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Hiperreal numbers

1. Sep 9, 2005

eljose

Hello,there i would like to know what are hiperreal numbers (i think that are numbers including infinitesimals and infinite but i am not sure) i have read that are numbers there including the infinitesimals (numbers smaller than any positive number) and infinite numbers (numbers bigger than any positive number) but i am not sure..could someone provide informatio?..thanks.

2. Sep 9, 2005

HallsofIvy

Staff Emeritus
3. Sep 9, 2005

Hurkyl

Staff Emeritus
Let's construct a theory of analysis of real numbers in the following manner:

I first define the superstructure over the reals as follows:

Let S(0) = R.
Then, let S(n+1) = S(n) U P(S(n))

That is, at each step I add in all subsets of what I've seen so far, so S(1) consists of the real numbers and all sets containing just real numbers, S(2) consists of the real numbers, and all sets containing just real numbers or sets of real numbers, et cetera.

Then I let S be the union of all of the S(n).

Inuitively, S simply consists of all the sets we would ever need to use when doing real analysis.

Now, I'll define a theory of real analysis as follows:

To define the language, we will take the usual symbols of first-order logic, whatever variable symbols we like, and each element of S gets its own constant symbol, and we will also use the symbol $\in$ for the set membership operation. Furthermore, we will only allow bound quantifiers. IOW, we can say things like $\forall A \in B: P(A)$, but we cannot say $\forall A: P(A)$.

The axioms of the theory will simply be every true statement one can make in the above language.

Now, we can do all sorts of ugly logic tricks to this, but this is the easiest way to explain it:

Consider the collection of statements:

0 < x
x < 1
x < 1/2
x < 1/4
x < 1/8
x < 1/16
...

As it turns out, in the above theory, you cannot prove that there does not exist an x satisfying all of these inequalities. (To prove that there is no x requires we take an external viewpoint) (In generally, first-order logic is only capable of talking about fintely many statements... there's the compactness theorem that says if any finite subset of statements is internally consistent, then the entire infinite collection of statements is internally consistent)

So, this means that there must be some model of the above theory in which all of the axioms of the above theory are true, and this collection of statements is true! If we use this model, then we have (externally) proven that this model contains an infinitessimal number. (There's a cool theorem that says any consistent collection of statements has a model)

There are lots of ways to go about actually "constructing" such a model, and you can get many inequivalent models. The sets of numbers in these models that corresponds to the reals are called hyperreals, but often one particular canonical construction is used, and its hyperrals are called the hyperreals.

Last edited: Sep 9, 2005


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# For the hydrogen atom, the Balmer Series is the series of line spectra produced when an election fal

For the hydrogen atom, the Balmer Series is the series of line spectra produced when an election falls from any higher energy level and has n=2 as its final energy state. The longest wavelength is the Balmer Series is 656.7 nm. what was the initial principal energy level (n) associated with this photon.
Will the energy value in the trasnistion be the geatest or smalles in the entie Balmer Series?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

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Ahmed Kirby
Since energy $E=\frac{hc}{\lambda }$
So, longest wavelength means minimum energy
For balmer series principal energy level will be from n = 3
Energy is smallest

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# The GiveWell Blog

Our goal with hosting quarterly open threads is to give blog readers an opportunity to publicly raise comments or questions about GiveWell or related topics (in the comments section below). As always, you’re also welcome to email us at info@givewell.org or to request a call with GiveWell staff if you have feedback or questions you’d prefer to discuss privately. We’ll try to respond promptly to questions or comments.

You can view previous open threads here.

• Martin Randall on March 10, 2022 at 6:35 pm said:

Any thoughts on Ukraine related giving opportunities?

• Miranda Kaplan on March 11, 2022 at 1:30 pm said:

Hi, Martin,

Thank you for raising this topic. The staff at GiveWell share many of our followers’ shock and sadness at the crisis unfolding in Ukraine, as well as the desire to help. GiveWell’s focus remains, as ever, on finding the most cost-effective ways to save and improve lives on a daily, ongoing, longer-term basis; we generally don’t investigate giving opportunities related to humanitarian crises, such as those caused by war and natural disasters, so we unfortunately don’t have specific recommendations for giving to help relief efforts in Ukraine.

That said, we have written about giving during a crisis in the past. That blog post isn’t strictly relevant to the situation in Ukraine, but you may find some of the content helpful as you vet options for support (or for further clarification of our views on giving to disaster relief in general).

• Mark Pearson on March 11, 2022 at 3:30 pm said:

About five years ago, one of your senior advisors did a deep dive into the research around deworming [1] [2] and concluded the main research study relied on made him lack confidence “in the generalizability of that finding [of long term effects] to other settings.” Per [3], GiveWell concluded there is “weaker evidence on the causal relationship between reducing worm loads and improved life outcomes”. Nevertheless, GiveWell funds these programs because it still thinks “the possibility that deworming children has a subtle, lasting impact on [children’s] development” [4]. GiveWell admits this is a distant possibility, as it discounts the estimated long-term effects by 87% due to concerns about generalizability [5].

This is a big lack of confidence.

– You’ve expanded (hired more analysts).
– You likely have five years of additional information since the last update, either through academic research or data collected by NGOs.

P.S. Even if you can’t get more confidence that a certain long-term effect is real versus not real, at least you could get a better estimate of the “replicability adjustment” factor. This 2019 doc provides some ideas and mentions “a deworming replicability adjustment in the range of 5-30% (i.e., 70-95% discount) seems appropriate. Where one falls in that range requires a number of judgment calls.” So even if you can’t use newer data/studies put a better estimate of effect size and replicability and thus not need a replicability adjustment, at least you can shrink the uncertainty around that adjustment. Per GiveWell’s work, the adjustment factor can easily be off by a factor of 2.5 (current 13% adjustment version reasonable range of 5%-30%). That obviously has a lot of impact on how cost effective deworming programs look.

• Miranda Kaplan on March 17, 2022 at 1:56 pm said:

Hi, Mark,

Thanks for your comment. First, I want to note that you submitted a similar comment on the December 2021 open thread post, which we didn’t approve until just a few days ago. That was an oversight on our part, and we’re sorry this response is coming belatedly!

We don’t believe that there is significant additional valuable information to be gained from continuing to explore the existing literature on the long-term impacts of deworming. However, we are interested in funding projects to collect new data that would allow us to refine both our estimate of the effect size from deworming treatment and our replicability adjustment. For example, though we haven’t yet published a page about this, we recently recommended additional funding to the University of Washington for the ongoing DeWorm3 research study. This funding will support the principal investigators of DeWorm3 to create a plan to track medium- and long-term impacts on the study participants.

Additionally, in 2020 we introduced a new model to improve the methodology with which we estimate worm burden (more here). This is an important input into our cost-effectiveness analysis, given that one of our key uncertainties about deworming is the greater infection intensity among study participants in Miguel and Kremer 2004 relative to communities likely to be treated in deworming campaigns today. This update resulted in substantial changes to our cost-effectiveness estimates for some, though not all, programs and locations.

We also hope to write more and more clearly about the decisions we made in creating our current CEAs for deworming programs. However, much of our focus these days is on finding new granting opportunities to keep pace with our funds raised (see this blog post for more), and given existing capacity specifically on the research team that makes decisions about our deworming CEAs, we’re unsure of when we’ll be able to get to this.

• Sheridan on March 21, 2022 at 1:37 am said:

Is GiveWell considering the use of smart contracts to enforce donation matching? Smart contracts are good at solving some thorny coordination and incentive problems.

