourier series. Similar approach can be used for other integral transforms; in particular, for the CosFourier and the SinFourier operators.

## Application of the Discrete Fourier

The discrete implementation of the Fourier transform is useful for the numerical analysis of solutions of linear equations and, in particular, those of the paraxial approximation in wave optics. [3][4][5].

## References

1. http://en.wikipedia.org/wiki/Hiragana
2. http://mizugadro.mydns.jp/PAPERS/josa1f0.pdf D.Kouznetsov, J.V.Moloney, E.M.Wright. Efficiency of pump in the double-clad fiber amplifiers. 1. Fiber with circular symmetry. -- J.O.S.A. B, June 2001, V.18, No.6, p.743-749.
3. http://mizugadro.mydns.jp/PAPERS/josa2f0.pdf D.Kouznetsov, J.V.Moloney. Efficiency of pump in the double-clad fiber amplifiers. 2. Broken circular symmetry. -- J.O.S.A. B, V.19, No.6, p.1259-1263 (2002).
4. http://mizugadro.mydns.jp/PAPERS/josa3f0.pdf D.Kouznetsov, J.V.Moloney. Efficiency of pump absorption in double-clad fibers amplifiers. 3. Calculation of modes -- J.O.S.A. B, V.19, No.6, p.1304-1309 (2002).


text:
# Question

Black and Blue, Inc., overlooked \$100,000 of inventory at the end of the current year because it was stored temporarily in a warehouse owned by another company. Before discovering this error, the company’s income statement showed the following:
Sales . . . . . . . . . . . . . . . . . . . . . . \$990,000
Cost of goods sold . . . . . . . . . . (560,000)
Gross profit . . . . . . . . . . . . . . . \$430,000
Restate these figures to reflect the inclusion of the overlooked inventory.

Sales0
Views88


text:
# Math Help - Trouble Finding Derivative

1. ## Trouble Finding Derivative

I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

$\frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]$

Also, in case it matters, $d=24$ and $w=6$, but in the example, the derivative is found with respect to the variables rather than their respective values.

My attempt:

$[(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]$

$[(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

$[(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

From this point, I could do stuff like factor out a $2(x+d)$, but it looks like mine won't be very easy to solve for zero.

This is what the derivative is supposed to be, which is easier to solve for zero:

$(x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)$

What am I doing incorrectly?

Thanks!

2. ## Re: Trouble Finding Derivative

Originally Posted by Algebrah
I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

$\frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]$

Also, in case it matters, $d=24$ and $w=6$, but in the example, the derivative is found with respect to the variables rather than their respective values.

My attempt:

$[(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]$

$[(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

$[(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

From this point, I could do stuff like factor out a $2(x+d)$, but it looks like mine won't be very easy to solve for zero.

This is what the derivative is supposed to be, which is easier to solve for zero:

$(x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)$

What am I doing incorrectly?

Thanks!
1. You have introduced and extra factor of two in the first term erroneously.

2. The second term should be:

$(x+d)^2 \frac{d}{dx}\left( \frac{w^2}{x^2}+1\right) =(x+d)^2 \frac{(-2)w^2}{x^3}$

CB

3. ## Re: Trouble Finding Derivative

Originally Posted by Algebrah
I've been working on optimization problems and have been having problems with this one. The only thing that I'm confused about is what's going on when I find the derivative of the function:

$\frac{d}{dx}[(x+d)^2(\frac{w^2}{x^2} +1)]$

Also, in case it matters, $d=24$ and $w=6$, but in the example, the derivative is found with respect to the variables rather than their respective values.

My attempt:

$[(2)(x+d) \cdot (1+1) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx^2-[2xw^2]}{x^4})]$
Where did that "(1+1)"? Surely you didn't differentiate "(x+ d)" and get "(1+ 1)"? "d" is a constant. The derivative of x+ d with respect to x is 1. Also you appear to have tried to write $\frac{w^2}{x^2}+ 1= \frac{w^2+ x^2}{x^2}$ and use the quotient rule- but you have done that incorrectly. Better is to write $\frac{w^2}{x^2}+ 1= w^2x^{-2}+ 1$ so the derivative is $-2w^2x^{-3}= \frac{-2w^2}{x^3}$.

$[(2)(x+d) \cdot (2) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

$[(4)(x+d) \cdot (\frac{w^2}{x^2} +1)] + [(x+d)^2 \cdot (\frac{2wx-2w^2}{x^3})]$

From this point, I could do stuff like factor out a $2(x+d)$, but it looks like mine won't be very easy to solve for zero.

This is what the derivative is supposed to be, which is easier to solve for zero:

$(x+d)^2(-\frac{2w^2}{x^3})+(\frac{w^2}{x^2}+1)(2)(x+d)$

What am I doing incorrectly?

Thanks!

4. ## Re: Trouble Finding Derivative

Perhaps if I hadn't neglected the $\frac{d}{dx}$ in front, things would've come together a lot quicker! Lesson learned.

Thanks a lot for the help!


text:
# BarotropicQGQL

### Basic Equations

This module solves the quasi-linear quasi-geostrophic barotropic vorticity equation on a beta plane of variable fluid depth $H - h(x, y)$. Quasi-linear refers to the dynamics that neglect the eddy–eddy interactions in the eddy evolution equation after an eddy–mean flow decomposition, e.g.,

$$$\phi(x, y, t) = \overline{\phi}(y, t) + \phi'(x, y, t) ,$$$

where overline above denotes a zonal mean, $\overline{\phi}(y, t) = \int \phi(x, y, t) \, 𝖽x / L_x$, and prime denotes deviations from the zonal mean. This approximation is used in many process-model studies of zonation, e.g.,

As in the SingleLayerQG module, the flow is obtained through a streamfunction $\psi$ as $(u, v) = (-\partial_y \psi, \partial_x \psi)$. All flow fields can be obtained from the quasi-geostrophic potential vorticity (QGPV). Here, the QGPV is

$$$\underbrace{f_0 + \beta y}_{\text{planetary PV}} + \underbrace{\partial_x v - \partial_y u}_{\text{relative vorticity}} + \underbrace{\frac{f_0 h}{H}}_{\text{topographic PV}} .$$$

The dynamical variable is the component of the vorticity of the flow normal to the plane of motion, $\zeta \equiv \partial_x v - \partial_y u = \nabla^2 \psi$. Also, we denote the topographic PV with $\eta \equiv f_0 h / H$. After we apply the eddy-mean flow decomposition above, the QGPV dynamics are:

\begin{aligned} \partial_t \overline{\zeta} & + \mathsf{J}(\overline{\psi}, \underbrace{\overline{\zeta} + \overline{\eta}}_{\equiv \overline{q}}) + \overline{\mathsf{J}(\psi', \underbrace{\zeta' + \eta'}_{\equiv q'})} = \underbrace{- \left[\mu + \nu(-1)^{n_\nu} \nabla^{2n_\nu} \right] \overline{\zeta} }_{\textrm{dissipation}} , \\ \partial_t \zeta' &+ \mathsf{J}(\psi', \overline{q}) + \mathsf{J}(\overline{\psi}, q') + \underbrace{\mathsf{J}(\psi', q') - \overline{\mathsf{J}(\psi', q')}}_{\textrm{EENL}} + \beta \partial_x \psi' = \underbrace{-\left[\mu + \nu(-1)^{n_\nu} \nabla^{2n_\nu} \right] \zeta'}_{\textrm{dissipation}} + F . \end{aligned}

where $\mathsf{J}(a, b) = (\partial_x a)(\partial_y b) - (\partial_y a)(\partial_x b)$. On the right hand side, $F(x, y, t)$ is forcing (which is assumed to have zero zonal mean, $\overline{F} = 0$), $\mu$ is linear drag, and $\nu$ is hyperviscosity. Plain old viscosity corresponds to $n_{\nu} = 1$. The sum of relative vorticity and topographic PV is denoted with $q \equiv \zeta + \eta$.

