text:
• ### The PANDA DIRC Detectors at FAIR(1707.09269)

July 28, 2017 physics.ins-det
The PANDA detector at the international accelerator Facility for Antiproton and Ion Research in Europe (FAIR) addresses fundamental questions of hadron physics. An excellent hadronic particle identification (PID) will be accomplished by two DIRC (Detection of Internally Reflected Cherenkov light) counters in the target spectrometer. The design for the barrel region covering polar angles between 22 deg. to 140 deg. is based on the successful BABAR DIRC with several key improvements, such as fast photon timing and a compact imaging region. The novel Endcap Disc DIRC will cover the smaller forward angles between 5 deg. (10 deg.) to 22 deg. in the vertical (horizontal) direction. Both DIRC counters will use lifetime-enhanced microchannel plate PMTs for photon detection in combination with fast readout electronics. Geant4 simulations and tests with several prototypes at various beam facilities have been used to evaluate the designs and validate the expected PID performance of both PANDA DIRC counters.
• The exclusive charmonium production process in $\bar{p}p$ annihilation with an associated $\pi^0$ meson $\bar{p}p\to J/\psi\pi^0$ is studied in the framework of QCD collinear factorization. The feasibility of measuring this reaction through the $J/\psi\to e^+e^-$ decay channel with the PANDA (AntiProton ANnihilation at DArmstadt) experiment is investigated. Simulations on signal reconstruction efficiency as well as the background rejection from various sources including the $\bar{p}p\to\pi^+\pi^-\pi^0$ and $\bar{p}p\to J/\psi\pi^0\pi^0$ reactions are performed with PandaRoot, the simulation and analysis software framework of the PANDA experiment. It is shown that the measurement can be done at PANDA with significant constraining power under the assumption of an integrated luminosity attainable in four to five months of data taking at the maximum design luminosity.
• Simulation results for future measurements of electromagnetic proton form factors at \PANDA (FAIR) within the PandaRoot software framework are reported. The statistical precision with which the proton form factors can be determined is estimated. The signal channel $\bar p p \to e^+ e^-$ is studied on the basis of two different but consistent procedures. The suppression of the main background channel, $\textit{i.e.}$ $\bar p p \to \pi^+ \pi^-$, is studied. Furthermore, the background versus signal efficiency, statistical and systematical uncertainties on the extracted proton form factors are evaluated using two different procedures. The results are consistent with those of a previous simulation study using an older, simplified framework. However, a slightly better precision is achieved in the PandaRoot study in a large range of momentum transfer, assuming the nominal beam conditions and detector performance.
• The ALICE Collaboration has measured inclusive J/psi production in pp collisions at a center of mass energy sqrt(s)=2.76 TeV at the LHC. The results presented in this Letter refer to the rapidity ranges |y|<0.9 and 2.5<y<4 and have been obtained by measuring the electron and muon pair decay channels, respectively. The integrated luminosities for the two channels are L^e_int=1.1 nb^-1 and L^mu_int=19.9 nb^-1, and the corresponding signal statistics are N_J/psi^e+e-=59 +/- 14 and N_J/psi^mu+mu-=1364 +/- 53. We present dsigma_J/psi/dy for the two rapidity regions under study and, for the forward-y range, d^2sigma_J/psi/dydp_t in the transverse momentum domain 0<p_t<8 GeV/c. The results are compared with previously published results at sqrt(s)=7 TeV and with theoretical calculations.


text:
If you're seeing this message, it means we're having trouble loading external resources for Khan Academy.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Area defined by polar graphs

4 videos
1 skill
We'll now use the power of the integral to find areas defined by polar graphs!

### Area enclosed by polar graph formula intuition

VIDEO 6:31 minutes

### Area enclosed by cardioid

VIDEO 6:26 minutes

### Area between two polar graphs

VIDEO 6:09 minutes

### Evaluating definite integral with calculator

VIDEO 2:59 minutes

### Area enclosed by polar graphs

PRACTICE PROBLEMS

format: plain


text:
Publication
Title
Quantum chemical DFT study of 4-azatricyclo [$5.2.2.0^{2,6}$] undecane-3,5,8-trione
Author
Abstract
 Fourier-transform-Raman and infrared spectrum of 4-azatricyclo [5.2.2.02,6] undecane-3,5,8-trione were recorded and analyzed. The vibrational wavenumbers were examined theoretically using the Gaussian03 set of quantum chemistry codes, and the normal modes were assigned by potential energy distribution (PED) calculations. The first hyperpolarizability, predicted infrared intensities and Raman activities are reported. The calculated first hyperpolarizability is comparable with reported values of similar structures which makes this compound an attractive object for future studies of nonlinear optics. Optimized geometrical parameters of the compound are in agreement with similar reported structures. The red shift of the NH stretching wavenumber in the infrared spectrum from the computational wavenumber indicates the weakening of NH bond.
Language
English
Source (journal)
Spectrochimica acta: part A: molecular and biomolecular spectroscopy. - Oxford
Publication
Oxford : 2010
ISSN
1386-1425
Volume/pages
75:5(2010), p. 1559-1565
ISI
000277231700030
Full text (Publisher's DOI)
UAntwerpen
 Faculty/Department Research group [E?say:metaLocaldata.cgzprojectinf] Publication type Subject Affiliation Publications with a UAntwerp address


text:
Populating price data from Bloomberg in Excel with BDH via macro not returning time series

Though issue has been addressed before on stackoverflow and reddit, I was not able to find any useful answers.

I'm running a macro to populate price data with a macro via excel bloomberg API, say

=@BDH("AAPL US EQUITY"; "PX_LAST"; "ED-5AY"; BToday())

and

=@BDH("AAPL US EQUITY"; "PX_LAST"; "ED-3AY"; BToday())

My problem is that my macro is too fast and sometimes it does not return the proper time series. Sometimes it returns just one value in a cell or it will get stuck on #N/A Requesting Data.... I have been trying everything from trying to utilize Application.Run "RefreshEntireWorksheet" and Application.Run "RefreshAllStaticData" as well as all kinds of exotic waiting functions.

I could not figure out why this happens nor find a satisfying solution which would let me loop through all my securities and desired time frames. I found out that it seems to work better, if the function already knows the cols and rows, which can also be specified in BDH. But I am quite lost on this. So any advice on how to solve this annoying problem is much appreciated. You can use the macro I wrote.

Sub example()
Dim startRange As Range
Set startRange = Range("A1")

startRange.Formula = "=@BDH(" & Chr(34) & "AAPL US EQUITY" & Chr(34) & _
", " & Chr(34) & "PX_LAST" & Chr(34) & ", " & Chr(34) & "ED-5AY" & _
Chr(34) & ", BToday())"

startRange.Offset(0, 2).Formula = "=@BDH(" & Chr(34) & "AAPL US EQUITY" & Chr(34) & _
", " & Chr(34) & "PX_LAST" & Chr(34) & ", " & Chr(34) & "ED-5AY" & _
Chr(34) & ", BToday())"
End Sub

I was able to replicate my problem, by running my macro in two different worksheets within the same workbook.

• Best option is to request the data using the COM addin from VBA directly, you may find examples or libraries online. Alternatively you can try using Application.onTime to check at regular intervals if the data has been populated and continue your program once it's the case. Oct 13 '20 at 14:35
• I saw the Application.onTime method in the reddit link. I literally copied the routine and call but it did bring any improvements. I will take a look at the COM addin. Oct 13 '20 at 14:37

1 Answer

It would be better not to use (old) BDP/BDH, but to use either the COM wrapper (which can be synchronous) or the new BQL or BQL.query calls (HELP BQLX). Some time ago I wrote a a few lines of code (similar to https://stackoverflow.com/questions/33661436/ ) to get synchronous behavior from BDP/BDH. In VBA, call Excel to search the range for value "#N/A Requesting Data". If something is still found, then refresh the calls, sleep 1/2 second and search again. else (if the search fails, then) all your data is ready.

• I ended up using this approach mikejuniperhill.blogspot.com/2013/06/… Oct 15 '20 at 15:01
• I used this one too, it works well. Oct 15 '20 at 15:52


text:
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### Recent Submissions

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This dissertation considers the representation theory of Temperley-Lieb algebras, TLₙ , along with some related cellular algebras, over positive and mixed characteristic fields. The Temperley-Lieb algebras are planar ...
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This thesis will consist of three main chapters. It is a standard fact of linear algebra that every matrix with rank k contains a k ⨯ k submatrix with rank k. In Chapter 2 we generalise this fact asymptotically to a class ...
• #### Modern Methods for Variable Significance Testing

This thesis concerns the ubiquitous statistical problem of variable significance testing. The first chapter contains an account of classical approaches to variable significance testing including different perspectives on ...


text:
# Mathematics 1010 online

## Complex Numbers

Recall how we built the number system : We started with the set of natural numbers, expanded it to the integers to make subtraction always possible, expanded them to rational numbers to make division always possible (except division by zero), and then expanded them to real numbers to account for the existence of numbers that manifestly are not rational.