For example, a donor who wishes to give to a non-recommended charity A could set up a contract where they will give $X to charity A, unless another donor commits to give$X to GiveWell, in which case the contract forces the first donor to give the $X to GiveWell as well. • Sam Dyson on March 23, 2022 at 4:05 am said: I think it’s great that we should focus on treating malaria as the cost per life saved is relatively low and it has such a high yearly death toll. I was wondering, however, if there are any charities that you would recommend who are working on ways to eradicate malaria long term. I’m relatively new to the idea of EA and was wondering what the thinking is on diverting funds towards preventative measures rather than treatment. Apologies if this has been covered already. • Miranda Kaplan on March 24, 2022 at 2:09 pm said: Hi, Sheridan, We haven’t considered this before; thanks for bringing it to our attention. Though the idea is interesting, enforcing matches between donors seems to fall outside of our remit. We see our responsibility to donors as primarily to provide information to help them direct their donations, and not to facilitate coordination among them. However, we support donors exploring ways to achieve their goals for supporting effective giving. • Miranda Kaplan on March 29, 2022 at 12:39 pm said: Hi, Sam, Thanks for your comment! No worries about the retreading of any old ground; we’re happy to make our views more legible to readers. To clarify, all the malaria-focused programs that we currently support work on malaria prevention rather than treatment. The Against Malaria Foundation distributes long-lasting insecticide-treated nets (LLINs) that protect against malaria-carrying mosquitoes, and Malaria Consortium protects children against malaria during the high season of transmission via its seasonal malaria chemoprevention (SMC) program. (Malaria Consortium operates other malaria programs besides SMC, and some of those do include treatment, but our recommendation is only for its SMC program. We’ve also directed funding to Malaria Consortium for LLIN distribution.) Other interventions we’ve investigated, like intermittent preventive treatment for malaria (IPTi) and the RTS,S malaria vaccine, are also preventive. We don’t have recommendations for programs specifically targeting malaria eradication over the longer term. In general, we see our strength in identifying opportunities to help people in low- and middle-income countries over a relatively short period of time, and there continue to be large unfilled funding gaps in shorter-term malaria prevention. However, we note that all of the malaria interventions named above can be part of a country’s long-term malaria eradication strategy. Additionally, Open Philanthropy, whose work includes identifying opportunities to explicitly help future generations, has made a couple of grants related to malaria eradication that you may be interested to read about, here and here. • Sam Dyson on April 3, 2022 at 7:51 am said: Fantastic, Thank you for getting back to me and for the links! Cheers Sam • I think a quick win for accessibility would be to change your all caps navigation bar and footer to sentence case. I struggled to correctly type my email address in the email subscription box because it appeared in (small) all caps. The same goes for other areas of the site – for example your “XYZ COMMENTED ON” lines in the comments sidebar. • Miranda Kaplan on April 11, 2022 at 8:34 am said: Hi, Al, Noted—thank you for the suggestion! I’ve passed it along to our web team. • Vasco on April 16, 2022 at 4:50 am said: Hi, Firstly, thanks for your work! Secondly, I would be interested to know whether GiveWell has conducted cost-effectiveness analyses of the current top charities which take into account the long-term effects of the interventions on economic prosperity, population size and politics. • Miranda Kaplan on April 20, 2022 at 1:04 pm said: Hi, Vasco, Thank you for following our work! The cost-effectiveness analyses for our top charities do not estimate effects on economic prosperity, population size, and political developments in settings where these programs operate. We haven’t done any empirical research on this, but we broadly believe that the improvements in health and increased consumption we attribute to our top-charity programs are likely to have positive effects on general economic development. This blog post on “flow-through effects” elaborates on our reasoning. We do not attempt to gauge life-saving interventions’ effects on population size, in part because we don’t believe that overpopulation and associated harm are a significant risk. More in this FAQ and in a longer blog post about a study we commissioned on this topic from David Roodman. • ANGELO TOMEDI on April 22, 2022 at 11:09 am said: A recent article in Lancet (Malaria in 2022: A Year of Opportunity, DOI:https://doi.org/10.1016/S0140-6736(22)00729-2) stated that malaria mortality and incidence rates have not changed appreciably since 2015, and in fact increased by 12% and 7%, respectively, compared with 2015. Does this mean that all of the efforts of by AMF and Malaria Consortium have not made even a perceptible dent in the incidence and mortality? Not that the efforts are not worthwhile, but that is very depressing. • Miranda Kaplan on April 29, 2022 at 12:59 pm said: Hi, Angelo, It’s true that the World Health Organization has reported increases in malaria cases and deaths in 2020 compared with 2019 (see the foreword and “This year’s report at a glance” sections of WHO’s 2021 World Malaria Report). We are also saddened by this loss of life, which continued to be concentrated in children under five. It’s not clear whether the 2020 increases are indicative of a larger trend. The WHO report attributes them largely to: – Disruptions in programs that work on preventing malaria due to COVID-19. According to the report, supply chain disruptions and the resulting higher costs of purchasing, shipping, and distributing the commodities that these programs rely on seem to have been especially problematic. These led to, e.g. delays in LLIN distribution campaigns that may have left some households inadequately protected against malaria (see figure 2.11 on page 14 of the malaria report). – A 2019 update in WHO’s calculation of the distribution of causes of death in children under five. According to the report, this meant that the estimated percentage of deaths attributable to malaria between 2000 and 2020 rose, but it doesn’t reflect a larger trend. We base our expectation of the impact of AMF and Malaria Consortium’s SMC program largely on past randomized controlled trials that demonstrated the effectiveness of these interventions at reducing malaria rates. We are also interested in supplementing this evidence by directly measuring the impact of the SMC and LLIN campaigns these organizations support. On a global scale, this impact is difficult to gauge: GiveWell directs a lot of funding to these programs, but that total is still a small fraction of global funding for malaria, and SMC and LLIN distribution are only two of several types of anti-malaria interventions. On a more local level, we have funded some research that we hope will help us accomplish this—for example, we recommended funding to Malaria Consortium for monitoring and evaluation of an LLIN distribution in Ondo state, Nigeria, and Malaria Consortium is using GiveWell-directed funding for research to measure the impact of SMC in Mozambique (more here). • Oscar on May 2, 2022 at 8:01 pm said: I can see the value in evaluating giving opportunities in terms of how good they are relative to direct cash transfers. But I’ve always wondered why GiveWell spends so much money on direct cash transfers when you expect there to be other giving opportunities that are, say, 8 times more effective. Thanks in advance! • Miranda Kaplan on May 4, 2022 at 1:23 pm said: Hi, Oscar, We think that GiveDirectly is an excellent charity, which is why it’s remained on our top charities list since 2012. We rate GiveDirectly highly on three of four of our traditional criteria—GiveDirectly is transparent with us, its program is backed by evidence, and it’s able to put large amounts of funding to use. But, according to our cost-effectiveness estimates (which are very important inputs into our funding decisions, though they’re not the only factor), it’s substantially less cost-effective than other programs we’ve funded recently. Because of that, we haven’t made or recommended grants for GiveDirectly, beyond the minimum incentive grant we’ve recommended to Open Philanthropy for each of the top charities on our list, since 2015. You can see a record of all our grants to GiveDirectly since 2014 on this page. (Note that the table on that page doesn’t include gifts made to GiveWell but restricted by donors to supporting GiveDirectly, or donations to GiveDirectly made directly by donors using our top charities list, but not on the basis of a specific recommendation from us.) We’ve continued to use GiveDirectly as a benchmark of comparison for other funding opportunities because it is among the best charities we’ve identified, and because it can use enough funding that we don’t expect to fund anything less cost-effective than GiveDirectly. But we realize that having it on our “recommended” list—which individual donors may use to guide their giving decisions—and not actually recommending funding for it is confusing. As grantmaking becomes a more significant part of GiveWell’s work, we plan to improve our communication around our recommendations to donors, and we’ll share further updates later this year. For more on this, you may want to read “Why aren’t you just directing the extra money to GiveDirectly?” and “Why is GiveDirectly still a Top Charity?” on this page. • Oscar on May 4, 2022 at 7:54 pm said: Wow, thanks for the detailed response Miranda! That makes a lot of sense • Treon verdery on May 22, 2022 at 3:35 am said: Your thoughtful analysis of vitamin A supplementation suggests that it can save a life for about$3000 of child doses. The vitamin A chemical is already very affordable, but chemically could become 100-1000 times more dose potent with adding a fluorine atom to it. This effect of halogenation on drug potency is well known to chemists. Perhaps there could be some advantage to a new cheaper vitamin A molecule. What do you think?

• Treon verdery on May 22, 2022 at 4:27 am said:

At first glance, there does not seem to be a way to give money directly to Helen Keller International’s VAS (Vitamin A Supplementation) program without the money being shared with two other HKI programs, changing the amount to save a life through HKI to some amount other than $3000. I’d like to donate a miniscule$300 during June to their VAS program alone. Do you know of a way to do that without costing them administration expenses?

• Miranda Kaplan on May 25, 2022 at 2:43 pm said:

Hi, Treon,

Thank you for the suggestion and for your support of our recommended programs!

Re: your first comment, according to our analysis, lowering the cost of the vitamin A capsules used for VAS would not change the program’s cost-effectiveness by much. As you allude to, the capsules are already very cheap ($0.06 each), and they represent a small proportion of program cost ($0.78-\$1.42 per child, per supplementation round, depending on location). We don’t know enough about the chemical process of halogenation to know whether it would confer other advantages beyond increasing potency, such as making the nutrient easier to absorb.

Re: your second comment, in the past, direct donors to Helen Keller have been able to restrict their online donations to VAS by writing “VAS” in the comments of the online form. It seems as though the form has changed and no longer includes a space for comment. We’re checking with Helen Keller to see whether they still offer the option to restrict, and will update here (and update this page on our site) when we have an answer.

In the meantime, as an alternative, you may donate to GiveWell in support of VAS using this page (by filling out the box labeled “Helen Keller International’s vitamin A supplementation program” under “Support GiveWell’s top charities”). We do not take any fees, donations through GiveWell are tax deductible in the United States and the Netherlands (to the extent permitted by law), and donating through GiveWell allows us to better track the impact of our program recommendations.


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# zbMATH — the first resource for mathematics

Continuity of the metric projection and local solar properties of sets. Continuity of the metric projection and local solar properties of sets, continuity of the metric projection and solar properties. (English) Zbl 1423.41050
Let $$M$$ be a nonempty closed subset of a real normed space $$X$$ and $$P_M$$ be the (set-valued) metric projection on $$M$$. The set $$M$$ is called a sun if for every $$x \in X \setminus M$$ there is a $$y \in P_M(x)$$ such that for all $$\lambda \geq 0$$, $$y \in P_M(y+\lambda (y-x))$$. If in addition, for any $$x \in X \setminus M$$, $$P_M(x) \neq\emptyset$$, and the condition from before holds for any $$y \in P_M(x)$$, then $$M$$ is called a strict sun. $$M$$ is called an LG-set if for any $$x \notin M$$, $$\varepsilon > 0$$ and $$y \in P_{M \cap B(y,\varepsilon)}$$, $$y \in P_M(x)$$.
Let $$Q$$ be a property that subsets of $$X$$ can have. Then we say that $$M$$ has property $$P$$-$$Q$$ if for all $$x \in X$$, $$P_M(x)$$ is nonempty and has property $$Q$$ and that it has property $$B$$-$$Q$$ if for all $$x \in X$$ and $$r > 0$$, $$M \cap B(x,r)$$ has property $$Q$$ or is empty.
The author obtains some implications and equivalences among these and similar notions, e.g. that any LG-set that is a $$B$$-sun is a strict sun and that a $$P$$-compact $$B$$-sun is a sun. In addition, he derives sufficient conditions for a closed locally Chebyshev set to be Chebyshev.

##### MSC:
 41A65 Abstract approximation theory (approximation in normed linear spaces and other abstract spaces) 54C65 Selections in general topology
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##### References:
 [1] Alimov, AR, Monotone path-connectedness of Chebyshev sets in the space C(Q), Sbornik Math., 197, 1259-1272, (2006) · Zbl 1147.41011 [2] Alimov, AR, Convexity of bounded Chebyshev sets in finite-dimensional asymmetrically normed spaces, Izvestiya Saratov Gos. Univ. Nov. Ser. Matem. Mekh. Inform, 14, 489-497, (2014) · Zbl 1361.46011 [3] Alimov, AR, Local solarity of suns in linear normed spaces, J. Math. Sci. (N.Y.), 197, 447-454, (2014) · Zbl 1314.46018 [4] Alimov, AR, Monotone path-connectedness and solarity of Menger-connected sets in Banach spaces, Izvestiya: Math., 78, 3-19, (2014) · Zbl 1303.41018 [5] Alimov, AR, A monotone path-connected set with radially continuous metric projection is a strict sun, Sib. Matem. Zh, 57, 16-21, (2016) [6] Alimov, AR, Selections of the metric projection operator and strict solarity of sets with continuous metric projection, Sbornik Math., 208, 915-929, (2017) · Zbl 1426.41039 [7] Alimov, AR; Tsar’kov, IG, Connectedness and other geometric properties of suns and chebyshev sets, J. Math. Sci., 217, 683-730, (2016) · Zbl 1361.46012 [8] Alimov, AR; Tsar’kov, IG, Connectedness and solarity in problems of best and near-best approximation, Russ. Math. Surv., 71, 1-77, (2016) · Zbl 1350.41031 [9] Amir, D.; Deutsch, F., Suns, moons and quasi-polyhedra, J. Approx. Theory, 6, 176-201, (1972) · Zbl 0238.41014 [10] Blatter, J.; Morris, PD; Wulbert, DE, Continuity of the set-valued metric projection, Math. Ann, 178, 12-24, (1968) · Zbl 0189.42904 [11] Brosowski, B.; Deutsch, F., Some new continuity concepts for metric projections, Bull. Amer. Math. Soc, 6, 974-978, (1972) · Zbl 0255.41029 [12] Brosowski, B.; Deutsch, F., On some geometric properties of suns, J. Approx. Theory, 10, 245-267, (1974) · Zbl 0272.41020 [13] Brosowski, B.; Deutsch, F., Radial continuity of set-valued metric projections, J. Approx. Theory, 11, 236-253, (1974) · Zbl 0283.41014 [14] Flerov, A., Locally Chebyshev sets on the plane, Math. Notes, 97, 136-142, (2015) · Zbl 1329.46016 [15] Malykhin, YV, Relative widths of Sobolev classes in the uniform and integral metrics, Proc. Steklov Inst. Math, 293, 209-215, (2016) · Zbl 1362.46039 [16] Repovš, D., Semenov, P.V.: Continuous selections of multivalued mappings. In: Recent Progress in General Topology. III, pp. 711-749. Atlantis Press, Paris (2014) · Zbl 1304.54002 [17] Tsar’kov, IG, Bounded Chebyshev sets in finite-dimensional Banach spaces, Math. Notes, 36, 530-537, (1984) · Zbl 0602.41033 [18] Tsar’kov, IG, Continuity of the metric projection, structural and approximate properties of sets, Math. Notes, 47, 218-227, (1990) · Zbl 0703.46010 [19] Tsar’kov, IG, Bounded Chebyshev sets in finite-dimensional Banach spaces, Izv. Math, 80, 442-461, (2016) · Zbl 1356.46013 [20] Tsar’kov, IG, Continuous ε-selection, Sb. Math, 207, 267-285, (2016) · Zbl 1347.41047 [21] Vlasov, LP, Approximative properties of sets in normed linear spaces, Russ. Math. Surveys, 28, 3-66, (1973) · Zbl 0293.41031
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.