Quasi-linear dynamics neglect the term eddy-eddy nonlinearity (EENL) term above.

### Implementation

The equation is time-stepped forward in Fourier space:

$$$\partial_t \widehat{\zeta} = - \widehat{\mathsf{J}(\psi, q)}^{\textrm{QL}} + \beta \frac{i k_x}{|𝐤|^2} \widehat{\zeta} - \left ( \mu + \nu |𝐤|^{2n_\nu} \right ) \widehat{\zeta} + \widehat{F} .$$$

The state variable sol is the Fourier transform of vorticity, ζh.

The Jacobian is computed in the conservative form: \mathsf{J}(f, g) = \partialy [ (\partialx f) g]

• \partialx [ (\partialy f) g]. The superscript QL on the Jacobian term above denotes that

triad interactions that correspond to the EENL term are removed.

The linear operator is constructed in Equation

GeophysicalFlows.BarotropicQGQL.EquationFunction
Equation(params, grid)

Return the equation for two-dimensional barotropic QG QL problem with parameters params and on grid. Linear operator $L$ includes bottom drag $μ$, (hyper)-viscosity of order $n_ν$ with coefficient $ν$ and the $β$ term:

$$$L = - μ - ν |𝐤|^{2 n_ν} + i β k_x / |𝐤|² .$$$

Nonlinear term is computed via calcN! function.

source

and the nonlinear terms are computed via

GeophysicalFlows.BarotropicQGQL.calcN!Function
calcN!(N, sol, t, clock, vars, params, grid)

Calculate the nonlinear term, that is the advection term and the forcing,

$$$N = - \widehat{𝖩(ψ, q+η)} + F̂ .$$$
source

which in turn calls calcN_advection! and addforcing!.

### Parameters and Variables

All required parameters are included inside Params and all module variables are included inside Vars.

For decaying case (no forcing, $F = 0$), vars can be constructed with DecayingVars. For the forced case ($F \ne 0$) the vars struct is with ForcedVars or StochasticForcedVars.

### Helper functions

GeophysicalFlows.BarotropicQGQL.updatevars!Function
updatevars!(sol, vars, params, grid)
updatevars!(prob)

Update the vars of a problem prob that has grid and params with the solution in sol.

source

### Diagnostics

The kinetic energy of the fluid is obtained via:

while the enstrophy via:

Other diagnostic include: dissipation, drag, and work.

## Examples

• examples/barotropicqgql_betaforced.jl: A script that simulates forced-dissipative quasi-linear quasi-geostrophic flow on a beta plane demonstrating zonation. The forcing is temporally delta-correlated and its spatial structure is isotropic with power in a narrow annulus of total radius $k_f$ in wavenumber space. This example demonstrates that the anisotropic inverse energy cascade is not required for zonation.


text:
# Defintion of Curvature

1. May 14, 2008

### maverick280857

Hi.

I am reading "An Introduction to Modern Astrophysics" by Carroll and Ostlie, for a summer project. In section 27.3 (Relativistic Cosmology) the curvature of a sphere is given by

$$6\pi \frac{C_{exp}-C_{meas}}{C_{exp}A_{exp}}$$

The situation is as follows:

Consider a sphere of radius R. An ant is moving on the sphere at a fixed polar angle $\theta$. The ant measures the circumference $C_{meas} = 2\pi R\sin\theta$ whereas the expected value of circumference is $C_{exp} = 2\pi D$ where $D = R\theta$. The expected area of the circle is $A_{exp} = \pi D^2$.

I am not sure how the above expression leads to the curvature of the sphere...

Any thoughts?

Thanks.
Cheers
Vivek.

Last edited: May 14, 2008
2. May 14, 2008

### Mentz114

My understanding of this is that the ant walks in a circle, not a great circle. In this case the expression you quote gives the curvature of the sphere. Imagine the ant walking is a circle on a flat surface, then let the flat surface curve. The ratio of the diameter to the radius changes.

3. May 15, 2008

### maverick280857

Mentz114, thanks for your reply. I'm not clear about the way the curvature has been written in terms of the two circumferences and the area, and also the $6\pi$ factor. How does all that come in?

4. May 15, 2008

### pmb_phy

That relation is teh Gaussian Curvature for a two dimensional surface. Gaussian curvature is not defined for spaces higher than two.

Pete

5. May 15, 2008

### Mentz114

maverick280857 , if you want a derivation, check out Wiki on 'Gaussian curvature'. There's even an expression in terms of Christoffel symbols.

6. May 16, 2008

### maverick280857

Thanks, I'm still learning these things and I don't know much about them. The page you have referred to does not explain the origin of the particular definition of curvature used here. But I will keep looking.

7. May 16, 2008

### maverick280857

Ok, so I guess its a special case of the expression given there, for 2 dimensions.

8. May 16, 2008


text:
# zbMATH — the first resource for mathematics

On the large order asymptotics of general states in semiclassical quantum mechanics. (English) Zbl 0791.35113
Summary: We consider the limit $$\hslash\to 0$$ of the solution $$\Phi(t,x,\hslash)$$ of the Schrödinger equation: $i\hslash {{\partial\Phi(t,x,\hslash)} \over {\partial t}}=- {{\hslash^ 2} \over {2m}} {{d^ 2 \Phi(t,x,\hslash)} \over {dx^ 2}} +V(x) \Phi(t,x,\hslash).$ We prove that, for any integer $$l\geq 2$$ and any initial condition $$\Phi(0,x,\hslash)$$ that belongs to the Schwartz-class, a solution $$\Phi^*(t,x,\hslash)$$ of the semiclassical equation approximates $$\Phi(t,x,\hslash)$$ such as $\|\Phi^*(t,\cdot,\hslash)- \Phi(t,\cdot,\hslash)\|_{L^ 2} \leq C\hslash^{1/2} \qquad (\hslash\to 0)$ .