We have to expand the numbers system one more time, to include complex numbers. Unfortunately it is not easy to motivate this expansion as compellingly as the previous expansions. Complex numbers have a tremendous range of applications, that will become apparent as you move on in Mathematics, Science, and Engineering, but they may be hard to appreciate at this stage.

For our purposes we start with the observation that there is no real number such that

This follows from the fact that the square of a positive number is positive, and that of a negative number is positive also. Thus there's no positive or negative number whose square is negative (and the square of zero of course is also non-negative).

We now take a leap of faith the kind of which sometimes gives mathematics a bad name. We don't know of a number satisfying the above equation, so we stipulate that there is a solution and we give it a name. We call it .

There are only two things about we need to know and appreciate. One is the fact that

The second fact is that we define the arithmetic operations with such that the ordinary rules of arithmetic continue to hold.

Let's consider some consequences of this fact. We obtain in particular

It is clear how this process continues. The powers of with natural numbers as exponents are and so on forever. We also have, for example, , , , , etc. What about the sum of and a real number like ? In other words, what is ? There is no way to simplify that expression, and so we leave it as it is!

Let's take stock and introduce some language. The symbol is called the imaginary unit. A complex number is an expression that can be written in the form where and are real numbers (and multiplies ). That particular form is sometimes called the standard form of a complex number. is called the real part of , and is its imaginary part. If the real part of is zero, and the imaginary part non-zero, then is called an imaginary number. (If the imaginary part is zero, then is a real number.) If you want to emphasize that the real part is zero you can call the number pure imaginary or purely imaginary.

Note that the imaginary part of a complex number is in fact a real number (and not, as you might reasonably expect, an imaginary number).

Consider, for example, the complex number . Its real part is , and its imaginary part is . The number is also a complex number. Its standard form is , its real part , and its imaginary part . The number is imaginary, the number is real. Both numbers are complex. In fact, all real numbers and all imaginary numbers are complex. On the other hand, some complex numbers are real, some are imaginary, and some are neither.

(A small aside: The textbook defines a complex number to be imaginary if its imaginary part is non-zero. According to that definition an imaginary number may have a non-zero real part. For example, the number would be imaginary. That definition is non-standard. I expect it's just a mistake. In any case you should not adopt that definition, it is not useful.)

The fact that some numbers are imaginary does not mean that they are inferior or mysterious, or somehow less than real. It just means they are complex numbers whose real part is zero. The word real in this context emerged in the course of history, but its modern meaning in our context is quite different from the meaning of "realistic". As far as a mathematician is concerned, real, imaginary, and complex numbers are all equally "realistic".

Here is one of the principles that make understanding mathematics possible:

You work with complex numbers as you would with ordinary algebraic expressions containing the variable , except that you replace with wherever it occurs.

Let's look at the four basic operations. For the sake of illustration we'll combine the complex numbers and . We'll also give the general formulas but you don't have to memorize them since the principle given in the previous paragraph tells you (almost) everything you need to know.

In general we have that

### Subtraction

In general we have that

### Multiplication

In general we have that
That is a formula I definitely do not memorize, and you shouldn't burden your mind with it. Instead, when you multiply two complex numbers just figure out the product from the principle listed above.

### Division

Division is a little more complicated. We have
If you think about it, it's not even clear that that expression is a complex number. To see that in fact it is, and to figure out its standard form, we need the concept of a conjugate complex of a complex number. It is obtained by replacing the imaginary part with its negative. Conjugate complex numbers are usually denoted with a bar. For example, the conjugate complex of is and the conjugate complex of is . In general
The significance of the complex of a conjugate complex number is that the product of a complex number and its conjugate complex is real! For example,
In general
This fact is so important that I'll restate it:

We now return to the problem of dividing two complex numbers. We are bothered by the complex number in the denominator and we get rid of it by multiplying with its conjugate complex in the numerator and denominator. Hence

The real part of is and its imaginary part is . When I first saw this trick some time ago I was awed! Surely you will have something to talk about at your next party!

In general we have

That is another formula you don't want to memorize! But you should be easily able to carry out a complex division by multiplying numerator and denominator with the conjugate complex of the denominator.


text:
# How to fill in missing transitions when sampling an MDP transition table?

I have a simulator modelling a relatively complex scenario. I extract ~12 discrete features from the simulator state which forms the basis for my MDP state space.

Suppose I am estimating the transition table for an MDP by running a large number of simulations and extracting feature transitions as the state transitions.

While I can randomize the simulator starting conditions to increase the coverage of states, I cannot guarantee all states will be represented in the sample, i.e. states which are possible but rare.

Is there a rigorous approach to "filling in the gaps" of the transition table in this case?

For example:

1. For each state that was unrepresented in the sample, simply transition to all other states with equal probability, as a "neutral" way to fill in the gap?

2. As above, but transition only to represented states (with equal probability)?

3. Transition to the same state with probability 1.0?

4. Ignore unrepresented states during MDP solving entirely, and simply have a default action specified?

I assume you use the 12 discrete features as state variables, and, for each of these variables, you will have at least two values. So, the minimum number of states will be $$2^{12} = 4096$$, which gives $$(2^{12})^{2} = 16777216$$ possible transitions. In order to reach this, you will need a huge amount of simulations, also taking into account that this number is a minimum since you might have more values per state variable, and you probably have more than one action per state transition.


text:
# How do you factor 0.09y^2 - 0.81?

Nov 30, 2015

$0.09 {y}^{2} - 0.81 = \left(0.3 y + 0.9\right) \left(0.3 y - 0.9\right)$

#### Explanation:

The difference of squares formula gives us
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
(you can multiply this out to confirm)

$0.09 {y}^{2} = {\left(0.3 y\right)}^{2}$
and
$0.81 = {0.9}^{2}$

Thus, by the difference of squares formula

$0.09 {y}^{2} - 0.81 = \left(0.3 y + 0.9\right) \left(0.3 y - 0.9\right)$


text:
de.mpg.escidoc.pubman.appbase.FacesBean
Deutsch

Hilfe Wegweiser Datenschutzhinweis Impressum Kontakt

# Datensatz

DATENSATZ AKTIONENEXPORT

Freigegeben

Zeitschriftenartikel

#### Combinatorial Algorithms for Data Migration to Minimize Average Completion Time

##### MPG-Autoren

Mestre,  Julián
Max Planck Society;

##### Externe Ressourcen
Es sind keine Externen Ressourcen verfügbar
##### Volltexte (frei zugänglich)
Es sind keine frei zugänglichen Volltexte verfügbar
##### Ergänzendes Material (frei zugänglich)
Es sind keine frei zugänglichen Ergänzenden Materialien verfügbar
##### Zitation

Gandhi, R., & Mestre, J. (2009). Combinatorial Algorithms for Data Migration to Minimize Average Completion Time. Algorithmica, 54(1), 54-71. doi:10.1007/s00453-007-9118-2.

The \textit{data migration} problem is to compute an efficient plan for moving data stored on devices in a network from one configuration to another. It is modeled by a transfer graph, where vertices represent the storage devices, and edges represent data transfers required between pairs of devices. Each vertex has a non-negative weight, and each edge has a processing time. A vertex completes when all the edges incident on it complete; the constraint is that two edges incident on the same vertex cannot be processed simultaneously. The objective is to minimize the sum of weighted completion times of all vertices. Kim (\textit{Journal of Algorithms, 55:42-57, 2005}) gave an LP-rounding $3$-approximation algorithm when edges have unit processing times. We give a more efficient primal-dual algorithm that achieves the same approximation guarantee. When edges have arbitrary processing times we give a primal-dual 5.83-approximation algorithm. We also study a variant of the open shop scheduling problem. This is a special case of the data migration problem in which the transfer graph is bipartite and the objective is to minimize the completion times of edges. We present a simple algorithm that achieves an approximation ratio of \mbox{$\sqrt{2} \approx 1.414$}, thus improving the 1.796-approximation given by Gandhi~\etal\ (\textit{ACM Transaction on Algorithms, 2(1):116-129}, 2006). We show that the analysis of our algorithm is almost tight.


text:
# zbMATH — the first resource for mathematics

Infinite games, Banach space geometry and the eikonal equation. (English) Zbl 1163.91007
Summary: We study a class of infinite games which turns out to be closely related to Banach space geometry. Using one of these games, we construct a bounded, differentiable, almost everywhere solution of the eikonal equation $$\| {\nabla} u\| = 1$$ on $$\mathbb R^{d}$$, with $$d \geqslant 2$$.