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# What does 5,000,000 tonnes of CO2 look like?

Reading a news article Probe after Gorgon releases millions of tonnes of greenhouse gas, I am confused. CO2 is a gas, right? Gases are not very dense, right? So what does 5MT of the stuff look like? How big a tank do we need to hold it?

• Please deselect the answer you have chosen. It is very wrong. May 11 '18 at 5:38
• Ummm... How so? May 11 '18 at 5:42
• In general, it's polite to wait at least a few hours (or days) before accepting an answer, in case a much better one comes along :-) May 11 '18 at 6:40
• I think there may well be several best answers here. :)) May 11 '18 at 6:54
• @David Hammen: Read the comments to that answer. It is not (very) wrong GIVEN THE ORIGINAL INPUT. If you know how to write answers that automatically adjust themselves when the question is changed, please feel free to tell us lesser minds how you do it :-) May 12 '18 at 19:22

The question is hard to give a definitive answer to, as it depends on the pressure at which the gas is kept in the tank. But to try to give an intuitive sense of how much gas this is, let's work out two things: Firstly, how big the box would be at atmospheric pressure; and secondly, how many standard gas bottles it would take.

## At atmospheric pressure

The mass of one mole of CO2 is 44 grams. So 5 MT of CO2 is $\frac{5\times 10^{12}}{44} = 114\times 10^9$ moles of gas.

The ideal gas law tells us that

$PV = nRT$

where $P$ is pressure, $V$ is volume, $n$ is the number of moles of the gas, $R$ is a constant, and $T$ is the temperature in Kelvin. If we assume a pressure of 100,000 Pa (roughly atmospheric pressure) and a temperature of 300 K (roughly room temperature), then

$V = \frac{nRT}{P} = \frac{114 \times 10^9 \times 8.314 \times 300}{10^5} = 2.8\times 10^9$ m3.

That volume works out to a cube with sides of 1.4 km.

## In gas bottles

A standard large CO2 bottle of the type that is familiar in the UK looks like the one on the right of this picture:

and according to BOC contains 34 kg of gas, at a pressure of up to 50 bar. So that's 29.4 cylinders per tonne or, to contain 5 million tonnes of CO2, 147 million of these cylinders.

"Look like" is not quite right, since gaseous CO2 is transparent to visible light. To get a sense of the volume, we can do some simple math, which unless I've made a mistake somewhere gives an area 21 km/13 miles on a side, stretching from sea level to space.

Showing my work:

x = 5 Gt = 5000000000 * 2000 lbs/ton = 10000000000000 lbs

Sea level pressure = 14.7 lbs/in^2, so 680272108843 in^2

144 in^2 per ft^2, so an area of 4724111866 ft^2, or a square 68732 ft on a side

Convert to miles = 13 miles on a side Convert to km = 20.9 km

• OK, my bad. One day I'll get used to all those prefixes... Post will be edited to reflect Million rather than 1000 Million. However, I think it is fair to say we're looking at a Very Big 100Km-high (Karman Line) Box. May 11 '18 at 5:25
• This should not be the correct answer. It is very wrong. Major problem: 5 million is 5000000, not 5000000000. Another problem: A pound mass and a pound force are different units. It happens to be a wash in this case, but equating the two (pounds mass and pounds force) is a sign of sloppiness. Rather minor problem: The article talked about tonnes (1000 kg) rather than tons (2000 lb). May 11 '18 at 5:42
• @David Hammen - Apart from my complete ineptitude with Mega-Giga-, jamesqf's arithmetic looks OK to me. My own check using his logic but my corrected number shows a 222.14m square, which sanity-checks with removing excess zeroes. So the Very Big Box meta-answer is still good enough for me. May 11 '18 at 6:05
• About "proper" units. In Oz we have been saddled with the Pascal as the unit of pressure, While this is technically correct, it is in fact rather useless for the laity. We note that in most of Europe (and France is the Home of SI) pressure is routinely expressed as Kg/cm^2. This avoids the possiblity of introducing headaches like momentum and acceleration. May 11 '18 at 6:15
• Using this approach (but in SI units, and with 5Mt rather than 5Gt), I reckon it's a block of the atmosphere resting on a square of sides around 700m. Done in a very quick 'n dirty fashion, though, so might well be wrong. Also, kinda wrong in principle, since the mass of that volume of CO2 would be significantly more than the mass of that volume of air. May 11 '18 at 6:37


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# $A$ nilpotent and $A+c_iB$ is nilpotent then $B$ is nilpotent.

Let $$A$$ and $$B$$ be $$n \times n$$ matrices over some field with $$A$$ nilpotent. Now let $$c_1,\ldots,c_{n+1}$$ be $$n+1$$ distinct scalars such that $$A+c_i B$$ is nilpotent for all $$i=1, \ldots,n+1$$. Then how can I show that $$B$$ is also nilpotent?

Thanks

Let us define the matrix valued polynomial $$p(t) = (A+tB)^n = \sum_{k=0}^n C_k t^k.$$ We see that each $$(i,j)$$-th component $$p(t)_{ij}$$ of $$p(t)$$ is a polynomial in $$t$$ of degree at most $$n$$. By the assumption, $$p(t)_{ij}=0$$ has $$n+1$$ distinct roots $$c_1,c_2,\ldots, c_{n+1}$$ on the underlying field $$\mathbb{F}$$. (This is why: if $$n\times n$$ matrix $$T$$ is nilpotent, the it holds $$T^k=O$$ for some $$k\le n$$ and hence $$T^n=O$$. Thus, by the assumption, $$p(c_k) = O$$ for all $$k=1,2,\ldots, n+1$$.) This means $$p(t)_{ij} \equiv 0$$ and it follows that $$p(t) \equiv O$$. Since the coefficient $$C_n$$ of $$t^n$$ is $$B^n$$, it follows that $$B^n=O$$.
• Can you be more explicit. I don't see why $p(t)_{ij}=0$ has $n+1$ distinct roots. PS I saw that your answer was upvoted and then downvoted. I did not downvote. I am just seeking a clarification. – Kavi Rama Murthy Jan 7 at 8:07
• Do we even need the fact that $A$ is nilpotent tho? – EuxhenH Jan 7 at 8:12
• @EuxhenH Maybe we can assume that one of $c_k$ is 0. – Song Jan 7 at 8:13
• @Song That would mean that $A$ is nilpotent as a consequence, but we still don't need it in the problem statement. – EuxhenH Jan 7 at 8:16


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# GuessPolySequenceFormulas

A Computer Algebra Package for Polynomial Sequence Recognition. The software package developed in the MS thesis research implements functions for the intelligent guessing of polynomial sequence formulas based on user-defined expected sequence factors of the input coefficients. We present a specialized hybrid approach to finding exact representations for polynomial sequences that is motivated by the need for an automated procedures to discover the precise forms of these sums based on user guidance, or intuition, as to special sequence factors present in the formulas. In particular, the package combines the user input on the expected special sequence factors in the polynomial coefficient formulas with calls to the existing functions as subroutines that then process formulas for the remaining sequence terms already recognized by these packages. The factorization-based approach to polynomial sequence recognition is unique to this package and allows the search functions to find expressions for polynomial sums involving Stirling numbers and other special triangular sequences that are not readily handled by other software packages. In contrast to many other sequence recognition and summation software, the package not provide an explicit proof, or certificate, for the correctness of these sequence formulas -- only computationally guided educated guesses at a complete identity generating the sequence over all n. The thesis contains a number of concrete, working examples of the package that are intended to both demonstrate its usage and to document its current sequence recognition capabilities.

## Keywords for this software

Anything in here will be replaced on browsers that support the canvas element


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## approximation

Posts: 134
Joined: Sat Sep 14, 2019 12:17 am

### approximation

can someone explain to me the rules of when we can approximate when K is smaller than 10^-3?

AArmellini_1I
Posts: 107
Joined: Fri Aug 09, 2019 12:15 am

### Re: approximation

Basically if K is less than 10^3, meaning that the acid/base barely disassociates, then we can assume that the dissociation affects the initial condition of your acid/base so little that it's insignificant. This means that when calculating your x, or the change in concentration, instead of have your denominator being [inital] - x, you can just leave it as [initial] since the concentration does not change significantly.

Posts: 134
Joined: Sat Sep 14, 2019 12:17 am

### Re: approximation

is the cutoff 10^-3? or 10^3?

bellaha4F
Posts: 104
Joined: Wed Sep 18, 2019 12:20 am

### Re: approximation

10^-3

Vivianvelazquez_1J
Posts: 109
Joined: Thu Jul 11, 2019 12:17 am
Been upvoted: 1 time

### Re: approximation

Ghadir Seder 1G wrote:is the cutoff 10^-3? or 10^3?

10^-3 is the cutoff.

Julie_Reyes1B
Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

### Re: approximation

It's also important to double check that the x value is in fact less than 5% of the initial concentration or the approximation is invalid. I always do that at the end of every problem in which I approximate, especially when the K value is only slightly smaller than 10-3

Return to “Equilibrium Constants & Calculating Concentrations”


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rcmdcheck_process is an R6 class, that extends the callr::rcmd_process class (which in turn extends processx::process.

## Usage

cp <- rcmdcheck_process$new(path = ".", args = character(), build_args = character(), check_dir = NULL, libpath = .libPaths(), repos = getOption("repos")) cp$parse_results()

Other methods are inherited from callr::rcmd_process and processx::process.

Note that you calling the get_output_connection and get_error_connection method on this is not a good idea, because then the stdout and/or stderr of the process will not be collected for parse_results().

You can still use the read_output_lines() and read_error_lines() methods to read the standard output and error, parse_results() is not affected by that.

## Arguments

• cp: A new rcmdcheck_process object.

• path: Path to a package tree or a package archive file. This is the package to check.

• args: Command line arguments to R CMD check.

• build_args: Command line arguments to R CMD build.

• check_dir: Directory for the results.

• libpath: The library path to set for the check.

• repos: The repos option to set for the check. This is needed for cyclic dependency checks if you use the --as-cran argument. The default uses the current value.

• env: A named character vector, extra environment variables to set in the check process.

## Details

Most methods are inherited from callr::rcmd_process and processx::process.

cp\$parse_results() parses the results, and returns an S3 object with fields errors, warnnigs and notes, just like rcmdcheck(). It is an error to call it before the process has finished. Use the wait() method to wait for the check to finish, or the is_alive() method to check if it is still running.


