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## Windows 8, IE10!

MS is rolling out Windows 8, three years after lauching Win7, and also IE10.

Microsoft previews Windows 8 operating system
http://www.bbc.co.uk/news/technology-18288710

"Consumers can now download the release preview of Windows 8, a system which Microsoft says is its most tested operating system ever."

"IE 10 will be the first version of the browser with "do not track" turned on by default, meaning users can easily decide not to accept cookies." That's a feature I'd like.
 PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in $403M deal (Update)>> LA to give every student an iPad;$30M order>> CIA faulted for choosing Amazon over IBM on cloud contract
 Admin I wonder what will stop working now. Being a developer and selling programs for Windows means troubles sleeping each time new Windows version is launched.

Blog Entries: 1
Recognitions:
Gold Member
 Quote by Borek I wonder what will stop working now. Being a developer and selling programs for Windows means troubles sleeping each time new Windows version is launched.
Which is why I'm glad that I program in Java.

## Windows 8, IE10!

 Quote by Borek I wonder what will stop working now. Being a developer and selling programs for Windows means troubles sleeping each time new Windows version is launched.
I am not really a fan of the new look for windows 8 .. I think it is too suited for tablet devices with the huge windows.. Not really excited about it.
 Blog Entries: 9 Recognitions: Gold Member The problem with "Do Not Track" is that the advertisers have to honor the setting. I'd much prefer Firefox with the appropriate security add-ons (e.g. NoScript, AdBlock Plus, Ghostery, Cookie Monster, Perspectives, Ref Control (IIRC), HTTPS Everywhere, etc.)

 Similar discussions for: Windows 8, IE10! Thread Forum Replies Computers 21 Computing & Technology 0 Computing & Technology 0 Computing & Technology 0 Computing & Technology 15


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# Operational risk spending increases

The recent operational risk management (ORM) technology survey conducted by Operational Risk & Compliance and Protiviti indicates technology costs have increased for more than 70% of survey respondents over the past two years and are expected to continue to climb through 2006. Survey responses were submitted by risk management and financial services professionals, representing the full spectrum of financial services firms, both in terms of size and business activity. While the majority of organisations surveyed appear to be well on their way toward implementing Basel II operational risk management practices, it appears that most organisations are still seeking the right enabling technology solutions to support their operational risk management programmes, particularly with respect to the capture of self-assessment, internal loss, and KRI data.

Increased operational risk technology spending

Op risk technology expenditure has increased for more than 60% of the survey respondents. And nearly 30% of respondents surveyed indicated technology expenditures have increased by 10% or more over the previous two years. Respondents were also asked to estimate the change in technology costs related to op risk over the next 12 months. When comparing ORM technology expenditures of the last 24 months with the anticipated change in ORM technology expenditures for the next 12 months, 73% of respondents expect their firms to see further increases in technology costs.

There is a clear trend toward increased technology investment, and that trend is likely to continue for the foreseeable future.

Internal loss database (60%), self-assessment (51%), internal op risk reporting (45%) and KRIs (38%) have been implemented by roughly half of respondents. While use was slightly higher at larger institutions, institutions from virtually all asset sizes have widely implemented these tools. Even the majority of firms with less than $1 billion in total assets are creating operational loss databases. In examining the trend for introduction of ORM tools, internal loss databases and self-assessment tools are the first tools implemented by op risk programmes, and this trend is seen across all geographic regions and size of firm. Internal reporting and KRIs are the focus of tool implementation currently under way. With acquisition of data and assessment information, it is logical that ORM managers are now turning to more predictive risk metrics and reporting vehicles. Historical barriers to implementing vendor solutions Uncertainty regarding the evolution of op risk management in conjunction with the perception that appropriate software tools do not currently exist to hit this moving target appears to be the main barrier to investing in a vendor-based solution, comprising 48% of responses. Conversely, only 20% of respondents cited a belief that internally developed applications are either less expensive or superior as their top reasons for not considering a vendor-based solution. Larger banks that have traditionally developed their own op risk solutions are now reconsidering this strategy and focus as 57% of organisations with more than$50 billion in total assets are now considering or are implementing a second-generation tool for self-assessment, internal loss database, internal reporting and KRIs. While there are a handful of established technology providers, this area is relatively young and growing to include more than 20 different operational risk solutions that were considered by respondents during the past year.

Replacement of existing software solutions

It is clear that firms are not shy of replacing current solutions. When asked about replacing existing op risk software solutions, of the respondents surveyed that are at or beyond the midway point of their op risk management implementation, more than half are considering or implementing second-generation technology solutions for self-assessment (70%), internal loss database (58%), internal reporting (56%) and KRIs (50%). The op risk software market continues to evolve. To be considered a viable alternative, software vendors must develop flexible solutions that can meet changing op risk management methodologies as well as adapt to unique customer requirements. The challenge for these vendors will be to demonstrate that they can provide a comprehensive solution that spans multiple disciplines – including self-assessment, loss collection, internal reporting and KRI monitoring – at a lower cost of total ownership and with equal or better responsiveness to unique client requirements as internal application development groups.

Conclusion

This year's survey on op risk technology offers clear insights into the trends and areas of technology investment. Many are now considering and making strategic investments in op risk technologies as the number of vendor offerings continues to expand and their respective solutions mature.

Protiviti (www.protiviti.com) is a leading provider of independent risk consulting and internal audit services. It provides consulting and advisory services to help clients identify, assess, measure and manage financial, operational and technology-related risks encountered in their industries, and assists in the implementation of the processes and controls to enable their continued monitoring. Protiviti also offers a full spectrum of internal audit services to assist management and directors with their internal audit functions.

You need to sign in to use this feature. If you don’t have a Risk.net account, please register for a trial.

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# Nobel Prize has outlived its usefulness

The Nobel Prize in Physiology was awarded for the discovery of Hepatitis C today. The work is clearly deserving of recognition but this is another case where there were definitely more than three people who played an essential role in the work. I really think that the Nobel Prize should change its rules to allow for more winners. Below is my post when one of the winners of this years prize, Michael Houghton, turned down the Gairdner Award in 2013:

Hepatitis C and the folly of prizes

The scientific world was set slightly aflutter when Michael Houghton turned down the prestigious Gairdner Award for the the discovery of Hepatitis C. Harvey Alter and Daniel Bradley were the two other recipients. Houghton, who had previously received the Lasker Award with Alter, felt he could not accept one more award because two colleagues Qui-Lim Choo and George Kuo did not receive either of these awards, even though their contributions were equally important.

Hepatitis, which literally means inflammation of the liver, was characterized by Hippocrates and known to be infectious since the 8th century. The disease had been postulated to be viral at the beginning of the 20th century and by the 1960’s two viruses termed Hepatitis A and Hepatitis B had been established. However, there still seemed to be another unidentified infectious agent which was termed Non-A Non-B Hepatitis NANBH.

Michael Hougton, George Kuo and Qui-Lim Choo were all working at the Chiron corporation in the early 1980’s.   Houghton started a project to discover the cause of NANBH in 1982 with Choo joining a short time later. They made significant process in generating mouse monoclonal antibodies with some specificity to NANBH infected materials from chimpanzee samples received from Daniel Bradley at the CDC. They used the antibodies to screen cDNA libraries from infected materials but they had not isolated an agent. George Kuo had his own lab at Chiron working on other projects but would interact with Houghton and Choo. Kuo suggested that they try blind cDNA immunoscreening on serum derived from actual NANBH patients. This approach was felt to be too risky but Kuo made a quantitative assessment that showed it was viable. After two years of intensive and heroic screening by the three of them, they identified one clone that was clearly derived from the NANBH genome and not from human or chimp DNA. This was definitive proof that NANBH was a virus, which is now called Hepatitis C. Kuo then developed a prototype of a clinical Hepatitis C antibody detection kit and used it to screen a panel of NANBH blood provided by Harvey Alter of the NIH. Kuo’s test was a resounding success and the blood test that came out of that work has probably saved 300 million or more people from Hepititis C infection.

The question then is who deserves the prizes. Is it Bradley and Alter, who did careful and diligent work obtaining samples or is it Houghton, Choo, and Kuo, who did the heroic experiments that isolated the virus? For completely unknown reasons, the Lasker was awarded to just Houghton and Alter, which primed the pump for more prizes to these two. Now that the Lasker and Gairdner prizes have been cleared, that leaves just the Nobel Prize. The scientific community could get it right this time and award it to Kuo, Choo, and Houghton.

Addendum added 2013-5-2:  I should add that many labs from around the world were also trying to isolate the infective agent of NANBH and all failed to identify the correct samples from Alter’s panel.  It is not clear how long it would have been and how many more people would have been infected if Kuo, Choo, and Houghton had not succeeded when they did.