##### MSC:
 35Q40 PDEs in connection with quantum mechanics 81Q20 Semiclassical techniques, including WKB and Maslov methods applied to problems in quantum theory
##### Keywords:
Schrödinger equation; semiclassical equation
Full Text:
##### References:
 [1] G.A. Hagedorn , Semiclassical Quantum Mechanics, I: The h \rightarrow 0 Limit for Coherent States , Commun. Math. Phys. , T. 71 , 1980 , pp. 77 - 93 . Article | MR 556903 · minidml.mathdoc.fr [2] G.A. Hagedorn , Semiclassical Quantum Mechanics. III: The Large Order Asymptotics and More General States , Ann. Phys. , T. 135 , 1981 , pp. 58 - 70 . MR 630204 [3] G.A. Hagedorn , Semiclassical Quantum Mechanics, IV: The Large Order Asymptotics and more General States in more than One Dimension , Ann. Inst. H. Poincaré , T. 42 , 1985 , pp. 363 - 374 . Numdam | MR 801234 | Zbl 0900.81053 · Zbl 0900.81053 · numdam:AIHPA_1985__42_4_363_0 · eudml:76287 [4] S. Robinson , The Semiclassical Limit of Quantum Dynamics, I: Time Evolution , J. Math. Phys. , T. 29 , 1988 , pp. 412 - 419 . MR 927028 | Zbl 0647.46060 · Zbl 0647.46060 · doi:10.1063/1.528029 [5] S. Robinson , The Semiclassical Limit of Quantum Dynamics, II: Scattering Theory , Ann. Inst. H. Poincaré , T. 48 , 1988 , pp. 281 - 296 . Numdam | MR 969167 | Zbl 0666.35071 · Zbl 0666.35071 · numdam:AIHPA_1988__48_4_281_0 · eudml:76401
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Yaamini’s Notebook: Skyline Error Checking Round 2

## This time it only took 4 hours!

Due to a combination of not working late at night, using the same proteins and transitions as last time, and knowing what good and bad peaks look like, I was able to spot check 100 peptides in the revised Skyline document. Because I was in a rush to finish and head to the airport, I didn’t take any side-by-side comparison pictures. I do however, have revised error rates.

Ovearll, I think the new .blib improved Skyline peak selection. I haven’t compared error rates so I don’t have anything quantitative to base my assertion on, but I have a few observations:

• More instances of a peak being chosen correctly for all replicates
• Way less missing peaks
• Fewer noisy peak selections

The previous error checking document is here, revised error checking document can be found here.

After error checking, I re-exported my Skyline reports.

Proteins and peak areas only

MSstats document, replicate not pivoted

from the responsible grad student http://ift.tt/2sPPH59
via IFTTT


text:
## 37.5% of what number is 6? HELP PLS I NEED IT BAD SHOW WORK PLS!

Question

37.5% of what number is 6? HELP PLS I NEED IT BAD SHOW WORK PLS!

in progress 0
1 year 2021-08-29T20:32:32+00:00 2 Answers 16 views 0

16

Step-by-step explanation:

Let the number be x.

37.5% * x = 6

Convert percentage to decimal form:

$$0.375x=6$$

Multiply both sides by 1000:

$$375x=6000$$

Divide both sides by 375:

$$\frac{375x}{375} =\frac{6000}{375}$$

$$x=16$$

16

Step-by-step explanation:

Let the number be x.

37.5% * x = 6

Convert percentage to decimal form:

Multiply both sides by 1000:

Divide both sides by 375:

x= 16

Im literally copying but im okay with that.


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

The first ever RStudio conference was held January 11-14, 2017 in Orlando, FL. For anyone else like me who spends hours each working day staring into an RStudio session, the conference was truly excellent. The speaker lineup was diverse and covered lots of areas related to development in R, including the tidyverse, the RStudio IDE, Shiny, htmlwidgets, and authoring with RMarkdown.
This is not a complete list by any means — with split sessions I could only go to half the talks at most. Here are some noncomprehensive notes and links to slides and resources for some of the awesome things are doing with R and RStudio that I learned about at the RStudio Conference.

Hadley Wickham kicked off the meeting with a keynote on doing data science in R. The talk focused on the tidyverse, and the notion of splitting functions into commands that do something, as compared to queries that calculate something, and how it’s generally a good idea to keep these different functionalties contained in their own separate functions. (Contrast this to things like lm that both computes values and does things, like printing those values to the screen, making it difficult to capture (see broom).
I asked Hadley after his talk about strategies to reduce issues getting Bioconductor data structures to play nicely with tidyverse tools. Within minutes David Robinson released a new feature in the fuzzyjoin package that leverages IRanges within this tidyverse-friendly package for efficiently doing things like joining on genomic intervals.
Charlotte Wickham’s 2-hour purrr tutorial was awesome. Here’s a link to a shared dropbox folder with code, challenges, slides, data, etc. The purrr package is a core package in the tidyverse, and I’ll be replacing many of the base ?apply and plyr ??ply functions that I still use here and there. The map_* functions are integral to working with nested list-columns in dplyr, and I think I’m finally starting to grok how to work with these.
Jenny Bryan gave a great talk on list columns. You can see her slides here. Jenny also put together this excellent tutorial with lots of worked examples and code snippets. And if you need some example list data structures for more practice or for teaching that aren’t foo/bar/iris/mtcars-level boring, see her repurrrsive package. Related to this, for more on list columns and purrr map functions, start reading at the “Many Models” section of Hadley’s R for Data Science book.
Julia Silge, data scientist at Stack Overflow, gave a great introduction to tidy text mining with R. You can read Julia and David’s Tidy Text Mining with R book here online (the book was authored in Rmarkdown using bookdown!).
Andrew Flowers, data journalist and former writer at FiveThirtyEight gave the second day’s keynote address on finding and telling stories using R. He gave a series of examples illustrating six motivating features that make data stories worth telling, along with potential danger inherent to each one:
1. Novelty (potential danger: triviality)
3. Archetype (oversimplification)
4. Trend (variance)
5. Debunking (confirmation bias)
6. Forecast (overfitting)
Yihui Xie led a two-hour tutorial on advanced RMarkdown. You can see his slides here. The rticles package has LaTeX Journal Article Templates for R Markdown for various journals. The tufte package now supports both PDF and HTML output. See an example here. Yihui’s xaringan package ports the remark.js library for slideshows into R. Careful. Yihui warns that you may not sleep after learning about how cool remark.js is. Yihui showed an early version of the in-development blogdown package that can build blog-aware static websites using the blazing-fast and well-documented Hugo static site generator. Finally, the bookdown package is just awesome. It takes multiple RMarkdown documents as input and renders into multiple output formats (screen-readable ebook, PDF, epub, etc.). It looks great for writing books and technical documentation with pushbutton publishing to multiple output formats with some nice built-in styles out of the box. Some examples:
Finally, a few gems from other talks that I jotted down:
• Chester Ismay gave a great talk on teaching introductory statistics using R, with the open-source course textbook written in RMarkdown using bookdown.
• Bob Rudis talked about using pipes (%>%), and pipes within pipes, and best piping practices. See his slides here.
• Hilary Parker talked about the idea of an analysis development, (and analysis developers), drawing similarities to software development/developers. Hilary discussed this once before on the excellent podcast that she and Roger Peng host, and you can probably find it in their Conversations On Data Science ebook that summarize and transcribe these conversations.
• Simon Jackson introduced corrr package for exploring and manipulating correlations and correlation matrices in a tidy way.
• Gordon Shotwell introduced the easymake package that generates Makefiles from a data frame using R.
• Karthik Ram quickly introduced several of the (many) rOpenSci packages related to data publication, data access, scientific literature access, scalable & reproducible computing, databases, visualization, taxonomy, geospatial analysis, and many utility tools for data analysis and manipulation.