##### MSC:
 91A05 2-person games 46B20 Geometry and structure of normed linear spaces 26B05 Continuity and differentiation questions
Full Text:


text:
# [Tex/LaTex] Rotating a one-page float for pdf viewing

floatslandscapepdfrotating

I have a few long tables that should be rotated 90 degrees. These tables pretty much need to take up their own page. Rotating the table itself is not a problem, but I would also like to have an attribute set in the appropriate page that made pdf viewers view the page in landscape mode.

I have read this question and this question and there is plenty of information for doing what I want using the landscape environment from pdflscape. Except for one problem: the table produced in these ways is not really a float. The landscape environment is not itself a float, so my tables take up a whole page, and the page that precedes them is full of whitespace.

Is there any way to designate the page containing some float as landscape while letting the float continue to float? Can I make a landscape-float environment? I don't mind if such a thing necessarily has to take up a whole page. If I have to, I'll resort to using sidewaystable from the rotating package and post-processing the pdf on pages that need rotating, but it would be nice to not have to do that every time I need to recompile the final draft.

#### Best Answer

Since the rotated table will take up a full page anyway (not as complex to position as a partial-page float), you can use afterpage to avoid breaking up the text flow.

\documentclass{article}

\usepackage{pdflscape}
\usepackage{afterpage}
\usepackage{lipsum} % for filler text only

\begin{document}
\pagestyle{myheadings} % to show header behavior
\lipsum
\afterpage{%
\begin{landscape}
\begin{table}[p]
\caption{Rotated table caption.}
\lipsum[1]
\end{table}
\end{landscape}
}
\lipsum
\end{document}


text:
Smashmallow Cinnamon Churro, Mouth Cancer Symptoms Pictures, Mumbo Jumbo Hermitcraft 6 Ep 43, Gödöllő To Budapest, Kahulugan Ng Mapapabuyong, Sprite Bottle 1 Litre Price, Speech Writing Pdf, 2020 Toyota Prius Prime Msrp, " />

# hypolimnion has ph value

The loss of the riparian system essentially removes critical habitats and buffers for urban runoff, thus reducing the resilience of the urban lake to eutrophication. Particularly in terms of drinking water production in reservoirs and dams with respect to the existing regulations of the Drinking Water Ordinance,[2] these deteriorations in water status are a serious problem. pH is usually (but not always) between 0 and 14. (1) (iv) Calculate the concentration, in mol dm –3, of the Nitric acid. Le pH de l’eau pure à 25°C, qui est égal à 7, a été choisi comme valeur de référence d’un milieu neutre. An acid with a […] The gas escapes into the atmosphere, the oxygenated water flows back through the downstream pipe. , … The brain, circulatory system, nerves, muscles, respiratory system, digestive system, and reproductive system can all benefit from a proper pH level. The amount and mobility of iron species also influences the redox-controlled phosphorus household. At very high concentrations, a yellow water colour is noticeable. In such a case, destratification would mix the high nutrient concentration water with the surface layer, increasing the possibility of an algal bloom. We use cookies to help provide and enhance our service and tailor content and ads. Fig. In some instances, it is desirable to maintain the thermal stratification and yet increase the DO concentration in the hypolimnion. Le potentiel hydrogène, plus connu sous le nom de pH permet de mesurer l'acidité ou la basicité d'une solution. Heavy metal concentrations may also impact the health of higher level species (e.g., macroinvertebrates, crustaceans, and fish) (Scheffer et al., 2001; Melbourne Water, 2005; Taylor and Breen, 2006; Taylor et al., 2007), which may lead to widespread mortality within an aquatic ecosystem. Hypolimnion: The bottom layer of the lake. For example, some fish require cold water temperatures but high DO concentrations. Hydrogen sulfide creates odor problems and may cause downstream fish kills. While it's theoretically possible to calculate a negative pH, pH values should be between 0 and 14 in practice. This assumption certainly causes underestimations, because some decomposition of algae, littoral macrophytes, and sessile algae occurs in the warm epilimnetic and metalimnetic layers before reaching the relatively quiescent, cool hypolimnion. This can be achieved by manipulating the fish populations in a lake. A sample calculation is given in Table 11-4 (see Wetzel and Likens, 2000). The hypolimnion layer (cooler and denser) in degraded urban lakes is typically of poor quality, as it often acts as a sink for organic matter and sediment causing low levels of dissolved oxygen (DO). In urban lakes, organic carbon loads, particularly those sourced from floating macrophytes, add to nutrient loads that are already potentially high because of the impacts of urban runoff. Examples for each method are provided, conditions and limitations of their use are enumerated, positive and negative effects mentioned, and some indication of costs given. The DO concentration in the hypolimnion can be increased by the introduction of air or pure oxygen. pH indicators are indicators that show the pH value of substances, they change color over the small range of the scale. Many of the same metal cycling processes that occurred in the sediments also occur in the water column. Dezember 2012. Persistent stratification—Stratification is the formation of two distinct layers in a lake, the epilimnion and hypolimnion (Fig. As a result Fe and Mn cycling at redox boundaries in the water column may be much more dynamic than in the sediment and opportunities for metal losses are greater. Prevention of sludge formation, increased ammonium production and the formation of toxic hydrogen sulfide. Warning: The ph scale has no limits. Conservation of deep zones as aerobic habitat for fish and other higher organisms. Higher alkaline pH values (pH 9–14) in the presence of oxygen as an oxidant lead to so-called oxygen corrosion. For an aqueous solution with a temperature of 25°C with pH value, less than 7 is considered as acidic, whereas pH value greater than 7 is considered as alkaline or base. Methods based on artificial mixing and oxygenation: Destratification by total mixing of the water column, Mixing and re-aeration of the hypolimnion, Sediment oxygenation by the RIPLOX procedure, Phosphorus precipitation with sediment sealing, Hydraulic regulation (selective water withdrawal), Extraction of hypolimnetic water (hypolimnetic siphoning). By the pH-stabilizing effect of hypolimnetic aeration, the application of these buffer solutions can be reduced, thus reducing operating costs. Iron is reduced in anoxic sediments and in the water columns and is transported by eddy diffusion to the oxic/anoxic boundary. The CO2-accumulation method of calculating production usually leads to underestimation because of the confounding factors mentioned earlier. More generalized results to predict efficient bubble diameters for different devices and conditions are underway (Little and DelVecchio,1999). Procedure IV - Adding Base Solution to the Buffer Solution Complete the table below using your data from Procedure IV. More specifically the layers are defined as follows: Figure 13.1. Limnol. Loss of organic production from the trophogenic zone via the outlet is assumed to be small—a condition which, again, is highly variable in open lakes. High organic carbon loading—In natural lake systems, organic matter is critical to microbial processes as it provides a food source for heterotrophic organisms (Ewel et al., 2001; Bond et al., 2008). Bedrock There are several assumptions underlying calculations of hypolimnetic CO2 accumulation. Yes, the answer is any pH below 7 (at 25C), but it doesn't stop at 0. pH actually ranges from between -2 and 15. EAWAG, Dübenedorf-Zürich, Schweiz,1996. The principle is evident in Figure 4.11, which shows one of the aerator types. For concentrations of iron and manganese in drinking water, the threshold values are 200 µg/l and 50 µg/l respectively. A solution that has a pH of 1.0 has 10 times the [H +] as a solution with a pH of 2.0, which in turn has 10 times the [H +] as a solution with a pH of 3.0 and so forth. Chap. If CO2 accumulation is calculated for the entire hypolimnion, and this is divided by the volume of the epilimnion, the quantity termed the relative assimilation intensity results. This value is considered neutral—neither acidic or basic. Is difficult to mix a narrow water stratum near the bottom water layer phrases traduites contenant  a. Usually a consequence of poor quality of specific organic pigments, so-called color.. 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# Prime divisors of $4k - 1$

In this question, I have to prove that there exists infinitely many primes for $n = 4k - 1$. We can consider this sequence $3, 5, 7, 11, 19, 23, 31, \ldots$, which are all a form of $4k - 1$.

So, we start by considering the positive integers of the form $n=4k-1$ and their possible prime divisors.

1. Show that every prime divisor of $4k - 1$ is odd.

2. Show that an odd number is either of the form $4k - 1$ or $4k + 1$.

3. Show that a product of two numbers of the form $4k + 1$ again has that form (more precisely, given two numbers $4k + 1$, $4l + 1$, show that their product has the form $4m + 1$ for some integer $m$).

I need help up to this point to solve the rest of the question. Any pointers will be greatly appreciated!

1. Every divisor of an odd number is odd. As $4k$ is always even, $4k-1$ is therefore odd.

2.

Every number leaves a remainder of $0,1,2,$ or $3$ when divided by 4. If the number was originally odd, then it leaves a remainder of $1$ or $3$ when divided by 4.