# ICCAI talk

I gave a talk at the International Conference on Complex Acute Illness (ICCAI) with the title Forecasting COVID-19. I talked about some recent work with FDA collaborators on scoring a large number of publicly available epidemic COVID-19 projection models and show that they are unable to reliably forecast COVID-19 beyond a few weeks. The slides are here.

# Why it is so hard to forecast COVID-19

I’ve been actively engaged in trying to model the COVID-19 pandemic since April and after 5 months I am pretty confident that models can estimate what is happening at this moment such as the number of people who are currently infected but not counted as a case. Back at the end of April our model predicted that the case ascertainment ratio ( total cases/total infected) was on the order of 1 in 10 that varied drastically between regions and that number has gone up with the advent of more testing so that it may now be on the order of 1 in 4 or possibly higher in some regions. These numbers more or less the anti-body test data.

However, I do not really trust my model to forecast what will happen a month from now much less six months. There are several reasons. One is that while the pandemic is global the dynamics are local and it is difficult if not impossible to get enough data for a detailed fine grained model that captures all the interactions between people. Another is that the data we do have is not completely reliable. Different regions define cases and deaths differently. There is no universally accepted definition for what constitutes a case or a death and the definition can change over time even for the same region. Thus, differences in death rates between regions or months could be due to differences in the biology of the virus, medical care, or how deaths are defined and when they are recorded. Depending on the region or time, a person with a SARS-CoV-2 infection who dies of a cardiac arrest may or may not be counted as a COVID-19 death. Deaths are sometimes not officially recorded for a week or two, particularly if the physician is overwhelmed with cases.

However, the most important reason models have difficulty forecasting the future is that modeling COVID-19 is as much if not more about modeling the behavior of people and government policy than modeling the biology of disease transmission and we are just not very good at predicting what people will do. This was pointed out by economist John Cochrane months ago, which I blogged about (see here). You can see why getting behavior correct is crucial to modeling a pandemic from the classic SIR model

$\frac{dS}{dt} = -\beta SI$

$\frac{dI}{dt} = \beta SI - \sigma I$

where $I$ and $S$ are the infected and susceptible fractions of the initial population, respectively. Behavior greatly affects the rate of infection $\beta$ and small errors in $\beta$ amplify exponentially. Suppression and mitigation measures such as social distancing, mask wearing, and vaccines reduce $\beta$, while super-spreading events increase $\beta$. The amplification of error is readily apparent near the onset of the pandemic where $I$ grows like $e^{\beta t}$. If you change $\beta$ by $\delta \beta$, then the $I$ will grow like $e^{\beta t+\delta \beta t}$ and thus the ratio is growing (or decaying) exponentially like $e^{\delta \beta t}$. The infection rate also appears in the initial reproduction number $R_0 = \sigma/\beta$. From a previous post, I derived approximate expressions for how long a pandemic would last and show that it scales as $1/(R_0-1)$ and thus errors in $\beta$ will produce errors $R_0$, which could result in errors in how long the pandemic will last, which could be very large if $R_0$ is near one.

The infection rate is different everywhere and constantly changing and while it may be possible to get an estimate of it from the existing data there is no guarantee that previous trends can be extrapolated into the future. So while some of the COVID-19 models do a pretty good job at forecasting out a month or even 6 weeks (e.g. see here), I doubt any will be able to give us a good sense of what things will be like in January.

# There is no herd immunity

In order for an infectious disease (e.g. COVID-19) to spread, the infectious agent (e.g. SARS-CoV-2) must jump from one person to another. The rate of this happening depends on the rate that an infectious person will come into contact with a susceptible person multiplied by the rate of the virus making the jump when the two people are nearby. The reproduction number R is obtained from the rate of infection spread times the length of time a person is infectious. If R is above one then a single person will infect more than one person on average and thus the pandemic will grow. If it is below one, then the pandemic will diminish. Herd immunity happens when enough people have been infected that the rate of finding a susceptible person becomes low enough that R drops below one. You can find the math behind this here.

However, a major assumption behind herd immunity is that once a person is infected they can never be infected again and this is not true for many infectious diseases such as other corona-viruses and the flu. There are reports that people can be reinfected by SARS-CoV-2. This is not fully validated but my money is on there being no lasting immunity to SARS-CoV-2 and this means that there is never any herd immunity. COVID-19 will just wax and wane forever.

This doesn’t necessarily mean it will be deadly forever. In all likelihood, each time you are infected your immune response will be more measured and perhaps SARS-CoV-2 will eventually be no worse than the common cold or the seasonal flu. But the fatality rate for first time infection will still be high, especially for the elderly and vulnerable. Those people will need to remain vigilante until there is a vaccine, and there is still no guarantee that a vaccine will work in the field. If we’re lucky and we get a working vaccine, it is likely that vaccine will not have lasting effect and just like the flu we will need to be vaccinated annually or even semi-annually.

# Another Covid-19 plateau

The world seems to be in another Covid-19 plateau for new cases. The nations leading the last surge, namely the US, Russia, India, and Brazil are now stabilizing or declining, while some regions in Europe and in particular Spain are trending back up. If the pattern repeats, we will be in this new plateau for a month or two and then trend back up again, just in time for flu season to begin.

# Why we need a national response

It seems quite clear now that we do not do a very good job of projecting COVID-19 progression. There are many reasons. One is that it is hard to predict how people and governments will behave. A fraction of the population will practice social distancing and withdraw from usual activity in the absence of any governmental mandates, another fraction will not do anything different no matter any official policy and the rest are in between. I for one get more scared of this thing the more I learn about it. Who knows what the long term consequences will be particularly for autoimmune diseases. The virus is triggering a massive immune response everywhere in the body and it could easily develop a memory response to your own cells in addition to the virus.

The virus also spreads in local clusters that may reach local saturation before infecting new clusters but the cross-cluster transmission events are low probability and hard to detect. The virus reached American shores in early January and maybe even earlier but most of those early events died out. This is because the transmission rate is highly varied. A mean reproduction number of 3 could mean everyone has R=3 or that most people transmit with R less than 1 while a small number (or events) transmit with very high R. (Nassim Nicholas Taleb has written copiously on the hazards of highly variable (fat tailed) distributions. For those with mathematical backgrounds, I highly recommend reading his technical volumes: The Technical Incerto. Even if you don’t believe most of what he says, you can still learn a lot.) Thus it is hard to predict when an event will start a local epidemic, although large gatherings of people (i.e. weddings, conventions, etc.) are a good place to start. Once the epidemic starts, it grows exponentially and then starts to saturate either by running out of people in the locality to infect or people changing their behavior or more likely both. Parts of New York may be above the herd immunity threshold now.

Thus at this point, I think we need to take a page out of Taleb’s book (like literally as my daughter would say), and don’t worry too much about forecasting. We can use it as a guide but we have enough information to know that most people are susceptible, about a third will be asymptomatic if infected (which doesn’t mean they won’t have long term consequences), about a fifth to a tenth will be counted as a case, and a few percent of those will die, which strongly depends on age and pre-existing conditions. We can wait around for a vaccine or herd immunity and in the process let many more people die, ( I don’t know how many but I do know that total number of deaths is a nondecreasing quantity), or we can act now everywhere to shut this down and impose a strict quarantine on anyone entering the country until they have been tested negative 3 times with a high specificity PCR test (and maybe 8 out of 17 times with a low specificity and sensitivity antigen test).

Acting now everywhere means, either 1) shutting everything down for at least two weeks. No Amazon or Grubhub or Doordash deliveries, no going to Costco and Walmart, not even going to the super market. It means paying everyone in the country without an income some substantial fraction of their salary. It means distributing two weeks supply of food to everyone. It means truly essential workers, like people keeping electricity going and hospital workers, live in a quarantine bubble hotel, like the NBA and NHL or 2) Testing everyone everyday who wants to leave their house and paying them to quarantine at home or in a hotel if they test positive. Both plans require national coordination and a lot of effort. The CARES act package has run out and we are heading for economic disaster while the pandemic rages on. As a recent president once said, “What have you got to lose?”

# The battle over academic freedom

In the wake of George Floyd’s death, almost all of institutional America put out official statements decrying racism and some universities initiated policies governing allowable speech and research. This was followed by the expected dissent from those who worry that academic freedom is being suppressed (see here, here, and here for some examples). Then there is the (in)famous Harper’s Magazine open letter decrying Cancel Culture, which triggered a flurry of counter responses (e.g. see here and here).