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# Plotting Spectrum Analysis

I am working on a spectrum analyzing simulator, which is meant to compute and plot the FFT of a sine wave carrier signal of $16 Hz$.

From my practicals with a spectrum analyzer, I know that the spike shows up on the $16Hz$ mark. But after computing my own FFT and passing the magnitude of the FFT to a plotter in Matlab, I get two spikes, one at the beginning and the other at the end.

My sample rate is $2048$ samples per second, and what I plotted was the respective $abs(FFT)$ against a frequency step of $16/2048$. Doing the fftshift only moves the spike to the $8Hz$ mark and not $16Hz$.

I've tried computing FFT in different programming languages and with different tools (java jtransform, Matlab, C language, Excel, two java written classes for FFT by different universities).

I dont think my FFT result are wrong, I think it has to do with the plotting. I do not understand why my plot does not look like it would in real life. From my description and information, can you see any problems that would cause my plot to be wrong?

Generic FFTs use complex-valued input arrays and provide complex-valued output arrays, so they can differentiate positive and negative frequencies for complex-valued signals. For real-valued signals, which is what it sounds like your case may be, this means that the positive and negative frequencies are mirrors of each other, so the spike at 16Hz will be mirrored by a spike at -16Hz. Matlab's FFT function output array has the positive frequencies first followed by the negative frequencies, so one would expect two spikes: the first at 16Hz, and the second wrapped to -16Hz which will be almost at the end of the array.

fftshift should have "unfolded" the output array so that the negative frequencies plot on the left, with DC (0Hz) in the center, and positive frequencies on the right. You should still see two spikes, one at -16 Hz and one at +16Hz, symmetric about zero.

For a Spectrum Analysis using matlab FFT function you need get the first half values from your absolute FFT result and you leave to get two spikes, take a look in the Nyquist Shannon sampling theorem.

Maybe are you confusing about FFT size and Sample rate, when you say:

My sample rate is 2048 samples per second

Are you right ? for me 2048 is your FFT size and not the Sample rate

against a frequency step of 16/2048

If i understood well it is your frequency resolution, if it is true are you limited by your sample rate and you never will be able to find 16Hz from the FFT!

• Ok ederwander, here is more explanation. I took 2048 samples of my sine wave (the wave in its entirety is 1sec long), thats why i concluded that my sampling rate is 2048. computing the absolute of the FFT of thses 2048 samples also returned a 2048 sized array of values (thats my FFT size). Now, if im to plot the first half of these results (1024 of them), what precisely do i plot them against to get the 16hz. – Naz_Jnr Dec 26 '13 at 23:04
• Nice, are you in the right way, but we really need know your sample rate, if you tell me i can help you better ... – ederwander Dec 27 '13 at 0:22
• if the definition of sampling rate is the number of samples of a wave taken per second, then my sampling rate is 2048. If otherwise, let me know what u mean by sampling rate so i can be ask d question better – Naz_Jnr Dec 27 '13 at 2:09
• When you record one signal the sample rate are defined by the number of samples per second, I really don't believe that your signal are sampled in 2048hz are you confusing the sample rate with frame size analysis, use wavread function from matlab to get the sample rate from your audio – ederwander Dec 27 '13 at 2:27
• yes, I have checked it out, my sampling rate is 2048hz. I did this for a greater accuracy. is there a disadvantage in this, and wat do I do next. – Naz_Jnr Dec 27 '13 at 22:24


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# zbMATH — the first resource for mathematics

Dynamical systems. Differential equations, maps and chaotic behaviour. (English) Zbl 0754.34001
London: Chapman & Hall. x, 330 p. (1992).
This is an updated and extended version of the authors’ well-known textbook $$(*)$$ [Ordinary differential equations. A qualitative approach with applications. London (1982; Zbl 0481.34005); Russian translation: Moscow (1986; Zbl 0671.34001)]. As the authors state, “this book is aimed at second and third year undergraduate students who have completed first courses in Calculus of Several Variables and Linear Algebra. Our approach is to use examples to illustrate the significance of the results presented. The text is supported by a mix of manageable and challenging exercises that give readers the opportunity to both consolidate and develop the ideas they encounter. As in $$(*)$$, we wish to highlight the significance of important theorems, to show how they are used and to stimulate interest in a deeper understanding of them.
We have retained our earlier introduction and discussion of linear systems (Chapters 1 and 2). Our treatment of non-linear differential equations has been extended to include Poincaré maps and phase spaces of dimension greater than two (Chapters 3 and 4). Applications involving planar phase spaces (covered in Chapter 4 of $$(*)$$) appear in Chapter 5. Problems involving non-planar phase spaces and families of systems are considered in Chapter 6, where elementary bifurcation theory is introduced and its application to chaotic behaviour is examined. Although ordinary differential equations remain the driving force behind the book, a substantial part of the new material concerns discrete dynamical systems and the title Ordinary Differential Equations is no longer appropriate. We have therefore chosen a new title for the extended text that clarifies its connection with the broader field of dynamical systems”.
Reviewer: W.Müller (Berlin)

##### MSC:
 34-01 Introductory exposition (textbooks, tutorial papers, etc.) pertaining to ordinary differential equations 34A26 Geometric methods in ordinary differential equations 34C23 Bifurcation theory for ordinary differential equations 37D45 Strange attractors, chaotic dynamics of systems with hyperbolic behavior 34C05 Topological structure of integral curves, singular points, limit cycles of ordinary differential equations 34C25 Periodic solutions to ordinary differential equations 37G99 Local and nonlocal bifurcation theory for dynamical systems 34C15 Nonlinear oscillations and coupled oscillators for ordinary differential equations


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# Tag Info

45

I just wanted to give a more concrete idea of how we know these equations even though we have trouble proving analytical theorems about them. Stuff moving in space Consider any stuff (as in, any conserved quantity) distributed over space. We know that we can describe this with a time-dependent density field $\rho(x,y,z,t)$ such that any little volume $dV$ ...

32

None of the interesting equations in physics can be derived from simpler principles, because if they could they wouldn't give any new information. That is, those simpler principles would already fully describe the system. Any new equation, whether it's the Navier-Stokes equations, Einstein's equations, the Schrodinger equation, or whatever, must be ...

27

Turbulence is indeed an unsolved problem both in physics and mathematics. Whether it is the "greatest" might be argued but for lack of good metrics probably for a long time. Why it is an unsolved problem from a mathematical point of view read Terry Tao (Fields medal) here. Why it is an unsolved problem from a physical point of view, read Ruelle and Takens ...

24

Frank White's Viscous Fluid Flow book contains a good list of these "exact" solutions. I am not sure if it is complete though. I've provided links to a few of the solutions. Steady flow between a fixed and moving plate Axially moving concentric cylinders Flow between rotating concentric cylinders Hagan-Poiseuille flow Combined Couette-Poiseuille flow ...

24

They are derivable from classical mechanics using either the continuum or molecular points of view. Starting with a continuum view, one applies conservation of mass, momentum, and energy to a control volume and the result is the Navier Stokes equations. The Navier Stokes equations, in the usual form, apply to Newtonian fluids, that is fluids whose stress ...

20

As @QMechanic mentioned in a comment, the Navier-Stokes equations are just $F = ma$, but they look much scarier. Assuming an incompressible fluid, you have: $$\rho \frac{D u_i}{D t} = -\frac{\partial \sigma_{ij}}{\partial x_j} + f_b$$ where $\rho$ is the density (mass per unit volume), $D u_i /D t$ is the acceleration (written in the Lagrangian form, ...

19

I once asked Putterman after a similar colloquium what he meant by this statement, and his answer was "long time tails". Long time tails are fractional powers that appear in the long time behavior of correlation functions, see, for example, here and here. These fractional powers are seen in molecular dynamics (they are more difficult to see experimentally), ...

19

You are not missing anything. Rather I think you are placing too much emphasis on the scientific accuracy of something said for effect in a very chatty presentation. The spoken words almost instantly are not repeated in the headlines, whereas the rest of the quote is. And the presenter does not develop this idea of being able to smell thioacetone almost ...

16

If you go through the process of non-dimensionalizing the equations, the math becomes more clear. If you start with the momentum equation (ignoring viscous forces because they aren't important for the analysis): $$\frac{\partial u_i}{\partial t} + \frac{\partial u_i u_j}{\partial x_j} = -\frac{1}{\rho} \frac{\partial p}{\partial x_i} + g$$ Then introduce ...

13

This usually only applies to a wall bounded flow and is normally restricted to incompressible fluids. This result usually manifests in boundary layer theory and can be obtained through order of magnitude analysis of the Navier-Stokes equations. The steady, incompressible, and constant property momentum equation in the $y$ direction takes the form, $$u \frac{... 12 The Navier Stokes equations are a combination of Newton's 2nd law of motion (differential form) with the 3D version of Newton's law of viscosity (i.e., the mechanical constitutive equation for a Newtonian fluid). What you were taught in Physics at school was correct. Not many analytic solutions exist to interesting practical problems, and numerical ... 12 The NS equations do include the gravity term, see the Wikipedia entry where it's included as a body force term. Under some conditions, it can be neglected: large Froude number (as tpg2114 showed) or hydrostatics (in which the gravity force is balanced by the pressure gradient) or horizontal flows (g in z-direction but flow is in x,\,y plane). It can ... 10 1. Why is force linear in the velocity? On a conceptual level, the idea behind stokes flow is that it is a very overdamped system because the viscosity is very high. When I say "overdamped" I am trying to make an analogy with a damped harmonic oscillator. You gave the equation$$\ddot{x}=-k^2 x,$$but a damped harmonic oscillator has a drag term:$$\ddot{x}=...

10

The traditional derivation of the Navier-Stokes equations starts by looking at a fluid parcel and the different fluxes over the surface in the integral form. The integral form is preferred as it is more general than the differential form: For the latter one has to assume differentiability and thus it is not valid for flow discontinuities such as shocks in ...

9

It's all about conservation of momentum, $F=ma$. Fluid can only change velocity by experiencing a force, and the only force it can feel is a pressure difference. So if there's a velocity difference, there's a pressure difference, and vice-versa (neglecting other things, like gravity). Read this beautiful exposition.

9

It comes from the notion of the boundary layer and whether it stays attached to the wall or not. If you consider the momentum equation normal to the wall, the only way there can be a pressure gradient normal to the wall is if there is a velocity or acceleration normal to the wall. If that is the case, then the boundary layer is no longer attached. This ...

7

Turbulence is not one of the great unsolved problems in physics. Physics tells us exactly how turbulence emerges as a direct consequence of local mass and momentum conservation. We can create multiparticle computer models such as lattice gas automata that generate turbulence at large length and time scales. We can write down the equations that govern ...

7

It's not just a simplifying assumption. There are good reasons as to why the approximation is used for incompressible flows: Vorticity is introduced into incompressible flows only through surface boundaries. A flow that is initially irrotational and and does not interact with a surface will remain irrotational. You can see this from the equation for the ...

7

There are known solutions to the Navier-Stokes equations. A simple example would be laminar shear-driven flow between two moving plates. Just as in the case of Einstein's equations, the known solutions regard simple situations with particular boundary conditions; a general solution that covers all possible cases is not known in either case. One should not ...