If the number leaves a remainder of $1$ when it's divided by $4$, then it's clearly of the form $4k + 1$

If the number leaves a remainder of 3 when it's divided by 4, then it's of the form $4x + 3=4(x+1)-1$, which is of the form $4k-1$

3.

\begin{align}(4k+1)(4l+1)&=16kl+4k+4l+1 \\ &=4(4kl+k+l)+1 \end{align}

which is of the form $4m+1$

• Where are you going with this? I don't see that you reached a proof that infinitely many primes of the form $4k-1$ exist. – hardmath Sep 21 '17 at 14:53
• @hardmath The OP is unclear. it says “I need help up to this point”, which could be reasonably interpreted as “i don’t know how to do any of these steps” (which are necessary ingredients in the proof). – Erick Wong Sep 21 '17 at 15:51
• @ErickWong: Ok, I read the "I need help up to this point to solve the rest of the question" as (emphasis added) focusing on the rest of the question. Still, if Jazzachi does not wish to wait for clarification (that may never come), I'd advise adding at least a sketch of how to proceed. It is within the Answerer's control. – hardmath Sep 21 '17 at 15:57
• @hardmath I didn't finish the proof because RideTheWavelet has already done so with reference to the three questions. – user472341 Sep 21 '17 at 22:19

Assume the first three points have been proved. Then if $n=4k-1$ is not prime, it has only odd divisors (point 1), which must therefore all be of the form $4k'\pm 1$ (point 2). If all of these are of the form $4k+1,$ then $n$ itself would have to take this form (by point 3). Thus, if $n=4k-1$ is not prime, then it has a proper divisor of the form $4k'-1.$

Suppose by way of contradiction that there are only finitely many primes of the form $4k-1$, $p_{1},p_{2},\ldots,p_{n}.$ If $n$ is odd, then $\prod_{i=1}^{n}p_{i}$ is of the form $4k-1,$ and if $n$ is even, this product is of the form $4k+1,$ so let $p=\prod_{i=1}^{n}p_{i}+4$ if $n$ is odd, and otherwise let $p=\prod_{i=1}^{n}p_{i}+2.$ In either case, $p$ is of the form $4k-1.$ If $p$ were composite, it would have a proper divisor of the form $4k'-1,$ as we showed above. If this divisor is not prime, then it too has a divisor of the form $4k''-1,$ and we may repeat this argument to say that $p$ has a prime divisor $p'$ of the form $4k-1.$ But $p'$ must divide $\prod_{i=1}^{n}p_{i},$ since this is the product of all primes of the form $4k-1,$ and it cannot be the case that $p'|4$ or $p'|2,$ so in fact $p'\not|\,p,$ and thus $p$ is prime. Since $p$ is of the form $4k-1$, and strictly larger than any prime in the list $p_{1},\ldots,p_{n},$ this contradicts our assumption that this was the list of all of the primes of the form $4k-1.$ This contradiction completes the proof.

• One could also avoid the separate even/odd cases by taking $p=4\prod p_i - 1$. – Erick Wong Sep 21 '17 at 17:16


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dc.contributor.author Castro, C. dc.contributor.author Zuazua, E. dc.date.accessioned 2017-02-21T08:18:21Z dc.date.available 2017-02-21T08:18:21Z dc.date.issued 2011-12-31 dc.identifier.issn 0025-5718 dc.identifier.uri http://hdl.handle.net/20.500.11824/596 dc.description.abstract We consider the problem of flux identification for 1-d scalar conservation laws formulating it as an optimal control problem. We introduce a new optimization strategy to compute numerical approximations of minimizing fluxes. We first prove the existence of minimizers. We also prove the convergence of discrete minima obtained by means of monotone numerical approximation schemes, by a Γ-convergence argument. Then we address the problem of developing efficient descent algorithms. We first consider and compare the existing two possible approaches. The first one, the so-called discrete approach, based on a direct computation of gradients in the discrete problem and the so-called continuous one, where the discrete descent direction is obtained as a discrete copy of the continuous one. When optimal solutions have shock discontinuities, both approaches produce highly oscillating minimizing sequences and the effective descent rate is very weak. As a remedy we adapt the method of alternating descent directions that uses the recent developments of generalized tangent vectors and the linearization around discontinuous solutions, introduced by the authors, in collaboration with F. Palacios, in the case where the control is the initial datum. This method distinguishes descent directions that move the shock and those that perturb the profile of the solution away from it. As we shall see, a suitable alternating combination of these two classes of descent directions allows building more efficient and faster descent algorithms. dc.format application/pdf dc.language.iso eng en_US dc.rights Reconocimiento-NoComercial-CompartirIgual 3.0 España en_US dc.rights.uri http://creativecommons.org/licenses/by-nc-sa/3.0/es/ en_US dc.subject 1-d scalar conservation laws dc.subject Alternating descent method dc.subject Flux identification dc.subject Numberical approximation dc.subject Optimal control dc.title Flux identification for 1-d scalar conservation laws in the presence of shocks dc.type info:eu-repo/semantics/article en_US dc.identifier.doi 10.1090/S0025-5718-2011-02465-8 dc.relation.publisherversion https://www.scopus.com/inward/record.uri?eid=2-s2.0-79960620619&doi=10.1090%2fS0025-5718-2011-02465-8&partnerID=40&md5=1f7ee9fb125a361c8d6c7230b177a80a dc.rights.accessRights info:eu-repo/semantics/openAccess en_US dc.type.hasVersion info:eu-repo/semantics/publishedVersion en_US dc.journal.title Mathematics of Computation en_US

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Except where otherwise noted, this item's license is described as Reconocimiento-NoComercial-CompartirIgual 3.0 España


text:
# Some basic Consumption CAPM questions

Say we are in a world described by the consumption CAPM. All investors in this world have quadratic utility. Also, assume that consumption is as follows: $$c_{t+1} = (1+m_t)c_t + s_t c_t e_t$$ where the shocks $$e_t$$ are iid with $$E_t(e_{t+1})=0$$ and $$V_t(e_t) = 1$$. Now $$m_t$$ and $$s_t$$ depend on the business cycle so that in down times $$m_t$$ and $$s_t$$ are high, but in boom periods $$m_t$$ and $$s_t$$ are low. Assume that the risk-free rate is independent of time.

I have a few questions as follows and was wondering if my answers make sense or if someone can correct me.

Say there is a company (e.g., luxury goods) that is very sensitive to the business cycle. Will this company have high or low returns?

My answer: Investors don't like uncertainty about consumption. You consume more luxury good in good times and less in bad times. Since it varies positively with consumption, you will require a low price and hence luxury goods will have high returns. Is my explanation right here?

Now consider a security that provides insurance against bad consumption shocks so that in down periods it pays off more. What can we say about the price of this security? Will it have a high/low price when consumption is high? Will it have high/low price during a down period?

My answer: This security would smooth consumption since it varies negatively with consumption. It pays off good in bad times and does not pay off in good times. As a result during a recession/down period, this security would have a high price and during a boom (when consumption is high), it will have a low price. Is my explanation right here?

Using the consumption CAPM, rank the expected returns on the following types of companies: i) computer software; ii) luxury goods; iii) basic consumer products; iv) oil and gas; v) high-growth company.

We need to compare the consumption beta of each company. The higher the consumption beta, the higher the expected return, all else equal. But I'm not quite sure how to roughly compare the betas of the above companies. Could anyone help?

Now for the above ranking, consider other risk factors that could explain expected returns, e.g., Fama-French 3 factors etc. How would the rankings change?

Again, I know I need to think about the covariance between the returns of the companies and how they co-vary with the risk factors, but I am unsure of the direction of the covariance. Could someone help here?


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Delete Middle Node from Linked List

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In this problem, we will delete the Middle Node from a given Linked List. We have covered the algorithm (along with implementation) to delete the middle node from a Singly Linked List and Doubly Linked List.

We will go through some of the basics of Linked List and the delete operation to remove a given node in a Linked List and then, move to our main topic. The challenge is to identify the node to delete that is the middle node in the given Linked List.

Linked List is a one of the DataStructure used for storing collection of data. A Linked List has the following properties:

1. A Linked List is linear dynamic DataStructure.
2. The number of nodes in Linked List is not fixed, it can grow and shrink on demand.
3. Each node of the Linked List is made up of two items - the data and a reference to the next node.
4. The Last node has reference to null.
5. The entry point is called head
6. If the list is empty then the head is null reference.

Why we use Linked List Over Array-

• The size of array must be known in advance before using it in program.
• Increasing size of the array is a time taking process. It almost impossible and to increase the size of an array at runtime.
• As array stored in our memory contiguously , inserting and deleting an element in an array means shifting of all the elements.