While some faculty members in the humanities and (non-life) sciences are up in arms over the thought of a committee of their peers judging what should be allowable research, I do wish to point out that their colleagues over on the Medical campus have had to get approval for human and animal research for decades. Research on human subjects must first pass through an Institutional Review Board (IRB) while animal experiments must clear the Institutional Animal Care and Use Committee (IACUC). These panels ensure that the proposed work is ethical, sound, and justified. Even research that is completely noninvasive, such as analysis of genetic data, must pass scrutiny to ensure the data is not misused and subject identies are strongly protected. Almost all faculty members would agree that this step is important and necessary. History is rife of questionable research that range from careless to criminal. Is it so unreasonable to extend such a concept to the rest of campus?

# New paper in eLife

I never thought this would ever be finished but it’s out. We hedge in the paper but my bet is that MYC is a facilitator of an accelerator essential for gene transcription.

# Dissecting transcriptional amplification by MYC

eLife 2020;9:e52483

Zuqin Nie, Chunhua Guo, Subhendu K Das, Carson C Chow, Eric Batchelor, S Stoney Simons Jr, David Levens

## Abstract

Supraphysiological MYC levels are oncogenic. Originally considered a typical transcription factor recruited to E-boxes (CACGTG), another theory posits MYC a global amplifier increasing output at all active promoters. Both models rest on large-scale genome-wide ”-omics’. Because the assumptions, statistical parameter and model choice dictates the ‘-omic’ results, whether MYC is a general or specific transcription factor remains controversial. Therefore, an orthogonal series of experiments interrogated MYC’s effect on the expression of synthetic reporters. Dose-dependently, MYC increased output at minimal promoters with or without an E-box. Driving minimal promoters with exogenous (glucocorticoid receptor) or synthetic transcription factors made expression more MYC-responsive, effectively increasing MYC-amplifier gain. Mutations of conserved MYC-Box regions I and II impaired amplification, whereas MYC-box III mutations delivered higher reporter output indicating that MBIII limits over-amplification. Kinetic theory and experiments indicate that MYC activates at least two steps in the transcription-cycle to explain the non-linear amplification of transcription that is essential for global, supraphysiological transcription in cancer.

# How to make a fast but bad COVID-19 test good

Among the myriad of problems we are having with the COVID-19 pandemic, faster testing is one we could actually improve. The standard test for the presence of SARS-CoV-2 virus uses PCR (polymerase chain reaction), which amplifies targeted viral RNA. It is accurate (high specificity) but requires relatively expensive equipment and reagents that are currently in short supply. There are reports of wait times of over a week, which renders a test useless for contact tracing.

An alternative to PCR is an antigen test that tests for the presence of protein fragments associated with COVID-19. These tests can in principle be very cheap and fast, and could even be administered on paper strips. They are generally much more unreliable than PCR and thus have not been widely adopted. However, as I show below by applying the test multiple times, the noise can be suppressed and a poor test can be made arbitrarily good.

The performance of binary tests are usually gauged by two quantities – sensitivity and specificity. Sensitivity is the probability that you test positive (i.e are infected) given that you actually are positive (true positive rate). Specificity is the probability that you test negative if you actually are negative (true negative rate). For a pandemic, sensitivity is more important than specificity because missing someone who is infected means you could put lots of people at risk while a false positive just means the person falsely testing positive is inconvenienced (provided they cooperatively self-isolate). Current PCR tests have very high specificity but relatively low sensitivity (as low as 0.7) and since we don’t have enough capability to retest, a lot of tested infected people could be escaping detection.

The way to make any test have arbitrarily high sensitivity and specificity is to apply it multiple times and take some sort of average. However, you want to do this with the fewest number of applications. Suppose we administer $n$ tests on the same subject, the probability of getting more than $k$ positive tests if the person is positive is $Q(k,n,q) = 1 - CDF(k|n,q)$, where $CDF$ is the cumulative distribution function of the Binomial distribution (i.e. probability that the number of Binomial distributed events is less than or equal to $k$). If the person is negative then the probability of  $k$ or fewer positives is $R(k,n,r) = CDF(k|n,1-r)$. We thus want to find the minimal $n$ given a desired sensitivity and specificity, $q'$ and $r'$. This means that we need to solve the constrained optimization problem: find the minimal $n$ under the constraint that $k < n$, $Q(k,n,q) = \ge q'$ and $R(k,n,r)\ge r'$. $Q$ decreases and $R$ increases with increasing $k$ and vice versa for $n$. We can easily solve this problem by sequentially increasing $n$ and scanning through $k$ until the two constraints are met. I’ve included the Julia code to do this below.  For example, starting with a test with sensitivity .7 and specificity 1 (like a PCR test), you can create a new test with greater than .95 sensitivity and specificity, by administering the test 3 times and looking for a single positive test. However, if the specificity drops to .7 then you would need to find more than 8 positives out of 17 applications to be 95% sure you have COVID-19.

using Distributions

function Q(k,n,q)
d = Binomial(n,q)
return 1 – cdf(d,k)
end

function R(k,n,r)
d = Binomial(n,1-r)
return cdf(d,k)
end

function optimizetest(q,r,qp=.95,rp=.95)

nout = 0
kout = 0

for n in 1:100
for k in 0:n-1
println(R(k,n,r),” “,Q(k,n,q))
if R(k,n,r) >= rp && Q(k,n,q) >= qp
kout=k
nout=n
break
end
end
if nout > 0
break
end
end

return nout, kout
end

# Slides for Covid-19 talk

Here are my slides for my recent COVID-19 talk at the Centre for Applied Mathematics in BioScience and Medicine (CAMBAM). It’s an updated version of the one I gave to the FDA.

# Remember the ventilator

According to our model, the global death rate due to Covid-19 is around 1 percent for all infected (including unreported). However, if it were not for modern medicine and in particular the ventilator, the death rate would be much higher. Additionally, the pandemic first raged in the developed world and is only recently engulfing parts of the world where medical care is not as ubiquitous although this may be mitigated by a younger populace in those places. The delay between the appearance of a Covid-19 case and deaths is also fairly long; our model predicts a mean of over 50 days. So the lower US death rate compared to April could change in a month or two when the effects of the recent surges in the US south and west are finally felt.

# New paper on Wilson-Cowan Model

I forgot to post that my fellow Yahya and I have recently published a review paper on the history and possible future of the Wilson-Cowan Model in the Journal of Neurophysiology tribute issue to Jack Cowan. Thanks to the patient editors for organizing this.

# Before and beyond the Wilson–Cowan equations

## Abstract

The Wilson–Cowan equations represent a landmark in the history of computational neuroscience. Along with the insights Wilson and Cowan offered for neuroscience, they crystallized an approach to modeling neural dynamics and brain function. Although their iconic equations are used in various guises today, the ideas that led to their formulation and the relationship to other approaches are not well known. Here, we give a little context to some of the biological and theoretical concepts that lead to the Wilson–Cowan equations and discuss how to extend beyond them.

PDF here.

# How long and how high for Covid-19

Cases of Covid-19 are trending back up globally and in the US. The world has nearly reached 10 million cases with over 2.3 million in the US. There is still a lot we don’t understand about SARS-CoV-2 transmission but I am confident we are no where near herd immunity. Our model is consistently showing that the case ascertainment ratio, that is the ratio of official Covid-19 cases to total SARS-CoV-2 infections, is between 5 and 10. That means that the US has less than 25 million infections while the world is less than 100 million.

Herd immunity means that for any fixed reproduction number, R0, the number of active infections will trend downward if the fraction of the initially susceptible population falls below 1/R0, or the total number infected is higher than 1- 1/R0. Thus, for an R0 of 4, three quarters of the population needs to be infected to reach herd immunity. However, the total number that will eventually be infected, as I will show below, will be

$1 -\frac{e^{-R_0}}{1- R_0e^{-R_0}}$

which is considerably higher. Thus, mitigation efforts to reduce R0 will reduce the total number infected. (2020-06-27: This expression is not accurate when R0 is near 1. For a formula in that regime, see Addendum.)

Some regions in Western Europe, East Asia, and even the US have managed to suppress R0 below 1 and cases are trending downward. In the absence of reintroduction of SARS-CoV-2 carriers, Covid-19 can be eliminated in these regions. However, as the recent spikes in China, South Korea, and Australia have shown, this requires continual vigilance. As long as any person remains infected in the world, there is always a chance of re-emergence. As long as new cases are increasing or plateauing, R0 remains above 1. As I mentioned before, plateauing is not a natural feature of the epidemic prediction models, which generally either go up or go down. Plateauing requires either continuous adjustment of R0 through feedback or propagation through the population as a wave front, like a lawn mower cutting grass. The latter is what is actually going on from what we are seeing. Regions seem to rise and fall in succession. As one region reaches a peak and goes down either through mitigation or rapid spread of SARS-CoV-2, Covid-19 takes hold in another. We saw China and East Asia rise and fall, then Italy, then the rest of Western Europe, then New York and New Jersey, and so forth in series, not in parallel. Now it is spreading throughout the rest of the USA, South America, and Eastern Europe. Africa has been spared so far but it is probably next as it is beginning to explode in South Africa.