With split sessions I missed more than half the talks. Lots of people here are active on Twitter, and you can catch many more notes and tidbits on the #rstudioconf hashtag. The meeting was superbly organized, I learned a ton, and I enjoyed meeting in person many of the folks I follow on Twitter and elsewhere online. A few days of 80-degree weather in mid-January didn’t hurt either. I’ll definitely be coming again next year. Kudos to the rstudio::conf organizers and speakers!
All the talks were recorded and will supposedly find their way to rstudio.com at some point soon. I’ll update this post with a link when that happens.


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# Why is 600 out of 1000 more convincing than 6 out of 10?

Look at this excerpt from "The study skills handbook", Palgrave, 2012, by Stella Cottrell, page 155:

Percentages Notice when percentages are given.

60% of people preferred oranges; 40% said they preferred apples.

This looks convincing: Numerical quantities are given. But is the difference between 60% and 40% significant? Here we would need to know how many people were asked. If 1000 people were asked of whom 600 preferred oranges, the number would be persuasive. However, if only 10 people were asked, 60% simply means 6 people preferred oranges. "60%" sounds convincing in a way that "6 out of 10" does not. As a critical reader, you need to be on the lookout for percentages being used to make insufficient data look impressive.

What is this characteristic called in statistics? I would like to read more about it.

• sample size matters – Aksakal Jul 28 '17 at 17:57
• I pick two people at random, they are both male, and therefore I conclude that 100% of Americans are male. Convincing? – Casey Jul 28 '17 at 19:06
• It's the "Don't compare apples with oranges" principle – wolfies Jul 28 '17 at 19:31
• To approach that question from a different angle you may consider digging the literature of the framing effect. However, it is an example of cognitive bias and is a psychological topic, not statistical. – Larx Jul 30 '17 at 2:46
• You can imagine a difference of 1 how much it will impact the estimated quantity. 7/10 is much further relatively from 6/10 than 601/1000 is from 600/1000. – mathreadler Jul 31 '17 at 15:22

I would like to list another intuitive example.

Suppose I tell you I can predict the outcome of any coin flip. You do not believe and want to test my ability.

You tested 5 times, and I got all of them right. Do you believe I have the special ability? Maybe not. Because I can get all of them right by chance. (Specifically, suppose the coin is a fair coin, and each experiment is independent, then I can get all rights with $0.5^5\approx0.03$ with no super power. See Shufflepants's link for a joke about it).

On the other hand, if you tested me large number of times, then it is very unlikely that I can get it by chance. For example, if you tested $100$ times, the probability of me getting all of them right is $0.5^{100}\approx 0$.

The statistical concept is called statistical power, from Wikipeida

The power of a binary hypothesis test is the probability that the test correctly rejects the null hypothesis (H0) when the alternative hypothesis (H1) is true.

Back to the super power on coin flip example, essentially you want to run a hypothesis testing.

• Null hypothesis (H0): I do not have the super power
• Alternative hypothesis (H1): I have the super power

Now as you can see in the numerical example (test me 5 times vs test me 100 times), The statistical power has been affected by the sample size.

More to to read here. (more technical and based on t-test).

An interactive tool to understand statistical power can be found here. Note, the statistical power changes with the sample size !

• Obligatory XKCD – Shufflepants Jul 28 '17 at 20:57
• This does not actually answer the question. The question has two parts: "Why is [it] more convincing…" and "What is this characteristic called in statistics?" You provided an example that asks the reader if they are more convinced by the example, but you have not explained the phenomenon, nor attempted to address why it's more convincing (just relying on the reader's intuitive reasoning on a single example). In addition, you haven't touched on the second question: If you believe this phenomenon doesn't have a name, then explicitly state that. – Makyen Jul 29 '17 at 17:48
• I can't help but feel like 100% is a special case and this answer doesn't really support my intuition that 600/1000 is more convincing than 6/10. Even if we take it to not be a special case, this is essentially just restating the problem with a different percentage. – NotThatGuy Jul 30 '17 at 23:14
• @Juya it is a web based tool built by someone. You can try it here – Haitao Du Aug 8 '17 at 15:59

Think about it in terms of proportions. Let's say that preferring an orange is a success, while preferring an apple is a failure. So your mean success rate is $\mu = \frac{\text{# of sucesses}}{n}$ or in this case .6

The standard error of this quantity is estimated to be $\sqrt{\frac{\mu(1-\mu)}{n}}$. For a small sample size (i.e. 10), the standard error is $\approx .155$ but for a sample size of 1000, the standard error is $\approx .0155$. So basically, as was mentioned in the comments, "sample size matters."

This concept is a consequence of the law of large numbers. From Wikipedia,

According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.

Results from a small sample may be farther from the expected value than that from a larger sample. And so, as stated in the question, one should be cautious of results calculated from small samples. The idea is also explained pretty well in this youTube video.

• There are several well-known theorems known in statistics as "laws of large numbers," but none of them make statements that look like the one in the question. How, then, do you establish the connection? – whuber Jul 28 '17 at 18:28
• From wikipedia en.wikipedia.org/wiki/Law_of_large_numbers, "According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed". Results from a small sample may be farther from the expected value than that from a larger sample. And so, as stated in the question, one should be cautious of results calculated from small samples. – Evan Phibbs Jul 28 '17 at 18:50
• That's a good explanation, thank you. Although you are correct to write that one should be cautious about just how many a "large number of trials" needs to be for the law to be trusted, the application is intuitively sound. I would like to suggest that your explanation belongs in your answer, where it would be more widely read and appreciated, rather than buried in a comment. Link-only answers (such as your link to Youtube) without explanation don't last here. – whuber Jul 28 '17 at 19:02

We're in the situation of estimating some population quantity by some sample quantity. In this case, we're using sample proportions to estimate population proportions, but the principle is considerably more general.

If you think of all the observations in your sample taking the value $1$ when they have the characteristic of interest ("preferred oranges to apples" in the example) and $0$ when they don't, then the proportion of $1$'s is the same as the mean of the set of $0$ and $1$ values -- so you can readily see that a sample proportion is actually a mean.

As we take larger and larger samples (using random sampling), the sample means will tend to converge to the population mean. (This is the law of large numbers.)

However what we really want to have some idea of is how far out we might be (such as might be represented by the width of a confidence interval for the proportion, or by the margin of error, which is normally half of such a width).