7

For a single-component fluid, the conservation of mass follows $$\left(\begin{array}{c}\text{mass of fluid } \\ \text{in volume }\Delta V\end{array}\right)=\left(\begin{array}{c}\text{flux of fluid } \\ \text{in/out of volume }\Delta V\end{array}\right)+\left(\begin{array}{c}\text{sources or} \\ \text{sinks in }\Delta V\end{array}\right)$$ In terms of a ...

7

The Reynolds number, with $\rho$ the density, $u$ the velocity magnitude, $\mu$ the viscosity and $L$ some characteristic length scale (e.g. channel height or pipe diameter) is given by $$\text{Re}=\frac{\rho~u~L}{\mu}.$$ This is a dimensionless relation of the ratio of inertial forces ($\rho u u$) to viscous forces ($\mu\frac{u}{L}$). It therefore signifies ...

7

To add to tpg124's answer and to answer the implict question in your statetent: How were the equations discovered in the first place if we can't solve them? simplicity and clarity of an equation's meaning and difficulty of working out the implications of that equation are two wholly different things. For the Navier-Stokes equation, the meaning is crystal ...

6

Solutions of the form $$cos(x_i)e^{-x_j}$$ are common specific solutions of the Navier-Stokes equations in simplified (not simple) problems. These are however problems where inertia is ignored, which you include. (Please note that I am using index notation, with $i,j\in\{1,2,3\}$). $x_j$ is then the wall normal direction. This is actually quite well ...

5

MS Mohamed et al begin with a "standard vector calculus formulation of the NS equations", $$\frac{∂ \bf u}{∂ t} - \mu ∆ {\bf u} + ({\bf u} \cdot \nabla) {\bf u} + \nabla p = 0$$ $$\nabla \cdot {\bf u} = 0$$ and use the rotational form, $$\frac{∂ \bf{u}}{∂ t} + \mu \nabla \times \nabla \times {\bf u} - {\bf u}\times(\nabla \times {\bf u}) + \nabla p^d ... 5 My answer will focus just on the mathematical parts pertaining to partial differential equations. Scale invariance is the fact that some partial differential equations stay the same if you appropriately scale the variables. For example the heat equation (where \boldsymbol{x} is the position vector in 1, 2 or 3D, doesn't matter)$$ \partial_t u(\...

5

The Navier-Stokes equation to which you refer is more generally the first moment of velocity of the Boltzmann equation. In order to get a proper connection to heating, you need a second-velocity-moment Navier-Stokes equation. The Boltzmann equation keeps track of distributions of particles. This changes the question from "What is the density and flow of a ...

5

I'm far from expert and my advice is basically to read Terry Tao's piece linked in John Rennie's comment. (Read Terry Tao is always good advice.) But I would like to make a couple of points. Singularities are not impossible or unphysical Mainly, I wanted to clear up the idea that, if we find smooth solutions that became singular in finite time that would ...

5

I belive you have it pretty much settled already. If I was to change anything, I would shrink instead of adding more items: Identify the relevant quantities of your system: Energy, Momentum, entropy, electric charge, mass ... Which may or may not be conserved. If you have boundary conditions, most probably you don't have energy and/or momentum conservation ...

5

First, Navier-Stokes governs the fluid in your setup. So, anything apart from the fluid will be an external force in N-S equation. Body-force means an external force that applies in the bulk of the fluid, like gravity or a magnetic force. Interaction with a "body", as a wing, which is external to the fluid domain, is done through boundary conditions : the ...

5

IANAFD but I'll stick my neck out and say this: resolving the Clay problem one way or another won't cause people doing CFD to lose any more sleep than they already do. First of all, Jean Leray proved the existence of weak solutions to Navier Stokes in $R^3$ way back in the 1930s, and that is pretty much what matters for the task of getting numerical ...

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


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# Find the value of the product of $\sqrt 3$ and $\sqrt[3]{2}$.

Last updated date: 18th Mar 2023
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Hint: write the roots in terms of exponents.
We are supposed to find the product of $\sqrt 3$ and $\sqrt[3]{2}$. So, the first step is to express roots in terms of exponents.
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {3^{\dfrac{1}{2}}} \times {2^{\dfrac{1}{3}}}$
Now, the step is to make the powers same,
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {3^{\dfrac{1}{2} \times \dfrac{3}{3}}} \times {2^{\dfrac{1}{3} \times \dfrac{2}{2}}}$
$\sqrt 3 \times \sqrt[3]{2} = {3^{\dfrac{3}{6}}} \times {2^{\dfrac{2}{6}}}$
We can take $\left( {{3^3}} \right)$ and $\left( {{2^2}} \right)$ in one bracket,
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {\left( {{3^3}} \right)^{\dfrac{1}{6}}} \times {\left( {{2^2}} \right)^{\dfrac{1}{6}}}$
We can write $\left( {{3^3}} \right) = 27$ and$\left( {{2^2}} \right) = 4$,
Therefore,
$\sqrt 3 \times \sqrt[3]{2} = {27^{\dfrac{1}{6}}} \times {4^{\dfrac{1}{6}}}$
We can take${\left( {27 \times 4} \right)^{\dfrac{1}{6}}}$ by the formula ${a^m} \times {b^m} = {(ab)^m}$
$\sqrt 3 \times \sqrt[3]{2} = {108^{\dfrac{1}{6}}}$
Now we can write it as,
$\sqrt 3 \times \sqrt[3]{2} = \sqrt[6]{{108}}$
Answer $\Rightarrow \sqrt[6]{{108}}$
Note: make sure in the end while writing the answer, you express it in terms of root.


text:
### Junior94's blog

By Junior94, 6 years ago,

Code Jam Round 1A starts in a few hours.

GL & HF.