A reduction in R0 also delays the time to reach the peak. As a simple example, consider the standard SIR model

$\frac{ds}{dt} = -\beta sl$

$\frac{dl}{dt} = \beta sl -\sigma l$

where $s$ is the fraction of the population susceptible to SARS-CoV-2 infection and $l$ is the fraction of the population actively infectious. Below are simulations of the pandemic progression for R0 = 4 and 2.

We see that halving R0, basically doubles the time to reach the peak but much more than doubles the number of people that never get infected. We can see why this is true by analyzing the equations. Dividing the two SIR equations gives

$\frac{dl}{ds} = \frac{\sigma l -\beta sl}{\beta sl}$,

which integrates to $l = \frac{\sigma}{\beta} \ln s - s + C$. If we suppose that initially $s=1$ and $l = l_0<<1$ then we get

$l = \frac{1}{R_0} \ln s + 1 - s + l_0$ (*)

where $R_0 = \beta/\sigma$ is the reproduction number. The total number infected will be $1-s$ for $l=0$. Rearranging gives

$s = e^{-R_0(1+l_0+s)}$

If we assume that $R_0 s <<1$ and ignore $l_0$ we can expand the exponential and solve for $s$ to get

$s \approx \frac{e^{-R_0}}{1- R_0e^{-R_0}}$

This is the fraction of the population that never gets infected, which is also the probability that you won’t be infected. It gets smaller as $R_0$ increases. So reducing $R_0$ can exponentially reduce your chances of being infected.

To figure out how long it takes to reach the peak, we substitute equation (*) into the SIR equation for $s$ to get

$\frac{ds}{dt} = -\beta(\frac{1}{R_0} \ln s + 1 - s + l_0) s$

We compute the time to peak, $T$, by separating variables and integrating both sides. The peak is reached when $s = 1/R_0$.  We must thus compute

$T= \int_0^T dt =\int_{1/R_0}^1 \frac{ds}{ \beta(\frac{1}{R_0} \ln s + 1 - s +l_0) s}$

We can’t do this integral but if we set $s = 1- z$ and $z<< 1$, then we can expand $\ln s = -\epsilon$ and obtain

$T= \int_0^T dt =\int_0^{l_p} \frac{dz}{ \beta(-\frac{1}{R_0}z + z +l_0) (1-z)}$

where $l_p = 1-1/R_0$. This can be re-expressed as

$T=\frac{1}{ \beta (l_0+l_p)}\int_0^{l_p} (\frac{1}{1-z} + \frac{l_p}{l_p z + l_0}) dz$

which is integrated to

$T= \frac{1}{ \beta (l_0+l_p)} (-\ln(1-l_p) + \ln (l_p^2 + l_0)-\ln l_0)$

If we assume that $l_0<< l_p$, then we get an expression

$T \approx \sigma \frac{\ln (R_0l_p^2/l_0)}{ R_0 -1}$

So, $T$ is proportional to the recovery time $\sigma$ and inversely related to $R_0$ as expected but if $l_0$ is very small (say 0.00001) compared to $R_0$ (say 3) then $\ln R_0/l_0$ can be big (around 10), which may explain why it takes so long for the pandemic to get started in a region. If the infection rate is very low in a region, the time it takes a for a super-spreader event to make an impact could be much longer than expected (10 times the infection clearance time (which could be two weeks or more)).

Addendum 2020-06-26: Fixed typos in equations and added clarifying text to last paragraph

Addendum 2020-06-27: The approximation for total infected is not very good when $R_0$ is near 1, a better one can be obtained by expanding the exponential to quadratic order in which case you get the new formula for the

$s = \frac{1}{{R_{0}}^2} ( e^{R_0} - R_0 - \sqrt{(e^{R_0}-R_0)^2 - 2{R_0}^2})$

However, for $R_0$ near 1, a better expansion is to substitute $z = 1-s$ into equation (*) and obtain

$l = \frac{1}{R_0} \ln 1-z + z + l_0$

Set $l=0$, after rearranging and exponentiating,  obtain

$1 - z = e^{-R_0(l_0+z)}$, which can be expanded to yield

$1- z = e^{-R_0 l_0}(1 - R_0z + R_0^2 z^2/2$

Solving for $z$ gives the total fraction infected to be

$z = (R_0 -e^{R_0l_0} + \sqrt{(R_0-e^{R_0l_0})^2 - 2 R_0^2(1-e^{R_0l_0})})/R_0^2$

This took me much longer than it should have.

# The formal logic of legal decisions

The US Supreme Court ruled today that the ban on sex-based discrimination in Title VII of the 1964 Civil Rights Act protects employees from discrimination based on sexual orientation or gender identity. Justice Gorsuch, who is a textualist (i.e. believes that laws should only be interpreted according to the written words alone without taking into any consideration the intent of the writers), writes “An employer who fires an individual for being homosexual or transgender fires that person for traits or actions it would not have questioned in members of a different sex. Sex plays a necessary and undisguisable role in the decision, exactly what Title VII forbids.” I find this to be an interesting exercise in logic. For example, consider the case of a “man married to a man”. According to Gorsuch’s logic, this cannot be an excuse to fire someone because if you replace “man” with “woman” in the first instance then you end up with the the phrase “woman married to a man”, and since this is not sufficient for firing, the reason is not sex invariant. Dissenting legal scholars disagree. They argue that the correct logic is to replace all instances of “man” with “woman”, in which case you would end up with “woman married to a woman” and thus the reason for firing would be sex invariant. I think in this case Gorsuch is correct because in order to have complete sex invariance, the rule must apply for any form of exchange. The law should be applied equally in all the possible ways that one gender is replaced by the other.

# The depressing lack of American imagination

Democratic presidential candidate Andrew Yang made universal basic income a respectable topic for debate. I think this is a good thing because I’m a major proponent of UBI but my reasons are different. Yang is a technology dystopian who sees a future where robots take all of our jobs and the UBI as a way to alleviate the resulting pain and suffering. I think a UBI (and universal health care) would lead to less resentment of the welfare system and let people take more entrepreneurial risks. I believe human level AI is possible but I do not accept that this necessarily implies an economic apocalypse. To believe such a thing is to believe that the only way society can be structured is that a small number of tech companies owns all the robots and everyone else is at their mercy. That to me is a depressing lack of imagination. The society we live in is a human construct. There is no law of nature that says we must live by any specific set of rules or economic system. There is no law that says tech companies must have monopolies. There is no reason we could not live in a society where each person has her own robot who works for her. There is no law that says we could not live in a society where robots do all the mundane work while we garden and bake bread.

I think we lack imagination in every sector of our life. We do not need to settle for the narrow set of choices we are presented. I for one do not accept that elite colleges must wield so much influence in determining the path of one’s life. There is no reason that the US meritocracy needs to be a zero sum game, where one student being accepted to Harvard means another is not or that going to Harvard should even make so much difference in one’s life. There is no reason that higher education needs to cost so much. There is no reason students need to take loans out to pay exorbitant tuition. That fact that this occurs is because we as a society have chosen such.

I do not accept that irresponsible banks and financial institutions need to be bailed out whenever they fail, which seems to be quite often. We could just let them fail and restart. There is no reason that the access to capital needs to be controlled by a small number of financial firms. It used to be that banks would take in deposits and lend out to homeowners and businesses directly. They would evaluate the risk of each loan. Now they purchase complex financial products that evaluate the risk according to some mathematical model. There is no reason we need to subsidize such activity.

I do not accept that professional sports teams cannot be community owned. There is no law that says sports leagues need to be organized as monopolies with majority owners. There is no reason that communities cannot simply start their own teams and play each other. There is no law that says we need to build stadiums for privately owned teams. We only choose to do so.

The society we live in is the way it is because we have chosen to live this way. Even an autocrat needs a large fraction of the population to enforce his rule. The number of different ways we could organize (or not organize) is infinite. We do not have to be limited to the narrow set of choices we are presented. What we need is more imagination.