Typically, the more data you have, the less uncertainty you will have about some quantity like a mean -- because the standard deviation of the distribution of the sample mean decreases as you take larger samples. [Imagine taking means of many different samples of size 4. The distribution of those means is less variable than the distribution of the original observations-- the standard deviation should tend to be about half as big. Now if you take means of many different samples of size 400, the standard deviation of that should be much smaller again (about $\frac{_1}{^{20}}$th of the standard deviation of the original observations).

The standard deviation of the distribution of the sample mean is one way to measure the typical distance a sample mean is from the population mean, which is decreasing (it decreases as $\frac{_1}{^\sqrt{n}}$, as in the examples above).

As a result, we're more confident about the accuracy of our estimate when the sample is large -- if we repeated our experiment again, other such means would be close to the current one -- they cluster together more and more tightly, and because (in this case) our estimate is unbiased, they're clustering together around the values we're trying to estimate. A single sample mean becomes more and more informative about where the population mean might be.

A rule of thumb for "counting" statistics, like counting the number of people that like oranges, or counting the number of "clicks" in a Geiger counter due to radioactive decay, is that the margin of error for the count is roughly the square-root of the expected count value. Counting statistics are known are Poisson statistics.

The square root of 6 is 2.4-ish, so the margin of error is about 40% (2.4/6). The square root of 600 is 24-ish, so the margin of error is about 4% (24/600). That is why having counted 600 is more significant that counting 6. The relative error is one-tenth.

I'm being a little sloppy about the definition of margin of error. It's really the 1-sigma value, and is not a hard cut-off, but it is the range where you expect most (68%) of the measurements to lie. So if you expect 6 orange eaters, you would expect a series of polls to give you mostly numbers in the 4 to 8 range, like 6,6,5,6,7,2,4,6,3,5,6,6,7,6,10,8,6,5,6,6,9,3,7,8.

I don't have the name you're looking for, but the issue isn't statistical. Psychologically, the way humans process numbers in our brains give greater weight (authority) to larger numbers than it does to smaller numbers because the magnitude (physical size) is visually as important as the representative value. Thus, 600/1000 appears more credible than 6/10. This is why shoppers prefer to see "10% Off!" for values less than 100 and "Save $10!" for values over 100 (called the "Rule of 100"). It's about how our brains react to perception. An amazing look into this and similar kinds of phenomena are discussed by Nick Kolenda in his online treatise, "An Enormous Guide to Pricing Psychology". • Though other answer are not incorrect, I do not see them actually correctly addressing the quoted text as this is. The text is addressing the impact of presenting the numbers and the perceived accuracy, not if the numbers actually are more accurate. That is, that you can make a person perceive that your information is more significant by saying 600 of 1000 or 60% or you can cause is to seem less significant by saying 6 of 10 even though each mean the same thing by implying a smaller sample size without if actually being stated, or even possibly true. – dlb Jul 28 '17 at 22:17 • This example seems fundamentally flawed in my opinion. For example a purchase of 100 gets a %10 discount for a total of 90 while a 100.01 purchase gets a discount of 25 for a total of 75.01 which results is a vastly different price. In fact there is a difference in values until you get up to 250.05(or 250.10 depending on rounding or truncating). The question is talking about sample size and standard error while your example is more about an actual difference because of our perception. – Joe W Jul 29 '17 at 14:19 • @Joe W, while I can't speak to the specifics of the "rule of 100" (not being a psychologist), the point my answer is making is that psychologists have determined that people trust a larger number, not because of the larger sample size it represents, but because of the perception of the greater importantance given to a larger number. As an engineer, I'd prefer it not be this way --- but it's the way the human mind works. Should you wish to challenge this, I'd recommend reading the underlying details of Mr. Kolenda's presentation. – JBH Jul 29 '17 at 14:53 • You are missing my point, there is a bigger difference in the 10% to 100 and 25 it is not just a matter of perception. The simple fact is if you took both values at 100 there is a 15% difference which is a value that is much simpler to measure then standard error when talking two different samples one of 10 and one of 1000. – Joe W Jul 29 '17 at 15:18 • I changed my question to use 10% and$10 to help you out. The point of the "rule of 100" is that people perceive larger numbers as having greater relevance and perceive monetary value as more imporant than a mere percentage. It's an issue somewhat outside the scope of the discussion for the OP's purposes. – JBH Jul 29 '17 at 15:20

While the actual margin of error is important, the reason it sounds more convincing is because of a more heuristic (rule of thumb) experience with people. The actual margin of error confirms this heuristic has merit.

If the sample is 6 for, and 4 against, this could be 50/50 if a single person changes their vote, or a single person was recorded in error. There is only two more people on the 6 side. Everybody knows two flakes, everybody knows the sample could be cherry-picked: You only asked waitresses and nobody else. Or you only polled 10 college professors in the halls of a university. Or you asked 10 wealthy people outside of Saks Fifth Avenue.

Even the mathematical margin of error presumes true randomness and doesn't account for selection bias, or self-selection bias, or anything else, people can intuitively understand that.

In contrast, the 600 vs. 400 result has 200 more people on one side than the other, and 100 people would have to change their mind. Those numbers are very hard to come by (but not impossible) by some accident of where you were polling, how you got people to agree, how individuals understood or interpreted the question, etc.

It is more convincing not because of a mathematical proof that it should be, but because we know from experience that crowds of 1000 are much more likely to be diverse in their opinions (on anything) than a group of 10. (unless you secretly did your polling at a political party convention or a KKK rally or something else likely to draw a one-sided crowd).

The math only precisely quantifies what we already know by intuition; that it is easier to randomly encounter one or two stray votes out of 10, than it is to randomly encounter 100 or 200 stray votes out of 1000.

Something that has not been mentioned is to look at the problem from a Bayesian point of view.

In a Bayesian setting, a natural approach to this problem would be to use a Beta-Binomial distribution. You can assume that the probability of someone preferring oranges over apples is $p$, which Beta distributed, and that the observations are binomially distributed with parameter $p$: $$p \sim \mathrm{Beta}(\alpha, \beta)\\ n_o|p \sim \mathrm{Bin}(n,p).$$

Let's assume that you have no a-priori reason to believe that more people prefer oranges over apples or vice-versa ($\beta=\alpha$) but also that you have no strong opinion about this (weak prior: $\beta=\alpha=1$). The prior distribution of $p$ is therefore uniform $\mathrm{U}(0,1)$.

After collecting answers from $n$ questionnaires about people's preferences, you note that $n_o$ respondents prefer oranges and $n_a=n-n_o$ of them prefer apples.

The posterior distribution of $p$ is: $$p|n_o,n_a \sim \mathrm{Beta}(n_o+1, n_a+1).$$

While the mode of the posterior of $p$ (i.e. the maximum-a-posteriori) is $n_o/(n_o+n_a)$ regardless of the number of respondents, the distribution itself is very different: it is much more peaked for large $n$ than for small ones.