• +36

 » 6 years ago, # | ← Rev. 2 →   +20 Are you that only person from Mozambique? That's cool.http://www.go-hero.net/jam/15/regions
•  » » 6 years ago, # ^ |   +3 Yeah and you seem to be in the same situation :)
 » 6 years ago, # |   +10 How to solve large input version of C (Logging)?
•  » » 6 years ago, # ^ |   +8 For each point, you can sort all other points around it and look for a line that passes through that point (+- collinearities) and cuts off as few other points as possible. You can just use something like 2 pointers, rotate the line and recalculate the number of points on either side of it. It runs in .
 » 6 years ago, # | ← Rev. 4 →   +8 Task C — is my approach incorrect or I just couldn't find all the bugs in an hour?To find an answer for a fixed point X: sort all vectors by polar angle from the basis in X, then considering every other point Y, count how many points there are strictly on the left side of X -> Y and on the right side. The minimum value for all Ys and all sides is the answer for the given X. If Y are considered in sorted order, this sides can be found by sliding window.UPD: Yep, my solution is exactly the same as Xellos described and I have also finally found a bug (long long f(); int x = f();). Lesson learned and all the extra flags from this were added alongside -Wall.
•  » » 6 years ago, # ^ |   +29 Sounds right, maybe you have an issue with corner case N = 1? That was my bug on which I spent almost the entire contest.
•  » » » 6 years ago, # ^ |   0 Thanks for suggestion, I had that covered, turns out it was a standard int overflow bug which I managed to avoid spotting because it was on a line with no arithmetic computations.
•  » » 6 years ago, # ^ |   +123 You should have used Excel.
•  » » 6 years ago, # ^ |   0 Maybe it's easier if you coded a quick O(N3) solution and compare what's wrong from both output? At least you can get 18 points in like 10 mins.
 » 6 years ago, # |   +64 I spent 10 minutes at start, 30 in the middle and 30 at the end of the contest to understand statement of the task A. But i didn't understand it yet. I just want to know, i alone with this problem?:)Only thing that i understand it is how to get number 25.
•  » » 6 years ago, # ^ |   +8 Same here, I still have no idea what that problem asked me to do...
•  » » 6 years ago, # ^ |   +8 I read the statement 3 to 4 times at the start and found it very, very ambiguous.
•  » » 6 years ago, # ^ |   -13 Code Jam has done it again. I'm impressed at the ability of the problem setters to make such HORRIBLE contests. This one was even worse than the Qualification Round, and that's quite a feat.I read Problem A statement like 5 times at the beginning of the contest, and I didn't understand what the hell it was asking. I asked for clarification, and some asshole jury replied with "Please read more carefully". If the problem was just that I was reading hastily, I wouldn't have contacted you, you stupid moron!So I decided to move on to Problem B, and solved it in around 15 minutes for both small and large. Then moved on to A again, and after seeing that there was no way I could understand the hyeroglyph written there, I tried to solve (without success) Problem C.So, anyway, can anyone please explain what the hell was Problem A asking?
•  » » » 6 years ago, # ^ |   +3 It's quite dumb.Basically there's a person who's eating mushrooms. They give you the # of mushrooms at 10 second intervals, and, through a process of eating and gaining mushrooms, what is the least number of mushrooms the person could have eaten using the two methods (note, for the second method, rate can be non-integral).
•  » » » » 6 years ago, # ^ |   +9 You don't need rate per second, you only need rate per 10 seconds, which is always integral.
•  » » » » » 6 years ago, # ^ |   +5 And that confused me for quite long time I also asked the question about that during the contest and the reply was: "Please read the problem statement more carefully. Consider looking at the limits and the example cases if those might help."My question was: "Can the speed of eating be decimal number like 3 mashrooms in 2 seconds => 1.5 per second?"The answer didn't help me. Probably the description for first test case "We can determine that she must eat mushrooms at a rate of at least 1 piece per second." confused me, trying to calculate rate per second...
•  » » » » » 6 years ago, # ^ |   -16 In what kind of multi-dimensional universe you can only eat integral number of mushrooms every 10 seconds however you can eat non-integral number of mushrooms per second???
•  » » » » » » 6 years ago, # ^ |   +1 Well, for example, 90 kilometers per hour is 1.5 kilometers per minute. That's in our universe, not sure about yours :)
•  » » » » » » » 6 years ago, # ^ |   0 I know, I guess I got frustrated because like I said this ruined the contest for me. Of course it's partially my fault too. It would have been much clearer in my opinion if the interval was always 1 second.
•  » » » 6 years ago, # ^ | ← Rev. 2 →   +55 I wholeheartedly agree that the task statement was exceedingly confusing. Hopefully this will clear it up:Every time the number of mushrooms on Kaylin’s plate decreases, it means she has eaten some of them. Bartholomew can add arbitrary mushrooms at Bartholomew times, so when the number increases, that’s his doing. What’s the smallest number of mushrooms Kaylin could’ve eaten? It doesn’t matter what the exact numbers are except that she can’t eat more than she has; what really matters is the difference between every two consecutive numbers. In particular, she doesn’t have to eat the mushrooms that are left on her plate at the end.The first example works like this:10 5 15 5 She had 10 mushrooms at time 0. She had 5 mushrooms at time 10. This means she must’ve eaten 5. She had 15 mushrooms at time 20. That’s fine, Bartholomew must’ve added some, but we don’t care. She had 5 mushrooms at time 30. This means she must’ve eaten 10. She left the 5 mushrooms on her plate and walked away. That’s it, she has eaten 5 + 10 = 15.Now, if she eats at a constant rate (in the second case we’re interested in), that rate must be at least 5 per 10 seconds because she ate 5 between times 0 and 10, and it must be at least 10 per 10 seconds because she ate 10 between times 20 and 30. Overall, the rate is at least 10 per 10 seconds. With this, we get: She had 10 mushrooms at time 0. She had 5 mushrooms at time 10. We know from our minimum speed that she must’ve eaten at least 10. Hence these 5 mushrooms must’ve been newly added by Bartholomew. She had 15 mushrooms at time 20. We know from our minimum speed that she must’ve eaten the 5 mushrooms she previously had. Bartholomew could’ve added some new mushrooms (at least 15) at any point between times 10 and 20. If he added them before time 20, then Kaylin would’ve already eaten some of them, but we want to minimize the amount she eats, so we can assume Bartholomew added exactly 15 new mushrooms exactly at time 20. She had 5 mushrooms at time 30. We know from our minimum speed that she must’ve eaten 10 out of the 15 mushrooms she previously had. She then left the 5 remaining mushrooms on her plate and walked away. So, if Kaylin ate at a constant rate, she must’ve eaten at least 10 + 5 + 10 = 25 mushrooms.
•  » » » » 6 years ago, # ^ |   +2 Thanks. I got it after reading ""we'll look at how many pieces of mushroom are on her plate", but sadly, that was after the contest.You should have been the one in charge of writing the statement, not the one who actually did it.
•  » » » » 6 years ago, # ^ |   +5 Thanks (:
•  » » » » 6 years ago, # ^ |   +5 Thanks a lot I understood the problem after reading your comment!
•  » » » » 6 years ago, # ^ |   0 she wouldn't walk away just like that, she probably ate Batholomew at the end cos she is a monster :D
•  » » 6 years ago, # ^ | ← Rev. 2 →   +6 I really didn't think the statement was unclear at all: you had to calculate the number of mushrooms the girl had eaten supposing he followed each of the two given strategies.The first strategy was pretty straightforward, so, I guess the problem was with the second one.In the second one, you assume that, as long as there are mushrooms in her plate, she eats them at a constant rate, that is, (number of shrooms eaten / second) is constant.
•  » » » 6 years ago, # ^ |   +5 I didn't have a problem understanding the second strategy. The first strategy was the problem. What the hell does it mean with "any number of mushroom pieces at anytime"? Just for the record, I still don't understand. If she eats "any number of pieces", I could just assume she never eats anything, and so the answer is always 0.
•  » » » » 6 years ago, # ^ |   0 Not if a[i + 1] < a[i]
•  » » » » 6 years ago, # ^ | ← Rev. 2 →   0 If she never eats anything, how can you account for the decrease in mushrooms between any 2 consecutive intervals? Which means, she has to eat that difference.
•  » » » » 6 years ago, # ^ | ← Rev. 2 →   0 Then, how would you explain an input like210 5if the only way mushrooms disappear from the plate is by being eaten?In this case, she can't eat 0 mushroom, she has to eat at least 5.
•  » » » » » 6 years ago, # ^ | ← Rev. 2 →   -42 That's because the problem statement was written by a monkey.It wasn't clear that the numbers given in the input were the amount of mushrooms on the girl's plate every 10 seconds. I interpreted the numbers were the amount of mushrooms Bartholomew puts on her plate every 10 seconds, because the statement is such a mess that by the time you get to the actual input, you're already confused.Wouldn't it be easier to just write a few more words like "If the given input is 10 5 15 5, then initially there are 10 mushrooms on her plate. Then she eats 5, and 5 remains. After that 10 more are put on her plate, then she eats 10, and there remains the final 5 mushrooms". With something like that, no one would have gotten confused.Seriously, the challenge has to be coming up with a solution and coding it, not deciphering the problem statement.I wouldn't be surprised if next time, they write a statement in latin. Everyone would have to use a translator, and that would add to the extra difficulty of the problem.
•  » » » » » » 6 years ago, # ^ |   0 "That's because the problem statement was written by a monkey." Lool :DI too couldn't understand the problem for a long time :P
•  » » » » » » 6 years ago, # ^ |   +32 You say "it wasn't clear that the numbers given in the input were the amount of mushrooms on the girl's plate every 10 seconds". Yet, the input description says that m_i are exactly "the number of mushrooms on Kaylin's plate at the start, and at 10-second intervals".I agree it could be clearer, but you should stop blaming it all on the problemsetter (especially with such anger) and take your part of the guilt. That's a way to become not only a better competitive programmer, but a better reader overall :)
•  » » » » 6 years ago, # ^ | ← Rev. 2 →   0 if m[i-1] > m[i], then she must have eaten at least m[i-1]-m[i] mushrooms in that time interval. There is nothing more to it :)
•  » » 6 years ago, # ^ |   +21 I spent about 20 minutes to understand. After n-th rereading statement, I saw "we'll look at how many pieces of mushroom are on her plate" and understood what we have and how solution can be implemented...
•  » » 6 years ago, # ^ |   +8 Yes, A was very difficult to understand. I just guessed what I should do, fortunately I guessed correctly.
•  » » 6 years ago, # ^ |   0 i could understand statement of the task A only in last 20 minutes of contest. really bad problem.
•  » » » 6 years ago, # ^ |   -26 Yes, you're not the only one. A few friends of mine got only 15 points because they couldn't understand that statement. And apparently, a lot of Codeforces users had serious trouble understanding it as well.What happened to Code Jam? Last year was wonderful, this year, so far, it's a mess.
•  » » 6 years ago, # ^ |   +53 Yes, I can solve task B + C under 16 minutes , but have spent another 16 minutes to solve task A (including 10+ minutes of guessing the meaning of this task).When I read the sentence after: For example, if the input is 10 5 15 5: I thought: "what the hell? did you miss to write the 'Input Format' part?".So the big secret is hidden in this line:  we'll look at how many pieces of mushroom are on her plate at 10-second intervals And in order to decrypt it you need to understand what means "10-second intervals", which is much beyond my language skill..
•  » » » 6 years ago, # ^ |   +8 I think it is better to say:we'll look at how many pieces of mushroom are on her plate every 10 seconds.
•  » » » » 6 years ago, # ^ |   -8 i also think it is better to say:we'll look at how many pieces of mushroom that are on here plate on every intervalthe 10 seconds should be ignored IMHO.
•  » » 6 years ago, # ^ |   +3 Problem A ruined my whole contest. What was the purpose for that stupid 10 second interval?? Even after the contest ended I was banging my head because I could not figure out how the result for the last sample case could be 244 (using the second strategy). Very poor, ambiguous, trite description unnecessarily prolonged.
•  » » » 6 years ago, # ^ | ← Rev. 6 →   0 The problem wording itself is confusing for method 2. But i figured the secret from their example.with input 10 5 15 5 what you need to do is to get the difference between the maximum difference among each eating interval to find her eating rate (forget about the 10 seconds thingy) and then sum every other digit that is less than, equal or has up to that value.i.e the difference between 15 and 5 is 10, so we start from index 0 to sum 10 + 5 + (10 out of the 15) {cos her eating rate is 10} and that gives us 25.
 » 6 years ago, # | ← Rev. 4 →   0 I got WA for task B (Small). My soln:if N <= B: then answer is N.else: Consider time instants from 1 to Max(B[i]). Find the seats that become available at time[i]. Add those seats to a list in increasing order of index. Let final list size be K. Ans will be (N % K)th element in the list.I basically thought that the values will repeat cyclically after a point. Is my method incorrect?Code with Max(B[i])EDIT: Got Correct Answer, after considering time instants upto LCM(B[i]).Code with LCM(B[i])
•  » » 6 years ago, # ^ | ← Rev. 2 →   +1 They will repeat starting from lcm(B[i]) (lowest^Wleast common multiple), not max(B[i])
•  » » » 6 years ago, # ^ |   +5 *least
•  » » » 6 years ago, # ^ |   0 I changed my code from Max(a[i]) to LCM(a[i]), still getting WA.Can someone please have a look at my code and tell me where I am wrong? Pastebin link — Code
•  » » » » 6 years ago, # ^ |   0 You missed a[2] in LCM
•  » » » » » 6 years ago, # ^ |   0 Thanks a lot slycelote :) It feels good to see "Correct!", although in Practice :P
 » 6 years ago, # |   0 What is the solution to B large?
•  » » 6 years ago, # ^ |   0 Can someone explain the Binary Search solution please?
•  » » » 6 years ago, # ^ | ← Rev. 2 →   +11 Notice as more time passes, more customers get served i.e. with increasing time customers served keep increasing. So you can do a binary search on the maximum time till which less than n customers have been served.Let this time be t and the customers served till t be n1. Therefore, n1 < n and customers served till t + 1 is greater than or equal to n.Now iterate through all the barbers. Those barbers for which Mi divides t will take a new customer at time t. Whenever, you encounter such a barber increase n1. The barber for which n1 becomes equal to n is the answer.Note: When I say served I mean either served or being served
•  » » » » 6 years ago, # ^ |   0 Thanks!
•  » » » » 6 years ago, # ^ |   0 Thanks for a great explanation akulsareen
•  » » 6 years ago, # ^ |   +14 You can perform a binary search on the time when the n th customer gets served. Once you have found correct time then finding the barber is also a simple matter. Let me know in case more explanation is needed.
 » 6 years ago, # |   +1 In B, the most obvious way(that led to a TLE [O(N*BlogB)]) was: n = N - B; for i in range(1,n+1): sort(M[]); // M=Original=pair, M[i].first = time, M[i].second = index of that barber ans = M[0].second; M[0].first += Original[M[0].second].first; return ans; I realize that the order of the indices will be cyclical. But how do you find that pattern and solve this quickly?
•  » » 6 years ago, # ^ | ← Rev. 2 →   +1 For finding the cycle, I just found out which seats were getting empty at time[i], and did that for all time instants uptil Max(M[i]).However, that gave WA, because we should consider LCM(M[i]), not Max(M[i]).Check out my implementation — Link to code
•  » » 6 years ago, # ^ |   0 It'll cycle every lcm(m[0], m[1], ...). But this solution will pass the small data only.
•  » » » 6 years ago, # ^ |   0 Can you please explain this in more detail?
•  » » » » 6 years ago, # ^ | ← Rev. 2 →   +2 For every i the i-th barber will be free after n seconds if n % m[i] = 0 so you need the least number that divides them all which is the least common multiple.Then you can reduce n to n % lcm(...) and iterate through the rest.
•  » » 6 years ago, # ^ |   0 I don't think it would work for Large — it will be cyclical, but with so many barbers it can be too long. I solved it with binary search — first determining the minute when last customer before n-th customer will get in (not get served, just gets in) and from there you can find which barber gets n-th customer.
 » 6 years ago, # |   0 What was the approach to solve Problem C small input.I got the solution for large input but lets say we would like to solve small input particularly what would be the approach ?
•  » » 6 years ago, # ^ |   +1 Try all possible sets of trees being left and find convex hull for each of them?
•  » » 6 years ago, # ^ | ← Rev. 2 →   +1 You can also go with the same approach as large input, but instead of using 2 pointers just iterate over all O(N2) lines connecting the points and for each line look how many points we can remove from either side of the line. Overall complexity is O(N3).
•  » » » 6 years ago, # ^ |   +17 This could actually pass the large input too if you have a fast enough computer. I was downloading the top solutions for C-large and I saw quite a few of those.
•  » » » 6 years ago, # ^ |   0 Can you explain how to reduce the O(N3) solution to O(N2) using two pointers?
•  » » » » 6 years ago, # ^ |   +21 As soon as you fixed the first point you sort the rest of the points by their polar angle relative to the first point. Then basically you will do a sweeping line and you will rotate it around that fixed point. This line will separate the plane into two halfplanes, also it will separate your set of points into two contiguous subsets (if you imagine your list being circular). So you introduce two pointers to keep track of first and last element in "left" halfplane for example. Then you rotate the line and move both pointers which will take O(N) to rotate it 360 degrees. And because of sorting it will be O(N2logN), not O(N2).
•  » » » » » 6 years ago, # ^ |   0 Got it. Thanks a lot.
 » 6 years ago, # |   +1 Can anyone tell me when will this round be added to the gym ?
•  » » 6 years ago, # ^ |   +8 You can practice on codejam webpage.https://code.google.com/codejam/contest/4224486/dashboard
•  » » » 6 years ago, # ^ |   0 Yes I know, but I wanted to solve some of the questions in codeforces too, just like the qualification round. Anyways thanks.
 » 6 years ago, # |   0 I had to read problem A like 5 times to understand it and I didn't even understood it because of the text per se but because of the test cases trying to figure out a way to produce them.
 » 6 years ago, # |   0 Can anyone help me how the question B is done? i mean the algorithm..? Thanks
•  » » 6 years ago, # ^ |   0 My solution was to find the time, when the Nth customer is served. This can be binary searched -> guess the time and count how many customers can be served by individual barbers and sum it together. When you have exact time T -> you have to find which one is going to serve Nth customer. This can be done in numerous ways. I counted the number of served customers in time T-1. And then I searched for barbers which were serving new customers in time T. And choose the appropriate one. The complexity is some logarithm from maximum time for binary search multiplied by the number of barbers. So it is something like O(log N * B).
•  » » » 6 years ago, # ^ |   0 What i did took long for it to finish running. I made an array of the barbers minutes and reduced it one by one and subtracted 1 from the N for each time the minutes became zero. then i repeated it until (while N>0) the array became the same as in the beginning. then i made N=N%(number deducted) and made it continue from there. Please find my mistake.
•  » » » » 6 years ago, # ^ |   +6 If the lengths to cut would have very big least common multiply your solution would be same as simulation, thus very slow. Good example are prime numbers — test case: 1 11 1000000000 2 3 5 7 11 13 17 19 23 29 31 The length have lcm much bigger than N, thus the array would be same later than N will reach 0.
•  » » » » » 6 years ago, # ^ |   0 Sorry I accidentally voted it less :( . Thanks I got the idea! Thanks very much :D :)
•  » » 6 years ago, # ^ | ← Rev. 4 →   +1 Note that, after T minutes, you can easily calculate how many people have been assigned a barber (including people that have finished). It is simply summing over the amount of people that each barber has, i.e. where ⌈·⌉ represents the ceiling function.Clearly, this sum grows as T grows, hence you can do a binary search for the highest time Tmax when less than N people are assigned a barber (i.e. such that cnt(Tmax) < N).At that exact time, you know that a barber will be available to service the N-th person. All you need to do is go through all available barbers (the k-th barber is available at this point if Tmax ≡ 0 (mod Mk)). Whenever encountering an available barber, we increment a variable X, which is initially set to X = cnt(Tmax). Once X = N, we have assigned a barber to the N-th person, and we only need to return that barber's id.Here's my code: http://hastebin.com/vigefudase.vala
 » 6 years ago, # |   +13 Just noticed — among the people who passed thus far into Round 2:http://www.go-hero.net/jam/15/languagesPython is placed second after C++ with 106 people vs 81 people using Java. Before it was always C++, Java, Python. Is it statistical quirk or a new trend? I bet Java will regain its second place in Round 2, but still worth noting I think.
•  » » 6 years ago, # ^ |   0 yeah.. java will take over in the next two rounds.
•  » » 6 years ago, # ^ |   0 It's a trend. After 3 (1A, 1B, 1C) rounds we have:C++ : 2260 Python: 456 Java: 336.So summing up: Among approximately 3,000 best active competitive programmers, without running speed restrictions, Python is more popular then Java. Would be interesting to see statistics for 500 best in three weeks.
 » 6 years ago, # |   0 Got 0/100
•  » » 6 years ago, # ^ |   -16 me too. lets do our best in the next round. :)
•  » » 6 years ago, # ^ |   +16 Maybe you'd do better if you trained more and shitposted less?
 » 6 years ago, # |   0 Contest Analysis for Round 1A: Analysis