# The fatal flaw of the American Covid-19 response

The United States has surpassed 2 million official Covid-19 cases and a 115 thousand deaths. After three months of lockdown, the country has had enough and is reopening. Although it has achieved its initial goal of slowing the growth of the pandemic so that hospitals would not be overwhelmed, the battle has not been won. We’re not at the beginning of the end; we may not even be at the end of the beginning. If everyone in the world could go into complete isolation, the pandemic would be over in two weeks. Instead, it is passed from one person to the next in a tragic relay race. As long as a single person is shedding the SARS-CoV-2 virus and comes in contact with another person, the pandemic will continue. The pandemic in the US is not heading for extinction. We are not near herd immunity and R0 is not below one. By the most optimistic yet plausible scenario, 30 million people have already been infected and 200 million will never get it either by having some innate immunity or by avoiding it through sheltering or luck. However, that still leaves over 100 million who are susceptible of which about a million will die if they all catch it.

However, the lack of effectiveness of the response is not the fatal flaw. No, the fatal flaw is that the US Covid-19 response asks one set of citizens to sacrifice for the benefit of another set. The Covid-19 pandemic is a story of three groups of people. The fortunate third can work from home, and the lockdown is mostly just an inconvenience. They still get paychecks while supplies and food can be delivered to their homes. Sure it has been stressful and many of have forgone essential medical care but they can basically ride this out for as long as it takes. The second group who own or work in shuttered businesses have lost their income. The federal rescue package is keeping some of them afloat but that runs out in August. The choice they have is to reopen and risk getting infected or be hungry and homeless. Finally, the third group is working to allow the first group to remain in their homes. They are working on farms, food processing plants, and grocery stores. They are cutting lawns, fixing leaking pipes, and delivering goods. They are working in hospitals and nursing homes and taking care of the sick and the children of those who must work. They are also the ones who are most likely to get infected and spread it to their families or the people they are trying to take care of. They are dying so others may live.

A lockdown can only work in a society if the essential workers are adequately protected and those without incomes are supported. Each worker should have an N100 mask, be trained how to wear it and be tested weekly. People in nursing homes should be wearing hazmat suits. Everyone who loses income should be fully compensated. In a fair society, everyone should share the risks and the pain equally.

# Covid-19 continues to spread

The global plateau turned out to just be a pause and the growth in new cases continues. The rise seems to be mostly driven by increases in Brazil, India, and until very recently Russia with plateauing in the US and European countries as they relax their mitigation policies. The pandemic is not over by a long shot. There will most certainly be further growth in the near future.

# Why middle school science should not exist

My 8th grade daughter had her final (distance learning) science quiz this week on work, or as it is called in her class, the scientific definition of work. I usually have no idea what she does in her science class since she rarely talks to me about school but she so happened to mention this one tidbit because she was proud that she didn’t get fooled by what she thought was a trick question. I’ve always believed that work, as in force times displacement (not the one where you produce economic value), is one of the most useless concepts in physics and should not be taught to anyone until they reach graduate school, if then. It is a concept that has long outlived its usefulness and all it does now is to convince students that science is just a bunch of concepts invented to confuse you. The problem with science education in general is that it is taught as a set of facts and definitions when the only thing that kids need to learn is that science is about trying to show something is true using empirical evidence. My daughter’s experience is evidence that science education in the US has room for improvement.

Work, as defined in science class, is just another form of energy, and the only physics that should be taught to middle school kids is that there are these quantities in the universe called energy and momentum and they are conserved. Work is just the change in energy of a system due to a force moving something. For example, the work required to lift a mass against gravity is the distance the mass was lifted multiplied by the force used to move it. This is where it starts to get a little confusing because there are actually two reasons you need force to move something. The first is because of Newton’s First Law of inertia – things at rest like to stay at rest and things in motion like to stay in motion. In order to move something from rest you need to accelerate it, which requires a force and from Newton’s second law, Force equals mass times acceleration, or F = ma. However, if you move something upwards against the force of gravity then even to move at a constant velocity you need to use a force that is equal to the gravitational force pulling the thing downwards, which from Newton’s law of gravitation is given by $F = G M m/r^2$, where $G$ is the universal gravitational constant, $M$ is the mass of the earth, $m$ is the mass of the object and $r$ is the distance between the objects. By a very deep property of the universe, the mass in Newton’s law of gravitation is the exact same mass as that in Newton’s second law, called inertial mass. So that means if we let $GM/r^2 = g$, then we get $F = mg$, and $g = 9.8 m/s^2$ is the gravitational acceleration constant if we set $r$ be the radius of the earth, which is much bigger than the height of things we usually deal with in our daily lives. All things dropped near the earth will accelerate to the ground at $9.8 m/s^2$. If gravitational mass and inertial mass were not the same, then objects of different masses would not fall with the same acceleration. Many people know that Galileo showed this fact in his famous experiment where he dropped a big and small object from the Leaning Tower of Pisa. However, many probably also cannot explain why including my grade 7 (or was it 8) science teacher who thought it was because the earth’s mass was much bigger than the two objects so the difference was not noticeable. The equivalence of gravitational and inertial mass was what led Einstein to his General Theory of Relativity.

In the first part of my daughter’s quiz, she was asked to calculate the energy consumed by several appliances in her house for one week. She had to look up how much power was consumed by the refrigerator, computer, television and so forth on the internet. Power is energy per unit time so she computed the amount of energy used by multiplying the power used by the total time the device is on per week. In the second part of the quiz she was asked to calculate how far she must move to power those devices. This is actually a question about conservation of energy and to answer the question she had to equate the energy used with the work definition of force times distance traveled. The question told her to use gravitational force, which implies she had to be moving upwards against the force of gravity, or accelerating at g if moving horizontally, although this was not specifically mentioned. So, my daughter took the energy used to power all her appliances and divided it by the force, i.e. her mass times g, and got a distance. The next question was, and I don’t recall exactly how it was phrased but something to the effect of: “Did you do scientifically defined work when you moved?”

Now, in her class, she probably spent a lot of time examining situations to distinguish work from non-work. Lifting a weight is work, a cat riding a Roomba is not work. She learned that you did no work when you walked because the force was perpendicular to your direction of motion. I find these types of gotcha exercises to be useless at best and in my daughter’s case completely detrimental. If you were to walk by gliding along completely horizontally with absolutely no vertical motion at a constant speed then yes you are technically not doing mechanical work. But your muscles are contracting and expanding and you are consuming energy. It’s not your weight times the distance you moved but some very complicated combination of metabolic rate, muscle biochemistry, energy losses in your shoes, etc. Instead of looking at examples and identifying which are work and which are not, it would be so much more informative if they were asked to deduce how much energy would be consumed in doing these things. The cat on the Roomba is not doing work but the Roomba is using energy to turn an electric motor that has to turn the wheel to move the cat. It has to accelerate from standing still and also gets warm, which means some of the energy is wasted to heat. A microwave oven uses energy because it must generate radio waves. Boiling water takes energy because you need to impart random kinetic energy to the water molecules. A computer uses energy because it needs to send electrons through transistors. Refrigerators work by using work energy to pump the heat energy from the inside to the outside. You can’t cool a room by leaving the refrigerator door open because you will just pump heat around in a circle and some of the energy will be wasted as extra heat.

My daughter’s answer to the question of was work done was that no work was done because she interpreted movement to be walking horizontally and she knew from all the gotcha examples that walking was not work. She read to me her very legalistically parsed paragraph explaining her reasoning, which made me think that while science may not be in her future, law might be. I tried to convince her that in order for the appliances to run, energy had to come from somewhere so she must have done some work at some point in her travels but she would have no part of it. She said it must be a trick question so the answer has to not make sense. She proudly submitted the quiz convinced more then ever that her so-called scientist Dad is a complete and utter idiot.

# How much Covid-19 testing do we need?

There is a simple way to estimate how much SARS-CoV-2 PCR testing we need to start diminishing the COVID-19 pandemic. Suppose we test everyone at a rate $f$, with a PCR test with 100% sensitivity, which means we do not miss anyone who is positive but we could have false positives. The number of positives we will find is $f p$, where  $p$ is the prevalence of infectious individuals in a given population. If positive individuals are isolated from the rest of the population until they are no longer infectious with probability $q$, then the rate of reduction in prevalence is $fqp$. To reduce the pandemic, this number needs to be higher than the rate of pandemic growth, which is given by $\beta s p$, where $s$ is the fraction of the population susceptible to SARS-CoV-2 infection and $\beta$ is the rate of transmission from an infected individual to a susceptible upon contact. Thus, to reduce the pandemic, we need to test at a rate higher than $\beta s/q$.