The Linked List consists of series of structures called Nodes. We can think of each node as a record. The first part stores the data , and second part stores a pointer to the next node. So , conclusion is each node contains two field , a data field and and next field. Generally , Linked List means Singly Linked List. In Singly Linked List , the link of the last node in the list is NULL, which indicates the end of the list.
Each Node is allocated in the heap with call to malloc() , so node memory contimues to exists until it is explicitly deallocated.

The node which is called as head is the first node in the list. The last node's next pointer points to NULL value.

Type Declaration for a Linked List of integers

struct ListNode
{
// define ListNode in Linked List
int data; // the data
struct ListNode *next; //pointer to the next List Node.
};

Deletion Operation on a List-

In this case, the node to be removed is always located between two nodes. Head and tails links are not updated in this case. It is done in two steps -

1. Start Traversing the Linked List, Once we find the node to be deleted , change the previous node's next pointer to the next pointer of the node to be deleted.

1. Dispose the current node to be deleted.

How to find the middle element?

To find the middle element, we can take two approaches:

• Uisng 2 traversals: The first traversal to find number of elements and the second traversal to find the middle element
• Using 1 traversal with slow and fast pointers

As we just need to position of middle element, we can get it using one traversal to count the total number of elements:

// one traversal
{
int count = 0;

while(temp.next != null)
{
++ count;
temp = temp.next;
}
return count / 2;
}

Steps to delete middle node from Singly Linked List

1. Get the position of the middle node using the above techniques (say nth position).
2. Traverse to the nth node of the singly linked list and also keep reference of n- 1 th node in some temp variable say prevnode.

1. Reconnect the n-1 th node with the n+1 th node i.e. prevNode->next = toDelete->next (Where prevNode is n-1th node and toDelete node is the nth node and toDelete->next is the n+1th node).

1. Free the memory occupied by the n th node i.e. toDelete node.

Implementation in C

void *delete(struct listNode **head , int position)
{
int k=1;
struct listNode *p, *q;
printf("List Empty");
return;
}

if(position == 1){ //from beginning
free(p);
return;
}

else{
//transverse the list until arriving at the position from which we want to delete
while((p!=NULL) && k<position){
k++;
q=p;
p = p->next;
}
if(p==NULL){ //at the end
printf("position doesnt exist");
return;
}
else{ //from the middle
q -> next = p ->next;
free(p);
}
return;
}
}

Time Complexity - O(N) . In the worst case we may need to delete the node at the end of the list.
Space Complexity - O(1).

We can use the approach of slow and fast traversal to delete the middle node in just one traversal. The pseudocode will be like:

// one traversal
{

while(first_node != null && second_node != null)
{
first_node = first_node.next;
second_node = second_node.next.next;
}

delete(first_node);
}

Apart from Singly Linked List , we also have two others types of Linked List i.e. Doubly Linked List and Circular Linked List.

We will look the structures of both the types and brief about how we can delete an intermediate node from these two.

A Doubly Linked List contains an extra pointer, typically called previous pointer, together with next pointer and data which are there in singly linked list.

The advantage of Doubly Linked List is that given a node in the list, we can navigate in both direction. In Doubly Linked List we can delete a node even we dont have previous node address ( since we have previous pointer ).

The disadvantage of Doubly Linked List are - Eachnmnode requires an extra pointer, requiring more space. The insertion and deletion of a node takes a bit longer.

Deleting an intermediate Node in Doubly Linked List

In this case, the node to be removed is always located between two nodes, and heads and tails are not updated, the removal can be done in two steps-

1. Maintain the previous node while also traversing the list. Upon locating the node to be deleted , change the previous node's next pointer to the next node of the node to be deleted. Also, change the previous pointer of the next node to the node to be deleted to the point to the previous node of the node to be deleted.
2. Dispose the current Node to be deleted.

Time Complexity - O(N).

Circular linked list is a linked list where all nodes are connected to form a circle. There is no NULL at the end. A circular linked list can be a singly circular linked list or doubly circular linked list.


text:
# What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters?

Mar 14, 2016

#### Answer:

We use Boyle's law, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$. The pressure required for compression is approx. $7.5$ $a t m$

#### Explanation:

${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2$ $=$ $\frac{196.0 \cdot L \times 1 \cdot a t m}{26.0 \cdot L}$ $=$ ??atm


text:
# IPython Notebooks

So while I wasn't paying attention, the Python scripting language has turned into an incredible tool for scientific math. I was aware of the badass numerical package numpy, but [IPython] has integrated it with amazing math (LaTeX) and graphics (matplotlib) display packages into a powerful tool for mathematical exploration.

Because MathJax is used for rendering LaTeX, this post is mostly an experiment to see how much of a IPython notebook I can cram into my blogging platform (Staceyapp). I was hoping I could inline some Notebook charts'n'graphs but turns out I can't without hacking IPython's HTML (rendered with 1500 lines of css) into the Staceyapp templating engine, which is a task for another time...

What I can do is simply link to the converted notebook page: Here's a test notebook to check out.

But MathJax is pretty awesome, here's a little test of rendering some LaTeX. Let's try Schrödinger's Equation:

$$\left [ – \frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V \right ] \Psi = i \hbar \frac{\partial}{\partial t} \Psi$$

Holy moly, that seemed to work! Take a look at the page source to see how easy that was.

Tip for installing IPython on Windows: because I have another Python package manager because client, I found the usual "pip install" didn't work and I had to manually install a lot of dependencies. Fortunately Christopher Gohlke at at UCI has made a lot of precompiled Python binaries available and they are super handy.

Update 2020 Installation is a lot easier today! I recommend the Anaconda distribution. Even easier, if you have a Google account you can use IPython notebooks right in your Google drive.

Tagged: SW hacks


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# Arrow of time and entropy?

The arrow of time is usually defined by the direction in which entropy increases. In a closed system, if there's a max entropy that the system can reach, does that mean time stops or at least become undefined at the max entropy state?

One may still describe the behavior of such a system at maximal entropy using a theory that does use the concept of time, or the time coordinate $t$. But it is true that operationally speaking, the passage of time ceases to exist because the equilibrium associated with the maximum entropy is incompatible with the existence of thinking observers.

Note that whenever someone measures time, remembers, thinks, plans, or does things, the total entropy has to strictly increase. All the macroscopic observables are stationary in the maximum-entropy state which means that nothing macroscopic is happening.

It is important to distinguish between the time and the flow of time.

The time, $t$, is just a coordinate like $x$, $y$ and $z$ that we use to specify points in spacetime. The time coordinate doesn't have an arrow any more than $x$, $y$ or $z$ have arrows. The time axis has a negative and positive direction, just like the spatial coordinates, but at normal energies all the fundamental equations we use in physics are time symmetric i.e. they don't distinguish between the future and the past.

But of course we can move in any direction in space, while our experience tells us that we have to move through time at one second per second. This is the flow of time. There are endless arguments about whether the flow of time is a quirk of human perception or something fundamental, but we should perhaps pass over those here.

You posit a system that is at maximum entropy. Entropy is a statistical quantity, and if you look at the microstructure of your system you'll see it is still changing with time. Gas molecules are still whizzing around and colliding with other gas molecules, and their interactions are still described by functions of time. So in this sense time still carries on just as it did before.

However when you consider large scale properties, like pressure or density, if the system is at maximum entropy these cannot change significantly - or more precisely it is exceedingly improbable that they will change significantly. So for example no macroscopic clock can tick, because the operation of a (real i.e. non-reversible) macroscopic system is always accompanied by an increase in entropy.

But does this mean the flow of time has stopped? I think this is largely semantics rather than physics. At the microscale the system is still evolving with time even though macroscale systems like you and I could not experience any flow of time. My view would be that t is not useful to say the flow of time has stopped in such a system. To make that statement just means you have not looked at the system closely enough.

• If the system is still evolving at maximum entropy, is there a way to say which direction the system is evolving towards? Normally, we deduce it from the increase in entropy, but if entropy is at a maximum, there is no clue. It sounds like, the arrow of time based on entropy is only a clue for observers like us but not really fundamental to the nature of time. – iBe Apr 15 '15 at 18:37
• @iBe: yes, I agree. I am unconvinced that an arrow of time based on entropy is that useful a concept in most areas of physics. – John Rennie Apr 15 '15 at 19:22

This is my first answer here so please don't expect a lot. I watched the BBC show yesterday just about this question. It is called Wonders of the Universe, episode 1 - Destiny.

According to the Professor Brian Cox, yes, arrow of time will stop eventually when maximum entropy is reached. It's going to happen in unimaginable amount of time. Literally unimaginable. If you count every atom in our Universe as one year — it would be not enough atoms to describe this number. Eventually every star is going to die. Then every black hole. Entropy is always increasing. In the end it would be nothing but cold photons floating in the empty space. When this happens maximum entropy is reached and time as we know will no longer exist.