text:
Consider the configuration consisting a +q charge...

Tags:
1. Sep 16, 2016

• Member advised to use the homework template for posts in the homework sections of PF.
and two -q charges aligned along the x axis as follows...

-q__________q__________-q <--- the space between them being d

a) Suppose the +q charge is displaced perpendicularly by a small vertical distance dy. What is the total force F acting on it? Taylor expand your answer to leading order in dy

b) Next suppose instead the +q charge is displaced along the x axis by a small dx distance. What is the force acting on it?

--------------

So for a) I'm kind of stuck at the Taylor expansion on the force.

F = (q)(-q)/4piε0d2 yhat + (q)(-q)/4piε0d2

where d= sqrt(x^2+dy^2)

I'm at a point where I'm trying to simplify something but feel that I"m doing something wrong in the math

Same for b)

Thanks!

2. Sep 16, 2016

vela

Staff Emeritus
One obvious problem here is you're trying to add a vector to a scalar. A less obvious one is you're using the variable $d$ in two different ways. You essentially have the magnitude of the force between the pairs of charges right:
$$F = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{x^2+y^2}$$ What should $x$ equal here?

What are the x-component and y-component of the net force on the middle charge going to be?

Once you have those expressions, then use the Taylor expansion about $y=0$.

3. Sep 16, 2016

The second expression should also have a yhat - I was posting in a bit of a hurry my bad.

Net force in the x direction is going to be 0 due to symmetry. The restoring force would be pointing back down to the origin.
I guess I'm stuck at getting that expression.

I should have
F = -2(q^2/4piε0(1/dy^2)yhat)

EDIT: Pardon the bad formatting and all - trying to get used to how the forum works. Also, pardon the mad noob question - I guess I understand the logic of where the problem is going I just can't put it into math.

Thanks again.

4. Sep 16, 2016

vela

Staff Emeritus
You were closer before. When the middle charge is collinear with the others, there should be no vertical force. Is your latest expression for the force consistent with that?

5. Sep 17, 2016

I"m supposed to calculate the force once it's shifted in delta y.

The picture above is where I always get to before not knowing what to do anymore.

Last edited by a moderator: May 8, 2017
6. Sep 17, 2016

Staff: Mentor

From your image it doesn't look to me like you've added the vectors of the forces properly.

By symmetry only the y-components of the forces survive the vector summation, but you haven't shown how those y-components are determined from the force magnitudes. Your work seems to be taking the full force magnitudes and just calling them y-components. Unless there's something special about your $\hat{y}$ vectors?

What do you need to multiply the force magnitudes by in order to "extract" the magnitude of their y-components?

7. Sep 17, 2016

I should have called it something else but

$\hat{r}$ = rvector/|r| = (0, y)/sqrt(x^2+y^2)

8. Sep 17, 2016

vela

Staff Emeritus
The numerator of what you wrote suggests $\vec{r} = (0,y)$ while the denominator suggests $\vec{r} = (x,y)$.

The vector $\hat r = \frac{\vec{r}}{\|\vec{r}\|}$ has to have a magnitude of 1. The vector $\frac{(0,y)}{\sqrt{x^2+y^2}}$ clearly doesn't, so the two can't be equal.

Forget about vector notation for right now. By what do you need to multiply the magnitude of the force by to get just the y-component? Note that $x$ won't appear in the final expression because $x=0$ for the middle charge.

9. Sep 17, 2016

r sin theta would give me that magnitude of y?

I really appreciate the help guys - sorry that I"m being a bit dense at the moment.

10. Sep 17, 2016

Staff: Mentor

If r is the magnitude of the vector of interest and θ is the angle that r makes with the positive x-axis, then yes, r sin(θ) would give you the magnitude of its y-component.

Now, see if you can "construct" sin(θ) from the given geometry using the triangle sides.

11. Sep 17, 2016

I"m not quite sure what you mean by construct here, so I assumed this

rsin theta = y

sin theta = y/sqrt(d^2+y^2)
So then I taylor expanded 1/sqrt(d^2+y^2) and since it's to the highest order of y, I ended up with

sin theta approximately equals y/d

Which when I put that back into the original force equation I come out with -2q^2y/4piepsilon0d^3

12. Sep 17, 2016

vela

Staff Emeritus
I don't think it matters in this case, but I'd wait to Taylor expand until after you get the expression for the force because the magnitude of the force also depends on $y$.

13. Sep 17, 2016


text:
# Ph.D. Defense | Similarity Algorithms for Embeddable Objects

Sep 24 2019 03:00 PM - Sep 24 2019 05:00 PM

Abstract:

The need to measure similarity between two objects is everywhere. Similarity between two objects is generally defined as an inverse function to the distance between them.
First, we propose the first non brute force algorithm to calculate the Gromov hyperbolicity constant. We present several approximate and exact algorithms to solve this problem. For example, we  provide an exact algorithm to compute the hyperbolicity constant $\delta$ in time $\bigO{n^{3.688}}$ for a discrete metric space. We also show that hyperbolicity at a fixed base-point cannot be computed in $O(n^{2.05})$ time, unless there exists a faster algorithm for (max,min) matrix multiplication than
currently known.
Then, we present a new system to find proteins similar in functionality. We employ text mining techniques to map text similarity to similarity in functionality. We use manually curated data from Swiss-Prot to train and build our system. The result is a search engine that given a query protein, reports the top similar proteins in functionality with 99\% accuracy. The system is tested extensively using GO annotations. We used this system, that predicts similarity in function, to enhance protein annotations. In particular, we were able to predict that some GO annotations should be added to some proteins. After careful literature reviews we were able to confirm many of those predictions, for example, in one case, we have 96\% prediction accuracy.
We also present a new algorithm for measuring the similarity between GPS traces. Our algorithm is robust against subsampling and supersampling. We perform experiments to compare this new similarity measure with the two main approaches that have been used so far: Dynamic Time Warping (DTW) and the Euclidean distance and our algorithm outperforms both of them in most of the cases.


text:
# Does the Jerk (derivative of acceleration) play a significant role in relativity?

In particular, I'm thinking of spaghettification that occurs as an object falls into a black hole (as an extreme example). But what about tidal forces like the tidal heating taking place on Jovian Moons like Io and Titan?

$$F=-\frac{\text{d}F_g}{\text{d}r}R=\frac{2G M m}{r^3}R$$
$$R$$ is the radius of the affected body.


text:
# Thread: trigonometric multiple angle formula

1. ## trigonometric multiple angle formula

Hi could I get Help with this question really struggling I need to use angle formula to show cos^2x=1/2 (cos2x-1).