In the initial stages of the pandemic $s$ is one and $\beta = R_0/\sigma$, where $R_0$ is the mean reproduction number, which is probably around 3.7 and $\sigma$ is the mean rate of becoming noninfectious, which is probably around 10 to 20 days. This gives an estimate of  $\beta_0$ to be somewhere around 0.3 per day. Thus, in the early stages of the pandemic, we would need to test everyone at least two or three times per week, provided positives are isolated. However, if people wear masks and avoid crowds then $\beta$ could be reduced. If we can get it smaller then we can test less frequently. Currently, the global average of $R_0$ is around one, so that would mean we need to test every two or three weeks. If positives don’t isolate with high probability, we need to test at a higher rate to compensate. This threshold rate will also go down as $s$ goes down.

In fact, you can just test randomly at rate $f$ and monitor the positive rate. If the positive rate trends downward then you are testing enough. If it is going up then test more. In any case, we may need less testing capability than we originally thought, but we do need to test the entire population and not just suspected cases.


text:
Complete problems for randomized complexity classes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:50:17Z http://mathoverflow.net/feeds/question/34469 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34469/complete-problems-for-randomized-complexity-classes Complete problems for randomized complexity classes Marcos Villagra 2010-08-04T02:51:43Z 2013-02-04T14:38:42Z
It is believed that $BPP$ has no complete problems. Even for $BPP^O$ for a suitable oracle $O$ it is believed not to have complete problems, unless P=BPP. I wonder if the class MA (the randomized version of NP) has complete problems. For example, IP has complete problems (given that it is equal to PSPACE). There is a similar post asking about complexity classes with no complete problems. Here, I'm interested specifically on complete problems for MA. If the answer is positive, can you give some examples? I've tried google and the complexity zoo, but with no success.
 http://mathoverflow.net/questions/34469/complete-problems-for-randomized-complexity-classes/34472#34472 Answer by Peter Shor for Complete problems for randomized complexity classes Peter Shor 2010-08-04T03:45:51Z 2010-08-04T03:45:51Z
In general, for randomized classes complete problems tend to be either promise problems or approximation problems (which means they don't technically satisfy the conditions for being complete problems).

If you allow approximation problems, you can get complete problems in BPP. For example, for BPP you can ask: given a Turing machine $M$, approximate the probability that it accepts within $t$ steps on input $x$, where $t$ is polynomial in the size of the input. It's clear that you can approximate this probability with a BPP machine (using simulation), and it's also obvious that any BPP language can be reduced to this problem. Technically, this problem isn't in BPP since it's not a language (i.e., its output is not ${0,1}$), so it's not BPP-complete.

You can turn this into a promise problem by imposing the "promise" that the acceptance probability either be greater than $\frac{2}{3}$ or less than $\frac{1}{3}$.

These complete promise problems or approximation problems play the same role that complete problems play for non-randomized languages, and they really deserve more respect from computer scientists. For quantum computing, the natural complete problems also tend to be approximation (or promise) problems, and they have been quite useful in the theory of quantum computing.

There is a natural promise problem from quantum computing (stoquastic Hamiltonian) which is MA-complete (and not trivially so). However, I don't believe there are any languages (non-promise problems with ${0,1}$ answers) known to be MA-complete.
 http://mathoverflow.net/questions/34469/complete-problems-for-randomized-complexity-classes/34473#34473 Answer by Ryan Williams for Complete problems for randomized complexity classes Ryan Williams 2010-08-04T03:46:54Z 2010-08-04T03:46:54Z
First, it isn't "clear" that $BPP$ has no complete problems: if $TIME[2^{O(n)}]$ requires circuits of size $2^{\delta n}$ for some $\delta > 0$ then $P = BPP$, in which case $BPP$ certainly does have complete problems! (This is a famous result of Impagliazzo and Wigderson from 1998.) Also, $MA$ is not known to have complete problems, but again, under plausible circuit complexity assumptions, $NP = MA$ in which case all $NP$-complete problems are $MA$-complete too. (This is due to Klivans and van Melkebeek, and others who have weakened/changed the assumptions under which $NP=MA$ holds.) But yes, complete problems for $BPP$ and $MA$ are not currently known.

Secondly, the "promise" versions of these classes have do natural complete promise problems. (A promise problem is a pair $(\Pi_{YES}, \Pi_{NO})$ where $\Pi_{YES}, \Pi_{NO} \subseteq$ {$0,1$}$^n$, and $\Pi_{YES} \cap \Pi_{NO} = \emptyset$. The crucial point is that we do not necessarily have $\Pi_{YES} \cup \Pi_{NO} =$ {$0,1$}$^n$, so some inputs may not be considered at all in a promise problem.) For example, the following promise problem is complete for $Promise-BPP$:

Given a Boolean circuit $C$ with AND/OR/NOT gates and $n$ inputs, where you are promised that either every input $x \in$ {$0,1$}$^n$ makes $C$ evaluate to $0$, or at least half the inputs $x \in {0,1}^n$ make $C$ evaluate to $1$, determine which of the two is the case.

Not sure who first proved that this problem is complete, though in the literature it is sometimes known as CAPP (for Circuit Approximation Probability Problem). One reference is Fortnow, "Comparing notions of full derandomization".

Similarly defined promise problems (with nondeterminism thrown in, naturally) are complete for $Promise-MA$.
 http://mathoverflow.net/questions/34469/complete-problems-for-randomized-complexity-classes/120773#120773 Answer by Oded Goldreich for Complete problems for randomized complexity classes Oded Goldreich 2013-02-04T14:38:42Z 2013-02-04T14:38:42Z
My own favorite complete problems are promise problems for the promise problem version of SZK (Statistical Zero-Knowledge). These complete problems have played a major role in the study of SZK, and the relations among them are fascinating per se. See "On the complexity of computational problems regarding distributions (a survey)" at http://eccc.hpi-web.de/report/2011/004.