• Hawking made this statement once but later retracted it as not true. – Peter R Jul 4 '16 at 14:16
• @PeterR thanks. Do you know where I can read about this? I tried to search the retraction and didn't find any. – borisy Jul 4 '16 at 14:51
• He talks about time reversal at the end of this speech he gave. hawking.org.uk/the-beginning-of-time.html – Peter R Jul 5 '16 at 2:00


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# Question on preventing k from reducing too quickly during KMV intersection

This question considers KMV, an algorithm that is able to estimate the cardinality (unique item) from a stream of data.

The way it does it is to first map the stream of data to a space that almost guarantee uniform distribution and then only store the minimum k values. To estimate the cardinality it looks at the spacing between the k values and extrapolate the cardinality.

One of the benefit of KMV compare to hyperloglog is that KMV can perform set intersection that create better result than hyperloglog with inclusion exclusion principle

I am looking at a particular implementation of KMV call theta sketch and I am interest in its implementation in intersection

Specifically I am having a hard time understanding how its k size is reducing almost 50% slower than my implementation.

I expect k's reduction to match the probably of intersection between the 2 sets, which I do see in my code. However I am seeing k's reduction in theta sketch much less prominent.

I look at the code and I think the only major difference is that theta sketch is much more optimized and its bound mechanism is through storing any number less than a certain number (ie. theta) instead of storing the minimum k number. But I still expect that k to reduce as fast as my implementation.

Can someone point me a direction ?

• I doubt this is answerable without reviewing the code of those two implementations, and that task sounds outside of our scope here. Any community votes? – D.W. Apr 20 at 22:40


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# Relations, Set Theory, Reflexive, Symmetric, Transitive

1. Aug 24, 2012

### Cabinbreaker

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

Determine whether the relations on three sets are Reflexive, Irrelfexive, Symmetric,, Asymmetric, Antisymmetric, Transitive, and Intransitive.

The relation $\subseteq$ on a set of sets.

2. Relevant equations
3. The attempt at a solution

I am having trouble figuring out how this will work. I think for the set {{a},{b},{c}} that should mean

<{a} , {{a},{b},{c}}>
<{b} , {{a},{b},{c}}>
<{{a},{b}} , {{a},{b},{c}}>
<{{a},{b}} , {{a},{b},{c}}>
<{{a},{b}} , {{a},{b},{c}}>
<{{a},{b},{c}} , {{a},{b},{c}}>
<∅ , {{a},{b},{c}}>

But I do not know how to use this to answer the question.

2. Aug 25, 2012

### tiny-tim

Welcome to PF!

Hi Cabinbreaker! Welcome to PF!

(btw, you don't need those extra {} round most of your letters )
eg reflexive means: if {a,b} then {b,a}

transitive means: if {a,b} and {b,c} then {a,c}

3. Aug 25, 2012

### Cabinbreaker

Then on the relation ⊆ on a set of sets (considering the set {{a},{b},{c}}), relations should be nonrelfexive, nonsymmetric, transitive, and Antisymetric.

Relations are not irreflexive or intransitive.

I think they are also not asymmetric because of <{{a},{b},{c}} , {{a},{b},{c}}> where xRy ⇒ ¬yRx ?

4. Aug 25, 2012

### vela

Staff Emeritus
What you've written doesn't make much sense to me. Either I'm not understanding the question or you've misinterpreted it.

Here's my take:

You have S={{a}, {b}, {c}}, which is a set of sets, and let $x, y \in S$. You say x is related to y if $x\subset y$. For example, say x={a} and y={b}. It's not true that $\{a\} \subseteq \{b\}$, so you would not say that {a} is related to {b}.

In general, a relation R is a subset of S×S. In this case, S×S contains the nine ordered pairs

{{a}, {a}}
{{a}, {b}}
{{a}, {c}}
{{b}, {a}}
{{b}, {b}}
{{b}, {c}}
{{c}, {a}}
{{c}, {b}}
{{c}, {c}}

So those are the pairs you should be considering whereas you seem to be considering the power set of S. Can you state the problem exactly as given to you?

5. Aug 25, 2012

### Cabinbreaker

The Problem says:

My difficulty is I do not understand the question. I was able to complete the first two. I do not know what it means by "The relation ⊆ on a set of sets." I used the power set because I thought where A ⊆ B it was asking for the ordered pair {A,B}.

6. Aug 25, 2012

### vela

Staff Emeritus
The set {{a}, {b}, {c}} is already a set of sets. That is, the elements of the set are sets themselves. You have to have a set of sets otherwise the relation $\subseteq$ doesn't make sense. For instance, if you have the set {a, b, c}, then you might ask if $a \subseteq b$, which doesn't make sense because neither a nor b is a set. On the other hand, with the set {{a}, {b}, {c}}, one element is the set {a} and another is the set {b}, and it does make sense to ask if $\{a\} \subseteq \{b\}$.

I enumerated the nine ordered pairs in S×S. What you want to do first is determine if the first set in each pair is a subset of the second set in the pair. Then you want to determine if the relation is reflexive, etc. If the relation is reflexive, you should find that every pair of the form {{x}, {x}} satisfies the relation. And so on.

7. Aug 25, 2012

### HallsofIvy

Re: Welcome to PF!

This is "symmetric", not "reflexive". A relation on set X is "reflexive" if an only if it contains {x,x} for every x in X.

8. Aug 25, 2012

### HallsofIvy

Then you do NOT want to look at a specific set as you do below.
A relation on a given set is a collection of ordered pairs from that set. If we are given a set, X, A and B are two subsets the the pair (A, B) will be in the relation if and only if $A\subseteq B$.

"Reflexive" here means that for any subset, A, of X, it is true that $A\subseteq A$ which is true because of the "=" part.
"Irreflexive" simply means "not reflexive".

"Transitive" here means "if $A\subseteq B$ and $B\subseteq C$ then $A\subseteq C$".

"Intransitive" simply means "not transitive".

"Symmetric" here means "if $A\subseteq B$ then $B\subseteq A$". Of course, that is not true.

"Asymmetric" simply means "not symmetric".

"Anti-symmetric" is a little more complicated. It means "if $A\subseteq B$ then it is NOT true that $B\subseteq A$. Here that is not true because of the "=" part.


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# current measurement over time

#### abc14

Joined Oct 15, 2017
117
I have a RC servo towerPro sg90 connected to Arduino, to measure the current I have put 1 ohm shunt resistor between servo GND and power ground.
Then I did a analogRead on the resistor, when servo sweep between 0 -180. I am getting following values out of my analogRead.

So in total I have 180 readings from 0 -180 sweep, and I have 15ms delay the loop, all in all its takes 2700ms or 2.7 secs to complete the sweep.
Values from analogRead ranging from 20 - all the way upto 250, since servo is controlling a small flap.

Now I have two questions here..

1. Is this analogRead is my voltage, or do I have to convert it into voltage, like analogread *(3.3/1023).. Then its simply voltage/R is my current.

2. Total current, will it be sums of all the 180 values calculated in the step 1.

I am going to copypaste the raw serial output as well... The serial monitor is running at 9600 buad rate, if its useful to know.

#### Attachments

• 59.8 KB Views: 2

#### Alec_t

Joined Sep 17, 2013
10,918
do I have to convert it into voltage, like analogread *(3.3/1023)
Yes, assuming your Arduino's A/D converter is a 10-bit one.
By 'total current' I suspect you mean total charge, which is the integral of current*time.

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#### abc14

Joined Oct 15, 2017
117
Yes, assuming your Arduino's A/D converter is a 10-bit one.

yes its 10 bit... Now I have gone ahead and converted the values to actual voltage readings, so most am getting is .775V or 775mV.. which equates to roughly about 777mA ?

#### Attachments

• 60.9 KB Views: 0

#### abc14

Joined Oct 15, 2017
117
Yes, assuming your Arduino's A/D converter is a 10-bit one.
By 'total current' I suspect you mean total charge, which is the integral of current*time.
yes total charge, goal here is to estimate how long the battery is going to last, as an input to my model how much charge/current is consumed by servo for each opening operation.

So from this data I can conclude the peak current draw is 770mA ?

#### Alec_t

Joined Sep 17, 2013
10,918
Looks like it. But current will also be affected by the load the servo is trying to shift, so may be much higher in practice.