To give you an idea, this is the posterior with $n_o=6$ and $n_a=4$:

While this is the posterior with $n_o=600$ and $n_a=400$:

How do you read these plots? You can reason as follows: "I observe that 6 out of 10 people (randomly chosen from the population) prefer oranges over apples but could the true underlying probability (for the whole population) be 0.4 or 0.8 instead? Well, according to the first plot this is quite possible." If you do the same for the second plot (i.e. with 1000 respondents), you get that $p=0.4$ or $p=0.8$ are very very unlikely (again, I am assuming the 1000 are IID samples from the population).

Please note that although these plots look similar to david25272's, they represent something very different.

His plots ask the question: "assuming a given value of $p$ known, what is the probability of observing $n_o$ people responding that they prefer oranges over apples?"

My plots answer the question: "assuming that I observe $n_o$ people responding that they prefer oranges over apples, what is the probability distribution of $p$, the probability of people preferring oranges over apples?"

Basically it's more convincing to have 600 out of 1000 than six out of 10 because, given equal preferences it's far more likely for 6 out of 10 to occur by random chance.

Let's make an assumption - that the proportion who preferred oranges and apples are actually equal (so, 50% each). Call this a null hypothesis. Given these equal probabilities the likelihood of the two results are:

• Given a sample of 10 people, there is a 38% chance of randomly getting a sample of 6 or more people who prefer oranges (which is not all that unlikely).
• With a sample of 1000 people there is less than 1 in a billion chance of having 600 or more out of 1000 people prefer oranges.

(For simplicity I am assuming an infinite population from which to draw an unlimited number of samples).

## A simple derivation

One way to derive this result is to simply list out the potential ways in which people can combine in our samples:

For ten people it's easy:

Consider drawing samples of 10 people at random from an infinite population of people with equal preferences for apples or oranges. With equal preferences it's easy to simply list all the potential combinations of 10 people:

Here's the full list.

r   C (n=10)    p
10  1       0.09766%
9   10      0.97656%
8   45      4.39453%
7   120     11.71875%
6   210     20.50781%
5   252     24.60938%
4   210     20.50781%
3   120     11.71875%
2   45      4.39453%
1   10      0.97656%
0   1       0.09766%
1024    100%

r is the number of results (people who prefer oranges), C is the number of possible ways of that many people preferring oranges, and p is the resulting discrete probability of that many people preferring oranges in our sample.

(p is just C divided by the total number of combinations. Note that there are 1024 ways of arranging these two preferences in total (i.e. 2 to the power of 10).

• For instance there is only one way (one sample) for 10 people (r=10) to all prefer oranges. The same is true for all people preferring apples (r=0).
• There are 10 different combinations resulting in nine of them preferring oranges. (One different person prefers apples in each sample).
• There are 45 samples (combinations) where 2 people prefer apples, etc, etc.

(In general we talk about n C r combinations of results r from a sample of n people. There are online calculators you can use to verify these numbers.)

This list allows us to gives us the probabilities above using just division. There is a 21% chance of getting 6 people in the sample that prefer oranges (210 of 1024 of the combinations). The chance of getting six or more people in our sample is 38% (the sum of all the samples with six or more people, or 386 out of 1024 combinations).

Graphically, the probabilities look like this:

With larger numbers, the number of potential combinations grows rapidly.

For a samples of just 20 people there are 1,048,576 possible samples, all with equal likelihood. (Note: I have only shown every second combination below).

r    C (n=20)   p
20   1          0.00010%
18   190        0.01812%
16   4,845      0.46206%
14   38,760     3.69644%
12   125,970    12.01344%
10   184,756    17.61971%
8    125,970    12.01344%
6    38,760     3.69644%
4    4,845      0.46206%
2    190        0.01812%
0    1          0.00010%
1,048,576  100%

There is still only one sample where all 20 people prefer oranges. Combinations that feature mixed results are much more likely, simply because there are many more ways that the people in the samples can be combined.

Samples that are biased are much more unlikely, just because there are fewer combinations of people that can result in those samples:

With just 20 people in each sample, the cumulative probability of having 60% or more (12 or more) people in our sample preferring oranges drops to just 25%.

The probability distribution can be seen to become thinner and taller:

With 1000 people the numbers are huge

We can extend the above examples to larger samples (but the numbers grow too rapidly for it to be feasible to list all the combinations), instead I have calculated the probabilities in R:

r   p (n=1000)
1000    9.332636e-302
900     5.958936e-162
800     6.175551e-86
700     5.065988e-38
600     4.633908e-11
500     0.02522502
400     4.633908e-11
300     5.065988e-38
200     6.175551e-86
100     5.958936e-162
0       9.332636e-302

The cumulative probability of having 600 or more out of 1000 people prefer oranges is just 1.364232e-10.

The probability distribution is now much more concentrated around the center:

[

(For example to calculate probability of exactly 600 out of 1000 people preferring oranges in R use dbinom(600, 1000, prob=0.5) which equals 4.633908e-11, and the probability of 600 or more people is 1-pbinom(599, 1000, prob=0.5), which equals 1.364232e-10 (less than 1 in a billion).

This is because higher number ensures greater accuracy. For ex, if u would pick up 1000 random people from anywhere on the planet and 599 of them are male against 10 random people with 6 male, the former would be more accurate. Similarly, if you assume a population of 7 billion and calculate the number of males, you would get a more precise number which would obviously be more convincing than with just 1000 people.


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# Recent history for Johnny Assay

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# Complex sets of the form $|f(z)| = c$

I'm looking for information on sets of the form $|f(z)| = c$, possibly with additional restrictions (e.g. $|z| < d$). I can probably also assume that $f$ is as nice as you'd like.

I'm interested in things like:

Into how many regions does the curve split the plane?

When is $|f(z)| = c$ path connected?

When is $|f(z)| \leq c$ bounded?

When is $|f(z)| \leq c$ simply connected?

If $(f_n)$ is a sequence of functions, what is the limiting behavior of $|f_n(z)| = c$?

If you could point me toward any references I'd be very grateful. I'd also be interested in references where specific cases are treated, just so I can get an idea of the common techniques.

I know this question is pretty broad but I'm not looking for a particular answer. Any information you could provide would be a great help.

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An interesting case: $|\sin z| = 1/2$. –  GEdgar Aug 30 '11 at 23:58

Assume $c > 0$, and $f$ a nonconstant entire function. $\{z: |f(z)| \le c\}$ is bounded if and only if $f$ is a polynomial. Any bounded component of $\{z: |f(z)| \le c\}$ must contain a zero of $f$. Conversely, if $z_0$ is a zero of $f$, then for $c$ sufficiently small there is a bounded component of $\{z: |f(z)| \le c\}$ in a neighbourhood of $z_0$. On the other hand, $\{z: |f(z)| > c\}$ can't have any bounded components, so any component of $\{z: |f(z)| \le c\}$ is simply connected.

$\{z: |f(z)|=c\}$ is locally a smooth curve except at points where $f' = 0$. If $f'(p) = 0$ and $|f(p)| = c$, then at $p$ there are $k$ smooth curves intersecting, with angles of $\pi/k$ from one to the next, where $f^{(j)}(p) = 0$ for $1 \le j < k$ and $f^{(k)}(p) \ne 0$.