2. ## Re: trigonometric multiple angle formula

Which "angle formula" are you supposed to use? Are you talking about the double angle formula for cosine?

$\displaystyle \cos(2x) = 2\cos^2 x -1$

Is that the "angle formula" you are referring to? If it is, then add 1 to both sides. Then divide both sides by 2.

3. ## Re: trigonometric multiple angle formula

I haven't been given one I've been told to use one of the multiple angle formula as to which one to use I'm not quite sure

4. ## Re: trigonometric multiple angle formula

$\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Let $\alpha = x$ and $\beta=x$. See what you get on the left and right sides.

5. ## Re: trigonometric multiple angle formula

Originally Posted by Nadir
Hi could I get Help with this question really struggling I need to use angle formula to show cos^2x=1/2 (cos2x-1).
The question is written incorrectly
It should be $\cos^2(x)=(1/2) (\cos(2x)\large\color{red}+1)$

6. ## Re: trigonometric multiple angle formula

Yeah I thought so I asked my teacher and he said he made a mistake with the question and it was incorrect thanks for pointing it out


text:
### Poincare’s Square and Unbounded Gomoku

Poincare’s hyperbolic disk model.

Henri Poincar’e was masterful in presenting scientific concepts and ideas in an accessible way. To explain that the Universe might be bounded and yet infinite, he imagined that the temperature of space decreased from the centre to the periphery in such a way that everything contracted with the distance from the centre. As travellers moved outward from the centre, everything got smaller in such a way that it would take an infinite time to reach the boundary.

Poincar’e described a beautiful geometric model with some fascinating properties. He envisioned a circular disk in the Euclidean plane, where distances were distorted to give it geometric properties quite different from those of Euclid’s Elements. For the Poincar’e disk, the metric is

$\displaystyle \mathrm{d}s^2 = \frac{4(\mathrm{d}x^2 + \mathrm{d}y^2)}{(1-(x^2+y^2))^2} = \frac{4(\mathrm{d}r^2 + r^2\mathrm{d}\theta^2)}{(1-r^2)^2}$

Basically, distances shrink with distance ${r=\sqrt{x^2+y^2}}$ from the origin. A diagram of Poincare’s disk is shown in the figure above, showing a selection of geodesics. Each of the (distorted) triangles in the figure has the same area in this metric.

From Circle to Square

We will consider a metric similar in some ways to that of the disk, but with points confined within a square centered at the origin. The metric of the Poincare Square is

$\displaystyle \mathrm{d}s^2 = \frac{\mathrm{d}x^2}{(1-x^2)^2} + \frac{\mathrm{d}y^2}{(1-y^2)^2}$

Let us consider the distance from the origin ${O}$ to the point ${P}$ at ${(x,0)}$:

$\displaystyle s(O,P) = \int \mathrm{d}s = \int_0^x \frac{\mathrm{d}x}{(1-x^2)} = \log\sqrt{\frac{1+x}{1-x}} \,.$

Clearly, the distance from the origin to the boundary at ${(1,0)}$ is infinite. We can invert this expression to obtain ${x}$ as a function of distance ${s}$:

$\displaystyle x = \tanh s \,.$

Geodesics

The general equations for geodesics may be written

$\displaystyle \frac{\mathrm{d}^2 x^\lambda}{\mathrm{d}\tau^2} + {\Gamma^\lambda}_{\mu\nu}\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau} = 0$

where ${{\Gamma^\lambda}_{\mu\nu}}$ are the Christoffel symbols, involving derivatives of the metric coefficients ${g_{\mu\nu}}$. They can be computed manually, but this is tedious and error-prone. Software to compute them automatically is available in association with the textbook on relativity by Hartle (2003). The Riemann tensor, Ricci tensor and curvature can also be computed in this way.

For the Poincar’e square, the Riemann tensor vanishes identically and the curvature of the space is zero. There are only two non-vanishing Christoffel symbols,

$\displaystyle {\Gamma^1}_{11} = \frac{2x}{1-x^2} \qquad\mbox{and} \qquad {\Gamma^1}_{22} = \frac{2y}{1-y^2} \,,$

so the geodesic equations are uncoupled from each other:

$\displaystyle \frac{\mathrm{d}^2 x}{\mathrm{d}\tau^2} + \frac{2x}{1-x^2} \left(\frac{\mathrm{d}x}{\mathrm{d}\tau} \right)^2 = 0 \qquad\qquad \frac{\mathrm{d}^2 y}{\mathrm{d}\tau^2} + \frac{2y}{1-y^2} \left(\frac{\mathrm{d}y}{\mathrm{d}\tau} \right)^2 = 0 \,.$

These equations are nonlinear, but they can be solved analytically, with the help of Mathematica. For initial conditions ${x(0) = x_0}$ and ${\mathrm{d}x/\mathrm{d}t (0)= u_0}$, the solution is

$\displaystyle x(t) = \frac {1 - \exp\left\{ - \frac{2 u_0}{1-x_0^2} \left[t - \frac{1 - x_0^2}{u_0}\log\left( -\sqrt{\frac{1-x_0}{1+x_0}}\, \right) \right] \right\} } {1 + \exp\left\{ - \frac{2 u_0}{1-x_0^2} \left[t - \frac{1 - x_0^2}{u_0}\log\left( -\sqrt{\frac{1-x_0}{1+x_0}}\, \right) \right] \right\} }$

The solution for ${y(t)}$ is identical in form, with initial conditions ${y(0) = y_0}$ and ${\mathrm{d}y/\mathrm{d}t (0)= v_0}$ (these solutions were confirmed by substituting them back into the equations).

Left: geodesics through the origin ${(0,0)}$ of the Poincar’e square. Right: geodesics through the point ${(-0.5,0.3)}$ of the Poincar’e square.

In the figure (left panel) we show a set of 100 geodesics passing through the origin. They are all “straight lines” in the Poincar’e metric. With two exceptions, they all extend from ${(-1,-1)}$ to ${(+1,+1)}$ or else from ${(-1,+1)}$ to ${(+1,-1)}$. The two exceptions are the lines corresponding to the ${x}$-axis and the ${y}$-axis.

In the figure (right panel) we show 100 geodesics passing through a general point — in fact, the point ${(-0.5,0.3)}$. Once again, they all extend from a corner to the opposite corner, with the exception of two geodesics, one parallel to the ${x}$-axis, the other to the ${y}$-axis.

Noughts and Crosses Unbounded

Children love to play noughts and crosses — aka tic-tac-toe — but soon lose interest: as long as both players use optimal strategies, the result is a draw, so the game has no deep theoretical interest. The first player has nine options. There are eight cells remaining for the second move, and so on. So there are at most ${9!}$ or about a third of a million games. However, many of these are equivalent. For the first move there are essentially three choices. You might like to estimate the total number of essentially unique games.

Gomoku is a more interesting game of this sort. Players alternately mark a cell with black and white stones or noughts and crosses. The goal is to get five stones or marks in a row, horizontal, vertical or diagonal. The game can be played on a Go board or with pencil and paper. There are typically 15 by 15 cells but, in principle, the playing field can be unlimited.

Left: an infinite grid for Gomoku or infinite noughts and crosses. Right: Infinite chessboard; all squares have equal area.

Is there any way to fit an infinite number of cells on a finite board? In the figure above we show a grid of lines parallel to the axes and equidistant from each other in the Poincar’e Square metric. The left panel shows an “infinite gomoku board”. In the right hand panel is an infinite chess board. Despite appearances, all the squares are of equal area.

Sources

${\bullet}$ Hartle, James B., 2021: Gravity: an Introduction to Einstein’s General Relativity. Republished 1st Edition, Cambridge University Press. 602pp. ISBN: 9-781-3165-1754-3


text:
# What is your limiting reactant if 32 grams of CH4 reacts with 32 grams of oxygen gas?

Apr 3, 2018

Oxygen is our limiting reactant.

#### Explanation:

We must start by creating a balanced reaction between Methane and Oxygen. As methane is a hydrocarbon reacting with oxygen this will be a combustion reaction resulting in carbon dioxide and water.

The combustion reaction is:

$C {H}_{4} + 2 {O}_{2} \to C {O}_{2} + 2 {H}_{2} O$
Now we want to find how many moles of each reactant we have to find which one is limiting.

If we do a rough calculation for finding how many moles of $C {H}_{4}$ and how many moles of ${O}_{2}$ we have we get the following:
Using the molarity equation:

$n = \frac{m}{M}$
M=molar mass of the compound/element,
m=mass in grams of the compound/element
n=number of moles of the compound/element

n(Oxygen)=$\frac{32}{32} = 1$ mol

n(methane)=$\left(\frac{32}{16}\right) = 2$ mol

However, in the reaction equation, we need 2 moles of oxygen for every one mole of Methane, and we only have 1 mole of oxygen and 2 moles of Methane. That means methane is in excess and oxygen is thus our limiting reactant.

So in this reaction we would only be able to react 1 mol of ${O}_{2}$ with 0.5 mol of $C {H}_{4}$.


text:
# The Standard Model fermions as excitations of an ether

Authors
Type
Published Article
Publication Date
Nov 08, 2012
Submission Date
Dec 19, 2009
Source
arXiv
The three-dimensional complexified exteriour bundle $C \otimes \Lambda(R^3)$ is proposed as a geometric interpretation of electroweak doublets of Dirac fermions. The Dirac equation on this bundle allows a staggered discretization on a three-dimensional scalar complex lattice field $C(Z^3)$. The three-dimensional affine group $Aff(3)$ is proposed as a geometric interpretation for the matrix defined by the three generations, with the three quark doublets and a lepton doublet in each generation. The corresponding lattice space $(Aff(3) \otimes C)(Z^3)$ allows a simple condensed matter (ether) interpretation. A possibility to construct anticommuting fermion operators based on canonical quantization is given. A set of axioms of the gauge action is defined such that the gauge action of the SM gauge group $SU(3)_c x SU(2)_L x U(1)_{Y}$ is the maximal possible one. This includes preservation of Euclidean symmetry and the symplectic structure, anomaly freedom, ground state neutrality, and the possibility of realization on the lattice. Strong interactions are realized as Wilson-like gauge fields, weak gauge fields as effective gauge fields describing lattice deformations, and the EM field as a combination of above types. Some possibilities to explain the qualitative characteristics of the mass parameters are discussed.


text:
B Possible outcomes of reactions

1. Apr 22, 2017

Biker

Hello,

1) In beta- decay, Electrons can have zero kinetic energy. Do they mean that they initially had kinetic energy to overcome the coulomb attraction which manifests itself in the "Free" electron mass? So it can literally have any energy from zero up to the maximum energy and also there is no reason for why a neutrino can't have all the energy ( excluding recoil energy). It is all probabilities.

2)
Example 198Hg79 ---)198Hg80 ----) 198Hg80
^ Gamma emission.
Do you have to have gamma emission in certain reactions? or in every reaction, there is a probability that it can happen in one step through beta decay?

Last edited: Apr 22, 2017
2. Apr 22, 2017

Staff: Mentor

Yes.

Electrons can get even lower energies - then they have to be in some orbital. This is called bound-state beta decay.
I don't understand that part of the question.

Gamma emission is possible, it does not have to happen.