#### abc14

Joined Oct 15, 2017
117
Looks like it. But current will also be affected by the load the servo is trying to shift, so may be much higher in practice.
Yes, beside since I didn't have actual 1 ohm resistor, hence i put 10x10ohms resistor in parallel to make 1 ohm resistor, the power rating of individual resistor is 1/4 Watts. Not sure how power rating effects the current calculation

Joined Jan 15, 2015
5,140
Among the problems you might have is you are using the Arduino 10 bit ADC and really looking at low level signals. I have no idea how much you do this sort of stuff but I tried a ADS1115 16 Bit 16 Byte 4 Channel I2C IIC Analog-to-Digital Converter along with an Arduino Uno and got really good results. Really slick as you can use I2C to connect. Here is an Arduino example. Among the nice features is programmable Gain so while the default is +/-6.144 volts and dividing 6.144 volts by 32767 yields a scale factor of 0.1875 mV per bit. Roughly 26 times the Arduino native 10 bit. In another PGA setting, you can establish a full scale of +/- 2.048 volts. That provides us a resolution of 0.0635 mV. Some other nice features are programmable address. Like anything on Amazon you need to shop carefully. I got two for under $4.00 each. I send the data to a 4 x 20 LCD. Something else nice is it is a 4 channel unit ADC single ended input yopu can also use it in a differential mode and have 2 differential channels. Very useful when doing shunt measurements. Using 10 1/4 watt resistors in parallel assuming all are exactly 10 Ohms you would have yourself a 2.5 watt 1.0 Ohm resistor which is 10 * 0.250 watt. Your voltage drop across the shunt is also subtracted from the load. So for example starting with 5 volts to a load and I place a 1 Ohm shunt in my line and my Vshunt is 1,000 my load is only seeing 4.0 volts. Ron #### Reloadron Joined Jan 15, 2015 5,140 #### nsaspook Joined Aug 27, 2009 6,949 I should have mentioned that. I also should have included the data sheet so here is the data sheet. You will find it worded: Analog input voltage AIN0, AIN1, AIN2, AIN3 GND – 0.3 VDD + 0.3 V Under Absolute Maximum Ratings. Ron If there's only VDD and GND that's usually a good indicator of unipolar pseudo-differential. An example of a device (SPI interface) that can do this: The ADS1220 can be used to measure various types of input signal configurations: single-ended, pseudodifferential, and fully-differential signals (which can be either unipolar or bipolar). http://www.ti.com/lit/ds/symlink/ads1220.pdf #### Reloadron Joined Jan 15, 2015 5,140 If there's only VDD and GND that's usually a good indicator of unipolar pseudo-differential. An example of a device (SPI interface) that can do this: The ADS1220 can be used to measure various types of input signal configurations: single-ended, pseudodifferential, and fully-differential signals (which can be either unipolar or bipolar). http://www.ti.com/lit/ds/symlink/ads1220.pdf Pretty cool chip which I never knew existed. Then too, unlike 7 ~ 8 years ago I no longer have a need to really stay on top of the new stuff and actually before I retired I had new engineers right out of school and what was nice was those guys were all over the newest, latest and greatest like white on rice. Hanging out here and still messing with a few circuits at least keeps my mind somewhat functional. Thanks for sharing that. Ron Thread Starter #### abc14 Joined Oct 15, 2017 117 If there's only VDD and GND that's usually a good indicator of unipolar pseudo-differential. An example of a device (SPI interface) that can do this: The ADS1220 can be used to measure various types of input signal configurations: single-ended, pseudodifferential, and fully-differential signals (which can be either unipolar or bipolar). http://www.ti.com/lit/ds/symlink/ads1220.pdf Hence ADS1220 is needed to measure negative voltage across the shunt ? Thread Starter #### abc14 Joined Oct 15, 2017 117 Among the problems you might have is you are using the Arduino 10 bit ADC and really looking at low level signals. I have no idea how much you do this sort of stuff but I tried a ADS1115 16 Bit 16 Byte 4 Channel I2C IIC Analog-to-Digital Converter along with an Arduino Uno and got really good results. Really slick as you can use I2C to connect. Here is an Arduino example. Among the nice features is programmable Gain so while the default is +/-6.144 volts and dividing 6.144 volts by 32767 yields a scale factor of 0.1875 mV per bit. Roughly 26 times the Arduino native 10 bit. In another PGA setting, you can establish a full scale of +/- 2.048 volts. That provides us a resolution of 0.0635 mV. Some other nice features are programmable address. Like anything on Amazon you need to shop carefully. I got two for under$4.00 each. I send the data to a 4 x 20 LCD. Something else nice is it is a 4 channel unit ADC single ended input yopu can also use it in a differential mode and have 2 differential channels. Very useful when doing shunt measurements.

Using 10 1/4 watt resistors in parallel assuming all are exactly 10 Ohms you would have yourself a 2.5 watt 1.0 Ohm resistor which is 10 * 0.250 watt. Your voltage drop across the shunt is also subtracted from the load. So for example starting with 5 volts to a load and I place a 1 Ohm shunt in my line and my Vshunt is 1,000 my load is only seeing 4.0 volts.

Ron

Hence in my case the voltage drop across shunt is around 775mV and my starting voltage is 4.7V meaning my load is only seeing 4V ?

#### TeeKay6

Joined Apr 20, 2019
572
If there's only VDD and GND that's usually a good indicator of unipolar pseudo-differential.

An example of a device (SPI interface) that can do this: The ADS1220 can be used to measure various types of input signal configurations: single-ended, pseudodifferential, and fully-differential signals (which can be either unipolar or bipolar).

@nsaspook
I don't see what distinction you are making. The ADS1220 too can only measure between the power supply rails (Vdd to Vss, +/-0.3V). The ADS1220 allows the analog ground to be more negative than the power ground, but it's still a ground and measurements essentially cannot go more negative than that ground. What am I missing?

#### nsaspook

Joined Aug 27, 2009
6,949
@nsaspook
I don't see what distinction you are making. The ADS1220 too can only measure between the power supply rails (Vdd to Vss, +/-0.3V). The ADS1220 allows the analog ground to be more negative than the power ground, but it's still a ground and measurements essentially cannot go more negative than that ground. What am I missing?
You're missing this.

The ADS1220 has additional power pins on the analog section. AVSS and AVDD that require symmetrical supplies (i.e. +/- 2.5V) for bipolar operation.

Note that AVSS is -2.5 IRT DGND and that analog inputs can be -0.1 of AVSS in bipolar mode.

2.5V Zeners provide the analog bipolar voltages and a 3.3V LDO regulator provides the digital section voltage.

Last edited:

Joined Jan 15, 2015
5,140
Hence in my case the voltage drop across shunt is around 775mV and my starting voltage is 4.7V meaning my load is only seeing 4V ?
That's about it, that's what I'm telling you. This why shunt resistances are kept very low. Take a look at this common everyday 50 amp shunt.

That shunt, across the voltage sense terminals is exactly 0.001 Ohm. The idea is we don't want our shunt dropping a high voltage because whatever we drop across the shunt never makes it to the load. So yes, in your case if you start with 4.7 volts and drop about 0.7 volts your load is seeing 4.0 volts. Years ago people went through great efforts to amplify current shunt outputs but today's A/D converters with a PGA (Programmable Gain Amplifier), make life much easier. They give us the ability to come off a mV source sensor and be able to read them less the need for all sorts of external amplification.

Before I forget see the post above. While that chip requires a +/- power input it can go negative. Another nice feature is it is a 24 bit ADC and like what I mentioned allows programmable gain. There is also good reason to use a differential input. That allows us to get an accurate voltage pickoff from our shunt. That's important. Just using "common" is not a good idea or practice.

Ron

#### TeeKay6

Joined Apr 20, 2019
572
You're missing this.

View attachment 199683

The ADS1220 has additional power pins on the analog section. AVSS and AVDD that require symmetrical supplies (i.e. +/- 2.5V) for bipolar operation.

Note that AVSS is -2.5 IRT DGND and that analog inputs can be -0.1 of AVSS in bipolar mode.

View attachment 199685View attachment 199686

View attachment 199688
2.5V Zeners provide the analog bipolar voltages and a 3.3V LDO regulator provides the digital section voltage.
I did not miss what you cite; that was in fact what I did cite. However, I did indeed misread your comment "An example of a device (SPI interface) that can do this..." I apologize for my error in reading. The ADS1220 is indeed a versatile ADC. You also said "If there's only VDD and GND that's usually a good indicator of unipolar pseudo-differential." Yet, T.I. (and other mfrs) sells several unipolar ADC's (using only Vdd & Gnd) that offer fully differential conversions. There is no strict need for bipolar operation in order to support fully differential measurements, although bipolar operation such as that of the ADS1220 certainly eases fully differential measurements of signals having a common-mode voltage that can be negative with respect to ground.

Last edited:

#### nsaspook

Joined Aug 27, 2009
6,949

#### abc14

Joined Oct 15, 2017
117
Interesting readings, stuck a scope across my shunt resistor, I am not seeing similar values of voltage drop across shunt as compared to my Arduino ADC ?