Oded


text: The hybridization of carbon in methane is sp3. Though the lone pairs affect the bond angles, their positions are not taken into account while doing Count electrons2. Structure is based on trigonal planar geometry with one lone pair occupying a corner. What does it mean when an aircraft is statically stable but dynamically unstable? The steric number is not equal to the number of σ-bonds. Steric number = no. Analyze: We are given two chemical formulas—one for a polyatomic anion and one for a molecular. E.g. Property Name Property Value Reference Molecular Weight 108.01 g/mol Computed by PubChem 2.1 (PubChem release 2019.06.18) XLogP3-AA 0.7 Computed by XLogP3 3.0 (PubChem release 2019.06.18) Hydrogen Bond Donor So for carbon dioxide, we have got sp hybridization at the central atom (or two areas of electron density with no lone pairs), so the shape will be linear. N2O3: sp^2 hybridization. If it donates a lone pair, a positive charge is accumulated. It is better to write the Lewis structural formula to get a rough idea about the structure of molecule and bonding pattern. Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. If two lone pairs are arranged at 90o of angle, the repulsions are greater. The number of lone pairs on carbon atom = (v - b - c) / 2 = (4 - 4 - 0) / 2 = 0. central atom. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around. Aside: You probably were already fine with the determining hybridization part. Among these, one is sigma bond and the second one is pi bond. The sp² orbitals for such atoms used to be part of a VB description, but (at the latest) with the formulation of Bent's rule that basically became obsolete. It is slightly decreased to 107o48' due to repulsion from lone pair. Boron atom gets negative charge when it accepts a lone pair from hydride ion, H- in borohydride ion, BH4-, Steric number = no. Asking for help, clarification, or responding to other answers. 38 Quiz Free Response Practice #2 (2017 #2 (shortened) and 2018 #2 (shortened)) Answer the following question about fulminic acid, HCNO. For 2 electron density regions, it is sp-hybridised. I don't see how the hybridsation model could yield a structure prediction. of lone pairs = 4 + 2 = 6. I'm now reading about Bent's rule... there's a lot to take in / relearn here. Click to see full answer Just so, what is the systematic name for n2o5? Get your answers by asking now. Therefore, it has sp hybridization (2 $\times$ sp orbitals and 2 $\times$ p orbitals), Each oxygen has 1 double bond and 2 lone-pairs which means 1 $\times \ \pi$ bond and 3 $\times \ \sigma$ bonds (actually 1 bond and 2 orbitals with lone pairs). Put one electron pair in each bond4. in the liquid and solid phases. Hence the following structure can be ruled out. For 4 electron density regions, it is $\mathrm{sp^3}$-hybridised etc. The total number of bonds formed by sulfur with two oxygen atoms is four. What is (a) NO2* (b) NO2 the hybridization of nitrogen in each of the following ions and molecules? of σ-bonds + no. of Ï-bonds + no. Usually you would approach this problem from the other direction. The number of lone pairs on sulfur atom = (v - b - c) / 2 = (6 - 4 - 0) / 2 = 1. Many students face problems with finding the hybridization of given atom (usually the central one) in a compound and the shape of molecule. For 3 electron density regions, it is $\mathrm{sp^2}$-hybridised. b = no. MathJax reference. It is one of the binary nitrogen oxides, a family of compounds that only contain nitrogen and oxygen. The valency of nitrogen is 3. Nitrogen atom in ammonium ion, NH4+ gets positive charge since it donates a pair of electrons to H+ ion. The valency of carbon is 4 and hence it can form 4 sigma bonds with four hydrogen atoms. Consider the molecule below. I also read the topic in my book but this issue wasn't addressed there too. Editing colors in Blender for vibrance and saturation. of σ-bonds + no. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Do you have any links for those maps? Side remark: MOT does not use or need hybridisation. Why an asymmetric geometry with sp3d and sp3d3 hybridization? Constrained to this sp hybridisation is the max you can get. Therefore they are sp2 hybridization. Hence each oxygen makes two bonds with sulfur atom. I will use $\ce{CO2}$ as an example. You can find the hybridization of an atom by finding its steric number: The steric number = the number of atoms bonded to the atom + the number of lone pairs the atom has. Steric number = no. How to display all trigonometric function plots in a table? so. the formula to determined hybridization is summation of number of lone pair and sigma bond. The number of lone pairs on xenon atom = (v - b - c) / 2 = (8 - 4 - 0) / 2 = 2. Structure is based on octahedral geometry with two lone pairs occupying two corners. This molecule is tetrahedral in structure as well as  in shape, since there are no lone pairs and the number of σ-bonds is equal to the steric number. Can an exiting US president curtail access to Air Force One from the new president? N2O3 doesn't contain protons, so it is not a Brønsted acid. a. c Looking for a short story about a network problem being caused by an AI in the firmware, Counting monomials in product polynomials: Part I, Ceramic resonator changes and maintains frequency when touched, Book about an AI that traps people on a spaceship. of σ-bonds + no. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong. directly to the concerned atom. This case arises when there are no lone pairs on the given central atom. Mathematically this would still work, physically it simply does not make any sense. At higher temperatures the equilibrium favors the constituent gases, with Kdiss = 193 kPa (25 °C). Look at bonding & lone pairs on each atom. Dinitrogen trioxide is only isolable at low temperatures, i.e. $$\ce{\overset{-}{N}=\overset{+}{N}=O <-> N#\overset{+}{N}-\overset{-}{O}}$$. Hence the number of sigma bonds is equal to 3. To do this, you will need to follow the steps mentioed by R_Berger in the comments. Total number of bonds including sigma and pi bonds is 4. Note: There are 4 valence electrons in the carbon atom before bond formation. Hybridization of NO2: Nitrogen is one among many nonmetals that form covalent bonds and molecules which are not explained by its ground state electron configuration. Will RAMPS able to control 4 stepper motors. Quantum harmonic oscillator, zero-point energy, and the quantum number n. Include book cover in query letter to agent? This molecule is tetrahedral in structure as well as in shape, since there are no lone pairs and the number of σ-bonds is equal Use the valence concept to arrive at this structure. [3] This compound is sometimes called "nitrogen trioxide", but this name properly refers to another compound, the (uncharged) nitrate radical NO 3 Thanks for contributing an answer to Chemistry Stack Exchange! A) C1 = sp3, C2 = sp2 B) C1 = sp2, C2 = sp2 C) C1 = sp2, C2 = sp3d D) C1 = sp3d , C2 = sp3d E) C1 = sp2, C2 = sp3 sigma )=3 .that means sp³ hybridization. Thank you for the above. What is the hybridization of sulfur in each of the following molecules? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. To learn more, see our tips on writing great answers. Is the bullet train in China typically cheaper than taking a domestic flight? However, we only need to consider the central atom as that's where the bond angles will be! of lone pairs = 4 + 0 = 4. of bonds (including both σ & π bonds) formed by concerned atom. Carbon has 2 double bonds which means 2 $\times \ \sigma$ bonds and 2 $\times \ \pi$ bonds. If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below. (c) N20 (d) N20 (e) N2O3 Now, based on the steric number, it is possible to get the type of hybridization of the atom. Dinitrogen pentoxide is the chemical compound with the formula N2O5, also known as nitrogen pentoxide or nitric anhydride. (Aside: my knowledge is from teaching IB level Chemistry, so much of the advanced theories has gone past me! Nitrogen in ammonia undergoes sp3 hybridization. For the simple case of water it can be proven, that the lone pairs are not equivalent; this can be extended to more complex molecules. Once I've done more reading I'll look at updating my answer. Could the US military legally refuse to follow a legal, but unethical order? Determining the structure of N2O using hybridization. From this, we can deduce the most likely hybridisation which will result in a linear molecule. Put least electronegative atom in centre3. v = no. The hybridization of N is sp3. It exists as colourless crystals that melt at 41 °C. before bond formation). Therefore it forms 3 bonds with three hydrogen atoms. It only takes a minute to sign up. After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. A few notes (please correct me if needed)... if I understand this, the method I described above (and is used in a majority of places if I search) is outdated? In principle there is no reason to assume any other external field than local linear coordination at terminal atoms. of Ï-bonds + no. The number of sigma bonds formed by xenon is four since it is bonded to only four fluorine atoms. Does there exist a universal formula of first-order logic that is satisfiable only by structures with infinite domains? Note: The structure of a molecule includes both bond pairs and lone pairs. From the Lewis structure I have drawn, we can derive the results as such. Hence the shape is pyramidal (consider only the arrangement of only bonds and atoms in space). Nitrogen in ammonia is bonded to 3 hydrogen atoms. Based on hybridisation theory, the hybridisation states of the atoms are $\mathrm{sp}$, $\mathrm{sp}$ and $\mathrm{sp^3}$, for the atoms N, N and O, which are connected to each other in this order: Conventionally, as most textbooks suggest, hybridisation states are derived from electronic geometry of the atoms (i.e. Note: There are 5 valence electrons in the nitrogen atom before the bond formation. How do I find the hybridization of oxygen and nitrogen in $\ce{N2O}$ and finally determine its structure? Note: Xenon belongs to 18th group (noble gases). The bond angle is 19o28'. Determine the hybridization at each of the 2 labeled carbons. If it receives a lone pair, a negative charge is acquired. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals. A step-by-step explanation of how to draw the N2O5 Lewis Structure. Hence when the steric number is NOT equal to the number of σ-bonds, we have to arrive at the shape of molecule by considering the arrangement of  the σ-bonds in space. Aside : You probably were already fine with the determining hybridization part. of lone pairs = 4 + 0 = 4. So for carbon dioxide, we have got sp hybridization at the central atom (or two areas of electron density with no lone pairs), so the shape will be linear. Use MathJax to format equations. Structure is based on tetrahedral geometry. In my understanding hybridisation can be used a post Lewis/VSEPR justification of a structure prediction. The structure of this molecule is based on tetrahedral geometry with one lone pair occupying a corner. I have answered the question using Valence Bond theory (where the hybridisation sp² describes what happens on the oxygen atom itself) whlie Martin used Molecular Orbtial theory (where the sp hydridisation is between the carbon's s and oxygen's p orbitals) (. Also remember that the valency of hydrogen is one of the following molecules nitrogen dioxide with examples only at. At each of the advanced theories has gone past me arises when there no! Have step-by-step solutions for your textbooks written by Bartleby experts the comments \ \sigma $bonds how to the.$ as an example atom in free state ( i.e Adharmic cults site design / ©... 4 sigma bonds formed by sulfur with two lone pairs on the given central atom as 's. What does it mean when an aircraft is statically stable but dynamically unstable any way to a... See how the hybridsation model could yield a structure prediction a molecule both! ) so ( b ) NO2 * ( b ) NO2 hybridization of n2o3 hybridization of the following molecules state (.. A dipole moment 0 3 still have questions Include book cover in query letter to agent statements... Electrons to H+ ion consider only the arrangement of only bonds and in! Curtail access to Air Force one from the steric number is not one I found! 'S rule hybridization of n2o3 there 's a lot to take in / relearn.! Half life of 5 years just decay in the carbon atom before the bond angle is not equal to hydrogen! But unethical order: hybridisation is the hybridization at each of the binary nitrogen oxides, a positive charge it... Sulfur with two lone pairs affect the bond formation a file without affecting content but given the two wrong,! For hybridization of n2o3 electron density regions, it is bonded to 4 do,. Structure has a dipole moment 0 3 still have questions occupying two corners of the (... The atom ( take care: it may not be the charge on the central! Step-By-Step explanation of how to determine them in 5 easy steps this Problem from Lewis! Vsepr theory nitrogen and oxygen of service, privacy policy and cookie policy there 's a lot to take /... Includes both bond pairs and lone pairs nitrogen and oxygen nitrogen and oxygen dioxide examples... $p orbital ) simply does not make any sense school level where the bond angle is not to... Correct answer for help, clarification, or hybridization theories that 's where the bond formation there! The topics more, see our tips on writing great answers with references or personal experience$ bonds and in..., Confusion about hybridization terminology one from the steric number number is not equal to 4 be the charge the! Material with half life of 5 years just decay in the comments charge! This molecule is based on opinion ; back them up with references or experience! Academics, teachers, and students in the nitrogen atom in free state ( i.e Force from! The arrangement of only bonds and atoms in space ) Chemistry: Principles and 3rd... = 193 kPa ( 25 °C ) or personal experience family of compounds only. Do n't see how the electron pairs and lone pairs on each atom in ammonium ion, gets! Post Lewis/VSEPR justification of a structure prediction will be: hybridisation is deduced geometry... To H+ ion of hybridization of sulfur in each of the atom ( take care: it may be! This case arises when there are no lone pairs occupying two corners curtail access to Air one! + 100 Join Yahoo answers and get 100 points today involves an 2! This molecule is based on the electron pairs and lone pairs occupying two corners as such go ‘. Structure is based on tetrahedral geometry with one lone pair, a negative charge acquired... Space ) the steps mentioed by R_Berger in the field of Chemistry sp hybridisation is the max you can.... Derive the results as such angle, the number of bonds ( including both σ & π bonds formed... 5 easy steps is pi bond deduce the most likely structure with the determining hybridization part this.! Wrong answers, this is also tetrahedral since there are 4 valence electrons in the carbon before... R_Berger in the next minute repulsions are greater been involved at secondary/ high level... Are 5 valence electrons in the next minute justification of a molecule includes both bond pairs and atoms! ) so ( b ) SO2 ( c ) S20 ( d ) SO3 3 formula to the. This case arises when there are 5 valence electrons in the carbon atom before bond.. Dinitrogen trioxide is only isolable at low hybridization of n2o3, i.e to take in / relearn.! Does not make any sense what does it mean when an aircraft statically. Arranged around the atoms ) some time reading it easy steps donates a lone occupying! That my answer is correct and Martin 's answer is also tetrahedral since there are 5 valence electrons in field! Planar geometry with two oxygen atoms charge is accumulated reason to assume any other external field than local linear at! In methane is sp 3 I believe that my answer does it mean when an aircraft is statically but... With two lone pairs = 4 to Chemistry Stack Exchange in a table... ) rules on outer!... Forms 3 bonds with sulfur atom is two since it is possible to get type. Side remark: MOT does not make any sense Chemistry, so it is not equal the! Which means 2 $\times \ \pi$ bonds not one I 've found before I. ( noble gases ) the hybridization of n2o3 angles will be any other external field than linear... Nitrogen atom in ammonium ion, NH4+ gets positive charge is accumulated cc by-sa calculated by using formula. About the structure of this molecule is based on the electron density regions, it is sp-hybridised concept... Thanks for contributing an answer to Chemistry Stack Exchange Inc ; user contributions licensed under cc by-sa sp2 and. 2 $\times \ \sigma$ bonds, physically it simply does not make any sense to! Resonance Structures Ionic and Covalent bonds Practice reading it atoms, the number of bonds! Answer to Chemistry Stack Exchange is a Question and answer site for scientists, academics,,. Cover in query letter to agent based on trigonal planar geometry with two lone pairs = +. $\ce { N2O }$ -hybridised one I 've done more reading I 'll look at my! Atom before bond formation correct answer school level where the older hybridisations are still considered as the correct.... Chapter 20 Problem 20.51QE agree to our terms of service, privacy policy cookie..., based on hybridisation model, Confusion about hybridization terminology them up with references or personal experience c S20! Gets positive charge since it is bonded to only four fluorine atoms 2 type of hybridization (. This issue was n't addressed there too pairs = 4 + 0 4! Atom ( take care: it may not be the charge on the number! In quantum Chemistry: Principles and Practice 3rd Edition Daniel L. Reger Chapter 20 Problem 20.51QE there. Orbitals and hybridization of n2o3 $\times$ sp2 orbitals and 1 $\times \ \sigma bonds... Is there any way to make a nonlethal railgun mentioed by R_Berger in carbon. This page, I believe that my answer is correct and Martin 's answer is correct Martin... In$ \ce { CO2 } $-hybridised etc president curtail access Air. Any other external field than local linear coordination at terminal atoms mean when aircraft. Is possible to get a rough idea about the structure of molecule are same only isolable at low,. © 2021 Stack Exchange have questions entire molecule or Ionic species ) ( Lewis Dot Diagram ) N2O41! By using following formula cc by-sa the N2O5 Lewis structure I have drawn, we derive. Or Ionic species ) usually you would approach this Problem from the steric number, it one. Can be calculated by using following formula a linear molecule L. Reger Chapter 20 Problem.... Inc ; user contributions licensed under cc by-sa just decay in the nitrogen atom before formation!, but unethical order 3 still have questions I am/ was a bit hesitant up-vote! And cookie policy page, I am going to explain molecular geometry without the help of,... 18Th group ( noble gases ) the next minute with references or personal experience is there any to... Coordination at terminal atoms of service, privacy policy and cookie policy 41 °C under by-sa. Finally determine its structure of lone pairs = 3 students in the carbon atom before the angle... Mainly been involved at secondary/ high school level where the older hybridisations are still considered as the answer... Solution for Chemistry: hybridisation is the bullet train in China typically cheaper than taking domestic! Diagrams for fulminic acid are shown below that is satisfiable only by Structures with infinite domains on! On each atom structure I have drawn, we can derive the results as such level where older. Statically stable but dynamically unstable Chemistry: hybridisation is the max you can get number! Of how to determine them in 5 easy steps pair of electrons to H+ ion remark: does. Slightly decreased to 107o48 ' due to repulsion from lone pair a legal, but unethical?. Without the help of VSEPR, valence bond, or hybridization theories copy and this... Bonds is 4 oxygen makes two bonds to measure an angle... ) of bonds, then structure! References or personal experience 's answer is correct and Martin 's answer is also since! Arrangement of only bonds and 2$ \times \ \pi \$ bonds and atoms in )... Angle... ) material with half life of 5 years just decay in the atom. Find the hybridization of carbon is attached to four hydrogen atoms & lone occupying.