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Thanks Robert, this is great. Is there a reference I should see for properties like these? –  Antonio Vargas Aug 31 '11 at 2:14
@Robert: What is a bounded component? –  Weltschmerz Aug 31 '11 at 3:37
@Weltschmerz: a connected component that is bounded. –  Robert Israel Aug 31 '11 at 18:17

If $f$ is a polynomial of degree $n$, the number of components of $\{z: |f(z)| \le c\}$ or $\{z: |f(z)| = c\}$ is $n$ minus the number of zeros $p$ of $f'(z)$ such that $|f(p)| \le c$, counted by multiplicity. For example, consider $f(z) = z^4 + z^3 + 1$. $f'(z) = 4 z^3 + 3 z^2$ has a zero of multiplicity 2 at 0, where $|f(0)|=1$, and a zero of multiplicity 1 at $-3/4$, where $|f(-3/4)| = 229/256$. So there are 4 components if $0 \le c < 229/256$, 3 if $229/256 \le c < 1$, and 1 if $c \ge 1$.

Here's an animation of the curve $\{z: |f(z)| = c\}$ in this example, as $c$ varies from $1/2$ to $3/2$. Notice two components merging at $c = 229/256$ and three merging at $c = 1$.

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# B Does right/left spin reverse in tunneling particles?

1. Feb 15, 2017

### fredt17

As I understand it, a relativistic reversal can occur in a Lorentz
boost. A particle's forward angular momentum can become backward
angular momentum, if the observer accelerates sufficiently. But the
particle and its angular momentum are unchanged. The only change
is in the observer’s frame of reference.

Could tunneling particles similarly reverse right/left angular
momentum from a change in the observer’s frame of reference?

The basis for this question is that a spinning object passing through
a surface twice reverses its apparent direction of spin, at least to an
observer on the surface.

Quantum tunneling maybe a result of quantum uncertainty, or
a wormhole, a loop, a tunnel, or some other mechanism. But this
reversal effect does not depend on the mechanism. This effect is
determined solely by the observer’s frame of reference.

This effect can be seen by poking a spinning pencil through a
folded sheet of paper. At one intersection with the paper,
the pencil will be spinning clockwise. At the other intersection, the
pencil will be spinning counter-clockwise, relative to the surface.

But there is only one pencil spinning in one direction. The apparent
direction of spin depends solely on which end of the tunnel (or wormhole
or other mechanism) the observer is looking at. It is purely a relativistic
effect.

So perhaps a Lorentz-boost-like relativistic reversal could occur in
tunneling particles. Has an experiment ever been conducted to test for
such an effect?

2. Feb 15, 2017

### Staff: Mentor

Yes--and you have to be careful describing what the effect actually is. Note that this effect is not particular to relativity; it also occurs in Newtonian mechanics. The apparent reversal is of the particle's spin relative to its direction of motion. It's just a consequence of changing the observer's motion relative to the object: if the object is moving in the positive $x$ direction, for example, in the original frame, and spinning clockwise about an axis pointing along its direction of motion, then in the new frame, it will be moving in the negative $x$ direction, and spinning counterclockwise about an axis pointing along its (new) direction of motion. But the change is not in its spin, only in its direction of relative motion, because of the change in the observer's frame.

More precisely, relative to vectors normal to the surface on the same side of it. This is a different kind of effect than what you were describing before: it has nothing to do with changing frames, at least not as you're describing it. It has to do with the change in orientation of the surface from one intersection to the other.

Not if "relativistic" here is intended to mean the same thing it meant earlier in your post. See above.

Also, the "tunnel" here has nothing whatever to do with quantum tunneling; it's just a surface intersecting an object twice. So the real answer to your question is no, the kind of effect you are talking about is not going to appear in quantum tunneling.

3. Feb 16, 2017

### fredt17

Thanks for your response. Could you clarify one point?

The apparent reversal is of the particle's spin relative to its direction of motion.
It's just a consequence of changing the observer's motion relative to the object:

Wouldn't the effect occur independent of the observer's motion? If Bob
is at one intersection on the surface, and Alice is at the other, wouldn't
their observations be different?

4. Feb 16, 2017

### Staff: Mentor

Which effect? What you quoted from me (please use the PF quote feature, btw) was referring to your first example, which is an effect due to a change in relative motion. What you are asking about in the last paragraph of your latest post is referrring to your second example. The effect in your second example depends on the direction that Alice and Bob view as the "positive" $x$ direction. This is a change in frame of reference, just not one due to relative motion (it's due to a spatial rotation). Neither of your examples describes any change in the object itself or its angular momentum.

5. Feb 16, 2017

### fredt17

One question. Whatever is happening in quantum tunneling that creates the reversal in apparent angular momentum
is this similar to the opposing right/left angular momentum in entangled particles?
Is it possible that entangled particles are connected by the same mechanism that
allows particles to appear to tunnel?

6. Feb 17, 2017

### Staff: Mentor

None of your examples are examples of quantum tunneling doing this. Can you give an example of quantum tunneling doing this?

Entanglement and tunneling are both quantum processes, but they're not the same quantum process. So I think the answer to this is no.

7. Feb 18, 2017

### fredt17

Perhaps I am confused (certainly a definite possibility). My question related to the way tunneling and quantum entanglement may share
a common attribute of reversed right/left angular momentum . Maybe they don't, but if they do, this may be from a similar source,
namely that the two effects are a result of connections that pass through the surface of our world twice.

8. Feb 18, 2017

### Staff: Mentor

Tunneling does not reverse right/left angular momentum.

I'm not even sure what this means--what is "the surface of our world"? But in any case it doesn't appear to apply to either tunneling or entanglement. It doesn't even apply to the original case you brought up in your OP, of a Lorentz boost.

9. Feb 18, 2017

### fredt17

Well, how do entangled particles interact instantaneously? There appears to be some method of transmission. And if that interaction takes zero time,
then there may be zero distance between the particles, from the particles frame of reference. That could be a tunnel or a wormhole or some other mechanism,
but the effect on left/right angular momentum would be the same. The two intersections would display opposite directions of spin.

10. Feb 18, 2017

### Staff: Mentor

They don't.

11. Feb 19, 2017

### fredt17

I hope you will consider Maldacena's article on entangled black holes in the November 2016 Scientific American.
His theory is based on entangled particles and he suggests entangled black holes could be connected by
a wormhole, and thus could interact to some extent.
His illustration shows a wormhole looping beneath a surface and reconnecting with the surface
at a different location. If an object entered that wormhole spinning clockwise, it would emerge at the
other wormhole spinning counter-clockwise.

12. Feb 19, 2017

Staff Emeritus
After 11 messages, I still can't figure what this thread is about.

13. Feb 19, 2017

### ZapperZ

Staff Emeritus
This is highly speculative! You need to show the physics that support this.