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# zbMATH — the first resource for mathematics

##### Examples
 Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.

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The Fermat-Weber location problem revisited. (English) Zbl 0855.90075
Summary: The Fermat-Weber location problem requires finding a point in ${ℝ}^{N}$ that minimizes the sum of weighted Euclidean distances to $m$ given points. A one-point iterative method was first introduced by Weiszfeld in 1937 to solve this problem. Since then several research articles have been published on the method and generalizations thereof. Global convergence of Weiszfeld’s algorithm was proven in a seminal paper by Kuhn in 1973. However, since the $m$ given points are singular points of the iteration functions, convergence is conditional on none of the iterates coinciding with one of the given points. In addressing this problem, Kuhn concluded that whenever the $m$ given points are not collinear, Weiszfeld’s algorithm will converge to the unique optimal solution except for a denumerable set of starting points. As late as 1989, Chandrasekaran and Tamir demonstrated with counter-examples that convergence may not occur for continuous sets of starting points when the given points are contained in an affine subspace of ${ℝ}^{N}$. We resolve this open question by proving that Weiszfeld’s algorithm converges to the unique optimal solution for all but a denumerable set of starting points if, and only if, the convex hull of the given points is of dimension $N$.

##### MSC:
 90B85 Continuous location
##### References:
 [1] J. Brimberg and R.F. Love, ”Local convergence in a generalized Fermat–Weber problem,”Annals of Operations Research 40 (1992) 33–66. · Zbl 0787.90040 · doi:10.1007/BF02060469 [2] J. Brimberg and R.F. Love, ”Global convergence of a generalized iterative procedure for the minisum location problem withl p distances,”Operations Research 41 (1993) 1153–1163. · Zbl 0795.90037 · doi:10.1287/opre.41.6.1153 [3] R. Chandrasekaran and A. Tamir, ”Open questions concerning Weiszfeld’s algorithm for the Fermat–Weber location problem,”Mathematical Programming 44 (1989) 293–295. · Zbl 0683.90026 · doi:10.1007/BF01587094 [4] L. Cooper, ”Location–allocation problems,”Operations Research 11 (1963) 37–52. · Zbl 0113.14201 · doi:10.1287/opre.11.3.331 [5] H. Juel and R.F. Love, ”Fixed point optimality criteria for the location problem with arbitrary norms,”Journal of the Operational Research Society 32 (1981) 891–897. [6] I.N. Katz, ”Local convergence in Fermat’s problem,”Mathematical Programming 6 (1974) 89–104. · Zbl 0291.90069 · doi:10.1007/BF01580224 [7] H.W. Kuhn, ”A note on Fermat’s problem,”Mathematical Programming 4 (1973) 98–107. · Zbl 0255.90063 · doi:10.1007/BF01584648 [8] H.W. Kuhn and R.E. Kuenne, ”An efficient algorithm for the numerical solution of the generalized Weber problem in spatial economics,”Journal of Regional Science 4 (1962) 21–34. · doi:10.1111/j.1467-9787.1962.tb00902.x [9] W. Miehle, ”Link-length minimization in networks,”Operations Research 6 (1958) 232–243. · doi:10.1287/opre.6.2.232 [10] A.E. Taylor and W.R. Mann,Advanced Calculus (Xerox College Publishing, Lexington, MA, 2nd ed., 1972). [11] E. Weiszfeld, ”Sur le point par lequel la somme des distances den points donnés est minimum,”Tohoku Mathematics Journal 43 (1937) 355–386.


text:
Home > matlab, Olympiad > ICPC 2015 World Final Problem B

## ICPC 2015 World Final Problem B

This the the solution to Problem B from the International Collegiate Programming Contest. The list of problems can be found here. The idea is to find the maximal area of the intersection of two moving polygons. The inputs give the initial positions of two convex polygons and the vectors giving their speed in the plane.

One idea of a solution is as follows: take a maximal time ${T}$ and a discretization of the interval ${[0,T]}$ by dividing it into ${N}$ parts. Iterate over the translations at times given by this discretization, compute at each step the area of the intersection of the two polygons (if there is any intersection at all), and in the end find the time ${t_0}$ for which this area is maximized. Now, even if the discretization step ${T/N}$ is large (greater than the demanded precision of ${10^{-3}}$), we can conclude that ${t_0}$ is an approximation of the final time with an error smaller than ${T/N}$. This is due to the fact that the function representing the area of the intersection has two monotonicity intervals, as a consequence of the fact that the polygons are convex. On the first interval, the intersection area is increasing, on the second one it is decreasing. Thus, once we have a discrete maximum, we are close to the real maximum.

Now, all we are left to do in order to achieve any desired precision is to refine the search near this discrete maximum, find a new, better approximation, and refine again, until we have enough precision. Of course, one first problem with this algorithm is the initial search using a maximal time ${T}$. It is possible that if ${T}$ or ${N}$ are not large enough, then we do not detect any intersection of the two polygons. Thus, an initial guess, based on a mathematical argument is needed in order to reduce the search of the optimal time to an interval which is small enough to have enough initial precision to detect the discrete maximum.

The algorithm presented below, uses Matlab’s predefined function polybool which can compute the intersection of two polygons. Of course, this is an overkill, since dealing with intersections of convex polygons is not that complicated (but still, I didn’t have enough time to play with the problem, in order to provide a more optimized version). I do not treat the search for the initial time interval. As I think about it, I guess a n argument based on finding some line intersections should give us a narrow enough time interval (with some care for the case when the two speed directions are collinear). The algorithm presented below solves the sample cases, but could fail in more general situations.

```
function Prob1_2015(p1,p2)

% choose initial polygons; see the text of the problem
p1 = [6 3 2 2 4 3 6 6 6 7 4 6 2 2 2];
p2 = [4 18 5 22 9 26 5 22 1 -2 1];

%p1 = [4 0 0 0 2 2 2 2 0 1 1];
%p2 = [4 10 0 10 2 12 2 12 0 -1 1];

np1 = p1(1);
np2 = p2(1);
cp1 = p1(2:1+2*np1);
cp2 = p2(2:1+2*np2);

vp1 = p1(end-1:end);
vp2 = p2(end-1:end);

cp1 = cp1(:);
cp1 = reshape(cp1,2,np1);

xp1 = cp1(1,:);
yp1 = cp1(2,:);
cp2 = cp2(:);
cp2 = reshape(cp2,2,np2);

xp2 = cp2(1,:);
yp2 = cp2(2,:);

%set precision
prec = 1;

%here there should be a clever initial choice of
%the starting time and stopping time
%this choice works well in this case
start = 0;
stop  = 5;

while prec>1e-6
n = 100;

timex = linspace(start,stop,n);
areas = zeros(size(timex));

for i = 1:n
t = timex(i);
xt1 = xp1+t*vp1(1);
yt1 = yp1+t*vp1(2);
xt2 = xp2+t*vp2(1);
yt2 = yp2+t*vp2(2);

[x,y] = polybool('intersection',xt1,yt1,xt2,yt2);
areas(i) = polyarea(x,y);
end
[m,I] = max(areas);
if m<1e-6
% if no polygonal intersection is detected
% there is nothing further to do
display('never');
break
end

tapp = timex(I);
fprintf('Precision %f | Time %f\n',prec,tapp);
start = tapp-prec;
stop  = tapp+prec;
prec  = prec/10;
end
clf
%this draws the final position of the polygon
%as well as their intersection
fill(xt1,yt1,'blue')
axis equal
hold on
fill(xt2,yt2,'red')
fill(x,y,'green')
axis equal
hold off
```

Here’s what you get for the first sample test provided in the questions:

``` >> Prob1_2015
res = 4.193548
```


text:
What is the electron configuration of an element with atomic number 17?

The electron configuration is ${\text{1s"^2 "2s"^2 "2p"^6 "3s"^2 "3p}}^{5}$.
The electron configuration becomes ${\text{1s"^2 "2s"^2 "2p"^6 "3s"^2 "3p}}^{5}$.


text:
# How to prove graph isomorphism is NP?

I know that Graph Isomorphism should be able to be verified in polynomial time but I don't really know how to approach the problem. Any help would be appreciated.

• What is graph isomorphism for you? Please add to your post a formal description. – Yuval Filmus Dec 2 '19 at 23:11

You haven't explained what graph isomorphism means for you, so let me assume that you mean the language of all pairs of graphs $$(G_1,G_2)$$ which are isomorphic.
Two graphs $$G_1 = (V_1,E_1),G_2 = (V_2,E_2)$$ are isomorphic if there exists a bijection $$f\colon V_1 \to V_2$$ such that $$(x,y) \in E_1$$ iff $$(f(x),f(y)) \in E_2$$.


text:
## Chemistry: The Central Science (13th Edition)

$Al(OH)_3 + 3HCl$-->$AlCl_3+3H_2O$
For the balancing of the equation we will initially balance the coefficients of the Cl, adding 3 to the HCl. Note that hydrogen and oxygen are not yet balanced, for this the coefficient of water will be 3. Finally, our equation is balanced as follows: $Al(OH)_3 + 3HCl$-->$AlCl_3+3H_2O$