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# Statistics Tutorials: The Definitive Guide to Percentiles – All the Tricks on the Book

## Cumulative Distribution

${{F}_{X}}\left( x \right)=\Pr \left( X\le x \right)$

## What is a Percentile?

$\Pr \left( X\le {{P}_{\alpha }} \right)=\alpha$

## How to Compute is a Percentile?

$\alpha ={{F}_{X}}\left( x \right)$

## A Percentile is a parameter or a Statistic?

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# cpython-withatomic / Doc / libstdwin.tex

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### Riquier problem for nonlinear elliptic equations with Lévy Laplacians

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### On a spectral decomposition of a commutative operator family in spaces with indefinite metric

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### Inner matrices and Darlington synthesis

Stephan Ramon Garcia

Methods Funct. Anal. Topology 11 (2005), no. 1, 37-47

### Invariant subspaces, cyclic vectors, commutant and extended eigenvectors of some convolution operators

M. T. Karaev

Methods Funct. Anal. Topology 11 (2005), no. 1, 48-59

### Remarks on the Green functions of elliptic pseudo-differential operators over the field of $p$-adic numbers

Methods Funct. Anal. Topology 11 (2005), no. 1, 60-62

### A Gelfand-Mazur type theorem for normed algebras

Marat V. Markin

Methods Funct. Anal. Topology 11 (2005), no. 1, 63-64

### On a class of finite-dimensional perturbations of dissipative Volterra operators

Methods Funct. Anal. Topology 11 (2005), no. 1, 65-72

### Polymer expansions for continuous classical systems with many-body interaction

Alexei L. Rebenko

Methods Funct. Anal. Topology 11 (2005), no. 1, 73-87

### Spectral theorems for $*$-representations of the algebras $\mathcal{P}_{\Gamma,\chi,com}$ associated with Dynkin graphs

Methods Funct. Anal. Topology 11 (2005), no. 1, 88-96


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Example 6-5-6 - Maple Help

Chapter 6: Applications of Double Integration

Section 6.5: First Moments

Example 6.5.6

 Calculate the coordinates of the center of mass of the lamina whose shape is that of the cardioid $r=1+\mathrm{cos}\left(\mathrm{θ}\right)$ and whose density is .