Spin-dependent tunneling is a known process. It resembles nothing like you mentioned here.

Zz.

14. Feb 19, 2017

### fredt17

Sorry for the confusion. Let me try again.

As suggested in Maldacena's article in Scientific American, entangled black holes may connect two different locations in
our universe through a wormhole.

This concept might also be applicable to quantum particles as quantum entanglement was the foundation of Maldacena's theory.
If so, a particle passing through such a wormhole could reverse its angular momentum.

Evidence in support of this idea might be found in possible reversals in tunneling particles. Has an experiment
ever been conducted to test for this possibility?
.

15. Feb 19, 2017

### ZapperZ

Staff Emeritus
Sorry, what evidence is there to connect this to tunneling particles? The particles undergoing tunneling do NOT go through "a wormhole". Do a search on "inelastic tunneling" where I've posted rebuttals to the idea that the particles do not pass through the barrier.

Again, spin-dependent tunneling is so well-known, it is not even funny.

Zz.

16. Feb 19, 2017

### fredt17

You could be right. Quantum tunneling is not what we know as tunneling.
It is only a label.

But forward/backward quantum momentum reverses in a Lorentz boost without any interaction
with the particle or its angular momentum.

So however the particle is getting from one place to another, a change in the observer's frame
of reference (as occurs in a Lorentz boost) could produce a similar reversal in right/left
angular momentum.

This is very speculative (to put it mildly), but it seems simple enough to test for..
.
.

17. Feb 19, 2017

### ZapperZ

Staff Emeritus
Then the title of this thread and your starting premise is faulty. This has nothing to do with the tunneling phenomenon.

Zz.

18. Feb 19, 2017


text:
Square Coins

Time Limit: 1000 ms Memory Limit: 32768 KiB

Problem Description

People in Silverland use square coins. Not only they have square shapes but also their values are square numbers. Coins with values of all square numbers up to 289 (=17^2), i.e., 1-credit coins, 4-credit coins, 9-credit coins, ..., and 289-credit coins, are available in Silverland.

There are four combinations of coins to pay ten credits:

ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin.

Your mission is to count the number of ways to pay a given amount using coins of Silverland.

Input

The input consists of lines each containing an integer meaning an amount to be paid, followed by a line containing a zero. You may assume that all the amounts are positive and less than 300.

Output

For each of the given amount, one line containing a single integer representing the number of combinations of coins should be output. No other characters should appear in the output.

Sample Input

2
10
30
0

Sample Output

1
4
27

Hint

hdoj1398 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。

Source

Asia 1999, Kyoto (Japan)


text:
# Arrange the digits from 0 to 9 into a number which is divisible by every integer from 12 to 21

Arrange the digits from 0 to 9 into a number which is divisible by every integer from 12 to 21 inclusive.

The numbers are

2137458960 and 6179584320

I found these results by calculating the least common multiple $\operatorname{lcm}(12,\dotsc 21) = 21162960$, then checking all multiples of this value up to 9876543210 with a script: https://jsfiddle.net/qdnyLfx8/

There must be more elegant solutions that work without using a computer.

• Well then I won't post the 2nd number. – Joel Rondeau Aug 16 '16 at 4:11
• The prime factors of 12 to 21 are 24×3×5×7×13×17×19, therefore the two 10-digit pandigital numbers are a multiple of 7054320. 2137458960 [7054320×303] and 6179584320 [7054320×876]. – Jamal Senjaya Aug 16 '16 at 4:29
• @JamalSenjaya Based on GOTO 0's LCM calculation you seem to have missed a factor of 3 somewhere. – user253751 Aug 16 '16 at 8:46

If I take all the factors of 12-21 and multiply them together, I get

5 * 7 * 9 * 13 * 16 * 17 * 19, or
21162960

So now all that is left to do is to keep adding it to itself until I get a 10 digit number that satisfies the requirements. The first such number is

2137458960


text:
Step 4 make the decision and state the conclusion for

• 7

This preview shows page 4 - 6 out of 7 pages.

Step 4: Make the decision and State the conclusion for the test If the test statistic falls within the rejection region, we reject H 0 . We have sufficient/significant evidence to support the alternative hypothesis. If the test statistic does NOT fall within the rejection region, then we fail to reject H 0 . There is insufficient/no significant evidence to support the alternative hypothesis. STAT3502 CU 24 / 41
t test for one population mean Inference for one population mean when σ is Unknown Example – Fast Food A fast food restaurant claims that the average waiting time in their drive through is less than one minute. We record the drive-thru waiting times for a random sample of 31 customers. The sample average time is 54.9 seconds and the sample standard deviation is 17.6 seconds. Waiting times are known to follow a normal distribution. Conduct a hypothesis test to examine the significance of the restaurant’s claim at the 1% level of significance. Solutions: From the question, we have n = 31, x = 54 . 9, s = 17 . 6, and α = 0 . 01. The value of σ is given. Therefore, we use a t procedure. STAT3502 CU 25 / 41 t test for one population mean Inference for one population mean when σ is Unknown Example – Fast Food Let μ be the average waiting time in the drive though. step 1. Stating Hypotheses H 0 : μ = 60 vs. H 1 : μ < 60 step 2. Finding the comparative number Given α = 0 . 01 and n = 31, we look at Table A-3 with d.f. = n - 1 = 30 We found that - t α = - 2 . 457. STAT3502 CU 26 / 41 t test for one population mean Inference for one population when σ is Unknown Example – Fast Food Rejection Rule: We reject H 0 at level of significance 1% if t test statistic -2.457. We fail to reject H 0 at level of significance 1% if t test statistic > -2.457. Step 3: Calculate test statistic t = x - μ 0 s / n = 54 . 9 - 60 17 . 6 / 31 = - 1 . 61 Step 4: Make the decision and State the conclusion for the test Since - 1 . 61 > - 2 . 457, our t test statistic does NOT fall within the rejection region. Fail to reject H 0 at the level of significance 1%. Therefore, we have insufficient evidence to conclude that the true mean waiting time in the drive-thru is less than one minute. STAT3502 CU 27 / 41 t test for one population mean Inference for one population – Grooms Example z v.s. t In 1968, census results indicated that the age at which American men first married had a mean of 23.3 years and a standard deviation of 4.8 years. We take a random sample of 28 men who married for the first time in 2008. Their average age was 25.2 years and the standard deviation of their ages was 4.6 years. We assume that ages of first-time grooms are normally distributed. Conduct a hypothesis test to determine whether the mean age of first-time grooms has changed since 1968 at significant level α = 0 . 05. STAT3502 CU 28 / 41 t test for one population mean Inference for one population – Grooms Example z v.s. t Solution: Let μ be the mean age of first-time grooms in 2008. Let X be the age of American men married for the first time. From the question, we have X N ( μ = 23 . 3 , σ = 4 . 8 ) , n = 28, x = 25 . 2 and s = 4 . 6. Because σ is known , we choose a z procedure for this hypothesis testing using p-value method. STAT3502 CU 29 / 41 t test for one population mean